trigonometry
inequalities
probability
modular arithmetic
induction
LaTeX
geometry

i suggest start a math marathon on our privite forum.  the people who solve the first problem can give a second one, like a math marathon. It's recommanded that you can make up the question on ur own, not coping from other contests.

i would like to start with a really easy one:

Prove the three angle bisectors in a triangle intersect at one point.

[quote="rem"]Every point on the angle bisector is equidistant from the sides of the angle, then the point of intersection of two bisectors is equidistant from all the sides so the third bisector passes through this point too. QED

OK next problem:
Prove if $A,B,C$ are angles in a triangle then:
$cosA+cosB+cosC\leq \frac{3}{2}$[/quote]

Cos0 = 1, Cos90 = 0 and Cos180 = -1

In adding up the angles of the cosine of a triangle, the highest value we can achieve is when all the angles are of the same magnitude (60 degrees). Cos60 = 0.5, thus 1.5 or 3/2 is the highest value we can achieve, since if we increase one angles size, the other angles decrease in size or atleast one of them. As the angle increases, the cosine of this angle rapidly decreases until the angle reaches 120 where is begins to slow its decrease in cosine value. While this angle increases, the other angles are decreasing at the same rate, but in their case their cosine value is increasing much slower than the other angle is decreasing, thus the value will be below 3/2.

I get how it works, but I have a feeling my wording isnt as great, so if you could post the 'proper' proof, thatd be awesome :D

I'm pretty sure I have an answer for your proposal Andy, but I dont have a question to propose so Ill wait  :oops:

[quote="andy_tok"]:lol: use
$x^{2}+y^{2}+z^{2}\geq 2xycosA+2yzcosB+2zxcosC$ x,y,z are positive reals

ABC is a triangle. Find the maximum of the following
$sin\frac{A}{4}sin\frac{B}{4}sin\frac{C}{4}+cos\frac{A}{4}cos\frac{B}{4}cos\frac{C}{4}$[/quote]

Mmm, am-gm and then jensens should work. Since sin x and cos x are concave in the interval (0, pi/4)?

[quote="rem"]Proof:
On interval $[0,\pi]$ the $cos$ function has $f(x)'=-sin(x)\leq 0$ so by Jensen inequality:
$\frac{1}{3}cosA+\frac{1}{3}cosB+\frac{1}{3}cosC\leq cos(\frac{A+B+C}{3})=cos(\frac{\pi}{3})=\frac{1}{2}$
So $cosA+cosB+cosC\leq \frac{3}{2}$ QED[/quote]

$f(x)'=-sin(x)\leq 0$ is not equivalent to f(x) is concave. you should use f''(x).

For this question you can't use Jensen directly in fact unless ABC is an acute triangle, otherwise cosx is concave on $[0, \pi/2]$ whereas it's convex on $[\pi/2, \pi]$

Proof:
$x^{2}+y^{2}+z^{2}\geq 2xycosC+2yzcosA+2xzcosB$
let x=y=z=1, and we get that
$cosC+cosA+cosB\leq \frac{3}{2}$

...i was wondering why $\cos{x}$ was concave in the interval $[0,\pi]$

[quote="SM4RT"]...i was wondering why $\cos{x}$ was concave in the interval $[0,/pi]$[/quote]
it's not.
 :)

[quote="rem"]Sorry for my first proof. :oops:
OK here's a hardcore brute force way:
By Cosine Law get:
$cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}$
So we have to prove:
$\sum_{cyc}\frac{a^{2}+b^{2}-c^{2}}{ab}\leq 3$
So multiplying through by $abc$ get:
$a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b\leq 3abc+a^{3}+b^{3}+c^{3}$
which is true by Schur.[/quote]

nice solution  :)

[quote="rem"]Every number from $11111$ to $99999$  inclusive is written on a separate card. Then the cards are all put in one row in some random order to make one big number. Prove it can't be a power of $2$.[/quote]

Because the sum of the digits of all these numbers is a multiple of 9(1,2...9 has the same probability to occur, so the sum of all the numbers from 11111 to 99999 is a multiple of 45)

[quote="JingyuanMo"]yo...do you mean that a 88889*5-digit number is formed?  :|[/quote]
Yeah.

@andy_tok: I don't think the sum of digits is a multiple of $9$.
It is $\frac{99999*100000}{2}-\frac{11110*11111}{2}$ which is not a multiple of $9$.

[quote="JingyuanMo"]
assume PO=x,and the radius is 1.Then use x and the angle a between PO and the nearest line to PO to represent the shadow areas.Take the derivative of the function with domain 0<a<30 (x is a constant here)[/quote]
Would you mind posting the function of the shadow area?

[quote="JingyuanMo"]APMO i think... :lol:[/quote]
lol...yea...i took APMO this year...lol...definitely met u... :D

sometimes we have to think idealisticly :P

[quote="Mathgod"][quote="bryant124"]try dis one...
let $0\le x_{i}\le 1$ for $i=1,2,...,n$
prove that $x_{1}+x_{2}+...+x_{n}-(x_{1}x_{2}+x_{2}x_{3}+...+x_{n}x_{1})\le [\frac{n}{2}]$[/quote]

LS is a linear function in terms of xi, therefore extreme values occur when xi is 0 or 1.[/quote]
lol...thats the first thing that came to my mind when i saw diz problem...but my friend told me there was a much easier way...(this method is like EXTREMELY easy already...lol)

[quote="Mathgod"][quote="bryant124"]try dis one...
let $0\le x_{i}\le 1$ for $i=1,2,...,n$
prove that $x_{1}+x_{2}+...+x_{n}-(x_{1}x_{2}+x_{2}x_{3}+...+x_{n}x_{1})\le [\frac{n}{2}]$[/quote]

LS is a linear function in terms of xi, therefore extreme values occur when xi is 0 or 1.[/quote]
true but that doesnt finish it off completely

[quote="bryant124"]
lol...thats the first thing that came to my mind when i saw diz problem...but my friend told me there was a much easier way...(this method is like EXTREMELY easy already...lol)[/quote]
 :maybe: use $(x_{i}-1)(x_{j}-1)\geq 0$ ?

lol...here is a mathematical investigation
first we use the formula
$girls=time \times money$
then, using the theorem "time is money", we have
$girls=time \times money=money \times money=(money)^{2}$
using another theorem, "money is the root of evil", we have
$girls=(money)^{2}=(\sqrt{evil})^{2}=evil$
and we are done :rotfl:  :rotfl:

Another one similar to this starts the same way but the last step is:
"money is the root of all problems"
$\Leftrightarrow girls=problems$
 :D

sounds interesting but
 :arrow: not that interesting

[quote="Kileel"]To keep this forum interesting, I will revive this thread by posting a small problem:

In a building there are seven elevators and each elevator services at most six floors.  From each floor, if it is possible to reach any other floor by using only one elevator, then what is the maximum number of floors the building can have?[/quote]

Let $n$ be the number of floors. Each of the $\frac{n(n-1)}{2}$ pairs of floors must be connected by an elevator. However, each elevator can only connect $\frac{6*5}{2}=15$ pairs of floors. Since there are only $7$ elevators, all the elevators together can connect $7*15=105$ pairs of floors. Thus we have $\frac{n(n-1)}{2}=105$, or $n^{2}-n-210=0$, so $n=15$.

A physics teacher divides a class of students into $k$ non-empty groups for a physics project. Independently, a chemistry teacher divides the same class into $k$ non-empty groups for a chemistry project. Prove that if there is a student with a larger physics group than chemistry group, then there must be a student with a larger chemistry group than physics group.

[quote="shalevbd"] so $n=15$.

[/quote]

But is n=15 actually attainable?  I think this is the bulk of the problem.  :D

woops, i guess $15$ doesn't work  :blush:
Assume $15$ floors: since there are $6*7$ elevator stops, and $15$ floors, there will be a floor with only two stops by the pigeon principle. But each elevator connects that floor to only five other floors, so that floor is connected to only ten other floors. Contradiction.

Now I show that $14$ works. To simplify things, lets merge pairs of floors together, in such a way that if two floors are merged, each elevator that stops in one of the floors must stop in the other. Merging consecutive pairs leaves $7$ floors, with elevators that stop on three floors. Name the floors $A$ to $G$. Then elevators that stop on:
$A,B,C$
$A,D,E$
$A,F,G$
$B,D,F$
$B,E,G$
$C,D,G$
$C,E,F$
work.


Figure $ A$ is made from three unit squares, and will be called a $ \emph{trio}$. Figure $ B$ is a 4 by 4 grid of unit squares. Without overlap, what is the maximum number of trios that can be placed on the grid such that each trio covers exactly three of the unit squares in the grid?

[asy]size(130);
picture p;
draw(unitsquare^^shift(1*up)*unitsquare^^shift(1*right)*unitsquare);
label("Figure $A$",(1,-1));
for(int i = 0; i<4; ++i) for(int j = 0; j<4; ++j)
draw(p,shift(i*up+j*right)*unitsquare);
label(p,"Figure $B$",(2,-1));
add(shift(3*right+2*down)*p);[/asy]

I'm terribly sorry, but I don't have LaTeX, so I will do the best I can with rather primitive methods of explanation.

I will call the upper-left square "1", the square to the right of that "2", etc., from left to right for each row going down until the bottom-right square "16". So 1-2-6 means a trio covering the upper-left square, the square to the right of that, and the square below the second.

Many arrangements can fit 4 maximum, like [1-5-6,2-3-7,8-11-12,10-14-15] and [1-2-5,3-6-7,10-11-14,12-15-16].

However, if we fit four of the trios onto the perimeter of the figure (1-2-5,3-4-8,9-13-14,12-15-16), we could make room for a fifth (either 6-7-10,6-7-11,6-10-11, or 7-10-11), thus the answer is [b]5[/b].

You don't need to have $ \text{\LaTeX}$ installed in order to type $ \text{\LaTeX}$ on AoPS.

Just put dollar signs around your math, and you can also refer to the $ \text{\LaTeX}$ guide [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]here.[/url]


function
inequalities
algebra proposed
algebra

Function $ f: (0;\plus{}\infty)\to(0;\plus{}\infty)$ satisfies the inequality $ {f(3x)\ge f\left(\frac{1}{2}f(2x)\right)\plus{}2x}$ for every $ x>0.$ Prove that $ f(x)\ge x$ for every $ x>0.$

Let $ a$ be this  inferior bound : $ a\equal{}\inf_{x>0}(\frac{f(x)}x)$. We can write   $ \forall_{x>0},f(\frac 12f(2x))\geq \frac{af(2x)}2\geq a^2x$  so :

$ \forall_{x>0},f(3x)\geq (a^2\plus{}2)x \ \Longrightarrow \forall_{x>0},\frac {f(x)}x$$ \geq \frac {a^2\plus{}2}3 \ \Longrightarrow a\geq  \frac{a^2\plus{}2}3\ \Longrightarrow  \ 1 \leq a \leq 2$ (q.e.d)
 :cool:


inequalities
function
LaTeX
algebra proposed
algebra

Let  $m$ and $n$ are integers ($m>1$ and $n >1$) $m$ is even
$f$ is a fonction such:
A)  $ \frac {f(x_1^m+ \cdots +x_n^m} n)=\frac {f(x_1)^m+ \cdots +f(x_n^m)} n$
  for all $x_1,x_2 \cdots ,x_n \in \mathbb {R}^+$
B) $f(2004) \neq 2004$ and $f(2006)\neq 0.$

Find $f(2005)$
 :P

Edited by pbornsztein to make it readable. Come on use latex codes!

Here is the solution I gave on the exam, I am not sure it is totally correct :
We can easily show the following facts :
- $\forall x \geq 0 : f(x^m)=(f(x))^m$
- $\forall x \geq 0 : f(x) \geq 0$
- $f(0) \in \{0,1\}$ and $f(1) \in \{0,1\}$
Now suppose $f(0)=0$
take $x=x_1^m$ and $x_2=x_3=...=x_n=0$ in [b]A)[/b] we get $\forall x \geq 0 : f(\frac{x}{n})=\frac{f(x)}{n}$
then $f(\frac{x_1^m+...+x_n^m}{n})=\frac{f(x_1^m+...+x_n^m)}{n}=\frac{f(x_1^m)+...+f(x_n^m)}{n}$
which means that ${f(x_1^m+...+x_n^m)=f(x_1^m)+...+f(x_n^m)}$ for all $x_1,...,x_n \geq 0$
and by puting $x_1^m=x$ and $x_2^m=y$ and $x_3=x_4=...=x_n=0$ we get
$f(x+y)=f(x)+f(y)$ for all $x,y \geq 0$
And this is a the well known Cauchy functional equation. we have $\forall n \in \mathbb N : f(n)=nf(1)$
But $f(2004)=2004f(1) \neq 2004$ which means that $f(1) \neq 1$
and $f(2006)=2006f(1) \neq 0$ which means that $f(1) \neq 0$
So $f(1) \not\in \{0,1\}$ . Contradiction.
So $f(0)=1$
Now take $x=x_1^m$ and $x_2=x_3=...=x_n=0$ in [b]A)[/b] we get $\forall x \geq 0 : f(\frac{x}{n})=\frac{f(x)+n-1}{n}$
So $f(\frac{x_1^m+...+x_n^m}{n})=\frac{f(x_1^m+...+x_n^m)+n-1}{n}=\frac{f(x_1^m)+...+f(x_n^m)}{n}$
which means that $f(x_1^m+...+x_n^m)=f(x_1^m)+...+f(x_n^m)-(n-1)$
and by taking $x=x_1^m$ and $y=x_2^m$ and $x_3=x_4=...=x_n=0$
We get $f(x+y)=f(x)+f(y)+(n-2)-(n-1)=f(x)+f(y)-1$ for all $x,y \geq 0$
Put $g(x)=f(x)-1$ to get $g(x+y)=g(x)+g(y)$ for all $x,y \geq 0$
Then $\forall n \in \mathbb N : g(n)=ng(1)$
which means that $\forall n \in \mathbb N : f(n)=n(f(1)-1))+1$
If $f(1)=0$ then $\forall n \in \mathbb N : f(n)=1-n<0$. Contradiction.
So $f(1)=1$ and then $\forall n \in \mathbb N : f(n)=1$
Of course $f(2005)=1$  :P

It's exactly the same solution that I gave :lol: . I think that this exam was very easy.


AMC
USA(J)MO
USAMO
AIME
summer program
Mathcamp
PROMYS

I'd just like to get to know some of the younger USAMOers who may be here.  Please post here if you are somewhere close to 7th, 8th, or 9th grade and made USAMO or almost did.  I'd also like to know who the other 8th grader who scored 11 on AIME is...

not that i made it to USAMO but i know people from my school that did like the 8th grade guy that got 11/15 on the AIME, his user name is Dark_Shadow, another person (8th grader) from our school made it to USAMO: an931freak

Wow do you go to Longfellow or something?

nvm I found out he was from San Jose.  Only a few thousand miles off...

[quote="kool_dudy"]Man...I just don't get how you guys are so young and actually were qualified to take the USAMO...

The kids in Canada definetely aren't as bright as the ones in USA...[/quote]

I wouldn't be so sure.  If I am not mistaken (although I may be), Canada is smaller population-wise than the U.S. and did almost as well on IMO.

The Canadian population is 31.4 million, according to their 2002 census.  The population of the United States is 290,342,554 in a 2003 estimate, making us roughly 9 times as populous.

But still...there shouldn't just be a few that are brilliant in math although there were recent Canadian contest results and it showed a grade 8 that was 11 years old that got into the second best group in a grade 10 Canadian contest...

pretty good...i can't do that....let alone be 11 and get into grade 8...

[quote="bugzpodder"][quote="kool_dudy"]Man...I just don't get how you guys are so young and actually were qualified to take the USAMO...

The kids in Canada definetely aren't as bright as the ones in USA...[/quote]

I take offense in this statement.  this is discriminatory and should be removed[/quote]

Well, then you're being silly.  Suck it up.

The statement isn't true.  In fact, I would imagine that when you take a few things into consideration, Canadians actually are smarter per capita.  Though I may be wrong.

It's worth noting, that besides the population increase, which should hypothetically give us 9 times as many genius smart kids, you should also look at the population you're examining.  This forum is motivated greatly by the AMC, which is an American competition.  A lot of USAMOers post here, so the people here also seem really smart.  This is just where they go.  The Canadian super geniuses may have their own forum, or none at all.  There are some here, don't get me wrong.

[quote="bugzpodder"]in canada, I have to agree the education system is much weaker than most of US... (but there are reasons, one of them is because of public education) but that definately does not mean that "Canadians are not bright".  It is just not fair in saying that!!!  HOw many canadians do you know??  where are you basing your opinion from??  do you even know most of the contests in canada?[/quote]

yes I agree that one can not generalize from USAMO, but I think kool_dudy is a Canadian himself, and was just being a little pessimistic when he wrote his original message.  I doubt he really meant it, and now if we could go back to the original purpose of this thread, which was for 7th, 8th, and 9th grade USAMOers to get to know one another.

sorry, although I could have just looked at this thread and not said anything, but I feel I must go out of my place and make a stand here.  And definately I am not being silly... this comment he made might as well have been unintentional... but I still will not allow it!

I'm sorry if I offended anyone but I'm a Canadian myself and I'm pretty sure of the people in Canada.

I don't mean to say that Canadians are stupid...(why would I call myself stupid) but it can be clearly shown that Americans do indeed have many more intelligent minds.  For example, I have never seen any grade 9 students taking a grade 12 contest.  On the other hand, there are plently of grade 7,8, and 9 in USA that take the USAMO which is a very very very hard contest according to Canadian high school students.

I, myself, am not even able to solve the AIME.  I'm not comparing myself to other Canadians but there is definetely a difference that is clearly shown between Canadians and Americans.

Besides, I am CANADIAN.

PS...and like what poss. said, get back to gr. 7,8, and 9 USAMO discussion

[quote="bugzpodder"]personally I have taken grade 12 tests in grade 9... (yes, I am canadian also)  and you must not have heard about Jacob Tsimerman, who was grade 9 when he got gold medal in IMO last year...[/quote]

I heard about Jacob Tsimerman...who goes to UTS...

I didn't know about IMO though...

Maybe NOT THAT MANY students take grade 12 contest whereas in USA...

I hope this is the last reply to this completely unrelated discussion to the topic

BTW...bugs...what grade you in now?

Pretty much everyone knows I'm the seventh grader, so I don't really need to post. I got an (2 :rt2: ) :^2:  on AIME.

I'd like to know what everyone thought of the USAMO.  I mismanaged my time, but I thought that some problems were actually quite doable (e.g., #1, #2, #4).  What did you think?

Grade: 8
AIME: 7
USAMO: 7 points for #4 + partial credit (maybe?) on #2...

[quote="captcha000"]aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrggggggggggg[/quote]

If that's all you have to say...

Yeah, I know you're annoyed I beat you by a point, but...\

BTW, are you staying for the Math Jam in about 30 minutes?

9th grader, 11 AIME (missed #1!), USAMO 17 points (#s 1, 2partial, 4)

Well gee.. I lived in Canada for 7 years (ages 3 to 10), and then I moved to the US, living here for 4 years... so.. heh.. what category would I fit into? lol

[quote="question"]For those who didn't make into MOSP, mainly 8th graders, and for those who made into MOSP, where you go for this summer? MathCamp in Maine or PROMYS at BU?[/quote]

Bah. Stupid generalizations.

Neither. I'm basically taking a break from math this summer (apart from ARML, that is.) Oh crap, I just realized, my ARML permission form...and I have to leave for DC tomorrow...

I'm going to the Duke TIP program and taking a course on AI. But yeah, I'm bringing along math books. :P

[quote="probability1.01"]How did you 8th graders do?  I got 10--satisfactory... I guess... This years scores were [i]really[/i] low.  I wonder how that other 11-on-AIME-er did...[/quote]

10 is a pretty amazing score for an 8th grader...

And this year's scores were not necessarily low (depending on your definition of "low" and how it is calculated)...in fact, i think less ppl got 0's this year than last...and there were quite a few scores near the top.

I wouldn't describe this year's scores as "low".

My understanding of the USAMO is that it is designed the distinguish the top 12 students in the nation for purposes of selecting an IMO team (and also honoring all the hard work of the students).  To that end I think this year's test was very successful as the problems were just hard enough that only the top students solved 4 or more problems.

Such a test necessitates a "low average" if viewed as a percentage.  This must be the case for a good selection test.  If there were more than maybe one problem that most students were expected to solve then those problems would cease to be meaningful.  Problems 1 and 4 were easy enough that nearly every students taking the exam had some kind of shot at scoring points on those problems, but neither were so easy that they would be irrelevant to any but the top few students.

An average of just over one complete problem and a cutoff for the winners of just over three complete problems seems like an excellent distribution.

Though the top 250 or so math students in America might be used getting 70 - 100% of the problems correct on a given exam, they shouldn't judge the USAMO in the same light.

I would also recommend to the younger USAMO participants to use this experience in a very specific way.  The USAMO can be very intimidating at first.  I remember wondering how I could possible prepare for my first USAMO and how proofs should be written.

Now that you have taken the exam, be sure to look back at the Math Jams from time to time.  The first two problems from each day should look approachable.  When you start to realize that you too can solve these problems you have overcome the biggest challenge in taking the USAMO.  At that point you should be able to practice solving a large enough variety of Olympiad level problems that should once again be on a path that feels like progress (and it is).

Great comments from Mr. Crawford, and I entirely agree.

Credit for the creation of the contest goes to the problem posers and the USAMO committee which creates and formalizes the problems.  It is a
hard balancing act to create a contest that distinguishes, without the danger of "extinguishing" enthusiasm.

My congratulations to all the students who participated. Remember, just being selected to participate is an achievement in itself, and the USAMO is a great learning experience.   Turn that learning to your advantage!

Steve Dunbar
AMC Director


Dear Mashimaru and Mathlinkers,
your problem is the first step to the Gossard's theorem.
Sincerely
Jean-Louis

Oh, I did not know that theorem. Could you please give some reference to either this problem or the Gossard's theorem? :blush:

[quote="jayme"]Dear Mashimaru and Mathlinkers,
see for example : http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
Sincerely
Jean-Louis[/quote]

@[b]jayme[/b]: could you please give another reference to the proof of the original problem? Since I arrived in this one as a lemma when trying to prove the problem which is called Gossard's theorem :P .

[quote="April"][quote="Mashimaru"]Let the Euler's line of $ \triangle ABC$ intersects $ AB,AC$ at $ D,E$ respectively. Prove that the Euler's line of $ \triangle ADE$ is parallel to $ BC$.[/quote]
I think your problem is equivalent to http://www.mathlinks.ro/viewtopic.php?t=191643 . Sorry if I am mistaken  :blush:[/quote]

@[b]April[/b]: Sorry but could please you point out how they are equivalent to each other?


vector
analytic geometry
abstract algebra
linear algebra
linear algebra unsolved

Consider the set $ S$ generated by the two vectors in $ Z_N^2$ for some $ N \in N$, $ (f,g)$ and $ (h,h)$.

Note: $ S$ is the set of all integer linear combinations of $ (f,g)$ and $ (h,h)$ whose entries are in $ (0,....,N\minus{}1)$

$ 1)$ What is the cardinality of the set?

$ 2)$ How many elements have a coordinate equal to zero?

Use {\bf BLAH} or \mathbb{BLAH} to do things like $ \bf BLAH$ and $ \mathbb{BLAH}$ and avoid awful notational conflicts like $ N \in N$.  Then it will be clear that you are looking for the span of the vectors $ (f, g)$ and $ (h, h)$ in the module $ \left({\bf Z}/N{\bf Z}\right)^2$.  Obviously, though, this depends quite a lot on the particular values of $ f, g, h$ and $ N$ -- if $ N$ is prime and $ (f, g) \neq (0, 0) \neq (h, h)$ and $ f \neq g$ then they span the entire space; but in other situations there are many possible results.


solve the system
\[ \begin{array}{l}
 \frac{x}{{p \plus{} \lambda }} \plus{} \frac{y}{{q \plus{} \lambda }} \plus{} \frac{z}{{r \plus{} \lambda }} \equal{} 1 \\
 \frac{x}{{p \plus{} \mu }} \plus{} \frac{y}{{q \plus{} \mu }} \plus{} \frac{z}{{r \plus{} \mu }} \equal{} 1 \\
 \frac{x}{{p \plus{} \nu }} \plus{} \frac{y}{{q \plus{} \nu }} \plus{} \frac{z}{{r \plus{} \nu }} \equal{} 1 \\
 \end{array}
\]

oh i forgot to mention the variables are x,y,z

$ \lambda , \mu , \nu$ are roots of $ 1\minus{}\frac {x}{p\plus{}\alpha}\minus{}\frac {y}{q\plus{}\alpha}\minus{}\frac {z}{r\plus{}\alpha}\equal{}0$
 :wink:


geometry
geometric transformation
reflection
geometry proposed

Two circles $ k_{1}$ and $ k_{2}$ are given in a plane. Draw a common tangent $ AB$ to these circles,such that $ A$ is on first and $ B$ is on second circle.Denote by $ X$ and $ Y$ two intersection points of these circles, such that point $ Y$ lies inside triangle $ \triangle ABX$. Throgh $ X$ draw parallel to $ AB$ and let this parallel intersect first and second circle at points $ C$ and $ D$, respectively.

 If $ Z$ is an intersection point of lines $ AC$ and $ BD$ prove that lines $ ZX$ and $ CD$ are perpenicular.

The problem should be to prove that $ ZY$ and $ CD$ are perpendicular.
We have $ \angle{ACY}=\angle{ZAB},\angle{BDY}=\angle{ZBA}$ because $ AB\parallel{}CD$ and pts $ C,Y,D$ are collinear in this order.
Also, $ \angle{ACY}=\angle{BAY},\angle{BDY}=\angle{ABY}$ as $ AB$ is tangent to circles $ k_1,k_2$ at points $ A,B$, repsectvely.
So, $ \angle{ZAB}=\angle{BAY},\angle{ZBA}=\angle{ABY}$, also $ AB$ is common so triangles $ AZB$ and $ AYB$ are congruent. Then, point $ Z$ is teh reflection of point $ Y$ over line $ AB$ so $ ZY\perp AB$. Because $ AB\parallel{}CD$ then $ ZY\perp CD$ QED.
Now if we had $ ZX\perp CD$ then because $ ZY\perp CD$ we would have $ YX\perp CD$ so $ XY\perp AB$ which is not the case when circle $ k_1$ is larger than circle $ k_2$ or vice versa (for the obvious case, consider $ k_1$ much larger than $ k_2$).
So, $ ZY\perp CD$.

Thank you very much [b]rem[/b] I think that you have not considered that point $ Y$ lies inside triangle $ \triangle ABX$ but however I understand your solution.

(I had almost finished this problem but I have not noticed that triangles $ \triangle ZAB$ and $ \triangle ABX$ / in your solution $ \triangle ABY$  / are congruent :wallbash_red:


inequalities
real analysis
real analysis unsolved

Let $ 1\leq p,q,r\leq \infty$ such that $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}\frac{1}{r}\plus{}1$ and $ f\in L_p,g\in L_q$. Prove $ f*g\in L_r.$

Young's Inequality for Convolution. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101943]Link.[/url]

Hi Kent Merryfield,

Thank you very much. Your link really helps.


$ A$ and $ B$ two elements $ \in$ $ P(E)$
let f a function such that :
$ f: P(E)\equal{}\equal{}>P(A)$ x $ P(B)$
$ X\equal{}\equal{}\equal{}> (X\cap A,X\cap B)$
$ 1$ . prove that $ A\cup B\equal{}E$ $ \Leftrightarrow f$ is $ injective$
$ 2$ . prove that $ A\cap B\equal{}\{\}$ $ \Leftrightarrow f$ is $ surjective$

[hide="1"]Suppose $ A \cup B \equal{} E$. Then $ f(X) \equal{} f(Y)$ implies that $ X \cap A \equal{} Y \cap A$ and $ X \cap B \equal{} Y \cap B$. Let $ a$ be an arbitrary element of $ A$. Then $ a \in X \Leftrightarrow a \in Y$. Similarly for $ b$. But $ A \cup B \equal{} E$ so this is all elements of $ E$. Thus $ X \equal{} Y$.

Suppose $ f$ is injective. Now suppose $ A \cup B \neq E$. Then there is an element $ s$ of $ E$ such that $ s \notin A$ and $ s \notin B$. But then $ f(\{s\}) \equal{} (\emptyset, \emptyset) \equal{} f(\emptyset)$, and $ s \neq \emptyset$, contradicting $ f$ injective. Thus $ A \cup B \equal{} E$.[/hide]

[hide="2"]Suppose $ A \cap B \equal{} \emptyset$. Let $ S_A$ be an arbitrary subset of $ A$ and let $ S_B$ be an arbitrary subset of $ B$. Then $ f(S_A \cup S_B) \equal{} ((S_A \cup S_B) \cap A, (S_A \cup S_B) \cap B)$
$ \equal{}((S_A \cap A) \cup (S_B \cap A), (S_A \cap B) \cup (S_B \cap B))$
Notice that $ S_B \cap A \equal{} \emptyset$ and $ S_A \cap B \equal{} \emptyset$. Furthermore, since $ S_A \subseteq A$, $ S_A \cap A \equal{} S_A$. Similarly $ S_B \cap B \equal{} B$. Thus we get that the above expression is equal to $ (S_A, S_B)$. So $ f$ is surjective.

For the other direction, suppose $ f$ is surjective, then suppose that $ A \cap B \neq \emptyset$. Then there is an element $ s$ in both $ A$ and $ B$. By $ f$ surjective there is an $ X$ such that $ f(X) \equal{} (\{s\}, \emptyset)$. Then $ X \cap A \equal{} \{s\}$, so $ s \in X$. But $ X \cap B \equal{} \emptyset$, and $ s \in B$, so $ s \notin X$. Contradiction. Thus $ A \cap B \equal{} \emptyset$.[/hide]

[quote="MellowMelon"][hide="1"]Suppose $ A \cup B \equal{} E$. Then $ f(X) \equal{} f(Y)$ implies that $ X \cap A \equal{} Y \cap A$ and $ X \cap B \equal{} Y \cap B$. Let $ a$ be an arbitrary element of $ A$. Then $ a \in X \Leftrightarrow a \in Y$. Similarly for $ b$. But $ A \cup B \equal{} E$ so this is all elements of $ E$. Thus $ X \equal{} Y$.

Suppose $ f$ is injective. Now suppose $ A \cup B \neq E$. Then there is an element $ s$ of $ E$ such that $ s \notin A$ and $ s \notin B$. But then $ f(\{s\}) \equal{} (\emptyset, \emptyset) \equal{} f(\emptyset)$, and $ s \neq \emptyset$, contradicting $ f$ injective. Thus $ A \cup B \equal{} E$.[/hide]

[hide="2"]Suppose $ A \cap B \equal{} \emptyset$. Let $ S_A$ be an arbitrary subset of $ A$ and let $ S_B$ be an arbitrary subset of $ B$. Then $ f(S_A \cup S_B) \equal{} ((S_A \cup S_B) \cap A, (S_A \cup S_B) \cap B)$
$ \equal{} ((S_A \cap A) \cup (S_B \cap A), (S_A \cap B) \cup (S_B \cap B))$
Notice that $ S_B \cap A \equal{} \emptyset$ and $ S_A \cap B \equal{} \emptyset$. Furthermore, since $ S_A \subseteq A$, $ S_A \cap A \equal{} S_A$. Similarly $ S_B \cap B \equal{} B$. Thus we get that the above expression is equal to $ (S_A, S_B)$. So $ f$ is surjective.

For the other direction, suppose $ f$ is surjective, then suppose that $ A \cap B \neq \emptyset$. Then there is an element $ s$ in both $ A$ and $ B$. By $ f$ surjective there is an $ X$ such that $ f(X) \equal{} (\{s\}, \emptyset)$. Then $ X \cap A \equal{} \{s\}$, so $ s \in X$. But $ X \cap B \equal{} \emptyset$, and $ s \in B$, so $ s \notin X$. Contradiction. Thus $ A \cap B \equal{} \emptyset$.[/hide][/quote]


AMC
AIME
geometry
parallelogram
analytic geometry
rectangle
USA(J)MO

How did you do on the AIME I 2009?

Sorry, I didn't save [14, 15] as an option on the other poll.

10/15

1. Almost everyone else at my school got this one wrong...lol
2. Fine
3. Fine
4. Made the parallelogram a square, graphed it with a vertex at the origin and found equation of line connecting 2 pts
5. Mass points
6. Fine
X7. Made a silly mistake so I got it wrong but not that bad when you get the cancelling formula thing
8. Just added them up, there must be a better way
X9. Failed because I counted 3-3-1 and 3-2-2 as having 6 permutations each not 3...
X10. No idea, I hate counting
11. x coordinates of the points must have an even difference
12. My favorite problem from the test, found the final tangent segment by adding the angles (arctan of the sides) of the other two right triangles
13. Seemed way too simple...bit of a guess really since I just assumed that they had to be factors and the series would alternate every 2, did anyone prove it?
X14. System solving fail, there's probably a nice simple way that I didn't see
X15. What the hell

For #8, $ 2^1 \minus{} 2^0 \neq 0$  :wallbash: haha oh well
I did [1, 7] and [9, 13]. Should have gotten 8 and 14.

And for that one cevian one, did everybody use mass points or something? You could've just drawn a parallelogram and used similar triangles... oh well.

P.S. I used a 1000 x 2009 rectangle instead of a square >.> haha we all did that and the alternating series one in a rather cheap way xP

[quote="BenM"]
13. Seemed way too simple...bit of a guess really since I just assumed that they had to be factors and the series would alternate every 2, did anyone prove it?[/quote]

Yeah, 13 was guessable... its not that easy to prove though, and the proof is nice.

And waht? 15 was WAY TOO EASY TO BE A 15. Interestingly, I solved 15, but I couldn't solve 5.

DANG I EPICLY FAILED!!!!!!!!!! HOLY CRAP.

I couldn't even do #5!!!!!!! :mad:  :mad:  :mad:

Elaborating on my previous post:

Wow I'm pissed. I left the test fairly happy, thinking I got about 9 right. I guess I didn't get any of the details right, since I was fairly confident I knew how to do 1-8, 11, and 14, and even had plenty of time to check my answer. Turns out I only got 4 right. WTF? How did I miss 1, 2, 4, 6, and 7??? Whatever, guess that blows my hopes for USAMO, even though it would have been close anyways. I doubt I did anything wrong except computation, too. -.

Guess I should start working on those Putnams for next year. Wink

Index: 154?

----

I looked back at what I got wrong, and it was the most stupidest (irony?) thing ever. I got 3, 5, 8, and 11 right. I had the right method/idea for 1-8, 11, and 14. I even checked my answers, but I get them wrong on trivialities. I'm not even sure how I got some wrong, I had the exact same equations as everyone else did in the solution pages. Just ridiculous, and I'm just ranting now.

So, should I study Spivak or Abstract Algebra now?

I got a 6
I missed #2, because I thought 164/4=42.

So you got an AMC 10/12 score of 138...which AMC did you take? 10A, 10B, 12A, or 12B?

Sadly, I phailed AMC 10A with 129 :(
So my index is 129+10(6)=189 :mad:
For USAMO selection, do they take the top middle schooler from each state, or what? :|

just the top person who didnt already make usamo. its always .5 less than the cutoff

"The goal is to select about 500 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.

Selection will be based on the USAMO index which is defined as 10 times the student's AIME score plus the student's score on the AMC 12 or the AMC 10.

The first selection will be the approximately 300 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.

The lowest AIME score among those 300 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.

The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.

To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.

The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests."


At an average speed of 60 miles per hour, how many miles does
a car travel in six hours and 30 minutes?

If the average speed is 60 miles an hour, then one mile is traveled a minute.

So 6 hours and 30 minutes is 390 minutes, or $ \boxed{390}$ miles.


function
calculus
derivative
algebra
polynomial
complex analysis
complex analysis unsolved

Let $f$ be an entire function such that $f(z)f'(z)f''(z)$ is nonzero for all $z$. Then $f(z)=\exp(az+b)$ for some $a,b$. Actually, one can show that it is enough $f(z)f''(z)\ne 0$ for all $z$.

If $f$ is of finite order, then apply factorization theorem for entire functions of finite order and remember that derivative of a polynomial of degree $\geq 2$ always has a complex root. If $f$ is not of finite order, then... hmm...  :roll:

The solution really depends on what we know... Assuming we know a variant of "Polya-Saxer": $f$ entire, $f\ne0$, and the kth derivative of $f$: $f^{k}\ne1\implies$ f constant.  We can find the desired result by considering $F=f/f' \implies F\ne0$ and $F' = \frac{(f'^{2}-f*f')}{(f')^{2}}=1-f(f'')/(f'^{2})$, where the second term never equals zero since neither $f$ nor $f'' = 0$.  Then we have $F$ constant and infinite radius of convergence $\implies$ $f=c*f'$.  Then, by considering f as a power series with infinite radius of convergence, we have a recurrence formula and find that the only solutions are $e^{az+b}$.  I believe that there are variants of this problem for meromorphic functions, as well (the argument  is identical, up to $f=c*f'$.  Then, it becomes more nuanced from there.

If we can't assume "Polya-Saxer", then we have to prove it... I assume the proof is a variation on Picard's, which I can't prove either...  In other words, you need to have a solid handle on nevalinna theory to prove it.


Prove that, $(x-y)^{5}+(y-z)^{5}+(z-x)^{5}$ is divisible by $5(x-y)(y-z)(z-x)$, where $x,y$ and $z$ are integers, that are not equal in pairs.

It's from old USSR olympiads. (Before 1970)

come on Nekruz coldnt you solve this problem yourself??
by the way how are you, what are you doing??
I will in Dushanbe inshallah after somedays, ok!
I liked that you are posting here :) please go on like this, bye Nekruz!
Say my hello to the others, see you

Davron


A set $M$ consists of $n$ elements. Find the greatest $k$ for which there is a collection of $k$ subsets of $M$ such that for any subsets $A_{1},...,A_{j}$ from the collection, there is an element belonging to an odd number of them.

[quote="watchcatno1"]Does the null set count too?[/quote]

Yes, it does


abstract algebra
group theory
linear algebra
matrix
geometry
geometric transformation
rotation

let O(3) be the group of orthogonal linear transformations in three dimensional euclidean space
find all morphisms to R*, the multiplicative group of all nonzero reals

the trivial one and the determinant are obvious choices, are there any others?

thx

Since elements of finite order (which must be mapped to $\pm1$) are dense in $O_3(\mathbb{R})$, those are the only continuous homomorphisms. Do you care about anything else?

Actually, I'm pretty sure those are the only ones without any assumption of continuity. The idea would be to prove that the kernel must contain $SO_3(\mathbb R)$.

$f$ such a morphism.
$\frac{O_3}{Ker(f)}$ is isomorphisc to a subgroup of $(R^*,.)$. Hence $\frac{O_3}{Ker(f)}$ is abelian.
This is why the derived group of $O_3$, and thus $SO_3$ is included in $Ker(f)$.
Conclusion follows.

suppose that $u$ is an orthogonal transformation. You know (well known), that  there exists a basis where the matrix of $u$ is bloc diagonal, with blocs of size at most $2$. And each bloc is a rotation. Then you just have to prove that the rotation of finite order are dense in $SO_2$ ..

cimabue, I don't understand what you mean. Could you clarify the situation for someone who knows nothing to Lie algebra ?  :)


This post is most likely a suggestion, so I put it here. :)
Does Mathlinks know that the biggest national olympiad in China is the Chinese Mathematical Olympiad? Why can't I find anything about it in the Resources part? Can Valentin help to add these stuff, please?
Thank you very much. ;)

You should try to google it.

I just did it and got the problems from 1998 to 2003.

[quote="leepakhin"]Of course I am able to find the problems everywhere. ;)
But if they are all put onto Mathlinks, it would be very good. :)[/quote]

Yes it will eventually be up there. If anyone has all Chinese Math Olympiad problems (or great number) in English, email me and I'll try to post some. Make sure you label the problems with year and what year they began.

PM me for my email.

You need to break it up so that each post has only one problem.

Thanks.

[quote="leepakhin"]I am posting the problems year by year [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=392055#p392055]here[/url]. :)
Could Valentin, administrator or any moderators please help to put the problems in the Resources part? Thank you very much. :)[/quote]

Oh okay.

EDIT: Read the below post.

Hold on. I think there was an issue back about this with Kalva..

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=CMO&t=5304

So, unless Valentin says something, I don't think we should do anything.

By the way, I thought that the math problems can't really be "copyrighted" like other literary work.  :?


geometry
rectangle
analytic geometry
geometry proposed

What the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle
$(a)$ of size $6\times 3$?
$(b)$ of size $5\times 3$?

Could anyone please post a solution to this question? :maybe:

For a):

(I'm assuming we also have to cover the edges of the rectangle).
Consider a $2\times 2$ square. The distance from the vertex common to all $4$ unit squares to one of the outer vertices is $\sqrt{2}$ and hence a circle with radius $\sqrt{2}$ with its centre coinciding with the centre of the square would cover the whole $2\times 2$ block. We can cover a $6\times 3$ rectangle in $6$ different $2\times 2$ blocks (an example would be if the bottom left vertex of the $6\times 3$ rectangle is the origin, then the $6$ circles would have their centres with coordinates as $(1,1),(1,2),(3,1),(3,2),(5,1),(5,2)$). So the upper bound is $6$. Now suppose we could cover a $6\times 3$ rectangle with $5$ circles with radius $\sqrt{2}$. Consider the points $(0,0),(0,3),(3,0),(3,3),(6,0)$ and $(6,3)$. The distance between any of the two is $3$ or more. Since the largest chord of a circle with radius $\sqrt{2}$ is $2\sqrt{2}\approx 2.8284<3$, it follows a circle cannot contain two of the aforementioned points. Then the lower bound is $6$ as well. Hence the final answer of $6$.

For b):

It follows from from part a) that the upper bound is 6. Like we did in a), consider the $5$ points $(0,0),(0,3),(5,0),(5,3)$ and $\left(\frac{5}{2},\frac{3}{2}\right)$. The distance between any two of these points, although tighter than last time, is at least $\frac{\sqrt{34}}{2}\approx 2.9154>2.8282\approx 2\sqrt{2}$ and the lower bound of $5$ follows, since $4$ circles with radius $\sqrt{2}$ would be incapable of covering more than $4$ of these points. Hence the answer is either $5$ or $6$. The $5$ circles with centres $\left(\frac{5}{4},\frac{1}{2}\right),\left(\frac{15}{4},\frac{1}{2}\right),(1,2),(4,2)$ and $(5,2)$ do the trick (boiling down to the fact that $\sqrt{2}>1.346$).


im staying at the dorms....im the only girl to stay at the dorms on my team heh..

yea wish i could stay at dorms but teachers you know

so hotels is my answer I guess

[quote="quantum leap"]So about how many people are going to be staying at hotels and how many at MIT frats/sororities?  Our two teams are split about 50-50.  The ones staying at the hotel are basically paying to [i]not[/i] stay with "nerdy" MIT students (their words, not mine).[/quote]

Yeah, we don't want to let all those cool kids [i]that go to math competitions[/i] corrupt us nerdy MIT students.

Anyway, are they staying in actual frats/sororities or dorms?  I would have thought that they would have tried to put you up in dorms first.  Then again, it's probably easier to put you all in the same place with a frat.

[quote="ThAzN1"](that's) what she said ^

dorms all the way: cheaper, more fun!  :lol:[/quote]

uhh, you're the only girl to stay at the dorms on your team?

this is news to me for SEVERAL REASONS

Dorms, definitely.  I'm on an indie team, so nobody really cares that we don't stay in hotels.

[quote="joml88"]
Yeah, we don't want to let all those cool kids [i]that go to math competitions[/i] corrupt us nerdy MIT students.[/quote]

 :wink:

[quote="joml88"]Anyway, are they staying in actual frats/sororities or dorms?  I would have thought that they would have tried to put you up in dorms first.  Then again, it's probably easier to put you all in the same place with a frat.[/quote]

We'll actually be staying in a frat (Theta Chi) and the girl in a sorority.  It's what was assigned to us when we asked for places to stay.

i like how im referred to as "the girl" :-p

and why is it surprising that im the only girl on my team staying at a dorm?

I thought he said "the girl" was staying at a sorority?   :maybe:

hotel

no icky dorms

btw, do they give us breakfast at the competition? i mean we do have to get up pretty early  :wink:

I wilbe in a hotel room, but I might just show up in one of your dorms saturday after the competition...

The schedule is [url=http://web.mit.edu/hmmt/www/datafiles/schedules/2007.shtml]here[/url].


[quote="2424"][quote="arqady"][quote="Pain rinnegan"]Show that for all non-negative $ a$, $ b$, and $ c$, we have

$ 13\left(a^4 \plus{} b^4 \plus{} c^4\right) \plus{} \left(a \plus{} b \plus{} c\right)^4 \ge 20\left(a^3b \plus{} a^3c \plus{} b^3c \plus{} b^3a \plus{} c^3a \plus{} c^3b\right)$[/quote]
Easy SOS. :([/quote]
I can prove this ineq by the Shur ineq.[/quote]

SOS can slove it :P
Inq
$ <\equal{}>14 \sum a^4\plus{}12abc(a\plus{}b\plus{}c)\plus{}6\sum a^2b^2 \ge 16 \sum ab(a^2\plus{}b^2)$
$ <\equal{}>\sum (a\minus{}b)^2[(a\minus{}b)^2\plus{}6(a^2\plus{}b^2\minus{}c^2)] \ge 0$(*)
$ VT(*) \ge \sum (a^2\plus{}b^2\minus{}c^2)(a\minus{}b)^2 \ge 0$
Done!
@2424[hide]Anh có phải tên Vượng ko ạ..Cho em xin nik chat em hỏi cái này :D[/hide]

[quote="caubetoanhoc94"][/quote][quote="2424"][quote="arqady"][quote="Pain rinnegan"]Show that for all non-negative $ a$, $ b$, and $ c$, we have

$ 13\left(a^4 + b^4 + c^4\right) + \left(a + b + c\right)^4 \ge 20\left(a^3b + a^3c + b^3c + b^3a + c^3a + c^3b\right)$[/quote]
Easy SOS. :([/quote]
I can prove this ineq by the Shur ineq.[/quote][quote="caubetoanhoc94"]

SOS can slove it :P
Inq
$ < = > 14 \sum a^4 + 12abc(a + b + c) + 6\sum a^2b^2 \ge 16 \sum ab(a^2 + b^2)$
$ < = > \sum (a - b)^2[(a - b)^2 + 6(a^2 + b^2 - c^2)] \ge 0$(*)
$ VT(*) \ge \sum (a^2 + b^2 - c^2)(a - b)^2 \ge 0$
Done!
@2424[hide]Anh có phải tên Vượng ko ạ..Cho em xin nik chat em hỏi cái này :D[/hide][/quote]
$ (*) < = > (a - b)^4 + (b - c)^4 + (c - a)^4 + 12(a^4 + b^4 + c^4 + 12abc - a^3(b + c) - b^3(c + a) - c^3(a + b)) \geq 0$! :lol:
@caubetoanhoc94[hide]uh.sao e bit a?nic chat cua a la vuakhtn.co j cu hoi,a san sang giai dap neu nam trong kha nang cua a.[/hide]


1. Certain list of 100 data has an average of b and a standard deviation of d, where d is positive.
Which of the following pairs of data, when added to the list, must result in a list of data with 102 elements with SD less than d.

a  -6   0
b   0   0
c   0   6
d   0   12
e   6   6

2. What does x>0=b mean? That x > 0 and x = b, or that x>0 and 0=b ?

3. On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between? (Im not sure what is 11.5 rounded to?)

4. If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t?

a  2
b  4
c  8
d  20
e  45

Please explain.

[quote="dear_xavier"]

2. What does x>0=b mean? That x > 0 and x = b, or that x>0 and 0=b ?

[/quote]

I believe $ x > 0 \equal{} b$ will mean $ x > 0$, so $ x$ is more than 0.  $ 0 \equal{} b$ means that $ b$ is simply 0.

[hide="4"]
You have $ \frac{s}{t}\equal{}64.12$
$ \longrightarrow s\equal{}64.12t$
And you have $ s\equal{}64t\plus{}r ($ $ r:$ remainder$ )$
$ \longrightarrow 64.12t\equal{}64t\plus{}r$
$ \longrightarrow \frac{3}{25}t \equal{}r$
Try $ r\equal{}2,4,8,20,45$. Get $ r\equal{}45$ because for $ r\equal{}2,4,8,20$ you get non-integer $ t$.
[/hide]

2 i think x>0=b means x>0 and b=0 and x>b

3. On a recent trip, Cindy drove her car 290 miles, rounded to the nearest 10 miles, and used 12 gallons of gasoline, rounded to the nearest gallon. The actual number of miles per gallon that Cindy's car got on this trip must have been between? (Im not sure what is 11.5 rounded to?)

For the miles to be rounded to the nearest 10 miles and be rounded to 290, her miles are between $ 85<x<95$. He gallons are between $ 4.4<y<5.4$. Do you get this. So the miles per gallon would be between $ 85/5.4$ (least m/p she could have gotten) and $ 95/4.4$ (greatest miles per gallon) which gets you that her miles per gallon is about: $ 15.74<x<21.59$


trigonometry
inequalities proposed
inequalities

Let F be the Fermat's point of $\triangle ABC$, where $\max\{A,B,C\}<2\pi/3$. Prove that \[\sin^2({\angle ABF+\angle ACF})(\frac{1}{\sin^2\angle ABF}+\frac{1}{\sin^2\angle ACF})+\sin^2({\angle BAF+\angle BCF})(\frac{1}{\sin^2\angle BAF}+\frac{1}{\sin^2\angle BCF})+\sin^2({\angle CAF+\angle CBF})({\frac{1}{\sin^2\angle CAF}+\frac{1}{\sin^2\angle CBF})\ge18}\]

[quote="warut_suk"]I have no idea. Can you post the solution please?[/quote]
Guy, this has been posted for 5 days only. If before the end the September there is still no solution, then I will post mine.

No one wanna give a try? :?
It's not really that hard ;)


For questions like these, can you just get rid of the squares?

[quote="blahblahblah"]It has a root of multiplicity 4 at $x=1$, and the rest of the stuff is not hard to factor.[/quote]
It isn't?! ${\frac{1}{72}}\, \left( {z}^{8}+4\,{z}^{7}+10\,{z}^{6}+4\,{z}^{5}-2\, {z}^{4}+4\,{z}^{3}+10\,{z}^{2}+4\,z+1 \right) \left( z-1 \right) ^{4}$ ?

Never mind; I thought there was a squared sign on the LHS. Its actually a cubed.

Yes, but how does that help?

Kalle, I did it with a computer algebra system, but the fact that the roots are at $z=1$ is not hard to see, so you can continue differentiating, and the remaining term is easily seen to be $\geq 0$.

I believe that this lemma was used before to prove a very difficult inequality.

[quote="blahblahblah"]I believe that this lemma was used before to prove a (very) difficult inequality.[/quote]
Indeed,thanks for interest ,proposer

For a similar inequality, see my solution to problem 6 of the inequalities posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=kiran+kedlaya&t=43486]here[/url]

[quote="zanttrang"][quote="blahblahblah"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=13183]This one[/url][/quote]
Interesting solution - out of curiosity, how did you guys actually do the factoring out of the $(x - 1)^4$ term?  It seems like it would be extremely tedious to do by hand.[/quote]
Do you mean expanding it?  If so, binomial theorem works. Hope that was what you were lookign for.

[quote="Altheman"]$z^8+4z^7+10z^6+4z^5-2z^4+4z^3+10z^2+{\mathbf 4z}+1 >0$  [/quote]

[hide]To solve the posted question it's sufficient to [u][b]justify the inequality  for[/b][/u] $z\ge 0 .$ Try to show that
\[ \begin{array}{c} \displaystyle z^8+4z^7+10z^6+4z^5-2z^4+4z^3+10z^2+4z^2+1 = \\ \\ =\displaystyle z^8+4z^7+10z^6+3z^5+z^3(z-1)^2 +3z^3+10z^2+4z+12z^4+4z^3+10z^2+{\mathbf 4z}+1 \ge 0 \end{array}  \]
[b][u]for [/u] $z\ge 0$ . [/b]
[/hide]

[quote="Altheman"]well, you could just use synthetic division on $x=1$ 4 times and stack your divisions, but how did you show that

$z^8+4z^7+10z^6+4z^5-2z^4+4z^3+10z^2+4z^2+1 >0$ $\forall z \in \mathbb{R}$?

is this just an easy application of the trivial inequality? i tried doing upper and lower bounds, but that did not really work[/quote]

We need only consider $z\in [0,\infty)$.  To deal with the negative term, $z^8+1\geq 2z^4$, and the rest of the terms are positive.


AIME
AIME I
AIME II
USA
Sets
modular arithmetic

A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer.  Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is the greatet number of elements that $\mathcal{S}$ can have?

[hide="Solution"]Call the ordered sequence of the members of $S$, ascending, $a_1, a_2, a_3 ... a_n$, where $n$ is the size of the set.  Denote $N = a_1 + a_2 + ... + a_n$.  Then $(n-1) | (N - a_k)$ for any $1 \le k \le n$.  The problem gives us $a_1 = 1, a_n = 2002$, so because you know that $N \equiv a_k \bmod (n-1)$ for every $a_k \in S$, you also know that $a_i - a_j \equiv 0 \bmod (n-1)$ for any $a_i, a_j \in S$.  This gives us $2001 \equiv 0 \bmod (n-1)$.  We prime-factorize $2001 = 3 \times 23 \times 29$, so our only possible values for $n$ are $n = 2, 4, 24, 30, 70, 88, 668, 2002$.

Note that $a_2 - a_1 \equiv 0 \bmod (n-1)$.  This gives us a method of finding a generic form for every member of $S$; we rewrite $a_2 \equiv a_1 \bmod (n-1)$, and we will say that $a_2 = a_1 + d_1 (n-1) = 1 + d_1 (n-1)$, for $d_k$ an integer.  We can repeat this procedure for every member of $S$ until we arrive at $a_n = 2002 = 1 + \left( \displaystyle\sum_{k=1}^{n-1} d_k \right) (n-1)$.  Because each $d_k \ge 1$ (to avoid adding zero, which would produce non-distinct members of $S$), we know that $D = \displaystyle\sum_{k=1}^{n-1} d_k \ge (n-1)$.  We come to the conclusion that $D = \displaystyle\frac{2001}{n-1}$, and the maximum value of $n$ that allows $D \ge n-1$ is $n = \boxed{030}, D = 69$.
[/hide]

[quote="t0rajir0u"][hide="Solution...?"]Call the ordered sequence of the members of $S$, ascending, $a_1, a_2, a_3 ... a_n$, where $n$ is the size of the set.  Denote $N = a_1 + a_2 + ... + a_n$.  Then $(n-1) | (N - a_k)$ for any $1 \le k \le n$.  The problem gives us $a_1 = 1, a_n = 2002$, so because you know that $N \equiv a_k \bmod (n-1)$ for every $a_k \in S$, you also know that $a_i - a_j \equiv 0 \bmod (n-1)$ for any $a_i, a_j \in S$.  This gives us $2001 \equiv 0 \bmod (n-1)$.  We prime-factorize $2001 = 3 \times 23 \times 29$, so our only possible values for $n$ are $n = 2, 4, 24, 30, 70, 88, 668, 2002$.

[/hide]

Edit:  Answer sheet gives a different answer, I'll work on it.[/quote]

Well, if you think about it, all elements satisfy $N \equiv a_k \mod{(n-1)}$ so if you have $n$ elements, the largest element has to be at least $n(n-1)$ (to create $n$ distinct elements with the same residue $\mod{(n-1)}$). So $n(n-1) < 2002 \Rightarrow n = 30$ is the largest one on your list that works.

[quote="t0rajir0u"][hide="Solution"]Call the ordered sequence of the members of $ S$, ascending, $ a_1, a_2, a_3 ... a_n$, where $ n$ is the size of the set.  Denote $ N \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$.  Then $ (n \minus{} 1) | (N \minus{} a_k)$ for any $ 1 \le k \le n$.  The problem gives us $ a_1 \equal{} 1, a_n \equal{} 2002$, so because you know that $ N \equiv a_k \bmod (n \minus{} 1)$ for every $ a_k \in S$, you also know that $ a_i \minus{} a_j \equiv 0 \bmod (n \minus{} 1)$ for any $ a_i, a_j \in S$.  This gives us $ 2001 \equiv 0 \bmod (n \minus{} 1)$.  We prime-factorize $ 2001 \equal{} 3 \times 23 \times 29$, so our only possible values for $ n$ are $ n \equal{} 2, 4, 24, 30, 70, 88, 668, 2002$.

Note that $ a_2 \minus{} a_1 \equiv 0 \bmod (n \minus{} 1)$.  This gives us a method of finding a generic form for every member of $ S$; we rewrite $ a_2 \equiv a_1 \bmod (n \minus{} 1)$, and we will say that $ a_2 \equal{} a_1 \plus{} d_1 (n \minus{} 1) \equal{} 1 \plus{} d_1 (n \minus{} 1)$, for $ d_k$ an integer.  We can repeat this procedure for every member of $ S$ until we arrive at $ a_n \equal{} 2002 \equal{} 1 \plus{} \left( \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} d_k \right) (n \minus{} 1)$.  Because each $ d_k \ge 1$ (to avoid adding zero, which would produce non-distinct members of $ S$), we know that $ D \equal{} \displaystyle\sum_{k \equal{} 1}^{n \minus{} 1} d_k \ge (n \minus{} 1)$.  We come to the conclusion that $ D \equal{} \displaystyle\frac {2001}{n \minus{} 1}$, and the maximum value of $ n$ that allows $ D \ge n \minus{} 1$ is $ n \equal{} \boxed{030}, D \equal{} 69$.
[/hide][/quote]

But how can we get $ a_2 \equal{} a_1 \plus{} d_1 (n \minus{} 1) \equal{} 1 \plus{} d_1 (n \minus{} 1)$ from $ a_2 \minus{} a_1 \equiv 0 \bmod (n \minus{} 1)$?
I don't get the last part of $ a_2 \equal{} a_1 \plus{} d_1 (n \minus{} 1) \equal{} 1 \plus{} d_1 (n \minus{} 1)$

We are given in the problem that $ a_1 \equal{} 1$.

I'm reviving only because I have a different solution.

[hide]Let S be the sum of the elements in S, and let n be the number of elements in S.
Let A=n-1
Let the elements in S be denoted by a_k.

S-1=0 (mod A) from the given
since 1 is the least term, a_k=1+j
S-a_k=0 (mod A)
S-1=A-j (mod A)
So A-j=0 (mod A)
So j=0 (mod A)
So j=mA for some constant m
Let a_k= 1 + b_k*A

S= {1, 1+b_1*A, 1+b_2*A, ..., 1+ b_(n-1)*A}, for b_1<b_2<...<b_(n-1)

Since the largest term is 2002, b_(n-1)*A=2001=3*23*29
Since b_1<b_2<...<b_(n-1), b_n-1>=A
Clearly, b_k & A are integers
So the greatest A can be is 29.
So n=29+1=[b]30[/b][/hide]

Here's a quick way to do this problem:

[hide]
Let $S=\{a_1,...,a_n\}$ with $1=a_1\le...\le a_n=2002$. Define $X=a_1+...+a_n$ and $f_i=\frac{X-a_i}{n-1}$. The given condition is that $f_i$ is an integer for $1\le i\le n$. Specifically, $n-1|X-1$ and $n-1|X-2002$, so $n-1|2001$. Since all the elements of $S$ are distinct, the $f_i$'s are also distinct; if we let $X-2002=k(n-1)$ and $X-1=r(n-1)$, then $r$ is at least $k+n-1$. Eliminating $X$ in these equations, we get $2001=(n-1)(r-k) \to 2001 \ge (n-1)(n-1)=(n-1)^2$, so $n\le45$. Noting that $2001=3*23*29$,  the largest divisor of $2001$ that is less than $45$ is $29$, which gives $n=30$ as our answer. It is easy to construct a set $S$ which works. [/hide]

Say there are $k$ elements with sum $p$. Then the means upon deleting $1,2002$ are both integers, so $\frac{p-1}{k-1}, \frac{p-2002}{k-1}$ are both integers. This means $p - 1 \equiv 0 \pmod {k-1}, p-2002 \equiv 0 \pmod {k-1}$. Then $p \equiv 1 \pmod {k-1}$ and $p \equiv 2002 \pmod {k-1}$, or $1 \equiv 2002 \pmod {k-1} \implies 2001 \equiv 0 \pmod {k-1}$. Now note that if $p-a \equiv 0 \pmod {k-1}$, then $p \equiv a \pmod {k-1}$, or $a \equiv 1 \pmod {k-1}$. Therefore, there needs to be at least $k$ values of $a$ satisfying this under $2002$ and above $1$. After $29 \implies k = 30$ which clearly works, the next largest factor of $2001$ is $87 \implies k = 88$, which clearly does not work. Hence, the answer is $\boxed{30}$.

[b]Solution:[/b] Let $X$ be the total sum of all elements of $S$ and $(E+1)$ be the number of elements in $S$. Then, we have that $X \equiv 2002 \pmod E$ and $X \equiv 1 \pmod E$ as well. This implies that $E|2001$, or in other words that $E$ is a factor of $2001$. Notice that, however, $\frac{2001}{E} > E$ by definition. This implies that $E < 45$, so the largest possible number of elements is $E = 29 \Rightarrow 29+1 = \boxed{030}$.


calculus integration LaTeX function trigonometry logarithms algebra

Calculate:

1. $ \int\frac {cos^2x }{ (1 \plus{} sinxcosx)^2 }dx$

2. $ \int\frac {x^2 \plus{} 1 }{ x^4 \plus{} 1 }dx$

3. $ \int\frac {\sqrt {sinx} }{ 1 \plus{} sin^2x }dx$

4. $ \int\frac {x \plus{} 1 }{ (x^2 \plus{} 1)^2\sqrt {x^2 \plus{} 2} }dx$

5. $ \int\frac {\sqrt {x^2 \plus{} 2x \minus{} 3} }{ (x \plus{} 1)^2(x \plus{} 3) }dx$

6. $ \int\frac {\sqrt {{sin}^{3}2x} }{ {sin}^{5}x}dx$

7. $ \int arcsin^{2}x dx$

8. $ \int\frac {2tgx \plus{} 3 }{ sin^{2}x \plus{} 2cos^{2}x }dx$

9. $ \int\frac {cosx \plus{} sinx }{\sqrt {sin2x} }dx$

P.S. I'm sorry for adding new integrals. Here are posted some of undefined integrals there are 3-10 more undefined integrals given at the olympiad and on the prerequisite competition. I posted the integrals "as is" in the book I have.

[hide="2"]

Since $ x^{4} \plus{} 1 \equal{} (x^{2} \plus{} \sqrt {2}x \plus{} 1)(x^{2} \minus{} \sqrt {2}x \plus{} 1)$, it seems we could use partial fractions and then complete the square of each denominatior to get arctans.

[/hide]

I can't read hidden text on my computer but by clicking Quote I can read the Latex and it seems that kabi has done the same as I suggested so I'll stop for a while.

[quote="borislav_mirchev"]kabi, are you sure about integral 5?[/quote]
Sorry i just read it wrong .Editing it in a min

$ \int\frac{cosx\plus{}sinx }{\sqrt{sin2x}}dx$
=$ \int \frac{cosx\plus{}sinx }{\sqrt{1\minus{}(1\minus{}sin2x)}}dx$
=$ \int \frac{cosx\plus{}sinx }{\sqrt{1\minus{}(sin(x)\minus{}cos(x))^{2}}}dx$
put $ sin(x)\minus{}cos(x)\equal{}t$

Borislav and kabi: your posts would look better if you would use the proper $ \text{\LaTeX}$ code for named functions like $ \sin x$ and $ \tan x.$ Put a backslash before the function name and then a space between the end of the name and the next letter you type - for instance, \sin x. Hence

$ \int\frac {\cos x + \sin x}{\sqrt {\sin 2x}}\,dx = ,$ etc.

Thank you for the recommendation. I'll use proper functions in the future. I knew the proper way but I'm too lazy.

(5) Let $ \sqrt{\frac{x\minus{}1}{x\plus{}3}}\equal{}y$

$ \int \frac{\sqrt{x^{2}\plus{}2x\minus{}3}}{(x\plus{}1)^{2}(x\plus{}3)}dx$

$ \equal{}\int \sqrt{\frac{x\minus{}1}{x\plus{}3}}\frac{1}{\left( x\plus{}1\right) ^{2}}dx$

$ \equal{}\allowbreak \int \,\frac{2y^{2}}{\left( 1\plus{}y^{2}\right) ^{2}}dy$

>$ \equal{}\allowbreak \tan ^{\minus{}1}y\minus{}\frac{y}{1\plus{}y^{2}}\plus{}C$

$ \equal{}\allowbreak \tan ^{\minus{}1}\sqrt{\frac{x\minus{}1}{x\plus{}3}}\minus{}\frac{\sqrt{\left( x\plus{}3\right) \left( x\minus{}1\right) }}{2\left( x\plus{}1\right) }\plus{}C$

7)

$ \int \left( \sin ^{\minus{}1}x\right) ^{2}dx$

$ \equal{}x\left( \sin ^{\minus{}1}x\right) ^{2}\minus{}\int \frac{2x\sin ^{\minus{}1}x}{\sqrt{1\minus{}x^{2}}}\,dx$

$ \equal{}\allowbreak x\left( \sin ^{\minus{}1}x\right) ^{2}\plus{}2\left( \sin ^{\minus{}1}x\right) \sqrt{\left( 1\minus{}x^{2}\right) }\minus{}2x\plus{}C$

8) Let $ \tan x\equal{}u$

$ \int \frac{2\tan x\plus{}3}{\sin ^{2}x\plus{}2\cos ^{2}x}dx$

$ \equal{}\allowbreak \int \frac{2u\plus{}3}{u^{2}\plus{}2}du$

$ \equal{}\allowbreak \ln \left( \tan ^{2}x\plus{}2\right) \plus{}\frac{3}{\sqrt{2}}\arctan \frac{\tan x}{\sqrt{2}}\plus{}C$

1) Let $ \tan x\equal{}u$

$ \int \frac{\cos ^{2}x}{\left( 1\plus{}\sin x\cos x\right) ^{2}}dx$

$ \equal{}\allowbreak \int \frac{1}{\left( 1\plus{}t^{2}\plus{}t\right) ^{2}}dt$

$ \equal{}\allowbreak \int \left[ \frac{2}{3\left( 1\plus{}t\plus{}t^{2}\right) }\plus{}\frac{1}{3}\frac{1\minus{}2t\minus{}2t^{2}}{\left( 1\plus{}t\plus{}t^{2}\right) ^{2}}\right] dt$

$ \equal{}\allowbreak \frac{2t\plus{}1}{3\left( 1\plus{}t\plus{}t^{2}\right) }\plus{}\frac{4}{3\sqrt{3}}\arctan \frac{2t\plus{}1}{\sqrt{3}}\plus{}C$

$ \equal{}\frac{2\tan x\plus{}1}{3\left( 1\plus{}\tan x\plus{}\tan ^{2}x\right) }\plus{}\frac{4}{3\sqrt{3}}\arctan \frac{2\tan x\plus{}1}{\sqrt{3}}\plus{}C$

4) Rewritting the integrand,

$ \int \frac{x\plus{}1}{(x^{2}\plus{}1)^{2}\sqrt{x^{2}\plus{}2}}dx$

$ \int \left( \minus{}\frac{x}{2\left( \sqrt{x^{2}\plus{}2}\plus{}1\right) \left( \sqrt{x^{2}\plus{}2}\minus{}1\right) \sqrt{x^{2}\plus{}2}}\allowbreak \plus{}\frac{1}{\left( x^{2}\plus{}1\right) ^{2}\sqrt{x^{2}\plus{}2}}\plus{}\frac{x\left( x^{2}\plus{}3\right) }{2\left( x^{2}\plus{}1\right) ^{2}\sqrt{x^{2}\plus{}2}}\right) dx$

$ \equal{}\allowbreak \frac{1}{4}\ln \frac{\sqrt{x^{2}\plus{}2}\plus{}1}{\sqrt{x^{2}\plus{}2}\minus{}1}\plus{}\frac{x\sqrt{x^{2}\plus{}2}}{2\left( x^{2}\plus{}1\right) }\minus{}\frac{\sqrt{x^{2}\plus{}2}}{2\left( x^{2}\plus{}1\right) }\plus{}C$

hello for 1) rewriting your integrand in the form
$ \frac{1}{\frac{1}{\cos(x)^2}+\sin(x)^2+\frac{2\sin(x)}{\cos(x)}}$
and using the formulas
$ \sin(x)^2=\frac{\tan(x)^2}{1+\tan(x)^2}$
$ \cos(x)^2=\frac{1}{1+\tan(x)^2}$
and setting $ t=\tan(x)$
we get the integral
$ \int\frac{1}{t^4+2t^3+3t^2+2t+1}\,dt$
which is equal to
${ \frac{1}{3}\frac{2t+1}{t^2+t+1}+\frac{4}{9}\sqrt{3}\arctan(\frac{\sqrt{3}}{3}(2t+1)})+C$
Sonnhard.

hello 3) leads to an elliptic integral.
Sonnhard.

3) $ \int\frac {\sqrt {sinx}}{ 1 + sin^{2}x }dx$

multiplying and dividing by $ 2 \sqrt {\sin x }$  we have  $ 2 \int\frac {\sin x }{( 2 \sqrt {sinx} ) ( 1 + \sin^{2}x ) }dx$

put $ \sqrt {sinx}  = t , \frac {1}{ 2\sqrt {sinx} } \cos x dx = dt$

therefore the integral becomes $ 2 \int \frac {t }{ ( t^4 + 1)(\sqrt {1 - t^4  )
}  dt$

put $ t = \frac {1}{y}$

the integral becomes  $ - 2 \int \frac {y^3 }{ ( y^4 + 1)(\sqrt {y^4 - 1 } ) } dy$

put $ \sqrt {y^4 - 1} = x$ and proceed
.


floor function
number theory proposed
number theory

Let be $ a\geq5$ an odd given natural number. Solve in $ \mathbb{Q}$ :

$ x^{\left\lfloor x\right\rfloor } \equal{} \frac {a}{2}$

$ x\equal{}\sqrt[n]{\frac{a}{2}}>n\equal{}[x].$ But for $ a\equal{}5$ no positive solutions.

$ x \equal{} \frac { \minus{} 2}{a}$ is unique negative solution.

If $ 0\le x\le 2$, then contradition.
Therefore positive solution exist only if $ n^n < \frac a2 < (n \plus{} 1)^n, 2\le n\in N$.

x = -2/a can't be a solution because a > 0

In fact, there isn't solution for $ a \geq 5$

[hide]See $ v_2$[/hide]

Suppose that $ x\in\mathbb{Q}$. We have :

$ v_{2}\left(x^{\left\lfloor x\right\rfloor }\right) \equal{} \left\lfloor x\right\rfloor v_{2}(x)$
$ v_{2}\left(\frac {a}{2}\right) \equal{} \minus{} 1$

Because $ x^{\left\lfloor x\right\rfloor} \equal{} \frac {a}{2}$, we have : $ \left\lfloor x\right\rfloor v_{2}(x) \equal{} \minus{} 1$, so $ \left\lfloor x\right\rfloor \equal{} \pm 1$

If $ \left\lfloor x\right\rfloor \equal{} 1$, then $ 1\leq x < 2$, so $ x^{\left\lfloor x\right\rfloor} < 2$, which is impossible because $ \frac{a}{2}>2$
If $ \left\lfloor x\right\rfloor \equal{} \minus{} 1$, then $ x < 0$, which is also impossible because $ a > 0$.

Indeed, there is no solutions in $ \mathbb{Q}$.


geometry
3D geometry
sphere
radical axis
cyclic quadrilateral
geometry unsolved

[quote="Klaas"]Six points A,B,C,D,E,F lies on one surface.Points: A,B,C,D; C,D,E,F;  E,F,A,B lies on one circle (three different circles). Prove, that points A,B,C,D,E,F lies on one circle.[/quote]

What do you mean by "surface"? Surely not a plane, since this would make the result wrong... I guess your problem anyway follows from the radical center theorem, as in http://www.mathlinks.ro/Forum/viewtopic.php?t=64923 .

  darij

I think that this problem has a meaning (because states about three different circles and also not about the kind of the quqdrilaterals), if the wanted conclusion is that the six points A, B, C, D, E, F, lie on a sphere, instead of a circle. But, if I am not mistaken, we can easy to prove a such that result (If the six points lie on the same plane, then the sphere has it's center at infinity).

Kostas Vittas.

[quote="darij grinberg"]So, any progress on understanding what the problem is supposed to mean?
darij[/quote]

Well, I don’t remember what I thought five months ago, but now I see that we can to define six points $A,$ $B,$ $C,$ $D,$ $E,$ $F,$ on a given sphere, such that the quadrilaterals $ABCD,$ $CDEF,$ $EFAB,$ to be cyclic and not on the same plane.

Let $ABCD$ be a cyclic quadrilateral, as the intersection shape of a given sphere from an arbitrary plane [[b][size=109]p[/size][/b]].

We denote as $S,$ the intersection point of the sidelines $AB,$ $CD$ $($ in my drawing $B,$ between $A,$ $S$ $).$

Through the point $S,$ we draw an arbitrary line which intersects the given sphere at two points, so be it $E,$ $F$ $($ $F,$ between $E,$ $S$ $).$

It is easy to show that the quadrilaterals $CDEF$ and $EFAB$ ( non-convex ), are cyclic as the intersection shapes of the given sphere from the planes [$SDE$] and [$SAE$], respectively.

We have now, the configuration of six points lie on a surface ( sphere ), as the problem states ( three different circles ), but not necessary on the same plane and hence, not on the same circle.

So, I don’t understand what the problem is supposed to mean, different than a sphere, as the wanted conclusion.

Kostas Vittas.

[quote="Davidd"]Work backwards, and assume all [b]are[/b] in the [u]same[/u] circle. Then you can not have three [u]distinct[/u] circles as described at the start, since three points define a unique circle.  If a fourth [or more] is on that circle, it can not be on another distinct from it.[/quote]

You are right about the circle as the wanted conclusion (it is impossible). But the problem states of course about three different circles, as for the cyclic quadrilaterals.

I think that Klaas knows that it is impossible to have three different circles on the same surface (plane or other) and simultaneously, the six points to be lie on the same circle.

So, we are searching what exactly may be supposed to mean, this problem.

I think that if we rewrite the last word of the notations, as ''sphere'' instead of ''circle'', it’s OK.

Kostas Vittas.


calculus integration function limit trigonometry derivative Gauss

how would i approx a integral of the form   (sin x)/square root x?

Oh and by the way the interval is from 0 to 1.
 It doesnt make sense since square root of 0  is not defined.

It does make sense, since the function is still limited in x=0 (the limit is zero). Since it also is continuous the function is Riemann-integrable in $(0,1]$. However, since a point has zero measure you can just add whatever value to the function in zero and integrate over the compact interval $[0,1]$, and the integral will remain the same. The most reasonable choice is to define the function as zero in the point x=0. I hope that answers your question.

edit: I hope you meant that division with $\sqrt{0}$ is not defined, because $\sqrt{0}$ is most definitely defined :)

[quote="Kent Merryfield"]But $\lim_{x\to\0}\frac{\sin x}{\sqrt{x}}=0.$

Think of this as the integral of the function $f(x)=\begin{cases}\frac{\sin x}{\sqrt{x}}&x>0\\ 0&x=0\end{cases}.$

$f$ is continuous on $[0,1].$ Use $0$ for the value at $x=0.$

On the other hand: your textbook probably has an error estimate for the error in the trapezoidal rule that involves a worst case estimate of $|f''(x)|.$ That's a problem in this case, as this $f(x)$ does not have a bounded second derivative - in fact, it doesn't have a bounded first derivative. Oh, the trapezoidal rule will converge anyway, but slower than the rate suggested by the theorem in question.

One way out of that: make a substitution in the integral. Let $u=\sqrt{x}$ so that $x=u^2$ and $dx=2u\,du.$

$\int_0^1\frac{\sin x}{\sqrt{x}}\,dx=2\int_0^1\sin(u^2)\,du.$

Now we have a $C^{\infty}$ integrand. Apply the trapzoidal rule, or the midpoint rule, for a method with an error proportional to the square of the step size, or Simpson's rule for an error proportional to the fourth power of the step size.[/quote]

But then when would I substitute back in the problem. After I have done the trapezoidal rule on it?

[quote="Kent Merryfield"]There is no "substitute back." $\int_a^bf(x)\,dx$ is a single real number. If we make a change of variables in that integral, it will still be that real number. If we approximate that integral by a numerical method, we'll have an approximation of that real number.

The specific approximation we get from a certain number of steps in the $du$ version will be different from (probably better than) the approximation from the same number of steps in the $dx$ version, but they are both approximations of the same number.[/quote]

Got it thanks!!!
the answer was .6277.

Now I need to compute it with gauss's rule but dont what the formula is.
Do happen to have the formula for gauss integration?
thanks

Do you mean "Gaussian quadrature?"

The "formula" is pretty complicated.

First, find a set of polynomials that are orthogonal on the interval of integration. Either that, or change variables to force the integral to be on your favorite interval. Let's assume that this in an integral over $[-1,1].$

The orthogonal polynomials are the Legendre polynomials. One way to define them:

$P(n)(x)=c_n\left(\frac{d}{dx}\right)^n(x^2-1)^n.$

(There are various conventions for the normalizing constant $c_n;$ we don't need to bother because all we care about are the zeros of $P_n$).

We can prove that $P_n(x)$ has $n$ distinct real zeros in the interval $(0,1).$ Let these zeros be $x_1,x_a,\dots,x_n.$ Find coeffients $a_j$ that satisfy the following system of equations:

For each $k,\,0\le k\le n-1,$ such that $\sum_{j=1}^na_jx_j^k=\int_{-1}^1x^k\,dx.$

[Note that these numbers - the $x_j$ and the $a_j$ - can be computed once and for all time and stored as a table of values.]

Then we approximate $\int_{-1}^1f(x)\,dx$ by $\sum_{j=1}^na_jf(x_j).$

To do this dynamically, we compute this for an increasing sequence of values of $n,$ watching to see how much the result changes.

The $n=2$ case would be $\int_{-1}^1f(x)\,dx\approx f\left(\frac{-1}{\sqrt{3}}\right)+ f\left(\frac1{\sqrt{3}}\right)$

The $n=3$ case would be $\int_{-1}^1f(x)\,dx\approx \frac89f(0)+\frac59\left( f\left(-\sqrt{\frac35}\right)+ f\left(\sqrt{\frac35}\right)\right)$

[quote="hpe"]Perhaps a better way to compute $\int_0^1 \sin(x^2) dx$ numerically consists in expanding the integrand into a power series and integrating term by term. Since $\sin(x^2) = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+2}}{(2n+1)!}$, one obtains
$\int_0^1 \sin(x^2) dx = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!(4n+3)} = \frac{1}{3} - \frac{1}{42}+\frac{1}{1320} - + \cdot$.
It's an alternating series, so there is an immediate error estimate. For example, the three terms given above are accurate to about $2 \cdot 10^{-5}$.[/quote]

Is there a set formula to get the power series?
I am having trouble seeing how you got that result.

Look here http://mathworld.wolfram.com/TaylorSeries.html .

[quote="hpe"][quote="numero1"]Is there a set formula to get the power series?
I am having trouble seeing how you got that result.[/quote]
The Taylor series for the sine is $\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$ for all $x$, hence $\sin(x^2) = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+2}}{(2n+1)!}$ and therefore $\int \sin(x^2)dx = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+3}}{(4n+3)(2n+1)!} + C$. The series converges uniformly for all $x$. Evaluate the primitive between $x=0$ and $x=1$, and you'll get the result.
Hope this helps :)[/quote]

Thanks but the question was to evaluate the integral of sin x/square root x.
How would i apply the power series method to that one?

thanks

[quote="numero1"]Thanks but the question was to evaluate the integral of sin x/square root x.
How would i apply the power series method to that one? [/quote]
1. That [b]is[/b] the same integral, except for a factor of 2. See post #3 in this thread.

2. Or you could do this:

$\int_0^1\frac{\sin x}{\sqrt x}\,dx=\int_0^1\frac1{\sqrt{x}}\left(x-\frac{x^3}{3!} +\frac{x^5}{5!}+\cdots\right)\,dx$

$=\int_0^1\left(x^{-1/2}-\frac16x^{5/2}+\frac1{120}x^{9/2}+\cdots\right)\,dx$

$=\left.\left(2x^{1/2}-\frac27\cdot\frac16x^{7/2}+ \frac2{11}\cdot\frac1{120}x^{11/2} +\cdots\right)\right|_0^1$

$=2-\frac2{42}+\frac2{11\cdot 120}-\frac2{15\cdot 7!}+\cdots$

Except for the easily-explained factor of 2, this is the same series as the one hpe derived - and that's not a coincidence.

3. It's really kind of wasteful to quote entire posts. Please edit them down to the minimum required to make clear what it is you're responding to.

thanks.
I could go as far as I like in the terms right?
I just used the first 3 terms.

[quote="numero1"]...how approx a integral of the form $\int\limits_{0}^{1} \frac{ \sin x}{\sqrt{ x}}\; dx$ ? ...  It doesnt make sense since ...[/quote]
Let $a<b\; ,\; f\in C[a,b]$  and  $w$ be a (positive) weight on $(a,b)$, that is a function $w: (a,b)\to [0,\infty)$ such that :
i) $\int\limits_{a}^{b}w(t)t^k\; dt \; ,\; k\in {\mathbb N}$ exist as proper (or improper) integrals ,
ii)  $\int\limits_{a}^{b}w(t)\; dt > 0 .$
  For instance $w(t): (0,1)\to [0,\infty)$ with $w(t)=t^{\alpha}(1-t)^{\beta}$ is a weight if and only if $\alpha>-1 ,\; \beta>-1 .$

Denote by $L(f;\cdot )\equiv L_1(a,b; f|\cdot)$ the Lagrange  polynomial , of degree $\le 1$ , which interpolates $f$ at the "knots" $a$ and $b$ .That is
\[ \left\{\begin{array}{c} L(f;a)=f(a)\; \; \; ,\; \; \; \; L(f;b)=f(b)\\ \\ L(f;t)=\displaystyle \frac{t-b}{a-b}f(a)+\frac{t-a}{b-a}f(b)\; . \end{array}\right.  \]
We have $\; \; \;(*)\; \; \; \; f(t)\approx L(f;t)$ for $t \in [a,b]$ or ( in case of functions $g\in C^2[a,b]$ )   the exact formula
\[ g(t)=L(g;t)+\varepsilon(g;t) \; \; \; ,\; \; \; \varepsilon(g;t)=(t-a)(t-b)\frac{g^{\prime\prime}(\theta)}{2!}  \]
with $\theta=\theta(a,b)\in [a,b]\; .$ Because $(*)$ enables us to write $w(t)f(t)\approx w(t)L(f;t)\; \; ,\; \; t \in (a,b)\; ,$  I appreciate that in this case the "trapezoidal formula" is
 \[ \begin{array}{|c|} \hline \int\limits_{a}^{b}w(t)f(t)\; dt = \int\limits_{a}^{b}w(t)L(f;t)\; dt + R[f;a,b]\\ \hline \end{array}\; ,  \]
$R[f;a,b]$ being the remainder term.   In our case $\begin{array}{|c|}\hline \displaystyle 0\le a < b<\infty \; ,\; \bar{w}(t)=\displaystyle \frac{1}{\sqrt{t}} \\ \hline \end{array} \; .$ Because  \[ \begin{array}{c} \int\limits_{a}^{b}\bar{w}(t)L(f;t)\; dt =\displaystyle \frac{2(\sqrt{b}-\sqrt{a}) }{3(\sqrt{a}+\sqrt{b})}\left[ (2\sqrt{b}+\sqrt{a}) f(a) + (\sqrt{b}+2\sqrt{a}) f(b)\right] \end{array}  \]
we give the "weighted" trapezoidal formula
\[ (1)\; \; \; \begin{array}{|c|} \hline\displaystyle \int\limits_{a}^{b}f(t)\frac{dt}{\sqrt{t}}=\\ =\displaystyle \frac{2(\sqrt{b}-\sqrt{a}) }{3(\sqrt{a}+\sqrt{b})}\left[ (2\sqrt{b}+\sqrt{a}) f(a) + (\sqrt{b}+2\sqrt{a}) f(b)\right]+ r[f] \\ \hline\end{array}  \]
where $0\le a <b \; ,\; f\in C[a,b]$ , $r[f]=r[f;a,b]$ being the  remainder term. It may be shown that for $g\in C^2[a,b]$ there exists  $\xi ,\; \xi=\xi(g;a,b)\in [a,b] ,$ such that
\[ (\#)\; \; \; \; \begin{array}{|c|}\hline \\ \displaystyle r[g]=-\; \displaystyle \frac{2(\sqrt{b}-\sqrt{a})^3(b+8\sqrt{ab}-4a)}{15}g^{\prime\prime}(\xi)\\ \\ \hline \end{array}\; .  \]
Let us remark that when $(a,b) : =(0,1)$ we have  \[ (@)\;\; \int\limits_{0}^{1}\frac{h(t)}{\sqrt{t}}\; dt=\frac{2}{3}\left(2h(0)+h(1)\right)-\frac{2}{15}h^{\prime\prime}(\eta)\; \; ,\; \; h\in C^2[0,1]  \]
with $\eta=\eta(f)\in [0,1] .$  Thus we  obtain  the approximation  $(\%)\; \; \; \; \int\limits_{0}^{1}\frac{\sin{t}}{\sqrt{t}}\; dt\approx \frac{2}{3}\sin{1}$ or
\[ \int\limits_{0}^{1}\frac{h(t)}{\sqrt{t}}\; dt=\frac{2}{3}\sin{1} +\frac{2}{15}\sin{\eta}\; \; ,\; \eta \in [0,1] .  \]
However $(\#)$ show us the in case when $\sqrt{b}-\sqrt{a}$ is  large, then perhaps the approximation  is not small.
In the following let us try to "improve"  formulae $(@)-(\%)$ .  Select a positive integer $N$ and consider the points $x_k=\frac{k^2}{N^2}\in [0,1]\; ,\; k\in \{0,1,...,N^2\}.$ For $g\in C^2[0,1],$  from (1) we give
\[ \begin{array}{c} \int\limits_{x_{k-1}}^{x_k}g(t)\frac{dt}{\sqrt{t}}=\displaystyle \frac{2}{3N}\left(\frac{3k-1}{2k-1}f(x_{k-1})+ \frac{3k-2}{2k-1}f(x_k)\right) +\\ + \displaystyle \frac{2}{15N^5}\left(-3k^2+8k-4+8\sqrt{k(k-1)}\right) g^{\prime\prime}(\zeta_k)\; , \; k\in \{1,2,...,N^2\}. \end{array}  \]
Further  sum above equalities. Perhaps one find a better result... [Sorry for possible erorrs,typos]

[quote="flip2004"]
Let $a<b\; ,\; f\in C[a,b]$  ... For $g\in C^2[0,1],$  from (1) [/quote]This is all very nice for general $f$ or $g$. But here $g$ is analytic, and that fact should be exploited.


Let $ ABC$ is triangle with $ \angle BAC \equal{} 90^0$, $ AB \equal{} AC$. Let $ M, N, P$ are three point lie on $ AB, BC, CA$ respectively and $ \frac{AM}{AB} \equal{} \frac{BN}{BC} \equal{} \frac{CP}{CA}$.
Prove that $ AN \perp PM$ and $ AN \equal{} PM$.

This is easy. AM = PC, and BN / AM = \sqrt 2, so just look at ractanlge with three vertexes A, M and P, and rectangle with diagonal vertexes A and N with sides on AB and AC. This rectagnles are congruent and you get one by rotating another for $ 90 ^o$.
Bye


Let

(sinx+siny)/sin(x+y)=m
(cosx+cosy)/sin(x-y)=n

Express sin(x+y) in terms of m and n

Have fun solving it~

Is this question considered competition material? If yes, at what standard?

hello, from the addition formulas we get
$ m\equal{}\frac{2\sin(\frac{x\plus{}y}{2})\cos(\frac{x\minus{}y}{2})}{2\sin(\frac{x\plus{}y}{2})\cos(\frac{x\plus{}y}{2})}\equal{}\frac{\cos(\frac{x\minus{}y}{2})}{\cos(\frac{x\plus{}y}{2})}$.
Similiarly we have $ n\equal{}\frac{\cos(\frac{x\plus{}y}{2})}{\sin(\frac{x\plus{}y}{2}}$, from this we have $ \frac{1}{mn}\equal{}\tan(\frac{x\minus{}y}{2})$. By squaring both equations
$ \sin(x)\plus{}\sin(y)\equal{}m\sin(x\plus{}y)$
$ \cos(x)\plus{}\cos(y)\equal{}n\sin(x\minus{}y)$
and adding we get:
$ 2\plus{}2\cos(x\minus{}y)\minus{}n^2(1\minus{}\cos(x\minus{}y)^2)\equal{}m^2\sin(x\plus{}y)^2$.
Using the well-known formula
$ \cos(x\minus{}y)\equal{}\frac{1\minus{}\tan(\frac{x\minus{}y}{2})^2}{1\plus{}\tan(\frac{x\minus{}y}{2})^2}$
we get
$ \cos(x\minus{}y)\equal{}\frac{m^2n^2\minus{}1}{m^2n^2\plus{}1}$
by the formula above.
By substituting all these we have last but not least
$ \sin(x\plus{}y)\equal{}\pm\sqrt{\frac{2\minus{}n^2}{m^2}\plus{}\frac{2}{m^2}\left(\frac{m^2n^2\minus{}1}{m^2n^2\plus{}1}\right)\plus{}\frac{n^2}{m^2}\left(\frac{m^2n^2\minus{}1}{m^2n^2\plus{}1}\right)^2}$
Sonnhard.

good job doc thx~is this question considered hard?what standard would it be at in the US?competition material?

It's somewhat hard, but it won't be at a competition.

I’d like to suggest the exponentials .I’m not sure that be useful or lead to a simpler solution.
sin(x) = (exp(ix)  -  exp(-ix))/ (2i)              i^2 = -1
cos(x) = (exp(ix) + exp(-ix))/ 2

t = exp(ix)        u = exp(iy)
substituting these variables yields :
u*t^2 – u + u^2*t - t = m*(t^2*u^2 - 1 )
u*t^2 + u + u^2*t + t = n*(u^2 -  t^2)*i

sin(x+y) = Im{ t*u}


Calculati primitivele functiei $f: (0,\infty)\rightarrow R$
\[f(x)=1/(x \sqrt{x^{4}+x^{2}+1)}\]

[quote="radac_mail"]Calculati primitivele functiei $ f: (0,\infty)\rightarrow R$
\[ f(x)=1/(x \sqrt{x^{4}+x^{2}+1) }\]
[/quote]

Fie $ J=\int{\frac{xdx}{\sqrt{x^{4}+x^{2}+1}}}$ si $ I=\int{\frac{dx}{x\sqrt{x^{4}+x^{2}+1}}}$, atunci $ J-I=\int{\frac{x-\frac{1}{x}}{\sqrt{x^{4}+x^{2}+1}}dx}=\int\frac{1-\frac{1}{x^{2}}}{\sqrt{\left(x+\frac{1}{x}\right)^{2}-1}}dx$

Introducand notatia $ x+\frac{1}{x}=t$ avem $ 1-\frac{1}{x^{2}}dx=dt$ si astfel
$ J-I=\int \frac{dt}{\sqrt{t^{2}-1}}=ln(t+\sqrt{t^{2}-1})$ deci
$ I=J-ln(x+\frac{1}{x}+\sqrt{x^{2}+\frac{1}{x^{2}}+1})$
Pe de alta parte $ J=\int{\frac{xdx}{\sqrt{x^{4}+x^{2}+1}}}=\int\frac{xdx}{\sqrt{\left(x^{2}+\frac12\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}}$ si introducand notatia $ x^{2}+\frac12=t$ avem $ xdx=\frac{dt}{2}$ deci $ J=\frac12\int\frac{dt}{\sqrt{t^{2}+\frac34}}=\frac12\ln(t+\sqrt{t^{2}+\frac34})$

Deci $ I=\frac12\ln(x^{2}+\frac12+\sqrt{(x^{4}+x^{2}+1})-ln(x+\frac{1}{x}+\sqrt{x^{2}+\frac{1}{x^{2}}+1})+C$


1)In how many ways can twelve loaves of bread be given away to 5 food banks? What is the number of ways if each food bank should get at least one loaf of bread?

2)From a group of five married couples three people are to be selected. How many ways if husband and wife cannot both be selected?

[quote]1) In how many ways can twelve loaves of bread be given away to 5 food banks?
What is the number of ways if each food bank should get at least one loaf of bread?[/quote][hide]
I'll assume that the 5 food banks are distinguishable (have distinct names)
and that the loaves of bread are [u]not[/u] distinguishable.

(a) For the first loaf, you have a choice of 5 food banks to donate it to.
For the second loaf, you have a choice of 5 food banks to donate it to.
. . . and so on.

Therefore, there are $5^{12} = 244,140,625$ ways.

(b) First, give each food bank one load of bread.
Then we will distribute the other seven loaves.

For the first loaf, you have a choice of 5 food banks to donate it to . . . etc.

Therefore, there are $5^7 = 78,125$ ways.

[/hide]
[quote]2) From a group of five married couples three people are to be selected.
How many ways if husband and wife cannot both be selected?[/quote][hide]
The first person select can be any of the [b]10[/b] people (either gender).

The second must not be the spouse of the first person.
So there are [b]8[/b] choices for the second member.

The third must not be the spouse of the first or second person.
So there are [b]6[/b] choices for the third member.

It seems that there are: $10\cdot8\cdot6 = 480$ ways.

But this selection procedure includes [u]ordering[/u] the 3 members.
Since the order of their selection makes no difference, we divide by 3!

Answer: $\frac{480}{3!} = 80$ ways.

[/hide]

Sorobaon  - Check the answer (both answers) of the first question ..


How many people do you need to make a community? Are there any Wisconsinites on here besides me? If not, could there be a community for the Midwest U.S. Area? I feel left out.

of course [i]i'm[/i] from wis. I saw Tim Black on here also. :wink:
we're the only 3 i saw that had a good number of posts(25 or so)

Iris made an account a few years ago. I saw that when I was randomly looking for wisconsinites on the member list, but she hasn't done anything in the past year or so. Has Tim been posting lately? What is his screenname?

picounts

i thought i saw one of his posts but i can't be sure... i'll get right on that.

EDIT: most of his posts were in spring of '04. His last was on 4/9/06. It seemed that most of his posts were mainly about ARML, MOP, anything AMC related. I guess the 76 posts caught my eye... falsly :(

So.... looks like it's just the two of us? How about a midwest forum in that case?

hmmm... I guess that would work since the closest one I can find is in Alabama. What would the Midwesterners talk about that people from the same states/countries talk about? ARML is the only thing I can think of and there is already a thread for that. I don't know :huh: We'll see what happens.

I'm from Wisconsin...

I'm not sure that we would have much to discuss in a Wisconsin community. There are too few state-wide competitions.

I'm from KS. The current KS community forum has pretty much nobody.

This is great! Encourage more people to head on over here :D

Well, uh, it looks like this is all we have. So how about if we all say where we're from and what age group we are in. I'll start. I live in Central Wisconsin, in Mosinee. (between Wausau and Stevens Point)I am an 8th grader at Mosinee Middle School.

Hey, long time no post, but I'd be in favor of a Wisconsin community.

Alabama's forum is highly inactive.

It would be fun to have a wisconsin community muhahaha. One that I could post something in.

 :)

obviously from my screenname, i'm from Indiana, and so is fanatic, kheg_g, slamchess, usaha, stupidkid, art of owana, and jeffdshen(but he never posts anything).

I'm from Wisconsin.  I'm from Appleton, most of you should know where that is.  It's 30 miles? south of Green Bay..

Wisconsin...

community=coolness

Midwest, also cool...

If your forum is dead, you should see Upstate NY!(even though NYSML is just a few days away)

The Wisconsin Forum is need of serious life support, though.

maybe if you made it a Wisconsin/Minnesota/North Dakota forum?

 :lol:

w/e. i guess you're free to post there anyhow.


If $ xyz\equal{}1$,proov that $ \frac{xy}{xy\plus{}x\plus{}1 }\plus{}\frac{yz}{yz\plus{}y\plus{}1 }\plus{}\frac{xz}{xz\plus{}z\plus{}1 }\equal{}1$

hello, we have
$ \frac{xy}{xy\plus{}x\plus{}1}\plus{}\frac{yz}{yz\plus{}y\plus{}1}\plus{}\frac{xz}{xz\plus{}z\plus{}1}\minus{}1\equal{}$
$ {\frac { \left( 2\,yxz\plus{}xz\plus{}xy\plus{}x\plus{}1\plus{}y\plus{}z\plus{}yz \right)  \left( yxz\minus{}1 \right)
}{ \left( xy\plus{}x\plus{}1 \right)  \left( yz\plus{}y\plus{}1 \right)  \left( xz\plus{}z\plus{}1
 \right) }}\equal{}0$ because of $ xyz\equal{}1$.
Sonnhard.

Using xyz =1, we get $ \frac{xy}{xy+x+1} = \frac{1}{1+xz+z} = \frac{y}{1+y+yz}$

$ \therefore \ \frac{yz}{1+y+yz} = \frac{1}{xy+x+1}$ and $ {\frac{xz}{1+xz+z} = \frac{x}{1+x+xy}}$

Thus the given expression equals $ \frac{xy}{xy+x+1} + \frac{1}{xy+x+1} + \frac{x}{xy+x+1} = 1$

Let  $ x \equal{} \frac {a}{b},y \equal{} \frac {b}{c},z \equal{} \frac {c}{a}$

$ \frac {xy}{xy \plus{} x \plus{} 1} \equal{} \frac {\frac {a}{c}}{\frac {a}{c} \plus{} \frac {a}{b} \plus{} 1} \equal{} \frac {ab}{ab \plus{} ac \plus{} bc}$

$ \frac {yz}{yz \plus{} y \plus{} 1} \equal{} \frac {\frac {b}{a}}{\frac {b}{a} \plus{} \frac {b}{c} \plus{} 1} \equal{} \frac {bc}{ab \plus{} ac \plus{} bc}$

$ \frac {yz}{yz \plus{} y \plus{} 1} \equal{} \frac {\frac {c}{b}}{\frac {c}{b} \plus{} \frac {c}{a} \plus{} 1} \equal{} \frac {ca}{ab \plus{} ac \plus{} bc}$

$ \frac {xy}{xy \plus{} x \plus{} 1}$ + $ \frac {yz}{yz \plus{} y \plus{} 1}$ + $ \frac {yz}{yz \plus{} y \plus{} 1}$ = $ \frac {ab}{ab \plus{} ac \plus{} bc}$ + $ \frac {bc}{ab \plus{} ac \plus{} bc}$ + $ \frac {ca}{ab \plus{} ac \plus{} bc}$= $ 1$


[hide]
Rewrite it as \[S = 75  \left( \frac{1}{9\times 10}+\frac{1}{10\times 11}+\cdots+\frac{1}{99\times 100}\right)\] The sum inside the parentheses telescopes, \[S = 75 \times \frac{100-9}{900}= \frac{91}{12}\] [/hide]

Was this meant as a pun?

[hide]Anyways.

I used telescoping series and got $\frac{91}{12}$[/hide]

Use the formula of the sum of the serie 1/((n)(n+1)), be careful, because the series begins at n=9 not n=1.

[hide]I get a 6.75[/hide]

[hide]It equals $75\times(\frac{1}{9\times10}+\frac{1}{10\times11}+\cdots+\frac{1}{99\times100})$
which telescopes to $\frac{91}{12}$.[/hide]

Sorry I made a mistake in the first calculation the answer is
[code]91/12[/code]

[hide]Factor out 75

Realize that frations are in terms $\frac{75}{(n)(n+1)}$

Play around with fractions of this type, and you find...

$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+\frac{1}{(n)(n+1)}= \frac{n}{n+1}$

so therefore you can take full sequence, and subtract what you don't want:

$75(\frac{99}{100}-\frac{8}{9}) = \frac{91}{12}$

>>$\frac{91}{12}$<<
[/hide]

Wasn't a varient of this problem on a MC target once?


Let $a, b, c>0$ such that their sum is $1.$ Prove that:

                      \[ a\sqrt [3] {1+b-c} +b\sqrt [3] {1+c-a} +c\sqrt [3] {1+a-b}\leq 1. \]

Why did you open a new topic for this? It was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=31799 .

  darij

I'm soory Darij. I didn't see it. Anyway, my solution was by using Bernoulli's inequality. ;)


This question was derived from AMC-10 2003.

If the 5-digit integer m is divided by 199, we call its quotient q and remainder r. For how many of m such that q+r is divisible by 198?

 :) Hope wording is okay.

If the 5-digit integer m is divided by 199, we call its quotient q and remainder r. How many of integers such that q+r is divisible by 198?

[hide]

Hmm....

I guess you would have

1(199) + 197

2(199) + 196

3(199) + 195

So it's just 396 + n(198). Now, you need to find the values of n from 10,000 to 99,999. There are 150.

Then, you'd have

1(199) + 397.

But there are too many for me to do this way. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!!!

There has to be an easier method.

[/hide]

Close but not necessarily right, Rep.

99999 isn't divisible by 198 ;)

Silverfalcon, [hide]$\displaystyle [99999/198]$ means 99999/198 without the remainder.[/hide]

Oh, yea. ;) Should've seen it.

That's correct answer.


algebra
function
domain
inequalities
algebra unsolved

For $ \phi: \mathbb{N} \mapsto \mathbb{Z}$ let us define \[ M_{\phi} \equal{} \{f:  \mathbb{N} \mapsto \mathbb{Z}, f(x) > f(\phi(x)), \forall x \in \mathbb{N} \}.\] Prove that if $ M_{\phi_1} \equal{} M_{\phi_2} \neq \emptyset,$ then $ \phi_1 \equal{} \phi_2.$ Does this property remain true if \[ M_{\phi} \equal{} \{f:  \mathbb{N} \mapsto \mathbb{N}, f(x) > f(\phi(x)), \forall x \in \mathbb{N} \}?\]

Well, in order to have $M_{\phi} \neq \emptyset$, we first need $\phi(\mathbb{N}) \subseteq \mathbb{N}$, otherwise no function $f \in M_{\phi}$ may exist, since some $\phi(x)$ will be outside the domain of $f$. Furthermore, denoting by $\phi^n$ the $n$-th iterate of $\phi$, no orbit $\mathcal{O}(x) = \{x,\phi(x),\ldots,\phi^n(x),\ldots\}$ of an element $x$ may be finite, since for  $f \in M_{\phi}$ we must have $f(x) > f(\phi(x)) > \cdots > f(\phi^n(x)) > \cdots$.

Precisely this sequence of strict inequalities shows that the answer to the last question is trivially yes, the property being [b]vacuously true[/b], since then such a decreasing sequence of integers cannot exist in $\mathbb{N}$, hence $M_{\phi} = \emptyset$ for all $\phi$.

Let $\phi \colon \mathbb{N} \to \mathbb{N}$, and for any $n \in \mathbb{N}$ denote by $\phi^n$ the $n$-th iterate of $\phi$ (with $\phi^0 = \textrm{id}_{\mathbb{N}}$). For any $x \in \mathbb{N}$ define $\mathcal{O}_{\phi}(x) = \{\phi^n(x) \mid n \in \mathbb{N}\}$, the orbit of $x$. In the sequel we will only consider $\phi$ with the property that for any $x \in \mathbb{N}$ the orbit $\mathcal{O}(x)$ is infinite (if even for one $x \in \mathbb{N}$ we would have $\mathcal{O}(x)$ finite, then $M_{\phi} = \emptyset$, since having $f(x) > f(\phi(x)) > \cdots > f(\phi^n(x)) > \cdots$ leads to a contradiction).

[b]The graph[/b]. Define a directed graph $\overrightarrow{G}_{\phi} = (V_{\phi}, E_{\phi})$ with $V_{\phi} = \mathbb{N}$ and $E_{\phi} = \{(x,y) \equiv \overrightarrow{xy} \mid y = \phi(x)\}$. The fact that $\mathcal{O}_{\phi}(x)$ is infinite for all $x \in \mathbb{N}$ ensures that $\overrightarrow{G}_{\phi}$ is acyclic. Now, $V_{\phi}$ partitions in the (connected) components of the subjacent undirected graph $G_{\phi}$. We will denote by $\mathcal{C}_{\phi}(x)$ the component containing $x$; clearly $\mathcal{O}_{\phi}(x) \subseteq \mathcal{C}_{\phi}(x)$, and moreover, for any $y \in \mathcal{C}_{\phi}(x)$ we will have $\mathcal{O}_{\phi}(y) \cap \mathcal{O}_{\phi}(x) \neq \emptyset$. Clearly, $\mathcal{C}_{\phi}(x)$ is a tree (induced) subgraph of $G_{\phi}$.

[b]The equivalence relation[/b]. The relation $\sim_{\phi}$ defined on $\mathbb{N} \times \mathbb{N}$ by $x \sim_{\phi} y$ if and only if $y \in \mathcal{C}_{\phi}(x)$ is an equivalence relation, with $\hat x = \mathcal{C}_{\phi}(x)$.

[b]The order relation[/b]. The relation $\prec_{\phi}$ defined on $\mathbb{N} \times \mathbb{N}$ by $x \prec_{\phi} y$ if and only if $x \in \mathcal{O}_{\phi}(y)$ is seen to be a (partial) order relation (it is here that $\mathcal{O}(x)$ is infinite is of paramount importance).

We can now define a function $f \colon \mathbb{N} \to \mathbb{Z}$. On each component $\mathcal{C}_{\phi}$ select an arbitrary $x \in \mathcal{C}_{\phi}$ and arbitrarily define $f(x) \in \mathbb{Z}$. For any $y \in \mathcal{C}_{\phi}$ we will have $\mathcal{O}_{\phi}(y) \cap \mathcal{O}_{\phi}(x) \neq \emptyset$; take any $z$ from it, thus $z = \phi^{n}(x) = \phi^m(y)$ for some $n,m \in \mathbb{N}$. Take then $f(y) = f(x) - n + m$. It is seen that $f$ is coherently defined, and $f(\phi(y)) = f(y) - 1$, thus $f(y) > f(\phi(y))$ for all $y \in \mathcal{C}_{\phi}$, and thus for all $y \in \mathbb{N}$, and so $f \in M_{\phi}$, therefore $M_{\phi} \neq \emptyset$.

Assume now $\phi_1 \neq \phi_2$ for some $\phi_1, \phi_2$ as above, i.e. $\phi_1(x) \neq \phi_2(x) = y$ for some $x \in \mathbb{N}$. If $\mathcal{C}_{\phi_1}(y) \neq \mathcal{C}_{\phi_1}(x)$, we can define a function $f \in M_{\phi_1}$ as above, but with $f(x) < f(y) = f(\phi_2(x))$, thus $f \not \in M_{\phi_2}$, hence $M_{\phi_1} \neq M_{\phi_1}$. If $\mathcal{C}_{\phi_1}(y) = \mathcal{C}_{\phi_1}(x)$, then $\mathcal{O}_{\phi_1}(y) \cap \mathcal{O}_{\phi_1}(x) \neq \emptyset$. We can now define a function $f \in M_{\phi_1}$ in a similar manner as above, but with $f(x) < f(y) = f(\phi_2(x))$, thus $f \not \in M_{\phi_2}$, hence $M_{\phi_1} \neq M_{\phi_1}$. Let $n \in \mathbb{N}$ be minimal such that $z = \phi_1^n (x) \in \mathcal{O}_{\phi_1}(y)$. If $n=0$, the above construction will do; if $n\geq 1$, use different step-values on the branches passing through $x$ and through $y$, so that $f(x) < f(y)$.

The problem is too much a set-theoretical issue (the reason probably it was not included on the ShortList). What in fact it refers to can be seen as follows

[color=#0000FF]The set $M_{\phi}$ is the set of isotone functions $f \colon (\mathbb{N}, \prec_{\phi}) \to (\mathbb{Z}, \leq)$. The problem states that this set of isotone functions determines the partial order relation $\prec_{\phi}$ induced by $\phi$, since the equality of two such sets, for $\phi_1$ and $\phi_2$, leads to the equality of $\phi_1$ and $\phi_2$.[/color]


Let $n$ be a positive integer. A number $A$ consists of $2n$ digits, each of which is 4; and a number $B$ consists of $n$ digits, each of which is 8. Prove that $A+2B+4$ is a perfect square.

[quote="peeta"]let $x$ be a number with n digits which are all 1's.
then
$A=4x(10^n+1)$
$B=8x$
so
$A+2B+4 = 4x(10^n+1)+16x+4 =4x(10^n-1)+8x +16x+4 = 4x\cdot 9x +24x +4= (6x+2)^2$[/quote]

Could you please explain how you got $A=4x(10^n+1)$?  Thanks. :)

[hide=Solution]
We want the following to be a perfect square, $$2^2 \left(\underbrace{111...1}_{2n}\right)+2\cdot 2^3\left(\underbrace{111...1}_{n}\right)+2^2$$
Notice the geometric sequences of powers of $10,$
$$2^2\left(\frac{10^{2n}-1}{9}\right)+2 \cdot 2^3 \left(\frac{10^n-1}{9}\right)+2^2$$
Factoring,
$$2^2\left(\frac{\left(10^n+1\right)\left(10^{n}-1\right)}{9}\right)+2^4\left(\frac{10^n-1}{9}\right)+2^2$$
Factoring out $10^n-1,$
$$2^2\left(10^n-1\right)\left(\frac{\left(10^n+1\right)+4}{9}\right)+2^2$$
Notice that we have,
$$2^2\left(\frac{\left(10^n+2\right)^2-3^2}{9}\right)+2^2$$
This is equal to
$$\frac{2^2(10^n+2)^2}{9}-\frac{2^2 \cdot 3^2}{9}+2^2=\frac{2^2\left(10^n+2\right)^2}{3^2}$$
This is always a integer because $10^n+2 \equiv 0 \mod 3.$ Notice this is a square number since the exponent is divisible by $2.$