Find all the integers $ n \ ,\ n\ge 2$ such that in the interval $ (n^3,2n^3)$ exists less than $ n$ perfect squares .

[hide]
If you list some beginning values $ n, n^3, 2n^3,$ and their
respective number of perfect squares in their open intervals, we see:

$ n$ | $ n^3$ | $ 2n^3$ | # of perfect squares
---------------------------------------------
2 | 8 ...| 16 ...| 1
---------------------
3 | 27 .| 54 ...| 2
---------------------
4 | 64 ..| 128 | 3
---------------------
5 | 125 | 250 | 4
---------------------
6 | 216 | 432 | 6

(The periods are used for spacing.)

I claim that the only integers for which there exist less than n
perfect squares in the interval $ (n^3, 2n^3)$ are
n = 2, 3, 4, 5 . . . (as listed in the table).

I claim that from n = 6 (inclusive), on up, that the number
of perfect squares is equal to or greater than n.

Example)

Look at three consecutive perfect square numbers such as
9, 16, and 25, and relate differences.

16 - 9 = 7, or $ 2\sqrt {9} \plus{} 1 \equal{} 7.$ . . . #
(Also) 25 - 16 = 9, or $ 2\sqrt {16} \plus{} 1 \equal{} 9.$

I'll make use of line # below for this example:

$ 25 \equal{}$
$ 9 \plus{} (7) \plus{} (9)$
$ 9 \plus{} (2\sqrt {9} \plus{} 1) \plus{} (2\sqrt {9} \plus{} 3)$, or
$ 9 \plus{} 2(2)\sqrt {9} \plus{} 1 \plus{} 3$

End of example - - -

We want a count of the number of potential perfect squares in the
interval.  Let's suppose the first perfect square is as far over to the
left as possible just after $ n^3$, that is, it is $ n^3 \plus{} 1.$

How many integers are in the interval $ (n^3, 2n^3)?$

$ 2n^3 \minus{} n^3 \minus{} 1 \equal{}$
$ n^3 \minus{} 1$

Look at the first number in the interval, $ n^3 \plus{} 1.$  As similar to an
example, you may have to go through a rounded version of 2$ \sqrt {n^3 \plus{} 1} \plus{} 1$
number of integers to get to the next perfect square.

As in the example, the gap from the first perfect square to the
second perfect square would then be 2$ \sqrt {n^3 \plus{} 1} \plus{} 3$,
or a rounded version.

Continuing in this manner, for n perfect square gaps added together,
that should be less than the number of integers in the interval,
$ n^3 \minus{} 1.$ This begets:

$ (2\sqrt {n^3 \plus{} 1} \plus{} 1) \plus{} (2\sqrt {n^3 \plus{} 1} \plus{} 3)$ + $ [remaining]$ $ [terms] < n^3 \minus{} 1$

$ (2\sqrt {n^3 \plus{} 1}$ $ \plus{}$ $ 2\sqrt {n^3 \plus{} 1}$ $ \plus{}$ ... a total of n times$ )$ $ \plus{}$ $ (1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9 \plus{} . . .) < n^3 \minus{} 1.$

$ 2n\sqrt {n^3 \plus{} 1} \plus{} n^2 < n^3 \minus{} 1$

In the table it was already noted that at n = 6 is the first time for a
positive n, n $ \ge 2$, that the number of perfect squares in the interval
is not less than n.

The left side of this inequality is asymptotic to $ 2n^{\frac52}$, and
the right side of the inequality is basically $ n^3$, so after checking
that these curves cross each other shy of n = 6 one time on the
positive x portion, I know that the left-hand side curve will continue
to have smaller values than the right-hand side curve for n = 6 and
on up.

Therefore, n = 2, 3, 4, 5 are the only integers that exist for which
there are less than n perfect squares in the interval $ (n^3, 2n^3)$.

[/hide]

[hide=" not too sure about this"]
Let $ k$ be the smallest integer such that $ n^3 < k^2$

We need to find the values of $ n$ such that $ n^3 < k^2 < (k \plus{} n)^2 < 2n^3$

The inequality gives $ (k \plus{} n)^2 < 2n^3$ and $ \minus{} k^2 < \minus{} n^3$. Summing these gives $ 2kn \plus{} n^2 < n^3$

$ \therefore 0 < n^3 \minus{} n^2 \minus{} 2kn \Leftrightarrow 0 < n^2 \minus{} n \minus{} 2k$ ($ \because n > 0$)

So the question becomes: for which $ n$ does there exist a $ k$, $ : k^2 > n^3$ such that $ 0 < n^2 \minus{} n \minus{} 2k$

We know that $ n^3 < k^2 \Rightarrow k \equal{} \lceil n\sqrt {n}\rceil$
but $ \lceil n\sqrt {n} \rceil \le n\lceil \sqrt {n} \rceil$

$ \therefore 0 < n^2 \minus{} n \minus{} 2\lceil n\sqrt {n}\rceil$

$ 0 < n \minus{} 1 \minus{} 2\lceil \sqrt {n}\rceil \Leftrightarrow 2 \plus{} 2\sqrt {n} < n$, ($ \because \lceil \sqrt {n} \rceil \le \sqrt {n} \plus{} 1$)

$ \therefore (n \minus{} 2)^2 > 4n \Longrightarrow n^2 \minus{} 8n \plus{} 4 > 0 \Longrightarrow n > 7$

Now we have shown that the statement is true for $ n > 7$ and we can check the other cases.

After checking cases $ n \equal{} 2,3,..,7$ we find that there are less that $ n$ perfect squares in the range $ (n^3,2n^3)$ for $ n\equal{}2,3,4,5$.

[/hide]

Edit: Okay, Arrange your tan, you were right. My case checking was a bit hasty and "interval holds" was poorly worded. I have fixed it now, cheers.
@kcn2rivers; True, but this made the numbers easier to work with, and it is fine because we were proving a stronger result.

Edit:

[quote="ocha"]not too sure about this  <-- I acknowledge this, and some of my own.

After checking cases $ n \equal{} 1,2,..,7$ we find that the interval holds for all integers $ n\ge 7$.

[/quote]

I don't see you starting checking cases with n = 1 because $ n \ge 2$ (as given).

I'm not sure in what context you meant "the interval holds."
If you meant in the original statement that there are fewer than n perfect squares
in the interval $ (n^3, 2n^3)$ for certain values of n, and you stated them as being
$ n \ge7$, then  I disagree.  In my table I show that n = 2, 3, 4, and 5 work.

However, if "the interval holds" is PARTICULAR to YOUR work, then I have no comment.
Also, I am not commenting on the other content of your post; I don't have enough
expertise in it, for one thing.

[quote="ocha"]
We need to find the values of $ n$ such that $ n^3 < k^2 < (k \plus{} n)^2 < 2n^3$.
[/quote]
Wouldn't this be $ n\plus{}1$ perfect squares in the interval? The problem asks for less than $ n$ perfect squares, so shouldn't the inequality be $ n^3 < k^2 < (k \plus{} n\minus{}2)^2 < 2n^3$ ?

[quote="kcn2rivers"][quote="ocha"]
We need to find the values of $ n$ such that $ n^3 < k^2 < (k \plus{} n)^2 < 2n^3$.
[/quote]
Wouldn't this be $ n \plus{} 1$ perfect squares in the interval? The problem asks for less than $ n$ perfect squares, so shouldn't the inequality be $ n^3 < k^2 < (k \plus{} n \minus{} 2)^2 < 2n^3$ ?[/quote]

I think it should be $ n^3 < k^2 < 2n^3 < (k \plus{} n \plus{} 1)^2$

[hide="New hint"]Prove that $ n< 3 \plus{} 2\sqrt {2}$[/hide]


MATHCOUNTS
videos
probability
geometry
AMC
AIME
AoPS Books

1) Do old mathcounts tests
2) Find out which ones you got wrong
3) Learn how to do them
4) If there's a specific type of problem, like say counting problems, that you're getting wrong, ask someone for help

it depends....what are you averagin on state tests and what do you need to get in order to make the countdownround(aproximatley)

If you just need to get better, get AoPS vol 1.(yes i know EVERYBODY says it but its just THAT good). if you are looking for a certain subjecct try the intro series. the intermediate will be almost useless for mathcounts.

its a bit late to sign up for the mathcounts problem series course

theres a good link [url=http://www.yashar.biz]here[/url] with tests

ummm, make up problems for your friends. thats what i do.

if your teachers still havent read you this, a study showed that you learn 10% of the things you read, 20% of the things you hear, 30% of the things you see, 40% of the things you do like an activity, and you learn 90% of the things you teach. something like that. my point is, if your school is involved in mathcounts a lot, then ask if you can teach something during a session. im just saying from my perspective and i dont know what you do(your school) for mathcounts so somethings may sound confusing

My tips:
1. Do past tests.
2. Take lessons w/ your coach or parents, seek help if needed.
3. Use AoPS books, like Introduction to _______ and The All-Time Greatest MATHCOUNTS Problems.
4. Do problems from the School Handbooks.
5. Do the Problems of the Week that are posted weekly on mathcounts.org
6. Practice a lot!

I'm sure that helps.

Tip #1! The most important tip of all.
DO NOT STUDY BECAUSE OF SOME OTHER REASON! IT WON'T WORK! YOU HAVE TO ACTUALLY MEAN WHAT YOU ARE DOING IN ORDER TO SUCCEED!
Which means...
Stop the video games.
And study.
For real.
If you just study because you want to play games, it's not going to work.

Well I personally think I won't make it to the team.
 :(

Yeah, like they all said, try to focus on improving yourself, not regaining ur video game privileges or beating other ppl.  In my 6th grade yr, I got too competitive and didn't do so well at chapter.  Try to do problems.  As you do more problems, you'll recognize the styles that mathcounts has.  Do all the handbook, old problems, anything you can find.  And, use the AoPS books too!

I placed in the top 40 in CA last year, my first year doing mathcounts. Basically, what people before me said, do practices. I sorta died on the harder probability ones but my friend who almost made nationals did too :P I don't have vol.1 so I dunno how it is, but at state they handed out a little book with excerpts from it and it has the basic probability, etc. stuff, if you know all that you can probably score like over 30 which is okay. I bought a vol.2 recently and its great so I'm guessing vol.1 must be too :D

[quote="darkmatter47"]Really? I placed 13th in New Jersey (say, is that unlucky?). Will vol 2 really help? I heard that a lot of the topics are like algebra 2 and such.[/quote]

it's not a particular subject area like "ALGEBRA 2". it picks various topics from various areas of competition math. i have it, and Im absolutely in love with it.
i would say its more for like AIME, but it should help with national mathcoutns stuff, if anything

What I meant was that MathCounts focuses on standard difficulty levels of algebra and geometry. A friend who bought said it barely helps with MathCounts, I suppose it's different with Nationals.

u guys are too good...... Idk if i'll make school team in 7th grade.(right now)

Anyways, I think ZhangPeijin is right. You have to love what you do if you wanna get good at it.

yeah...

everyone's right here

you got to study like you mean it

refer to this example:

There's a big difference if you actually try to do it rather than I'm doing this for my video game priveleges

(ignore the spellings)

[quote="alanchou"]volume 2 relies very much on knowledge of just about every concept in volume 1.  If you don't place top 20, you probably should go with volume 1.  But don't just practice with AoPS texts, although they're great, it's not enough.[/quote]
Well not really, mathcounts can't determine THAT much. In CA we had like 4 way ties for every single score from 42 down or something. I'd say AoPS books help a good deal, but they're not everything. The thing they do a great job on is they let you gain a greater degree of mastery of the topics because you actually get to prove all the stuff, and do practices, etc. Volume 2 is just a bunch of topics you see in high school but to a greater depth. Mathcounts is GREATLY speed biased, and has a lot of those annoying counting/probability problems (that I suck at) and geometry so if you're good on probability and geometry, then computation is what should be worked on (speed that is). Mathcounts actually cut my computation speed in like 1/2 lol.

Try working through as many MATHCOUNTS problems as possible. In hindsight, I wish I didn't waste my time on AOPS books. It's usually too easy or too much usless knowledge for MATHCOUNTS. Most people on this site would disagree. I know the AOPS books are good math resources, but they didn't help me in MATHCOUNTS. MATHCOUNTS is for using almost common knowledge creatively. At least, this is my view.

AoPS books are NOT designed to help you on Math Counts, the AMCs, etc. They certainly do, but they aren't specifically made for one competition. Just working through them helps you gain a better array of problem solving strategies.


How many integers are in the solution set of $ |x\minus{}2|\leq5.6$

this is 2 equations.
$ x\minus{}2 \leq 5.6$ and $ x\minus{}2 \geq \minus{}5.6$
$ \minus{}3.6 \leq x \leq 7.6$
giving 7 positive, 3 negative, and 0 as solutions, making the answer $ \boxed{11}$


factorial
search
number theory solved
number theory

find smallest n such that the decimal representation of n! (n factorial) begins with 1994

1. solve in positive integers:
a^2+5b^2=p^2
a^2-5b^2=q^2

problem number 1 is it a system or two dio equations? :D

The sysitem p*p+5 m*m = square
                  p*p- 5 m*m = square Have infinitely many coprime solutions.
     I dont remember the first example, but it was something like 13,40 or 7, 19. Well at least this number where involved in some way and the solution were p and m less  than 100.
    Then for infinitely many you find more from the preciding ones, like
p2= p*p*p*p - 5 m*m*m*m and etc..
  I dont remember exactly the example and solution but try to relate each solution of the sistem with one of
  a*a*a*a -b*b*b*b = 5*x*x ( wich have infinitely may solutions

charly you can use the shyntax a^3 or a3 (with HTML enabled in the post) to express a*a*a.

i did a search using computer and only one pair of (a,b) satisfy the condition.  but then again, the solutions to pell's equation grows large pretty fast...

problem 2 is a system of equations.  i swear i have answered Lagrangia's post before but i do not see my answer?

yes i know some about 100 posts were lost at some point due to a failure in the database system.

I am changing the provider now, hope that up to Saturday I'm done with it :)


Here's a thought for the day.. What [i]are[/i] numbers?

[quote="MysticTerminator"]wow! can we have a thought for the day? as in the getting started forum?[/quote]
but this is the "Advanced Problems" forum...

Thought for the day: If "walking" represents addition and "walking backwards" represents subtraction, then what does "moon-walking backwards" represent?
I think Syntax Error would be good on this one :)

And here's a more concrete thought of the day.. what does f(x) = x^x look like? Does it have any local minima or maxima?

What about f(x) = x^(1/x)?

...or how about f(x) = x^x^x^x... (x times)

[quote="Ragingg"]Moon walking backwards is clearly an arithmetic regression.  A smooth one.

15,13,11,9,7,5,3,1[/quote]
Um, you're incorrect. Sorry, think again.

[quote="ComplexZeta"]x^(1/x) has a local maximum at (e, e^(1/e)), which everyone from SF Bay Area should know since that was the theme of the 2003 SFBA ARML T-shirt. x^x has a minimum at x=1/e.

x^x^x^... is an interesting function. It converges for e^-e <= x <= e^(1/e) if I remember correctly.[/quote]

Can you prove these things?

And your x^x^x... function is technically different than Fiery's, but is much more interesting. Fiery's is only defined for positive integers and basically just grows really really fast.

Sure, I can prove the first two by taking derivatives. I missed that Fierytycoon wrote x times for the x^x^x^... function. I'm not sure how to prove the convergence issue for the x^x^x^... function. I'll work on it, but it looks fairly difficult.

or how about f(x)=x^(1/x)^(x)^(1/x).....where there are
x (x^(1/x))'s

Why not infinitely many? It's hard to define your function when x isn't a positive (or at least nonnegative) integer.

[color=cyan]Well, you could ask:  what is a continuous, differentiable function that shares the values of fierytycoon's function at positive integers?  That might be interesting.[/color]

darkquantum, your function is just f(x^x^-1), where f(x) = x^x^x^...

[quote="ComplexZeta"]No it isn't because his is finite.[/quote]
Naturally, I meant your version of his.

x^x^x^... = y = sqrt(2Li(x))/ln x, I think.

Um, I know this sounds stupid, but isn't this something of a "thought of the week"? :)

[quote="ComplexZeta"]In that case you are correct.[/quote]

But am I?

NO!

(x^x^-1)^(x^x^-1) = x^x^(x^-1 - 1) != x^(x^-1)^x^(x^-1). How embarrassing! So darkquantum's function with an infinitely long power tower is a whole function.


triangle ABC
Squares ABDE and BCGH are constructed on the sides AB and BC of a triangle ABC. The midpoints S and I of the segments AC and DH together with the centers of the original squares F and N then form another square FINS.

Consider the quadrilateral $ACHD$.

$S,N,I,F$ are the midpoints of its sides.

So $FI,SN,AH$ are parallel and $FS,IN,CD$ are parallel.

Moreover, $FI=SN=\frac{AH}{2}$ and $FS=IN=\frac{CD}{2}$

Consider the triangles $ABH$ and $DBC$.

Because $AB=DB,\angle ABH=\angle DBC,BH=BC$

so they are congruent. We get $AH=DC$.

Hence, $FI=SN=FS=IN$.

Now $\Box FINS$ is a rhombus.

If we rotate $A$ and $H$ with pivot $B$ for $90^\circ$, they go to $D$ and $C$

That means $AH$ are rotated to $DC$, so $AH$ makes $90^\circ$ angle with $DC$.

This makes every angle of $\Box FINS$ be $90^\circ$.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 48
Sincerely
Jean-Louis


calculus
integration
limit
logarithms
calculus computations

Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \ln \left(1\plus{}\frac{k^a}{n^{a\plus{}1}}\right).$

We can write:

$ \lim_{n\to\infty}\sum^{n}_{k \equal{} 1}\ln\left(1 \plus{} \frac {k^a}{n^{a \plus{} 1}}\right) \equal{} \minus{} \lim_{n\to\infty}\sum^{n}_{k \equal{} 1}\sum^{\infty}_{i \equal{} 1}\frac {( \minus{} 1)^i}{i}\left(\frac {k}{n}\right)^{a\,i} \frac {1}{n^i} \equal{}$

$ \equal{} \minus{} \sum^{\infty}_{i \equal{} 1}\frac {( \minus{} 1)^i}{i} \lim_{n\to\infty} \frac {1}{n^i} \sum^{n}_{k \equal{} 1}\left(\frac {k}{n}\right)^{a\,i} \equal{}$

$ \equal{} \minus{} \sum^{\infty}_{i \equal{} 1}\frac {( \minus{} 1)^i}{i} \lim_{n\to\infty} \frac {1}{n^{i \minus{} 1}} \int^1_0 x^{a \, i} \,dx$

Now, we see that

$ \lim_{n\to\infty} \frac {1}{n^{i \minus{} 1}} \int^1_0 x^{a \, i} \,dx$

does not equal zero only for $ i \equal{} 1$, so we have

$ \lim_{n\to\infty}\sum^{n}_{k \equal{} 1}\ln\left(1 \plus{} \frac {k^a}{n^{a \plus{} 1}}\right)  \equal{} \minus{} \frac {( \minus{} 1)^1}{1} \lim_{n\to\infty} \frac {1}{n^{1 \minus{} 1}} \int^1_0 x^{a\cdot 1} \,dx \equal{} \frac {1}{a \plus{} 1}$

Since we've used the assumption that the logarithm expansion is absolutely convergent, we need $ a > 0$ for this to be OK.

Also an approach that boils down to Riemann integral:

Define

$ f_n(a)\equal{}\sum^{n}_{k\equal{}1} \ln\left(1\plus{}\frac{k^a}{n^{a\plus{}1}}\right)$

We have:

$ \frac{d f_n(a)}{da} \equal{} \sum^{n}_{k\equal{}1} \frac{1}{1\plus{}\frac{n^{a\plus{}1}}{k^a}}\ln \frac{k}{n} \equal{}  \sum^{n}_{k\equal{}1} \frac{1}{n}\frac{1}{\frac{1}{n}\plus{}\left(\frac{n}{k}\right)^a}\ln \frac{k}{n} \sim  \sum^{n}_{k\equal{}1} \frac{1}{n}\frac{1}{\left(\frac{n}{k}\right)^a}\ln \frac{k}{n} \equal{}$

$ \equal{} \sum^{n}_{k\equal{}1} \frac{1}{n}\left(\frac{k}{n}\right)^a \ln \frac{k}{n}$

So

$ \lim _{n \to \infty} \sum^{n}_{k\equal{}1} \ln\left(1\plus{}\frac{k^a}{n^{a\plus{}1}}\right) \equal{}\int \left(\lim _{n \to \infty}\frac{d f_n (a)}{da}\right) \,da \equal{}\int \int^1_0 x^a \ln x \, dx \,da \equal{}$

$ \equal{}\int^1_0 \int x^a  \,da \ln x \, dx \equal{} \int^1_0 x^a \, dx \equal{} \frac{1}{1\plus{}a}$


I basically put a comic strip with a couple word missing, and you guys have to figure out what word was it.

[img]http://i822.photobucket.com/albums/zz142/xiaosanfelix/GuesstheMissingWord1.gif[/img]

[hide="First One"]

TIGER[/hide]

[hide="Second One"]

RIGGED[/hide]


Rita, who is a long distance runner, runs at an average velocity of 9 miles per hour. Three hours after Rita leaves your house, you leave in your car and follow the same route. If your average velocity is 45 miles per hour, how long will it be before you catch up to Rita?

This question was taken from a chapter titled Parametric Equations.

A friend said the answer is 45 minutes but he did not explain how he got his answer.  I also disagree the answer is 45 minutes because he answered the question as though it is a regular distance word problem but is it?

Why would it be in a parametric equation chapter?

[quote="MellowMelon"]Your friend is correct. What is your progress/answer?

I have no idea why it is in the parametric equations chapter. Using those is overkill.  Everything in this problem is linear.[/quote]

Textbooks often give you a simple problem and force you to solve it with a difficult method.  Authors seem to like it.


The following Ptolemy-like criterion for the concyclicity of four points, in spite of its apparent simplicity, appears to have been only very recently discovered, and (no, I'm not joking) it is incredibly difficult to prove.

[b]Theorem. [/b] The convex quadrangle $ABCD$ is inscribed in a circle if and only if
\[
AB \cdot BC \cdot CA + AC \cdot CD \cdot DA = BC \cdot CD \cdot DB + AB \cdot BD \cdot DA.
\]

--Vesselin

Hmm.. This definitely doesn't look right. Are you sure you weren't referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25372]this[/url]?

Ah, yes, sorry.  :blush:  This has also been discussed here ... It is very strange since in [url=http://arxiv.org/pdf/math.GM/0410234]this[/url] article (in Russian) has the following abstract: [i]A convex quadrilateral with sides $a,b,c,d$ and diagonals $p,q$ is cyclic iff $abp-bcq+cdp-daq=0$. This condition, in spite of its simplicity, appears to be unnoted and unexpectedly proof-resilient. We employ advanced methods of computer algebra and nonlinear analysis. [/i] (!!!)  :?

 :D  :D  :D

Isn't it funny?   :lol:  Haha, it's indeed trivial, after thinking a while! I didn't think on it initially, just posted it directly here after seeing it in the mentioned article  :) . Check the article out and tell us what do you think!  ;)

--Vesselin


integration
calculus
calculus computations

\[ I(\epsilon) = \int_1^\infty x^{-1-2\epsilon} \left[ \int_{1/x}^\infty u^{-\frac{1}{2}-\epsilon  \frac{du}{1+u} \right] dx.}\]
\[ 0< \int_0^{1/x} u^{-\frac{1}{2}-\epsilon} \frac{du}{1+u} < \int_0^{1/x} u^{-\frac{1}{2}-\epsilon} du = \frac{x^{-\frac{1}{2}+\epsilon}}{\frac{1}{2}-\epsilon}.\]
\[ I(\epsilon)=\int_1^{\infty}x^{-1-2\epsilon}\left\{
  \int_0^{\infty}u^{-\frac{1}{2}-\epsilon} \frac{du}{1+u}
\right\}dx+O \left( \int_1^{\infty}x^{-\frac{3}{2}-\epsilon}dx \right)\]

I can not understand `$ O$' here. Why?
thanks

An $ O$ should come with an understood "as a certain variable tends to a certain place." That context is not clear here. Since we have an expression written as $ I(\epsilon),$ and since it is $ \epsilon$ with the usual connotations of that, my default assumption would be that this should be as $ \epsilon\to 0.$

But it that's the case, then that last integral tends to a finite nonzero limit as $ \epsilon\to0$ and could thus be written as $ O(1).$

So I don't understand this, either.

Yes , it is when $ \epsilon \to 0$.
I just can not understand how to get the last formula.
Thanks


geometry
incenter
trigonometry
geometry unsolved

The incircle with center $ I$ of a triangle $ ABC$ touches $ BC,CA,AB$ at $ D,E,F$ respectively. The circle with $ BC$ as diameter cut $ EF$ at $ P,Q$, where $ P$ is nearer to $ B$ than $ Q$. Calculate angles $ \angle PCA$ and $ \angle PIA$, in terms of angles $ A,B,C$ of $ \triangle ABC$

Nice problem :)

Assume that $ P'$ is intersection of lines $ CI$ and $ EF$. In triangle $ \triangle CEP'$ we get that $ \measuredangle EP'C = \frac{\beta}{2}$. From sine theorem in same triangle we get that $ \frac{CP'}{\sin {\frac{\pi}{2} + \frac{\alpha}{2}}} = \frac{CE}{\sin {\frac{\beta}{2}}}$.

In triangle $ \triangle CIB$ we have $ \frac{CI}{\sin {\frac{\beta}{2}}} = \frac{BC}{\sin {\frac{\beta}{2} + \frac{\gamma}{2}}} = 4R \sin \frac {\alpha}{2}$. Finally, $ CI = 4R \sin \frac {\alpha}{2} \sin \frac {\gamma}{2}$. In right-angle triangle $ \triangle CEI$ we get $ CE = CI \cos \frac{\gamma}{2}$.

Thus, $ CP' = 4R \cos \frac{\alpha}{2} \sin \frac {\alpha}{2} \sin \frac {\gamma}{2} = BC \cos \frac{\gamma}{2}$, and triangle $ \triangle CP'B$ has right angle at $ P$'. This means that $ P = P'$ and points $ C$, $ I$ and $ P$ are collinear.

Obviously, ${ \measuredangle PCA = \frac{\gamma}{2}}$ and $ \measuredangle PIA = \frac{\pi}{2} - \frac{\beta}{2}$.

From a well-known property, $ I\in CP\cap BQ$ .


it's easy:
$(2a^{2}+2b^{2}-c^{2})(2c^{2}+2a^{2}-b^{2})\leq (2a^{2}+bc)^{2}$ which is after multiplication equivalent to $a\leq b+c$. Do the same thing with $(2b^{2}+ac)$ and $(2c^{2}+ab)$. Done! :P

[quote="arqady"][quote="eagerkill"]Let $a,b,c$ is the side of a triangle. Prove that
$(2a^{2}+2b^{2}-c^{2})(2b^{2}+2c^{2}-a^{2})(2c^{2}+2a^{2}-b^{2})\leq (2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab)$[/quote]
I think, it's true for all non-negative $a,$ $b$ and $c.$[/quote]

really? can you prove it? :maybe:

just wondering with that sos thing is there some kind of well known manipulation how to find the required form $S_{c}(a-b)^{2}+S_{a}(b-c)^{2}+S_{b}(c-a)^{2}$


articles
number theory
prime numbers
LaTeX

I've never seen anything like that. Post your code - the problem is probably in something you've done there.

now that I re-compiled it only some lines (like the one ending with world- renowned) is within the margins. i dont like the way its justifying everything, the right margin is really sloppy.

the source code:

\documentclass[20pt]{article}
\usepackage{amssymb}
\linespread{1.6}
\pdfpageheight 11in
\pdfpageheight 8.5in
\usepackage{graphicx}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\rhead{\ttfamily{Complete Sequences of Positive Integers} \hfill \thepage}

\begin{document}

\noindent \textbf{\LARGE{\ttfamily{Abstract}}}

\indent \indent \ttfamily{
\large{
\textit{Complete sequences} of positive integers have been studied since the sixteenth century. However, intrigue in this family of sequences has begun to grow only in the last century. In fact, these sequences are rapidly gaining world-wide attention for their potential use in many fields, including cryptography and quantum mechanics \cite{mathworld}.\\
\indent \indent Of the many conjectures proposed about these sequences, one conjecture proposed a decade ago has perplexed many \\world-renowned mathematicians. \textit{Blank}, named after the Romanian mathematician who proposed it, is the the most famous conjecture about these sequences that has yet to be proved \cite{primenumbers}.In this paper, research is conducted on this family of sequences by utilizing math-based computer programs such as \textit{MatLab} and \textit{Mathematica}. Analysis of the results generated by the computer programs led to proofs of many interesting properties. The graphs of the sequences, for example, provided insight on the differences and similarities between separate families belonging to the sequence.\\
\indent \indent We culminate the paper, utilizing the lemmas proved through the research, by proving \textit{blank}. We also discuss several applications of this sequence.

\begin{thebibliography}{99}

\bibitem{mathworld} J. L. Brown Jr., ``Unique Representations of Integers as Sums of Distinct Lucas Numbers,'' \textit{Fibonacci Quarterly}, no. 7, pp. 243-252, 1969.

\bibitem{primenumbers} R. Crandall and C. Pomerance, \textit{Prime Numbers: A Computational Perspective.} New York: Springer-Verlag, 2001.

\end{thebibliography}
       }
}

\end{document}

Try this out:

[code]
\documentclass[20pt]{article}
\usepackage{amssymb}
\linespread{1.6}
\pdfpageheight 11in
\pdfpageheight 8.5in
\usepackage{graphicx}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\rhead{\ttfamily{Complete Sequences of Positive Integers} \hfill \thepage}

\begin{document}

\noindent \textbf{\LARGE{\ttfamily{Abstract}}}

\large{\textit{Complete sequences} of positive integers have been studied since the sixteenth century. However, intrigue in this family of sequences has begun to grow only in the last century. In fact, these sequences are rapidly gaining world-wide attention for their potential use in many fields, including cryptography and quantum mechanics \cite{mathworld}.

Of the many conjectures proposed about these sequences, one conjecture proposed a decade ago has perplexed many world-\-renowned mathematicians. \textit{Blank}, named after the Romanian mathematician who proposed it, is the the most famous conjecture about these sequences that has yet to be proved \cite{primenumbers}. In this paper, research is conducted on this family of sequences by utilizing math-based computer programs such as \textit{MatLab} and \textit{Mathematica}. Analysis of the results generated by the computer programs led to proofs of many interesting properties. The graphs of the sequences, for example, provided insight on the differences and similarities between separate families belonging to the sequence.

We culminate the paper, utilizing the lemmas proved through the research, by proving \textit{blank}. We also discuss several applications of this sequence.

\begin{thebibliography}{99}

\bibitem{mathworld} J. L. Brown Jr., ``Unique Representations of Integers as Sums of Distinct Lucas Numbers,'' \textit{Fibonacci Quarterly}, no. 7, pp. 243-252, 1969.

\bibitem{primenumbers} R. Crandall and C. Pomerance, \textit{Prime Numbers: A Computational Perspective.} New York: Springer-Verlag, 2001.

\end{thebibliography}
}

\end{document}[/code]

Separate paragraphs with a blank line in your TeX source, not \\ \indent \indent.  Also, I changed your text formatting a little bit, which made the document look nicer.


One hundred people are standing in a line and they are required to count off in fives as 'one, two, three, four, five, one, two, three, four, five,' and so on starting from the first person in the line. Those who count off 'five' walks out of the line. Those remaining repeat this procedure until only four people remain in the line. What was the original position in the line of the last person to leave?

The problem seems ambiguous, and the choices stated are:
(A) 94
(B) 96
(C) 97
(D) 98
(E) 99

wow was I stupid...I had that...but then I thought that I had a counter example...that was pretty dumb...lol

[quote="dream23hope"]People are removed in this manner:
round 1: every 5th person
round 2: every 6th person since person 5 and 10 have left, so person 6 and 12 leave.
round 3: every 7th person
...
so last person to leave would be largest prime smaller than 100.

Ans: 97[/quote]

I don't understand how you made the conclusion it has to be the largest prime, can you explain?

[quote="JavaMan"]I don't understand how you made the conclusion it has to be the largest prime, can you explain?[/quote]

In every round [i]n[/i], the people who leave are of multiple of ([i]n[/i]+4) as shown above. Thus people who leave do not occupy numbers which are prime.

But in round 2, number 31 is removed, and 4+2 = 6 doesn't divide 31.

My mistake, I didn't realize the numbering-off is continuous i.e. if the last person says 3, the first  person says 4. Sorry once more.


We can choose $2^k$ different numbers from $0,1,2,...,3^k-1$, so that three numbers in arithmeticprogression will not occur.

For example we can take numbers only with digits 0 and 1 in the 3-adic system.


Diophantine equation
number theory
AUT
Austria

Determine all integers $ a$ and $ b$, so that $ (a^3\plus{}b)(a\plus{}b^3)\equal{}(a\plus{}b)^4$

The following couples $ (a,b)$ are the only solutions: $ (0,k),\,(k,0),\,k\in \mathbb{Z},\,(\pm1,\mp1),(\pm 3,\pm 5),\,(\pm 5,\pm3)$  :ninja:

can you explain how you found the answers hence solution

Solutions $ (0,k),\,(k,0),\,k\in \mathbb{Z}$ are trivial. Now, let $ a$ and $ b$ be different from $ 0$. The above equation can be rewritten as

$ a^4\plus{}a^3b^3\plus{}ab\plus{}b^4\equal{}a^4\plus{}4a^3b\plus{}6a^2b^2\plus{}4ab^3\plus{}b^4$, namely $ ab(a^2b^2\plus{}1)\equal{}ab(4a^2\plus{}4b^2\plus{}6ab)$.

Since $ ab\ne 0$, we have $ a^2b^2\plus{}1\equal{}4a^2\plus{}4b^2\plus{}6ab$. But $ a^2b^2\plus{}1\equal{}4a^2\plus{}4b^2\plus{}6ab\plus{}2ab \minus{}2ab$, namely $ a^2b^2\plus{}2ab\plus{}1\equal{}4a^2\plus{}4b^2\plus{}8ab$.

This can now be written as $ (ab\plus{}1)^2\equal{}(2a\plus{}2b)^2$ $ \to$ $ (ab\plus{}1)^2\minus{}(2a\plus{}2b)^2\equal{}0$ $ \to$ $ (ab\plus{}1\plus{}2a\plus{}2b)(ab\plus{}1\minus{}2a\minus{}2b)\equal{}0$.

Now, two cases

1) $ ab\plus{}1\minus{}2a\minus{}2b\equal{}0$ $ \to$ $ a\equal{}\frac{2b\minus{}1}{b\minus{}2}\equal{}2\plus{}\frac{3}{b\minus{}2}$, thus $ b\in \{\minus{}1,1,3,5\}$ and $ a\in \{1,\minus{}1,5,3\}$

2) $ ab\plus{}1\plus{}2a\plus{}2b\equal{}0$ $ \to$ $ a\equal{}\frac{\minus{}2b\minus{}1}{b\plus{}2}\equal{}\minus{}2\plus{}\frac{3}{b\plus{}2}$, thus $ b\in \{\minus{}1,1,\minus{}3,\minus{}5\}$ and $ a\in \{1,\minus{}1,\minus{}5,\minus{}3\}$

 :D

If you don't see the factorization, try the substitution

a+b=x
ab=y

And solving for x.

[hide=Solution]
Expanding, we can rewrite the equation as $a^3b^3+ab=4a^3b+6a^2b^2+4ab^3$, from which all pairs $(a, b)$ for which $ab = 0$ come out. Now assume $ab \ne 0$. Then we can rewrite the equation as $a^2b^2+1=4a^2+6ab+4b^2$. Adding $2ab$ to both sides, we get $(ab+1)^2 = (2a+2b)^2$.
What remains is to solve the equations $ab+2a+2b+1=0$ and $ab-2a-2b+1=0$, which we can use SFFT to get $(a+2)(b+2)=3$ and $(a-2)(b-2)=3$. These are all the rest of the solutions.

In conclusion, the solutions to the equation are $\boxed{\{(0, k), (k, 0), (-1, 1), (1, -1), (-3, -5), (-5, -3), (3, 5), (5, 3)\}}$, where $k$ is any integer.
[/hide]


trigonometry
inequalities proposed
inequalities

The sum of positive real numbers $ a,b,c$ equals $ 1$. Prove that:

$ \sqrt{\frac{1}{a}\minus{}1}\sqrt{\frac{1}{b}\minus{}1}\plus{}\sqrt{\frac{1}{b}\minus{}1}\sqrt{\frac{1}{c}\minus{}1}\plus{}\sqrt{\frac{1}{c}\minus{}1}\sqrt{\frac{1}{a}\minus{}1} \ge 6.$

let $ p\equal{}\sqrt{\frac{1}{a}\minus{}1}$ and so on
then since $ a\plus{}b\plus{}c\equal{}1$ we have $ p^2q^2\plus{}q^2r^2\plus{}r^2p^2\plus{}2\equal{}p^2q^2r^2$ so let $ pq\equal{}\frac{1}{\cos A}$ and so on.
it suffices to prove $ \sum \frac{1}{\cos A} \ge 6$.

[quote="moldovan"]The sum of positive real numbers $ a,b,c$ equals $ 1$. Prove that:

$ \sqrt {\frac {1}{a} \minus{} 1}\sqrt {\frac {1}{b} \minus{} 1} \plus{} \sqrt {\frac {1}{b} \minus{} 1}\sqrt {\frac {1}{c} \minus{} 1} \plus{} \sqrt {\frac {1}{c} \minus{} 1}\sqrt {\frac {1}{a} \minus{} 1} \ge 6.$[/quote]
$ \sqrt {\frac {1}{a} \minus{} 1} \equal{} \sqrt {\frac {b \plus{} c}{a}} \ge \frac {\sqrt {2}\sqrt [4]{bc}}{\sqrt{a}}$. Using this everywhere, it becomes:
$ \sum_{cyc} \sqrt [4]{\frac {c^2}{ab}} \ge 3$ which is obvious by AM-GM..


6. 	A cat is trying to catch a mouse in each of the labyrinths (a), (b), and (c): The cat starts at vertex C and the mouse at vertex M. The cat and the mouse take turns moving, with the cat moving first; each move is from a vertex to an adjacent vertex. If the cat ever gets to the same vertex as the mouse, it catches the mouse. In which (if any) of the three labyrinths can the cat catch the mouse, assuming the mouse makes no stupid moves? (A "stupid move" is a move after which the cat has a strategy that guarantees it can catch the mouse, where it didn't have one before.)

This was a fun problem :D I got (b) for my answer.  The mouse can't be caught in (a), because he starts two lengths away from the cat, and the cat moves first. Since the cat moves 1 length and then the mouse does the same, the cat is an even distance away on the start of his turn. Poor Mr. Cat will alway remain at least two lengths away (since zero would mean the mouse walks right into the cat, which it will never do :lol:).

For (b), the cat can go through the diagonal in the top right to switch around their positions (by moving two lengths in one turn). Now cat starts his turn an odd distance away and for the mouse, even. The mouse has to stay at least 3 lengths away, or be caught. Since he can't circle around the cat to get to the diagonal and change the positions again, the cat eventually traps him in a corner and now...suffice to say he'll learn at lot about the internal composition of a cat.

For (c), whenever the cat tries to use one of the diagonals to switch, the mouse can counter by using the diagonal on the other side. Thus, he'll never get caught.  :P

hm for part a i got that no matter how stupid either of them are the mouse wins anyway...you can color the vertices alternating black and white, and in order for the cat to catch the mouse, it needs to be on a different color when it is the cat's turn. but we see that this is impossible.

for c, the mouse basically 'mirrors' the cat, for example if the cat moves up, the mouse moves down, if the cat moves right, the mouse moves left, etc.

For (a) I just made this picture:
[img]http://i23.photobucket.com/albums/b378/ycz6/quiz-catmouse-a.gif[/img]
and a parity argument follows.

(b) was a brute force of sorts, though obviously by the above picture the cat has to go through the triangle first.

(c) is just symmetric.
[img]http://i23.photobucket.com/albums/b378/ycz6/quiz-catmouse-c.gif[/img]

More elaborate but overprecise proof:

When “distance” is used in this solution, it will always mean Manhattan distance. Unless stated or suggested otherwise, the distance referred to is on the cat’s turn to move.

At the beginning in all three labyrinths, the cat has a distance of 2 from the mouse – an even number. However, to catch the mouse, the cat must have a distance of one, an odd number, on its turn to move. An orthogonal move always results in a change of distance of 1; if both the cat and the mouse move orthogonally the total distance change will be either 0 or 2, both even numbers. Since an even number plus or minus an even number is always an even number, the distance between the cat and the mouse when only orthogonal moves are used will always remain even. Since Labyrinth A only has orthogonal moves, the only way for the cat to catch the mouse (that is, for the distance between the two to become 0) is for the mouse to move onto the cat’s vertex. However, due to the no stupid move rule, the mouse may only move onto the cat’s vertex if that is the only move option. However, since every vertex in Labyrinth A neighbors at least two other vertices, that can never occur. Therefore, the cat cannot catch the mouse in Labyrinth A.

However, labyrinths B and C allow diagonal moves. A diagonal move will change the distance between the cat and the mouse by either 0 (if the non-mover is equidistant from both vertices of the diagonal move) or 2 (if it is not). Coupled with an orthogonal move by the other, a diagonal move can switch the distance between the cat and the mouse on the cat’s turn to move between odd and even. However, to determine whether the cat can force the mouse to be caught, we must further analyze the dynamics between the two.

As the mouse cannot do anything to stop or hinder the cat, the cat will inevitably be able to make the distance between them odd so that it is capable of capturing the mouse. From there, any move the mouse makes can be said to be either towards the cat or away from the cat.

<Comment not in original: Next paragraphs can be summarized as "The mouse cannot circle around the cat at all, whether indefinitely or get to the diagonal.">

If the mouse moves away from the cat, the cat can mimic that move, and their relative positions will stay the same. If repeated, the mouse will eventually run into a boundary, and then a corner. From there, the mouse must move toward the cat, which will result in either immediate capture, or a closing of distance.

If the mouse moves toward the cat, the distance between the two will obviously decrease. Therefore, the mouse moving toward the cat is obviously unsustainable and can only be attempted in pursuit of a place to make a diagonal move to revert the distance between the two to even. Thus, the mouse’s only hope for survival is to revert the distance between the two to odd.

Say the mouse is moving toward some goal point. To catch the mouse, both components of the distance between the two entities must be 0 after the cat’s move. As, once the component that the mouse is trying to reduce reaches 0, the mouse will begin moving away from the cat, and their relative position will be constant, the cat must move to maintain parity between the components if it hopes to catch the mouse. As that is trivial to do, the cat can easily move in the most efficient path to catch the mouse. <Yep, kinda took a leap here.>  From there, we can easily say that the cat will only be able to catch the mouse if it can be within one square of a point on the path before the mouse can. Thus, we can say that the mouse will be able to avoid interception if the difference between the cat’s distance to the goal point and the mouse’s distance to the goal point is greater than 1.

In the case of trying to make a diagonal move, the vertex in question is the second vertex of the diagonal move, and, rather than depending on its Manhattan distance to the vertex, the mouse depends on the number of moves required to complete the diagonal move.

In Labyrinth B, after switching the distance between the two entities to odd, the cat will have a distance of either 0 or 2 from the final vertex; since the mouse is an odd number of spaces from the cat, it must requires at least two moves to make the diagonal move. Therefore, it will not be able to revert the distance between the two entities back to even, and the cat will be able to catch the mouse in Labyrinth B.

On the other hand, in Labyrinth C, since, unless the mouse makes a stupid move and moves toward the cat after its diagonal move, there will be at least three spaces between the two, and since the cat will be as far away from the other diagonal as possible, the distance between the cat and the mouse will remain at least three as the mouse moves toward the other diagonal until it makes the diagonal move. As the cat must lose distance if it attempts to make another diagonal move, this can repeat indefinitely, and the cat will not be able to catch the mouse in Labyrinth C.


vector
geometry
geometric transformation
reflection
ratio
projective geometry
angle bisector

Lemma 1: $ m\angle{BPC}\equal{}90^{\circ}$.
Proof: Reflect $ C$ over $ P$.  Because $ P$ is on the midsegment of the triangle, the image, $ C'$, of this reflection must be on $ AB$.  We have $ CP\equal{}PC'$ and $ m\angle{CBP}\equal{}m\angle{PBC'}$.  It then follows $ BP \perp CC'$, as desired.

Now drop a perpendicular from $ A$ to $ BK$.  Let this line intersect $ BK$ at $ Q$, and $ MN$ at $ R$.  Notice $ \triangle{KCP} \sim \triangle{KAQ}$, and $ \triangle{BCP} \sim \triangle{BAQ}$.  Moreover, because $ \frac{CK}{AK}\equal{}\frac{CB}{AB}$, the ratio of similitude is the same.  Thus, $ B, K, P,$ and $ Q$ are in harmonic division, so lines $ AB, AK, AP,$ and $ AQ$ are harmonic.

Now let $ AP$ intersect $ MN$ in a point $ P'$.  Then $ M, N, P',$ and $ R$ are in harmonic division.  Because $ P'$ and $ R$ are harmonic conjugates wrt $ M$ and $ N$ which are on the circle, we have that the polar of $ P'$ passes through $ R$.  But the polar of $ P'$ passes through $ A$, because $ P' \in MN$.

Thus, the polar of $ P'$ is $ AR$.  Because $ BK$ is perpendicular to $ AR$ and passes through the center of the circle, we must then have $ P' \in BK$, so $ P'\equal{}P$, as desired.

Thank [b]K81o7[/b]. Can you slove it by vector?

Haven't tried, but maybe?  Not really sure how...

Mathangel, could you elaborate?

Dear Mathlinkers,
this is the Lascases's theorem ; I think that we have see this kind of problem several time on Mathlinks.
Sincerely
Jean-Louis

[quote="K81o7"]Lemma 1: $ m\angle{BPC} \equal{} 90^{\circ}$..[/quote]
right !

we can used $ PE \parallel{} AB so \angle PBE \equal{} \angle BPE$ we obtain $ PE \equal{} EB \equal{} EC .$ -> $ \angle BPC \equal{} 90^0$
let$ P'$ is a point $ MN$ cut $ BK$ .In viewing lemma 1 we have result

Suppose that $ MN$ intersects $ BK$ at $ P'$. We need to prove that $ P \equiv P'$.

Now, note that from Menelaus theorem, we have:

$ \frac {MA}{MB} \frac {P'B}{P'K} \frac {NK}{NA} \equal{} 1$ (Note that $ MA \equal{} NA$) and $ \frac {EC}{EB} \frac {PB}{PK} \frac {DK}{DC} \equal{} 1$ (Note that $ EC \equal{} EB$)

We need to prove that $ \frac {P'B}{P'K} \equal{} \frac {PB}{PK}$ which is equivalent to $ \frac {MB}{NK} \equal{} \frac {DC}{DK}$

The final equation could be derived with algebra.

I think that my solution is more elementary but [b]K81o7[/b]'s one gives us a deeper insight to the problem.

[img]http://i209.photobucket.com/albums/bb172/Ineq/hay-7.jpg[/img]
$ H \equal{} AB\cap CP$
Easy to prove $ ONPC,HAOC$ is inscribed quadrilateral by add angles
so $ OAHC$ is inscribed quadrilateral,$ OM \perp AH,ON \perp AC,OP \perp HC$ so $ M,N,P$ is collinear (Simson line)

[quote][color=darkred][b][u]An equivalent original enunciation.[/u][/b] The in circle $ w \equal{} C(I,r)$ touches the sides $ [BC]$ . $ [CA]$ , $ [AB]$

in the points $ D$ , $ E$ , $ F$ respectively. Denote the midpoints $ M$ , $ N$ of the sides $ [BC]$ , $ [CA]$ respectively.

Prove that  the lines $ EF$ , $ BI$ , $ MN$ are concurrently (in the point $ P$ ) and $ PB\perp PC$ [/color].[/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ P\in EF\cap BI$ . Observe that $ m(\widehat {PEC}) \equal{} m(\widehat {AEF}) \equal{} 90^{\circ} \minus{} \frac A2 \equal{} \frac {B \plus{} C}{2}$ and $ m(\widehat {PIC}) \equal{} \frac {B \plus{} C}{2}$ .

Thus, $ \widehat {PEC}\equiv\widehat {PIC}$ $ \implies$ $ PEIC$ is cyclically, i.e. the points $ D$ , $ E$ , $ P$ belong to the circle with diameter $ [CI]$ .

Particularly, $ PB\perp PC$ . Since in the right triangle $ BPC$ the point $ M$ is the middle of the hypotenuse $ [BC]$ obtain

$ m(\widehat {PMC}) \equal{} 2\cdot m(\widehat {CBP}) \equal{} B$ , i.e. $ PM\parallel AB$ $ \implies$ $ P\in MN$ . In conclusion, the lines $ EF$ , $ BI$ , $ MN$ are concurrently.

[b]Otherwise[/b], the ray $ [PB$ is the bisector of the angle $ \widehat {DPF}$ , i.e. $ m(\widehat {IPD}) \equal{} m(\widehat {IPF}) \equal{}$ $ m(\widehat {AFP}) \minus{} m(\widehat {ABP}) \equal{}$

$ \frac {B \plus{} C}{2} \minus{} \frac B2 \equal{} \frac C2 \equal{} m(\widehat {ICD})$ . Thus,  $ \widehat {IPD}\equiv\widehat {ICD}$ , i.e. the quadrilateral $ IPCD$ is cyclically a.s.o.[/color]

Oh! It's nice solution too, thank Virgil Nicula very much.

[quote="mathangel"]I think that my solution is more elementary but [b]K81o7[/b]'s one gives us a deeper insight to the problem.[/quote]

Wow perhaps, but now look at the other solutions posted...

We can slove $ \vec {MN}\equal{}k.\vec {MP}, k \in R$ $ \Longrightarrow M, N, P$ are collinear.

[quote="thanhnam2902"]let $ \triangle ABC$. Let $ BK$ is bisector of $ \angle B, K \in AC$. Let $ D, E$ are midpoint of $ AC, BC$ respectively. Let $ BK$ meet $ DE$ at $ P$. Let $ (O)$ is inscribed circle of $ \triangle ABC$ and $ AB, AC$ tangent with $ (O)$ at $ M, N$ respectively. Prove that three point $ M, N, P$ are collinear.

-----------------------------------------------------------------------------------------------------------------------------------------------
Real, I sloved it by vector. But I don't like it, because it's very long. I want you help me slove it by vector and it's short. Or you can slove it by other way.[/quote]
[hide="Solution"]
Let the incircle of $ \triangle ABC$ meet $ BC$ at $ K$.  First of all, realize that $ B$, $ I$, $ K$, and $ P$ are collinear since they all lie on the bisector of $ \angle ABC$.  Notice that $ DE$ is a midline of $ \triangle ABC$, so $ AB\parallel DE$, from which we obtain that $ \angle EPB \equal{} \angle ABI \equal{} \angle IBE$ and so $ PE \equal{} BE \equal{} EC$ and we see that $ \angle IPC \equal{} 90$.  Notice that $ \angle IKC \equal{} 90 \equal{} 180 \minus{} \angle IPC$, so $ IPCK$ is cyclic.  Also, $ \angle INC \equal{} 90 \equal{} \angle IPC$, so $ INPC$ is cyclic, which means that $ INPCK$ is cyclic.  It follows that $ \angle INP \equal{} 180 \minus{} \angle ICP \equal{} 180 \minus{} (\angle PCK \minus{} \angle ICB)$.  Notice that $ \angle PCK \equal{} 90 \minus{} \angle IBC \equal{} 90 \minus{} \frac {\angle ABC}{2} \equal{} \frac {\angle BAC \plus{} \angle ACB}{2}$ and we also have that $ \angle ICB \equal{} \frac {\angle ACB}{2}$, which means that $ \angle ICP \equal{} \frac {\angle BAC}{2} \equal{} \angle IAM$.  Yet, $ \angle IMA \plus{} \angle INA \equal{} 90 \plus{} 90 \equal{} 180$, so $ AMIN$ is cyclic, and it follows that $ 180 \minus{} \angle INP \equal{} \angle ICP \equal{} \angle IAM \equal{} \angle INM\implies \angle INM \plus{} \angle INP \equal{} 180$, so $ M$, $ N$, and $ P$ are collinear, as desired. [/hide]

Dear Mathlinkers,
now, for a proof with historical note and a lot of figures
see: http://perso.orange.fr/jl.ayme   vol. 4  An unlikely concurrence, revisited and generalized.
Sincerely
Jean-Louis

Let $ CP$ intersect $ AB$ at $ C'$. It's easy to see that $ P$ is the midpoint of $ CC'$, and using that $ BP$ is the bisector of $ \angle B$, we get that $ C'BC$ is isosceles.

Now, let $ MN$ and $ PM$ intersect $ BC$ at $ X$ and $ X'$, respectively. Let the incircle of $ ABC$ touch $ BC$ at $ Q$.

It is known that $ CM$, $ BN$ and $ AQ$ concurr, and this implies $ (C, Q, B, X) \equal{} \minus{}1$

Also, by simmetry, $ CM$, $ BP$ and $ C'Q$ concurr. This implies $ (C, Q, B, X') \equal{} \minus{}1$

Then we get that $ X \equal{} X'$. So $ N$ and $ P$ both lie on line $ MX$, and the result follows.

Here is the Menelaus solution:
We can assume WLOG N is between A and K.
We have to show $ \frac{PK}{PB} \cdot \frac{BM}{MA} \cdot \frac{AN}{NK}\equal{}1$. But by Menelaus,
$ \frac{PK}{PB} \cdot \frac {BE}{EC} \cdot \frac{CD}{DK}\equal{}1$
So we have to show
$ \frac{BM}{MA} \cdot \frac{AN}{NK} \cdot \frac{CE}{BE} \cdot \frac{DK}{CD}\equal{}1$
But $ AN\equal{}MA$ and $ CE\equal{}BE$, so we have to show
$ \frac{BM}{NK} \cdot \frac{DK}{CD}\equal{}1$
Let $ BM\equal{}BR\equal{}x$ (R is the point of tangency of the incircle to BC), $ AM\equal{}AN\equal{}y$, $ CN\equal{}CE\equal{}z$.
We know $ BM\equal{}x$.
By the angle bisector theorem $ AK\equal{}AC \cdot \frac{AB}{AB\plus{}BC}\equal{}\frac{(x\plus{}z)(y\plus{}z)}{2x\plus{}y\plus{}z} so NK\equal{}AK\minus{}y\equal{}\frac{x(y\minus{}z)}{2x\plus{}y\plus{}z}$. So $ \frac{BM}{NK}\equal{}\frac{2x\plus{}y\plus{}z}{z\minus{}y}$
We know $ CD\equal{}\frac{1}{2} \cdot AC\equal{}\frac{y\plus{}z}{2}$.
But $ DK\equal{}CK\minus{}CD\equal{}\frac{(x\plus{}z)(y\plus{}z)}{2x\plus{}y\plus{}z}\minus{}\frac{y\plus{}z}{2}\equal{}\frac{(y\plus{}z)(z\minus{}y)}{2(2x\plus{}y\plus{}z)}$
So $ \frac{BM}{NK} \cdot \frac{DK}{CD}\equal{}1$ and we are done.


calculus
integration
number theory unsolved
number theory

provwe that for all positive integers $ n$ the equation $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 59^n$ has integral solutions

A positive integer n can represent in the form $ x^2\plus{}y^2\plus{}z^2$ if and only if it is not in the form : $ 4^l(8k\plus{}7)$

For this particular case here are the examples:
$ n\equal{}2k$
$ (x,y,z)\equal{}(59^k,0,0)$
$ n\equal{}2k\plus{}1$
$ (x,y,z)\equal{}(59^k 5 ,59^k 5 ,59^k 3)$


Which books do you think I should read in preparing for AMC10/AMC12 and AIME?

I have so far AoPS1, AoPS2, Geometry Revisited. Any other suggestions?

Um, do the first two, but don't do Geometry Revisited unless you want to get started on Olympiad Geometry for some reason. Geometry Revisited is THE GOD of all geometry.

EDIT: Apart from Mr. Rusczyk. Heheh...  :heli:

When they come out, I would recommend the Intermediate Algebra and Intermediate Counting & Probability books (and the rest of the intermediate books).  The classes (took NT & CP) were great, so the books probably will be also.


function
integration
geometric series
calculus
calculus computations

The function $ f$ is defined by the power series:
$ f(x) \equal{} 1 \plus{} (x\plus{}1) \plus{} (x\plus{}1)^2 \plus{} ... \plus{} (x\plus{}2)^n \plus{} ... \equal{} \sum_{n\equal{}0}^{\infty} (x\plus{}1)^n$ for all read numbers x for which the series converges.

(a)  Find the interval of convergence of the power series for $ f$.

(b) The power series above is the Taylor series for $ f$ about $ x\equal{}\minus{}1$.  Find the sum of the series for $ f$.

(c) Let $ g$ be the function defined by $ g(x) \equal{} \int_{\minus{}1}^x f(t) dt$.  Find the value of $ g(\frac{\minus{}1}{2})$.

(d) Let h be the function defined by $ h(x) \equal{} f(x^2\minus{}1)$.  Find the first three nonzero terms and the general term of the Taylor series for $ h$ about $ x\equal{}0$, and find the value of $ h(\frac{1}{2})$.

Here's a line I've used for my class. If you forget nearly everything you know about series and can only hang onto one important fact, let the thing you remember be the geometric series: when it converges and what its sum is.


I want to make a large table with two columns that splits itself equally in three, er... meta-columns across the page, if you know what I mean, so that there wouldn't be a large empty white space on the right.

My attempt:
[code]\begin{multicols}{3}

\begin{tabular}[]{r r}
  1  &  1  \\
  2  &  4  \\
  3  &  9  \\
  4  &  16 \\
  5  &  25 \\
  6  &  36 \\
  7  &  49 \\
  8  &  64 \\
  9  &  81 \\
 10  &  100 \\
 11  &  121 \\
 12  &  144 \\
 13  &  169 \\
 14  &  196 \\
 15  &  225 \\
 16  &  256 \\
 17  &  289 \\
 18  &  324 \\
 19  &  361 \\
 20  &  400 \\
 21  &  441 \\
 22  &  484 \\
 23  &  529 \\
 24  &  576 \\
 25  &  675 \\
\end{tabular}

\end{multicols}[/code]

(The multicols package was used.) That did not work.

Thanks in advance.

I'm not sure that I completely understand what you wan. Centering the table between \begin{center}...\end{center} presumably is not what you want.
If you want to spread the columns out you would normally use p{[i]width[/i]} instead of r, but that would give left-aligned columns which you wouldn't want with integer content.
So I suggest you use something like [code]\begin{tabular}{r@{\hspace{3cm}}r}[/code]to move the second column across or [code]\begin{tabular}{@{\hspace{3cm}}r@{\hspace{3cm}}r}[/code] to move both columns.
Obviously you can replace 3cm by any suitable width.

No, stevem, he wants this:
[code]
1 1               5 25            9 81
2 4               6 36           10 100
3 9               7 49           11 121
4 16              8 64           12 144
[/code]
but he wants the splitting into 3 "metacolumns" to be automatic and depending on the length of the list.

OK then, in that case let's try something as close to leoxnlin's code as we can. The tabbing environment works with multicol where tabular doesn't.
For example,
[code]\setlength{\columnsep}{2cm}
\begin{multicols}{3}
\begin{tabbing}
  1  \`  1 \\
  2  \`  4 \\
  3  \`  9  \\
  4  \`  16 \\
  5  \`  25 \\
  6  \`  36 \\
  7  \`  49 \\
etc
\end{tabbing}
\end{multicols}[/code]
There's a few details [url=http://www-h.eng.cam.ac.uk/help/tpl/textprocessing/teTeX/latex/latex2e-html/ltx-58.html]here[/url] on the tabbing environment, though it is not as well documented or as easy to use as the tabular environment.

It does work if the number of rows is divisible by 3, otherwise it can't do that. So
[code]\setlength{\columnsep}{2cm}
\begin{multicols}{3}
\begin{tabbing}
  1  \`  1 \\
  2  \`  4 \\
  3  \`  9  \\
  4  \`  16 \\
  5  \`  25 \\
  6  \`  36 \\
  7  \`  49 \\
  8  \`  64 \\
  9  \`  81 \\
  10 \` 100 \\
  11 \` 121 \\
  12 \` 144
\end{tabbing}
\end{multicols}[/code]gives 3 equal columns.
Do note that you must not have a \\ on the last line otherwise that introduces another row and then there will be 3n + 1 rows.

[quote="stevem"]Do note that you must not have a \\ on the last line otherwise that introduces another row and then there will be 3n + 1 rows.[/quote]

Oooooohhhhh...  :oops: Thanks stevem and fedja.  :)


How many regular hexagons appear in the figure below?

1+2+3+4+5+6 hexagons of side length 1

1+2+3 hexagons of side length 2

This gives 27 hexagons


So I started to prepare for my next year math league stuffs, I really need your help for the next several weeks :) Thank you so much!

20.	Solve the equation $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}= 3$.

[hide="Well..."] Symmetry suggests $x = a+b+c$, which checks out.  You could just expand and do a lot of algebra. [/hide]

[hide="very little algebra"]the equation can be rewritten as $(\frac{x-a-b}{c}-1)+(\frac{x-b-c}{a}-1)+(\frac{x-c-a}{b}-1)=0$
$\iff (x-a-b-c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=0$
so $x=a+b+c$
[$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq 0$ need to be assumed, because it doesn't contain the variable $x$][/hide]

If you want to be fully correct, you need to give cases. The answer should be:

$x=a+b+c$ if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ne 0$.

$x$ is anything if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= 0$.

[where $a,b,c\ne 0$]

This is a nice problem.


geometry
vector
geometric transformation
reflection
inradius
geometry proposed

The incircle of triangle ABC has centre I and touches the sides AB and BC at points X and Y respectively. The line through X and I meets the incircle at T. Let X' be the intersection of CT and AB. The point L is located on the segment X'C s.t. X'L=CT. Prove A, L and Y are collinear iff AB=AC.

JAJA! I've got it :D:D, give me a little time for me to copy it, cuz is a little long (I used vectors :D)

Ok, the solution is as follows:
Let $S$ be the incircle of triangle $\triangle{ABC}$. Draw the tangent to $S$ trough $T$ and let  $V$ and $W$ be the intersections with $AC$ and $BC$ respectivly. We know that $VW$  is parallel to AB since $T$ is diametrically oposed to $X$. Let $S_1$ be the incircle of the triangle $\triangle{VCW}$, and $X''$ be the tangency point with $VW$. We notice that circle $S$ is the excircle of triangle $\triangle{VCW}$, so that the point $T$ is the reflection of $X''$ trough the midpoint of $VW$.
Althouhg, triangle $\triangle{ABC}$ is a dilatation of triangle $\triangle{VCW}$, so we get that the point $X'$ is the reflection of $X$ trough the midpoint of the segment $AB$.
Let $Y', Z'$ be the reflections of $Y, Z$ trough the midpoint of $BC$ and $AC$ respectivley, let $L'$ be the point of intersection of the lines $CX'$ and $AY'$, and let $Q$ be the point diametrically oposed to $Y$. If we show that $L'=L$ then we are done,
because when $AC=AB$ we get that $Y'=Y$ and, it would follow that $A, L \mbox{ and } Y$ are colinnear. For this we will show that $\frac{\mid\vec{AL'}\mid}{\mid{AY'}\mid}=\frac{\mid\vec{QY'}\mid}{\mid\vec{AY'}\mid}$
We will make this by Vectors :D:
Let $C$ be the origin. Tthe value of $\vec{Y'}$, and $\vec{X'}$ are:\[\begin{array}{cc}\vec{Y'}=\vec{B}\frac{\mid\vec{CY'}\mid}{\mid\vec{CB}\mid}=B\vec{B}\frac{\mid\vec{BY'}\mid}{\mid\vec{CB}\mid}=\vec{B}\left(\frac{a+c-b}{2a}\right)\\\vec{X'}=\vec{A}+(B-A)\frac{\mid\vec{AX'}\mid}{\mid\vec{AB}\mid}=\vec{A}+(B-A)\frac{\mid\vec{BX}\mid}{\mid\vec{AB}\mid}=\vec{A}\left(\frac{b+c-a}{2c}\right)+\vec{B}\left(\frac{a+c-b}{2c}\right)\end{array}\]
Also we know that \[\vec{L'}=\vec{X'}\alpha=\vec{A}\left(\frac{b+c-a}{2c}\right)\alpha+\vec{B}\left(\frac{a+c-b}{2c}\right)\alpha=\vec{A}\left(\frac{b+c-a}{2c}\right)+\vec{Y'}\left(\frac{a}{c}\right)\alpha\] where $\alpha$ is a constant, and the representation is unique (because $\vec{A} \mbox{ and } \vec{Y'}$ are lineary independent). But since $A, L\mbox{ and } Y'$ lie on a line, then:\[\begin{array}{cc}\displaystyle\left(\frac{b+c-a}{2c}+\frac{a}{c}\right)\alpha=1\\\Rightarrow\alpha=\displaystyle\frac{2c}{a+b+c}\end{array}\]
From this we get that:\[\vec{L'}=\vec{A}\left(\frac{b+c-a}{a+b+c}\right)+\vec{Y'}\left(\frac{2a}{a+b+c}\right)\]
Then the value of the vector \[\begin{array}{cc}\vec{AL'}=L'-A=\left(\frac{2a}{a+b+c}\right)(Y'-A')\\\Rightarrow\left(\frac{L'-A}{Y'-A'}\right)=\frac{2a}{a+b+c}\end{array}\].
And since the vectors $\vec{AL'} \mbox{ and } \vec{AY'}$ are on the same line, we can define $\frac{L'-A}{Y'-A'}$ as the complex division, and is equal to $\frac{\mid\vec{AL'}\mid}{\mid\vec{A'Y'}\mid}$
In the other hand, since $Q$ is diametrically oposite to $Y$ we have that $\frac{Q-Y'}{A-Y'}=\frac{2r}{h_a}$ (where $r$ and $h_a$ are the inradius, and the lenght of the hight from vertex $A$ respectively)
From now on, easy. We just need to show that $\displaystyle\frac{2a}{a+b+c}=\frac{2r}{h_a}$
And this is obvoiusly true, hence point $L'=L$ and we are finished.:D
Personally I think it's interesting problem. I just wait to see if anybody shows it in a more elegant way  :lol:  :cool:

If you need I made a little drawing, so the solution is kinda more easy to understand :)

Take ${D} \equiv TX \cap CA$, ${E} \equiv AI \cap BC$, apply Cristea’s theorem (transversal) for the $\Delta ABC$ with transversal $XID$, find $CD/AD$, next, apply Menelaos for $\Delta ACX'$ with transversal $XTD$, find $\frac{CT}{TX'}(=\frac{X'L}{LC})$ and, finally, from Menelaos in $\Delta BCX'$ with the transversal $ALE$ one should get $b=c$!

Best regards,
sunken rock


The sum of the areas of all triangles whose vertices are also vertices of a $1\times 1 \times 1$ cube is $m+\sqrt{n}+\sqrt{p}$, where $m$, $n$, and $p$ are integers.  Find $m+n+p$.

[hide]
a cube has 8 vertices, so we have $\binom 83=56$ possible triangles. On each face of the cube we have 4 possibility, so 24 triangles with area
$=\frac{1}{2}$. Then we can take 2 consecutive vertices and one on the opposite edge, with 2 possibilities for each edge, so 16 triangles with area
$=\frac{1*\sqrt{2}}{2}$. The remanent 16 triangles are equilateral with side $=\sqrt{2}$ and area $\frac{\sqrt{3}}{2}$. So the total area is
$8\sqrt{2}+12+8\sqrt{3}=12+\sqrt{128}+\sqrt{192} \rightarrow m+n+p=332$[/hide]

i understand yor reasoning but i am overcounting the $\frac{\sqrt{2}}{2}$ case, i am getting 2 triangles for each edge and since there are 12 edges i count 24.  I understand all the other numbers of triangles and the total number there should be but i just don't get where i am going wrong.

[hide]There are $\binom{8}{3}=56$ total triangles.

Case 1:  All vertices are on the same face.  There are six faces, and one of the four vertices of the face is not a vertex of the triangle.  Therefore, there are $6\cdot 4=24$ triangles for this case.  Each triangle has an area of $\frac{1}{2}$, so the sum of the areas of the triangles in this case is $12$.

Case 2:  Two vertices are on the same edge, and the third vertex is on the opposite edge.  There are twelve edges, and there are two vertices on the opposite edge.  Therefore, there are $12\cdot 2=24$ triangles for this case.  Each triangle has an area of $\frac{\sqrt{2}}{2}$, so the sum of the areas of the triangles in this case is $12\sqrt{2}$.

Case 3:  No vertices are on the same edge.  It is difficult to count the possibilities for this case.  However, we can construct such a triangle by choosing a vertex $A$ of the cube, and taking the three vertices that share an edge with $A$.  We can construct $8$ triangles in this way.  Since $24+24+8=56$, we have accounted for all possible triangles.  Each triangle in this case is equilateral with sides $\sqrt{2}$, so each triangle has an area of $\frac{\sqrt{3}}{2}$.  The sum of the areas of the triangles in this case is $4\sqrt{3}$.

The sum of the areas of all the triangles is $12+12\sqrt{2}+4\sqrt{3}=12+\sqrt{288}+\sqrt{48}$.  The answer is $\boxed{348}$.[/hide]

Yes matt276eagles is right, because for the $\sqrt{2}$ equilatal triangle, there are only 8 of these, my reasoning is that each of the sides is a diagonal of a face and so it overcounts it 3 times. So there are infact 8.

(note - I know its the same as other solutions, writing it up so I remember it better)

There are three possible types of triangles that we could get, which are $1 \cdot 1 \cdot \sqrt2, 1 \cdot \sqrt2 \cdot \sqrt3, \sqrt2 \cdot \sqrt2 \cdot \sqrt2$. To get the first type, we can just select a face and there are four such triangles, so there are $6 \cdot 4 = 24$ ways to get this triangle, which has area $\frac{1}{2}$. To get the second type, we take both points on one of the twelve edges, and then select one of the two vertices which differ in height, which gives again $12 \cdot 2 = 24$ ways to get this triangle, which has area $\frac{\sqrt2}{2}$. Lastly, to get the third type, there must be $56 - 24 - 24 = 8$ such triangles, which have area $\frac{\sqrt3}{2}$. The answer is $12+12\sqrt2+4\sqrt3 = 12 + \sqrt{288} + \sqrt{48} \implies \boxed{348}$.


If $ x$, $ y$, $ z$ are positive reals, then prove that:
\[ 3(x \plus{} y)^{2}(y \plus{} z)^{2}(z \plus{} x)^{2}(x \plus{} y \plus{} z)\ge 16(x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} xy \plus{} yz \plus{} zx)^{2}xyz
\]

Thanks  :)
And if $x,\,y,\,z$ are the side lengths of a triangle, is it true?


linear algebra
matrix
Putnam
algebra
polynomial
superior algebra
superior algebra solved

A,B \in M_3(Z)such that A+kB are inverttible for k \in {1;2;3;4;5;6}
and for the same  k \in {1;2;3;4;5;6} (A+kB)^(-1) \in M_3(Z)
Prove that A+2000B are invetible, its inverse is in M_3(Z)

Dorinel Anca, Targoviste

[quote="Moubinool"]A,B \in M_3(Z)such that A+kB are inverttible for k \in {1;2;3;4;5;6}
.....[/quote]

sorry A+kB are inverttible for k \in {0;1;2;3;4;5;6}

this is actually a stolen putnam problem. Don't remember the year now, but I'll come back later.

the solution is based on the fact that A is invertible in M_n(Z) if and only if det A = 1.

-I_3 in M_3(Z) invertible det(-I_3)= - 1

Yes Valentin Putnam 1994

http://www.kalva.demon.co.uk/putnam/psoln/psol944.html

Do you have another solution of Targoviste 2000

Let P(x)=det(A+xB). As Valentin said, A+kB invertible, so P(k)=1 or -1 for all k from 0 to 6. P is a 3'rd degree polynomial. Among the 7 numbers 0,1,2,3,4,5,6 there are either 4 roots of P(x)-1 or of P(x)+1 (Dirichlet). Let's assume we have 4 different roots of P(x)-1. Since p(x)-1 is a 3'rd degree polynomial, it must be 0 for all x, so P(x)=1 for all x, which means that det(A+xB)=1 for all x, thus the conclusion.

yes that is the year :) the only difference is that this is a 3x3 matrix :) so basically it's the same problem.

Valentin,

[ Vn + V(n+1) ] = [ V(4n+1) ] is another Putnam problem (or Dorinel Anca problem!)

Vx denotes square root of x

what does it have to do with matrices? :) I also saw this particular problem with many authors on our district  olympiads :)

cristi is it a problem with square root matrices ?


The official answer list has been posted on the [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-2-1012-Answers.shtml]AMC web site[/url].

Omg. thats exactly the reason i requested the quetsion.  I dont have the bookelt (test people wont let ou keep), but on the test i remember there was a function question and it was 7(A) (dont know which problme-must be 9), but there was no 7 on the answer key so i was stressing out, and tryin to get the test..

[quote="pieterminate"]Is number 9 wrong? It says on the website that it's 2, but A is 7... :|[/quote]

I would assume that it is a typo, since 2 isn't even an option.  The correct answer is $(A)7$, so the letter is correct but the number is wrong.


Let ABC be a triangle with AB=AC. If the bisector of the angle B intersects AC in D such that BC=AB+2AD.
 Determine the angle A.

Answer is $\sin^{-1} \frac{3}{4}$, but wrong forum.

can you post your solution in more details?

why is this in the round table?  :what?:

Do your own homework and read the forum descriptions before you post in them.


[quote="GameBot"]Alex, Bing, Cara, Dan and Earl are people playing marbles. Alex holds 15 marbles, Bing holds 2 marbles, Cara holds 3 marbles, Dan holds 7 marbles, and Earl holds 8 marbles. What is the least possible number of marbles that must change holders so that each player will hold an equal number of marbles?[/quote]
total marbles = $ 14\plus{}2\plus{}3\plus{}7\plus{}8\equal{}35$ # marbles each person must have = $ 7$ Alex must give away $ 8$, and Earl must give away $ 1$, Bing must recieve $ 5$, and Cara must recieve $ 4$. $ 8\plus{}1\plus{}5\plus{}4\equal{}18$ Divide by 2 since each marble is recieved and given away $ \frac{18}{2}\equal{}9$
NOTE: You only need to just add up number needed to be given away, or add up the number you must recieve, not both.


Poate aceste materiale pentru unii nu erau accesibile, acum le au in lb.romana!

[quote="Moubinool"]can we have a PDF version ?

thank you[/quote]

 here it is !


Quincy and Celine have to move 16 small boxes and 10 large boxes. Celine can move a small box in 2 min and a large box in 6 min. Quincy can move a small block in 3 min and a large block in 5 min. They start moving the boxes at 9 A.M. What is the earliest time at which they can be finished moving all of the boxes?

[hide="solution"]
Since Celine is faster at moving small boxes and Quincy is faster at moving large boxes, we can first start by making them move the boxes they're faster at.
For Celine to move 16 small boxes, it will take $ 16*2\equal{}32$ minutes. In that time, Quincy can move 6 large boxes and will be moving the 7th one.
When Celine finishes the small boxes, she will start moving the remaining large boxes.
At 35 minutes, 7 large boxes would have been moved.
At 38 minutes, 8 large boxes would have been moved.
40 minutes-9 large boxes.
44 minutes-10 large boxes.
So the answer is $ \boxed{9: 44   A.M.}$[/hide]
I'm not that sure of my answer, though.


1. Solve for the given real number $ a$ the following euqation in the real numbers $ x$:
    $ (a\minus{}1)\cdot(|x|\plus{}|x\plus{}2|)\equal{}3a\minus{}4$

first case: $ x \le \minus{} 2$:

$ (a \minus{} 1)( \minus{} 2x \minus{} 2) \equal{} 3a \minus{} 4$
$ \minus{} 2x \minus{} 2 \equal{} \frac {3a \minus{} 4}{a \minus{} 1}$
$ \minus{} 2x \equal{} \frac {5a \minus{} 6}{a \minus{} 1}$
$ x \equal{} \frac {6 \minus{} 5a}{2a \minus{} 2}$

$ \frac {6 \minus{} 5a}{2a \minus{} 2} \le \minus{} 2 \implies a < 1$ or $ a \ge 2$

second case: $ \minus{} 2 \le x \le 0$:

$ (a \minus{} 1)( \minus{} x \plus{} x \plus{} 2) \equal{} 3a \minus{} 4$
$ 2a \minus{} 2 \equal{} 3a \minus{} 4$
$ a \equal{} 2$

third case: $ x \ge 0$:

$ (a \minus{} 1)(2x \plus{} 2) \equal{} 3a \minus{} 4$
$ x \equal{} \frac {a \minus{} 2}{2a \minus{} 2}$

$ \frac {a \minus{} 2}{2a \minus{} 2} \ge 0 \implies a < 1$ or $ a \ge 2$.

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

Unless I made a computational error (easy to do in such a problem), the solution is:

For $ 1 \le a < 2$ , there are no solutions.

For $ a \equal{} 2$, every $ x \in \mathbb{R}$ such that $ \minus{} 2 \le x \le 0$ is a solution.

For all other values of $ a$, the solutions are $ \frac {a \minus{} 2}{2a \minus{} 2}$ and $ \frac {6 \minus{} 5a}{2a \minus{} 2}$.


Is there any Expresion in $n$ denoted by $E(n)$ so that \[ E(n)=E(n+1)=n \]

??? I guess this expresion use floor or ceil function.

As understand it, $E(n) = E(n+1) =n$ for all $n$. But that's clearly not possible...
So can you please explain what you want¿

$E(n)=E(n+1)=n\ \forall n$ $\implies E(n+1)=E(n+2)=n+1$ $\implies n = n+1$ $\implies 0=1$ ... so if u find one tell me because that would prolly be another nice proof for $0=1$ ...


trigonometry
geometry
geometric transformation
homothety
circumcircle
power of a point
radical axis

Be ABC triangle with NPC and the excircles (Ia),(Ib),(Ic). If X,Y,Z are the touch points between NPC and (Ia),(Ib),(Ic) also (L,M);(E,K);(S,Q) are the touch points between (Ia)-(AB,AC);(Ib)-(BC,AB);(Ic)-(AC,BC) respectively. Prove that: XM.YK.ZQ=XL.YE.ZS

Here are a few empirically discovered facts:

The pairs of lines $(ZQ,YE),\ (XM,ZS),\ (YK,XL)$ meet on the NPC. Call the intersection points $P_1,P_2,P_3$, respectively. Even more can be said: if $N$ is the nine-point center, then $ZQ,YE$ concur on the perpendicular from $N$ to $BC$.

The lines $XP_1,YP_2,ZP_3$ are concurrent.

Note that these two results solve the problem, as we can use trig Ceva in the NPC. I'll try to prove these, but I'm really tired and it will probably have been solved by Darij by the time I come back :D:D.

My first statement can be derived from a configuration which generalizes this one (I’m using these notations so that the relations with the problem are clear):

Given non-intersecting circles circles $(I_b),(I_c)$ (the non-intersecting part can be dropped, I think) with external tangnent $EQ$ ($E\in (I_b),Q\in (I_c))$), and a circle $n$ tangent to both at $Y,Z$ respectively, then $EY,QZ$ intersect on $n$ and on the radical axis of $(I_b),(I_c)$. Call this point of intersection $P_a$. Moreover, the line $P_aN\perp EQ$, where $N$ is the center of $n$.

Thos follows easily from some playing around with the cyclic quadrilateral $EQZY$.

There's a more general setting for this, so let me write this in detail (sorry if it's going to turn out really long :D).

Given are three pairwisely non-intersecting circles (this can be dropped, but I'm keeping it just to be on the safe side :)) called $(I_a),(I_b),(I_c)$ (I'm keeping the notations to highlight the connection). Let $(N)$ be a circle externally tangent to all three of them in $X,Y,Z$ respectively. The Radical axis of $(I_b),(I_c)$ cuts $(N)$ in a point $P_a$ opposite to $X$ wrt $YZ$. In the same way we define $P_b,P_c$. Then $XP_a,YP_b,ZP_c$ are concurrent.

I'm sure it's clear how this solves the problem.

In order to prove that thing up there, I'll use a well-known problem: We are given triangle $ABC$ andthe points where its incircle touches the sides, $D,E,F$, and a point $P$. Then, if $PA,PB,PC$ cut the incircle at $P_a,P_b,P_c$, then $DP_a,EP_b,FP_c$ are concurrent (it was even solved on the board, I think). We can now apply this for $ABC=$ the triangle formed by the tangents to $(N)$ in $X,Y,Z$, $P=$ the radical center of $(I_a),(I_b),(I_c)$, and $D,E,F=X,Y,Z$ respectively.

I hope there are no errors.

I'm really tired of this Feuerbach configuration, so I'll just comment about one of your first observations, Grobber:

[quote="grobber"]The pairs of lines $(ZQ,YE),\ (XM,ZS),\ (YK,XL)$ meet on the NPC. Call the intersection points $P_1,P_2,P_3$, respectively. Even more can be said: if $N$ is the nine-point center, then $ZQ,YE$ concur on the perpendicular from $N$ to $BC$.[/quote]

In fact, here is how I prove this observation (I've already sent the proof to Juan Carlos):

Let $U_1$, $V_1$, $V_1$ be the midpoints of the arcs $M_bM_c$, $M_cM_a$, $M_aM_b$ on the nine-point circle, where $M_a$, $M_b$, $M_c$ are the midpoints of the sides BC, CA, AB of triangle ABC. Then I will prove that the point $U_1$ lies on the lines ZQ and YE and on the perpendicular from the point N to the line BC. This will, of course, yield your result and the fact that $U_1 = P_1$.

Let $U_2$, $V_2$, $W_2$ be the midpoints of the arcs $M_bM_aM_c$, $M_cM_bM_a$, $M_aM_cM_b$, respectively, on the nine-point circle.

Let R be the point where the C-excircle of triangle ABC touches the side AB. Then, it is easy to prove (by angle chasing) that QS || $U_1V_1$, SR || $V_1W_2$ and RQ || $W_2U_1$. In other words, the corresponding sidelines of triangles QSR and $U_1V_1W_2$ are parallel; hence, these triangles are homothetic. Let $Z_1$ be the homothetic center of these two triangles, i. e. the center of the homothety z which maps the triangle QSR to the triangle $U_1V_1W_2$. This homothety z must also map the circumcircle of triangle QSR to the circumcircle of triangle $U_1V_1W_2$. But the circumcircle of triangle QSR is the C-excircle of triangle ABC, and the circumcircle of triangle $U_1V_1W_2$ is the nine-point circle. Hence, the homothety z maps the C-excircle of triangle ABC to the nine-point circle; this means that the center $Z_1$ of this homothety is a center of similitude of the C-excircle and the nine-point circle, and actually it is the internal center of similitude, since the homothety z has a negative factor. But the internal center of the similitude of the C-excircle and the nine-point circle is the point of tangency of these circles, i. e. the point Z.

Thus, the point $Z_1$ coincides with the point Z. It follows that the point Z is the homothetic center of triangles QSR and $U_1V_1W_2$, and this yields that the point Z lies on the lines $QU_1$, $SV_1$ and $RW_2$. Consequently, the line $QU_1$ passes through Z, or, in other words, the point $U_1$ lies on the line ZQ. Similarly, the point $U_1$ lies on the line YE.

Now it only remains to show that the point $U_1$ lies on the perpendicular from the point N to the line BC. But this is easy: The point $U_1$ is the midpoint of the arc $M_bM_c$, hence it lies on the perpendicular bisector of the segment $M_bM_c$. But this perpendicular bisector is clearly perpendicular to the line BC (since it is perpendicular to the line $M_bM_c$, and the line $M_bM_c$ is parallel to the line BC), and it passes through the point N (since the points $M_b$ and $M_c$ lie on the nine-point circle whose center is N). Thus, the perpendicular bisector of the segment $M_bM_c$ is the perpendicular from the point N to the line BC. And thus, we see that the point $U_1$ lies on this perpendicular from the point N to the line BC.

Proof complete.

I'm sorry but I have not thought about your generalizations yet: too much geometry for me...

  Darij


What does the brightness of a bulb in a circuit depend on (a copper wire is used as conductor)? Is it the current, voltage (potential difference), or power?

Any help would be appreciated.

I believe it has to do with the power. But power is the product of voltage and current. So i guess it depends on all three. Well, there are several equations for power.

P = VI = I^2(R) = (V^2)/R

but these equations are for power produced by a single resistor (i believe), so the total brightness depends on the total power created by the circuit. So i believe power is the answer you are looking for.

Ok, suppose you have a battery, two identical bulbs, and connecting wires.

If the bulbs were connected to the bulb in series, then the current passing through both bulbs is the same but the voltage across each bulb is half the voltage of the battery (omit internal resistence of battery).

Now if the bulbs are connected to the battery in parallel, the current passing through each bulb is half of the current in the main circuit, but the voltage across each bulb is the same as the voltage of the battery.

Suppose the current through the circuit was $A$ amperes and the voltage provided by the battery was $V$ volts. In each case each bulb receives power equivalent to $\frac{1}{2}AV$. How is it then that the bulbs connected in parallel are brighter than the bulbs connected in series?

But the currents will not be the same in both cases! Let's be a bit quantitative: suppose both bulbs have the same resistance $R$ and are connected to an EMF source $E$ with an internal resistance $r$.

(1): Parallel connection: $V = E\left(1-\frac{r}{r+R/2}\right) = E\frac{R}{2r+R}\Rightarrow P = \frac{V^{2}}{R}= \frac{E^{2}R}{(2r+R)^{2}}$.

(2): Serial connection: $I = \frac E{2R+r}\Rightarrow P = I^{2}R = \frac{E^{2}R}{(r+2R)^{2}}$, so the parallel connection is brighter if $R > r$ (and certainly for $r = 0$).

Thanks.. this helps a lot. I think I will print this out for my physics teacher.. he never taught me this.  :)

One thing you have to be careful that "brightness"  need not be equal to power put in.

A 1000 watt  heater coil (which radiates  most of the power as heat)  may be barely glowing red  and  be less bright than a 100 watt bulb and and ordinary 100 watt bulb (which gets hot and thus wastes energy ) may be less bright than a 40 watt tube-light.

So the value you get above  has to be modified by "efficiency" or percentage of power which is going as light.

That why a LED light which uses same "power" as others is much bright becuase almost all the energy is transmitted as light. (A simple  light bulb is fairly inefficient (less than 20%)

Hope this helps.


Find the remainder when $ x^n$ is divided by $ x^2\minus{}x\minus{}1$ where $ n$ is a positive integer greater than or equal to $ 2$.

[hide="Solution and discussion"] Write $ x^n \equal{} (x \minus{} \phi)(x \minus{} \varphi) Q(x) \plus{} ax \plus{} b$.  We have $ \phi^n \equal{} a \phi \plus{} b$ and $ \varphi^n \equal{} a \varphi \plus{} b$, so (recalling that $ \frac{1}{\phi} \equal{} \minus{} \varphi$)

$ a \equal{} \frac{\phi^n \minus{} \varphi^n}{\phi \minus{} \varphi} \equal{} F_n$
$ b \equal{} \frac{\phi^{n\minus{}1} \minus{} \varphi^{n\minus{}1}}{\phi \minus{} \varphi} \equal{} F_{n\minus{}1}$.

Hence our remainder is $ F_n x \plus{} F_{n\minus{}1}$.  Note that the more interesting result is that

$ x^2 \minus{} x \minus{} 1 | x^n \minus{} F_n x \minus{} F_{n\minus{}1}$.

I may or may not write a blog post about identities of this type.  Note that we could've proven this by induction:  $ x(F_n x \plus{} F_{n\minus{}1}) \equal{} F_n x^2 \plus{} F_{n\minus{}1} x \equiv F_{n\plus{}1} x \plus{} F_n \bmod (x^2 \minus{} x \minus{} 1)$. [/hide]


Let  $x,y,z$ be nonnegative numbers such that $x^{2}+y^{2}+z^{2}=1$, then

$\frac{2}{\sqrt{3}}\sum\frac{x}{y^{2}+z^{2}}\ge \frac{3}{x+y+z}\sum{2x^{3}+xyz \over (2x+y)(2x+z)}\geq 1$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=132530


is it possible to get LaTeX to work on my phpbb forum?

Hi,
Have a look at this topic [url]http://www.mathlinks.ro/Forum/topic-70632.html[/url].

A+


$ ABCD$ is a square. How many points $ P$ are there in the plane of the square that have the property that $ PAB$, $ PBC$, $ PCD$, $ PDA$ are all isosceles triangles?

Is it one? The most obvious point would be the center.

The problem said nothing about the triangles being [i]generate[/i]. May we count the four vertices as well as the nine points that luisgometria and wizintraining mentioned?

To prove that there are 9, any point forming an isoceles triangle with a segment is either on it's perp bisector or on one of the circles with center at one endpoint passing through the other. So draw all four circles and the two perpendicular bisectors to the sides of the square, we find 9 of the intersections work.


If
$ c \geq b\geq a>0$
then
$ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2\geq3(\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c})$

$ (\frac{a}{b})^2\plus{}(\frac{b}{c})^2 \ge 2.\frac{a}{c}$
similarly for other fractions and we r done!


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{1\plus{}\sin \theta \minus{}\cos \theta}\ d\theta$

Put $ t \equal{} tan frac{\theta}{2}$
so the integral reduces to $ \int_{\sqrt{3}}^{\infty}\frac{2dt}{t^2 \plus{} 2t}  \equal{} (log(\frac{t}{t\plus{}2} )$ from $ {\sqrt{3}}$ to $ {\infty}$

so the answer $ \equal{} log(\frac{\sqrt{3} \plus{} 2}{\sqrt{3} }$

hello kunny, setting $ t=\tan(\frac{\theta}{2})$ we have $ \cos(\theta)=\frac{1-t^2}{1+t^2}$,$ \sin(\theta)=\frac{2\theta}{1+t^2}$ and $ d\theta=\frac{2\,dt}{1+t^2}$, then we get
${ \int\frac{1}{1+\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}}\cdot\frac{2 dt}{1+t^2}}=\int\frac{dt}{1+t}=\ln|1+t|=\ln|1+\tan(\frac{\theta}{2})|$ and our result is $ -\frac{1}{2}\ln(3)+\ln(\sqrt{3}+3)-\ln(2)$
Sonnhard.

incorrect ?? :oops:    what is wrong can u please tell me??

$ \int_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{1\plus{}\sin \theta \minus{}\cos \theta }d\theta$

$ \vspace{1pt}\equal{}\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{2\sin \frac{\theta }{2}\left( \sin \frac{\theta }{2}\plus{}\cos \frac{\theta }{2}\right) }d\theta$

$ \equal{}\int_{\frac{\pi }{3}}^{\frac{\pi }{2}}\allowbreak \left( \frac{1}{\tan \frac{\theta }{2}}\minus{}\frac{1}{\tan \frac{\theta }{2}\plus{}1}\right) d\left( \tan \frac{\theta }{2}\right)$

$ \equal{}\allowbreak \ln \frac{\sqrt{3}\plus{}1}{2}$

[quote="vijaymenon"]Sorry made a mistake its actually the following.

the integral reduces to $ \int_{\sqrt {3}}^{\infty}\frac {dt}{t^{2} \plus{} t} \equal{} (log(\frac {t}{t \plus{} 1})$ from  $ \sqrt {3}$to $ \infty$

sop answer $ \equal{} log(\frac {\sqrt {3} \plus{} 1}{\sqrt {3}}$[/quote]

The interval of Integration is wrong. :(

[quote="IndoChina"]$ \int_{\frac {\pi }{3}}^{\frac {\pi }{2}}\frac {1}{1 \plus{} \sin \theta \minus{} \cos \theta }d\theta$

$ \vspace{1pt} \equal{} \int_{\frac {\pi }{3}}^{\frac {\pi }{2}}\frac {1}{2\sin \frac {\theta }{2}\left( \sin \frac {\theta }{2} \plus{} \cos \frac {\theta }{2}\right) }d\theta$

$ \equal{} \int_{\frac {\pi }{3}}^{\frac {\pi }{2}}\allowbreak \left( \frac {1}{\tan \frac {\theta }{2}} \minus{} \frac {1}{\tan \frac {\theta }{2} \plus{} 1}\right) d\left( \tan \frac {\theta }{2}\right)$

$ \equal{} \allowbreak \ln \frac {\sqrt {3} \plus{} 1}{2}$[/quote]

Yes. That's exactle same as my solution. :lol:

sorry i got it iput the limits wrongly it was from $ \frac{1}{\sqrt {3}}$ to $ 1$

so ans is indeed $ \equal{} \allowbreak\ln\frac {\sqrt {3} \plus{} 1}{2}$

O.K. :lol:

Oh, it's high time I went to bed, it's 3:21 AM in Japan. Good Night!

hello, yes, this is the same result as mine.
Sonnhard.


Find All functions $f: \mathbb{R} \to \mathbb{Q}$ such that:

1- $\forall x \in \mathbb{R}$:  $(x-1) \lneq f(x) \le x$

2- $\exists p \in \mathbb{Z}$ : $\forall x \in \mathbb{R}$, $f(x+p)= p+f(x)$

3- $\forall n \in \mathbb{Z}$: $f(n) = n$

[quote="lordWings"]Sorry, I'm not used to this notation. What does "$a \lneq b$" mean? I can't decide between these two options.
1) $a<b$ and not $a=b$, so it's strictly $a<b$.
2) not $a \leq b$, so it's $a>b$[/quote]
$a\leq b$ means $a$ is less or equal to $b$
examples: $2\leq 3$ &  $3\leq 3$

There's a little line over the sign. I'm asking about $\lneq$ (lneq), and not about $\leq$ (leq).
Is it "is less and not equal" or "is not less nor equal"?

Sorry i did not see the little line, it's this
[quote="lordWings"]
1) $a<b$ and not $a=b$, so it's strictly $a<b$.
[/quote]


geometry
3D geometry
tetrahedron
sphere
trigonometry
geometry proposed

Find the length of a side of regular tetrahedron which is inscribed in the sphere with radius 1.

Let $ABCD, R$ be a regular tetrahedron and the radius of its circumscribed sphere. Denote $\vartheta  = \measuredangle ABH,{\rm{ }}AB = \ell $
and let $H,K$ be the midpoints of  $CD, AB$ respectively (see attached figure). Cosine theorem to $\triangle ABH$ yields:

$\cos \vartheta  = \frac{{AB^2  + BH^2  - AH^2 }}{{2 \cdot AB \cdot BH}} = \frac{{AB}}{{2 \cdot BH}} = \frac{\ell }{{\sqrt 3 \ell }} = \frac{1}{{\sqrt 3 }}$

Thus we have $\sin \vartheta  = \sqrt {1 - \cos ^2 \vartheta }  = \sqrt {1 - \frac{1}{3}}  = \frac{{\sqrt 6 }}{3}$ and consequently

$R = OA = \frac{{AK}}{{\sin \measuredangle AOK}} = \frac{{AK}}{{\sin \measuredangle ABH}} = \frac{{{\ell  \mathord{\left/
 {\vphantom {\ell  2}} \right.
 \kern-\nulldelimiterspace} 2}}}{{\sin \vartheta }} = \frac{{\sqrt 6 }}{4}\ell $

If $R=1$ then $\ell  = \frac{4}{{\sqrt 6 }} = \frac{2}{3}\sqrt 6 $. $\blacksquare$

Thank you,Leon.You are right. :)

Comment:Let $E$ be a interception point of two lines of $AO$ and $BH,$we have $AO:OE=3:1$.

For a regular tetrahedron with a side of $l$,

The hight :$h=\frac{\sqrt{3}}{4}l$.

The radius of inscribed sphere:$r=\frac{1}{4}h=\frac{\sqrt{6}}{12}l$

The radius of circumscribed sphere:$R=3r=\frac{3}{4}h=\frac{\sqrt{6}}{4}l$

The area of $\triangle ABH:S=\frac{\sqrt{2}}{4}l^2$

The volume :$V=\frac{\sqrt{2}}{12}l^3$

kunny


LaTeX
number theory unsolved
number theory

consider n is a natural number and k+1 is the least prime number that K+1>n.
prove that 1,2,...,k are divisors of n!.

really there is nobody to solve this problem?

Let me try to explain why none is replying:

The later ones are more important:

1) It's not in Latex format. It doesn't matter much though in this case, but it comes useful in many mathematical things you need to express here. And learning is pretty easy and worth it.

2)Totally stupid title. If few has the interest to open the topic then chances are low someone will answer it. See point (a) [url=http://www.mathlinks.ro/viewtopic.php?t=135914]here[/url].

3)It is not clear if "k" and "K" are the same thing. It is not conventional to define a new prime as "k+1". Generally "p" is used. For all these one might think he has misunderstood the problem.

4)If everything is as I understood ("k"="K" and all), then it is so easy (at most [url=http://en.wikipedia.org/wiki/Bertrand%27s_postulate]Bertrand's postulate[/url] is needed and then it follows directly) that one might think that he has misunderstood the problem or wouldn't simply want to reply.


An integer $ a$ has just two digits. When the digits are reversed the resulting number $ b$ is $ p\, \%$ larger than $ a$. Given that $ p$ is an odd integer find the largest value of $ p$

[hide="Solution"]The percentage change between a and b is $ \frac{100(b-a)}{a}$

Write $ a$ as $ 10m+n$

Then $ b$ is $ 10n+m$

Thus our expression for the percentage becomes:

$ \frac{900(n-m)}{10m+n}$

Where $ n,m$ are integers less than 10 and $ m \ne 0$.

In order for our expression to be an odd integer, then if $ (n-m)$ is even, $ (10m+n)$ must be a multiple of 8 but not 16. If $ (n-m)$ is odd, $ (10m+n)$ must be a multiple of 4 but not 8.

With these restrictions in place, and systematically checking values of $ m,n$ which satisfy the restrictions on the denominator, we find that the largest possible percentage difference is $ 75%
$

In fact, this occurs at 4 distinct values of $ a$.

$ (a,b)=(12,21),(24,42),(36,63),(48,84)$[/hide]


You were assigned homework from Ms. B, but you don't know how to do it. You have no friends, and the only place you could go is the supermarket.
There is 1/2 chance of meeting Ms. A, 1/3 chance of meeting Ms. B, and so on, so there is 1/26 chance of meeting Ms. Z. If no two teachers were in the supermarket at the same time, what are the possibilities of meeting Ms. B?

First of all, there is a mistake here, I think. If there is a 1/2 chance of meeting Ms. A, and so on, where the chance of meeting a teacher is the reciprocal of her place in the alphabet +1, shouldn't the chance of meeting Ms. Z be 1/27? And even then, do the probabilities of meeting a teacher take into account the fact that no 2 teachers are there at once?

As for the first point, you're absoluterly right, I'll go and fix it right now.
The second point, however, might need some thoughts. Think about case in which there's only Ms. A and Ms. B. (A is 1/2 and B is 1/3 chance) You must meet Ms. B but not Ms. A (since they can't be both there) so (1-1/2)*1/3=1/6.

Uh Tare, I don't think thats quite right.. in the just A,B case, you are *given* that no two teachers are there at the same time, so the probability of meeting Ms B. is (1/6)/(1/6 + 2/6 + 2/6) = 1/5.


Let $a,b,c>0$, prove that
\[ \frac a{\sqrt{b^2+ca}}+\frac b{\sqrt{c^2+ab}}+\frac c{\sqrt{a^2+bc}}\geq\frac3{\sqrt2}. \]

[quote="hungkhtn"]Also a wrong problem, too. Let $a=0$.[/quote]

You mean $a \rightarrow 0$ ??

 :)


group theory
abstract algebra
vector
linear algebra
number theory
superior algebra
superior algebra unsolved

I'm having some trouble in intuitively and rigorously understanding what a generator is.

So what is given a group $ G$, what is $ <S>$? I'd really appreciate some examples, as well as theory.
Also, for g in G, why is <g> = {g^n}?

Let $ G$ - group, $ S\subseteq G$. Group generated by $ S$ (denoted by $ \left < S\right >$) is subgroup of $ G$ and it contains all elements that can be made by using finite numbers of multiplying and inversing elements from $ S$.
\[ \left<S\right> : \equal{} \{ g_1^{\varepsilon_1}\ldots g_k^{\varepsilon_k} | g_1, \ldots, g_k \in S, \varepsilon_1, \ldots, \varepsilon_k \in \{1,\minus{}1\}, k\in \mathbb{N} \}\]
Or equivalently we can say that's smallest subgroup of $ G$ containing $ S$.
I can use some analogy from linear algebra, let $ X$ - vector space, $ A\subseteq X$. You remember notion of linear span of $ A$? It was denoted by $ \textrm{span} A$, $ \textrm{lin} A$ or something like that :) It's very similar notion to those of $ \left < S\right >$.

You mean that
\[ \left < g\right > \equal{} \{ g^n | n\in \mathbb{Z}\},
\]
right? When you multiply $ g$ or get inverse of $ g$ and repeat that process on results, you always get something like $ g^n$.

On the end, I advice you to get some textbook of algebra :)

I assume $ S$ is a subset of $ G$. $ <S>$ represents the set of all elements of $ G$ that are expressible as the product of elements or inverses of elements in $ S$.

Example: Let's take the integers under addition as $ G$, and let $ S \equal{} \{4,6\}$. Then $ <S>$ is every integer that can be written as a linear combination of $ 4$s and $ 6$s (since groups have inverses, this is with respect to both addition and subtraction). Using number theory you can determine this to be the even integers pretty easily.

You probably know what subgroups are if you're learning generators. The way I think about $ <S>$ is the smallest possible subgroup of $ G$ containing all elements of $ S$. In a more concrete way, while $ S$ can be any subset of $ G$ so it doesn't have to be a group, $ <S>$ "completes" $ S$ and makes it into a group.

Some theory related to generators: Cyclic groups are groups that can be generated by one element. For instance, the integers under addition can be generated by $ 1$. Cyclic groups have a number of important properties. If $ G$ is a cyclic group, the following are true.
1. $ G$ is abelian.
2. If $ G$ is infinite, $ G$ is isomorphic to the integers under addition.
3. If $ G$ is finite, $ G$ is isomorphic to the integers under addition modulo $ n$, where $ n$ is the number of elements in $ G$.
Also, all finite groups with a prime number of elements are cyclic.

One way to think about $ <S>$ is the set of all "words" using the elements of $ S$ and their inverses for an alphabet. In most groups, a large number of these words will be the same thing. In our example with the integers and $ S \equal{} \{4,6\}$ above, $ 4\plus{}4\plus{}4 \equal{} 6\plus{}6$, so the words $ 444$ and $ 66$ are the same. Free groups, however, constitute the unique group that has no overlap between any words (except in cases with elements and their inverses adjacent). I haven't studied free groups much, so I'll stop here before I say something stupid and ask you to look somewhere else for more info.

OK, thanks for your replies! So why is $ D_{2n}\equal{}<r,s>$?

The dihedral group of order $ 2n$ contains elements that are all of the form $ r^us^t$.  Do not forget to write down the relations as well, for accuracy.


This is complex to me anyways....

The solution to the inequality $\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}\ge1$ is the union of a finite number of disjoint intervals, each of the from $a<x\le b$. If the length of the interval is defined as $b-a$, what is the sum of the lengths of all these intervals?

The wording also confuses me...

[quote="Thales418"]${x=u+\frac{9}{3}}$[/quote]

It's right here.

Can someone finish up Thales418's work for the real root? or maybe do it a different way?
I really want to know how to do this problem. Also, what do all the weird words mean?

you can check the graph of the LHS. Then you can check that the graph meets the graph of $y=1$ three times. Let the meeting points $(a,1),(b,1),(c,1)$ where $1<a<2<b<3<c$. then the range of $(LHS) \geq 1$ is $\{ x : 1<x\leq a \}$, $\{ x : 2<x\leq b \}$, and $\{ x : 3<x\leq c \}$. so what we want is $(a-1)+(b-2)+(c-3)=a+b+c-6$.

note that $a,b,c$ are the solutions of $(LHS)=1$ or $x^{3}-9x^{2}+23x-17=0$. there fore the sum of the solutions, namely $a+b+c$, is equal to 9. hence, the needed value is $a+b+c-6=3$.

I still don't understand this, and your graph doesn't seem to match your equation...

Can someone give me a full explanation of the original problem

Much Appreciated!


group theory
abstract algebra
vector
induction
invariant
linear algebra
linear algebra unsolved

Suppose we have $G$ a subgroup in $GL_n(C)$ such that all its elements have all eigenvalues of modulus 1. Prove that for any non-zero vector $X$ we have $inf_{A\in G}{||A(X)||}>0$ where $||$ is a norm on the matrices.

Can someone help me with this problem? :?

I think one can deduce the solution from the Molien's theorem from
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=303812#p303812]www.mathlinks.ro/Forum/viewtopic.php?p=303812#p303812[/url] .

 Let's prove it by induction on the dimension. The case $ n=1 $ is
clear. Now suppose $ n > 1 $. Suppose first, that the algebra
generated by elements of $ G $ is the whole $ M_{n}( \mathbb{C} )
$. Then take elements $ A_{1},A_{2}, \ldots , A_{n^{2}} \in G $,
such that they form a basis for $ M_{n}( \mathbb{C} ) $. Then we
have for every $ B \in G $ that $ | tr(A_{j}B) | \leqslant n $ ,
for $ j=1,2, \ldots , n^{2} $ ( because $ A_{j}B \in G $, and
hence has all eigenvalues of modulus $ 1 $ ). Therefore we see
that $ G \subset M_{n}( \mathbb{C} ) $ is precompact ( i.e.
bounded ) , so it's closure $ \overline{G} $ is compact and
moreover does not contain non-invertible matrices, because
determinant of each element of $ G $ has modulus $ 1 $, so it is
true also for elements of $ \overline{G} $. Now let us show that
it is impossible that $ inf_{A \in G} ||Ax|| = 0 $ for some $ x
\neq 0 $. Indeed, if we have that $ lim_{k \rightarrow \infty}
||A_{k}x || = 0 $, for some sequence $ A_{1},A_{2}, \ldots \in G
$, then we can take some subsequence of it, which converges to
some $ B \in \overline{G} $, so we will have that $ Bx = 0 $, so $
B \in G $ will be not invertible, which is not true.
 Now consider the case, when the algebra generated by elements of
$ G $ is not the whole $ M_{n}( \mathbb{C} ) $. Then by Molien's
theorem, there exists non-trivial invariant subspace $ 0
\subsetneqq W \subsetneqq M_{n}( \mathbb{C} ) $ of $ G $. Then we
have an action of $ G $ on $ W $ and on $  \mathbb{C}^{n} / W $.
Note also that eigenvalues of every $ A \in G $, restricted on $ W
$ are also eigenvalues of $ A $, and every $ A \in G $, acting on
$ \mathbb{C}^{n} / W $ also has all eigenvalues containing on the
set of eigenvalues of $ A $, so all of them are of modulus $ 1 $.
Therefore we can use the induction hypothesis for action of $ G $
on $ W $ and on $ \mathbb{C}^{n} / W $. Take some $ x \neq 0 $.
Then if $ x \in W $, by induction hypothesis, $ inf_{A \in G}
||Ax|| > 0 $. If now $ x \notin W $, then it's $ [x] \in
\mathbb{C}^{n} / W $ is non-zero , so $ inf_{A \in G} ||A[x]||_{
\mathbb{C}^{n} / W } > 0 $, where $ || y ||_{ \mathbb{C}^{n} / W }
$ on $ \mathbb{C}^{n} / W $ is induced from $ ||x|| $ on $
\mathbb{C}^{n} $. But of course $ ||Ax|| \geqslant ||A[x]||_{
\mathbb{C}^{n} / W } $, so $ inf_{A \in G} ||Ax|| > 0 $.

I hope it is correct.

Exactly, this is how I (finally!!!) solved this damn problem! I still wonder about the difficulty of that exam at X...