I think you can remove the latest option... those people won't get what you're asking in the poll :D

Most people whose native language is English do not write or speak perfectly.

They don't, but they would be able of doing so if they wanted to... especially people intelligent enough to visit this place :)

I'm not convinced at all. One must be taught how to speak and write properly, and many schools do not teach such things.

Perfect doesn't mean you're a specialist. Perfect just means that you build your sentences without errors, you know how to spell most words and you can write fluently in a way everyone understands you :)

To these norms, you're perfect in english ComplexZeta :)

I tend to think that ``perfect'' means ``without errors.'' Having only a few errors does not constitute perfection.

Since I am NOT a perfect writer, I did not use the word "perfect" as I should have. Its intended meaning is the one Peter presented ;)

English can be quickly learnt in such a way that you can have a small talk with other people. But writing and speaking perfectly is a real huge step I think.  :)

English ain't  ;)  my mother's tongue but I'm not so bad I suppose. I prepare for CAE  :)

[quote="RC-7th"]Perhaps the last one should be "I do not speak english, but can read/write." Otherwise, how could you understand the poll?[/quote]It may be curious, but almost all people that I have encountered (and I do not refer only to Romanians here) that didn't knew to speak english, knew to say something like "no speak(ing) english" ;) so they might understand :D

[quote="Valentin Vornicu"]It may be curious, but almost all people that I have encountered (and I do not refer only to Romanians here) that didn't knew to speak english, knew to say something like "no speak(ing) english" ;) so they might understand :D[/quote]

Well umm... I can imagine those people can say that... but I cannot imagine that they'd visit an english site like this then :P

You don't need to know English to use this site!  Use Babelfish!

Oooww yeah

Prove integers that a,b,c sufficing $a^2-2bc<b^2+c^2$ with sides of a triangle :roll:

Good writing and speaking skills. :blush:

I disagree with the 'mother language.' (I correct my mother on grammar all the time, lol!)

Well, my English is relatively good, but people from other langauges could use Google translator to translate a webpage you know.

Bien, mi inglés es relativamente bueno, pero la gente de otros langauges podría utilizar el traductor de Google para traducir un Web page que sabes.

Bien, mon anglais est relativement bon, mais les gens d'autres langauges pourraient utiliser le traducteur de Google pour traduire une page Web que vous savez.

Bene, il mio inglese è relativamente buono, ma la gente da altri langauges potrebbe utilizzare il traduttore di Google per tradurre un Web page che conoscete.

حسنا ، يود الانكليزيه جيدة نسبيا ، ولكن الناس من سائر لانغاوغيس يمكن استخدام غوغل مترجم ترجمة صفحة الويب تعلمون.

Translated by Google translator.

English may be my mother tongue, but I am also competent in French.

(I write all kinds of essays and whatnot)

I love English. I can't really say how good I am but I just love it.

P.S. I mean English as a subject not just speaking/writing but studying texts etc.

WELL...
I speak english...
and I started learning about 6 years ago....
(I'm Korean...)
and I'm fluent...
and my friends are like, "OMG UR SO GOOD HOW CAN U BE LIVING HERE FOR ONLY 5 YEARS???"
and I reply... "oh. I studied"

:)
sooo
i'm pretty good, but sometimes, i don't bother to have correct grammar...

Average. I feel so... AVERAGE.

Especially for an American.

Eh. 12 year old geniuses that are always at a loss of words like me I guess ARE average.

Going by what Peter said on the first page (and what Valentin meant), I do speak and write English perfectly.

However, I do use phrases like "a lot" that some formal writers frown upon, and I will not necessarily follow the more "advanced" rules used in the sentence correcting portion of the SAT.  Nevertheless, I can effectively communicate with very few errors, and people understand what I say.

[quote="hunter34"][quote="kimmystar94"]whole bunch of my teachers are like, "you have potential. you should use better vocab."
once during lunch, my friend and i spoke in SAT vocab the whole time, and my other friends didn't understand..
it was pretty funny...
my sister's response..
"that's so nerdy"[/quote]

I find this statement grammatically incoherent.[/quote]

Incorrect grammar, punctuation, and capitalization, to be precise. :P

But we are not here to criticize people of their errors, no matter what the errors may be.

Stop rhapsodizing about peripheral errors! And blah blah blah.  :P

I think I am "perfect" in the sense provided... yeah, and my mom always asks me to correct her grammar and spelling. She really needs me. She can't spell "cotton". Actually it's probably just her typing skills.

Anyway I chose the second because both my parents are pure Asian. Which part of Asia I'm concealing for now, or as long as somebody rediscovers the critical piece of information embedded within the Counting Game in G&FF.

P.S. When bored I randomly generate words from my dictionary for pure recreation.

IMPORTANT EDIT: If you feel offended or anything, please read my new signature. Thank you.

I put down bad writing/speaking skills :huh:


Prove that if $3\leq d\leq 2^{n+1}$, then $(a^{2^n}+1)$ is not divisible by $d$ for all positive integers $a$

Let $d \geq 3$ a prime such that $d \mid a^{2^n}+1$.
We can assume that $(a,d)=1$.
We have $a^{2^{n+1}}\equiv1[d]$ so the order $\alpha$ of $a$ in $\left(\mathbb{Z}/d\mathbb{Z}\right)^*$ divides $2^{n+1}$. But $a^{2^n}\equiv-1[d]$, so $\alpha=2^{n+1}$. By Fermat little theorem we obtain $\alpha \mid d-1 \Rightarrow 2^{n+1}<d$.


A bowling lane is $ 3\frac12$ feet wide and 60 feet long. How many square feet are in its area?

We just multiply to get $ 210$ square feet.

[hide="Solution"] All you have to do with this problem is to multiply!! Which is extremely easy.

$60\times3=180$, and $60\times\frac{1}{2}=30$, then we have $180$ and 30 left over, and $180\times30=5400$ which is way to big, and this next you just add $180+30=\boxed{210}$

The reason I did 60x3, is because of 60 feet and the 3 wasn't in the fraction?? Or watever, then whatever I do to one problem I have to do the other, so then I do 60x1/2 because the rest of the fraction and 60 feet, and 60x3=180 and 60x1/2=30, so we have some number left over so you just add them together and get our answer[/hide]


How do you solve problems like this one:

A number leaves a remainder of 2 when divided by 5, leaves a remainder of 1 when divided by 4, and leaves a remainder of 2 when divided by 3. Find the smallest possible number that fits these conditions.

and this one:

If there are 2 chairs to a table, there will be one chair left over. If there are 3 chairs to a table, there will be a table with no chairs. What is the number of chairs plus the number of tables?

I can get by with guess and check, but is there a faster way to do it? Guess and check probably won't work with larger numbers.

Modular arithmetic.  [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206]Introduction to Number Theory[/url] book explains it (I think you may be able to borrow from Mr. Lomas), although I found that the [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]Introduction to Number Theory[/url] course was more useful in teaching HOW it works.

Chinese Remainder theorem:

Notation:
$ x \equiv a$ (mod b) means $ x\minus{}a$ is divisible by $ b$.
Then we can see the $ x\equiv a$ mod b is the same a saying that $ x$ leaves a remainder of $ a$ upon dividing through by $ b$.

Therefore, we want to solve the system:
$ x \equiv 2$ mod 5
$ x \equiv 1$ mod 4
$ x \equiv 2$ mod 3

Now, Chinese Remainder Theorem guarantees that if 5,4,3 are relatively prime (which they are), then this system has a unique solution mod 5*4*3=60. The cool thing about this though, is you do not have to try all 60 possibilities, instead we take 2 of the equations, and replace them by a new equation, which we can do by Chinese Remainder Theorem.

For example, we take $ x \equiv 1$ mod 4 and $ x \equiv 2$ mod 3. By C.R.T. this has a unique solution mod $ 3*4\equal{}12$, so we write out the 2 sequences which satisfy each independently

1,5,9,13 <-- too big, so we stop
2,5,8,11,14 <-- too big, so we stop.

Here, 5 belongs to both sequences, so we can replace the 2 equations with $ x \equiv 5$ mod 12.

Now, we have the 2 equations $ x \equiv 2$ mod 5, and $ x \equiv 5$ mod 12.

For the mod 12, we have:
5,17,29,41,53

Now, we don't have to write all of the numbers mod 5, we just have to check which of these numbers is $ \equiv 2$ mod 5.

So subtracting 2 from each member of the sequence,
3,15,27,39,51

we see only the second is divisible by 5. Therefore, 17 is the unique solution mod 60, and hence the smallest solution.

This is the general method one should use to do these types of problems without going into more advanced number theoretic mathematics dealing with multiplicative inverses and the extended euclidean algorithm.

For the second, this isn't even number theory, we just have the system: $ 2t\plus{}1\equal{}c$ and $ 3(t\minus{}1)\equal{}c$

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

$ c\equal{}2t\plus{}1\equal{}2*4\plus{}1\equal{}9$

Therefore, there are 4 tables and 9 chairs.

For the first problem, you can simply add three to both modulars:

$ x$ mod $ 5$ $ \equiv{2}$ becomes $ x\plus{}3$ mod $ 5$ $ \equiv{0}$
$ x$ mod $ 4$ $ \equiv{1}$ becomes $ x\plus{}3$ mod $ 4$ $ \equiv{0}$

x is three less than the LCD of 4 and 5. x= $ \boxed{17}$


In a circle, both foci and the center are at the same point.

just remember that a circle is just a very round ellipse, just like how a square is a very square rectangle

aka, meaning,a circle is a special case of an ellipse

ALL CIRCLES ARE ELLIPSES


I watched this movie yesterday. It has quite a few things to say about nuclear weapons. The locked Korea topic just reminded me of this. I wonder if you people have watched it?

I didn't. Is Chuck Chuck Norris? Why can he divide by 0? i have always been mad at the biased math people who try to tell me I can't. Whay can Chuck Norris divide by 0?

[url]http://en.wikipedia.org/wiki/Chuck_Norris_Facts[/url]

Wikipedia is our friend. :)

stop spamming :mad: chuck is of course chuck murdoch


Let $ \huge x, y, z>0$ and $ \huge x^3\plus{}y^3\plus{}z^3\equal{}3$
Prove that !

$ \huge P\equal{}3xy\plus{}3yz\plus{}3zx\minus{}xyz \le 8$

[quote="bigbang195"]Let $ \huge x, y, z > 0$ and $ \huge x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$
Prove that !

$ \huge P \equal{} 3xy \plus{} 3yz \plus{} 3zx \minus{} xyz \le 8$[/quote]

we have $ (\sum x^3) \plus{} 6 \equal{} \sum(x^3 \plus{} 1 \plus{} 1) \ge 3\sum x$ (AM-GM)
so $ \sum x \le 3$

$ 3(1 \minus{} xyz) \equal{} \sum x^3 \minus{} 3xyz \equal{} (\sum x)(\sum x^2 \minus{} \sum xy) \le 3(\sum x^2 \minus{} \sum xy)$

$ \to (\sum xy)\minus{}xyz \le (\sum x^2) \minus{}1$

$ P \le \sum x^2 \plus{} 2\sum xy \minus{} 1 \equal{} (\sum x)^2 \minus{} 1 \le 9 \minus{} 1 \equal{} 8$
:)

[quote="bigbang195"]Let $ \huge x, y, z > 0$ and $ \huge x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$
Prove that !

$ \huge P \equal{} 3xy \plus{} 3yz \plus{} 3zx \minus{} xyz \le 8$[/quote]
The following much more stronger inequality is also true.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x^3\plus{}y^3\plus{}z^3\equal{}3.$ Prove that
\[ 2(xy\plus{}xz\plus{}yz)\leq3(1\plus{}xyz)\]


For each positive integer $m>1$, let $P(m)$ denote the greatest prime factor of $m$. For how many positive integers $n$ is it true that both $P(n)=\sqrt{n}$ and $P(n+48)=\sqrt{n+48}$?

A. 0

B. 1

C. 3

D. 4

E. 5

[hide="Hint"] $P(x) = \sqrt{x}$ iff $x$ is the square of a prime. [/hide]

I think the only solution is $1$, which gives $P(49)=7$. This leads to an answer of $B$.

[hide="Really?"] Actually, the only solution is $n = 121$.  The problem is very tricky - because $1$ is not a prime, $P(1)$ is not defined (edit: this is why the problem specified $m > 1$). [/hide]

[hide]it is evident that if P(m)=\sqrt{m}, then $m$ is a square of a prime, that means that $n=p^{2}$ and $n+48=q^{2}$ where $p$ and $q$ are primes, then we get the equation:

$(q-p)(q+p)=48$, and we can easily find solutions to this, both factors have the same partity, if even, getting rid of 4, we have =12, then no solution, if odd, then q-p=1, and q+p=49, q=25, p=24, n=576, 1 solution[/hide]

I think you should look over that second paragraph a little more carefully...


:?: PA,PB are tagent to circle O ,C is a point on PA;AH is perpendicular to BC ;prove that :∠HEC=2∠PCB. Please help me with this problem . THANKS. :blush:


Hi. I hate it when books only tell you the 'what' and not the why. If I dont know the why, why should I accept it. I hope some of you can help me learn the why.

What proof is there that shows an atom exists?
How do we know/find the shape of an atom?
How do we measure the radius of an atom?
How do we find the mass of an atom?
How do we find the charge of a subparticle?
What keeps the protons together? How do we know this?
How do we know protons, electrons, and neutrons are held together by quarks and gluons?

Most of these are the results of experiments, and then equations are refined to fit those results and predict new ones.  The radius of an atom is not a well-defined number but it is a result of, among other things, the Schrodinger equation.  Pick up an introductory book to particle physics and quantum mechanics if you want to understand these things.

[b]What proof is there that shows an atom exists?[/b]
The classic demonstration of this is [url=http://en.wikipedia.org/wiki/Geiger-Marsden_experiment]gold foil experiment[/url]
[b]How do we know/find the shape of an atom?
How do we measure the radius of an atom?[/b]
These may be derived with the [url=http://en.wikipedia.org/wiki/Schrodinger_equation]Schrödinger equation[/url].
[b]How do we find the mass of an atom?
How do we find the charge of a subparticle?[/b]
The elementary charge of a subatomic particle was determined by the [url=http://en.wikipedia.org/wiki/Oil_drop_experiment]oil drop experiment[/url].  The mass-to-charge ratio of the [url=http://en.wikipedia.org/wiki/Proton#History]proton[/url] and [url=http://en.wikipedia.org/wiki/J.J._Thomson#Third_experiment]electron[/url] were used to determine the masses of those particles.  The massof the neutron may then be determined using conservation of energy and momentum of atomic nuclei.
[b]What keeps the protons together? How do we know this?
How do we know protons, electrons, and neutrons are held together by quarks and gluons?[/b]
I know less about this, but see [url=http://nobelprize.org/nobel_prizes/physics/laureates/1990/press.html]the 1990 Nobel prize in physics[/url].

:lol:  Thanks!

Ill pick up a book on particle physics and quantum mechanics, and ask some more questions later.


Suppose the cevians AP, BQ, CR meet at T. Prove that TP/AP+TQ/BQ+TR/CR=1

Note: $ \frac{TP}{AP}=\frac{[TBC]}{[ABC]}$

Summing cyclically, we must show that:

$ \frac{[TBC]+[TCA]+[TAB]}{[ABC]}=1$

But this is obviously true.


function
real analysis
real analysis unsolved

Let $ X$ and $ Y$ be compact spaces. Show that for each continuous real-valued function $ f$ on $ X \times Y$ and each $ \varepsilon>0$, there exist continuous functions $ f_{1},\ldots,f_{n}$ on $ X$ and continuous functions $ g_{1},\ldots,g_{n}$ on $ Y$ such that $ |f(x,y)\minus{}\sum f_{i}(x)g_{i}(y)|<\varepsilon$ for all $ (x,y) \in X \times Y$.

It seems the Stone-Weierstrass theorem can be applied but how?

[quote="chungyc"]Let $ X$ and $ Y$ be compact spaces. Show that for each continuous real-valued function $ f$ on $ X \times Y$ and each $ \varepsilon > 0$, there exist continuous functions $ f_{1},\ldots,f_{n}$ on $ X$ and continuous functions $ g_{1},\ldots,g_{n}$ on $ Y$ such that $ |f(x,y) \minus{} \sum f_{i}(x)g_{i}(y)| < \varepsilon$ for all $ (x,y) \in X \times Y$.

It seems the Stone-Weierstrass theorem can be applied but how?[/quote]
Functions of the form f(x)g(y) (where $ f\in C(X)$ and $ g\in C(Y)$) separate the points of $ X \times Y$. The set of finite sums of functions of that form is an algebra. By the S–W theorem the closure of this set is the whole of $ C(X \times Y)$.


vector
geometry
analytic geometry
graphing lines
slope

Let x,y be real numbers and 4x+3y=4. Find the minimum of x*x+y*y.

[hide="More methods"]
This is equivalent to finding $r^{2}$ for the circle $x^{2}+y^{2}=r^{2}$ tangent to the line $4x+3y=4$, which can be done many ways. Vectors work, as well as an area equality (since the radius will be the altitude to the hypotenuse of the triangle formed by the line and the axes), as well as just taking the line from the origin with slope $\frac{3}{4}$ and finding its intersection with the other line.  [/hide]

[hide="Yet more methods"]Plug $y={4-4x\over 3}$ and find the minimum of the quadratic.[/hide]


This is another question i came across in the Hwa Chong Challenge (Singapore) sample papers.

Evaluate sin 20° sin 40° sin 60° sin 80° sin 100° sin 120° sin 140° sin 160° .

The answer this time is [hide]9/256[/hide], not as nice a number as that of my first post (See Trigo (multiplication of cotangent).

It is a competition question and so needs to be done pretty fast (in probably 6 minutes or less).  My teacher managed to solve it using some long method involving complex nos. but that definitely cannot be done within the time given (1 hour for 14 questions--this is one of the hardest ones inside though).

Does anyone know a way to solve this type of problems?


articles
number theory
prime factorization

A mom of 3 kids was talking to a math professor. She told him that the product of the ages of her kids was 72. She also told him that the sum of the ages of her kids was ___ (she told him a number, we just dont know the number).
Then the professor immediately said, "That is not enough info.".
Then, the mom said "the oldest has an iguana" . Then, the professor immediately got the ages of her kids.
What are the ages of her kids? Assume that somebody who is n years old is exactly n years old, where n is a positive integer.

[hide]Of all the possible natural triplets $(a,b,c)$ such that $a\leqslant b\leqslant c$ and $abc=72$, only two have the same sum: $(2,6,6)$ and $(3,3,8)$. (Which is the point of the professor's "There's not enough info.") Since the sentence "the oldest has an iguana" implies that there's ONE which is the oldest, the kids have $3$ years, $3$ years and $8$ years.[/hide]

We can see that the word oldest is used. This means that the oldest age has no age equal to it.
You gave the fact that all ages are integers.
Thus, all integer factors of 72, where the greatest factor is not repeated, work.
Prime Factorization of 72
$72 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$
$72 ={2^{3}}\cdot 3^{2}$

There are many answers.
However, if the sum and the product of a given set of three integers is not enough to make it distinct from another set, there must be one such set that does not meet this.
Thus there must be a $2^{3}$ person and two $3$ year olds.

EDIT: Beat by a second.

[quote="Ihatepie"]It can be 9,4,2     9,8,1       2,2,18    36,1,1   4,1,18...    its impossible to know how much it is without knowing how much it is when you add them up.[/quote]
No, because after the mom tells the professor the sum, it will be obvious which triplet it is. But there are 2 triplets of the same sum, one of which is the answer. Read Farenhejt's above post. :D

I do think asdf4 has got it!!! ;)
Rob
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[hide]
We can see that during every hour $h$ the clock chimes $h+1$ times. So we can figure that the clock chimes $2(2+3+4...13)$ times per day. We can use the formula for sum of first $n$ numbers to calculate this quickly and find that the clock rings $180$ times in a day. We can also see that the clock will chime $91$ times in the remainder of the 26th. So now we can set up an equation

$91+180x=2003$

Solving this, we find that $x=10.62$ Thus, we can round up and say that the 2003rd chime will occur 11 days after the 26th. There are 28 days in February, the 2003rd chime will occur on $\boxed{March 9, 2003}$ Was this one of the choices??[/hide]

[quote="now a ranger"]A clock chimes once at 30 minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February 26, 2003, on what date will the 2003rd chime occur?[/quote]

[hide="solution"]Every 12 hours there are $12+(1+2+\cdots+11+12)=90$ chimes so every day there are 180 chimes.

$\frac{2003}{180}$ is about $11.13$ so it is 11 days away.  The decimal part is less than $.5$ so it does not pass over to the next day.

11 days from the 26th is March 9th, 2003.[/hide]


search
real analysis
real analysis theorems

I am a 1s yr student on maths honours & i am given an impossible task on giving a lecture on Schroeder-Bernstein theorem.I know 2 proofs of it -1 by birkhoff,mclane & the other by whom I dont know.So i want some proof of the theorem which is very elegant,not very long (otherwise i wll be confused while presenting it)But i dont want a proof using axiom of choice or things like that.Could you suggest some article bcoz i think i have to give some examples to show its real power.Plz give me a reply bcoz I dont hav much ime.

What is Schroeder-Bernstein theorem :idea:

It states that if $A,B$ are two sets and there exist injections from $A$ into $B$ and from $B$ into $A$, then there is a bijection from $A$ to $B$.

We call Cantor-Bernstein in China.  :blush:

If my memory is good :?:
1-We suppose first that $B$is in $A$ thus if there exist an injection $f$ from $A$ to $B$ then there is also a bijection$g$ from $A$ to $B$ .
2-take:$f:A-->B$and $h:B-->A$ then there composition is injective, and apply 1)
 :?

Hmm rt this result was 1st proposed by cantor & its 1st proof was given by Dedekind.
Can you tell me why is this considered to be such a deep theorem?

As I was saying Dedekind gave its 1st proof.His proof is the most elegant of all proofs published so far.This proof is there in the complete works of Dedekind & its in german.The problem is neither do I have that book nor do I know German.Can any of you find that proof & translate it to me in English please.

I know two proofs of the "Cantor-Bernstein Theorem", both very elegant (at least in my opinion).
I'm posting the books where i read them, above :
[1] Introductory real analysis, Kolmogorov and Fomin (This book has some problems that you can use as examples).
[2] PROOFS from the book, Aigner and Ziegler, where they present the proof from a book of Paul Cohen, which is realy straight from the "book". I strongly recommend you to search for this proof.

If you can't find these books, let me know and i'll try to post them when i have time.


geometry
circumcircle
3D geometry
prism
geometry proposed

In triangle $ABC$ common tangents of circumscribe and excircle (wrt A) of $\triangle ABC$ touches Circumscribe at $X$ and excircle at $Y$.
let $Z$ be intersection point of these circle (the one which is not nearer to $XY$) prove that$XZ$ is tangent to excircle.

An "external" version of [url=http://mathworld.wolfram.com/PonceletsPorism.html]Poncelet's poism[/url] works here:

If $P$ is a point on the circumcircle of $ABC$, and the tangents from $P$ to the $A$-excircle cut the circumcircle in $M,N$ (different from $P$), then $MN$ is tangent to the excircle.

It shouldn't be hard to modify the proof for the usual porism (for triangles) in order to get a proof for this version. Now we can apply the above to our particular case.Take $P=X$. Then two of the sides of $XMN$ coincide, and this can only happen if one of the vertices is a common point of the two cicles (the other two vertivces coincide with $X$). It’s easy to see now that this common point must be the one farthest from the line $XZ$ (and not the other one).

Let $D,E,F$ be tangency points of the $A$-excircle of $ABC$ with sides $BC,CA,AB$. Let $M,N,P$ be midpoints of $EF,DF,DE$. Let $I,O$ be the centers of circumscribed circles of $\triangle DEF$($k(I,r)$) and $\triangle MNP$. Consider the inversion $\phi$ WRT $k$. It maps $A,B,C$ to $M,N,P$ so $M,N,P,Z$ are con-cyclic on a circle with radius $\frac {r}{2}$. Now let $S$ be the midpoint of $IZ$ $L$ is the symmetric point of $Z$ WRT $ON$. Since $OL=\frac {r}{2}=OZ=ZS=SL$ we have that $L$ is on circle $MNP$. Let $ZL$ meet $k$ again at $G$ then let $R$ be the center of $IG$. Since $G,R,I$ circle $IGL$ is tangent to $k$.Since $\angle RLG=\angle RGL=\angle IGZ=\angle IZL=\angle OZL=\angle OLZ$ then $R,L,O$ are collinear so circle $IGL$ is tangent to circle $MNP$. So $\phi(\odot ILG)$ is a line that is tangent to circles $\phi(\odot MNP)=\odot ABC$ and $\phi(k)=k$. so $\phi(\odot ILG)$ is a common tangent to $k$ and circle $ABC$ touching $k$ at $G$ where $G,Z$ are  on opposite sides of $IO$. So $G=Y$ and $\phi(L)=X$ since $S$ is the circumcenter of $ILZ$ and $Z,L,I$ are collinear $\odot IZL$ touches $k$. so $\phi(\odot IZL)$ touches $k$ but $\phi(\odot IZL)$ is line $XZ$ so $XZ$ touches $k$


quadratics
conics
parabola
analytic geometry
function

...Really tough for me anyways. It was in the nationals 90-91 sprint round. Give it a try, it's the only one in the round I can't figure out:

Points A and B are selected on the graph of y=(-1/2)x 2 so that triangle ABO (O is the origin in the diagram) is equilateral. Find the length of one side of triangle ABO.

I can't scan the diagram, so I'll describe it as best I can: there are the two axes (x and y) with the graph of the parabola shown. Obviously, the graph for y=(-1/2)x 2 is a parabola below the x axis with the tip tangent to the x axis at the origin. The origin is point O. Points A and B are on the parabola on opposite sides of the y axis equally distant from both the x axis and the y axis. Triangle ABO is drawn; it is equilateral as the problem states.

I can't figure it out. Help! :wacko:

[hide]I made point B on the equilateral triangle on the positive side of the x-axis. If I take line OB I can find its equation easily, $y = -x \sqrt 3$ because the angle that the x-axis makes with line OB is a 60 degree angle and the therefore for every unit that the line goes up, it goes $\sqrt 3$ to the left. To find where these the porabola and this line intersects, we set the two equal and get $-\frac {1}{2} x^2 = - \sqrt 3 x$. Solving, we get that $x$ can equal $2 \sqrt 3$, so the entire side must be $2* 2 \sqrt 3 = 4 \sqrt 3$.[/hide]

I've got it:

[hide] Simply draw y=(-1/2)x^2 (doesn't have to look good) and the equilateral triangle on a coordinate graph. Notice that the y value (the height of the triangle) is equal to sqrt(3) times x (half the length of a side the triangle). Now replace y in terms of x, giving sqrt(3)x=(-1/2)x^2. This simplifies to 2sqrt(3)x=-x^2, or x=-2sqrt(3). But this is actually the coordinate value of x of one of the vertices of the triangle on the graph, so the actual length of half the side of the triangle is 2sqrt(3), so the side length is 4sqrt(3).

Answer: 4sqrt(3)

Side note: I could explain this better with pics and diagrams, but I can't, so this is the best I can do with text.

[/hide] By the way, 09husbri, I'm on your national team. Guess who I am!

Well, first you draw a 30-60-90 triangle with the 30 degree angle at the origin, then extend the hypotenuse until it hits the parabola...

[hide="myanswer"]

i dont know if this is the eziest way but for it to be equilateral, (this'll be harde to explain w/o graph but neways) 2x=the sqrt of (x^2+((-1/2)x^2)^2) this is because one leg from the origin is (-1/2)x^2 (based on the function givent at the beginning) and the other leg will obviously be x
solving the equation above by squaring both sides u get
3x^2=x^4/4 so:
12x^2=x^4
x^2=12
so x=sqrt12
and one side fo the triangle will be 4sqrt3
[/hide]

Ahhhh, I've got it now. Thanks, everybody!

And confidential to Monkey....let's see...partial to animals with funny names...art thou the one with thy rubber duck, [i]Henry[/i]? :D


find the smallest  $k$ such that $2(x^3+y^3+z^3)+3xyz>=(x^k +y^k+z^k)(x^{3-k}+y^{3-k}+z^{3-k})$ for all non-negative $x,y,z$.

Obviously true for k=2 due to Schur's. But I am not sure Schur's is the sharpest though.

I think you wanted to say $k=1$ (he asks for the smallest $k$ ;) ).
And all $k<1$ don't work. Indeed, put $y=z=1$, then we obtain $x^3+3x\geq 2x^{k}+2x^{3-k}$. Assuming $x\to 0$ we obtain contradiction.


ratio
inequalities proposed
inequalities

That $\phi$ is related to golden ratio, Fibonacci sequence and here it makes the problem interestingly special.

Excuse me:
[quote="hungkhtn"]We have
\[ \sum_{sym} \left(\frac{a}{b+c}\right)^k \ge \min(2,\frac{3}{2^k}).  \]
So we are done.[/quote]
symmetric?

[quote="arqady"]$a>0,$ $b>0,$ $c>0,$ $\phi=\frac{1+\sqrt5}{2}.$
Prove that $\left(\frac{\phi a}{b+c}\right)^\frac{1}{\phi}+\left(\frac{\phi b}{a+c}\right)^\frac{1}{\phi}+\left(\frac{\phi c}{a+b}\right)^\frac{1}{\phi}\geq\phi^2.$[/quote]
cyclic!
 :?

[quote="hungkhtn"]We have
\[ \sum_{sym} \left(\frac{a}{b+c}\right)^k \ge \min(2,\frac{3}{2^k}).  \]
So we are done.[/quote] (should be cyc not symm) It seems to me this has been proven only for $k\ge\frac23$, disproved for $k\le\frac13$, open for the rest. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=12763]here[/url] unless there are more recent results. :?:

of course it is true for all k>0 and it has been proved in this forum.
I have proven also the case of four variables by hunghtn.
just use the IPT for reducing it to the case of two equal variables and then Vasc's RCF theorem! :D
It is also mentioned in the theorems and formulas section here
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]www.mathlinks.ro/Forum/viewtopic.php?t=64933[/url]

post 8  ;)

No,It's Nebitt extention, I had post in this forum and only use Mixing Variable. Not any other advance.

yes this is nesbitt's extention here isn't it? :)

(bu the way your proof with mixing variables must be quite long, especially for the case with four variables discussed before)

[quote="akhrif"]we can use $\phi^2=\phi+1????$[/quote]
It follows what?

[quote="hungkhtn"]We have
\[ \sum_{sym} \left(\frac{a}{b+c}\right)^k \ge \min(2,\frac{3}{2^k}).  \]
So we are done.[/quote]
According to hungkhtn:
\[ \frac{3}{2^\frac{1}{\phi}} = {\phi}^{2-\frac{1}{\phi}}  \]
 :?

[quote="arqady"]$a>0,$ $b>0,$ $c>0,$ $\phi=\frac{1+\sqrt5}{2}.$
Prove that $\left(\frac{\phi a}{b+c}\right)^\frac{1}{\phi}+\left(\frac{\phi b}{a+c}\right)^\frac{1}{\phi}+\left(\frac{\phi c}{a+b}\right)^\frac{1}{\phi}\geq\phi^{2}.$[/quote]
Unsolved in 2006! :wink:

[quote="arqady"]$a>0,$ $b>0,$ $c>0,$ $\phi=\frac{1+\sqrt5}{2}.$
Prove that $\left(\frac{\phi a}{b+c}\right)^\frac{1}{\phi}+\left(\frac{\phi b}{a+c}\right)^\frac{1}{\phi}+\left(\frac{\phi c}{a+b}\right)^\frac{1}{\phi}\geq\phi^{2}.$[/quote]
I think it solved (the post of hungkhtn).
We need only prove $min \left(2;\frac{3}{2^{\frac{1}{\phi}}}\right) \ge \frac{\phi^{2}}{(\phi)^{\frac{1}{\phi}}}$. Indeed.

Arqady,

happy new year!
Don't tell me you have a solution that uses the special nature of $\phi$  :P


The elementary school is packing everything away for the summer.  All that is left to pack are 9 foam playground balls of equal density.  One ball, with diameter 4, is bright red. The rest are green.  The volume of the red ball is equal to the volumes of the green balls put together.
These balls are perfectly packed into a cubical box (with lid) so that each green ball is pushed up into each corner (touching 3 sides of the box) while the red ball is nestled snugly between the green ones.  (No balls are squished.)
Find the shortest distance from the red ball to the box, in simplest radical form.

[quote="13375P34K43V312"]Nice problem. Here's a hint.

[hide="bomb"]Once you find the radius of the big ball, consider the long diagonal of the cube, and consider the "circumcube" of the smaller ball. [/hide][/quote]

Why the circumcube of the smaller ball?  Shouldnt it only be the circumcube of half of the smaller ball since the other half touches the larger ball.

Oh well, I have a horrible time visulizing this, but I get the radius of the small ball is 1, the bigger is 2, and the circumbue of half of the smaller ball's long diagonal is $\sqrt{3}$ so its $3x^{2}=(1+2+\sqrt{3})^{2}$, so we have $x^{2}=4+2\sqrt{3}$ (from the cube formed with the long diagonal being half of the long diagonal of the larger cube) so $x=\sqrt{3}+1$?

Im not positive thats right.....but it might be right since it came out nice..
EDIT: (look at below post of mine below the spam) Its $x=\sqrt{3}-1$ if its anywhere on the ball (subtract radius 2) to the cube, my initial answer is frmo center of ball to cube.

[quote="da ban man"][quote="13375P34K43V312"]Nice problem. Here's a hint.

[hide="bomb"]Once you find the radius of the big ball, consider the long diagonal of the cube, and consider the "circumcube" of the smaller ball. [/hide][/quote]

Why the circumcube of the smaller ball?  Shouldnt it only be the circumcube of half of the smaller ball since the other half touches the larger ball.

Oh well, I have a horrible time visulizing this, but I get the radius of the small ball is 1, the bigger is 2, and the circumbue of half of the smaller ball's long diagonal is $\sqrt{3}$ so its $3x^{2}=(1+2+\sqrt{3})^{2}$, so we have $x^{2}=4+2\sqrt{3}$ (from the cube formed with the long diagonal being half of the long diagonal of the larger cube) so $x=\sqrt{3}+1$?

Im not positive thats right.....but it might be right since it came out nice..[/quote]
Actually $x=\sqrt{3}-1$ might be the answer..my first answer was from the center of the ball to the side of the cube.  My second (just subtract the radius 2) from it is from the ball (anywhere on it) to the box as the shortest distance.

[quote="da ban man"]
Actually [hide="this"]$\sqrt{3}-1$[/hide] might be the answer..my first answer was from the center of the ball to the side of the cube.  My second (just subtract the radius 2) from it is from the ball (anywhere on it) to the box as the shortest distance.[/quote]

What you got is right, I believe.  Yes, I meant from anywhere on the ball, not the center of the ball.

[quote="13375P34K43V312"]
Here's a hint.
[hide="bomb"]Once you find the radius of the big ball, consider the long diagonal of the cube, and consider the "circumcube" of the smaller ball.  [/hide][/quote]
Yeah, please make sure you're able to find the radius of the big red ball. :P

I think he meant
[hide="HINT"]Once you find the radius of the green balls, consider the long diagonal of the cube, and consider the "corner radius cube" of the small ball. [/hide] if that makes more sense.

[quote="archimedes1"]
Yeah, please make sure you're able to find the radius of the big red ball. :P[/quote]

well, that step's a little easier...


How has the lack of a calculator impacted your results/success in your AMC experiance?

I tend to like to use a lot of space, so that I can write real big and organize my scrap paper into piles and stuff. The calculator just seemed to get in the way. Now it's not there, and I can use my space more effectively.

I'm also glad because before, I only really used them for arithmetic and stuff. Occasionally I'd graph something or something like that, but I really didn't use them much. But there were a lot of people that did; people that might write a program for one problem and let the calculator crank it out while they work on the next, which would save them time [i]and[/i] get them a correct answer. For me, relatively speaking, the exclusion of the calculator has helped.

I don't know how to program anything, i just need a calculator with arithmetic stuff so that i won't make stupid mistakes like on the A test and save much time on the B.

Obviously, my result would be better if I had it.

It didnt really affect me because there werent any rigourous calculations on the AMCs this year...well atleast not for the first 20 problems on the 12A (probalby something like number 24 a calculator wouldve helped on the 12A) or the 10B test.

Really, the only problem that I missed on last years 12B is the one that I attempted to program on my calculator. The AMC did a great job at making sure that the test this year was doable without a calculator. I felt that not having a calculator did not hinder me at all on the AMC.

Meh, I only took the AMC8 last year anyways, so It didn't really matter.

I don't think the lack of a calculator affected me much, and it allowed cool problems like 24 on the 12B, which would have been bashable with a calculator. It also forces you to do less brute force on the combinatorics problems. In retrospect, though I was skeptical at first, it was a good decision.

[quote="The Zuton Force"]You have to keep in mind that the test this year was designed with the exclusion of calculators in mind, whereas in previous years it was not. If someone were to take a no calculators test with a calculator, naturally, they would ace it.

That being said, I think there was a whole lot less computation this year than in previous years:
13, 18, 19, 21, 24, and 25 were the only problems I might have asked for a calculator on on this year's 12B. And for 24, only because I cranked it with finite differences, the much better solution is to interpolate it, or even better, to legitimately derive the formula.

For the 12A, there was significant computation in 6, 11, 15, 16, 19, 20, but it wasn't bad at all because most of those were easier problems, and there wasn't much room for screwing up.

24 had a good deal of algebraic manipulation if you didn't do it by sticking in a circle, but a TI-89 would completely trivialize it: solve for tangent in terms of x, set derivative equal to 0, substitute.

You know where I would just love a calculator? The AIME's. Some of them just have horrendous computation. I mean they give more time, but still.[/quote]

I tend to think they design the previous years with the inclusion of calculators in mind. All of the AMCs are designed to be doable without a calculator.

[quote="123456789"]I think we should be allowed a four-function calculator, because I usually mess up on calculations.[/quote]

that is somewhat the purpose to test people who have weak calculations while testing the people how well they know and can apply concepts. they don't want to test how well someone can program something on to their calculators. As for me, even though my score has improved somewhat, i think i am not doing as well without the calculator. but the score increase has to do with the fact that last year, i was in the middle of pre-algebra and i didn't know squat, and this year i am learning algebra 2 and precalculus, so my mathematical abilities have improved along with my score from last year.

Lacking the calculator didn't affect me much because I'm not too experienced with the complex functions in the first place, and I got stuck on the easy questions (probably due to nerves), so having a calculator wouldn't be much better. I think this year's computations were fair enough. to be done by hand quickly.

I think it was designed for no calculators since problems like the one about the postman would have required no work at all.

Hard to say, really. I've improved a lot in math since last year, so I can't really compare myself. I probably would have gotten #25 on the AMC 10A (which I took unofficially anyway) this year if I had a calculator. It wouldn't have made a difference on the 10B, because it wouldn't help at all on the problem I missed (I made a logic mistake as opposed to a computation mistake.)


trigonometry limit geometry function inequalities search calculus

[quote="M4RI0"]Well I'm my country we use both notations, that's why I wrote it that way but the problem is fine  :P[/quote]

I do know the difference of notation. What I want to say is that $\lim_{x\rightarrow 2}\left(tg2 \pi x+cos \frac{ \pi x}{2}+tg \frac{\pi x}{2}\right)\neq 0$

In other words, the difference is $\cos$ versus $\sin$.

I'm tired, the answer is there don't exists the limit of the problem!

[quote="kunny"][quote="M4RI0"]Compute: $\lim_{x\to{2}}\frac{tg2 \pi x+cos \frac{ \pi x}{2}+tg \frac{\pi x}{2}}{x^{2}+4x-12}$[/quote]

This problem is wrong.

The right problem is $\lim_{x\rightarrow 2}\frac{\tan 2\pi x+\sin \frac{\pi x}{2}+\tan \frac{\pi x}{2}}{x^{2}+4x-12}$.[/quote]

$\lim_{x\rightarrow 2}\frac{\tan 2\pi x+\sin \frac{\pi x}{2}+\tan \frac{\pi x}{2}}{x^{2}+4x-12}$

$=\lim_{y\rightarrow 0}\frac{\tan 2\pi y-\sin(\frac{\pi y}{2})+\tan(\frac{\pi y}{2})}{y^{2}+8y}$
(using $y=x-2$)

$=\lim_{y\rightarrow 0}\frac{\frac{2\pi}{\cos^{2}{(2\pi y)}}-\frac{\pi}{2}\cos{(\frac{\pi y}{2})}+\frac{\frac{\pi}{2}}{\cos^{2}{(\frac{\pi y}{2})}}}{2y+8}$
(using hospital/clinic... whatever :D )

$=\boxed{\frac{\pi}{4}}$

yes. misan is right. (also you can solve it by this substitution $x-2: =y$ wher $y\to 0$ + Taylor.)
kunny's comment is true. because if you put instead $\sin{\pi{x}/2}$ this $\cos{\pi{x}/2}$, then your limit will be equal to $-oo$ which also proved by Taylor

Here is a solution for Japanese high school students who aren't permitted to use l'Hopital's theorem in university entrance exam.

$\lim_{x\rightarrow 2}\frac{\tan 2\pi x+\sin \frac{\pi x}{2}+\tan \frac{\pi x}{2}}{x^{2}+4x-12}$

$=\lim_{x\rightarrow 2}\frac{\tan 2\pi (x-2)-\sin \frac{\pi}{2}(x-2)+\tan \frac{\pi}{2}(x-2)}{(x-2)(x+6)}$

${{=\lim_{x\rightarrow 2}\frac{1}{x+6}\{\frac{\tan 2\pi (x-2)}{2\pi (x-2)}\cdot 2\pi-\frac{\sin \frac{\pi}{2}(x-2)}{\frac{\pi}{2}(x-2)}\cdot \frac{\pi}{2}+\frac{\tan \frac{\pi}{2}(x-2)}{\frac{\pi}{2}(x-2)}\cdot \frac{\pi}{2}}}\}$

$=\frac{1}{8}(1\cdot 2\pi-1\cdot \frac{\pi}{2}+1\cdot \frac{\pi}{2})$

$=\frac{\pi}{4}$, which is misan's answer.

[quote="M4RI0"]Compute: $\lim_{x\to{2}}\frac{\tg2 \pi x+\sin\frac{ \pi x}{2}+\tg \frac{\pi x}{2}}{x^{2}+4x-12}$[/quote]
[color=darkblue][u][b]Sorry, M4R10[/b][/u], this limit isn't interesting. Come on with mine to geometry Section !

After usage $x^{2}+4x-12=(x-2)(x+6)$ and the substitution $t=x-2$ it reduces to

$\lim_{t\to 0}\left(\sum_{k=1}^{n}f_{k}(t)\right)=\sum_{k=1}^{n}\lim_{t\to 0}f_{k}(t)$, where for any $k\in\overline{1,n}$, the $\lim_{t\to 0}f_{k}(t)$ exists and is finite,

while each limit uses same technique, i.e. $\lim_{t\to 0}\frac{\sin at}{t}=a$ or $\lim_{t\to 0}\frac{\tan at}{t}=a$.

[b][u]Exercise.[/u][/b] I present here an interesting limit : $\lim_{x\to 0}\frac{x-\sin x}{x^{2}}$ [u][b]without the l"Hospital's rules. [/b][/u][/color]

[quote="Virgil Nicula"]
[color=darkblue][b][u]Exercise.[/u][/b] I present here an interesting limit : $\lim_{x\to 0}\frac{x-\sin x}{x^{2}}$ [u][b]without the l"Hospital's rules. [/b][/u][/color][/quote]

[hide]
The limit is $\boxed{0}$. Since the function $\frac{x-\sin x}{x^{2}}$ is odd, it suffices to prove $\lim_{x\rightarrow 0^{+}}\frac{x-\sin x}{x^{2}}=0$. Note that this limit is equal to $\lim_{x\rightarrow 0^{+}}\frac{x-\sin x}{\sin^{2}x}$ since $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$.

Now use the inequality $\sin x<x<\tan x$ for $0<x<\frac{\pi}{2}$. Then $\frac{x-\sin x}{\sin^{2}x}>0$ but also $\frac{x-\sin x}{\sin^{2}x}<\frac{\tan x-\sin x}{\sin^{2}x}$. By squeeze theorem, it is enough to prove the upper bound tends to 0.

The upper bound is in fact equal to $\frac{\tan \frac{x}{2}}{\cos x}$, so it indeed tends to 0.
[/hide]

However, it's not strong enough to prove the stronger $\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}$.

[quote="scorpious119"]$\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}$.[/quote]

We are to induct the inequality $\frac{x^{3}}{3!}-\frac{x^{5}}{5!}<x-\sin x<\frac{x^{3}}{3!}$.

We can prove the inequality directly without using l'Hopital's theorem or Taylor expansion.

[quote="Kunny"].... $\frac{x^{3}}{3!}-\frac{x^{5}}{5!}<x-\sin x<\frac{x^{3}}{3!}$. We can prove these inequalities directly without using l'Hopital's theorem or Taylor expansion.[/quote]
Please Kunny, how prove you ? Thanks.

Please refer to the following links.

1.[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=1967&p=131909[/url]

2.[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1223044444&t=19548[/url]

and you could also use intergral by parts.


Express $ \frac{5}{8} \plus{} \frac{1}{4}$ as a decimal to the nearest thousandth.

For your fraction way, why change the fractions to denominators of $ 24$?  You can keep 'em at $ 8$...

$ \frac{5}{8} \plus{} \frac{1}{4} \equal{} \frac{7}{8} \equal{} 0.875$


logarithms
limit
Euler
algebra unsolved
algebra

Is there a simpler expression for sum(1/n)? If so how is it.

No, there isn't.
However, for n sufficiently large, we have
       $\sum_{k=1}^{n}\frac1{k}\approx \ln n$.
More precisely,

       $\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\frac1{k}}{\ln n}=1$.

Or, even more precisely,

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n = \gamma \approx 0.577215665...$, a constant known as the Euler-Mascheroni constant.

We can state even a bit more: $|\sum_{n=1}^x \frac{1}{n} - ln(x) - \gamma | < \frac{1}{x}$.


How do I show that if n=a^2+b^2 for integers a,b then n cannont be congruent to 3 mod 4 (n does not equal 4m+3 for all integers m)  Thanks!

Consider $ a^2$ and $ b^2 (mod) 4$ Clearly they are $ 0,1$ Since the largest is $ 1\plus{}1\equal{}2$, $ 3$ cannot be achieved

It's not a matter of largeness when you're dealing with mods.  The exact values $ a^2 \plus{} b^2 \bmod 4$ can take are $ 0, 1, 2$.


floor function
ceiling function
logarithms

Determine all couples (a,b) ( $ a,b \in R $ ) such that :
 a[bn] = b[an]  for all $ n\in N $

This is probably wrong. It's also not rigorous at all. Oh well.

[hide]If $a=b$ or if $a$ and $b$ are both integers or if at least one of $a$ and $b$ is $0$, it's obviously true. So, assume $a,b\not\in\mathbf{Z}$ and $a\neq{b}$. Now let's define $0\le\alpha,\beta<1$ s.t. $a=\lfloor{a}\rfloor+\alpha$ and $b=\lfloor{b}\rfloor+\beta$.

The conditions set on $a$ and $b$ can be written as $\frac{a}{b}=\frac{\lfloor{an}\rfloor}{\lfloor{bn}\rfloor}$. The fraction on the left is clearly a constant, so the one on the right must be as well.

I've been trying to figure out how to prove this but I don't think it's possible for the fraction on the right to be constant because eventually $n\alpha$ and/or $n\beta$ will exceed $1$ (I believe this will occur for $\alpha$ when $n=\lceil{-\log_{10}{\alpha}\rceil}$ or even earlier), at which point the fraction will have a different value. Maybe someone can build on what I have?

Edit: I just realized that I left out the case that one of $a,b$ is a nonzero integer and the other is not an integer. In this case the fraction on the right will not be constant (using similar reasoning as above).[/hide]

[quote="seamusoboyle"]If you tell me what [x] is i'll give it a go...[/quote]the greatest integer not exceeding x


I have found somewhere in a book a link : The Silow theorem., but I don't know it...
Could sb help me?
...and..
It would be great to describe the fobrenius theorem to.
Thanks

Look here: http://mathworld.wolfram.com/SylowTheorems.html
and here: http://mathworld.wolfram.com/FrobeniusTheorem.html

Is this what you wanted¿

I've seen the Sylow theorem in Fundamentals of Abstract Algebra...


trigonometry
geometry
circumcircle
trig identities
Law of Sines
geometry unsolved

Points $ P_1,P_2,...,P_7$ respectively lie on the sides $ BC,CA,AB,BC,CA,AB,BC$ of triangle $ ABC$, and are such that

$ \angle P_1P_2C\equal{}\angle AP_2P_3\equal{}\angle P_3P_4B\equal{}\angle CP_4P_5\equal{}\angle P_5P_6A\equal{}\angle BP_6P_7\equal{}60^{\circ}$.

Prove that $ P_1\equal{}P_7$.

here a horrible trigonometric solution:

call $ \overline{CP_1} \equal{} x$ so for some law of sines

$ \displaystyle \overline{CP_7} \equal{} a \minus{} \frac{\left ( c \minus{} \frac{\left ( b \minus{} \frac{\left ( a\minus{} \frac{\left ( c \minus{} \frac{\left ( b \minus{} \frac{x \cdot \sin (60\plus{}\gamma)}{\sin60} \right ) \cdot \sin60}{\sin (60\plus{}\alpha)} \right ) \cdot \sin (60\plus{}\beta)}{\sin60} \right ) \cdot \sin60}{\sin (60\plus{}\gamma)}\right ) \cdot \sin (60 \plus{} \alpha)}{\sin60}\right ) \cdot \sin60}{\sin (60 \plus{} \beta)}$

denote $ \delta \equal{} 60 \plus{} \alpha\$, $ \ \theta \equal{} 60 \plus{} \beta \$ and $ \ \sigma  \equal{} 60 \plus{} \gamma$

$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} \equal{}$

$ \equal{}a [\sin{\theta} \sin{\sigma} \minus{} \sin{\delta}\sin60] \plus{} b [\sin{\delta} \sin{\sigma} \minus{} \sin{\theta}\sin60 ] \plus{} c [ \sin{\delta} \sin{\theta} \minus{} \sin{\sigma}\sin60 ] \plus{} x \cdot \sin{\theta} \sin{\sigma}$

but $ \sin{\theta}\sin{\sigma} \minus{} \sin{\delta}\sin60 \equal{}$

$ \equal{} \frac{1}{2} [\cos(\beta \minus{} \gamma) \minus{} \cos (60 \plus{} \alpha) \minus{} \cos {\alpha} \plus{} \cos (120 \plus{} \alpha)] \equal{}$

$ \equal{} \frac{1}{2}\cos(\beta \minus{} \gamma) \minus{} \cos {\alpha}$ and equally for the others, so

$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} \equal{}$

$ \equal{} a[\frac{1}{2}\cos(\beta \minus{} \gamma) \minus{} \cos {\alpha}] \plus{}b[\frac{1}{2}\cos(\gamma \minus{} \alpha) \minus{} \cos {\beta}] \plus{} c [\frac{1}{2}\cos(\beta \minus{} \alpha) \minus{} \cos {\gamma}] \plus{}x \cdot \sin{\theta} \sin{\sigma}$

but $ \frac{1}{2}\cdot a \cdot \cos(\beta \minus{} \gamma) \equal{} R \cdot \cos(\beta \minus{} \gamma)\sin {\alpha}$ and equally for the others, where R is the circumradius of ABC

and $ a \cos {\alpha} \equal{} R \sin {2 \alpha}$ and equally for the others, then

$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} \equal{} R[\cos(\beta \minus{} \gamma)\sin {\alpha} \minus{} \sin {2 \alpha} \plus{} \cos(\gamma \minus{} \alpha)\sin{\beta} \minus{} \sin {2\beta} \plus{} \cos(\beta \minus{} \alpha)\sin{\gamma} \minus{} \sin {2\gamma}] \plus{} x \cdot \sin{\theta} \sin{\sigma}$

but $ \cos(\beta \minus{} \gamma)\sin {\alpha} \equal{} \sin(\beta \plus{} \gamma)\cos(\beta \minus{} \gamma) \equal{} \frac{1}{2}\sin{2\beta}\plus{}\frac{1}{2}\sin{2\gamma}$ and equally for the others, so

$ \displaystyle \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} \equal{} R[\frac{1}{2}\sin{2\beta}\plus{}\frac{1}{2}\sin{2\gamma} \minus{} \sin {2 \alpha} \plus{}\frac{1}{2}\sin{2\alpha}\plus{}\frac{1}{2}\sin{2\gamma} \minus{} \sin {2\beta} \plus{} \frac{1}{2}\sin{2\beta}\plus{}\frac{1}{2}\sin{2\alpha} \minus{} \sin {2\gamma}] \plus{} x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow$

$ \Longleftrightarrow \overline{CP_7} \cdot \sin{\theta} \sin{\sigma} \equal{} x \cdot \sin{\theta} \sin{\sigma} \ \Longleftrightarrow\ \boxed{CP_7 \equal{} x} \Longleftrightarrow \boxed{\boxed{P_1 \equiv P_7}}$

 :D  :D

$ \angle P_3P_4P_5\equal{}60^{\circ} \Longleftrightarrow P_5P_2P_4P_3$ is cyclic.$ (1)$
$ \angle P_1P_2C\equal{}\angle P_4P_5C\equal{}60^{\circ} \Longleftrightarrow P_5P_2P_1P_4$ is cyclic.$ (2)$
$ \angle AP_6P_5\equal{}\angle P_5P_2P_3\equal{}60^{\circ} \Longleftrightarrow P_6P_5P_2P_3$ is cyclic.$ (3)$
$ (1),(2),(3) \Longleftrightarrow P_1P_4P_3P_6P_5P_2$  lie on a circle and it is obvious that $ \angle BP_6P_1\equal{}60^{\circ}\Longleftrightarrow P_1\equal{}P_7$


find the remainder when 2^2008 is divided by 2008


On a balance scale, 3 green balls ballance 6 blue balls, 2 yellow balls balance 5 blue balls, and 6 blue balls balance 4 white balls. How many blue balls are needed to balance 4 green, 2 yellow and 2 white balls?

[hide] first I wrote this out:
3g=6b
2y=5b
4w=6b

so we already know 2w=3b
and 2y=5b
now all we need is how many g=b
voila 1g=2b
so you have=8b

8b+3b+5b=16b
therefore the answer is 16 blue balls.[/hide]

[hide]3g=6b
2y=5b
6b=4w

so we can say
4g=8b
2y=5b
2w=3b

8b+5b+3b=16b[/hide]


geometry
power of a point
radical axis
analytic geometry
perpendicular bisector
geometry theorems

[quote="dondigo"]The red line is the radical axis of the two circles. Prove the equality of the two segments ($a=a$)[/quote]

How do you define "radical axis"? With the standard definition - the radical axis of two circles is the locus of all points whose powers with respect to these two circles are equal -, the problem is trivial: In fact, more generally, let k and k' be two circles, and P a point on their radical axis lying outside of both of these circles. Let a tangent from the point P to the circle k touch k at a point T, and let a tangent from the point P to the circle k' touch k' at a point T'. Then, we have PT = PT', because $PT^2$ is the power of the point P with respect to the circle k, while $PT^{\prime}\ ^2$ is the power of the point P with respect to the circle k', and now, since the point P lies on the radical axis of the circles k and k', its powers with respect to these circles must be equal, i. e. we have $PT^2=PT^{\prime}\ ^2$ and thus PT = PT'.

If you are using a different definition of the radical axis, then you first have to show that it is equivalent to the standard definition.

  Darij

Yes, indeed it was trivial. Firstly I tried to prove it using analytic geometry, but it was like a nighmare and I didn't think of using "normal" geometry.
Anyway. I discovered, that the same situation is when the second circle is considered to be a point. But, where is the radical axis of two circles, when one of them is inside the other. When they are tangent to each other, it's obvious, but I don't know how to do the second case. Analogous question: when is the radical axis of a circle and a point , which is inside the circle?

[quote="dondigo"]But, where is the radical axis of two circles, when one of them is inside the other.[/quote]

See this picture (the radical axis is red):

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=539[/img]

[quote="dondigo"]Analogous question: when is the radical axis of a circle and a point , which is inside the circle?[/quote]

See here:

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=540[/img]

Generally, if you have two circles k and k', then the radical axis of these circles is the perpendicular bisector of the segment SS', where S is the image of the center of the circle k under the inversion with respect to k', and S' is the image of the center of the circle k' under the inversion with respect to k. If one of the circles k and k' degenerates to a point - say, the circle k' -, then the point S is simply the center of the circle k'. This way, you can easily construct radical axes.

  Darij


Determine all integers k, n1, n2, ..., nk such that
n1|2(n2) - 1
n2|2(n3) - 1
n3|2(n4) - 1
...
nk|2(n1) - 1.
Nice, isn't it?

If k \geq 2 and onw of the ni is 1, then all of them are 1 by induction, so suppose ni>1 for all i.

Suppose k>2.
We have all ni odd and positive.
If 3|ni then 3|2ni+1-1, and ord3(2)=1, so 2|ni+1, but all ni where odd.

Let p be the smallest prime that divides one of ni. By what I said earlier, p>3.
Then p|ni|2ni+1-1.
But p|2p-1-1, so p|2gcd(p-1,ni+1)-1 (2 n-1 is a Mersene sequence) . Since p>3 we get
2<gcd(p-1,ni+1)<p. Let q be a prime divisor of gcd(p-1,ni+1). Then q|ni+1 and q<p. But p was the smallest prime divisor of some ni. This is a contradiction.

So k=1. Then n|2 n-1.
Lets find the solutions to this one.
n is odd. Let p be the smallest prime divisor of n. Then p>2, and n=p r *m with gcd(m,p)=1.
We have p|2 p r *m-1 and p|2 p-1-1, so p|2 gcd(p r *m,p-1)-1=2 gcd(m,p-1)-1. But m has only prime divisors larger that p, so gcd(m,p-1)=1, so p|1.
This shows that the only solution for k=1 is n=1.

Yes, the problem is quite nice.


Can somebody post series for \[ \tan(x) \] ? I couldn't find it in my book  :)

$\tan{x}+0+0+0+0+0+\ldots$  :D

I don't think this is a question for the forums..

It's ugly:

http://mathworld.wolfram.com/Tangent.html


Let $p=x^{2}+y^{2}+z^{2}$ ,$q=\sqrt{2x-\frac{2}{y}}+\sqrt{2y-\frac{2}{z}}+\sqrt{2z-\frac{2}{x}}-\sqrt{3x+3y+3z}$ and $K=\{u\mid{q(u)=0}\}$. $Inf(p(u))$, when $u$ goes through $K$, is 6 and is reached only for $x=y=z=\sqrt{2}$. Thus the unique solution is $(\sqrt{2},\sqrt{2},\sqrt{2})$.


[quote="gemath"]Given triangle $ABC$ with arbitrary point $P$ prove that
$\min\{\frac{bc}{MB\cdot MC},\frac{ca}{MC\cdot MA},\frac{ab}{MA\cdot MB}\}\le 3$[/quote]

You mean point M?


function
advanced fields
advanced fields unsolved

Let $ A$,$ B$ be commutative C*-algebras and $ \phi: A\rightarrow M(B)$ (multiplier algebra of B) an involutive homomorphism for which $ \phi(A)B$ is dense in $ B$. Then $ \phi$ is the dual of a continuous function that is not necessarily proper.

Since $ A$,$ B$ commutative, $ A$,$ B$ are $ C_0(X)$,$ C_0(Y)$ for some locally compact spaces $ X$,$ Y$. Now, $ M(C_0(Y))\cong C_b(Y)$. I don't understand how to go from bounded functions to duals of continuous functions.

what do you mean by involutive here?
$ C_b(Y) \cong C(\beta Y)$, where $ \beta$ denots the stone-cech-compactification. $ \phi : A \to M(B)$ corresponds to a unital homomorphism $ A^\plus{} \to M(B)$, and thus to a continuous map $ \beta(Y) \to X^\plus{}$. this corresponds to a continuous map $ f : Y \to X^\plus{}$. now you have to translate the hypothesis $ \overline{\phi(A)B}\equal{}B$ to the fact, that the image of $ f$ lies in $ X$.


Prove that for all positive real numbers a, b, c we have the inequality:
a^5c^8/(a^3b^2c^4 + 2b^6)  + b^7c^5/(a^4b^5c^3 + 2a^13) + a^8/(a^6c^4 + 2bc^13) ≥ 1

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249871[/url]


complex analysis
complex analysis unsolved

For any language $ L, L^* \equal{} (L^*)^*$

I know in order to proof this I have to show that $ L,L^* \subset (L^*)^*$ and vice versa, but I am stuck inside that step.

This probably belongs in [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]computer science[/url], or maybe combinatorics, but certainly not complex analysis.

$ L^*\subset (L^*)^*$: Given an $ x\in L^*$, it is true by definition that $ x\in (L^*)^*$, and therefore $ L^*\subset (L^*)^*$.

$ (L^*)^*\subset L^*$: It suffices to show that for each $ z\in (L^*)^*$ we have $ z\in L^*$.  Given a $ z\in (L^*)^*$, there is by definition a concatenation $ y_1y_2...y_n \equal{} z$ with $ y_k\in L^*$ for $ 1\leq k\leq n$.  Furthermore, since each $ y_k\in L^*$ for each $ y_k$ there is by definition a concatenation $ x_{k,1}x_{k,2}...x_{k,m_k} \equal{} y_k$ with $ x_{k,j}\in L$.  But this means $ z \equal{} x_{1,1}x_{1,2}...x_{1,m_1}x_{2,1}x_{2,2}...x_{n,1}x_{n,2}...x_{n,m_n}$, so $ z\in L^*$.


search
inequalities proposed
inequalities


HMMT
geometry
power of a point
Pythagorean Theorem

Prove that the Radical Axis of two circles is a straight line.

[quote="Altheman"][quote="computer_nuke"][quote="Altheman"]man...this is so in the oly. geo transcripts...

anyway...you want to use the definition of power of a point as (point to center)^2-(radius)^2...just use analytic from here[/quote]

(point to center)^2-(radius)^2? I thought it was if a line through P goes through the circle at A and B, and the other circle at C and D it would be to do:

$(PA)^{2}(PB)^{2}=(PC)^{2}(PD)^{2}$?

O BTW, Alex congrats at HMMT for placing 3rd in Algebra  :clap2:[/quote]

1) thanks

2) you can always do it the hard way...

the power of a point with respect to a circle is the length of the tangent segment from teh point to the circle squared, using the pythagorean theorem, this is equivalent to what i gave[/quote]

Right, thanks.


Given that in the expansion of $(1+2x-10x^{2})(1+x)^{10}$, the coefficient of $x^{k}$ is $0$. Find the value of $k$.

[quote="A M Y"][hide="well..."]
$-10 C_{n+2}^{10}+2 C_{n+1}^{10}+1 C_{n}^{10}=0$ where $n$ is $10-k$

Dividing through by $10!$ and multiplying through by $n!(8-n)!$ results in:

$\frac{-10}{(n+1)(n+2)}+\frac{2}{(10-n)(n+1)}+\frac{1}{(9-n)(10-n)}=0$
$\implies n=6 \implies \boxed{k=4}$
[/hide]

[b]Ooohhh I'm P vs NP now?  :o [/b][/quote]
The equation you wrote is only work for $0\leq n\leq8$ (i.e. $2\leq k\leq10$). And so you need to note that the coefficients of $x,\ x^{11},\ x^{12}$ are not 0.


induction
number theory unsolved
number theory

We call a sequence of integers a [i]Fibonacci-type sequence[/i] if it is infinite in both ways and $a_{n}=a_{n-1}+a_{n-2}$ for any $n\in\mathbb{Z}$. How many [i]Fibonacci-type sequences[/i] can we find, with the property that in these sequences there are two consecutive terms, strictly positive, and less or equal than $N$ ? (two sequences are considered to be the same if they differ only by shifting of indices)

[i]Proposed by I. Pevzner[/i]

It is obviosly $\frac{N(N-1)}{2}$. For any (k,k+1) we have Fibonaccy tipe sequence with $a_{0}=k,a_{1}=k+1,k\ge 1$. In this sequence has not another terms with $a_{i+1}=a_{i}+1,a_{i}\ge 1$. It proved needed result.

Here's the solution:
Let those two consecutive terms be $a_{1},a_{2}$. we have:
$a_{2}-a_{1}\le 0$, so $a_{2}\le a_{1}$
$a_{2}+a_{1}>N$, so $a_{1}+a+2\ge N+1$
All we have t do is count the number of pairs $(a_{1},a_{2})$.
Case 1: $N$ is even.
Because $a_{1}\ge a_{1}$ and $a_{2}+a_{1}\ge N+1$, obtain $2a_{1}\ge N+1$, so $a_{1}\ge \frac{N}{2}+1$. So, we can write $a_{1}=\frac{N}{2}+k$, where $k$ is a positive integer, $k\le \frac{N}{2}$, because $a_{1}\le N$. We can also write $a_{2}=\frac{N}{2}+p$, $p$ is an integer, such that $-\frac{N}{2}+1\le p\le \frac{N}{2}$, $p\le k$, because $a_{1}\ge a_{2}$, $0<a_{2}\le N$.
We have $k\ge p$ and $k+p\ge 0$. Then for every value of $k$, $1\le k\le \frac{N}{2}$, there are $2k$ values of $p$, therefore the total number of $(p,k)$ (which is equal to the number of $(a_{1},a_{2})$), is:
$\sum_{i=1}^{\frac{N}{2}}(2i)=(\frac{N}{2})(\frac{N}{2}+1)$
Case 2: $N$ is odd.
Because $a_{1}\ge a_{1}$ and $a_{2}+a_{1}\ge N+1$, obtain $2a_{1}\ge N+1$, so $a_{1}\ge \frac{N+1}{2}$. So, we can write $a_{1}=\frac{N-1}{2}+k$, where $k$ is a positive integer, $k\le \frac{N+1}{2}$, because $a_{1}\le N$. We can also write $a_{2}=\frac{N-1}{2}+p$, $p$ is an integer, such that $-\frac{N-1}{2}+1\le p\le \frac{N+1}{2}$, $p\le k$, because $a_{1}\ge a_{2}$, $0<a_{2}\le N$.
We have $k\ge p$ and $k+p\ge 0$. Then for every value of $k$, $1\le k\le \frac{N+1}{2}$, there are $2k-1$ values of $p$, therefore the total number of $(p,k)$ (which is equal to the number of $(a_{1},a_{2})$), is:
$\sum_{i=1}^{\frac{N+1}{2}}(2i-1)=(\frac{N}{2}+1)^{2}$
So the answer is $(\frac{N}{2}+1)^{2}$ if $N$ is odd and $(\frac{N}{2})(\frac{N}{2}+1)$ if $N$ is even.

[quote="maky"]the correct answer is $\frac{N(N+1)}{2}$...[/quote]

We construct a bijection from the set $F$ of all fibonacci sequences with the
given properties to the set $M : = \{(a,b) \in \mathbb{N}^{2}: 0<b \le a \le N\}$.
Clearly $M$ has $\frac{N(N+1)}{2}$ elements, so we are done.

First let $a_{n}$ be a sequence in $F$. By assumption there are two consecutive
positive terms $a_{n}$ and $a_{n+1}$, and we easily see $a_{k}>0$ for $k \ge n$.
On the other hand by the explicit formula for fib. sequences there has to be some
$k$ s.t. $a_{k}\le 0$. By shifting of indices we therefore get a unique normal form of
the fib.  sequence, which is characterized by : $a_{0}\le 0$, $a_{k}>0$ for all $k \ge 1$.

By induction it can easily be seen that for every fib. sequence in normal form we have:
(a) For $k \le-1$ at least one of $a_{k}$ and $a_{k+1}$ is $\le 0$.
(b) For $k \ge 3$ we have $a_{k}\ge a_{1}+a_{2}$.

We conclude that  a fib. sequence in normal form is in $F$ iff $a_{1},a_{2}\le N$. (c)

Now the rest is easy: The bijection $f: F \to M$ is given by
$f((a_{n})_{n}) : = (a_{1},a_{2})$, where $(a_{n})_{n}$ is the unique normal form of the fib. sequence.
This is well defined as $0 < a_{1},a_{2}\le N$ by (c) and $0 \ge a_{0}= a_{2}-a_{1}$
which implies $a_{2}\le a_{1}$. Injectivity and surjectivy are clear by construction.

that is the official solution, too.
good job, solyaris !
the solution i gave in the contest was somewhat different.
i defined the relation $\sim$ between elements of $\mathbb{Z}^{2}$. two pairs of integers are similar if they generate the same fibonacci sequence. it is very easy to see that $\sim$ is an equivalence relation, and now we are interested in the number of classes with at least one representant with both components in $[1,N]$. we try to find a recurrence : (denote by $x_{k}$ the number of classes with at least one representant with both components in $[1,k]$)
$x_{1}=1$, that is $(1,1)$.
$x_{2}=3$, that is $(1,1)\sim(1,2)$, $(2,1)$, $(2,2)$.
then when we go from $n$ to $n+1$ we get that the pairs $(i,n+1)$ with $1\leq i\leq n$ are similar to ones already counted (with both components in $[1,n]$), and pairs $(n+1,i)$ with $1\leq i\leq n+1$ form new classes.
thus $x_{n+1}=x_{n}+n+1$ and we are done.
i missed a lot of details but this is the main idea

[quote="maky"]the correct answer is $\frac{N(N+1)}{2}$...[/quote]
I do not quite understand. Take $N=3$, you have $4$ pairs: $(3,3),(3,2),(3,1),(2,2)$.
By "two consecutive terms" does the problem mean exactly two terms or at least two terms? :maybe:

for $N=3$ there are also $(2,1)$ and $(1,1)$.
for any sequence that is requiered, we know that it has two consecutive terms both in the interval $[1,N]$. the sequences are the same if they differ by shifting of indices.

Sorry, I misunderstood the problem. i thought there ar eexactly two terms in the range. :(
Then here's my solution:
Look at the two consecutive terms $a_{1},a_{2}$ such that $a_{1}+a_{2}>N$, $a_{1},a_{2}\le N$. Therefore $a_{1}+a_{2}$ is in the range $[N+1,2N]$; for $a_{1}+a_{2}=N+k$, there are $(N-k)+1$ values for $a_{1}$, because $a_{1},a_{2}$ are integers that belong to $[1,N]$.   Also, every value of $a_{1}$ yields a distinct value of $a_{2}$ if the sum $a_{1}+a_{2}$ is given. Threfore the total number of sequences which is the total number of pairs $(a_{1},a_{2})$(because evry distinct sequence has a distinct pair $(a_{1},a_{2})$) is:
$\sum_{i=1}^{N}(N-k)+1=N^{2}-\frac{N(N+1)}{2}+N=\frac{N(N+1)}{2}$


[quote="xcenario"]$ 4a^2c^2 \minus{} (c^2 \minus{} b^2 \plus{} a^2)^2$[/quote]

Hint:
[hide]You might use $ X \equal{} 2ac, Y \equal{}$ $ (c^2 \minus{} b^2 \plus{} a^2)$ in $ X^2 \minus{} Y^2 \equal{} (X \minus{} Y)(X \plus{} Y)$.[/hide]

Solution:
[hide]$ [2ac \minus{} (c^2 \minus{} b^2 \plus{} a^2)][2ac \plus{} (c^2 \minus{} b^2 \plus{} a^2)] \equal{}$

$ [2ac \minus{} c^2 \plus{} b^2 \minus{} a^2][2ac \plus{} c^2 \minus{} b^2 \plus{} a^2] \equal{}$

$ \minus{} 1[a^2 \minus{} 2ac \plus{} c^2 \minus{} b^2][a^2 \plus{} 2ac \plus{} c^2 \minus{} b^2] \equal{}$

$ \minus{} 1[(a \minus{} c)^2 \minus{} b^2][(a \plus{} c)^2 \minus{} b^2] \equal{}$

$ \minus{} 1[(a \minus{} c) \minus{} b][(a \minus{} c) \plus{} b][(a \plus{} c) \minus{} b][(a \plus{} c) \plus{} b] \equal{}$

$ \minus{} 1(a \minus{} b \minus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)(a \plus{} b \plus{} c)$

{This form has all of the coefficients of variable a positive, has half of the coefficients
of variable b positive, and has half of the coefficients of variable c positive,
for the purposes of balance/aesthetics.}
[/hide]


pigeonhole principle
combinatorics unsolved
combinatorics

[quote="Johan Gunardi"]There are 1990 points inside a square with side length 12. Prove that there exists an equilateral triangle with side 11 which contains at least 498 of these points.[/quote]
Yes.
[hide="Solution"]
Divide the square into 4 squares of side length 6. Since $ 1990 > 4 \cdot 497$ we have that in one square there is more than $ 497$ points(in fact at least $ 498$..)
Then just prove that a equilateral triangle with side 11 can contain such a square..
[/hide]

[quote="Akashnil"][quote="Mathias_DK"]Then just prove that a equilateral triangle with side 11 can contain such a square..
[/quote]How? Please explain.[/quote]
Was too lazy to do it myself. But just draw one. It is proof enough.

Btw. There is a really nice problem just like this one from the first danish math competition ever ^^
It goes like this:
A circle with radius 2, have 15 points in it. Prove the existence of a circle with radius 1, which have 3 points in it.:)

[quote="Mathias_DK"][quote="Akashnil"][quote="Mathias_DK"]Then just prove that a equilateral triangle with side 11 can contain such a square..
[/quote]How? Please explain.[/quote]Was too lazy to do it myself. But just draw one. It is proof enough.[/quote]I still cant get it. I think it can't be drawn. Post a pic... may be that will explain.

[quote="Akashnil"][/quote][quote="Mathias_DK"][quote="Akashnil"][quote="Mathias_DK"]Then just prove that a equilateral triangle with side 11 can contain such a square..
[/quote]How? Please explain.[/quote]Was too lazy to do it myself. But just draw one. It is proof enough.[/quote][quote="Akashnil"]I still cant get it. I think it can't be drawn. Post a pic... may be that will explain.[/quote]
You're right.. It can't.. We'll just have to place the triangles in right positions:)


Let $ABC$ be a right triangle,$<A=90^{o}$, with inradius $r_{0}$. Take $D$ on $BC$ such that the incircles of $ABD$ and $ADC$ have inradii equals to $r$. Prove that: $\frac{1}{r}=\frac{1}{r_{0}}+\frac{1}{AD}$

We have the inradii relation for 2 equal incircles $\frac{2}{r}-\frac{r_{0}}{r^{2}}= \frac{2}{h_{a}}.$ Dividing it by $r_{0}$ and subtracting both sides from $\frac{1}{r_{0}^{2}},$

$\left(\frac{1}{r}-\frac{1}{r_{0}}\right)^{2}= \frac{1}{r_{0}^{2}}-\frac{2}{h_{a}r_{0}}= \frac{1}{r_{0}^{2}}\left(1-\frac{2r_{0}}{h_{a}}\right)$

Substituting $sr_{0}= |\triangle ABC| = \frac{h_{a}a}{2},\ \frac{2r_{0}}{h_a}= \frac{a}{s},$

$\left(\frac{1}{r}-\frac{1}{r_{0}}\right)^{2}= \frac{1}{r_{0}^{2}}\cdot \frac{s-a}{s}= \frac{s(s-a)}{s^{2}r_{0}^{2}}$

Since $r < r_{0},\ \frac{1}{r}> \frac{1}{r_{0}},$

$\frac{1}{r}-\frac{1}{r_{0}}= \frac{\sqrt{s(s-a)}}{sr_{0}}= \frac{\sqrt{s(s-a)}}{|\triangle ABC|}= \frac{1}{\sqrt{(s-b)(s-c)}}$

Not that we need the last 2 expressions for the given problem, but they hold for any triangle, not necessarily right. Substituting $r_{0}= s-a$ valid for the right angle $\angle A$ and $AD = \sqrt{s(s-a)}$ valid for any triangle (see solution of the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113632]Three equal incircles[/url]),

$\frac{1}{r}= \frac{1}{r_{0}}+\frac{1}{\sqrt{s(s-a)}}= \frac{1}{r_{0}}+\frac{1}{AD}$

Note that for the right angle $\angle A,$ $AD^{2}= s(s-a) = sr_{0}= |\triangle ABC|.$

[quote="juancarlos"][color=darkred]Let $ABC$ be a right triangle,$AB\perp AC$, with inradius $r$. Take $M$ on $BC$ such that the incircles of $ABD$ and $ADC$ have inradii equals to $\delta$. Prove that: $\frac{1}{\delta}=\frac{1}{r}+\frac{1}{AM}$[/color][/quote]

[color=darkblue][b]Proof.[/b] Denote the area $S$ of the triangle $ABC\ .$ Thus, $AB\perp AC\Longleftrightarrow$ $S=p(p-a)\ .$ From the my proof (remark) at the topic

http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 obtain (generally)  $\delta =\frac{r}{a}\cdot\left[p-\sqrt{p(p-a)}\right]$ and $AM^{2}=\sqrt{p(p-a)}\ .$

Therefore (in the our case), $AM=\sqrt{pr}$ and $\frac{1}{\delta}=\frac{1}{r}+\frac{1}{AM}$ $\Longleftrightarrow$ $\frac{p+\sqrt{p(p-a)}}{pr}=\frac{1}{r}+\frac{1}{\sqrt{p(p-a)}}\ ,$ what is truly.

[b]Remark.[/b] Generally, i.e. in any triangle $ABC$ exists the relation $\boxed{\ \frac{1}{\rho}=\frac{1}{r}+\cot \frac{A}{2}\cdot\frac{1}{AM}\ }\ .$[/color]


geometry
geometric transformation
LaTeX
geometry solved

Points $D,E,F$ are on the sides $BC, CA$ and $AB$, respectively which satisfy $EF || BC$, $D_1$ is a point on $BC,$ Make $D_1E_1 || D_E, D_1F_1 || DF$ which intersect $AC$ and $AB$ at $E_1$ and $F_1$, respectively. Make $\bigtriangleup PBC \sim \bigtriangleup DEF$ such that $P$ and $A$ are on the same side of $BC.$ Prove that $E, E_1F_1, PD_1$ are concurrent.

[color=red][Edit by Darij: See my post #4 below for a [b]possible correction[/b] of this problem. However, I am not sure that it is in fact the problem given at the TST... Does anyone have a reliable translation?][/color]

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
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etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

[quote="orl"]Prove that $E, E_1F_1, PD_1$ are concurrent.[/quote]

I don't understand.

A hardy attempt to reconstruct the correct problem statement:

[color=blue][b]Problem.[/b] Let D, E, F be points on the sides BC, CA, AB of a triangle ABC such that EF || BC. Let $D_1$ be a point on the side BC, and let the parallels to the lines DE and DF through the point $D_1$ meet the lines AC and AB at the points $E_1$ and $F_1$, respectively. Let the parallels to the lines DF and DE through the points B and C intersect at a point P. Prove that the lines EF, $E_1F_1$ and $PD_1$ are concurrent if and only if the point $D_1$ is the midpoint of the segment BC.[/color]

[hide="Changes in the problem statement"]Here is what I changed in the problem statement:
- Made the definition of the points $E_1$ and $F_1$ clearer.
- Replaced the definition of the point P through the similarity $\bigtriangleup PBC \sim \bigtriangleup DFE$ (not $\bigtriangleup PBC \sim \bigtriangleup DEF$, as Orl wrote!) and the condition that P and A lie on the same side of the line BC by the (equivalent, but more appropriate for this problem) definition of P as the point of intersection of the parallels to the lines DF and DE through the points B and C (in fact, both definitions are equivalent: the condition $\bigtriangleup PBC \sim \bigtriangleup DFE$ yields < PBC = < DFE, while EF || BC yields < ABC = < AFE, so that < PBA = < PBC + < ABC = < DFE + < AFE = < DFA, and thus PB || DF, so that the point P lies on the parallel to the line DF through the point B, and similarly the point P lies on the parallel to the line DE through the point C, so the point P is the point of intersection of these two parallels).
- Replaced the "line E" by the "line EF".
- Added the condition that "the point $D_1$ is the midpoint of the segment BC". Without this condition, the problem is plainly wrong.[/hide]

[i]Solution.[/i] Since both of the lines BP and $D_1F_1$ were constructed as parallels to the line DF, we get a triple of parallel lines: $BP\parallel D_1F_1\parallel DF$. Similarly, $CP\parallel D_1E_1\parallel DE$.

Let the lines BP and CP meet the line $E_1F_1$ at the points U and V, respectively.

Let the lines $PD_1$ and $E_1F_1$ intersect at R. Then, since $BP\parallel D_1F_1$, by Thales we have $\frac{UR}{F_1R}=\frac{PU}{D_1F_1}=\frac{PR}{D_1R}$, and since $CP\parallel D_1E_1$, Thales yields $\frac{VR}{E_1R}=\frac{VP}{E_1D_1}=\frac{PR}{D_1R}$. Thus, $\frac{UR}{F_1R}=\frac{PU}{D_1F_1}=\frac{VR}{E_1R}=\frac{VP}{E_1D_1}$. This equation splits into $\frac{UR}{F_1R}=\frac{VR}{E_1R}$, what is equivalent to $\frac{UR}{F_1R}=\frac{RV}{RE_1}$ or to $\frac{UR}{RV}=\frac{F_1R}{RE_1}$, and into $\frac{PU}{D_1F_1}=\frac{VP}{E_1D_1}$, what becomes $\frac{VP}{PU}=\frac{E_1D_1}{D_1F_1}$.

Since BC || FE, CP || ED and PB || DF (these are just the relations EF || BC, CP || DE and BP || DF, slightly rewritten), the triangles BCP and FED are homothetic. Thus, $\frac{BP}{PC}=\frac{FD}{DE}$. In other words, $\frac{BP}{PC}=\frac{DF}{ED}$.

Now we will use a fact which, if not known, is a nice little exercise for the reader (it [i]should[/i] appear in good books, alas it doesn't...):

[color=blue][b]Menelaos theorem for quadrilaterals.[/b] Let ABCD be a quadrilateral, and let X, Y, Z, W be four collinear points on its sidelines AB, BC, CD, DA, respectively. Then, $\frac{AX}{XB}\cdot\frac{BY}{YC}\cdot\frac{CZ}{ZD}\cdot\frac{DW}{WA}=1$.[/color]

Applying the Menelaos theorem for quadrilaterals to the quadrilateral UVCB with the collinear points R, P, $D_1$, P on its sidelines UV, VC, CB, BU, respectively, we get

$\frac{UR}{RV}\cdot\frac{VP}{PC}\cdot\frac{CD_1}{D_1B}\cdot\frac{BP}{PU}=1$.

In other words:

$\frac{UR}{RV}\cdot\frac{VP}{PU}\cdot\frac{BP}{PC}\cdot\frac{CD_1}{D_1B}=1$.

Using the relations $\frac{UR}{RV}=\frac{F_1R}{RE_1}$, $\frac{VP}{PU}=\frac{E_1D_1}{D_1F_1}$ and $\frac{BP}{PC}=\frac{DF}{ED}$, this becomes

$\frac{F_1R}{RE_1}\cdot\frac{E_1D_1}{D_1F_1}\cdot\frac{DF}{ED}\cdot\frac{CD_1}{D_1B}=1$.

Now, again, this rewrites as

$\frac{F_1R}{RE_1}\cdot\left(\frac{E_1D_1}{ED}: \frac{D_1F_1}{DF}\right)\cdot\frac{CD_1}{D_1B}=1$.

Equivalently,

[color=green][b](1)[/b][/color] $\frac{F_1R}{RE_1}\cdot\left(\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}\right)\cdot\frac{CD_1}{D_1B}=1$.

Now, since $D_1E_1\parallel DE$, Thales yields $\frac{E_1D_1}{ED}=\frac{CE_1}{E_1E}$. In other words, $\frac{E_1D_1}{ED}=\frac{CE_1}{E_1E}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}$. Another application of Thales using $D_1E_1\parallel DE$ shows that $\frac{CE_1}{E_1E}=\frac{CD_1}{D_1D}$, so this becomes

$\frac{E_1D_1}{ED}=\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}$.

Similarly,

$\frac{F_1D_1}{FD}=\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\cdot\frac{AF}{BF}$.

Thus,

$\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}=\left(\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\cdot\frac{AE}{CE}\right): \left(\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\cdot\frac{AF}{BF}\right)$.

Since EF || BC, by Thales we have $\frac{AE}{CE}=\frac{AF}{BF}$, so that this simplifies to

$\frac{E_1D_1}{ED}: \frac{F_1D_1}{FD}=\left(\frac{CD_1}{D_1D}\cdot\frac{E_1E}{AE}\right): \left(\frac{BD_1}{D_1D}\cdot\frac{F_1F}{AF}\right)$
$=\frac{CD_1}{BD_1}\cdot\frac{E_1E}{AE}\cdot\frac{AF}{F_1F}=\left(-\frac{CD_1}{D_1B}\right)\cdot\left(-\frac{E_1E}{EA}\right)\cdot\left(-\frac{AF}{FF_1}\right)$
$=-\frac{CD_1}{D_1B}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}$.

Hence, the equation [color=green][b](1)[/b][/color] becomes

$\frac{F_1R}{RE_1}\cdot\left(-\frac{CD_1}{D_1B}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}\right)\cdot\frac{CD_1}{D_1B}=1$.

In other words,

$-\left(\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}\right)\cdot\left(\frac{CD_1}{D_1B}\right)^2=1$.

Thus,

[color=green][b](2)[/b][/color] we have $\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}=-1$ if and only if $\left(\frac{CD_1}{D_1B}\right)^2=1$.

But according to the Menelaos theorem, applied to the triangle $AE_1F_1$ with the points R, E, F on its sidelines $E_1F_1$, $AE_1$ and $AF_1$, the equation $\frac{F_1R}{RE_1}\cdot\frac{E_1E}{EA}\cdot\frac{AF}{FF_1}=-1$ is equivalent to the assertion that the points R, E, F are collinear, i. e. that the point R lies on the line EF; since the point R is defined as the point of intersection of the lines $PD_1$ and $E_1F_1$, this assertion is thus equivalent to stating that the lines EF, $E_1F_1$ and $PD_1$ are concurrent.

On the other hand, the equation $\left(\frac{CD_1}{D_1B}\right)^2=1$ is equivalent to $\frac{CD_1}{D_1B}=1$ (since $\frac{CD_1}{D_1B}=-1$ is impossible, unless the point $D_1$ is infinite, what we don't want to allow); and this is actually just the statement that the point $D_1$ is the midpoint of the segment BC.

Hence, the result [color=green][b](2)[/b][/color] rewrites as follows: The lines EF, $E_1F_1$ and $PD_1$ are concurrent if and only if the point $D_1$ is the midpoint of the segment BC.

And the problem is solved...

  Darij

I do not see why D1 has to be the midpoint of BC.

  The configuration is projective. There should be no limitations on position of D1.

  The only changes needed is correcting typo to "line EF",  and not "line E" and

  orientation of PBC as Darij did.

 ( Please, do not think that I translated from original Chineese  text!)

   Thank you.

    M.T.

Here is my solution(?) :

  Using central projection transform triangles ABC and PBC into  triangles with a

   point D into middle of BC. This will put point P

  onto A angle median.Then move line EF until it becomes midline, P will be in

  the A angle vertex.

  Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They

  intersect on the  mid-line.

  Thank you.

   M.T.

[quote="armpist"]I do not see why D1 has to be the midpoint of BC.

  The configuration is projective.[/quote]

No, it isn't. There is a lot of parallels around in the problem statement.

[quote="armpist"]Here is my solution(?) :

  Using central projection transform triangles ABC and PBC into  triangles with a

   point D into middle of BC. This will put point P

  onto A angle median.Then move line EF until it becomes midline, P will be in

  the A angle vertex.

  Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They

  intersect on the  mid-line.[/quote]

Not that I would understand it, but at least it is nice that you call it a "solution(?)", not a "solution". And as I said above, you can't just project everything expecting the parallel lines to remain parallel after the projection, and particularly you can't force the triangle DEF to be the medial triangle of the triangle ABC.

  Darij

Dear Darij,

  In  your wonderful post

            http://www.mathlinks.ro/Forum/topic-59919.html

  you wrote about a line through midpoint:

    "..... It is not undefined. The midpoint between an Euclidean point and an infinite point is the latter infinite point (the Euclidean point is "insignificant"). Only the midpoint between two infinite points is undefined. And yes, assertion (c) very well holds with the point X being infinite.  ......  "

   So  it is possible to send BC to infinity (preserving parallels)  and EF (chasing

   BC in parallel fashion) only half way there?  Some experts have found it being

   hysterically funny.

   But really, where are all the lines that are parallel to the line being on it's way to

   infinity?

   Well,  I think differently: the more unexplained phenomena - the better.

   People tend to like it because of the amusement and entertainment

   components of it...

   Thank you.

   M.T.

I have just read the official version of this problem. Here is my translation:
Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. $D_1$ is a point on side BC(different from B,D,C), through $D_1$ draw $D_1E_1//DE,D_1F_1//DF$, which intersect AC and AB at points $E_1,F_1$ respectively. Join $E_1F_1$. Construct,on the same side of A, triangle PBC, such that $\triangle PBC\sim\triangle DEF$. Join $PD_1$. Prove that $EF,E_1F_1,PD_1$ are concurrent.

I know very well this must be the worse translation in the world...

[quote="mecrazywong"]I have just read the official version of this problem. Here is my translation:
Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. $D_1$ is a point on side BC(different from B,D,C), through $D_1$ draw $D_1E_1//DE,D_1F_1//DF$, which intersect AC and AB at points $E_1,F_1$ respectively. Join $E_1F_1$. Construct,on the same side of A, triangle PBC, such that $\triangle PBC\sim\triangle DEF$. Join $PD_1$. Prove that $EF,E_1F_1,PD_1$ are concurrent.[/quote]

That's funny, since this means the official version is wrong...

  darij

Hello guys,

  I think I see what is wrong.  Darij put us in thinking that the triangle PBC can be

  constructed  only by lines through B and C parallel to DE and DF.

  There is another triangle with vertex P still on the same line.  And it has proper

  orientation.  The problem is correct with only typo line E should be line EF.

   Now it all should work out, I just reduced the number of parallel lines for Darij.

   Thank you.

   M.T.

[quote="armpist"]Hello guys,

  I think I see what is wrong.  Darij put us in thinking that the triangle PBC can be

  constructed  only by lines through B and C parallel to DE and DF.

  There is another triangle with vertex P still on the same line.  And it has proper

  orientation.  The problem is correct with only typo line E should be line EF.[/quote]

You mean that triangle PBC should be similar to triangle DEF (with corresponding letters for corresponding vertices)? Try it, it won't hold.

  Darij

Dear Darij,

    PBC  corresponds to FDE for direct similarity.   I guess the problem statement

    just states similarity to make it harder to find the proper triangle.

    In my sketch it works out fine. The 3 lines intersect . Line  EF  if rather close to

    BC  for angle D to be obtuse.

    If D and D1 on the same half of BC  then  PBC ~ EDF.

    Thank you.

    M.T.

[u]A vectorial solution ([i]orientated segments[/i]).[/u]

The fixed points: $D$, $E$, $F$, $P\Longrightarrow$ the constant values $d$, $k$.

The mobile points: $D_1$, $E_1$, $F_1$, $R\Longrightarrow$ the variable value $x$.

$\frac{AE}{EC}=\frac{AF}{FB}=k\Longrightarrow \underline {\overline {\left| E=\frac{A+kC}{1+k},\ F=\frac{A+kB}{1+k}\right| }}.$

$\frac{BD}{DC}=d\Longrightarrow \underline {\overline {\left| D=\frac{B+dC}{1+d}\right| }}.$

$\frac{BD_1}{D_1C}=x\Longrightarrow \underline {\overline {\left| (1+x)D_1=B+xC\right| }}\ (1).$

$\triangle PBC\sim \triangle DFE\ (D\in AP)\Longrightarrow \frac{BP}{FD}=\frac{AB}{AF}=\frac{k+1}{k}\Longrightarrow$

$P=B+\frac{k+1}{k}(D-F)\Longrightarrow\underline {\overline {\left| k(1+d)P=-(1+d)A+(1+k)B+d(1+k)C\right| }}\ (2).$

$\frac{EE_1}{E_1C}=\frac{DD_1}{D_1C}=\frac{D_1-D}{C-D_1}=\frac{\frac{B+xC}{1+x} -\frac{B+dC}{1+d}}{C-\frac{B+xC}{1+x}}=\frac{x-d}{1+d}\Longrightarrow$

$E_1=\frac{E+\frac{x-d}{1+d} C}{1+\frac{x-d}{1+d}}=\frac{(1+d)E+(x-d)C}{1+x}\Longrightarrow$

$(1+x)E_1=(1+d)\frac{A+kC}{1+k}+(x-d)C\Longrightarrow$

$\underline {\overline {\left| (1+x)(1+k)E_1=(1+d)A+(k+x+kx-d)C\right| }}.$

$\frac{FF_1}{F_1B}=\frac{DD_1}{D_1B}=\frac{D_1-D}{B-D_1}=\frac{\frac{B+xC}{1+x}-\frac{B+dC}{1+d}}{B-\frac{B+xC}{1+x}}=\frac{d-x}{x(1+d)}\Longrightarrow$

$F_1=\frac{F+\frac{d-x}{x(1+d)} B}{1+\frac{d-x}{x(1+d)}}=\frac{x(1+d)\frac{A+kB}{1+k}+(d-x)B}{d(1+x)}\Longrightarrow$

$\underline {\overline {\left| d(1+k)(1+x)F_1=x(1+d)A+(kdx+kd+d-x)B\right| }}.$

$R\in EF\cap E_1F_1\Longleftrightarrow R=uE+(1-u)F=vE_1+(1-v)F_1\Longleftrightarrow$

$\frac{u}{kx+k+x-d}=\frac{1-u}{kdx+kd+d-x}=\frac{1}{k(1+d)(1+x)},$ $\frac v1=\frac{1-v}{d}=\frac{1}{1+d}\Longleftrightarrow$

$\underline {\overline {\left| (1+k)(1+d)(1+x)R=(1+d)(1+x)A+(kdx+kd+d-x)B+(kx+k+x-d)C\right| }}\ (3).$
==========================================================================
$D_1\in PR\Longleftrightarrow \left| \begin{array}{ccc} 0 & 1 & x \\ -(1+d) & 1+k & d(1+k) \\ (1+d)(1+x) & kdx+kd+d-x & kx+k+x-d \ \end {array}\right| =0$

$\Longleftrightarrow k(1+d)(x^2-1)=0\Longleftrightarrow x=1$, i.e. $D_1$ is the midpoint of the side $[BC]$.

[u]Remark.[/u] If the points $P$, $A$ are on the same side of the line $BC$ and $\triangle PBC\sim \triangle DEF$,

($PB\parallel DE$, $PC\parallel DF$) then the official version is correctly !

[quote="mecrazywong"]I have just read the official version of this problem. Here is my translation:
Points $ D$, $ E$, $ F$ are on the sides $ BC$, $ CA$, $ AB$ of an acute-angled $ \triangle{ABC}$ respectively, which satisfy $ EF\parallel BC$. $ D_1$ is a point on side $ BC$(different from $ B$, $ D$, $ C$), through $ D_1$ draw $ D_1E_1 \parallel DE, D_1F_1 \parallel DF$, which intersect $ AC$ and $ AB$ at points $ E_1$, $ F_1$ respectively. Join $ E_1F_1$. Construct,on the same side of $ A$, triangle $ PBC$, such that $ \triangle PBC\sim\triangle DEF$. Join $ PD_1$. Prove that $ EF,E_1F_1,PD_1$ are concurrent.
[/quote]
Yes, this is the official version appeared in the published book by the Chinese coach teams.


algebra
polynomial
limit
real analysis
real analysis unsolved

In the following  $P(X): =\sum\limits_{k=0}^{p}a_kX^{p-k} \; ,\; Q(X): =\sum\limits_{k=0}^{p+1}b_kX^{p+1-k} \;$ are such that   $a_0> 0 \; , \; b_0 >0$  and
 $P(x)> 0 \; ,\; Q(X)>0$  for all $X \in {\mathbb R} .$ Denote  $\displaystyle A_n(P)=\displaystyle \frac{1}{n}\sum\limits_{k=1}^{n}P(k)\; \; ,\; \; \displaystyle G_n(Q): = \displaystyle \sqrt[n]{P(1)P(2)\cdots P(n)}\; ,\; (n\ge 2) ,$
 and consider the sequence $\left(X_n\right)_{n=2}^{\infty}$  having the general term  $\displaystyle \begin{array}{|c|} \hline \\ \displaystyle X_n\equiv X_n(P,Q): =\displaystyle \frac{(p+2)}{(p+1)}\frac{G_n(P)A_n(Q)}{G_n(Q)A_n(P)} \\ \\ \hline \end{array}\; ,\; n\in \{2,3,...\}.$
[color=green][b]Prove or disprove that $\begin{array}{|c|} \hline \lim\limits_{n\to \infty}X_n(P,Q)= e\\ \hline \end{array}\; ,$   $e$   being Napier's constant.  [/b] [/color]

Asymptotic expansion technic give

$A_n(P)\sim a_0 \frac{n^p}{p+1}$
$A_n(Q)\sim b_0 \frac{n^{p+1}}{p+2}$

$G_n(P) \sim a_0{n!}^{\frac{p}{n}}$
$G_n(Q) \sim b_0{n!}^{\frac{p+1}{n}}$

$X_n \sim \frac{n}{n!}$ wich tends to $e$


calculus
integration
logarithms
calculus computations

Evaluate: [hide]$ \int \frac {log (x \plus{} \sqrt {1 \plus{} x^2})}{\sqrt {1 \plus{} x^2}}dx$[/hide]

I prefer step-by-step solution along with explanation; instead of just hinting or giving the straight answer.

Thank you.

I think it's a typo.

$ \int\frac{\log(x\plus{}\sqrt{1\plus{}x^2})}{\sqrt{1\plus{}x^2}}\,dx$ is much easier- the substitution $ x\equal{}\sinh t$ solves it immediately.
$ \int\frac{\log(\sinh t\plus{}\cosh t)}{\cosh t}\cosh t\,dt\equal{}\int t\,dt\equal{}\frac12 t^2\plus{}C$, and
$ \int \frac{\log(x\plus{}\sqrt{1\plus{}x^2}}{\sqrt{1\plus{}x^2}}\,dx\equal{} \frac12\left(\sinh^{\minus{}1}(x)\right)^2\plus{}C\equal{} \frac12\left(\log(x\plus{}\sqrt{1\plus{}x^2})\right)^2\plus{}C$

The question as you wrote it is probably impossible.

[quote="jmerry"]I think it's a typo.

$ \int\frac {\log(x \plus{} \sqrt {1 \plus{} x^2})}{\sqrt {1 \plus{} x^2}}\,dx$ is much easier- the substitution $ x \equal{} \sinh t$ solves it immediately.
$ \int\frac {\log(\sinh t \plus{} \cosh t)}{\cosh t}\cosh t\,dt \equal{} \int t\,dt \equal{} \frac12 t^2 \plus{} C$, and
$ \int \frac {\log(x \plus{} \sqrt {1 \plus{} x^2}}{\sqrt {1 \plus{} x^2}}\,dx \equal{} \frac12\left(\sinh^{ \minus{} 1}(x)\right)^2 \plus{} C \equal{} \frac12\left(\log(x \plus{} \sqrt {1 \plus{} x^2})\right)^2 \plus{} C$

The question as you wrote it is probably impossible.[/quote]

Maybe you have a mistake when do $ \int\frac {\log(\sinh t \plus{} \cosh t)}{\cosh t}\cosh t\,dt \equal{} \int t\,dt$
The result can be   $ \frac {ln10\left[log\left(x\plus{}\sqrt{1\plus{}x^2}\right)\right]^2}{2} \plus{} C$

it is just a matter of notation , somebody treat $ \log$ as base 10 but somebody treat $ \log$ as base e. And I believe jmerry is the latter .

Maybe you're right. It's just a matter of notation.

I think better is substitution: $ t \equal{} \ln (x \plus{} \sqrt {1 \plus{} x^2})$, then $ dt \equal{} \frac {dx}{\sqrt {1 \plus{} x^2}}$, hence we calculate:
$ \int t dt \equal{} \frac {t^2}{2} \plus{} C \equal{} \frac {1}{2}\ln^2(x \plus{} \sqrt {1 \plus{} x^2}) \plus{} C$

[quote="CyanSnow"]Evaluate $ \int \frac {\ln \left(x + \sqrt {1 + x^2}\right)}{\sqrt {1 + x^2}}\ \mathrm {dx}\ .$[/quote]
$ \int\frac {1}{\sqrt {1+x^2}}\ \mathrm {dx}=\ln\left(x+\sqrt {1+x^2}\right)+\mathcal C$ $ \Longleftrightarrow$ $ \left[\ln\left(x+\sqrt {1+x^2\right)\right]'=\frac {1}{\sqrt {1+x^2}}\ .}$

Denote $ \boxed {f(x)=\ln \left(x + \sqrt {1 + x^2}\right)}\ .$ Therefore, $ f'(x)=\frac {1}{\sqrt {1+x^2}}$ and

$ \int \frac {\ln \left(x + \sqrt {1 + x^2}\right)}{\sqrt {1 + x^2}}\ \mathrm {dx}=\int f(x)f'(x)\ \mathrm {dx}=$ $ \frac 12\cdot f^2(x)+\mathcal R\ .$


logarithms
calculus
calculus computations

Given a prime p. Find the maximum value (if it exists) of :

  $ f(x) \equal{} (x \plus{} 1)p$^$ ( \minus{} x/2)$

?? what seems to be the problem ??

$ f'(x) = p^{-\frac{x}{2}} - \frac{1}{2} (x+1) p^{-\frac{x}{2}} \ln (p)= 0$

since $ p^{-\frac{x}{2}} > 0$ we have:

$ 2 = (x+1) \ln(p)$

$ x = \frac{2}{\ln(p)}-1$

and the maximum value is:

$ f\left(\frac{2}{\ln(p)}-1\right) = \left(\frac{2}{\ln(p)}-1+1\right)p^{\frac{1}{2}-\frac{1}{\ln(p)}} =$

${ = \frac{2\sqrt{p}}{\ln(p)}p^{\log(p, \frac{1}{e})}} =$

$ = \frac{2 \sqrt{p}}{e \ln(p)}$

Am I missing something - why prime?

oh sorry my question is this :

 Let p_i(x) = be the maximum of f(x) when p=the i-th prime. wich you already gave. if we consider the product

  of p_i(x) of all primes ie  p_1(x)p_2(x)... to infinity. ¿does there exist a constant c such that :

  c> p_1(x)p_2(x)....?


well,dont u ppl think that the age limit of 40 for teh fields medal is a bit too harsh,imagine ppl like andrew wiles and dan voiculescu not getting the deserved medal.
dont see the reason behind keeping such a restriction :?

to encourage young people. unfortunately those two were bad lucky, however they got all the credit they deserved :D

It's a travesty that there's an age limit to such a prestigious award. But for the same reason that there continues to be no Nobel Prize award for mathematics, this rule too will probably never change. Traditions are very hard to break. Though some serious discussion was given to changing it after the Andrew Wiles fiasco, nothing ever came of it.

The problems with the current set up of the Fields Medal are, 1) that people live longer now than ever before, so 40 is no longer considered so late in life, 2) even if the award really is to promote future promise (as they say it is), then they should give them even younger -- before the recipients have already made their careers with great achievements, 3) there are already sufficiently many awards to promote future promising careers - they are the postdoctoral fellowships and grants.

I agree with all of you people, but I guess we can't change the situation... However, I really don't understand the logic behind such restrictions, except for trivial stories like this one: I've heard that the reason why there is no Nobel for mathematics is that.. Nobel's wife cheated on him with a mathematician :D. I don't think it's true, of course.

No, he didn't have.

His girlfriend finally married to a mathematician.. So it is said that he didn't like mathematics... that's why there is no math prize in Nobel Prizes.

why in the world did she have to do that ? :) couldn't she have chosen a chemist or something? :D

lol. Unfortunately Nobel was a chemist, so I don't think that would have applied if it had been with one of his own kind.

But actually this story is all just part of mathematical folk-lore. Nobel did not have a wife, and more than likely this had nothing to do with Mittag-Leffler. No one really knows for certain why he excluded math from the Nobel Prize categories - it is all a matter of speculation.

Here are two good sources of information about the founding of the Field's Medal - http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Fields.html

And the mythology behind the Nobel Prize - http://almaz.com/nobel/why_no_math.html

By the way, here is another interesting page I found about controversies in the Nobel Prize - http://www.pbs.org/kqed/nobel/sttimeline.html

And for those who don't know, there is also another very prestigious award given for Mathematics which does not have an age limit. It is called the Wolf Prize. It's not as old as the Field's Medal, but I speculate that in some years it will become as significant an award. The past winners of the Wolf Prize can be found here - http://www.aquanet.co.il/wolf/wolf5.html

Yeah, I think it started being awarded in 2002:

[url]http://www.abelprisen.no/index_english.html[/url]

indeed, it was big news then.

Well nobel's right, math is not a sort of science...

But it's the basics for all other sciences!!

Next Fields Medals will be in ICM International Congress of Mathematicians 2006 in Barcelona. Every 4 years a big meeting of about
3000 mathematicians from all countries, plenary lecture, conference,
poster session...

http://www.icm2006.org/#


function
inequalities
abstract algebra
real analysis
real analysis solved

If a,b are from (0,+00(, a<b prove that the eqs:

((a+b)/2)^(x+y)=a^x*b^y has at least one solution in (a,b).

Is there another way of proving this without applying Lagrange's theorem for funtions: f(q)=lnq on the interval [a,(a+b)/2] and then the same thing on the intervall [(a+b)/2,b].

cheers!

hmm... forgot to add that this problem was given for the 11th grade in the Republic of Moldavia 2003.

I would also like to add that the contest in Moldavia wasn't very hard.. at least it seemed to me... :D

cheers!

By soln in (a,b) you mean x and y should be in (a,b)? What is the unknown of the eqn? x, y, or both?

Yes x and y should be in (a,b), so the unknown stuff from the eqs are x and y!

cheers!

In case you mean a solution in x,y s.t. x and y are in (a,b) here it is (it's similar to Lagrange though, it uses continuous functions)

First we make x=y=a and we get ((a+b)/2)^2a > (ab)^a from AM-GM. By the same way ((a+b)/2)^2b > (ab)^b if x=y=b. Now we make x=a, y=b. We should prove ((a+b)/2)^(a+b) < a^a*b^b. We take ln of this and we must prove f((a+b)/2) < (f(a)+f(b))/2, where f(x)=xlnx. It's easy to prove that f is convex, so ((a+b)/2)^(a+b) < a^a*b^b. Now let's assume we take x an y inside (a,b). We have 2 cases:

Case 1 ((a+b)/2)^(x+y) > a^x*b^y. Then we make x->a and y->b independently. Because at the limit we have ((a+b)/2)^(a+b) < a^a*b^b we must "pass" through a case when ((a+b)/2)^(x+y) = a^x*b^y. I can elaborate on that, if you think this "independent variation" isn't clear enough.

Case 2 ((a+b)/2)^(x+y) < a^x*b^y. we do sth similar to case 1, just make x->a and y->a.

hmm... I was thinking more of a non analisys solution..

some that doesn't uses Lagrange, continuos functions.. Fermat, rolle things like that!!could there be found let's say a junior solution?

cheers! :D  \varepsilon

I think an actual soln can't be found, so a proof must be based on inequalities. But if you prove the existence of solutions based on inequalities it means that you are using Darboux, so..

So i guess that settles it!! no elementary solution for this one!!

cheers! and g'night... kinda late here in Romania.. 3 in the morning.. shish..! :D :D .. talking about math fanatics! :D

[quote="Lagrangia"]

Is there another way of proving this without applying Lagrange's theorem for funtions: f(q)=lnq on the interval [a,(a+b)/2] and then the same thing on the intervall [(a+b)/2,b].

cheers![/quote]

Can you elaborate on applying Lagrange's theorem method?
Thanks.