How important is your nationality ? This question might be silly but i just want to know this .

To me it's nothing. I was born in London, but I only lived in the UK for a year and half, so I don't feel very British. I've lived in the USA most of my life, and my parents were born here, but it wasn't many generations ago that my mom's family was Irish and my dad's family was Austrian and Israeli. So unlike (for example) my Chinese friends whose grandparents immigrated here, I don't have any clear "heritage" either.

Last year in my writing class, I wrote an essay (developing the ideas of Benedict Anderson) about nationalism and communities. Anderson wrote about the development of nationalism to replace the religious and social class communities that have been slowly decaying for the past few hundred years. My argument was that globalization is now beginning to undermine these national communities, and non-geographic social communities are starting to replace them. People use our advanced communication and traveling technology to build strong communities based on cultural interests. Nationalism isn't dead, but the Internet is eroding it.

i'm korean
but yeah well so was cho seung-hui

so i'm actually american

i'm taiwanese but i was born here but i still say im taiwnaese

im chinese (mainland)
so many taiwanese...
cmon mainland chinese people unite   :lol:

[quote="Ignite168"][quote="1=2"]He's obviously american because he won last years Mathcounts.

USA, but I went to Canada(my parents made me) for some sightseeing.[/quote]

Stop posting in topics you don't understand at all for god's sake.[/quote]

agreed

I think nationality is important.  But in a different way.  There are people who care about their nationality, in that sense that they want to change it.
In Europe, obvious examples are the Scots, the Basques, the Catalonians, the Flemings (like me).  Other exampes are the Chechens, the Kurds,...

[quote="Xevarion"]
Last year in my writing class, I wrote an essay (developing the ideas of Benedict Anderson) about nationalism and communities. Anderson wrote about the development of nationalism to replace the religious and social class communities that have been slowly decaying for the past few hundred years. My argument was that globalization is now beginning to undermine these national communities, and non-geographic social communities are starting to replace them. People use our advanced communication and traveling technology to build strong communities based on cultural interests. Nationalism isn't dead, but the Internet is eroding it.[/quote]
Or the internet can change it, rather than eroding it!

[quote="Treething"][quote="Ignite168"][quote="1=2"]He's obviously american because he won last years Mathcounts.

USA, but I went to Canada(my parents made me) for some sightseeing.[/quote]

[b]Stop posting[/b] [size=50]in topics you don't understand at all for god's sake[/size][b].[/b][/quote][/quote]

Lmao.

[quote="Ignite168"][quote="Zakary"]I'm American, and I'm glad to be in a very good country.  However, I find it funny (as in strange and unsettling) that there is hate for Americans in other parts of the world.[/quote]

This is exactly why I hate Americans.

I hate them because they have no idea why we hate them.[/quote]

Is that why you are from Michigan?

[quote="Treething"][quote="Ignite168"][quote="1=2"]He's obviously american because he won last years Mathcounts.

USA, but I went to Canada(my parents made me) for some sightseeing.[/quote][size=150]Stop posting[/size][size=59] in topics you don't understand at all for god's sake.[/size][/quote][/quote]

Shut up. :P

[quote="1=2"][quote="Ignite168"][quote="Zakary"]I'm American, and I'm glad to be in a very good country.  However, I find it funny (as in strange and unsettling) that there is hate for Americans in other parts of the world.[/quote]

This is exactly why I hate Americans.

I hate them because they have no idea why we hate them.[/quote]

Is that why you are from Michigan?[/quote]

I'm not from Michigan idiot, if I was, I'm pretty sure all the Michiganders would know.

This is why you should STOP POSTING ABOUT THINGS YOU HAVE NO IDEA ABOUT, AND SINCE THAT'S EVERYTHING, STOP POSTING.

Originally your location thingy said "Michigan".

[quote="Ignite168"][quote="noneoftheabove"][quote="Treething"]sheesh taiwan isn't even a country

(jk? idk)[/quote]

...Excuse me "Stop posting about things you have no idea about" applies to you also =P[/quote]

Taiwan is not a country[/quote]

actually it is..

[quote="Art of Owna"][/quote][quote="Ignite168"][quote="noneoftheabove"][quote="Treething"]sheesh taiwan isn't even a country

(jk? idk)[/quote]

...Excuse me "Stop posting about things you have no idea about" applies to you also =P[/quote]

Taiwan is not a country[/quote][quote="Art of Owna"]

actually it is..[/quote]

Actually, it isn't.

Everyone thinks Taiwan is a country, it's so stupid. Saying Taiwan is a country and not part of China is like saying Hawaii isn't part of the US.

Wow, finally someone noticed that UN doesn't recognize Taiwan.

[quote="kohjhsd"]Taiwan was member of the United Nations (UN) until 1971, when it was replaced by the current "China" by the United Nations, since there can't be two Chinas in the UN.[url=http://en.wikipedia.org/wiki/United_Nations_General_Assembly_Resolution_2758]*[/url]

It is still a [url=http://en.wikipedia.org/wiki/Political_status_of_Taiwan]controversial issue[/url] though. Taiwan is still considered a country by many people, but at the same time, many people consider Taiwan not an official country.[/quote]
It's funny that they both consider the other be an illegitimate ruler over the other part.  But ...let's be realistic, take a look at the number of people on the mainland and in Taiwan. :huh:
I don't know what the status of Taipei is right now, but for a long time Taiwan still considered Nanjing their capital, even though they had no control over it at all.

[quote="noneoftheabove"][quote="Treething"]HEY I CAN BE PRETENTIOUS AND USE BIG WORDS TOO

United Nations doesn't recognize Taiwan as a country; I'm surprised that so many of you argue so vehemently about Taiwan's status.[/quote]

...Since when was being recognized by the United Nations part of a definition of a country?[/quote]
It isn't. :wink:
[img]http://images.acclaimimages.com/_gallery/_SM/0195-0512-2623-3108_SM.jpg[/img]

Well, then there isn't any possible proof that Taiwan is a country.
What, did you have a vision that Taiwan is a country?

[quote="noneoftheabove"]Sigh, i would rather not debate about this especially in the f&g forum but here goes...

Definition of a country:
Defined Land-check
Population-check
Government-check
Soverignty-check(China cannot change how taiwan operates unless you mean by military threats...then US owns the entire world)

Yes, there are countries that do recognize Taiwan as a country which is proof.
There are also other countries who would recognize Taiwan as a country but the threat economically from China to that country would be too much of a risk.
It is not recognized in the UN because China is part of the UN Security Council which basically allows for any country in that council to veto any law from being passed.[/quote]

Since when do you define a country?

Since when does "countries" overrule an entire body governed by almost every country in the world?

[quote]Definition of a country:
Defined Land-check
Population-check
Government-check
Soverignty-check(China cannot change how taiwan operates unless you mean by military threats...then US owns the entire world)
[/quote]
[url]http://en.wikipedia.org/wiki/Country[/url]
I think the word you're looking for is
[url]http://en.wikipedia.org/wiki/State[/url]

Taiwan IS recognized as a country, but NOT as a state.
edit: actually it is(by some), but i meant not as a nation
or something like that
i dont really know what im talking about

[quote="noneoftheabove"]Sigh, i would rather not debate about this especially in the f&g forum but here goes...

Definition of a country:
Defined Land-check
Population-check
Government-check
Soverignty-check(China cannot change how taiwan operates unless you mean by military threats...then US owns the entire world)

Yes, there are countries that do recognize Taiwan as a country which is proof.
There are also other countries who would recognize Taiwan as a country but the threat economically from China to that country would be too much of a risk.
It is not recognized in the UN because China is part of the UN Security Council which basically allows for any country in that council to veto any law from being passed.[/quote]

Those 20 or so nations only recognize Taiwan because Taiwan is paying them millions.

[quote="noneoftheabove"]Sigh, i would rather not debate about this especially in the f&g forum but here goes...

Definition of a country:
Defined Land-check
Population-check
Government-check
Soverignty-check[/quote]

So my school counts as a country? (If you want to argue semantics about the word "population", go replace "my school" with "the nearby college".)

I'm Russian-American, and i'm proud of both my Rusian and American heritage. And i think Taiwan is a country.

[quote="Ignite168"]ohhh the flosssssy flosssssssssy [/quote]

Did you really have to do that? I hate that song.  :mad:


inequalities
symmetry
SFFT
number theory
relatively prime
algebra
system of equations

What is the best way to solve following problem?

The product of two positive integers minus their sum is 39.
The integers are relatively prime, and each is less than 20.
What is the sum of the two integers?

[hide="Solution"]Let the two integers be $ x$ and $ y$.

Then, we have $ xy\minus{}x\minus{}y\equal{}39\implies xy\minus{}x\minus{}y\plus{}1\equal{}(x\minus{}1)(y\minus{}1)\equal{}40$

Now all we have to do is easy brute force.

Answer: $ (x,y)\equal{}(5,11), (11,5)$[/hide]

i would use my own way the fourmula or guess and check either way i get the answer
fourmula is easier
list the factors and use guess and check if you are not comfortable with the fourmula

[quote="FantasyLover"][hide="Solution"]Let the two integers be $ x$ and $ y$.

Then, we have $ xy \minus{} x \minus{} y \equal{} 39\implies xy \minus{} x \minus{} y \plus{} 1 \equal{} (x \minus{} 1)(y \minus{} 1) \equal{} 40$

Now all we have to do is easy brute force.

Answer: $ (x,y) \equal{} (5,11), (11,5)$[/hide][/quote]

so the answer is $ 5\plus{}11\equal{}\boxed{16}$ :P

[quote="The Pink Panther"]You didnt have to post that 3 times :wink:[/quote]

Sometimes the server is too slow and so people get impatient, and click the submit button mutiple times.

[quote="AIME15"][quote="FantasyLover"][hide="Solution"]Let the two integers be $ x$ and $ y$.

Then, we have $ xy \minus{} x \minus{} y \equal{} 39\implies xy \minus{} x \minus{} y \plus{} 1 \equal{} (x \minus{} 1)(y \minus{} 1) \equal{} 40$

Now all we have to do is easy brute force.

Answer: $ (x,y) \equal{} (5,11), (11,5)$[/hide][/quote]

so the answer is $ 5 \plus{} 11 \equal{} \boxed{16}$ :P[/quote]
Gah, misread the problem :(

@Math Champion: Yes, I was calling that a brute force  :P

[quote="Math Champion"]You don't have to brute force at the end. You could list the factors of 40 and try them.

Edit: Well, I guess you could call this brute force.  :)[/quote]
You [i]could[/i] list them out, but that would take longer, y'know.

Also, couldn't you solve this by a system of equations?

You would have 1 equation and 1 inequality. How would you do systems of equations with that?

...which still results in case checking, except by then you've already wasted a minute.

Sorry! lol  :)  Guess there is no good way to solve it then.

Casework is a perfectly legitimate and "good" way to solve problems.

[quote="FantasyLover"][hide="Solution"]Let the two integers be $ x$ and $ y$.

Then, we have $ xy \minus{} x \minus{} y \equal{} 39\implies xy \minus{} x \minus{} y \plus{} 1 \equal{} (x \minus{} 1)(y \minus{} 1) \equal{} 40$

Now all we have to do is easy brute force.

Answer: $ (x,y) \equal{} (5,11), (11,5)$[/hide][/quote]

Remark the symmetry of $ x,\ y$, Without Loss of Generality, $ 1\leq x\leq y < 20$, we have $ (x \minus{} 1,\ y \minus{} 1) \equal{} (1,\ 40), (2,\ 20),\ (4,\ 10),\ (5,\ 8)$

$ \Longleftrightarrow (x,\ y) \equal{} (2,\ 41),\ (3,\ 21),\ (5,\ 11),\ (6,\ 9)$. Since $ x,\ y < 20$ are relatively prime, yielding $ (x,\ y) \equal{} (5,\ 11)$.

The answer is $ x \plus{} y \equal{} 16$.

[quote="myyellowducky82"][quote="Math Champion"]Replace a variable in the inequality by finding it in terms of the other in the equation.[/quote]
Wouldn't this make the equations easier, though?  :|[/quote]

What do you mean? That's the point of the replacement.

Just find out what SFFT is and this will be hecka ez.

If you read any of the previous posts, the idea of using SFFT has already been posted :wink:

[url=http://www.artofproblemsolving.com/Wiki/index.php/SFFT]Simon's Favorite Factoring Trick[/url]

So....I guess we sitll haven't come up with a good way to do this problem?

The skill is not so new, because Japanese junior high school students study in private school.

I think the guy who posted this got h\is answer so mods can lock

No...We still haven't found a quick way.

[quote="Math Champion"]Does that help in this problem though?[/quote]
Yep. But it's not very "fast."

[quote="FantasyLover"][hide="Solution"]Let the two integers be $ x$ and $ y$.

Then, we have $ xy \minus{} x \minus{} y \equal{} 39\implies xy \minus{} x \minus{} y \plus{} 1 \equal{} (x \minus{} 1)(y \minus{} 1) \equal{} 40$

Now all we have to do is easy brute force.

Answer: $ (x,y) \equal{} (5,11), (11,5)$[/hide][/quote]

I think that takes less than 5 minutes.

Seriously, if you even check every possible case without even thinking after factoring, you only have 8 cases to check. If you just let $ x>y$ by symmetry, that leaves only 4. I'm pretty sure that does not take much time.

Some problems can only be solved by casework; compared to several other problems, this requires relatively almost no casework at all.

Wow, guys.

Do we need like 40 posts to solve a simple problem?

SFFT is a perfectly fast way to do this problem, and if you are a decent (not even 'good') mathlete, you will be able to do that problem in less than a minute.

Plus, SFFT is better for problems with big numbers.

So i guess mods can lock this thread now.


Determine the number of 100-digit numbers whose all digits are odd, and
in which every two consecutive digits differ by 2.

We have that we can use the digits, $1,3,5,7,9$
Group these into two groups, $1,5,9$, and $3,7$

Clearly, if were building up this number, if we have a number from one set, the next is from the other set. Lets pretend that the first number is from the first set, we list every two digit combination, and what 2 digit combination can go after it.

$13, 53, 57, 97$

$13 \rightarrow 13, 53, 57$
$53 \rightarrow 53, 57, 13$
$57 \rightarrow 53, 57, 97$
$97 \rightarrow 53, 57, 97$

So every two digit number can be followed by 3 other numbers.

total combinations=$4(3^{49})$
But, if a number began in the second set, reading it backwards would give a number in the first set, So the total combinations is:

 $8(3^{49})$


LaTeX
superior algebra
superior algebra unsolved

[color=olive]Let n be a positive integer. Define f : Z ! Zn by f(a) = [a]. Is f a homomorphism? One-to-one? Onto? An isomorphism? Explain.[/color]

Can any one help? Thanks very much.
 :lol:

um....I guess Z denotes the group of integer.

1. You guess¿
2. Please use $\LaTeX$, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 for an introduction.


Daca M e o multime finita, si "$\cdot$" este o lege de compozitie asociativa pe M si daca exista $a\in M$ astfel ca $f(x)=axa$ este o functie bijectiva, atunci $(M,\cdot )$ e monoid.

Nu este necesara finititudinea multimii $M$, si nici injectivitatea functiei $f$.

De fapt, problema initiala zicea ca axa e injectiva si M finita. De aici am dedus ca de fapt e si surjectiva. Nu am postat problema in forma ei initiala. Totusi, poti da te rog si rezolvarea?

O problema foarte tare si intr-un context mult mai general ar fi sa demonstrati ca:

        " Pe orice multime se poate introduce o structura de grup."

 Problema asta mi-a fost comunicata de domnul Marius Vladoiu, Fac. Matematica-Informatica, Univ. Bucuresti. :lol:

Văd că nimeni nu vrea să posteze soluţia...
Problema: pe mulţimea $M$ se defineşte o lege asociativă astfel încât există $a\in M$ cu proprietatea că funcţia $f: M\to M,f(x)=axa$ este surjectivă. Atunci $M$ este monoid.
Soluţie: exista $b\in M$ astfel încât $aba=a$. Fie $e=ab$ şi $e'=ba$. Fie $x\in M$, arbitrar. Atunci $ex=x$ şi $xe'=x$. Intr-adevăr, există $y\in M$ astfel încât $x=aya$. Atunci $ex=ab(aya)=(aba)ya=aya=x.$ Analog, $xe'=(aya)ba=x.$
Luând $x=e,x=e',$ rezultă $ee'=e',ee'=e,$ deci $e=e'$. Aşadar există element neutru.

Am scris greşit o relaţie...scuze! Am corectat!


Hello,
My name is akech, 23 years old, from Sudan. Because of the civil war in Sudan, i did not get an opportunity to go to school until i was fifteen. I immigrated to the US five years ago, but was too old to go to high school so I skipped high school, but went to college and majored in Mathematics and Theology. I will graduate with a double major this coming May.

It is very nice to be here with you all.


geometry
trigonometry
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral, inscribed in a circle $ \omega$. Let $ A'$, $ B'$, $ C'$, $ D'$ be the points where the tangents at $ A$ and $ B$, at $ B$ and $ C$, at $ C$ and $ D$ and at $ D$ and $ A$, respectively, intersect. Prove that the lines $ AC$, $ BD$, $ A'C'$ and $ B'D'$ are concurrent, that is, they intersect at one point.

This is just Brianchon's theorem applied to the degenerate hexagons $ A'AD'C'CB'$ and $ A'BB'C'DD'$.

However, this is called …Newton’s Theorem “In a circumscribed quadrilateral the diagonals and the lines joining the points of contact of opposite sides with the inscribed circle are concurrent”. (A’B’C’D’ is the quadrilateral, A’C’ and B’D’ its diagonals, AC and BD the lines joining the contact points of opposite sides with the inscribed circle) and has the following proof as well:

For faster writing: let M, N, P and Q be the contact points of AB, BC, CD and DA with the incircle and, let {R} = AC∩MP.
Apply sine theorem to DARM and DCRN and, seeing that ÐAMR=180º-ÐCPR (the 2 angles count half of the measure of the arcs MQP and PNM) and ÐARM=ÐPRC, we get AR/CR=AM/CP ( 1 ).
Let {R’} = AC∩NQ. After similar considerations we get  AR'/CR'=AQ/CN ( 2 ), but AQ=AM and CN=CP, and from (1) and (2) we get R≡R’.
Similarly we shall get that MP and NQ intersect BD at the same point, thence AC, BD, MP and NQ are concurrent.

Best regards,
sunken rock

[quote="Cassi"]Let $ ABCD$ be a cyclic quadrilateral, inscribed in a circle $ \omega.$ Let $ A',$ $ B',$ $ C',$ $ D'$ be the points where the tangents at $ A$ and $ B,$ at $ B$ and $ C,$ at $ C$ and $ D$ and at $ D$ and $ A,$ respectively, intersect. Prove that the lines $ AC,$ $ BD,$ $ A'C'$ and $ B'D'$ are concurrent, that is, they intersect at one point.[/quote]
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265111[/url]


function
floor function
algebra proposed
algebra

Find all functions, (or conditions on a function) so that

$f(\lfloor x \rfloor )=\lfloor f(x)\rfloor$

for all x

f(n) is in Z
f(x)=f([x])+g(x)
where g(x)is in [0,1) :)

Excuse me, i ment polynomials. Its not a terribly hard problem

then as f(x) would be continuous f(x+1)-f(x) would be 1,0 or -1
so f(x)=x+m,-x+m or m
where m is in Z


function
Ring Theory
superior algebra
superior algebra unsolved

Let $ A \equal{} C([0,1])$ be the ring of continuous real functions defined on $ [0,1]$. If $ M$ is a maximal ideal of $ A$ show that there exist $ a \in [0,1]$ such that $ M \equal{} \{ f \in A: f(a) \equal{} 0\}$.

Find my topic in this box, please!  :wink:


Show that for all positive $ x$ and $ y$ such that $ x < y$, $ \displaystyle (1 \plus{} \frac{1}{x} \displaystyle )^x < \displaystyle  (1 \plus{} \frac{1}{y} \displaystyle )^y$.

[hide="solution"]
Let $ f(x)\equal{}\left(1\plus{}\frac{1}{x}\right)^x$ I'll show that $ f$ is increasing.

$ f'(x)\equal{}\left(1\plus{}\frac{1}{x}\right)^x\left[\ln\left(1\plus{}\frac{1}{x}\right)\minus{}\frac{1}{x\plus{}1}\right]$

Now I want to show that $ g(x)>0$ where $ g(x)\equal{}\ln\left(1\plus{}\frac{1}{x}\right)\minus{}\frac{1}{x\plus{}1}$
Notice that $ \lim_{x\to 0^\plus{}}g(x)\equal{}\plus{}\infty$, $ \lim_{x\to\infty}g(x)\equal{}0$ and $ g'(x)\equal{}\minus{}\frac{1}{(x\plus{}1)^2x}<0$, hence $ g$ is decreasing, so for all positive value of $ x$ inequality $ g(x)>0$ holds.
Q.E.D.[/hide]

Thanks for the reply. Unfortunately, I was trying to use that fact to prove that the limit $ \lim_{x \to \infty} \left ( 1 \plus{} \frac{1}{x} \right )^x$ exists (once I show that $ f(x) \equal{} \left ( 1 \plus{} \frac{1}{x} \right)^x$ is nondecreasing, I can take any sequence of reals that grows without bound positively; if I can show that it converges, the limit of $ f(x)$ as $ x$ approaches infinity therefore exists). So, no using natural logs for this proof. :P Sorry I didn't specify.

Can you think of any algebraic solution to this?

Thanks.


function
Asymptote
analytic geometry
trigonometry
calculus
calculus computations

The original problem is : Sketch the graph of this function f (x) = (x+2) / (x-2) (just the beginning  :D )

This is my real concern. [b]From the sketch above, draw the sketch of f ^2 (x)[/b].

Sorry, I don't know how to type the normal formula :(
Thx guys.

I am assuming your notation means $ (f(x))^2$.

The first thing to note is that this function lies entirely above the x-axis.
Any vertical asymptotes are the same.

As an (optional) intermediate step, you may wish to consider the graph of $ |f(x)|$. Anything below the x-axis is reflected to appear above the x-axis.

From here, it is a simple matter to square the function. Anything less than 1 becomes smaller (i.e. same x-coordinate, but from a visual perspective, the y-coordinate is 'lower' on the graph, when compared with y = |f(x)|), and anything greater than 1 becomes bigger (i.e. same x-coordinate, but the y-coordinate is 'higher' on the graph).

I hope this makes sense.

I think she means $ f^{2}(x)$ which means its iterated within itself.

Jshen: Yeap, million thanks  ^^
ckck: Sorry, perhaps I did not make myself clear enough. Are you thinking of f(f(x))  :lol:  ? I assume that both f^{2}(x) and (f(x))^2 denote the same function . [size=75]Anw, I'm a girl [/size]:D

In general, $ f^{n}(x)\equal{}\underbrace{f(f(f(\ldots f(x)\ldots )}_{n\ \text{times}}$ and $ f(x)^n\equal{}(f(x))^n$.

(This is different for trig functions though)


Q1.  $x^{4}-y^{4}=5$
       $x+y=1$
Solve for reals
Q2. $c$ is the 1st term of a sequence and is not equal to 1. The 2nd term of the sequence is 3 more than the previous divided by 1 less than the previous term.
a) For what values of c will the sequence recur like c,c,c etc..
b) Can any term in the sequence be equal to 1

Q3. In a triangle the base is divided by an altitude in the ratio 18:7. If another line is drawn parallel to the altitude as shown in the diagram, such that it halves the area of the triangle. Then find the ratio by which this line divides the base.

Q4. In the diagram the circle has a radius of 2. If the grey area is Y and X is the area of black. Find Y-X. The circle touches the square as shown in the diagram.

Q5. There are x stations in a ralway network. Every train ticket made for these stations says the starting station and the destination. So, for 2 stations - there would be 2 combinations- from 1st to 2nd, and 2nd to 1st. If there were 3, then 1st to 2nd, 2nd to 1st, 1st to 3rd, 3rd to 1st, 2nd to 3rd, 3rd to 2nd. 6 tickets made. If y stations are added to x stations then 200 new tickets need to be issued. How many combinations of x,y satisfy this?.. Sorry for the poor wording...

from some uk math competition...

[hide="1"]Let $x-y = a$. Then $x^{2}+y^{2}= \frac{1}{2}\left((x+y)^{2}+(x-y)^{2}\right) = \frac{1}{2}(a^{2}+1)$. Therefore, $10 = 2(x^{4}-y^{4}) = (x+y)(x-y)2(x^{2}+y^{2}) = a(a^{2}+1)$. The RHS is strictly increasing (because for $f: \mathbb{R}\to\mathbb{R}: x\mapsto x^{3}+x$ we get $f'(x) = 3x^{2}+1 > 0$), so the solution is unique. We see that $a=2$ gives a correct equation, so $x-y = 2$ and $x+y = 1$, i.e. $\left(x,y\right) = \left(\frac{3}{2},\frac{-1}{2}\right)$.[/hide]

nice solution. Could someone please solve 5. I just need to verify my solution is correct.

[quote="1=2"][hide="2a"]$c=\frac{4c-3}{c-1}$

$c^{2}-c=4c-3$

$c^{2}-5c+3=0$

$c=\frac{5\pm\sqrt{13}}{2}$[/hide]

[hide="2b"]Sicne the series is infinite, the next term of the sequence is (1/0)+3. So 1 is impossible.[/hide]

[hide="5."]There are a total of x(x-1) tickets. After you add on y more tickets, the total is (x+y)(x+y-1)=x^2+2xy-x+y^2-y

So $2xy+y^{2}-y=(2x+y-1)y=200$.

y must be a factor of 200. Let's list the factors out.

1,2,4,5,8,10,20,25,40,50,100,200

If y=200, x is negative.

If y=100, x is negative again.

All values of y from 20 through 200 will make x negative. If y is 10, x isn't an integer.

If y=8, x=9.

If y=5, x=18.

If y=4, x is not an integer.

If y=2, x isn't an integer.

If y=1, x=100.

(x,y)=(9,8), (18,5), (100,1)[/hide][/quote]
Your 2nd answer is wrong. It's c+3 not 4c-3. Dam, i wrote only 2 solutions for the 5th.

Solving the remaining questions

[hide="3"]The height of the triangle is irrelevant. If not convinced, let $h$ be the height, $x$ be the distance from the altitude to the parallel line dividing the triangle into two equal areas, $t$ is the length of the segment made by the parallel line. If you draw a figure, and assuming WLOG that the parallel line is on the right side of the altitude dividing the base into $7: 18$, we have by similar triangles,

$\frac{18-x}{18}=\frac{t}{h}\Rightarrow t=h\frac{18-x}{18}$

The area of the whole triangle is $\frac{25}{2}h$ and since the parallel line divides it into two equal areas,

$\frac{25}{4}h=\frac12(18-x)t=\frac12(18-x)h\frac{18-x}{18}$

$15^{2}=(18-x)^{2}\Rightarrow x=3,33$

33 is a line outside of the triangle, so we take $x=3$ and we have the ratio $10: 15$ or $2: 3$[/hide]

[hide="4"]Note the 3 points of contact between the circle and the square. This is enough information to determine the length of the side of the square. Draw the radii $r$ from the center of the circle to the 3 points of contact. Then, the length from the center of the circle to the nearest side of the square is $\frac{r}{\sqrt{2}}$. So, the length of the side is $s=r+\frac{r}{\sqrt{2}}$.

Let $W$ be area of the white part in the figure. Then, area of square is $X+W$ and area of circle is $Y+W$. So,

$Y-X=\text{circle}-\text{square}=\pi r^{2}-s^{2}$

$Y-X=r^{2}\left(\pi-\frac32-\sqrt{2}\right)=4\pi-6-4\sqrt{2}$[/hide]

[hide="comment"]You could do number 1 by brute force with rational roots theorem (lower and upper bounds make it significantly quicker, as does the fact that it is obvious there are no negative solutions when you see the equation), but I like Kurt's solution, even with the slight bit of calculus. How did you come up with that?[/hide]


calculus
derivative
limit
calculus computations

Hello.

How many real roots exist to $ x^{100} \minus{} (4^x)x^{98} \minus{} x^2 \plus{} 4^x \equal{} 0$ ?

Thanks!

you  write it [code]-x^{98}(4^{x}-x^{2}) +4^{x}-x^{2}=(4^{x}-x^{2})(1-x^{98})=(2^{x}-x)(2^{x}+x)(1-x^{98})[/code] and then you can study each factor with derivatives or equalities

hello
[hide="solution"]
$ x^{100} \minus{} (4^x)x^{98} \minus{} x^2 \plus{} 4^x \equal{} 0$
$ (x^{98}\minus{}1)(x^2\minus{}4^x)\equal{}0$
$ (x^{98}\minus{}1)\equal{}0$ will give $ x\equal{}^\plus{}_\minus{}1$
for $ f(x)\equal{}(x^2\minus{}4^x)$
look its derivative is always -ve.
and $ \lim_{x\to\minus{}\infty}f(x)\equal{}\plus{}\infty$
 $ \lim_{x\to\infty}f(x)\equal{}\minus{}\infty$
this shows f(x) meets x-axis at a point other than $ ^\plus{}_\minus{}1$
therefore
$ x^{100} \minus{} (4^x)x^{98} \minus{} x^2 \plus{} 4^x \equal{} 0$ has three roots.
[/hide]
thank u


function
number theory unsolved
number theory

Suppose that $n$ has exactly 3 distinct prime factors and $\phi (n)|n-1$. Find all $n$.

[quote="vardanch"]
$x\leq 3$ and $x\geq 2$
[/quote]
why?


In one day, US citizens spend 8 million dollars on taxi fares. The United States has approximately 250 million people. Using this estimate, find the number of dollars spent per person per day on taxis to the nearest tenth of a cent.

$ \frac {8000000}{250000000} \equal{} \frac {8}{250} \equal{} \frac {4}{125} \equal{} \boxed{.032}$ dollars, which is to the nearest tenth of a cent.


Galois Theory
superior algebra
superior algebra theorems

what text book do you use in your university?
what do you recommend for reading?

Look up authors like Artin and Dummit & Foote. The latter are supposed to be better. Also, A Course in universal algebra by Burris and Sankappanavar

It depends for what.
As an introductions you can try:
"Abstract Algebra" by Beachy and Blair.

If you want more advanced material I would go for (that's what I wanna buy):
"Field and Galois Theory (GTM)" by Morandi.

Algebra (GTM) by S. Lang or Jacobson N. Lectures in abstract algebra, vol 1, 2, 3 (GTM , Springer).

I want something with interesting and challenging problems.

I am just a sophomore in the introductory class but we use "Contemporary Abstract Algebra" 5th Edition by Gallian.


geometry
algebra
system of equations
geometry unsolved

Consider the triangle $ ABC$ and a point $ P$, found inside the triangle $ ABC$, that satisfies $ m(\angle PBA)\equal{}m(\angle PCB)\equal{}m(\angle PAC)$. It is known that:
$ PA\equal{}4, PB\equal{}6$ and $ PC\equal{}9$. Find the area of $ \triangle ABC$.

$ PA \equal{} \frac {bc^2}{\sqrt{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}}$
In order to find S(ABC), we solve the system of equations with variables a,b,c.

Could you prove your formula, mr.danh?
Thank you very much.

The following file will show the formula I posted. I'm quite busy, so I'm sorry if my proof has some typo mistakes.

I proved that the value of the angles PBA, etc is tan (<PBA)=4S/(a^2+b^2+c^2), where S is the area of ABC.
Maybe that could be used to find directly the S.

Best regards,
sunken rock


superior algebra
superior algebra unsolved

let L is a finite extension of K. Let $ V$ be a linear rep. of G over L with character $ \chi$; by restricting scalars we can consider V as a linear rep. of G over K. Prove that the character of this rep. is equal to $ Tr_{L/K}(\chi)$?

$ GL_n(L) \equal{} GL_{n[L: K]}(K)$. Then use transitivity of the trace.

[quote="-oo-"]how do you define such a restriction of scalars here? you cannot get a homomorphism $ G \to GL(n,K)$ from a homomorphism $ G \to GL(n,L)$. it's vice versa.[/quote]... by $ Aut_L(V) \rightarrow Aut_K(V)$.


geometry
circumcircle
geometric transformation
reflection
analytic geometry
calculus
vector

[color=darkred]Let $C(I,r)$ be the incircle of $\triangle ABC$ and denote $M\in AB\cap CI$, $N\in AC\cap BI$. Prove that $OI_{a}\perp MN$, where $O$- the circumcenter and $I_{a}$- the $A$- exincenter of $\triangle ABC$.[/color]

[color=darkblue][b]Remark.[/b] Using the property "If $H$- the orthocenter of $\triangle ABC$, then the circumcircle $(O')$ of $\triangle BHC$ is the reflection of the circumcircle $(O)$of $\triangle ABC$ w.r.t. the line $BC$and the Euler's point of $\triangle ABC$ belongs to $AO'$" and appling the proposed problem to the orthic triangle of $\triangle ABC$ we get the problem which was posted by [u][b]N. T. Tuan[/b][/u] at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=139222 [/color]

[quote="N.T.TUAN"]Trivial by coordinate method.[/quote][color=darkblue]Sorry, by "coordinate method" isn't trivially ! Can you post this proof ?
In the my opinion, you"ll make many calculus and write many rows, so that ... "abandon du travail".[/color]

I don't know a lot of on plane geometry, so I allway use cordinate, complex, vector, etc. Codinates of $I_{a}$ and $O$ maybe find from

1) http://en.wikipedia.org/wiki/Circumcircle#Coordinates_of_circumcenter
2) http://mathworld.wolfram.com/Excircles.html

Codinates of $M,N$ is easy find!

[quote="Virgil Nicula"][color=darkred]Let $C(I,r)$ be the incircle of $\triangle ABC$ and denote $M\in AB\cap CI$, $N\in AC\cap BI$. Prove that $OI_{a}\perp MN$, where $O$- the circumcenter and $I_{a}$- the $A$- exincenter of $\triangle ABC$.[/color][/quote]
[color=darkblue]$1.\blacktriangleright$[b][u]A metrical method.[/u][/b]. Denote the power $p(X)$ of the point $X$ w.r.t. $C(O,R)$ of $\triangle ABC$. Thus,

$\{\begin{array}{c}p(M)=-MA\cdot MB=OM^{2}-R^{2}\\\ p(N)=-NA\cdot NC=ON^{2}-R^{2}\end{array}\|$ $\Longrightarrow$ $\boxed{\ OM^{2}-ON^{2}=NA\cdot NC-MA\cdot MB\ }$.

Denote the points $U$, $V$ where the $A$- exincircle touches the lines $AB$, $AC$ respectively.

$\{\begin{array}{ccc}I_{a}U\perp AB\Longrightarrow I_{a}A^{2}-I_{a}M^{2}=UA^{2}-UM^{2}\\\\ I_{a}V\perp AC\Longrightarrow I_{a}A^{2}-I_{a}N^{2}=VA^{2}-VN^{2}\\\\ UA=VA=p\end{array}\|$ $\Longrightarrow$

$I_{a}M^{2}-I_{a}N^{2}=UM^{2}-UN^{2}$ $\Longrightarrow$ $\boxed{\ I_{a}M^{2}-I_{a}N^{2}=(MA-NA)\cdot (MA+NA-2p)\ }$.

Using the relations $\{\begin{array}{c}(a+c)AN=(a+b)AM=bc\\\ (a+b)MB=ac\\\ (a+c)NC=ab\end{array}$ prove easily that

$OM^{2}-ON^{2}=I_{a}M^{2}-I_{a}N^{2}=$ $\frac{abc|b-c|[(a+b+c)^{2}-bc]}{[(a+b)(a+c)]^{2}}$, i.e. $OI_{a}\perp MN$.

[b]Remark.[/b] $OI_{a}\perp MN$ $\Longleftrightarrow$ $OM^{2}-ON^{2}=I_{a}M^{2}-I_{a}N^{2}$ $\Longleftrightarrow$

$NA\cdot NC-MA\cdot MB=(MA-NA)(MA+NA-2p)$ $\Longleftrightarrow$

$\frac{a}{c}\cdot NA^{2}-\frac{a}{b}\cdot MA^{2}=MA^{2}-NA^{2}+2p(NA-MA)$ $\Longleftrightarrow$

$ab\cdot NA^{2}-ac\cdot MA^{2}=bc\cdot (MA^{2}-NA^{2})+2bcp(NA-MA)$ $\Longleftrightarrow$

$c(a+b)\cdot MA^{2}+2bcp\cdot NA=b(a+c)\cdot NA^{2}+2bcp\cdot MA$ $\Longleftrightarrow$

$bc^{2}\cdot MA+2bcp\cdot NA=b^{2}c\cdot NA+2bcp\cdot MA$ $\Longleftrightarrow$

$c\cdot MA+2p\cdot NA=b\cdot NA+2p\cdot MA$ $\Longleftrightarrow$

$(a+c)\cdot NA=(a+b)\cdot MA\ (=bc)$, what is truly.[/color]

This problem was on Tournaments of Towns, and has realy beautiful synthetic solution.

Let $I_{b},I_{c}$ be rest of excicrcles centers. Points $C,I,I_{c}$ and points $B,I,I_{b}$ are collinear. Let $O_{1}$ be circumcenter(circle $k_{1}$) of triangle $II_{a}Ib$. Point $I_{a}$ is orthocenter of $II_{b}I_{c}$ and point $O$ center of Euler's circle ($k_{2}$)of triangle $II_{b}I_{c}$. Because points $O_{1},O,I_{a}$ are collinear it's enough to prove that $MN \perp O_{1}O$. Because $AI_{c}BI$ is cyclic then is $MA\cdot MB=MI\cdot MI_{c}$. Power of the point of $M$ w.r.t. $k_{1}$ and $k_{2}$ are excaltly $MA\cdot MB$ and $MI\cdot MI_{c}$, hence $M$ lies on radical axe of these circle. Similiar $N$ lies on this line, from which we conclude $O_{1}O\perp MN$ and $OI_{a}\perp MN$.


inequalities
trigonometry
geometry
BPSQ
function
area of a triangle
inequalities theorems

I think it is just Schur for $t=2$, that is, $a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-b)\geq 0$

Anyways, shall we try a solution that doesn't involve multiplying out (too much?).  Letting $a=x+y$ which is acceptable by the triangle condition and so on, we easily transform the inequality into

$\sum xy\geq \sqrt {3}\cdot \left(\sum x^2yz\right)^{\frac {1}{2}}$

Squaring, we get:

$\sum x^2y^2+2\sum x^2yz\geq 3\sum x^2yz$ which is obvious by Muirhead or AM-GM

So not too hard, overall.

No, it's not too hard...it's just that the book gave a messy trig solution...so  :)

But you are absolutely right.

our ineq is equivalent to:

$(a-b+c)(a-c+b)+(b-c+a)(b-a+c)+(c-a+b)(c-b+a)\geq4\sqrt{3}s$

let $ x=b+c-a, y=c+a-b, z=a+b-c$

we get:

$xy+yz+zx\geq\sqrt{3(x+y+z)xyz}$

which is true by equivalent translation ;)

is that you want?? :D

Well, the inequality $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S$ does not originate from Klamkin; it's the hyper-wellknown [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14388]Weitzenböck inequality[/url]. Its refinement, $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S+Q$, with $ Q=\left(b-c\right)^{2}+\left(c-a\right)^{2}+\left(a-b\right)^{2}$, is the so-called Hadwiger-Finsler inequality. Both inequalities are proven in Arthur Engel's "Problem-Solving Strategies", New York 1998, chapter 7, example E21. Here is the proof of the Hadwiger-Finsler inequality:

Since the angles A, B, C of triangle ABC lie in the interval ]0°; 180°[, their halves $ \frac{A}{2}$, $ \frac{B}{2}$, $ \frac{C}{2}$ lie in the interval ]0°; 90°[. Since the function $ x\to\tan x$ is convex on this interval, we thus have by Jensen's inequality

$ \tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\geq 3\tan\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3}=3\tan\frac{A+B+C}{6}$
$ =3\tan\frac{180^{\circ}}{6}$     (since A + B + C = 180° by the sum of angles in a triangle)
$ =3\tan 30^{\circ}=3\frac{\sqrt3}{3}=\sqrt3$.

But by the half-angles theorem, $ \tan\frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}$, where $ s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. In other words,

$ \tan\frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}=\sqrt{\frac{\left(s-b\right)^{2}\left(s-c\right)^{2}}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}=\frac{\left(s-b\right)\left(s-c\right)}{\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}$.

Since, by the Heron formula, the area S of triangle ABC equals $ S=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, this becomes

$ \tan\frac{A}{2}=\frac{\left(s-b\right)\left(s-c\right)}{S}$.

Since $ s-a=\frac{a+b+c}{2}-a=\frac{b+c-a}{2}$ and similarly $ s-b=\frac{c+a-b}{2}$ and $ s-c=\frac{a+b-c}{2}$, this rewrites as

$ \tan\frac{A}{2}=\frac{\frac{c+a-b}{2}\cdot\frac{a+b-c}{2}}{S}=\frac{\left(c+a-b\right)\left(a+b-c\right)}{4S}$
$ =\frac{\left(a-\left(b-c\right)\right)\left(a+\left(b-c\right)\right)}{4S}=\frac{a^{2}-\left(b-c\right)^{2}}{4S}$.

Similarly,

$ \tan\frac{B}{2}=\frac{b^{2}-\left(c-a\right)^{2}}{4S}$;
$ \tan\frac{C}{2}=\frac{c^{2}-\left(a-b\right)^{2}}{4S}$.

Hence, the inequality $ \tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\geq \sqrt3$ becomes

$ \frac{a^{2}-\left(b-c\right)^{2}}{4S}+\frac{b^{2}-\left(c-a\right)^{2}}{4S}+\frac{c^{2}-\left(a-b\right)^{2}}{4S}\geq \sqrt3$.

Multiplying this with 4S, we get

$ \left(a^{2}-\left(b-c\right)^{2}\right)+\left(b^{2}-\left(c-a\right)^{2}\right)+\left(c^{2}-\left(a-b\right)^{2}\right)\geq 4\sqrt3\cdot S$.

The left hand side of this rewrites as

$ \left(a^{2}-\left(b-c\right)^{2}\right)+\left(b^{2}-\left(c-a\right)^{2}\right)+\left(c^{2}-\left(a-b\right)^{2}\right)$
$ =\left(a^{2}+b^{2}+c^{2}\right)-\left(\left(b-c\right)^{2}+\left(c-a\right)^{2}+\left(a-b\right)^{2}\right)=\left(a^{2}+b^{2}+c^{2}\right)-Q$;

thus, we have $ \left(a^{2}+b^{2}+c^{2}\right)-Q\geq 4\sqrt3\cdot S$, so that $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S+Q$, and the Hadwiger-Finsler inequality is proven.

  Darij

yeah, i remember seeing something like that in my book... :D

nevertheless, thanks Darij.  :)

A nice and short solution for this inequality, can be found in my book of inequalities, problem GD. 42 ;)  :D

Please a take a look.

[url=http://planetmath.org/encyclopedia/ProofOfHadwigerFinslerInequality.html]planetmath.org/encyclopedia/ProofOfHadwigerFinslerInequality.html[/url]

kunny

I saw, thanks. I know this proof. It appear in very many old and new books.

You are nicely, [i]Kunny[/i] and [i]Mekrazywong[/i] ! Thanks you very much ! I wish to both ... sweet sleep and pleasant dreams ! [hide="Here is the my (?) solution for the Hadwiger-Finsler inequality."]From the well-known identities $r_ar_b+r_br_c+r_cr_a=p^2,\ r_a+r_b+r_c=4R+r$
results: $3(r_ar_b+r_br_c+r_cr_a)\le (r_a+r_b+r_c)^2\Longleftrightarrow\boxed {\ p\sqrt 3\le 4R+r\ }\ .$
From the well-known identities $ab+bc+ca=p^2+r^2+4Rr,\ S=pr$
results: $\boxed {\ \sum a^2\ge 4S\sqrt 3+\sum (b-c)^2\ }\Longleftrightarrow \sum bc\ge pr\sqrt 3 +p^2\Longleftrightarrow$
$p^2+r^2+4Rr\ge pr\sqrt 3+p^2\Longleftrightarrow \boxed {\ p\sqrt 3\le 4R+r\ }\ .$
[/hide]

[color=darkblue][b]A proof of the [u]Hadwiger-Finsler's inequality[/u][/b] : $\boxed{\ \sum a^{2}\ge 4S\sqrt 3+\sum (b-c)^{2}\ }\ .$

$\sum a^{2}\ge 4S\sqrt 3+\sum (b-c)^{2}$ $\Longleftrightarrow$ $\sum \left(b^{2}+c^{2}-2bc\cos A\right)\ge 4S\sqrt 3+\sum \left(b^{2}+c^{2}\right)-2\sum bc$ $\Longleftrightarrow$
$\sum bc(1-\cos A)\ge 2S\sqrt 3$ $\Longleftrightarrow$ $\sum bc\sin^{2}\frac{A}{2}\ge S\sqrt 3$ $\Longleftrightarrow$ $\sum (p-b)(p-c)\ge S\sqrt 3$ $\Longleftrightarrow$
$r(4R+r)\ge pr\sqrt 3$ $\Longleftrightarrow$ $p\sqrt 3\le 4R+r$, what is truly.

[b]Remark.[/b] From the well-known identities $r_{a}+r_{b}+r_{c}=4R+r$, $r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}$
and the inequality $3\cdot (xy+yz+zx)\le (x+y+z)^{2}$, for $x: =r_{a}\ ,\ y: =r_{b}\ ,\ z: =r_{c}$, we obtain
$3\cdot\left(r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}\right)\le (r_{a}+r_{b}+r_{c})^{2}$, i.e. $3p^{2}\le (4R+r)^{2}\Longleftrightarrow$ $\boxed{\ p\sqrt 3\le 4R+r\ }\ .$[/color]

[b]Problem:[/b]Prove that:
$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq 4\sqrt{3}S\plus{}(a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}$
[i]My solution:[/i]
$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq 4\sqrt{3}S\plus{}(a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}$
<=>$ 2(ab\plus{}bc\plus{}ca)\geq 4\sqrt{3}S\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}$
<=>$ 4S(\frac{1}{sinA}\plus{}\frac{1}{sinB}\plus{}\frac{1}{sinC})\geq 4\sqrt{3}S\plus{}4S(cotgA\plus{}cotgB\plus{}cotgC)$
<=>$ (\frac{1}{sinA}\minus{}cotgA)\plus{}(\frac{1}{sinB}\minus{}cotgB)\plus{}(\frac{1}{sinC}\minus{}cotgC)\geq\sqrt{3}$
<=>$ \frac{1\minus{}cosA}{sinA}\plus{}\frac{1\minus{}cosB}{sinB}\plus{}\frac{1\minus{}cosC}{sinC}\geq\sqrt{3}$
<=>$ tg(\frac{A}{2})\plus{}tg(\frac{B}{2})\plus{}tg(\frac{C}{2})\geq\sqrt{3}$
This inequality is true
Done!

[quote="al.M.V."]
The following one is obviously much stronger: $ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4\sqrt3\cdot S \plus{} Q$, where $ Q \equal{} \left(b \minus{} c\right)^{2} \plus{} \left(c \minus{} a\right)^{2} \plus{} \left(a \minus{} b\right)^{2}$.
This is called the [b]Hadwiger-Finsler inequality[/b].[/quote]
The following inequality is more stronger:
$ \sum_{cyc}\frac{a}{h_a\plus{}r_a}\geq\sqrt3.$ :wink:

Please don't diccuss this inequality . It is a problem from M&Y-issue 9

[quote="arqady"]
The following inequality is more stronger:
$ \sum_{cyc}\frac {a}{h_a \plus{} r_a}\geq\sqrt3.$ :wink:[/quote]

Wait for certain months, please!

[quote=sqing]Prove that $ab+bc+ca\geq4S^2\left(\frac{1}{s(s-a)}+\frac{1}{s(s-b)}+\frac{1}{s(s-c)}\right)$.[/quote]

$$\iff  \sum_{cyc} ab \geq 4\sum_{cyc} (s-a)(s-b) \iff \sum_{cyc} ab \geq \sum_{cyc} (c-a+b)(a-b+c)\iff 2\sum_{cyc}(a-b)^2\geq 0$$

I get this once:

\[a^2+b^2+c^2\geq 4\sqrt{3}\Delta+\frac{1}{2}\sum{(\frac{a}{b+c}+\frac{b+c}{a})(b-c^2)}\]

BQ

[quote=YESMAths]We also have the easy one: :wink:
\[ \sum_{cyc}a^2\cos\cfrac{B-C}{2}\ge 4\sqrt{3}\triangle. \][/quote]

\[ \sum_{cyc}a^2\cos^3\cfrac{B-C}{2}\ge 4\sqrt{3}\triangle. \]

This my problem, XinQun lin proved it.

BQ

[quote=al.M.V.]One of Klamkin's Quickies (it wasn't too quick for me at first.... :)) reads:
Prove the inequality $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S$, where $ a$, $ b$, $ c$ are the sidelengths of a triangle $ ABC$, and $ S$ is its area.
This is the so-called [b]Weitzenböck inequality[/b].

The following one is obviously much stronger: $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S+Q$, where $ Q=\left(b-c\right)^{2}+\left(c-a\right)^{2}+\left(a-b\right)^{2}$.
This is called the [b]Hadwiger-Finsler inequality[/b].[/quote]

I get this once:

\[ a^{2}+b^{2}+c^{2}\leq 4\sqrt3\cdot S+\sum{\frac{b+c}{a}(b-c)^2}\leq 4\sqrt3\cdot S+3\sum{(b-c)^2}\]

BQ

[quote=xzlbq]

\[ \sum_{cyc}a^2\cos^3\cfrac{B-C}{2}\ge 4\sqrt{3}\triangle. \]

[/quote]
After using Ravi's substitution we need to prove that
$\sum_{cyc}(x+y+2z)^3\sqrt{x^3y^3(x+y)}\geq4\sqrt{3xyz(x+y+z)\prod\limits_{cyc}(x+y)^3}$.
By Holder $\left(\sum_{cyc}(x+y+2z)^3\sqrt{x^3y^3(x+y)}\right)^2\sum_{cyc}(x+y)^2\geq\left(\sum_{cyc}(x+y)(x+y+2z)^2xy\right)^3$.
Hence, it remains to prove that
$\left(\sum_{cyc}(x+y)(x+y+2z)^2xy\right)^3\geq96xyz(x+y+z)\prod_{cyc}(x+y)^3\sum_{cyc}(x^2+xy)$, which is true but ugly.

[url=https://math.stackexchange.com/questions/2143589/imo-problem-1961-different-proofs-of-the-weitzenbock-inequality]IMO problem 1961, different proofs of the Weitzenbock inequality[/url]

[quote=al.M.V.]
Prove the inequality $ a^{2}+b^{2}+c^{2}\geq 4\sqrt3\cdot S$, where $ a$, $ b$, $ c$ are the sidelengths of a triangle $ ABC$, and $ S$ is its area.
This is the so-called [b]Weitzenböck inequality[/b].[/quote]
$$ a^{2}+b^{2}+c^{2}\geq 4\sqrt3 \left(\frac{a^2+b^2+c^2}{ab+bc+ca} \right)^{\frac{2}{3}} S.$$
[size=50]（ Adil Abdullayev）[/size]

[quote=sqing]Prove that $ab+bc+ca\geq4S^2\left(\frac{1}{s(s-a)}+\frac{1}{s(s-b)}+\frac{1}{s(s-c)}\right)$.[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2006025p14040671]Macedonian Mathematical Olympiad 1998:[/url]
$$\frac{ab+bc+ca}{4S} \ge  \sqrt3$$
In  triangle $ABC$. Prove that
$$a(b+c-a)cos\frac{B-C}{4}+b(c+a-b)cos\frac{C-A}{4}+c(a+b-c)cos\frac{A-B}{4}\geq4\sqrt 3 S $$
$$a^2cos^2\frac{B-C}{2}+b^2cos^2\frac{C-A}{2}+c^2cos^2\frac{A-B}{2}\geq4\sqrt 3 S $$


If $2x^{2}-3x-5=a(x-3)^{2}+b(x-3)+c$ is true for all values of x, compute a+b+c.


[quote="alex2008"]Let be $ x\ge y\ge z > 0$ such that $ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y} \equal{} 1$ . Prove that :

$ x^3y \plus{} y^3z \plus{} z^3x\le \frac {1}{3}$[/quote]
Overkill maybe :P

$ 1 \equal{} \sum_{cyc} \frac {x^2y}{z} \ge \sum_{cyc} x^2$ can be proved using $ x \ge y \ge z > 0$ :)
[hide="Proof"]
Equivalent to $ (y\minus{}z)(x^3y\minus{}y^3z) \plus{} (x\minus{}y)(z^3x\minus{}y^3z) \ge 0$ (after multiplying with $ xyz$)
But $ (y\minus{}z)(x^3y\minus{}y^3z) \ge yz(y\minus{}z)(x\minus{}y)(x\plus{}y)$ by the condition and
$ (x\minus{}y)(z^3x\minus{}y^3z) \ge \minus{}yz(x\minus{}y)(y\minus{}z)(z\plus{}y)$.
So $ (y\minus{}z)(x^3y\minus{}y^3z) \plus{} (x\minus{}y)(z^3x\minus{}y^3z) \ge yz(y\minus{}z)(x\minus{}y)(x\minus{}z) \ge 0$ :)
[/hide]

And by Vasc's ineq $ x^3y \plus{} y^3z \plus{} z^3x \le \frac{(x^2\plus{}y^2\plus{}z^2)^2}{3} \equal{} \frac{1}{3}$. Done :)


induction
number theory unsolved
number theory

Prove that  A=(2n)!:(n!xn!) is a common mutiple of B=n+1

I think the question is:

Show that $ (n\plus{}1)| \frac{2n!}{n!n!}$

But $ \frac{1}{n\plus{}1} \binom{2n}{n}$ are always integers (the Catalan sequence)

In fact, a beautiful proof is to consider the following sequence, $ a_{0} \equal{} 1$, and $ a_{n} \equal{} \sum_{i\equal{}0}^{n\minus{}1}a_{i}a_{n\minus{}1\minus{}i}$. You can easily check by induction that $ a_{n} \equal{} \dfrac{\binom{2n}{n}}{n\plus{}1}$ which is an integer because of being a linear combination of integers.


For n real positive numbers and n>2 $ a1,a2,...,an$ prove that

$ \frac {a1}{a1 \plus{} a2} \plus{} \frac {a2}{a2 \plus{} a3} \plus{} ... \plus{} \frac {an}{an \plus{} a1} > 1$

Also, you probably meant real numbers  $ > 0$.

probably, malino means $ a_1, a_2,.., a_n >0$ with $ n\ge 3$

Edit: OK scratch that. It (my solution)'s false.

[hide="solution"]First, if there are 2 or more ascending pairs among $ (a_1, a_2), (a_2, a_3), ... (a_n, a_1)$, say at indices $ 1, k$, then $ \frac {a_1}{a_1 \plus{} a_2} \plus{} \frac {a_k}{a_k \plus{} a_{k \plus{} 1}} > \frac {a_1}{2a_1} \plus{} \frac {a_k}{2a_k} \equal{} 1$.  Thus if we assume $ a_1$ is the greatest, then we must only prove it for the ordering $ a_1 > a_n > a_{n \minus{} 1}\ldots a_2$.

We show that $ \frac {a_1}{a_1 \plus{} a_2} \plus{} \frac {a_n}{a_n \plus{} a_1} > 1.$  This will obviously prove the question, because the rest of the terms are all positive.  We have $ \frac {a_1}{a_1 \plus{} a_2} \plus{} \frac {a_n}{a_n \plus{} a_1} \equal{} 1 \plus{} ( \minus{} \frac {a_2}{a_1 \plus{} a_2} \plus{} \frac {a_n}{a_n \plus{} a_1})$.  Now consider the function $ f(x) \equal{} \frac {x}{x \plus{} a_1}$.  We have that if $ x > y$, then:

$ x \plus{} a_1 > y \plus{} a_1$

$ \frac {a_1}{x \plus{} a_1} < \frac {a_1}{y \plus{} a_1}$

$ 1 \minus{} \frac {a_1}{x \plus{} a_1} > 1 \minus{} \frac {a_1}{y \plus{} a_1}$

$ \frac {x}{x \plus{} a_1} > \frac {y}{y \plus{} a_1}$

$ f(x) > f(y)$.

So inside the parenthesis above is a positive number, because $ a_n > a_2$.  Thus the whole expression is greater than 1 and thus it is proven.

This fails for $ n \equal{} 2$ because $ a_n \equal{} a_2$ so the term in the parenthesis is equal to 0.[/hide]

Edit: So close! but ninjad.

Edit 2: Yongyi, doesn't Chebychev's inequality require an ordering?  The problem doesn't say anything about ordering the variables.

Additional restriction:

$ a_i \geq 0$ for all $ i$. (If negatives are allowed consider $ a_1 \equal{} \minus{} 1,a_2 \equal{} 2,a_3 \equal{} \minus{} 3,\dots$)

I think I have a correct (and elegant) solution now.

[hide]
By Cauchy-Schwartz we have
\[ \left(\sum_{\text{cyc}}\frac {a_i}{a_i \plus{} a_{i \plus{} 1}}\right)\left(\sum_{\text{cyc}}a_i(a_i \plus{} a_{i \plus{} 1})\right)\geq (a_1 \plus{} a_2 \plus{} \dots \plus{} a_n)^2,\]
or
\[ \sum_{\text{cyc}}\frac {a_i}{a_i \plus{} a_{i \plus{} 1}}\geq\frac {(a_1 \plus{} a_2 \plus{} \dots \plus{} a_n)^2}{\sum_{\text{cyc}}a_i(a_i \plus{} a_{i \plus{} 1})}.\]
However, it is clear that
\[ (a_1 \plus{} a_2 \plus{} \dots \plus{} a_n)^2 \equal{} \sum a_i^2 \plus{} \sum_{i > j}a_ia_j > \sum a_i^2 \plus{} \sum_{\text{cyc}} a_ia_{i \plus{} 1} \equal{} \sum_{\text{cyc}}a_i(a_i \plus{} a_{i \plus{} 1})\]
for $ n\geq 3$, hence $ \sum_{\text{cyc}}\frac {a_i}{a_i \plus{} a_{i \plus{} 1}} > 1$ and we are done.[/hide]

If we notice from the very first time that manlio is making a hidden message in this problem

The following inequality is valid for all $ a_1, a_2,..., a_n > 0$

$ \frac {a_1}{a_1 \plus{} a_2} > \frac {a_1}{a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n}$

similarly, we got the idea for other terms, summing them up and the result appears unarguably.

Sorry, I edited my problem. :blush:

Great math's solution is simpler. :)


What has happened to some of the specific forums ...it is coming zero posts ...zero topics !!!!

same problem here i see 0  topics in inequalities proposed and own problems :o
is the site getting a new look or is some forum being added? or is my PC wrong?...

This should be fixed now.  It was a caching error on our end, not a hack.

I don't know if this is right place to post this, but, I used Nmap to port scan aops, it seems that you have a lot of ports aside from 80 or 8080 open. Just saying.


symmetry
trigonometry
geometry
circumcircle
geometry proposed

Let AM,AH,AD are median, height and bisector of ABC triangle. Consider AK is symmetry of AM toward AD. Prove that:
\[ \frac{MH}{AM} . \frac{\sin BAK}{\sin KAM}\equal{}\frac{AB}{BC}\]

Let the tangents of the circumcircle $\odot(ABC)$ at $B,C$ meet at $P.$ $AP \equiv AK$ is the A-symmedian of $\triangle ABC.$

Substituting $\frac{BK}{KM}=\frac{ \sin \widehat{BAK}}{\sin \widehat{KAM}} \cdot \frac{AB}{AM}$ and $\frac{MH}{KM}=\frac{PA}{PK}$ $\Longrightarrow$ $\frac{PA}{PK}=\frac{AB^2}{BC \cdot BK} \ (\star).$

On the other hand, we get $\frac{PA}{PK}=\frac{AC}{CK} \cdot \frac{ \sin \widehat{PCA}}{\sin \widehat{PCK}}=\frac{AC}{CK} \cdot \frac{\sin \widehat{B}}{\sin \widehat{A}}=\frac{AC^2}{CK \cdot  BC}.$

Hence, combining with the expression $(\star)$ gives $\frac{CK}{BK}=\frac{AC^2}{AB^2},$ which is well-known identity of the A-symmedian.


AMC
AIME
algebra
polynomial
modular arithmetic
AMC 10

I would like to start a marathon containing [b]original[/b] problems.  If you have a good problem that you made up, this is a good place to post it.  The difficulty of the problems can be anywhere from an AMC 10 level problem to a 13-15 AIME problem, but the problems should be original.

I'll start with an interesting problem:

Let $ S_1 \equal{} \{1,2\}$.  For all $ n > 1$, let $ S_n$ equal the set of sums of one or more distinct elements of $ S_{n \minus{} 1}$.  For example, $ S_2 \equal{} \{1,2,1 \plus{} 2\} \equal{} \{1,2,3\}$.  How many distinct elements does $ S_6$ have?

Does the question mean to ask how many [b]possible[/b] distinct elements does $ S_6$ have?

Also does the problem count as original if it's made up but typical?  Which I guess is kind of contradicting.  :roll:  :P

Sorry, that was a typo.  I meant to say that $ S_n$ is the [i]set[/i] of all sums of one or more distinct elements of $ S_{n\minus{}1}$. (I'll fix the problem)
As long as a problem is not a complete copy of another problem it's fine.  Basically, if you don't just switch numbers around to "change" it, I consider it original.

If $ S_n \equal{} \{1,2,\dots,k\}$, then $ S_{n \plus{} 1}$ consists of the $ \frac {k(k\plus{}1)}2$ distinct elements:
\[ 1,2,\dots,k,k \plus{} 1,k \plus{} 2,\dots,k \plus{} (k \minus{} 1), k \plus{} (k \minus{} 1) \plus{} 1, k \plus{} (k \minus{} 1) \plus{} 2, \dots, k \plus{} (k \minus{} 1) \plus{} (k \minus{} 2) \plus{} \dots \plus{} 1,\]
so $ S_{n \plus{} 1} \equal{} \left\{1,2,\dots,\frac {k(k \plus{} 1)}2\right\}$.

So in general, $ |S_{n \minus{} 1}| \equal{} \frac {|S_n|(|S_n| \plus{} 1)}2$. By computation, $ |S_2| \equal{} 3,|S_3| \equal{} 6,|S_4| \equal{} 21,|S_5| \equal{} 231$, and $ |S_6| \equal{} \boxed{26796}$.

[b]New problem[/b]
There exist constants $ c_1,c_2,\dots,c_{2010}$ such that:
\[ c_1x^{2009} \plus{} c_2(x \plus{} 1)^{2009} \plus{} \dots \plus{} c_{2010}(x \plus{} 2009)^{2009} \equal{} (x \plus{} 2010)^{2009}\]
for all $ x$. Find the remainder when $ |c_1| \plus{} |c_2| \plus{} \dots \plus{} |c_n| \plus{} |c_{n \plus{} 1}|$ is divided by $ 1000$.

[hide]I think you meant until $ c_{2010}$ instead of $ c_{n\plus{}1}$, right? If so, we can use the identity found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]here[/url]
\[ \sum_{i \equal{} 0}^{n}(\minus{}1)^{i}{n\choose i}P(x\plus{}n\minus{}i) \equal{} 0\]
for $ \deg P<n$. For this problem we have $ n\equal{}2010$ and the polynomial $ x^{2009}$ has degree $ 2009$.
So the sum is just

\[ \sum_{k\equal{}0}^{2010}\binom{2010}{k}\minus{}1\equal{}2^{2010}\minus{}1\equiv (2^{10})^{201}\equiv 24^{201}\minus{}1\equiv (\minus{}176)^{67}\minus{}1\equiv(\minus{}24)^{33}\cdot(\minus{}176)\minus{}1\equiv(\minus{}176)^{11}\cdot176\minus{}1\equiv\minus{}176^{12}\minus{}1\equiv\minus{}(24)^6\minus{}1\equiv\minus{}(\minus{}176)^2\minus{}1\equiv24\minus{}1\equiv23\pmod{1000}\][/hide]
If this is correct here's the next problem (though not really original):
Let $ a,b \in \mathbb{N}_0$ and $ p \ge 5$ a prime. Show that $ \binom{pa}{pb} \equiv \binom{a}{b} \pmod{p^3}$


function
factorial
algebra
binomial theorem
calculus
calculus computations

I encountered the following series when trying to compute something about a random walk using a generating function with $ p \plus{} q \equal{} 1$ and $ s \in \mathbb{R}$ chosen such the series converges. Let
\[ z(s) \equal{} \sum^{\infty}_{n \equal{} 0} \binom{2n}{n} (p \cdot q)^{n} \cdot s^{2n}
\]
The result is supposed to be something along the lines
\[ z(s) \equal{} (1 \minus{} 4pqs^2)^{ \minus{} 0.5}.
\]

The function you want is $ \frac {1}{\sqrt {1 \minus{} 4pqs^2} }$.  (This relation implies the generating function for the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].)

[quote="obor"]The result is supposed to be something along the lines
\[ z(s) \equal{} (1 \minus{} pqs^2)^{ \minus{} 0.5}.
\]
[/quote]
Suggestion: start there - or better, put in the $ 4$ from t0rajir0u's post - and expand that using the binomial series. Where do you wind up? (You would probably have discovered you needed the $ 4$ anyway.)

By expanding both side directly, we find that $ \binom{2n}{n} \equal{} 4^n \cdot \frac{(2n\minus{}1)(2n\minus{}3)\cdots(3)(1)}{(2n)(2n\minus{}2)\cdots(4)(2)} \equal{} (\minus{}4)^n \binom{\minus{}1/2}{n}$. So

$ z(s) \equal{} \sum_{n\equal{}0}^{\infty} \binom{\minus{}1/2}{n} (\minus{}4pqs^2)^n \equal{} (1 \minus{} 4pqs^2)^{\minus{}1/2}$

[quote="sos440"]So $ z(s) \equal{} \sum_{n \equal{} 0}^{\infty} \binom{ \minus{} 1/2}{n} ( \minus{} 4pqs^2)^n \equal{} (1 \minus{} 4pqs^2)^{ \minus{} 1/2}$[/quote]

Thanks, I see I should have been working backwards and setting $ x \equal{} 4pqs^2$ and $ \alpha \equal{} \minus{} \frac{1}{2}$ in the binomial series $ (1\plus{}x)^{\alpha} \equal{} \sum^{\infty}_{k \equal{} 0} \binom{\alpha}{k} \cdot x^k.$ But assumed I would not have known the final closed form in advance how will you proceed to derive a solution generally?

[quote="sos440"]By expanding both side directly, we find that $ \binom{2n}{n} \equal{} 4^n \cdot \frac {(2n \minus{} 1)(2n \minus{} 3)\cdots(3)(1)}{(2n)(2n \minus{} 2)\cdots(4)(2)} \equal{} ( \minus{} 4)^n \binom{ \minus{} 1/2}{n}$. [/quote]

I see that this has to be the same. But what again is the definition of a non-integer factorial as used in the binomial coefficient?

[quote="Kent Merryfield"]Define $ \binom{\alpha}{k}$ for $ k$ a nonnegative integer and arbitrary $ \alpha$ by

$ \binom{\alpha}{k} \equal{} \frac {\alpha(\alpha \minus{} 1)\cdots(\alpha \minus{} k \plus{} 1)}{k!}.$

This is really a definition of convenience inspired by (Newton's) binomial theorem itself.[/quote]

Thanks Prof. Merryfield for this definition. But I still would like to have a reply to my other question, i.e. how to build intuition for solving series problems. I mean if you don't know the final answer how to approach those problems? Maybe you also have some references for series calculations. Thanks.


trigonometry
integration
calculus
calculus computations

Help me to check the solution:

$ \cos x\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k}}}{\left( 2k \right)!}}{{x}^{2k}}\Rightarrow \cos \left( {{e}^{\minus{}x}} \right)\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k}}}{\left( 2k \right)!}}{{\left( {{e}^{\minus{}x}} \right)}^{2k}}$

$ \text{*  }\int\limits_{0}^{\plus{}\infty }{\cos \left( {{e}^{\minus{}x}} \right).dx}\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k}}}{\left( 2k \right)!}}\int\limits_{0}^{\plus{}\infty }{{{\left( {{e}^{\minus{}x}} \right)}^{2k}}.\left( \minus{}1 \right).d\left( \minus{}x \right)}\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k\plus{}1}}}{\left( 2k\plus{}1 \right)!}}\left[ {{\left( {{e}^{\minus{}x}} \right)}^{2k\plus{}1}} \right]_{0}^{\plus{}\infty }$

$ \equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k\plus{}2}}}{\left( 2k\plus{}1 \right)!}}\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k}}}{\left( 2k\plus{}1 \right)!}}\equal{}\sin 1$

$ \Rightarrow \int\limits_{0}^{x}{\cos \left( {{e}^{\minus{}x}} \right).dx}\equal{}\sum\limits_{k\equal{}0}^{\plus{}\infty }{\frac{{{\left( \minus{}1 \right)}^{k\plus{}1}}}{\left( 2k\plus{}1 \right)!}}\left[ {{\left( {{e}^{\minus{}x}} \right)}^{2k\plus{}1}} \right]_{0}^{x}\equal{}\minus{}\sin \left( {{e}^{\minus{}x}} \right)\plus{}\sin 1$

$ \text{but }\minus{}{{\left[ \sin \left( {{e}^{\minus{}x}} \right) \right]}^{'}}\equal{}\minus{}\cos \left( {{e}^{\minus{}x}} \right).{{\left( {{e}^{\minus{}x}} \right)}^{'}}\equal{}\cos \left( {{e}^{\minus{}x}} \right).{{e}^{\minus{}x}}$

There are problems with swapping the sum and the integral. Why can you do perform that?

Yes I can repeat it once more: why can you swap the sum and the integral?

E.g., for $ k\equal{}0$ you have $ \int_0^\infty 1 dx \equal{} \infty$.

Also, you integrate $ e^{\minus{}2kx}$ incorrectly.

In fact, the integral itself is divergent, you can see it by making a simple change of variable.


function
limit
real analysis
real analysis unsolved

Considering the continuous function $f: (0,\infty)\to\mathbb{R}$ with $f(x)=f\left(\frac{2x}{1+x^{2}}\right) \ \forall x>0$. Compute $f'(2006)$.

Since constant functions are obviously solutions, the problem is to show that $f$ must be constant.

[hide="Hint"]$\frac{2x}{1+x^{2}}=\frac2{\frac1x+x}$[/hide]

Well.. $\frac{2x}{1+x^{2}}=\frac{2}{\frac{1}{x}+x}\leq 1$ and $>0$ ... Can you please give a further hint?

$f(x)=f\left(\frac2{\frac1x+x}\right)=f\left(\frac1x\right)$, so $f(x)=f\left(\frac{x+\frac1x}2\right)$. Iterate.

By request, here's the full version:

We have $f(x)=f\left(\frac{x+\frac1x}2\right)$ by the preceding posts. Let $x_{0}=x,x_{n+1}=\frac{x_{n}+\frac1{x_{n}}}2$. As a special case of the Babylonian square root iteration, the $x_{n}$ converge to 1 for all $x>0$. (It's easy to show that $1<x_{n+1}<x_{n}$ for $n>1$, and 1 is the only fixed point)

Since $f(x)=f(x_{n})$ and $f$ is continuous, $f(x)=f(\lim x_{n})=f(1)$ and $f$ is constant.


In triangle $ ABC$, sides $ a,b$ and $ c$ are opposite angles $ A,B$ and $ C$ respectively. $ AD$ bisects angle $ A$ and meets $ BC$ at $ D$. Then if $ x \equal{} \overline{CD}$ and $ y \equal{} \overline{BD}$ the correct proportion is:
$ \textbf{(A)}\ \frac {x}{a} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(B)}\ \frac {x}{b} \equal{} \frac {a}{a \plus{} c} \qquad\textbf{(C)}\ \frac {y}{c} \equal{} \frac {c}{b \plus{} c} \\
\textbf{(D)}\ \frac {y}{c} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(E)}\ \frac {x}{y} \equal{} \frac {c}{b}$

[hide]
The angle bisector theorem says $ x: b\equal{}y: c$, so the correct proportion is E [/hide]

[quote="sunehra"]That's actually wrong. $ \frac {x}{b} \equal{} \frac {y}{c}$ does not imply $ \frac {x}{y} \equal{} \frac {c}{b}.$

[hide="My Solution"]
By the Angle Bisector Theorem and ratio relations, we have $ \frac {x}{b} \equal{} \frac {y}{c} \equal{} \frac {x \plus{} y}{b \plus{} c} \equal{} \frac {a}{b \plus{} c} \boxed{\text{(D)}}$.
[/hide][/quote]

Yes, yes it implies $ \frac{x}{y}\equal{}\frac{b}{c}$!!!

No complete Solution ever posted. Can someone please solve this?

Thanks for the revive :) I like new problems.
[hide = Hint]
Angle Bisector Theorem (its very useful). If you don't know this I can explain.
[/hide]
[hide = Solution]
Using the Angle Bisector Theorem, we get the ratio between $\overline{CD}$ and $\overline{BD}$ as $\frac{\overline{CD}}{\overline{BD}} = \frac{x}{y} = \frac{b}{c}$. Let's add 1 to both sides to get $\frac{x + y}{y} = \frac{b + c}{c}$. Recall the $x + y = a$ because $x + y = \overline{CD} + \overline{BD}  = \overline{BC} = a$. Then, we have $\frac{a}{y} = \frac{b + c}{c}$ which we can rearrange to get our answer: $\boxed{\textbf{(D) } \frac{y}{c} = \frac{a}{b + c}}$


function
real analysis
real analysis unsolved

Could you give an example of a continious function from $L^{\infty}(\mathbb{R})$ which is not bounded everywhere.

[quote="alekk"]what does "not bounded everywhere" mean here ?[/quote]

I meant, not bounded in the whole $\mathbb{R}$.

What do you mean? :? If it is in $L^\infty$ and continuous, it is bounded by some constant outside a set of measure $0$ and, therefore, by continuity, on that set too...


I need to make a small program that does integer partions by solving

                                                   P(n,k)= P(n,n-k)  + P(n,k-1)

Where n is an integer, and k is the higest integer used to sum up N.

Can I make something like this in excel.  If not, what about maple?  I've never done coding before so if you have any good sites or advice, that would be great.

I think you want $ P(n, k) \equal{} P(n \minus{} k, k) \plus{} P(n, k \minus{} 1)$ (with $ P(0, k) \equal{} P(n, 1) \equal{} 1$), right?  I don't know much about either Excel or Maple, but Maple should certainly have the capacity to do simple recursions like this, and searching for "recursion in maple" on your favorite search engine will bring up some useful-looking links.  ([url=http://www.math.sunysb.edu/~scott/Book331/Recursion_making_Koch.html]This[/url] is a good start, for example, but you'll need to figure out how to modify it for a two-variable recursion.)

Thanks for the help, anyone know anything else

what about excel?


Draw a pentagon with each of its five diagonals. How many triangles are contained in the figure?

Is there a formula for this?

[hide]I got 30.[/hide]

[asy]size(150); pen p=fontsize(8); pair O=(0,0), A=O+dir(90), B=O+dir(162), C=O+dir(234), D=O+dir(306), E=O+dir(18); draw(A--B--C--D--E--cycle^^A--C--E--B--D--cycle); dot(A^^B^^C^^D^^E); label("A",A,N,p); label("B",B,WNW,p); label("C",C,WSW,p); label("D",D,ESE,p); label("E",E,ENE,p);[/asy]
[hide]I can't think of a combinatorical way, but I get $ 35$ just counting.[/hide]

Luckily, this is only a pentagon...
[hide]
[asy]size(150); pen+p=fontsize(8); pair+O=(0,0),+A=O+dir(90),+B=O+dir(162),+C=O+dir(234),+D=O+dir(306),+E=O+dir(18); draw(A--B--C--D--E--cycle^^A--C--E--B--D--cycle); dot(A^^B^^C^^D^^E); label("A",A,N,p); label("B",B,WNW,p); label("C",C,WSW,p); label("D",D,ESE,p); label("E",E,ENE,p);[/asy]
we just count them and get 35, and I believe this was an old mathcounts problem...
[/hide]
@DonkeyKong-I hope there is, but probably not.  Especially cause the problem never says it was a regular pentagon...

[quote="alanchou"]Especially cause the problem never says it was a regular pentagon...[/quote]
The fact that it has 5 diagonals is more important than it being regular in this problem.

Nice try at copying my code... :P

i get 35 also. just counts by 5s since its a pentagon

I counted 6 unique triangles and using symmetry i multiplied it by 5 and got 30.
I cant seem to get 35.

@ I_like_pie: How do you make that picture?

[quote="Fraenkel"]
@ I_like_pie: How do you make that picture?[/quote]

It's asy.
It's a graphing system.
you start it off by [code][asy]and end with[/asy][/code]
[quote="alanchou"]@DonkeyKong-I hope there is, but probably not. Especially cause the problem never says it was a regular pentagon...[/quote]

So there might be a formula if it's regular pentagon, right?

uh, regular really does little to the problem...it just makes many of the triangles congruent. its simply that the pentagon is convex that is important.

[quote="Fraenkel"]@ I_like_pie: How do you make that picture?[/quote]
[url=http://asymptote.sourceforge.net/]Asymptote[/url]. There are several informative topics about it in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]this[/url] forum, especially [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149650]this[/url] one.

Another method:

Since it is a regular pentagon that has five lines of symmetry....find out how many different types triangles there are and multiply by 5.  There 5 of each type. There are 7 different types times 5 is 35.

Question:  will this work on other types of regular polygons?


Prove that the sequence defined by $ y_0\equal{}1$ and $ y_{n\plus{}1}\equal{}\frac{1}{2}(3y_{n}\plus{}\sqrt{5{y_{n}}^2\minus{}4})$ for $ n\geq0$ consists only of integers>

Try induction:
$ y_{n\plus{}2}\equal{}3y_{n\plus{}1}\minus{}y_{n}$ :wink:


You know what I support? I support people doing their OWN homework! Do your own homework! You're not learning to  be a creative thinker at all like this.

[quote="brady00n"]Hey, I'm doing a project on school vouchers and current educational policy.  What would be some good ideas to start out the speech?  I support educational reforml

Thanks![/quote]

The 1º step for a winner is to loose, so do sth by yourself, maybe doing wrong things sometimes helps you to improve and grow yourself

You might want to start thinking about how the market system's normal corrections and refinements don't work because your school and teachers have a monolopy enforced by the man with a gun (try not attending in most states).

The voucher program allows competitive forces to work, not only on the schools that get the vouchers, but the existing schools that for the first time might loose money if they don't keep their customers (parents) happy.

I think the other two responders are a bit harsh.  You didn't ask us to do the work, just share our initial thoughts.  I don't know anyone who hasn't learned or "borrowed" ideas from others.


Given two matricies, $A,B$, show that if:

$\text{trace}A^{n}= \text{trace}B^{n}$ for all $n \in \mathbb{N}$ then $A$ and $B$ are similar.
By similar i mean $A =P^{-}1BP$  for some $P$

You need {} around any exponent with more than one character, such as $-1$.

The problem is false; consider $A=0$ and $B$ nilpotent but nonzero.


function
AMC
USA(J)MO
USAMO
algebra
polynomial
analytic geometry

Let R denote the set of all real numbers. Find all functions f from R to R satisfying:
(i) there are only finitely many s in R such that f(s) = 0, and
(ii) f(x^4 + y) = x :^3: f(x) + f(f(y)) for all x, y in R.

Where is this problem from?

A pretty good beginning:
Setting x = 1, y = 0, so f(1) = f(1) + f(f(0)), so f(f(0)) = 0.
Setting x = 0, y = 0 gives us f(0) = f(f(0)) = 0, so f(0) = 0.
Setting y = 0 gives us f(x4) = x3f(x) for all x.
Setting x = 0 gives us f(y) = f(f(y)), so we can reduce our given statement to the statement that f(x4 + y) = x3f(x) + f(y) for all x, y.

Thus, if s is a zero, so is s4, s16, and so on, giving us infinitely many zeroes if f has any zeroes other than 1, -1 and 0.

We see that for x /= 0, x3f(x) = f(x4) = (-x)3f(-x), from which we get that f(x) = -f(-x) for all x (since it's also true of 0).

Well, let's see if I can't keep going.

We've reduced the given equation to f(x^4 + y) = x^3*f(x) + f(y), and we also know f(x^4) = x^3*f(x), so we have
f(x^4 + y) = f(x^4) + f(y).  We also know f(x) = -f(-x), so f(-x^4 + y) = f(-x^4) + f(y), so f has additivity.  This means that f is sort-of linear -- on any arithmetic sequence, it will act linearly.  So, in other words, on the set {... -3d, -2d, -d, 0, d, 2d, 3d, ...} we know that f acts as multiplication by a constant; however, this constant may differ for different such sets.  I know I'm very close to proving it's the identity function, but I haven't quite figured out how to tie it all together -- I know it's the identity on its range, I know that it acts the same on x and x^4, but I'm not sure how to stick all of that together to get continuity.

Whenever x=0, we have f(y) = f(f(y))

Doesn't this imply that if f has an inverse, then:

y = f(y)?

The possibility that f(y)=0 is excluded by the statement of the problem.

A little additional work:

Suppose f(x) = 0. By Joel's work,

x = 1, x = 0 or x = -1.

We show that

f(1) = 0, or f(-1) = 0,
leads to contradiction:

Suppose f(1) = 0.
since we have f(x^4 + y) = x^3f(x) + f(y),
let x = 1, and let y be some value such that f(y) = 0.
Then
f(1 + (y-1)) = f(1) + f(y-1)
=> f(y) = f(1) + f(y-1) => f(y-1) = 0.
We can continue infinitely from here, which leads to contradiction.

Suppose f(-1) = 0.
Then let x = -1, y be a value such that f(y) = 0.
Then
f(1 + (y-1)) = -f(-1) + f(y-1)
f(y) = -f(-1) + f(y-1)
f(y-1) = 0;

so we again get an infinite descent. Contradiction.

Therefore, the only value of x such that f(x) = 0 is x = 0.

[quote="JBL"][quote="blahblahblah"]Whenever x=0, we have f(y) = f(f(y))

Doesn't this imply that if f has an inverse, then:

y = f(y)?[/quote]

No, unfortunately: consider the function g: g(x) = 1 for all x.[/quote]

Ah, but that g has no inverse.  So if f is a bijection, then it must be the identity function, correct?

You're right that if f has an inverse, then f(y) = y.  But that's a big if.  How are you going to prove that f has an inverse?

So with our current progress, wouldn't (i) be satisfied if (ii) is (thus making (i) superfluous)?

No, because we had to use (i) to get to that stage.  It is true that we will not get any more useful information from (i), however.

[quote="blahblahblah"][quote="probability1.01"]Building on all previously established work...

By JBL, our function must satisfy f(x^4 + y) = x^3*f(x) + f(y)

Case y=0:

 f(x^4) = x^3*f(x) + f(0) = x^3*f(x) by Nukular
Thus, our function must satisfy f(x^4 + y) = f(x^4) + f(y).  In other words, f(x+y) = f(x) + f(y) for all non-negative x and real y.

Now I wish I had a nifty way to prove this algebraically, as in a similar USAMO problem, but we see that if x stays fixed, f(y) at y increases as quickly as at x+y.  Since we can play around with the numbers, the function should increase at a constant rate throughout (linear).  If f(x) = mx + b, then:

 mx + my + b = mx + b + my + b
b = 2b
b = 0

Therefore, it seems that f(x) = mx where m is any real constant.[/quote]

Unfortunately, your work is flawed because f(x^4) != x^3*f(x) (excuse my mix of C++ and accepted mathematical parlance there).[/quote]

C++ excused.  By JBL, though (sorry for citing inaccuracy), f(0) = 0 and he also shows that f(x^4) = x^3*f(x).

Proof that f(x+y)=f(x)+f(y) implies that f is a first degree polynomial with f(0)=0

f(x+0) = f(x)+f(0), which proves the second part:

Let a>b

Then f(x+a) = f(x)+f(a) and f(x+b) = f(x)+f(b)

So f(x+b)-f(x+a) = f(b)-f(a) for any a,b,x.
Significantly, [f(x+b)-f(x+a)]/(b-a) = [f(b)-f(a)]/(b-a)

This is saying that the slope is the same at any two points.  We can then place the points arbitrarily close together and obtain the result that:

f'(x) = C

so f'(x) = Cx+b --> f(x) = Cx because f(0)=0

f(x+y)=f(x)+f(y) does not imply that f(x)=kx.

why not? Isnt that just Cauchy's functional equation? I mean, we can prove it, but isnt it well known?

You can't prove it because it simply isn't true. You can prove it's true for rational x, f(x)=kx is the only solution, and you can prove it for all rational multiples of pi and e, etc, but you can't prove that the value of k is always the same unless you have one other bit of information (continuity, monotonicity, etc.).

In other words, assuming the axiom of choice, there are solutions to f(x + y) = f(x) + f(y) other than f(x) = kx.

Hm do I really need the Axiom of Choice? You can keep giving me numbers, and as long as each one you give me is linearly independent of the previous ones, I can assign it whatever value I choose. I think I might even be able to construct an explicit function f that is not linear, but it would probably be hard.

[quote="JBL"]Every single person whose written on this: you are all proving things which assume continuity.  You can't do this because we aren't given it!  (Also, I proved additivity for all x, y, by throwing in the negative condition in my second or third post, I think -- you didn't need to reprove it.)[/quote]

I am aware that continuity is required, I am sure that my little proof did not bring us any closer to a solution of the problem.

Furthermore, there are definitely discontinuous functions which satisfy f(x+y)=f(x)+f(y)

So, if I understand the previous posts, we have the following properties of f:

(1) f is additive: f(x + y) = f(x) + f(y).
(2) f is idempotent: f(f(x)) = f(x).
(3) The only zero of f is 0.

By (1) and (2), we have f(f(x) - x) = f(f(x)) - f(x) = 0.  In other words, f(x) - x is a zero of f.  Thus, by (3), we have f(x) - x = 0, which means f(x) = x.

Does that look okay?


Can someone tell me the steps to solve this?
x^2=-i
the answer is +-(1-i)/sqrt(2)
but I have no idea how to arrive at that answer and know I'm just missing something silly :oops: . Thanks!

try solving $(a+bi)^{2}=-i$ where $a$ and $b$ are reals... you get $2ab=-1$ and $a^{2}=b^{2}$

ok.. so solving those I get
2ab=-1 => a=-(1/2b)
substituting this into the second equation I get
(-1/2b)^2=b^2 => 1/sqrt(2)=b
Now, do I take the latter and substitute it back into the first equation to get a?
I guess I'm just not making the connection yet as to how I transform that into the real and imaginary parts of the answer....

? I don't understand kunny? :?:
I mean I understand that (1+i)^2 = (1,1)(1,1)=(0,2), but not the attachment to the problem

[hide]$(a+bi)^{2}=-i$
$a^{2}-b^{2}+2abi=-i$
Equating reals and imaginaries:
$a^{2}=b^{2}$ and $2ab=-1$
$ab=-1/2$ so $a=\frac{\sqrt{2}}{2}$ and $b=-\frac{\sqrt{2}}{2}$.
We have switch a and b to get both the "positive" and "negative" square root of $-i$
Plugging in $a$ and $b$ to get $a+bi$,
$\boxed{\pm\frac{\sqrt{2}(1-i)}{2}=x}$.[/hide]

oh jeez... I was thinking way too hard!!!! :rotfl:

$(1+i)^{2}=2i\Longleftrightarrow i=\left(\pm \frac{1+i}{\sqrt{2}}\right)^{2}$, thus
$x^{2}=-i\Longleftrightarrow (ix)^{2}=i$, yielding $ix= \pm \frac{1+i}{\sqrt{2}}$.


counting
distinguishability
combinatorics unsolved
combinatorics

[quote="pluristiq"]posted before, by another Romanian. coincidence? ;)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317040[/quote]

Not quite a coincidence. (Aalexandru is my colleague and we had an exam today.)

But my question is a little bit different
[quote]In how many ways can one distribute $ n$ (distinct) letters in $ k$ mailboxes such that one of the boxes receives [b]at least[/b] $ N$ letters?[/quote]

[quote="pluristiq"]Question:are the mailboxes distinguishable? if so, does a particular mailbox need to have at least $ N$ letters, or just *some* mailbox?[/quote]
Yes, the mailboxes are distinguishable. Some mailbox should have at least $ N$ letters.
In the case that one fixed mailbox must have $ N$ letters the answer is $ \sum\limits_{r \equal{} N}^n {\left( \begin{array}{l}
 n \\
 r \\
 \end{array} \right)} \left( {k \minus{} 1} \right)^{n \minus{} r}$

(we place $ N \leq r \leq n$ letters in the fixed box and the other $ n\minus{}r$ letters in $ k\minus{}1$ boxes). I was tempted to multiply this number by $ k$, but we end in an over counting.

yeah, that's why you need Inclusion Exclusion.

the number of ways that $ j$ particular mailboxes can each get at least $ N$ letters is

\[ S_j\equal{}\sum_{N\leq r_1, ..., r_j\leq n} \binom{n}{r_1}\binom{n\minus{}r_1}{r_2}\cdots \binom{n\minus{}r_1\minus{}\cdots r_{j\minus{}1}}{r_j}(k\minus{}j)^{n\minus{}r_1\cdots \minus{}r_j}\]

so by PIE the answer is $ \sum_{j\equal{}1}^k(\minus{}1)^{j\plus{}1}\binom{k}{j}S_j$.

as far as I know, other than for certain special cases, there's no closed-formula answer tothis problem; someone correct me if I'm wrong.


[quote="warut_suk"]Find all $x,y\in N$ such that
$\frac{x}{y} = \frac{(x^2-y^2)^{\frac{y}{x}}+1}{(x^2-y^2)^{\frac{y}{x}}-1}$[/quote]

Obvious that $x\neq y$ from the equation .

Simplify the original equation into

$(x+y)^{x-y}=(x-y)^{x+y}$

Now we can see that $x>y$ (if $x<y$ , LHS will be fraction while RHS is integer , contradict)

Now letting $x+y=a$ ; $x-y=b$ . Clearly $a\neq b$ or else this will result in $y=0$ which contradict with the question.

We wish to find the integral solution for $a^b=b^a$ for $a>b$ . Since $b|a$ , say $a=b^k$ for natural $k$ . Then we have

$b^{bk}=b^{b^k}$ :arrow: $bk=b^k$  :arrow: $k=b^{k-1}$

letting $b=k^p$ for integer $p$
 :arrow: $k=k^{pk-p}$ :arrow: $1=p(k-1)$ :arrow: $p=1$ ; $k-1=1$

Hence $k=2$ , $b=2$ , $a=2^2=4$

$x+y=4$ ; $x-y=2$  :arrow: $x=3,y=1$ :)

[quote="riddler"]How did you get
$(x+y)^{x-y}=(x-y)^{x+y}$
 :?
I think it should be....
$x(x^2-y^2)^{\frac{y}{x}}-x=y(x^2-y^2)^{\frac{y}{x}}+y$
$(x-y)(x^2-y^2)^{\frac{y}{x}}=(x+y)^1$
$(x-y)^{\frac{y+x}{x}}=(x+y)^{\frac{x-y}{x}}$[/quote]

keep on going by powering $x$ and you will see that the $\frac{1}{x}$ thing will gone  :P


I need help on these problems!..Thank you.

1. How many postive integer divisors of 2008^2008 are exactly divisible by 2008 positive integers?

2. How many positive integers n are there in such that n is an exact divisor of at least one of 10^40 and 20^30?

3. Compute the least positive prime p greater than 32 such that p^3 + 3p^2 is a perfect sqaure.

3. $ p^3\plus{}3p^2\equal{}p^2(p\plus{}3)\longleftrightarrow p\plus{}3\equal{}x^2, x\in \mathbb{N}$. The least $ p$ such that $ p\plus{}3$ is a square and $ p\ge 32$ is $ 61$.

[hide]2.  Let $ S$ be the set of factors of $ 10^{40}$ and let $ T$ be the set of factors of $ 20^{30}$.  We want to find the cardinality of the union of $ S$ and $ T$.  Note that $ 10^{40}\equal{}2^{40}*5^{40}$ and $ 20^{30}\equal{}2^{60}*5^{30}$.  Thus their greatest common factor is $ 2^{40}*5^{30}$.
Using the formula for the number of factors of a number, we know the cardinality of $ S$ is $ 41^2\equal{}1681$, the cardinality of $ T$ is $ 61*31\equal{}1891$, and the cardinality of their intersection is $ 41*31\equal{}1271$.  Using the Principle of Inclusion-Exclusion, the cardinality of their union is:
$ 1681\plus{}1891\minus{}1271\equal{} 2301$[/hide]

[hide]$ 2008^{2008}\equal{} 2^{3 \cdot 2008} \cdot 251^{2008}$. A divisor of this number is divisible by exactly $ 2008$ positive integers if it is in the form $ 2^a \cdot 3^b$ where $ \left(a\plus{}1\right)\left(b\plus{}1\right)\equal{}2008$. Then we have $ \left(a,b\right)\equal{}\{(1, 2008), (2, 1004), (4, 502), (8, 251), (251, 8), (502, 4), (1004, 2), (2008, 1)\}$. All of these are divisors of $ 2008$, so the answer is $ \boxed{8}$.[/hide]

[quote="123456789"][hide]$ 2008^{2008} \equal{} 2^{3 \cdot 2008} \cdot 251^{2008}$. A divisor of this number is divisible by exactly $ 2008$ positive integers if it is in the form $ 2^a \cdot 3^b$ where $ \left(a \plus{} 1\right)\left(b \plus{} 1\right) \equal{} 2008$. Then we have $ \left(a,b\right) \equal{} \{(1, 2008), (2, 1004), (4, 502), (8, 251), (251, 8), (502, 4), (1004, 2), (2008, 1)\}$. All of these are divisors of $ 2008$, so the answer is $ \boxed{8}$.[/hide][/quote]

Actually, the form of the divisor is $ 2^a 251^b$ (though this is trivial).  However, your solutions $ \{(1, 2008), (2, 1004), (4, 502), (8, 251), (251, 8), (502, 4), (1004, 2), (2008, 1)\}$ correspond to the values of $ (a \plus{} 1, b \plus{} 1)$, not $ (a,b)$, though each still corresponds to a divisor with $ 2008$ factors.

Another way to find that there are only $ 8$ solutions: Once we find that $ (a \plus{} 1)(b \plus{} 1) \equal{} 2008$, there exists a solution $ (a,b)$ for each pair of divisors of $ 2008$.  $ 2008 \equal{} 2^3 251$, so it has $ 4\cdot 2 \equal{} 8$ divisors, and thus $ 4$ pairs of divisors.  But each pair can be arranged in $ 2$ ways (ex, $ (2,1004)$ and $ (1004,2)$ are distinct), so there are a total of $ 8$ solutions for $ (a,b)$


geometry
power of a point
radical axis
geometry unsolved

The circles $\gamma_1$ and $\gamma_2$ intersect at the points $Q$ and $R$ and internally touch a circle $\gamma$ at $A_1$ and $A_2$ respectively. Let $P$ be an arbitrary point on $\gamma$. Segments $PA_1$ and $PA_2$ meet $\gamma_1$ and $\gamma_2$ again at $B_1$ and $B_2$ respectively.
a) Prove that the tangent to $\gamma_{1}$ at $B_{1}$ and the tangent to $\gamma_{2}$ at $B_{2}$ are parallel.
b) Prove that $B_{1}B_{2}$ is the common tangent to $\gamma_{1}$ and $\gamma_{2}$ iff $P$ lies on $QR$.

Let the centres of the circles be $O_1$, $O_2$ and $O$ respectively. Then:

(a). Connect $O_1B_1$, $O_2B_2$ and $PO$. Then $O_1B_1 || PO || O_2B_2$, which means the tangents at $B_1$ and $B_2$ are also paralel.

(b). Since P lies on QR, so $PB_1 \cdot PA_1 = PB_2 \cdot PA_2$. Thus $A_1, A_2, B_2, B_1$ lies on the same circle. Therefore
$\angle{PB_2B_1}+\angle{O_2B_2A_2}=\angle{B_1A_1A_2}+\angle{O_2A_2B_2}=$
$=\angle{O_2A_2B_2}+\angle{OA_2A_1}+\angle{OA_1B_1}=\angle{PB_1B_2}+\angle{O_1B_1A_1}$

Thus $\angle{B_1B_2O_2}=\angle{B_2B_1O_1}$. But by the conclusion of the part (a) we have that those two angles have the sum of $180^o$. So there must be $\angle{B_1B_2O_2}=\angle{B_2B_1O_1}=90^o$, which yields $B_1B_2$ is the common tangent of $\gamma_1$ and $\gamma_2$.

Mr [b]shobber[/b], part (b) of the problem require an iff :oops:

Here is my solution:

Consider the inversion $ \mathcal{H}$ through pole $ P$ with product $ k \equal{} \overline{PA_1}.\overline{PB_1}$ which takes $ A_1\mapsto B_1$ and $ A_2\mapsto A_2'$ then $ \mathcal{H}$ takes $ (O)\mapsto (B_1A_2')$, but since $ (O)$ is tangent to $ (O_1), (O_2)$, so is $ B_1A_2'$. Then $ B_1B_2$ is the external tangent of $ (O_1),(O_2)$ iff $ B_2\equiv A_2'$ or $ \overline{PA_2}.\overline{PB_2} \equal{} k \equal{} P_{P/(O_1)}$, which implies that $ P$ lies on the radical axis of $ (O_1),(O_2)$, which is $ QR$.


There are 3 pipes, one small, one medium, and one large.  They are all used to fill a pool.  If all 3 are used at the same time then it takes 3 hours to fill the entire pool.  It takes the largest pipe 1/3 the time of the smallest pipe to fill the pool.  It takes the medium twice the time of the large pipe.  How long does it take the smallest pipe to fill the entire pool.

The answer is 16.5.  I can't seem to get that.  Maybe the textbook is wrong?  Iam sooooo confused.

                                                      Help?

[hide="Explanation"]Let $ x$ be the rate of the small pipe. Thus, $ 3x$ is the rate of the large pipe and $ 1.5x$ the rate of the medium pipe. Amount pumped is equal to rate $ \times$ time, so time is equal to $ \frac{distance}{rate}$. We have that $ \frac{d}{3x\plus{}1.5\plus{}x} \equal{} \frac{d}{5.5x} \equal{} 3$, so $ \frac{d}{x} \equal{} \boxed{16.5}$.

[/hide]

For every $ 3$ litres the large pipe delivers, the medium one delivers $ 1.5$ and the small one $ 1$ litre.

The ratio of these is the same as $ 6: 3: 2$, so the small pipe delivers $ \frac{2}{6\plus{}3\plus{}2} \equal{} \frac{2}{11}$ of the total poolfull every $ 3$ hours and so $ \frac{2}{33}$ of the poolful every hour.

It therefore takes $ \frac{33}{2} \equal{} 16.5$ hours to fill the pool by itself.


AMC
USA(J)MO
USAMO
geometry
ARML
AMC 12
AIME

For those who already qualified USAMO or working towards it, how much time do you spend on math everyday?  And what kind of problems are you guys doing?

I've done a USAMO a day for the past few weeks but I ran out last week :(  So now I work on USA TSTs and some APMOs.  Some days I don't do anything at all, and some days I spend 6+ hours.  It all depends.

It depends a lot I agree. I've been doing random problems and also going through the intro section 101 Combo Problems by Titu Andreescu. I do like 4 problems a day in that book (or at least I did until they got impossible haha). I might do math all day. I suppose the least amount of math that I do daily would be like an hour because I have two math classes a day...

I'm kinda bad about really practicing that much in test conditions. I suppose I should do that more, but I spend more time just sorta messing around with interesting ideas and coming up with problems.

Mainly I just run around looking at random olympiads. I decided for the next few days I'll do a lot of geometry, and then kind of works on combinatorics. How much I spend on math each day depends on my mood and how busy I am that day. Luckily this weekend I'm able to work on math almost all day, but when the weekend is over...

Well, for over the past year, I just work on the problems in the upcoming contest.

I probably did limited math probelms (by myself) from like last year's ARML up to November.  But from like December-February, I worked on the AMCs because they were the upcoming contest (mainly AMC12s and the harder question so ill be also practicing for the AIMEs simultaneously).  Then I worked on the AIMEs for like February-March.  Now, I am doing some old USAMOs, USAMTS, and any kind of proof problems to prepare for the USAMO (probably not enough time to actually do well on them though)

I only really do math problems on the weekends: maybe 1-2 hour by myself, and then i have this math class every weekend where we do like 4 hours of math (2 for proofs and 2 for problem solving).

But like the week before the AIME, I probably did 3-4 hours every night and just bsed all my homework during school.

I do mostly contest problems (AMC, AIME, ARML, Mandelbrot, USAMTS (although im too lazy to do...),some state contests, anything math related you can find on the internet-power of google)
If you want to be very good at contest math, I dont think the school textbooks and what your teachers teach you will help that much...most of those problems are fomula oriendtated and really require a straight up process while these contest probelms requires the person to think and with harder problems, come up with very unique solutions.

cdymdcool, whats your goal in math?  Are you trying to pass the AMCs only or are you trying to make it to USAMO or something?

normally: 0 hours a day
2 days before the test: 8 hours a day

Yeah, I don't really have too much time because I spent a lot of time practicing piano, so doing USAMOs under test conditions is kinda out for me. But I mess around, look at random USAMOs, math books (courtesy of a friend who has a whole host of them) and some WOOT stuff. I'm not expecting to do well on the USAMO, as my main goal was to do well on AIME this year, but with any luck I should be able to pull off a 7.  :P

Pre-AIME I did an AIME every other day or so between the AMC B tests and the AIME I. And no prep for the AMC (probably why I bomb it every time it comes around).

[quote="davidyko"]Pre-AIME I did an AIME every other day or so between the AMC B tests and the AIME I. And no prep for the AMC (probably why I bomb it every time it comes around).[/quote]That was essentially my preparation for AIME. The weekend before AMC I did like 7 practice tests (we had a 4-day weekend).

For USAMO I've been doing old USAMO problems in spare time (at home, on the bus to and from school, in free time at school) and reading ACoPS. This time of year I probably do like 3+ hours a day

I don't often take tests in time either; like when preparing for the AMC I took my stash of problems from the AMC 12 Problem Series (basically the hard problems from the last 30 years) and worked on those for some time every day. I essentially didn't do any preparation for the AIME.

I do math when I feel like it, probably a little less than I should. When I'm in the mood for a problem, I just pick a random test and a random problem, or when I see a problem somewhere I may be inclined to work on it. I never want to feel like I'm forcing myself to do math.

practice some USAMOs, look at some books, etc. etc. 4-7 hours on weekend, 1-3 otherwise.

[quote="mathemonster"]do u guys think that doing an hour every day, finishing v1 of aops, taking the aops online amc 10 course, doing as much as i can of aops v2, and working through old amc 10s and AIMES will be enough to make the 2009 goals in my signature?[/quote]Depends on a lot of things. That really isn't an answerable question, it depends how good you are, how quickly you learn, how you're "doing" the AoPS books, etc.

i am doing almost all the problems in the aops books i think i learn fast and i could prbly score alot higher than my 106.5 on the 10 now.

Yea, it schould definitely, bring you up high enough.
I mean, I can get hi 130s to lo 140s without even finishing volume 1...
You can definitely get a 132+

Tho, as warning, don't make stupid mistakes!

I don't really put that much time in practice...I always happen to have something else to do.
So far I've tried (seriously trying, not just skimming) 2 USAMO-like problems.

[quote="Anaxerzia"]After a year on the forums and with AoPS, I went from not being MathCounts level to USAMO qualifier in 8th grade this year, so yeah even if you're just starting out don't be discouraged :)[/quote]

dang it ravi. i wish i could say the same thing. :D
after one year, i went from 16 on the AMC 8 to missing USAMO because of the stupid 2nd index crud (with a 7 on the AIME)

I went from 23 AMC 8 to 22 AMC 8. Woot!
On the other hand, I did better on every other competition.
Still failed AIME though.

[quote="Anaxerzia"]After a year on the forums and with AoPS, I went from not being MathCounts level to USAMO qualifier in 8th grade this year, so yeah even if you're just starting out don't be discouraged :)[/quote]
Same.

How much is needed to get to blue mop?

3 questions pythag, don't aim too high, since you're in 6th grade I think. Blue MOP is really hard, and ever getting an honorable mention is a really big achievement.

I'm starting to practice today. I'm going to spend some time trying past olympiad problems, and I'm going to also spend some time trying to do well-written proofs for problems that I've already solved.

[quote="mathemonster"]georgiamathdude how much aops v1 did u do until u could get a 140!?[/quote]

I am up to chapter 18 now. I do like a chapter every two weeks. And 140 is my record, not my everyday thing... Mid 130s for me.

I do ~2 hours during the week and ~4 on weekends. The only problem is half the time I am staring at the problem or not thinking...

Obviously most people don't actually meet their goals :P Including me.
On weekends I try to do tons of math, but there's not always much time. 3 hours a day average friday, saturday, sunday, then like 10 minutes on normal days, except when there's no homework (2 hours). But I still managed to fail so whatever.

Well, that's on topic. But anyways, "goals are set to be 'almost met' " - Some random person.

[quote="Elemennop"][quote="Ubemaya"]I set myself a goal of 2 hours a day, and ALWAYS I manage to get distracted by the internet or SSBB. :([/quote]

why would you play brawl? it sucks compared to melee[/quote]
Online opponents :wink:

I usually have a bunch of pointless and time-consuming homework and other random stuff, so most of my time is usually spent on weekends.  Maybe 3-4 hours on Saturday, and a little mre on Sunday.

i currently do about 45 minutes 4 times a week and 2 hours on the weekend

Basically all my AP classes that are currently in review are fair game for USAMO practice.... so that gives me about 3-4 hours during school ;)
It's a bit difficult to concentrate though, I tend to solve better at home, but I end up being more apathetic about it once I am home... doing nothing but math for 8 hours every day for a month does, believe it or not, get tiring after a while.

My goal is really simple: to get 13+ on the USAMO.  Of course, I would like to do better.  I'm really good at setting goals which I can usually achieve without too many problems.  Of course, you never know what questions you'll get, so maybe I'll fail my goal.

Anyway, I'm really working through tons of Olympiad problems.  The problem is that I keep on working on them in a very chopped up manner.  So if I'm doing a problem, especially a geometry problem, I have to keep on retracing my steps each time I go back to that problem, which wastes a lot of time.

Throughout the school year, since I have a private math class, I'm always doing at least 42 minutes just within school.  And when I get out, generally, I'll spend at least an hour, even two.  If I have no homework, like today, I'm bound to spend even more time.

I have a prediction:  I think I'm going to have a good day and get numbers 1, 2, 4, and 5 correct.  I've never done the USAMO under time pressure, so I hope I can do as well as I am on the past ones I'm going over.


Would it be possible to somehow make the transcripts more "printer-friendly"?

the suggestion of making binders for our classes came up earlier and I intend to actually try it...except these current transcripts are like avg 23 pages apiece...times 12 plus the Problem Sets and Solutions and thats a lotta paper and ink...is there a way we could (other than manually editing all 23 pages worth everytime) get a better version for our printers' sake? (and our parents' willingness to buy stacks of paper and ink cartridges)

Why not just keep a copy on your computer and not bother to print everything out?

Can you copy/paste into word & shrink font size and/or use a dual-column print format?

I usually do keep a copy on my comp, but its realli inconvient if im trying to look something up and im currently not using the computer...that's why im making binders in the first place.

I've used word to try to shrink stuff but then the images dont show up too well, and after a while text gets hard to read (after we go smaller than size 10). I haven't tried dual-columns though. Lemme see how that works.

or if one of your parents works in an office thingie they can print it out on a laser printer or something and not be charged for paper or ink

Or better yet use the school's computer/printer :twisted:

I doubt keeping a binder is helpful since you're not going remember where to look if you want to find the divisibility rule for 13, but if you keep it digital than you can just search for it.
If you must have it on paper I suggest you acquire a laser printer like my parents, because in the long run if you use it that often it will be worth it (try to write two graphs, one with time as variable and the rate of paper and ink being used up as the constant and one with just y=the cost of laser printer and see how long it'll take in order for them to intersect (and for the first to exceed the second))

i have a laser one...its juss...still inconvient as far as paper and ink goes...

exactly the reason i wish to get a printer-friendly version of the transcripts...

[quote="JBL"]I'm not sure what that graph you were suggesting was, Tare, but laser printer ink is crazy-expensive like.[/quote]
...you mean a toner, right? Exactly why you need a graph to compare the amount of money you spend total as time passes by writing two equations and solving for the intercept (which is the point in which a laser-jet printer actually saves money than a normal printer)

yeah, i was wondering this too.  When i try to print out transcripts, its still a lot of pages, even if i shrink the font.  But i dont know what this dual-column print format is.

I believe it makes it kind of like a newspaper...with 2 columns...but I hvae no clue how that would save space...

like most replies that people make do not fill up a line, so if you make it columns it basically cuts the the 20 or so pages in half

Well, not quite...but yes i do see how that would save a little bit of space...

You could even reduce the margins if you must.

reducing margins wouldn't do too much...

and putting it in pdf wouldnt be too much better...b/c i usually do some editing of repetitive posts (like when they show the first 3 ppl having the right answer). Now if the transcripts were to be edited beforehand that would make a world of difference.

If you know any simple scripting languages perhaps you could write a script that compresses transcripts by cleaning out whatever it is you don't want.

I guess you could delete everything not said by the host, but then it wouldn't make much sense...there's not much you can reduce from a transcript I believe, so I guess there's nothing drastic you can do about it...

Tos summarize:
a.) make the font size smaller
b.) reduce the space in between the lines
c.) reduce margin
d.) put it in pdf file
e.) leave it on the computer
f.) buy a laser-jet printer
g.) print it from somewhere else and make someone else pay for the electricity/ink
h.) all of the above

anyone try using the print 2-4 pages on one sheet and two-sided printing? that combined w/ the other things might help. :shock: