hello mathlinkers who will go to vietnam this summer.

what do you think of the idea that everyone can bring items like coins or something similar to the IMO to change them? especially those from countries that are a bit smaller ;)

Naphthalin

Sure, that's a good idea.

And we will bring Canmoo too, but not for exchange! :P

Canmoo will be ours!

But yes it's a good idea, I still have the coins the slovenian team in mexxico (2005) gave us, together with other stuff, as "preparation" for 2006. But doing so with all teams is more fun^^
But I will not be there :(

sorry for the question but what is Canmoo exactly?
even if i come from germany i can bring some nice money. e.g. old money of the german "democratic" republic^^

Naphthalin

I suspect Canmoo is the Canadian Moose :moose: that has been traveling around at the IMO for the past years. While previous attempts of getting posesion of the moose have failed, I have designed a master plan this year that will surely lead to the capturing of the moose :w00tb: beware moose guardians!

Oh guys very nice topic! During the last four years I have collected so many interesting things from many friends at IMO(particularly a lot of coins :) ,I haven't counted how much it is in total,but I guess I am a millioner :D ). Last year,besides Armenian coins and other stuff, I took with me two botteles of Armenian brandy :blush: ,the people liked it very much though they were drank in the end :oops: .

It's really a very nice idea !  :lol:

I'll bring some with me....... (of course, not only coins...  :wink: )

We always bring coins (we have a coin that has 1/2 printed on it, and we think it's funny).

@Valentin Vornicu:
Many teams had the moose for some time, including mine (We had a master plan last year).
But I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI.
I hope I'm wrong.

[quote="Fuaran"]But I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI.
I hope I'm wrong.[/quote]
That's quite wrong. :wink:

[quote]Fuaran wrote:
But I heard (from a reliable source) that the moose won't be at the IMO this year, just the IOI.
I hope I'm wrong.

That's quite wrong. Wink[/quote]

Yay!
Actually, I heard it from a former Canadian participant. Looking forward to see you all at Hanoi!

I don't know what to bring... I will bring the best spaghetti of  Italy! One spaghetto for each contestant  :D

I remember some team bringing food last year as well, Vietnam I think... But they got a HUGE box of spaghetti, everyone got a package  :lol: .


Show that for every prime number $ p>3$ there exist integer numbers $ x, y, k$ such that $ 0<2k<p$ and $ kp\plus{}3\equal{}x^{2}\plus{}y^{2}$

[EDIT: Sorry, I was reading the wrong thing.]

We consider to set :
$ A_1 \equal{} \{k^2 \minus{} 3| 0\leq k\leq \frac {p \minus{} 1}{2}\}$
$ A_2 \equal{} \{ \minus{} k^2|0\leq k\leq \frac {p \minus{} 1}{2}\}$
Because $ |A_1| \plus{} |A_2| > p$ so exist $ x,y\in\{0,...,\frac {p \minus{} 1}{2}\}$ such that
$ p|x^2 \plus{} y^2 \minus{} 3$
$ x^2 \plus{} y^2 \minus{} 3 \equal{} kp$
$ x^2 \plus{} y^2 \minus{} 3\leq 2(\frac {p \minus{} 1}{2})^2 \minus{} 3 \equal{} \frac {p^2 \minus{} 2p \minus{} 5}{2}$
So $ 2k\leq p \minus{} 1$
This problem claim.


calculus
derivative
linear algebra
matrix
function
integration
trigonometry

I don't necessarily need a proof of this, but more of a reason why I should believe it to be true. At this point I don't believe it at all.

Show that the diagonal elements of the matrix representation in any basis for $ H_1$ for the operator $ D\equal{}\frac{\partial}{\partial x}$ are purely imaginary.

$ H_1$ is defined to be the space of all functions such that $ \int_0^a \phi^*\phi \ dx<\infty$ (and * is complex conjugate).

Why I don't believe it: It seems as if you could use some sort of Fourier sine basis $ \{\sqrt{\frac{a}{2}}\sin\left(\frac{n\pi x}{a}\right)\}$ and get no imaginary numbers in the matrix representation at all.

With $ \sin,\cos$ basis you get zeros on the diagonal. And $ 0$ counts as a purely imaginary number (it has real part $ 0$).

Thanks, that makes sense. This is still a little strange in my mind. I've thought of a couple of approaches, none seem to be working.

Brute force: Let $ B\equal{}\{\phi_n \}$ be a basis. Then $ D$ has the matrix form $ D_{qn}\equal{}\int \phi^*_q D \phi_n dx$ so that the diagonal elements are $ D_{mm}\equal{}\int_0^a \phi_m^*\frac{\partial \phi_m}{\partial x} dx$. There doesn't seem to be an obvious reason from here that $ \text{Re}(D_{mm})\equal{}0$.

Alternate approach: Diagonalize the matrix via eigenvalues, and use some sort of similarity argument?

We have $ D_{mm} \equal{} \int_0^a \overline{\phi_m(x)}\phi_m'(x)\,dx$. Therefore, $ \overline{D_{mm}} \equal{} \int_0^a \phi_m(x)\overline{\phi_m}'(x)\,dx$. We want to show that $ D_{mm} \plus{} \overline{D_{mm}} \equal{} 0$. By the product rule for derivatives, $ D_{mm} \plus{} \overline{D_{mm}} \equal{} \int_0^a \frac {d}{dx}(\phi_m(x)\overline{\phi_m}(x))\,dx \equal{} |\phi_m(a)|^2 \minus{} |\phi_m(0)|^2$.

Therefore, the statement you want to prove is true if all elements of $ H_1$ attain the same value at $ 0$ and at $ a$. Most likely, your definition of $ H_1$ is incomplete.

(Moral of this story: when you have a differential operator acting on a Hilbert space of functions, be ready to integrate by parts.)

Wow, I'm sorry. I actually had gotten to where you ended, and completely forgot about the fact that I knew that $ \phi(a)\equal{}\phi(0)\equal{}0$ (hence why I only used $ \sin$ for my basis in the first post). Thanks a lot!


geometry
probability
geometric transformation
dilation
function
trigonometry
3D geometry

Well, it was inevitable that somebody had to start it, so why not me??? After I'm the one who always studies for school.

Schedule::

March 1 - Chemistry
March 7 - Physics
March 10 - English
March 15 - Biology
March 20 - Maths
March 31 - Computer Science

List of PPl who are writing this yr (and in AOPS)

[hide]
chapplikka
chemrock
Cyber-Shreyas
Eucleidan Geometry
hash_include
madness
nitishexpress
pardesi
shruthibaskaran
SOUMYASHANT NAYAK
srinitrs
valeriummaximum
vihang

ICSE:
AnitSahu

Other Boards:
Varunarasu
[/hide]
So get the ball rolling

Hmm.. Let the hilarity ensue, what else can I say? And I think it was Anit Sahu right?

include Raghav in CBSe shre, he is here as Euclidean Geometer

Have u all finished portions for chem..????
me not even once...

Dei...
Onakke idhellam toooo mucha thereela  ?
Naan exam ku munnadi oru dharava padichhu mudipena ne pureelai :!:
Any ways,
Have u all been asked to mug egs. of tranquillizers blah blah... diff kinds of drugs :?:
I mean the names
Its bugging and belting

naanum orutharava mudikanumnu romba try panituiruken

2 Good Questions

1. What do you mean by the functionality of a monomer?
2. How do the double bonds in natural rubber control its properties?

[quote="valeriummaximum"]the double bonds in the dont do anything actually.....Once natural rubber is stretched there is no restoring force for it to regain the original sahpe...
since double bond is functional site...so they help in interlinking the two chains of natural rubber thereby helping to establish the rubber like property[/quote]

Wrong!!

Clue:
[hide]Think isomerism[/hide]

cis and trans?????

how is it related to the intermolecular forces shre...

Natural rubber is the cis form isnt.... Cis 1, 4 something if I remember what I read for the preboards :D Functionality is the number of functional groups in a monomer?

the natural rubber's name is cis-1,4-polyisoprene.....

Common Shre, Explain how the configuration affects the intermolecular forces???

Ranga, he just said properties you know, never mentioned anything about intermolecular forces..  I think he meant wrong in a way that your answer that the double bonds are useless is wrong.. I maybe wrong, but who cares...

2 more questions:

1. During preparation of Nitrogen from ammonium chloride and sodium nitrite, what are the two impurities evolved? How can nitrogen be purified from them?
2. Two ways of preparing S2Cl2

elasti got that one it seems! Brutes, his friends had nved sumita arora downright n then they were discussing this morn before him so he got it Cha! :mad:  :|

[quote="bruteforce"][quote] Out due to inheritance[/quote] ur children took away ur marks  :rotfl:  :rotfl:

it was not multilevel , not multiple , so all that came into my mind was heirarchial  :D

what r we expected to write for that wrong sql question ? i didn't even notice it , so just assumed that it was customer.i_id and item.i_id . Isn't that enough? should we mention that it is wrong?[/quote]

Even i thought that they had made a mistake so i wrote it by assumption :)

I just stated the assumption and found the output.

Hierarchical please... Note the spelling  :P

[quote]it was not multilevel , not multiple , so all that came into my mind was heirarchial [/quote]

The spelling is wrong...It should be hierarchical....(anyway,who cares...)

A centum wouldn't exactly surprise me .... :P  :P
btw,
I had almost got my output wrong assuming 'R' comes after 'P'...
I guess I should revise my alphabets again... :rotfl:
The paper was a bit too easy...

EDIT: I see raghav has already mentioned that...

Another worst mistake i would have made and almost made was that i was considering the loop for the word MyString :D :o . Then i realised that there was no special chars or whatever in that so i checked again to find that the word was actually sth else and that was the name of the string :rotfl:

Great everybody has written hierarchical. :( :((P.S: Spell checker in Firefox is really nice ;))

I liked that question where you had to swap every odd and even position. There was just 1 trick in that question which is ofcourse not very easy to miss considering that his own example was illustrating the trick. Not like the INOI where he gives pathetic examples.

@brute: I wrote. Assumption: its I_id and not Item_id and wrote the answer.

I also liked that 'Write the equivalent POS form' question. A lot of people here just inverted the expression.

Yup the POS question was good...I tells how people just mugged up the K mapp and went for exam

He used have given a 10 size array ex. for the swapping..  Then most of the students wouyld have not thought of the trick

No, Rohit, I interpreted Nitish comments like this. We weren't really taught boolean in a very understanding or conceptual way. I'm sure that nobody in our class can really explain why a K-Map works and why do we base it on gray code.

The problem is that none of us really understand SOP and POS, Maxterm and Minterm so well enough to answer twisted questions on that.... The problem is not with you, it was with the teaching.

[quote="madness"][quote="Nitishexpress"]Yup the POS question was good...I tells how people just mugged up the K mapp and went for exam
[/quote]

Now that was both rude and unnecessary :huuh: :mad: Well there are plenty of instances i can show where you 'pros' have made mistakes. That means u are all bloody nvs and bloody bloody hypocrites too. :mad:

EDIT: Dont get me started, then a whole lot of unnecessary comments will appear without invitation[/quote]

COOL DOWN :coolspeak: Yes shre is right. KVKK didn't do us any great favours with his teaching. "Just NV my text to get 95" that's all he says.

[quote="Nitishexpress"]Thalla mmadness, i thought there was only one meaning for that... and that is what shre told... if u make a mistake it doesnt mean that question is a nv question...[/quote]

hello! :D Sorry da if i wasnt clear, i didnt call that question an nv ques at all, infact that was a good trick question as i told shre.Read my post properly da, even the best of persons make mistakes in logical questions :D thats what i meant. And ok, the msg got over and i cant agree more, esp the networking part :D


geometry
perimeter
trigonometry
linear algebra
matrix
integration
calculus

One a semicircle of radius 1, two points are selected at random and independently, along the perimeter of the semicircle. Together with the midpoint of the diameter, these three points form a triangle. Prove that the expected area of the triangle is $ (2\plus{}\pi)^{\minus{}1}$.

"The midpoint of the diameter" is a wonderful way of referring to the center of a circle :wink:

[hide="Possible approach"]

If I understand you correctly, $ \alpha$ and $ \beta$ are chosen uniformly at random in $ [0,\pi]$, and we are asking for the expected are of the triangle with vertices $ (0,0)$, $ (\cos\alpha, \sin\alpha)$, $ (\cos\beta, \sin\beta)$. If this is so, the area can be calculated as a 3x3 determinant

$ \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ \cos\alpha & \sin\alpha & 1 \\ \cos\beta & \sin\beta & 1 \end{vmatrix}$

we just need to be careful about signs - depending on the order of $ \alpha$ and $ \beta$ this could be positive or negative. This can be avoided by forcing, say, $ \beta \leq \alpha$, so calculating the expectation as

$ E \equal{} \frac{1}{2 \pi^2} \int_{\alpha\equal{}0}^\pi \int_{\beta\equal{}0}^\alpha (\cos\alpha \sin\beta \minus{}\sin\alpha \cos\beta) \, d\beta\, d\alpha$

which is straightforward. Unfortunately I'm getting a different answer from yours... need to check my calculations.

[/hide]

I think you're missing the cases when one of the points is on the diameter.

Ah. So I did misread the question. OK, that's easy to fix - just add another integral for the case of one point on the diameter. Also, my use of the determinant formula in the previous solution was just plain silly; the triangle with one vertex at the origin and two vertices at angles $ \alpha$, $ \beta$ on the unit circle obviously has area $ \frac{1}{2} |\sin (\alpha\minus{}\beta)|$: just rotate it so that $ \beta\equal{}0$ and then the base is 1 and the altitude is $ \sin(\alpha\minus{}\beta)$.

[hide="Correct solution"]
The probability for each point to end up on the circular arc is $ p\equal{}\frac{\pi}{\pi\plus{}2}$, and the probability to land on the diameter is $ q\equal{}\frac{2}{\pi\plus{}2}$. So we have

1. Probability $ p^2$: both points on the arc, expected area is

$ E_1 \equal{} \frac{1}{\pi^2}\int_0^\pi \int_0^\pi \frac{1}{2}|\sin(\alpha\minus{}\beta)|\, d\alpha \, d\beta$

(the coefficient $ \frac{1}{\pi^2}$ just normalizes the integral so that the total probability is 1).

2. Probability $ 2pq$: one point on the arc, one on the diameter, expected area is

$ E_2 \equal{} \frac{1}{2\pi} \int_{x\equal{}\minus{}1}^1 \int_{\alpha\equal{}0}^\pi \frac{1}{2} |x \sin\alpha| \, d \alpha \, d x$

3. Probability $ q^2$: both points on the diameter, area 0.

An easy calculation gives $ E_1 \equal{} \frac{1}{\pi}$ and $ E_2 \equal{} \frac{1}{2\pi}$ so

$ E \equal{} p^2 E_1 \plus{} 2pq E_2 \equal{} \frac{1}{2\plus{}\pi}$.

[/hide]

Thanks, I was hoping to see a non-calculus solution, but apparently not  :P


(From 2002 Belarussian Maths Olympiad)

Prove that there exist infinitely many positive integers which cannot be represented in the form

x[size=59]1[/size]^3+x[size=59]2[/size]^5+x[size=59]3[/size]^7+x[size=59]4[/size]^9+x[size=59]5[/size]^11

where x1,x2,x3,x4,x5 are positive integers.

The number of $k$-th powers $\leq n$ is $\leq n^\frac{1}{k}$.
Assume that two different sums of powers of your form are different, then there are at most
$n^\frac{1}{3}n^\frac{1}{5}n^\frac{1}{7}n^\frac{1}{9}n^\frac{1}{11}=x^r,r<1$
different numbers of that type $\leq n$.
But $\lim_{n \to \infty} \frac{x^r}{x} = 0$,
so only a very small part of all integers can be represented in this way, so there are infinitly many that cannot be represented.


Hi All,
I am looking for summer school specially in Europe for summer 2008.
Please help me on any thing about this kind of  programs (websites, name of university,..).

a summer school like a mathe camp? or a summer school that caches you up if you fail something? Also it would help to know what area of europe you live in.

Hm, you might be able to take an online long-distance [url=http://cty.jhu.edu/]CTY[/url] course. They cover many topics around the high school age, but might be a bit pricy.


[quote="XK-39"][quote="zheng"]Point P is selected at random from the interior of the pentagon with vertices A=(0,2) ,B=(4,0) ,C=(2pi+1,0) , D=(2pi=1,4) , and E=(0,4). What is the probability that angle APB is obtuse? Express your answer as a common fraction.[/quote]

Are you looking for the solution Alcumus provides or a clarification on this solution? I've attached Alcumus's solution for you to take a look out. If you don't understand something in the solution, raise your question in this thread.[/quote]

I was looking for clarification

[quote="PowerOfPi"]You are new, but I don't think this belongs here. I am not sure of the difficulty, so maybe High School Intermediate topics.[/quote]

Hey,
PowerOfPi let me say but it was an alcumus question. Plus i was only looking for clairification.

Yeah, it's fine here. Just one of the more difficult Alcumus problems. Plus, clarification on what?

[quote="pinkmuskrat"]Plus, clarification on what?[/quote]

On the solution(which i got)


algebra
polynomial
number theory open
number theory

The polynomials P(x) = x^2 and Q(x) = x^2-1 differ by one, they also factorise fully over the integers.

I want to know what are the highest degree polynomials that have this property? Here is a cubic example

P(x) = (x+1)(4x+1)(4x+1), Q(x) = x(4x+3)(4x+3) => P(x) - Q(x) = 1

I could go higher but I wont spoil the fun. I dont know a complete solution.

$ P(x) \equal{} x^{2^n}$ and $ Q(x) \equal{} x^{2^n} \minus{} 1 \equal{} (x \minus{} 1)(x \plus{} 1)(x^2 \plus{} 1) \cdots (x^{2^{n \minus{} 1}} \plus{} 1)$ satisfy the conditions for any natural number $ n$

EDIT: Nevermind I misunderstood the question

I should perhaps have been clearer but that's not what I menat by "fully" factorise. I meant they must factorise into linear factors
i.e. P(x) = (a1 x + b1)(a2 x + b2) ... (an x + bn) for integers ai, bi, and similarly for Q(x).


If you have 6 pizza toppings to choose from, how many different ways can you choose toppings to make a pizza with at least 3 toppings?
Is the answer 6C3

Using the digits 1 through 7 once each, how many numbers can are there where the digits alternate between odd and even?

[hide]If you want a pizza with atleast 3 toppings, it can have either 3,4,5, or 6 toppings.  Therefore your answer is 6C3+6C4+6C5+6C6

you must start and end with an odd.  4!3! [/hide]

[hide]If you have at least 3 toppings, then you could have 3,4,5, or 6 toppings.  This can be done in $\binom63+\binom64+\binom65+\binom66=\boxed{42}$ ways.

The digits must be arranged like:

oeoeoeo

There are $4!$ ways to rearrange the odd numbers and $3!$ ways to rearrange the even numbers.  That makes $4!\cdot3!=144$ total  combinations.[/hide]


I understand Google Chat is up and running. Anybody have any experience with it? Should I get it?

As I understand it, GC works with a Jabber-enabled client with server talk.google.com, username yourusername@gmail.com, and password yourpassword.

Can you post a link saying something about this, cause googled "Google Chat," but I didn't find anything.

http://www.spreadhello.com/

That's the closest thing I know of. Uh...but it's not really "Google Chat". It's more "Google's AIM-wannabe".

Yea, I found that, but I didn't think that was what solafidefarms meant.

It isn't *officially* released yet, but you can use it with GAIM or iChat. Here's where I heard about it: be warned slashdot has some bad language: http://it.slashdot.org/it/05/08/23/2316223.shtml?tid=217&tid=218

Also, for the direct link to the article: http://www.smashsworld.com/2005/08/im-on-google-talk-right-now.php

There are some reports that it isn't working any more; ... but it should tomorrow.

Ah yes. Hmm...I'm trying to get it to work. Trillian fails me.

trillian and gaim work for most ppl they can talk to msn and yahoo and aim

It is free, and is very good iff all your friends have gmail.

http://www.google.com/talk/index.html

It's alive! :-D And it also works on Trillian and Gaim now.

I had problems with Yahoo, I have no clue what gmail is. I like my MSN and AIM...


It is the ratio of a circle's circumference to its diameter, which I would think is the simplest definition. It can be found in many places, including nature. Some formulas can be found [url=http://mathworld.wolfram.com/PiFormulas.html]here[/url].

There is no need to revive the topic.
Your comment added nothing.
Please refrain from reviving in the future.

LOL countyguy thAts exactly how far i know.
At mathpath 11, someone recited 3321 digits.


I am taking USAMTS for the 1st time this year.
I know I shouldn't expect to do too well this year, but what is the average score?

I think the median is 40-50, according to the USAMTS webiste. Doing your best is the most important thing though :)


algebra function domain polynomial trigonometry calculus vector

$ f(f(x))\plus{}f(x)\equal{}2x$

$ f(f(f(x)))\plus{}f(f(x))\equal{}2f(x)$

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

$ 2(2x\minus{}f(x))\equal{}2f(x)$

$ 4x\minus{}2f(x)\equal{}2f(x)$

$ f(x)\equal{}x$

Hope it's correct!

[quote="milin"]$ f(f(x)) \plus{} f(x) \equal{} 2x$

$ f(f(f(x))) \plus{} f(f(x)) \equal{} 2f(x)$

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

$ 2(2x \minus{} f(x)) \equal{} 2f(x)$

$ 4x \minus{} 2f(x) \equal{} 2f(x)$

$ f(x) \equal{} x$

Hope it's correct![/quote]

How do you go from $ f(2x \minus{} f(x)) \plus{} 2x \minus{} f(x) \equal{} 2f(x)$   to   $ 2(2x \minus{} f(x)) \equal{} 2f(x)$  :?:
 :cool:

[quote="milin"]$ f(f(x)) \plus{} f(x) \equal{} 2x$

$ f(f(f(x))) \plus{} f(f(x)) \equal{} 2f(x)$

[/quote]

Why does $ f$ need to be distributive?

I had an explanation in my mind when I did this, but I see now that it is wrong.  :(

I think this looks better.

[hide="Solution."]
$ f(f(x)) \plus{} f(x) \equal{} 2x$

If $ x$ is in the domain of $ f$, then $ f(x)$ is also in the domain of $ f$. Thus we can substitute $ f(x)$ for $ x$. We are [i]not[/i] assuming that $ f(x)\equal{}x$:

$ f(f(f(x)))\plus{}f(f(x))\equal{}2f(x)$

Since $ f(f(x))\plus{}f(x)\equal{}2x\implies f(x)\equal{}2x\minus{}f(f(x))$:

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

$ 2x\equal{}4x\minus{}2f(f(x))$

$ f(f(x))\equal{}x$

Substitute this back into the original equation:

$ f(f(x))\plus{}f(x)\equal{}2x\implies x\plus{}f(x)\equal{}2x \implies \boxed{f(x)\equal{}x}$

This easily checks as the solution.
[/hide]

[quote="10000th User"][quote="jvenezuela716"]

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

[/quote]The flaw is in this line.[/quote]

The RHS should be 2x, which makes it an identity.  Instead, $ f(x)$ was substituted for $ x$, so it is no surprise we end up with $ x \equal{} f(x)$.

I think its obvious the function $ f(x) \equal{} x$ works.  The question is are there others?

[hide="ideas"]
$ f(f(x)) \plus{} f(x) \equal{} 2x$
$ f(x) \plus{} x \equal{} 2f^{ \minus{} 1}(x)$
Hence $ f(x) \equal{} 2f^{ \minus{} 1}(x) \plus{} x$ and therefore $ f(f(x)) \plus{} 2f^{ \minus{} 1}(x) \equal{} x$ so $ x \minus{} f(x) \plus{} 2f^{ \minus{} 1}(x) \equal{} 0$

Don't exactly know what to do from here...

If we let $ f$ be a polynomial of degree $ n$, $ f(f(x))$ has degree $ n^2$.  $ f(f(x)) \plus{} f(x)$ can only be less than degree $ n^2$ if the leading terms can cancel, which is only remotely possible if $ n \equal{} n^2$ for which $ n \equal{} 0$ or $ n \equal{} 1$, both trivial cases.

If $ n \equal{} 0$, $ f(x) \equal{} c \implies 2c \equal{} 2x$ which is not true for all $ x$'s.  In the other case, we know that $ n \equal{} 1 \implies f(x) \equal{} ax \plus{} b \implies a(ax \plus{} b) \plus{} b \plus{} ax \plus{} b \equal{} 2x \implies (a^2 \plus{} a)x \plus{} ab \plus{} 2b \equal{} 2x$ and equating coefficients $ a^2 \plus{} a \minus{} 2 \equal{} 0 \implies (a \plus{} 2)(a \minus{} 1) \equal{} 0$ so $ a \equal{} \minus{} 2$ or $ a \equal{} 1$.  Additionally, we have $ ab \plus{} 2b \equal{} 0 \implies b(a \plus{} 2) \equal{} 0$ so either $ b \equal{} 0$ (as in the case of $ a \equal{} 1$) or $ a \equal{} \minus{} 2$, as previously shown.

Hence the only polynomials that satisfy the condition are $ y \equal{} x$ and $ y \equal{} \minus{} 2x \plus{} b$ for all $ b$. Looking back on the conditions, for the second case we must require b>0 and restrict the domain such that $ 0 < \minus{} 2x \plus{} b \implies x < b/2$ so the domain is $ (0,b/2)$ so that the range is $ (0,b)$, while for $ y \equal{} x$ the domain and the range are $ (0,\infty)$.
[/hide]

thanhnam2902, no problem.  However, I think I should remind you that I didn't prove that those were the only solutions but the only polynomial solutions - a pretty important category of functions, but not the only category...

Just a thought: from Taylor series, we know that most functions can be written as polynomials with an infinite number of terms.  Does my work with degrees of polynomials say anything about a Taylor series?

If so, then we can rule out exponential, trig, logarithmic functions, functions with x (or functions of x) raised to rational powers, etc - all functions that can be written as Taylor series.

A less general generalization could deal with extending my work to include all rational functions - certainly doable, though much more complicated.

let a,b,c,d  >0  Find MIN
$ (a^2\plus{}b^2\plus{}10)(c^2\plus{}d^2\plus{}10)\minus{}(a\plus{}b)(c\plus{}d)$

[quote="facis"]
The RHS should be 2x, which makes it an identity.  Instead, $ f(x)$ was substituted for $ x$, so it is no surprise we end up with $ x \equal{} f(x)$.[/quote]

I suppose that is a mistake (still thinking about it), but I don't think your logic is any better:

[quote="facis"]
$ f(f(x)) \plus{} f(x) \equal{} 2x$
$ f(x) \plus{} x \equal{} 2f^{ \minus{} 1}(x)$[/quote]

Are you not also assuming that $ x\equal{}f^{\minus{}1}(x)$? In fact, you are making an even larger assumption that $ f$ has an inverse in the first place, which need not be the case. (Consider $ f(x)\equal{}(x\minus{}2)^2\implies f(1)\equal{}f(3)\equal{}1 \implies f^{\minus{}1}(1)$ does not exist.)

@jvenezuela716:  The point of the reasoning above is more like let $ x\equal{}f(y)$ but then it really doesn't matter, once we substitute this in everywhere in an equation, if we call the variable $ x$ or $ y$ (but it has caused quite some confusion).  I know I did try to make the assumption that $ f$ has an inverse, but i really didn't get anywhere doing it.  I was just trying not to get buried in $ f(f(f(...)))$'s but it didn't help at all.

The mistake is that somewhere in all the attempted proofs that claim $ f(x)\equal{}x$ as the only solution is that somewhere, someone accidentally mistook $ x$ for $ f(x)$ or vice -versa.  For instance, you took $ f(f(x)) \plus{} f(x) \equal{} 2x$ and, thinking of the case where $ f$ had been substituted in for $ x$, proceeded as if $ f(f(x)) \plus{} f(x) \equal{} 2f(x)$, thus replacing $ f(x)$ with $ x$ in one part of the equation but not the whole equation, which basically worked as if you have assumed $ x\equal{}f(x)$ and then showed that it worked. (Specifically, you have $ f(x) \equal{} 2x\minus{}f(f(x))$ and then $ f(f(x))\plus{}2x\minus{}f(f(x))\equal{}2(2x\minus{}f(f(x)))$ which implies $ f(f(x)) \plus{} f(x) \equal{} 2f(x)$, not the original equation).

I don't think it is possible to manipulate just notation with $ f$'s to solve for the function unless we know more about the function, but perhaps different types of functions that f could be - the more general, the better.  For instance, if we were given $ f$ is a polynomial, then $ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}2x\plus{}b$ iff $ b>0$ and the domain and range restrictions I previously showed are the only possiblities.

If there are more solutions besides the polynomial ones, my guess is that they are not continuous anywhere. I'm going to keep working on this problem.

MellowMelon that wouldn't surprise me much either, though I would only go as far to say not differentiable.  I guess the thing to do would be to stick this in a calculus thread with a link here and ask what they think about my Taylor series idea.  Taylor series require infinite differentiability (and must converge), so if no Taylor series is possible, it might (not quite sure what the convergence would mean) rule out infinitely differentiable functions.  Anyway, I'll post a link in one of those forums.

edit: Hope someone comes up with something: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1115411#1115411[/url]

In fact $ f(x) \equal{} x$ is the only solution.

[hide]
Fix $ x$ and just consider the sequence
$ x,f(x),f(f(x)),f(f(f(x))),\ldots$ of positive numbers.
What does the functional equation tell you about this sequence?
[/hide]

@solaris: I think you are missing something.  For instance, consider $ y \equal{} \minus{} 2x \plus{} b$ for all $ b > 0$ with domain $ (0,b/2)$ and range $ (0,b)$.  $ f(x) \equal{} x$ is certainly a solution, but not the only solution.  All the sequence with $ f$'s shows is that we have a sequence of positive numbers.  It is not necessary for this sequence to always be increasing, so I don't really see what's special here.

At the least, I'm going to attempt rational functions.

Let $ f(x) \equal{} p(x)/q(x)$ where $ p$ and $ q$ are polynomials with degrees $ m$ and $ n$ respectively.  As $ x$ appoachs $ \pm \infty$, $ f(x)$ approaches a 'polynomial' with degree $ m \minus{} n$ or a function that behaves like some multiple of $ x^{m \minus{} n}$ ($ n > m$).  Therefore, $ f(f(x)) \equal{} p(p/q)/q(p/q)$ approaches a degree $ (m \minus{} n)^2$ polynomial.  As long as $ m \ge n$ we can only have leading terms of $ f(f(x)) \plus{} f(x)$ cancel to give a linear term if $ (m \minus{} n)^2 \equal{} (m \minus{} n)$ for which $ m \minus{} n \equal{} 1$ or $ m \minus{} n \equal{} 0$ (no linear term here).  So we must consider rational functions such that $ m \equal{} n \plus{} 1$.  In the cases when $ n < m$ there cannot be a linear term in $ f(f(x)) \plus{} f(x)$ so we only need to worry about $ m \equal{} n \plus{} 1$. (The linear cases I proved earlier are given by $ n \equal{} 0$).

Let's define $ g(x) \equal{} \frac {f(x)}x$ . It's easy to see that $ g(x)\equal{}\frac 2{1\plus{}g(f(x))}$ . Let's define the inferior & superior bounds : $ I \equal{}\inf_{x>0}(g(x)) \leq S \equal{}\sup_{x>0}(g(x))$
$ I \leq g(f(x))\leq S\  \Longrightarrow \ \frac 2{1\plus{}I}\geq \frac 2{1\plus{}g(f(x))}\equal{}g(x) \geq \frac 2{1\plus{}S}$ $ \  \Longrightarrow\  \frac 2{1\plus{}I}\geq S\geq I\geq  \frac 2{1\plus{}S}$ $ \  \Longrightarrow\  I\plus{}IS \geq 2\geq   S\plus{}IS$
Finally $ I\equal{}S\equal{}1$ and $ f(x)\equal{}x$ ... do you follow me ?  :cool:

What the heck? Stop making bogus solutions that claim $ f(x)$ must be $ x$. There are other possibilities for $ f$, like $ \minus{} 2x$ plus any constant.

EDIT: Sorry, that was too harsh; the other posts in this topic had already irritated me. You have made some good progress there, but $ I \equal{} S \equal{} \minus{} 2$ is another solution to the last inequality. There are others too, since $ f(x) \equal{} \minus{}2x$ isn't the only other solution.

[quote="MellowMelon"]What the heck? Stop making bogus solutions that claim $ f(x)$ must be $ x$. There are other possibilities for $ f$, like $ \minus{} 2x$ plus any constant.

EDIT: Sorry, that was too harsh; the other posts in this topic had already irritated me. You have made some good progress there, but $ I \equal{} S \equal{} \minus{} 2$ is another solution to the last inequality. There are others too, since $ f(x) \equal{} \minus{} 2x$ isn't the only other solution.[/quote]

That's not possible because of $ f: R^\plus{}\rightarrow R^\plus{}$  :rotfl:

Oh geez, I probably could have solved it if I had seen that. :blush: Well, do you have any ideas if we extend it to $ \mathbb{R}$?

@facis:  well yes, but the given domain is $ (0,\infty)$ and the given range is $ (0,\infty)$, so you
are considering a different problem (which may be more interesting).

@Diogene: I guess my solution goes along similar lines: Fix $ x$ and consider the sequence
$ a_n : \equal{} f^n(x) : \equal{} f(f( \ldots f(x)  \ldots ))$, then the functional equation gives a linear recurrence
for $ a_n$, which can be solved explicitly (in terms of $ a_0$ and $ a_1$), and from the explicit
formula it is evident that the sequence has to contain negative reals if $ a_0 \neq a_1$.

@Mellowmelon: If  you consider the same problem on the whole set of reals,  you get much more
solutions: Consider a base $ B$ of the reals as a vector space  over the rationals, and define $ f$
on all 1-dimensional subspaces spanned by a $ b \in B$ separately (e.g. as $ f(x) \equal{} x$ or as
$ f(x) \equal{} \minus{}2x$). Then the resulting $ f$ satisfies the equation. But somehow I would guess that there
are a lot more solutions than these, so I am slightly pessimistic that one can find all solutions
to your problem if there are no further conditions on $ f$ like continuity or monotonicity.
However, if you have additional conditions a lot depends on what these conditions are.
E.g. if you restrict your attention to functions that are increasing, it is obivous that $ f(x) \equal{} x$
is the only solution. If you would rather like to focus on continuous functions
it is getting much harder, I guess ...

[quote="gagarin"]Find f:R+->R+ such that
f(f(x))+f(x)=2x[/quote]

Let's define $ a_0 \equal{} x$, $ a_1 \equal{} f(x)$, $ a_n \equal{} f(f(...f(x)...))$(n times).
Than the recursive relation holds: $ a_n \plus{} a_{n \minus{} 1} \equal{} 2a_{n \minus{} 2}$. The characteristic equation is $ \alpha^2 \plus{} \alpha \minus{} 2 \equal{} 0$, with solutions $ 1$ and $ \minus{} 2$. So we have $ a_n \equal{} c_1 \plus{} c_2*( \minus{} 2)^n$, with $ c_1,c_2$ being real contstants. But since the co-domain of our function is $ \mathbb{R}^ \plus{}$ obviously $ c_2 \equal{} 0$. So $ a_n \equal{} c_1, \forall n \in \mathbb{N}_0$. And since $ a_0 \equal{} x$, we get $ c_1 \equal{} x$, so $ f(x) \equal{} x$ is the only solution.

[quote="solyaris"]@facis:  well yes, but the given domain is $ (0,\infty)$ and the given range is $ (0,\infty)$, so you
are considering a different problem (which may be more interesting).

@Diogene: I guess my solution goes along similar lines: Fix $ x$ and consider the sequence
$ a_n : \equal{} f^n(x) : \equal{} f(f( \ldots f(x) \ldots ))$, then the functional equation gives a linear recurrence
for $ a_n$, which can be solved explicitly (in terms of $ a_0$ and $ a_1$), and from the explicit
formula it is evident that the sequence has to contain negative reals if $ a_0 \neq a_1$.
[/quote]

Solyaris, I interpreted f:R+->R+ as meaning for all real positive numbers in the range we get a real positive number, not that the domain and range are required to include all real positives.

Behemont, how do you go from $ a_n \plus{} a_{n \minus{} 1} \equal{} 2a_{n \minus{} 2}$ to $ a^2 \plus{} a \minus{} 2 \equal{} 0$?  I'm not very good with finding generating functions.  On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.

Diogene, I do not follow.  Perhaps I just can't think well right now but I do not see where you get your formula for $ g(x)$ based on $ g(f(x))$.

[quote="facis"]
Solyaris, I interpreted f:R+->R+ as meaning for all real positive numbers in the range we get a real positive number, not that the domain and range are required to include all real positives.
[/quote]

I see. Well, the usual interpretation of $ f: A \to B$  is that $ f$ has to be defined on all of $ A$ and
all values have to be in $ B$. (This is what I meant in my post.)

[quote="facis"]
Behemont, how do you go from $ a_n \plus{} a_{n \minus{} 1} \equal{} 2a_{n \minus{} 2}$ to $ a^2 \plus{} a \minus{} 2 \equal{} 0$?  I'm not very good with finding generating functions.  On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.
[/quote]

If you want to avoid the theory of recurrence equations, you can simply rewrite the equation as
$ a_n \minus{} a_{n \minus{} 1} \equal{} \minus{} 2(a_{n \minus{} 1} \minus{} a_{n \minus{} 2})$, and writing $ b_n \equal{} a_n \minus{} a_{n \minus{} 1}$ this gives
$ b_n \equal{} \minus{} 2 b_{n \minus{} 1}$, i.e. $ b_n \equal{} ( \minus{} 2)^{n \minus{} 1} b_1$, so
$ a_n \minus{} a_0 \equal{} b_n \plus{} \ldots \plus{} b_1 \equal{} (( \minus{} 2)^{n \minus{} 1} \plus{} \ldots \plus{} 2^1 \plus{} 2^0) b_1 \equal{} \frac {1 \minus{} ( \minus{} 2)^n}{3}(a_1 \minus{} a_0)$.
As $ a_n > 0$ for all  $ n$ and $ \frac {1 \minus{} ( \minus{} 2)^n}{3}$ can take arbitrarily large positive and negative values, this implies $ 0 \equal{} a_1 \minus{} a_0 \equal{} f(x) \minus{} x$.

There is no need to go "from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$". It is rather
that from a general property of the sequence (positive terms) you conclude something about the
starting values of the sequence $ a_0,a_1$.

solyaris, very well explained.

As for 'going from $ f^n$ back to $ f$' I meant that if we ignore the f:R+->R+ condition (as I had done, since I had never seen this specific notation before) one way to find all possibilities for $ f$ would be to find all functions such that $ f^n(x) \equal{} \frac {1 \minus{} ( \minus{} 2)^n}{3}(f(x) \minus{} x) \plus{} x$.  I see that the format of the answer requires that $ a_1\minus{}a_0\equal{}f(x)\minus{}x\equal{}0$ because the other factor diverges and would guaruntee a negative value for some sufficiently large $ x$ otherwise.

As for "avoiding the theory of recurrence equations" I'm not sure it's necessary, I just haven't heard of it before and I am not very good at finding generating functions (which I believe is what you are referring to) - probably more a lack of experience than inability.  My best guess is that $ a^2\plus{}a\minus{}2\equal{}0$ is being solved to find a partial fraction decomposition.  And then the decomposition is changed into an infinite geometric series, and the coefficients of each term equated to the values of the sequence.  But I have no clue how to get to the expression to do the decomposition on.

[quote="facis"]

Behemont, how do you go from $ a_n \plus{} a_{n \minus{} 1} \equal{} 2a_{n \minus{} 2}$ to $ a^2 \plus{} a \minus{} 2 \equal{} 0$?  I'm not very good with finding generating functions.  On a side note, it appears that your solution could find all possible answers, but we need to consider how to go from the sequence $ a_n$ which is $ f^{n}(x)$ back to $ f(x)$ - that is, how to remove the recursion.

[/quote]

its a known method for solving recurrence equations..for the second question, read my solution once again.. :)

@Behemont: As I said right above your post...

[quote="facis"]
As for 'going from $ f^n$ back to $ f$' I meant that if we ignore the f:R+->R+ condition (as I had done, since I had never seen this specific notation before) one way to find all possibilities for $ f$ would be to find all functions such that $ f^n(x) \equal{} \frac {1 \minus{} ( \minus{} 2)^n}{3}(f(x) \minus{} x) \plus{} x$.
[/quote]

I completely understand how this problem works out with the condition $ f: R\minus{}>R$ but what I meant was assuming we ignore this condition, how could we find all possible solutions?  Thoughts?


[u]xref4search [/u]: closed general form of sequence u_{n + 2} = u_{n + 1} + 6u_{n} + 6n[quote="nkht-tk14"]Given sequence u_{n} determined by
$ u_{0} \equal{} 6;u_{1} \equal{} 42$ and $ u_{n \plus{} 2} \equal{} u_{n \plus{} 1} \plus{} 6u_{n} \plus{} 6n$
Find the general formula of $ u_n$[/quote]

So $ (u_{n\plus{}2}\plus{}n\plus{}2\plus{}\frac 16)\equal{}(u_{n\plus{}1}\plus{}n\plus{}1\plus{}\frac 16)\plus{}6(u_n\plus{}n\plus{}\frac 16)$

So $ u_n\plus{}n\plus{}\frac 16\equal{}\frac{111}{10}3^n\minus{}\frac{74}{15}(\minus{}2)^n$

So $ u_n\equal{}\frac{111}{10}3^n\minus{}\frac{74}{15}(\minus{}2)^n\minus{}n\minus{}\frac 16$

Your answer is incorrect.

Why don't you make a confirmation for $ n\equal{}0$?

[quote="kunny"]Your answer is incorrect.

Why don't you make a confirmation for $ n \equal{} 0$?[/quote]

My anwer is correct.

Why dont you make a confirmation of your counter-examples ?

$ u_n \equal{} \frac {111}{10}3^n \minus{} \frac {74}{15}( \minus{} 2)^n \minus{} n \minus{} \frac 16$

So $ u_0 \equal{} \frac {111}{10} \minus{} \frac {74}{15} \minus{} \frac 16$ $ \equal{} \frac {333}{30} \minus{} \frac {148}{30} \minus{} \frac {5}{30}$ $ \equal{} \frac {333 \minus{} 148 \minus{} 5}{30}$ $ \equal{} \frac {180}{30} \equal{} 6$

I made my confirmations only for $ n\equal{}0$, $ n\equal{}1$, $ n\equal{}2$, $ n\equal{}3$ and $ n\equal{}4$ and considered that is was enough.

Sorry, my calculation iwas incorrrect. :blush:


geometry
circumcircle
trigonometry
geometry proposed

$\triangle ABC$ is acute with altitudes $h_a,h_b,h_c.$ $R,r,r_0$ denote the radii of its circumcircle, incircle and incircle of its orthic triangle. Prove synthetically the relation
\[h_a \plus{} h_b \plus{} h_c \equal{} 2R \plus{} 4r \plus{} r_0 \plus{} \frac{r^2}{R} \]

Hint: If $H$ is the orthocenter, show that $HA+HB+HC=2(R+r).$ Also, keep in mind that  $\odot(HBC),$ $\odot(HCA),$ $\odot(HAB)$ are congruent to $\odot(ABC).$

Hint for a trigonometric solution: Use the identity $r_0=2R \cos A \cos B \cos C.$

$ \|\ \sum h_a = 2R + 4r + r_o + \frac {r^2}{R}\ \|\ \odot\ 2R$ $ \Longleftrightarrow$ $ \|\ \begin{array}{c} \sum bc = 4R^2 + 8Rr + 2Rr_0 + 2r^2 \\
 \\
(\ \begin{array}{c} bc + ca + ab = p^2 + r^2 + 4Rr \\
 \\
r_0 = 2R\cos A\cos B\cos C\end{array}\ )\end{array}\ \|$ $ \Longleftrightarrow$

$ \boxed {\ p^2 = (2R + r)^2 + 2Rr_0\ }$ $ \Longleftrightarrow$ $ \cos A\cos B\cos C=\frac {p^2-(2R+r)^2}{4R^2}\ .$ [b]Nice relation ![/b]

$ \|\ \sum h_a = 2R + 4r + r_o + \frac {r^2}{R}\ \|\ \odot\ 2R$ $ \Longleftrightarrow$ $ \|\ \begin{array}{c} \sum bc = 4R^2 + 8Rr + 2Rr_0 + 2r^2 \\
 \\
(\ \begin{array}{c} bc + ca + ab = p^2 + r^2 + 4Rr \\
 \\
r_0 = 2R\cos A\cos B\cos C\end{array}\ )\end{array}\ \|$ $ \Longleftrightarrow$

$ \boxed {\ p^2 = (2R + r)^2 + 2Rr_0\ }$ $ \Longleftrightarrow$ $ \cos A\cos B\cos C = \frac {p^2 - (2R + r)^2}{4R^2}$ , what is truly.

[b][u]Remark[/u].[/b] A triangle $ ABC$ is nonobtuse $ \Longleftrightarrow\ \sum\sin A\ \ge\ 1+\sum\cos A\ \Longleftrightarrow\ p\ \ge\ 2R+r\ .$

Dear Mathlinkers,

[url= http://jl.ayme.pagesperso-orange.fr/Docs/Trigonometrie.pdf] here [/url]  P. 23...

Sincerely
Jean-Louis


Fifteen is twenty percent of what number?

if 15 is twenty percent of a number then we can write the equation 15 = $ \frac{20x}{100}$ you may simplify this to x/5 but either way you still arrive at x=75

Generally, to solve percent problems, we use this equation:

$ \text{PART} \equal{} \text{PERCENT} \times \text{WHOLE}$

To solve percents, we substitute the numbers we know. So let's use the problem as an example. $ 15$ is the part, and $ 20$ is the percent. We're solving for the whole.

$ 15 \equal{} 0.20 \times \text{WHOLE}$

(Remember that $ 20\%$ is $ 0.20$ or $ 0.2$.)

$ \frac{15}{0.20} \equal{} \text{WHOLE}$

$ 75 \equal{} \text{WHOLE}$

So our answer is $ \boxed{75}$.


Prove that there exists a infinite number of integers $ m$ satisfying $ m! \equal{} 2^p(2k \plus{} 1)$ $ (m,p,k\in \mathbb N)$ and $ m \minus{} p \equal{} 1^2 \plus{} 2^2 \plus{} \cdots \plus{} 2005^2$

Let $ c>0$ be an integer and $ m\equal{}\sum_{i\equal{}1}^c2^{n_i}$, where $ n_1>n_2>\cdots>n_c>0$ are integers.
We can show
$ \sum_{j\equal{}1}^{\infty}[\frac{m}{2^j}]\equal{}\sum_{i\equal{}1}^c(2^{n_i}\minus{}1)\equal{}m\minus{}c$.
Obviously we can let $ c\equal{}1^2\plus{}2^2\plus{}\cdots\plus{}2005^2$ and $ p\equal{}m\minus{}c$.

[quote="ZetaX"]Your titles are completely useless. Use titles that describe the problem! (No, "m,p,k" doesn't, thats why it got that new title by me; same for the other two such topics  :wink: )[/quote]
The current title is really funny. :P  :rotfl: . Do you want to spread consciousness for good naming by it? :D


If a six sided dice is rolled, how many times would we expect to roll it until all numbers appear at least once?What if we extend this to an n sided dice?

[hide="Hint"]
This can be broken into pieces. After rolling one number, on average how many turns will it take to roll a different number? After rolling two different numbers, how long on average will it take to roll a third different number? Continue this.[/hide]
[hide="Answer"]I get $ 6(\frac {1}{6} \plus{} \frac {1}{5} \plus{} \frac {1}{4} \plus{} \frac {1}{3} \plus{} \frac {1}{2} \plus{} \frac {1}{1}) \equal{} 14.7$ The pattern is obvious for the extended problem.[/hide]

Let's say, for example, you've already chosen four different numbers. You still need to choose one of 2, which you have a 2/6=1/3 chance of. Thus, it should take 3=6/2 turns to do this (the rigorous way to prove this is using an infinite series). So the answer is 6/6+6/5+6/4+6/3+6/2+6/1.

[quote="kl2836"]can you show the inifinite series proof?[/quote]

No, the rigorous way does not have to use infinite series. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228627]here[/url].

In fact, the infinite series method is inelegant - who would want to read it? The preferred method is recursion.

Here is the infinite series method anyways, after reading yongyi's post.
Ok, in the example I gave, there is a 2/6 chance it will take one turn, $ 2/6 \cdot 4/6$ chance it will take two, $ 2/6 \cdot (4/6)^2$ chance it will take 3, etc. So it should take $ 1(2/6) \plus{} 2(4/6)(2/6) \plus{} 3(4/6)^2(2/6) \plus{} 4(4/6)^3(2/6)....$ turns on average. Let this equal x. It is clear $ x \minus{} (4/6)x \equal{} x$, from which we find $ x \equal{} 6/2 \equal{} 3$.


Excuse me, anybody knows how to subscribe for Mathematics and Informatics Quarterly magazine, because I've already contacted 2 email address at the website http://olympiads.win.tue.nl but they haven't replied my email until now. If somebody have subscribe this magazine could you tell me how to subscribe that? Thanks

is m&ic a really good magazine? what's in it?

I don't know it, but this is a math magazine for high school student, of course, there's many fun and nice problem. Actually, Bulgaria is a good country in mathematics.


search
floor function
arithmetic sequence
number theory proposed
number theory

For real positive numbers $x,$ the set $A(x)$ is defined by
\[A(x)=\{[nx]\ | \ n\in{ \mathbb{N}}\},\] where $[r]$ denotes the greatest integer not exceeding real numbers $r.$Find all irrational numbers $\alpha >1$ satisfying the following condition.
Condition: If  positive real number $\beta$ satisfies $A(\alpha)\supset A(\beta),$ then $\frac{\beta}{\alpha}$ is integer.

I get a contradictory thing:

Let $\alpha$ be such number. There exist $x\in R$ such that
$\frac{1}{\alpha}+\frac{1}{x}=1$.
Take also irrational numbers $\beta_{n}$ and $\frac{x}{n}$ such that
$\frac{1}{\beta_{n}}+\frac{1}{x/n}=1$.
For any $n \in N$ we have sequence $[kx]$ is in sequence $[k\cdot x/n]$.
From Beatty Theorem both sequences cover $N$ with no overlapping,
therefore sequence $[k\beta_{n}]$ is in sequence $[k\alpha]$.
Thus $\frac{\beta_{n}}{\alpha}$ is an integer for any $n, \text{ }\Rightarrow$
$\frac{x-1}{x}\cdot \frac{x}{x-n}=\frac{x-1}{x-n}$ is an integer for any $n$. The only possibility for that $x=1$, it follows that there does not exist
such $\alpha$. Maybe I misunderstood something.

You have to ensure that $\frac{x}{n}>1$! (which is seldom a case, for any $\alpha >2$ and $n>1$ we have $\frac{x}{n}<1$). But your proof implies that if $\alpha$ exists at all then $\alpha >2$.

You are right Megus, let me try again.

First step:  Let $\alpha$ be such number. There exist $x\in R$ such that $\frac{1}{\alpha}+\frac{1}{x}=1$.
Take also irrational numbers $\beta_{n}$ and $\frac{x}{n}$ such that $\frac{1}{\beta_{n}}+\frac{1}{x/n}=1$.
For any $n \in N$ we have sequence $[kx]$ is in sequence $[k\cdot x/n]$.

From Beatty Theorem both sequences cover $N$ with no overlapping, therefore sequence $[k\beta_{n}]$ is in sequence $[k\alpha]$.
Thus $\frac{\beta_{n}}{\alpha}$ is an integer for any $n, \text{ }\Rightarrow$ $\frac{x-1}{x}\cdot \frac{x}{x-n}=\frac{x-1}{x-n}$ is an integer for any $n \geq 2$.
Contradiction, it follows that $\frac{x}{n}\leq 1$, for $n \geq 2$, thus $x \leq 2$ and therefore $\alpha > 2$.

Second step:  We search all $\alpha >2$, for which there exist $\beta$ such that $A(\beta) \subset A(\alpha)$ and $\frac{\beta}{\alpha}$ is not an integer.
Let $\beta=\alpha(k+x)$ where $k \in \mathbb{Z}$ and $x\in (0,1)$. Observe that for all $n$, $n\beta=kn\alpha+xn\alpha$

Thus $kn\alpha+[xn]\alpha\leq kn\alpha+xn\alpha \leq kn\alpha+([xn]+1)\alpha$ $\text{ }\text{ }\text{ (1)}\text{ }$

and $n\beta$ is between these two consecutive multiples of $\alpha$.

The main condition is not satisfied when we can find $n$ such that

$xn\alpha-[xn]\alpha \geq 1$ and $([xn]+1)\alpha-xn\alpha \geq 1$.

If $x$ is irrational then $\{xn\}$ is dence in $(0,1)$ and there exist $n$ such that $\frac{1}{2}\geq xn-[xn]\geq \frac{1}{\alpha}$
and it satisfies both conditions, as $\alpha > 2$.

Thus $x$ is rational and therefore if there exist such $\beta$, then $\beta/\alpha$ is rational number.

Third step: Let $x=\frac{p}{q}$, where $(p,q)=1$. We want to do not be captured by interval $[\frac{1}{\alpha},\frac{1}{2}]$.

If $q$ is even,  $q =2q_{1}$, then $xn-[xn]=\frac{p}{2q_{1}}-[n\frac{p}{2q_{1}}]$.
Taking $n=q_{1}$ we have thatt $\frac{1}{2}= n\frac{p}{2q_{1}}-[n\frac{p}{2q_{1}}]=\frac{p}{2}-[\frac{p}{2}]$, contradiction.

If $q$ is odd $q =2q_{1}+1$ then we can find $n$ such that $\{\frac{np}{q}\}= \frac{q_{1}}{2q_{1}+1}$.
As we do not want to be captured by the interval $[\frac{1}{2},\frac{1}{\alpha}]$ we need $\frac{1}{\alpha}> \frac{q_{1}}{2q_{1}+1}$, thus $\frac{2q_{1}+1}{q_{1}}>\alpha$.
Because we need the existance of such $q_{1}$ and $q_{1}\geq 1$, there might exist a construction for $\beta$ iff $\alpha < 3$.

Fourth step: $\frac{\beta}{\alpha}=\frac{u}{v}$, where $(u,v)=1$ and we know $v$ is odd, $\Rightarrow$ $v\geq 3 > \alpha$.

Consider $l$, inverse of $u$ modulo $v$, $0< l < v$ and $lu=mv+1$, where $m$ is fixed.
Consider numbers $n$ of the form $n=kv+l$, then
$n\beta=(kv+l)\frac{u}{v}\alpha=(k+m)\alpha+\frac{1}{v}\alpha$
As $\alpha$ is irrational and $\{x\alpha\}$, for $x \in \mathbb{Z}$ is dense in $(0,1)$ there exist $k$ such that $1-\frac{\alpha}{v}\leq \{(k+m)\alpha\}\leq 1$, then
$[(k+m)\alpha] < [(k+m)\alpha+\frac{1}{v}\alpha]=[n\beta]$
Also $[n\beta] < [(k+m+1)\alpha]$ because $(k+m+1)\alpha-\left((k+m)\alpha+\frac{1}{v}\alpha\right)=\frac{v-1}{v}\alpha \geq 1$, ($v \geq 2, \alpha > 2$).

It follows that all irrational numbers $\alpha > 2$ are good. It means that if there exist $\beta$ such that $A(\beta) \subset A(\alpha)$ then $\frac{\beta}{\alpha}$ is an integer.

I think there is a simpler way to show that for $\alpha>2$, $A(\beta)\subseteq A(\alpha)\implies\beta/\alpha\in\mathbb{Z}$.

Assume for the sake of contradiction that $\beta/\alpha\notin\mathbb{Z}$ but $A(\beta)\subseteq A(\alpha)$; then for every $n\in\mathbb{N}$, there exists $m$ such that $\lfloor{m\alpha}\rfloor=\lfloor{n\beta}\rfloor$, or equivalently, $[\lfloor{n\beta}\rfloor/\alpha,\lfloor{n\beta+1}\rfloor/\alpha)$ contains an integer for every $n$. (*)

Weakening (*) a bit, this means $((n\beta-1)/\alpha,(n\beta+1)/\alpha)$ has an integer for all $n\ge1$. If $\beta/\alpha$ is irrational, then since $2/\alpha<1$, Kronecker's density theorem yields a contradiction.

On the other hand, if $\beta/\alpha=p/q$ for some coprime positive integers $p,q$ with $q>1$, then $\beta$ is irrational, then by Bézout's identity we can find some arithmetic progression $P=\{qm+r\}_{m\ge0}$ such that for every $n\in P$, we have $\{n(\beta/\alpha)\}=1/q\le1/2<1-1/\alpha$. But $q\beta$ is irrational, so by Kronecker's density theorem we can get $\{qm\beta\}\in[\{-r\beta\},\{-r\beta\}+\epsilon)$ for any $\epsilon>0$ and thus $\{n\beta\}\in[0,\epsilon)$, whence there exists $n\in P$ such that $0<\{n(\beta/\alpha)-\{n\beta\}/\alpha\}\le1-1/\alpha$, contradicting (*).

Does anyone know which Bezout's identity math154 is referring to, or provide a link? Thanks.

Also, does anyone know any motivations as to math154 other than "experience", because I do not know how understanding the solution to this problem would help with solving problems on another math olympiad; i.e., I do not get the reasons for math154 to use Kronecker's density theorem and Bezout's theorem which I have yet to understand. To me, his solution just looks like a variable bash, and I do not see any pattern to his reasoning. Can someone explain any possible motivation for his solution? Thanks!


[quote="TripleM"]Let a,b,c be nonnegative real numbers. Prove that
ab+bc+ca :ge: :sqrt:(3abc(a+b+c)).

Let a,b,c be nonnegative real numbers such that a+b+c = 1. Prove that
a:^2:+b:^2:+c:^2:+:sqrt:(12abc):le:1.[/quote]

the first one looked intimidating but was really not that bad...
working backwards, we square it to get a:^2:b:^2:+a:^2:c:^2:+b:^2:c:^2:+2ab:^2:c+2abc:^2:+2a:^2:bc :ge:3ab:^2:c+3abc:^2:+3a:^2:bc , which works out to be a:^2:b:^2:+a:^2:c:^2:+b:^2:c:^2:-ab:^2:c-abc:^2:-a:^2:bc:ge:0, which is true because it is in fact weaker than expanding the trivial inequality (ab+bc-ca):^2:+(ab-bc+ca):^2:+(-ab+bc+ca):^2::ge:0

[quote="darkquantum"]For number2, using AM-GM, we get that max of abc is 1/27, attained when a=b=c=1/3.

This means that a^2+b^2+c^2<=1/3.
Since the max of the LHS is attained when max(abc) is attained, the max of a^2+b^2+c^2=1/3.
Thus, the original one holds.[/quote]

why must a^2+b^2+c^2:le:1/3? If a=1 and b=c=0, then the LHS is 1 which is bigger than 1/3.

....so? you're not trying to prove it for max abc, you're trying to prove it for all (a,b,c) as long as a+b+c=1.

You have now successfully confused me.

errr
perhaps some revision is in order...

for the 2nd one :

[hide]

squaring a+b+c=1 and using first inequality doesn't seem to be a bad idea

[/hide]

yup..you're right..that solves it. On the second one, I had gotten up to ab+bc+ca:ge::sqrt:(3abc) and then I sorta forgot all about the first inequality (which solves it since a+b+c=1).


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $ G$ is a finite group and H is a subgroup of G and [G:H]=n>1. Prove that either G has a non-trivial normal subgroup whose index divides $ n!$ or G is isomorphic to a subgroup of $ S_n$.

I think you want: ... whose index divides $ n!$ and is not equal to $ n!$ ...

consider the natural map $ G \to Sym(G/H)$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=19724]here[/url] for a proof that $ K\equal{}\bigcap_{g\in G}(gHg^{\minus{}1})$ is a normal subgroup, and $ G/K$ can be identified with a subgroup of $ S_n$.


How many three-digit odd integers are greater than 299?

the first three-digit odd integer greater than 299 is 301
the last three-digit odd integer greater than 299 is 999

therefore, we have a list of odd integers: 301,303,305...995, 997, 999

how do we count these???
well, we know that the interval between each of these numbers is two, so let's divide by two
however, if we do that, we will deal with decimals, and i don't like those in counting problems
so let's add one to each of them first

now we have the list 302,304,306....996,998,1000
now we divide by 2 to get the list 151, 152, 153....498, 499, 500

we can subtract 150 from each term to get: 1, 2, 3......348, 349, 350

therefore, there are $ \boxed{350}$ 3-digit integers greater than 299


Are there gift certificates/cards for the Aops bookstore?  nice present for math teachers...

Currently, you can only order these by phone.  Eventually, we'll have a system that allows you to order them online.


Find all prime numbers $p,q$ and even $n>2$ such that :

$p^n+p^{n-1}+...+p+1 = q^2+q+1$   .

See the topic Primes http://www.mathlinks.ro/Forum/viewtopic.php?t=29570


let $a_{i},b_{i}\in \mathbb{R}\ \ ,\ 1\le i\le n$ be constant numbers find Max and Min of :

$y=b_{i}|x-{a}_{i}|+b_{2}|x-{a}_{2}|+\dots+b_{n}|x-{a}_{n}|$ that $x,y \in \mathbb{R}$

Oh sorry, I did not pay close attention. :blush:

[quote="K81o7"]It seems as far as the maximum gose, you could make it as large as you wanted...
As for the minimum, it seems to be at the weighted arithmetic mean of the $a_{i}$, ie $\frac{a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}}{b_{1}+b_{2}+...+b_{n}}$?  Maybe I'm not understanding it...or I'm just wrong.[/quote]

You are wrong Consider $b_{i}<0\ ,\ i=1,2,...,n$ and you can not make  it as large as you want.

[quote="ali666"]let $f(x)=b_{1}|x-a_{1}|+...+b_{n}|x-a_{n}|$
and S={$f(-\infty),f(a_{1}),...,f(a_{n}),f(\infty)$} then
 Max(f)=Max(S) ,Min(f)=Min(S)[/quote]

can you prove it please?


geometry perimeter ratio trigonometry quadratics function algebra

At some point or another, we were all beginning problem solvers. Yet almost every post (even in this forum) contains questions that are inaccessible, at least for the time being, to true beginning problem solvers. Therefore I have decided to post some problems (in approximate increasing order of difficulty) that I feel will be of reasonable difficulty to a student of this level. I encourage all true beginners (especially those who may only frequent the non-math forums on this site) to attempt some of these.

1. If [tex]30x-18=72[/tex], [tex]xy+6y=72[/tex], and [tex]xyz=72[/tex], find [tex]x+y+z[/tex].

2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square.

3. What is the positive difference between the 2004th positive odd integer and the 2004th negative odd integer?

4. A side of a regular hexagon has length 1. Find the area of the hexagon.

5. (a) If [tex](x+1)(x-1)=80[/tex], find [tex]x^2[/tex].
(b) If [tex](x+1)(x-1)=22[/tex], find [tex]x^2[/tex].

6. Find the sum of the first 15 positive odd integers.

7. (more challenging) The ratio of a circle's circumference to its area is 1:7. What is the radius of the circle?

[quote="JSRosen3"]At some point or another, we were all beginning problem solvers.[/quote]

And I mostly still am a beginning problem solver, because I don't get in enough practice time. Thanks for sharing problems at a good level for beginners.

Anyone want to give these a shot? Even if you can't completely solve them, you can post your work so far and someone can give you a hint to help you finish it.

JSRosen3 wrote:2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square.

2. My solution will be in spoiler, of course.

[hide]A problem-solver learns to think "Pythagorean theorem" as soon as the term "right triangle" is mentioned. This problem is asking about a triangle with one right angle (that is, 90 degree angle) in it. The two sides next to that angle are called "legs." The longest side of the triangle, which is opposite the right angle, is called the hypotenuse (Greek for "thing hanging below"). The famous theorem of Pythagoras says that the area of the squares on the two legs adds up to the area of the square on the hypotenuse.

One leg has length 13. A square constructed on that leg will have area 13 times 13, or 169. The other leg has length 15, so the square has area 15 times 15, or 225. Adding 169 to 225, we find the area of the square on the hypotenuse is 394, which is our answer for part (b) of the question. The length of the hypotenuse, then, is the square root of 394, which is approximately equal to 19.8494 units. [/hide]

Is that process of solution clear? This is definitely within the scope of Getting Started Forum.

I'll try to insert a diagram here.

Yes, and it's also very thorough.  I think I'll start doing that too so that beginners looking at solutions can understand them.  Another thing to note:

[hide]An immediate result of the Pythagorean Theorem is that if a and b represent the lengths of the legs, and c represents the length of the hypotenuse, then a^2 + b^2 = c^2.  Thus, we do not have to think strictly geometrically.  If you look at the left of the screen below the navigation, there will sometimes be an animated geometric proof of the Pythagorean Theorem[/hide]

As a beginner, I'll give a shot. Just one easy one:

 [hide]6. Find the sum of the first 15 positive odd integers. Basically, it's 1+3+5+..... but rather than doing all these, we can simply write this as x+x+2+x+4+x+6 so, just find 2+4+6+ to +28 which is 210.  Then we have 11x's where x is 1. So, 11+210 = 221.  I hope this is right because it's kinda shame if I get this easy one wrong.  Plus, tell me if there's easier way.

[/hide]

You've got the right idea, but you made some calculation errors.  BTW, there's no shame whatsoever involved here.  I get Getting Started stuff wrong all the time due to small errors.  If you want an easier way, you can look at the little animated figure under the navigation on your left right here on AoPS.  Occasionally it will show a formula for the sum of the first so many odd numbers.

Good work so far, Silverfalcon. You're almost there. The only mistake you made was that [hide]you said there were 11 x's rather than 15.[/hide] Fix that and you should be good to go.

Another way to do this is [hide]to start small, namely by adding the first 2 odd integers, then the first 3 odd integers, etc. and looking for a pattern. You'll find one pretty quickly[/hide].

3.

[hide]

Since all even numbers are multiples of two, we can expect all even number to be in the form 2n.  For example, 1*2 = 2, 2*2 = 4, 3*2 = 6, etc.  All odds are one less than evens.  Thus the odds are 2n-1.  Note that we don't use 2n+1, because that would make the first odd number 3, which is wrong.

Thus the 2004th odd number is 4007.  The negative of the 2004 odd number is -4007.  4007-(-4007) = 4007+4007=8014

[/hide]

probability1.01 wrote:I get Getting Started stuff wrong all the time due to small errors.

One of my worst getting started mistakes was saying that 2/2 = 0.

1.

[hide]

30x - 18 = 72.  Adding 18 to both sides and dividing by thirty gives x = 3.

Plugging x = 3 into the next equation gives 9y = 72, or y =8.  Putting three and 8 into the last one gives 24z = 72.  Dividing by 24 gives z = 3.  3+8+3 = 14.

[/hide]

5.

[hide]

One very useful factorization is called the difference of squares.  If we have x^2-y^2, it can be factored as (x-y)(x+y)  We can expand (x-1)(x+1) and get x^2 - 1.

(a).  x^2 - 1 = 80

       x^2 = 81

(b).  x^2 - 1 = 22

       x^2 = 23

Recognizing  these factorizations is important because, as you see for part b, guessing and testing doesn't help since x was not a rational  number, but x^2 is, and x^2 was what the question asked for.[/hide]

I commend JSRosen3 for the wonderful level of these problems. Also, all of the solutions have been very clear and have avoided 'hand wavy' stuff (hand waving is bad, so try not to do it). I did find one problem though. Once you think you have solved a problem, [b]go back and look to see what was being asked[/b]. For #2 that tokenadult solved, he did everything correctly up until the very end. If he had went back and reread the problem, he would have seen that part (a) was asking for the [b]perimeter[/b] of the square, not the sidelength. I'm just trying to stress the importance of answering what was being asked. Good job and keep it up!

7.

[hide]

The circumference of a circle is given by the formula C = d*pi where d is the diameter, or C = 2*pi*r, where r is the radius.  The area is pi*r^2.  We want the ratio of these two quantities to be 1/7.  Let's start with (2r*pi)/(pi*r^2) and start simplifying.  The pi's on the numerator and the denominators cancel out.  2r/r^2 becoms 2/r because the r's cancel out.  Now we just solve for r in the equation 1/7 = 2/r.  Multiplying both sides by 7r gives r = 14.[/hide]

[quote="probability1.01"]I suggest some other people try these problems too.  So far, it's been nr1337, Silverfalcon, and tokenadult.  Can't think of any others, but BHorseMath agreed to do some problems in a post on the F&G forum :twisted:[/quote]

Sorry - I had to do something to fight my boredom today!

1. [hide]If 30x-18=72, x=3. So xy+6y=72 can be rewritten as 3y+6y=72, or y=8. That means xyz=72 is equal to 3*8z=72, or z=3. x+y+z=3+8+3=14.[/hide]

2. [hide]Using the Pythagorean Theorem, the hypotenuse is (13^2+15^2)^1/2, or  :sqrt: 394. The perimeter of the square is 4 :sqrt: 394, and the area is ( :sqrt: 394)^2=394.[/hide]

3. [hide]2004th positive odd integer=2004*2+1=4009

    2004th negative odd integer=-2004*2-1=-4009

    4009-(-4009)=4009+4009=8018[/hide]

5. [hide](x+1)(x-1) can be multiplied out into x^2-1. If x^2-1=80, x^2=81. If x^2-1=22, x^2=23.[/hide]

6. [hide][15(1+31)]/2=240[/hide]

7. [hide]2 :pi: r: :pi: r^2=1:7

    7*2 :pi: r= :pi: r^2

    14=r[/hide]

I don't know why but I cannot get number 4.

4.

[hide](Sorry, I don't have Geometer's Sketchpad so I'll explain as clearly as I can with text)

Divide the hexagon into six equilateral triangles by connecting the center of the hexagon with each vertex.  Now, focus on one of the triangles.  The central angle of the triangle (the angle containing the center of hexagon) is 60 degrees.  (Because there are 6 triangles, and the total number of degrees in the center is 360).

Now, draw the altitude of the triangle, which is also the apothem of the polygon.  The altitude/apothem divides the base of the equilateral triangle in two, forming two right triangles that have a base of 0.5.  Notice the right triangles are 30-60-90 triangles, so the apothem will be  :rt3: / 2

Then, following the formula of the area of a regular polygon, 1/2 aP, subsitute in the values.  (:rt3: / 2 for apothem, 6 for perimeter).  The area of the regular hexagon is (3 :rt3:) / 2.[/hide]

LynnelleYe,

On #3, the 2004th positive odd integer is not [hide]4009[/hide], because the nth positive odd integer is [hide]2n-1[/hide], not [hide]2n+1[/hide]. Same thing for #6. You had the right idea on both, though.

Revision for tolkenadult's number 2:

[hide]He has not finished the question because  :sqrt: 394 is only the side length. Therefore, correct answer is 4  :sqrt: 394.[/hide]

My way of doing number 4:

[hide]If you know the formula for the area of a hexagon, which is 6s^2 :rt3: /4 where s=side of the hexagon, you could just plug 1 in for s and get 3 :rt3: /2.  If you didn't know the formula, you could divide the hexagon up into 6 equilateral triangles like one of the posters above me did.[/hide][/hide]

Finally somebody posted what my teacher calls the 1-2-3-4 formula. It's for the area of an equilateral triangle, and it's a lot quicker than doing everything with the apothem.

1-1 variable
2-to the power of 2
3-multiplied by  :rt3:
4-divided by 4

so, the formula is, with s being the side of the eqiuilateral triangle, s:^2:*:rt3: /4

Then, for a regular hexagon, you just multiply that result by 6 for the 6 triangles that make up a hexagon.

Another useful formula is the general formula for the area of a regular n-gon with n sides and side length x:

[tex]A=\frac{nx^{2}}{4\tan \frac{180}{n}}[/tex]

So if you plugged in 6 for n, you'd get that the area of a regular hexagon is

[tex]\frac{6x^{2}}{4\tan \frac{180}{6}}=\frac{6x^{2}}{4\tan 30}=\frac{6x^{2}}{4(\frac{1}{\sqrt{3}})}=\frac{6x^{2}\sqrt{3}}{4}=\frac{3x^{2}\sqrt{3}}{2}[/tex].

The derivation of the general formula is a good exercise in trig identities. Anyone want to give it a shot?

Doesn't the trig seem a little non-beginner-ish?

The trig might be a little advanced, but if you know the identities it's definitely doable.

While we're talking about trig, what's tan[36]?  My brother has been wanting me to find the area of a regular pentagon for a while, but I need to know tan[36] (and NOT a deciimal approximation).

[tex]\tan{36}=\sqrt{5-2\sqrt{5}}[/tex]

I'm trying to think of a good way to derive this. Using quintuple-angle formulas is probably not such a good idea  :D

To find tan (or anything else) 36, consider

[tex]x^4+x^3+x^2+x+1=0.[/tex]

One way to find the roots is to divide by [tex]x^2[/tex] so that you get a quadratic in x+1/x. Another way is to multiply by x-1 to get [tex]x^5-1=0[/tex]. Find the roots each way and equate real and imaginary parts.

Okay, I haven't taken trig (I just finished geometry), but i have had a minute exposure to sin cos tan, but i have no idea what anything beyond that is. What are identities and quintuple-angle formulas, and how did you get that for tan 36? :shock:

Trig identities are just identities for trig functions; i.e., they are always true. Here are a few examples:

[tex]\sin{2\theta}=2\sin{\theta}\cos{\theta}[/tex]

[tex]\cos{(\alpha+\beta)}=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}[/tex]

So, if you wanted to find [tex]\cos{105}[/tex], you could write it as [tex]\cos{(60+45)}[/tex], and then use the identity above to obtain [tex]\displaystyle \cos{105} = \cos{(60+45)} = \cos{60} \cos{45} - \sin{60} \sin{45}[/tex] = [tex](\frac{1}{2} \times \frac{\sqrt{2}}{2}) - (\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}) = \frac{\sqrt{2} - \sqrt{6}}{4}[/tex].

Quintuple-angle formulas are expressions for things like [tex]\tan{(5\theta)}[/tex] in terms of only [tex]\tan{\theta}[/tex]. Since [tex]\tan{180}[/tex] can be expressed as [tex]\tan{(5 \times 36)}[/tex], we can use a quintuple-angle identity for the tangent function to find [tex]\tan{36}[/tex] in terms of [tex]\tan{180}[/tex], which we know to be 0. It would be very messy, though, so I'd look for an alternative (perhaps geometric?) method before I plowed into quintics.

Identities are things true for any angle.  For example,
[tex]
\sin^2 \theta + \cos^2 \theta = 1

\cos^2 \theta = \frac{1 + cos 2\theta}{2}
[/tex]

When he said quintuple ange formulas, that means a formula for (some trig function) of 5x.  For example,
[tex]
\cos 5\theta = \cos^5 \theta - 10\cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta
[/tex]

And now we are WAY beyond  getting started.

Yeah, we're a little beyond the realm of Getting Started, although I don't think that all trig is necessarily beyond it. Perhaps a mod would be willing to post an introduction to trig, just like the one for arithmetic sequences? The one for arithmetic sequences was excellent.

Just for fun,
[tex]
\displaystyle\tan 5\theta = \frac{5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}{\cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta}
[/tex]

Gah!  :o That's frightening. That's why I'm looking for a geometric way to find [tex]\tan{36}[/tex].

Did you look at my method? You can find it that way.

Using ComplexZeta's method:

[tex]x^4 + x^3 + x^2 + x + 1 = 0[/tex]

[tex]x^2 + x + 1 + \frac{1}{x} + \frac{1}{x^2} = 0[/tex]

Let [tex]y=x+\frac{1}{x}[/tex]. Then [tex]y^2 + y - 1 = 0[/tex], so [tex]y = \frac{-1 \pm \sqrt{5}}{2}[/tex].

Recall that [tex]y=x+\frac{1}{x}[/tex]. Rearranging:

[tex]x + \frac{1}{x} = \frac{-1 \pm \sqrt{5}}{2}[/tex]

[tex]2x^2 + 2 = -x \pm x\sqrt{5}[/tex]

[tex]2x^2 + (1 \pm \sqrt{5})x + 2 = 0[/tex]

Using the Quadratic Formula, the four solutions are:

[tex]\displaystyle x_1 = \frac{\sqrt{5}-1}{4} + \frac{\sqrt{2(\sqrt{5}+5)}}{4} \times i[/tex]

[tex]\displaystyle x_2 = \frac{\sqrt{5}-1}{4} - \frac{\sqrt{2(\sqrt{5}+5)}}{4} \times i[/tex]

[tex]\displaystyle x_3 = \frac{-\sqrt{5}-1}{4} + \frac{\sqrt{-2(\sqrt{5}-5)}}{4} \times i[/tex]

[tex]\displaystyle x_4 = \frac{-\sqrt{5}-1}{4} - \frac{\sqrt{-2(\sqrt{5}-5)}}{4} \times i[/tex]

Let [tex]P(x) = x^4 + x^3 + x^2 + x + 1[/tex], and let [tex]Q(x)= (x-1)P(x) = x^5-1[/tex]. The roots of [tex]P(x)[/tex], are the same as the roots of [tex]Q(x)[/tex], disregarding the root [tex]x=1[/tex], since we introduced it when we multiplied [tex]P(x)[/tex] by [tex]x-1[/tex]. The roots of [tex]Q(x)[/tex] are the fifth roots of unity, which are (excluding 1):

[tex]\cos{72}+i(\sin{72})[/tex]

[tex]\cos{144}+i(\sin{144})[/tex]

[tex]\cos{216}+i(\sin{216})[/tex]

[tex]\cos{288}+i(\sin{288})[/tex]

Now, set one of these roots equal to its corresponding root that was found above. Since we want [tex]\tan{36}[/tex], choose the second root (since 144 and 36 are supplementary). Since this root is in the second quadrant, we need to choose the root above that is also in the second quadrant, namely [tex]x_3[/tex]. So, evaluating real and imaginary parts seperately, we have:

[tex]\cos{144} = \displaystyle \frac{-\sqrt{5}-1}{4}[/tex].

[tex]\cos{36} = -\cos{144} = \displaystyle \frac{\sqrt{5} + 1}{4}[/tex].

[tex]i \sin{144} = \displaystyle \frac{\sqrt{-2(\sqrt{5}-5)}}{4} \times i[/tex]

[tex]\sin{36} = \sin{144} = \displaystyle \frac{\sqrt{-2(\sqrt{5}-5)}}{4} = \displaystyle \frac{\sqrt{2(5-\sqrt{5})}}{4}[/tex].

And finally:

[tex]\tan{36} = \displaystyle \frac{\sin{36}}{\cos{36}} = \frac{\frac{\sqrt{2(5-\sqrt{5})}}{4}}{\frac{\sqrt{5} + 1}{4}} = \frac{\sqrt{2(5-\sqrt{5})}}{\sqrt{5}+1}[/tex].

(also, if a mod wants to move this discussion to intermediate, feel free to do so--but please don't move the whole thread, just the recent discussion of identities, complex numbers, etc.)

[i]P.S. Thanks for showing me how to [tex]\TeX[/tex] the plus/minus, ComplexZeta.[/i]

I understand why the method of finding the roots of the polynomial in two different ways and equating them works. But how would we have thought of this in the first place? How can we train ourselves to think [tex]x^4 + x^3 + x^2 + x + 1[/tex] when we hear [tex]\tan 36[/tex]?

You might think about 5th roots of unity when you want to know (insert trig function here)(insert multiple of 18 here). Then you might consider [tex]x^5-1[/tex]. Then you might get rid of the root you already know, and voila!

Sadly, this method gets very messy when we have 360/some big number.  I tried to find cos 360/17 like Gauss did using this method and didn't get very far very quickly.

[spoiler]The ratio of a circle's circumference to its area is 1:7. What is the radius of the circle?

Area = pi.r^2
circumference = 2pi.r

1/7 = circumference/area (using ratios I guess)
1/7 = (2pi.r)/(pi.r^2)
1/7 = 2/r
r = 14?[/spoiler]

I liked that. Good problem, took some thinking because I haven't seen one like that before.

Here are my answers:

[hide]

1.If 30x - 18 = 72  xy + 6y = 72  xyz = 72 what is x+y+z?

30x -18 = 72

30x = 90

x = 3

3y + 6y = 72

9y = 72

y = 8

3*8*z = 72

24z = 72

z = 3

x+y+z = 3+3+8 = 14

2. A right triangle has legs 13 and 15. The hypotenuse of the triangle is also the side of a square. Find (a) the perimeter of the square, and (b) the area of the square.

a :^2:  + b:^2:  = c:^2:

13:^2:  + 15:^2:  = c:^2:

169 + 225 = c:^2:

c:^2:  = 394

c =  :sqrt: 394

a =  :sqrt: 394*4

a  :approx: 79.40

b = :sqrt: 394 :^2:

b = 394

3. What is the positive difference between the 2004th positive odd integer and the 2004th negative odd integer?

nth positive odd integer = 2(n-1) + 1

2004th positive odd integer = 4007

nth negative odd integer = -(2(n-1) +1) = -(nth positive odd integer)

2004th negatice odd integer = -4007

4007 - (-4007) = 4007*2 = 8014

8014

[/hide]

I'll try the rest in a couple days

Oh yeah, thanks for the easier problems.


Prove that \[ a^rb^s\plus{}b^rc^s\plus{}c^ra^s>a^sb^r\plus{}b^sc^r\plus{}c^sa^r\] unde \[ a>b>c>0, r>s>0\]

Let $ a=\alpha c , b=\beta c, \alpha>\beta>1$,and the inequality becomes  after simplyfing with $ c^{r+s}$, \[ \alpha^r \beta^s+ \beta^r +\alpha^s > \alpha^s \beta^r +\beta^s +\alpha^r\] equivalent to \[ \frac{\alpha^r -1}{\alpha^s -1}> \frac{\beta^r-1}{\beta^s-1}\] and if $ \alpha^s=u and \beta^s=v, u>v>1, r\s=t>1$.The inequality becomes \[ \frac{u^t-1}{u-1}> \frac{v^t-1}{v-1}\] and to prove this one we have to prove that $ f(x)=(x^t-1)/(x-1) \nearrow$ with $ x>1$.We have \[ f'(x)=\frac{tx^{t-1}(x-1)-(x^t-1)}{(x-1)^2}=\frac{(t-1)x^t-tx^{t-1}+1}{(x-1)^2}\] To have $ f'(x)>0$ we have to have \[ g(x)=(t-1)x^t-tx^{t-1}+1>0\] for any x>1.
But $ g'(x)=t(t-1)x^{t-2}(x-1)>0$ it results that $ g \nearrow$ and then $ g(x)>lim_{x \rightarrow 0,x>0}g(x)=0$ and the inequality is proved.


The sequence $ a_n$ is defined as follows: $ a_0\equal{}1, a_1\equal{}1, a_{n\plus{}1}\equal{}\frac{1\plus{}a_{n}^2}{a_{n\minus{}1}}$.
Prove that all the terms of the sequence are integers.

[quote="bambaman"]The sequence $ a_n$ is defined as follows: $ a_0 \equal{} 1, a_1 \equal{} 1, a_{n \plus{} 1} \equal{} \frac {1 \plus{} a_{n}^2}{a_{n \minus{} 1}}$.
Prove that all the terms of the sequence are integers.[/quote]
[hide="Solution"]
We proceed with induction.  $ a_{2}\equal{}2$, so $ a_2$ is an integer (as are $ a_0$ and $ a_1$).  Now, assume that $ a_i$ is an integer for $ 1\le i\le n$.  We have that $ a_{n\plus{}1}\equal{}\frac{1\plus{}(\frac{a_{n\minus{}1}^2\plus{}1}{a_{n\minus{}2}})^2}{a_{n\minus{}1}}$.  Upon multiplying out, we have that $ a_{n\plus{}1}\equal{}\frac{a_{n\minus{}2}^2\plus{}a_{n\minus{}1}^4\plus{}2a_{n\minus{}1}^2\plus{}1}{a_{n\minus{}2}^2a_{n\minus{}1}}$.  Now, $ a_{n\minus{}1}$ and $ a_{n\minus{}2}$ are coprime since $ a_{n\minus{}1}^2\plus{}1\equal{}a_{n\minus{}2}a_n$.  Thus, it suffices to show that $ a_{n\minus{}2}^2|(a_{n\minus{}2}^2\plus{}a_{n\minus{}1}^4\plus{}2a_{n\minus{}1}^2\plus{}1)$ (which is clearly true since $ a_{n\minus{}2}^2|(a_{n\minus{}1}^2\plus{}1)^2$) and $ a_{n\minus{}1}|(a_{n\minus{}2}^2\plus{}a_{n\minus{}1}^4\plus{}2a_{n\minus{}1}^2\plus{}1)$ (which is clearly true since $ a_{n\minus{}1}|(a_{n\minus{}2}^2\plus{}1)$).  Since both are proven, $ a_{n\plus{}1}$ is an integer, so induction is complete. [/hide]

Together,
\[ a_{n\plus{}1}a_{n\minus{}1}\equal{}1\plus{}a_n^2\wedge a_{n\plus{}2}a_{n}\equal{}1\plus{}a_{n\plus{}1}^2\implies \frac{a_{n\plus{}1}}{a_n}\equal{}\frac{a_{n\plus{}2}\plus{}a_n}{a_{n\plus{}1}\plus{}a_{n\minus{}1}}\]

Telescoping, we find that \[ a_{n\plus{}1}\equal{}\frac{a_{n\plus{}2}\plus{}a_n}{3}\iff a_{n\plus{}2}\equal{}3a_{n\plus{}1}\minus{}a_n\] hence the conclusion.


inequalities
integration
calculus
abstract algebra

yes, that's obviously true. But what's the joke? :|

Yes, the joke is not only that abc = 1 is a useless condition, but also that the inequality is completely trivial:

For every real number x, we have $\displaystyle \frac{x}{x^2+1} \leq \frac12$.

But I wonder why Maverick posts this in the "Unsolved Problems" section when it's a real joke.

  Darij

[quote="siuhochung"][quote="fuzzylogic"]Even a,b,c > 0 are unnecessary, right?[/quote]

I think so.[/quote]

Yes, because (x-1)^2 implies x/(x^2 + 1) <= 1/2 which obviously holds for any x.

[quote="Peter VDD"]This starts getting funny :D

Hey why don't we make a contest "Weakest inequality of the month" :maybe:[/quote]
 :D
that's interesting

[quote="Peter VDD"]This starts getting funny :D

Hey why don't we make a contest "Weakest inequality of the month" :maybe:[/quote]

Given a >= b

Prove 2a >= 2b

yeah, but that one is rather cheap.

One that takes you a few seconds before you see it :D

[quote="blahblahblah"]In triangle abc, prove that:
a^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.[/quote]

LOL??  :shock: :shock: :shock:

Is that a funny ineq for which conditions are too much??  I was going to send that one in for our national junior olympiad... didn't even know it was well-known or so :? but dEUS already told me the same...

[quote="blahblahblah"]In triangle abc, prove that:

a^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.[/quote]

Hmm, this actually rewrites as

$a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+c^{2}a-\left( a^{3}+b^{3}+c^{3}+2abc\right) \geq 0$,

or as

$\left( b+c-a\right) \left( c+a-b\right) \left( a+b-c\right) \geq 0$,

what is pretty clear for a triangle abc. Although I wouldn't call it a joke. Just one thing: If you say that abc is a "real" triangle, then you can replace the $\geq$ sign by a strict > sign.

  Darij

Yes, but that rewriting is pretty much impossible for pupils having rarely worked with algebra, so I found some more straightforward way to convert the thing to proving that c/(a+b) < 1, which is obvious. :)

Surprises me that this is soo all-round known :P I really never heard it anywhere before.

Speaking of jokes...
What's $\displaystyle{\int_1^{cabin}dx/x}$

Peter VDD wrote:Yes, but that rewriting is pretty much impossible for pupils having rarely worked with algebra, so I found some more straightforward way to convert the thing to proving that c/(a+b) < 1, which is obvious.

Surprises me that this is soo all-round known  I really never heard it anywhere before.

A neat problem would be, in triangle abc, prove:

a^3 + b^3 + c^3 + 3abc >= a^2b + ab^2 + b^2c + bc^2 + a^2c + c^2a >= a^3 + b^3 + c^3 + 2abc.

The leftmost inequality is handled by [hide]Schur[/hide], the rightmost is handled by the triangle inequality.

But there are factorizations for each part:

Clearly x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) >= 0, and clearly (x+y-z)(x+z-y)(y+z-x) >= 0 as well.

[quote="Peter VDD"]ln cabin? :)[/quote]

Grrrr.  That wasn't the answer I wanted and you know it!  ;)

[quote="JSRosen3"]Given: a < b

Prove: b > a[/quote]

WTF?!  That's what they have Pre-Olympiad and Advanced for.  Keep that stuff out of Intermediate!

Oh, sorry, I forgot to cite the problem. It's IMO 1983, Problem 6. Sorry about the difficulty, Magnara, but if you try it for a while you may find a very elegant solution!

mangara, what did you mean then? :) sorry I'm not getting it :P (me=non-english)

I don't even know that :P

As soon as it's not mathematics... :D

[quote="Peter VDD"]ln cabin? :)[/quote]
Old joke but the actual answer  is " House Boat"

(It is log cabin plus "sea")
(Of course you have to change the problem a little to make the integral indefinite)

(Now Another  classic one: " what is purple and commutes? ")

[quote="Gyan"][quote="Peter VDD"]ln cabin? :)[/quote]
Old joke but the actual answer  is " House Boat"

(It is log cabin plus "sea")
(Of course you have to change the problem a little to make the integral indefinite)[/quote]

Grooooaaaaan.

[quote="BenG1008"]Tell me if I'm not allowed to simply do this for the a<b/b>a proof:
Given:  a<b
Assume that b <or=a
You then have a contradiction, since a<b is given, and it is not possible for each to be smaller than the other.
Therefore b must be greater than a, since that is only possiblity left.[/quote]

I'd mention the law of trichotomy in there somewhere.


Given a convex quadrilateral $ABCD$ such that $AB=3,\ BC=4,\ CD=5,\ DA=6,\ \angle{ABC}=90^\circ.$ Find the area of $ABCD.$

[hide]Then $AC=5$ and $[ABC]=6$. $[ACD]=\sqrt{8(8-5)(8-5)(8-6)}=12$ by Heron's forumula, and it follows that $[ABCD]=18$. ($[ABC]$ denotes the area of polygon $ABC$.)[/hide]

[hide]$\triangle ABC$ is right, so $AC=5$. Drop an altitude from $\angle C$ in $\triangle ACD$ and obtain two more 3-4-5 triangles. Each of these three triangles has an area of $6$, so $[ABCD]=\boxed{18}$.[/hide]


geometry
circumcircle
perpendicular bisector
geometric transformation
cyclic quadrilateral
geometry proposed

Let $ (O_1),(O_2)$ be two circles intersect each other at $ A,B$. Let an arbitry line through $ A$ intersects $ (O_1),(O_2)$ respectively at $ C,D$. The tangents from $ C$ and $ D$ of $ (O_1),(O_2)$, respectively meet together at $ P$. Prove that the perpendicular bisector of $ PB$ always touches a fixed circle.

[b]p/s[/b]: I have solved this problem by [b]spiral similarity[/b] with some major hints. I just want to know wether this problem can be solved by using inversion?? Thanks in advance.

[b]mathVNpro[/b]

[quote="limes123"][hide="Solution"]Assume WLOG that $ \angle BDP\leq 90^{\circ}.$Let $ X, Y$ be foots of perpendiculars from $ O_1, O_2$ respectively on $ BC, BD$. Simple angle chasing shows that $ CBDP$ is cyclic quadrilateral and $ \angle BO_1C \equal{} \angle BO_2D$ (1). Let $ O$ be the circumcenter of circumcircle of $ CBDP$. It's obious that $ BYOX$ is cyclic, and using (1) we get that also $ BO_2OO_1$ is cyclic. Let $ M$ be the midpoint of $ PB$. We have $ \angle BOM \equal{} \angle BDP \equal{} \angle BO_2Y \equal{} \angle BO_2O$ which shows that $ OM$ is tangent to circumcircle of $ O_1BO_2$, which is fixed. [/hide][/quote]
this is exactly main idea in my solution. Can anybody offer a inversion one?


geometry
circumcircle
rhombus
Euler
angle bisector
geometry proposed

Let a triangle ABC, $ \angle A\equal{}60^o$, O is the circumcenter of ABC. BO meets AC at D, CO meets AB at E. Denote M,N,P be the midpoints of BC,BD,CE. Prove that the bisector of angle NMP passes though the nine-point center of triangle ABC.

N and P lie on midparallels from M, so <NMP  has the sides parallel to AB and AC, hence its angle bisector is parallel to AF, the angle bisector of BAC, F being the midpoint of the arc BC. If H is the orthocenter of ABC, see that AHFO is a rhombus, AF its diagonal, M the midpoint of OF, and the parallel to AF from M passes through a point P on OH such that PH = 3*PO.

Best regards,
sunken rock

[quote="mr.danh"]Let a triangle ABC, $ \angle A \equal{} 60^o$, O is the circumcenter of ABC. BO meets AC at D, CO meets AB at E. Denote M,N,P be the midpoints of BC,BD,CE. Prove that the bisector of angle NMP passes though the nine-point center of triangle ABC.[/quote]
I find Problem :
Let the Euler line of ABC meet line AB in E and line AC in F. Then the Euler line of AEF is parallel to BC

[quote="TSTUSA12345"][quote="mr.danh"]Let a triangle ABC, $ \angle A \equal{} 60^o$, O is the circumcenter of ABC. BO meets AC at D, CO meets AB at E. Denote M,N,P be the midpoints of BC,BD,CE. Prove that the bisector of angle NMP passes though the nine-point center of triangle ABC.[/quote][/quote]
I find Problem :
Let the Euler line of ABC meet line AB in E and line AC in F.
Any Body


modular arithmetic
decimal representation
combinatorics
counting
function
IMO Shortlist

Let $f(k)$ be the number of integers $n$ satisfying the following conditions:

(i) $0\leq n < 10^k$ so $n$ has exactly $k$ digits (in decimal notation), with leading zeroes allowed;

(ii) the digits of $n$ can be permuted in such a way that they yield an integer divisible by $11$.

Prove that $f(2m) = 10f(2m-1)$ for every positive integer $m$.

[i]Proposed by Dirk Laurie, South Africa[/i]

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

Let $ A$ be the set of integers $ n$ with $ 0\leq n < 10^{2m-1}$ and satisfying property 2. Let $ B$ be the set of integers $ n$ with $ 0\leq n < 10^{2m}$ and satisfying property 2. We will form a bijection from the elements in set $ A$ to the elements of set $ B$.
Take any element in set $ B$ -- a number with decimal representation $ a_{1}a_{2}...a_{2m-1}a_{2m}$. Then, there exists a permutation of the elements of set $ \{a_{1},a_{2},...,a_{2m}\}$ into a set $ \{b_{1},b_{2},...,b_{2m}\}$ so that the number $ b_{1}b_{2}...b_{2m-1}b_{2m}$ is divisible by $ 11$ so $ \sum_{i = 1}^{2m}(-1)^{i}b_{i}\equiv 0\pmod{11}$.
Then $ \sum_{i = 1}^{2m}(-1)^{i}(kb_{i}+l)\equiv\sum_{i = 1}^{2m}(-1)^{i}(kb_{i})\equiv k\sum_{i = 1}^{2m}(-1)^{i}(b_{i})\equiv 0\pmod{11}$, therefore the integer $ \sum_{i = 1}^{2m}(-1)^{i}(c_{i})\equiv 0\pmod{11}$ is also divisible by $ 11$ where $ c_{i}\equiv kb_{i}+l\pmod{11}$        (*)
Now, because for $ i = 1,2,...,2m$ $ a_{i}$ is one of $ 0,1,2,...,9$ then $ a_{i}+1$ is never divisible by $ 11$. Then, there exists an integer $ k$ from the set $ \{1,2,...,9,10\}$ such that $ k(a_{2m}+1)\equiv 1\pmod{11}$ so $ k(a_{2m}+1)-1\equiv 0\pmod{11}$. Also note that for $ i = 1,2,3,...,2m-1$ we have $ k(a_{i}+1)-1$ is one of $ 0,1,2,..,9$ mod $ 11$ and never $ 10$ mod $ 11$ because $ a_{i}+1$ is never divisible by $ 11$. Then, using the property (*) twice we get that $ \sum_{i = 1}^{2m}(-1)^{i}(d_{i})\equiv 0\pmod{11}$ where $ d_{i}\equiv k(b_{i}+1)-1\pmod{11}$. Then, if we take $ e_{i}= k(a_{i}+1)-1$ then $ e_{i}$ is an integer from $ 0$ to $ 9$ for $ i = 1,2,3,...,2m-1$ and $ e_{2m}= 0$. Then, because then $ d_{i}$ is an integer from $ 0$ to $ 9$ then because $ \sum_{i = 1}^{2m}(-1)^{i}(d_{i})\equiv 0\pmod{11}$, the integer $ d_{1}d_{2}...d_{2m}$ is divisible by $ 11$ and $ \{d_{1},d_{2},...,d_{2m}\}$ is a permutation of $ \{e_{1},e_{2},...,e_{2m}\}$ (because $ \{b_{1},b_{2},...,b_{2m}\}$ is a permutation of $ \{a_{1},a_{2},...,a_{2m}\}$). Therefore, the digits of $ e_{1}e_{2}...e_{2m}$ can be permuted to get an integer divisible by $ 11$, and we can interchange odd and even positions in order to get $ e_{2m}= 0$ being the right-most digit. Therefore, the digits of the number $ e_{1}e_{2}...e_{2m-1}$ can be permuted to get an integer divisible by $ 11$ (because the $ 0$ on the right end can be just removed). So we get a one-to-one bijection from $ a_{1}a_{2}...a_{2m}$ to $ e_{1}e_{2}...e_{2m-1}$.
Now we can reverse the procedure to go from every integer $ e_{1}e_{2}...e_{2m-1}$ in set $ B$ and add choose any $ a_{2m}$ from the set $ \{0,1,2,...,9\}$ and then find the required $ k$ such that $ k(a_{2m}+1)-1\equiv 0\pmod{11}$, and then reverse the process described above (solve for $ a_{i}$ in $ k(a_{i}+1)-1\equiv e_{i}\pmod{11}$ and then we get a unique solution $ \pmod{11}$ with $ a_{i}$ belonging to the set $ \{0,1,2,...,9\}$) and reconstruct the number $ a_{1}a_{2}...a_{2m}$ belonging to $ A$. There are ten ways to choose $ a_{2m}$ for every element in $ B$, and each yields a unique element from $ A$, and we will get every element in $ A$ because every element in $ A$ can be mapped on an element in $ B$ by the reverse procedure. So, for every element in $ B$ there are $ 10$ elements from $ A$ so $ f\left(2m\right) = 10 f\left(2m-1\right)$ QED.

Here is another solution which uses a different bijection.

[hide = Solution]

Let $A$ be the set of elements in $\{0, 1, 2, \cdots, 9\}^{2m}$ such that it satisfies condition $(ii)$ of the problem.

Let $B$ be the set of ordered pairs of elements in $\{0, 1, \cdots, 10\}^{2m-1}$ and integers from $1$ to $10$ inclusive such that for a pair $(b_1, b_2, \cdots, b_{2m-1}, x) \in B$, we have that $(b_1, b_2, \cdots, b_{2m-1})$ satisfies condition $(ii)$ of the problem and $x$ is not among the integers present in $(b_1, b_2, \cdots, b_{2m-1})$.

Observe that by counting the elements in $B$ based on casework on $x$, it's clear that $|B| = 10f(2m-1)$, since the number of elements of $B$ with every $x = 1, 2, \cdots, 10$ is equal by symmetry (only mod $11$ matters), and $f(2m-1)$ is simply counting the number of elements of $B$ where $x = 10$.

Also, it's obvious that $|A| = f(2m)$.

Therefore, it suffices to construct a bijection from $A$ to $B$.

To do this, we will biject $(a_1, a_2, \cdots, a_{2m}) \in A$ to the pair $(a_2 - a_1, a_3 - a_2, \cdots, a_{2m} - a_1) \in B$ and the number $10-a_1$. Note that this is in $B$ since $10-a_1$ is note among $a_i - a_1$ for $i \neq 1$ ( as otherwise we'd have $a_i = 10$).

Now, this is clearly an injection to the elements of $B$, as $10-a_1$ and $a_i - a_1$ for $i \neq 1$ clearly determine all of the $a_i$.

It suffices only to show that it is a surjection.

For a given pair in $B$, say $(b_1, b_2, \cdots, b_{2m-1})$ and $x$, it's clear that $(10-x, b_1 + 10-x, \cdots, b_{2m-1} + 10-x)$ is a $2m-$tuple in $A$ and that it maps to $(b_1, b_2, \cdots, b_{2m-1})$ and $x$, so we are done.

$\square$

[/hide]

Suppose that we allow the digit $A$, which has value $10$, inside our $k$-digit numbers. Also, define $f(k,d)$ to be the number of good strings of length $k$ which exclude the digit $d$. Hence, $f(k)=f(k,A)$. Now, note that the number of good strings in $f(2m,A)$ which end in a digit $d$ is equivalent to the number of good strings with $2m-1$ digits which can be split into two groups whose sums differ by $d$. However, subtracting $d\pmod{11}$ from these $2m-1$ digits, we see that there is a bijection between good strings in $f(2m-1,A-d)$ and good strings in $f(2m,A)$ which end in $d$. So, $f(2m,A)=f(2m-1,1)+\ldots +f(2m-1,A)$.

Now, we claim $f(2m-1,d)$ is constant for nonzero $d$. To see this, we can just map every digit $k$ in our string to $k\zeta\pmod{11}$, where $\zeta$ is a fixed primitive root. Then, $f(2m-1,d)=f(2m-1,d\zeta)=\ldots$, as desired. In particular, this means that $f(2m)=10*f(2m-1)$, as desired.

How about a solution that doesn't require that $11$ is prime? [url=https://anhngq.files.wordpress.com/2010/07/imo-2003-shortlist.pdf]A comment after the first official solution[/url] says that $10$ can be replaced by any even base $b$ (as long as $11$ and $9$ are replaced by $b+1$ and $b-1$), yielding $f\left(2m\right) = b f\left(2m-1\right)$.

We fix a positive integer $m$. Call a $2m$-digits number (with leading zeroes allowed) [i]tasty[/i] if and only if its digits can be permuted in a way that yield an integer divisible by $11$. Let $A=f(2m-1)$ and $B_i$ be the number of tasty numbers starting with digit $i$, where $i=0,1,\hdots,9$.
------
[color=red][b]Claim:[/b][/color] $A=B_0$

[i]Proof:[/i] See that $\overline{x_1x_2\hdots x_{2m-1}}\mapsto \overline{0x_1x_2\hdots x_{2m-1}}$ is indeed a working bijection. $\blacksquare$
-----
Thus, we restrict our attention to $B_0,B_1,\hdots,B_9$. It turns out that they are all equal. The critical observation to prove this is the following.
------
[color=red][b]Claim:[/b][/color] If we allow $10$ as a digit, then for any $a,b\in\mathbb{F}_{11}$, if $\overline{x_1x_2\hdots x_{2m}}$ is tasty, then its component-wise linear transformation
$$\overline{(ax_1+b)(ax_2+b)\hdots (ax_{2m}+b)}$$
is also tasty, where each number is treated as a residue modulo $11$.

[i]Proof:[/i] As $\overline{x_1x_2\hdots x_{2m}}$ is tasty, one can partition $x_1,x_2\hdots,x_{2m}$ to two sets $T_1, T_2$ such that $|T_1|=|T_2|=m$ and
$$\sum_{t\in T_1} t \equiv \sum_{t\in T_2} t \pmod{11}.$$
Thus,
$$\sum_{t\in T_1} at+b \equiv \sum_{t\in T_2} at+b\pmod{11}$$
which yields the desired partition of the transformed number. $\blacksquare$
-------
All that remains is to find a linear transformation in $\mathbb{F}_{11}$ which fix $10$ and maps $0$ to $k$. It's easy to see that the desired transformation is $x\mapsto (k+1)(x+1)-1$. This completes the problem.


In a ten-team league, each team plays every other team exactly twice. Find the total number of games played in the league.

The first team will play $ 18$ games against other teams ($ 9$ other teams, and play $ 2$ games against each). The next will play $ 16$ games not already counted (we already counted its $ 2$ games against the first team). Continuing in a similar path we have $ 18\plus{}16\plus{}14...\plus{}4\plus{}2\equal{}9\times{10}\equal{}\boxed{90}$.


Analise’s resting heart rate is 75 beats per minute. When she sleeps, it slows to 60 beats per minute. What percent decrease is this?

[hide] We have $ \frac{75\minus{}60}{75}\equal{}\frac15\equal{}\boxed{20\%}$.[/hide]


vector
integration
algebra
polynomial
partial fractions
linear algebra
real analysis

Prove that the $ n$th order linear homogeneous ordinary differential equation has exactly $ n$ fundamental sets of solutions.

EDIT: Of course by referring to the characteristic equation and from the fundamental theorem of algebra this should be pretty obvious, but how would one prove that the characteristic equation would yield all the fundamental sets?

A straightforward dimensionality argument is as follows:  the set of solutions to an $ n^{th}$ order linear homogeneous ODE is clearly a vector space.  Specifying $ y(0), y'(0), ... y^{(n\minus{}1)}(0)$ uniquely specifies a particular solution, so this vector space has dimension $ n$.  The characteristic equation method produces a solution space of dimension $ n$.

Taking a Laplace transform turns the initial value problem into a partial fractions problem, and we know partial fractions.

Define $ Lf(s)\equal{}\int_0^{\infty}f(x)e^{\minus{}sx}\,dx$
$ L(x^ke^{rx})(s)\equal{}\frac{k!}{(s\minus{}r)^{k\plus{}1}}$ (1)
$ L(f')(s)\equal{}s\cdot Lf(s)\plus{}f(0)$ (2)

By (2), the ODE $ f^{(n)}(x)\plus{}a_{n\minus{}1}f^{(n\minus{}1)}(x)\plus{}\cdots\plus{}a_1f'(x)\plus{}a_0f(x)\equal{}0$ becomes $ (s^n\plus{}a_{n\minus{}1}s^{n\minus{}1}\plus{}\cdots\plus{}sa_1\plus{}a_0)Lf(s)\plus{}q(s)\equal{}0$, where $ q(s)\equal{}f^{(n\minus{}1)}(0)\plus{}(s\plus{}a_{n\minus{}1})f^{(n\minus{}2)}(0)\plus{}\cdots\plus{}(s^{n\minus{}1}\plus{}a_{n\minus{}1}s^{n\minus{}2}\plus{}\cdots\plus{}a_1)f(0)$ is a polynomial of degree at most $ n\minus{}1$.
By (1), the terms $ \frac{1}{(s\minus{}r)^{k}}$ of the partial fractions decomposition of $ \frac{\minus{}q}{p}$ correspond exactly to standard solutions $ \frac{x^k}{k!}e^{rx}$.

You could also just look at the linear algebra directly.