geometry perimeter inradius search function

Find all right triangles which have integer side lengths such that the area is  equal to the perimeter.  Prove that you have found all of them.

[b]Please hide your solution.[/b]

[quote="LynnelleYe"]A Pythagorean triple can always be expressed as 2n, n^2-1, n^2+1[/quote]

That isn't true.  Otherwise, every Pythagorean triple would have two values which differ by 2.

[quote="LynnelleYe"][hide="Not_too_sure_but..."]A Pythagorean triple can always be expressed as 2n, n^2-1, n^2+1. So this is equivalent to
2n+n^2-1+n^2+1=(2n)(n^2+1)/2
Which simplifies to
n^3-2n^2-n=0
n(n^2-2n-1)=0
But n=0 results in a triangle with lengths of 0 and -1, and n^2-2n-1=0 doesn't have integer solutions, so there are no such triangles?
Unless not all Pythagorean triples can be expressed this way. I don't remember whether or not they can.[/hide][/quote]
[hide="Hint "]Label the sides a, b, and c[/hide]
No, the side lengths of a triangle are always positive.

I disagree with LynnelleYe.

[hide] Let $a,b,c$ be the sides of the right triangle,where $a\leqq b<c$

From the given condition, we have $a^2+b^2=c^2$ and $a+b+c=\frac{1}{2}ab$

Eliminating $c$ from these equation,we obtain $ab-4(a+b)+8=0\Longleftrightarrow (a-4)(b-4)=8$

Since $a,b$ are natural numbers, $-3\leqq a-4\leqq b-4$,

we have $(a-4,b-4)=(1,8),(2,4)\Longleftrightarrow (a,b)=(5,12),(6,8)$

Therefore the desired triangles are $5^2+12^2=13^2,6^2+8^2=10^2$ [/hide]

The general form for Pythagorean triples is $\{d(m^2 - n^2), 2dmn, d(m^2 + n^2)\}$ where m, n and d are positive integers with $m > n$.

I know I forgot to mention it before, but could you hide your solution kunny?

[quote="LynnelleYe"]Oh right...
How would you do this then?[/quote]
Do you understand kunny's solution?

Hi this question has been asked at a little lower level too:
[url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=right+triangle+perimeter+area&t=25481]Here.[/url]

A simple solution can be also be thought of as:

[hide]Area =  (1/2) * perimeter * inradius  = perimeter
so inradius = 2
But inradius = $\frac {a+b-c}{2}$
so we have $(a+b-c) = 4$
$a+b-4=c$
or $a^2+b^2+2ab-8a-8b+16 = c^2 = a^2+b^2$
which gives  $ab-4a-4b+8=0$
or $(a-4)(b-4)= 8$
Since 8 can be factored as  (1*8) and (2*4)  only answers for (a,b) are  $(5,13)$ and $(6,8)$ [/hide]

BTW the question has also been asked here  for any triangle (not   only right angle) too where area = perimeter.

[quote="Gyan"]
BTW the question has also been asked here  for any triangle (not   only right angle) too where area = perimeter.[/quote]
Do you know where? This was going to be one of the follow-up problems I was going to put out.

[quote="Gyan"]See Nene's solution in the thread posted above. You may also have luch using search function as I seem to recall a seperate thread for this too.[/quote]
Thanks

Deriving this formula is easy. Think of the circle: $x^2 + y^2 = 1$ and the line $y=k(x+1)$ with $0<k<1$ and $k=\frac{p}{q}$ where $p$ and $q$ are integers. So the solutions for the intersection of this line and the circle can be found:

$x^2 + (k(x+1))^2 = 1
\\
x^2 + k^2 x^2 + 2k^2 x + k^2 = 1
\\
(k^2 + 1)x^2 + (2k^2)x + (k^2-1) = 0
\\
x=\displaystyle \frac{-(2k^2) \pm \sqrt{(2k^2)^2-4(k^2 + 1)(k^2-1)}}{2(k^2 + 1)}
\\
x=\displaystyle\frac{-2k^2 \pm 2}{2(k^2 + 1)}=-1, \frac{1 - k^2}{k^2 + 1}
\\
y=k(\frac{1 - k^2}{k^2 + 1}+1)=\frac{2k}{k^2 + 1}$
Since our point is on the circle we know $x^2 + y^2 = 1^2$ and we have a triple $(x,y,1)$ that makes a right triangle. So, we can manipulate our triple to find the formula now:
$(x,y,1) \Rightarrow \displaystyle\left(\frac{1 - k^2}{k^2 + 1},\frac{2k}{k^2 + 1},1\right)
\\
\Rightarrow \displaystyle\left(1 - k^2,2k,1+k^2\right)
\\
\Rightarrow \displaystyle\left(1 - \left(\frac{p}{q}\right)^2,2\left(\frac{p}{q}\right),1+\left(\frac{p}{q}\right)^2\right) \Rightarrow \displaystyle\left(q^2 - p^2,2pq,q^2+p^2\right)$


who's a basketball-lover and would like to play this nice sport in IMO? (cuz we always played soccer in the past years, maybe this year will be different :P) ?

Besides the exam, what do the students usually do at the IMOs? Like the social aspects.

well in the last night everyone gets (or at least tries to :P) get drunk, people hang out together, doing different stuff (usually the americans and canadians play mafia :)), the south-american people play soccer, I remember in 2001 almost all the people were playing SET :P

also there are excursions (in the states they took us into a big rollercoaster park, which was pretty cool - for us people which don't have such amusament parks in our country) etc.

also board games are featured almost all the time, among the most popular are chess :P (with all sorts of different variations), monopoly, risk (probably now also Catan Settlers) and of course card games.

so do not worry about boring at IMO. the only thing you have to worry about at IMO is to get a medal :P

Rather the Germans are known for playing mafia all the night... BTW, is it true that the game was invented in Romania?

It is strange, but I have never been thinking of IMO nightlife... although at the IMO training camps at Bad Homburg we have spent one half of our spare time by gambling (we were housed in a youth hostel, and they had a couple of game machines to finance their house), and at Oberwolfach we usually hung around in the library until 2 o'clock (it is, BTW, a really cool experience to walk the way from the hill where the institute lies down to the river where we have our hotel at midnight, in complete darkness).

  Darij

[quote="darij grinberg"]Rather the Germans are known for playing mafia all the night... BTW, is it true that the game was invented in Romania?
Darij[/quote] John Clyde, former USA IMO Team member in 1998 (I think), and Putnam winner with Duke's Team in 2001, told me so in 2001 at MOP. But I remember him mentioning that in fact the game was invented by Romanians at IMO in 1996, but not in Romania :P
Anyway, the americans invented all sorts of variations, including Indian Mafia, which I like best.

[quote="darij grinberg"]What is Indian Mafia?

Here in Germany we play two kinds of Mafia only: the standard Mafia and the Bavarian Mafia (with the "Todesart" = kind of death, and lots and lots of special figures, like terrorists, sheriffs, mathematicians, deflorators, psychiatrists, freelancers, just to mention a few).

  Darij[/quote] Bavarian Mafia is one of the variation usually played when more than 10 people are playing. I am curious, what do the mathematicians do? :P

Indian Mafia is played with cards, and it's more of a "probability game". I'm also too lazy to explain exactly how it's played right now :P

Anyway this thread was about Basketball, so I guess I'll change the topic name into "games at IMO" :)

[quote="Valentin Vornicu"]I am curious, what do the mathematicians do? :P[/quote]

Well, a mathematician is a kind of generalized detective: He may point to a non-empty set of players and the administrator must say how many of them are "bad" (mafiosi, terrorists, etc.). However, he can do this only once in his life (after doing it he is dead).

  Darij

well... singapore is trying to field a good basketball team this year in addition to a good maths team... and the very same professors are training us (=... well maybe minus valiowk. unless we somehow get her to play (=

yup i like bball so i'll probably be playing.  i like soccer too!  though im quite rusty

in our singapore math camp we usually stay up to play cards everyday... generally bridge and tai tee.  sometimes variation appear here and there

i remember one year when we played soccer on the last night... till someone knocked into a post he didnt see.  then we started taking penalty kicks.  another year we played bball at 3 am.

looking for a basketball player? here i am! not a phenomenon, but i like playing, and i've been training 11 years, so i guess i can play my role ;)

we would also need a gym, which seemed to be a problem in the past years. maybe with the whole Olympic Sport Games taking place in Athens after the real Olympiad (the maths one that is :P:D)  this won't be a problem anymore, but who knows? :)

well I love playing soccer at the Imo, but basketball is fine....we have been training as a team for soccer matches but we can train a little of basketball....

Btw Mafia is also played here in colombia and in almost all latin countries (in the iberoamerican olympiad it was a very popular game) so i dont know how it is so well known all over the world but is cool!

And also i want to ask something....After the ceremony in imo 2003 we were ina hotel and some friends and I drank lots of beer (they didnt gave liquor to us so a co-leader gave us) we form a regular hexagon (or at least it looked regular) of empty bottles and i saw some people took pictures of it......Maybe some member of mathlinks took a picture and if somebody did i would like to see it! :)  ;)  ;)

[quote="Valentin Vornicu"]well in the last night everyone gets (or at least tries to :P) get drunk, people hang out together, doing different stuff (usually the americans and canadians play mafia :)), the south-american people play soccer, I remember in 2001 almost all the people were playing SET :P [/quote]

Yeah... and I remember the guide of Brasil in 2002. The most beautiful girl ever seen... Not even movies can replace her :(. Her name was Daniella, and that's all i know :(

Well, I think I'll have to practise swimming... As our team leader told us, if Mexico beats France, we cannot get on the airplane  :D
PS : What is mafia ?

maybe this year. i've been playing quite a bit of basketball this term so watch out (well at the moment, the safest place is still under the basket because i can't seem to hit it as yet - but soon!)

Playing MAFIA with canadians and estonians :))
Forever! [img]http://www.froster.org/forum/style_emoticons/default/icon_smile_blush.gif[/img]

[quote="kueh"]i don't like to mention this, but how about mao - some random version which no one knows so its fair :lol:[/quote]
How could I forget it %)
This game taught me ~10 English phrases.
Besides guessing the rules I had to guess their sence  :D

[quote="ondrob"]Mafia was known in Slovakia 25 years ago (I know from my older friends), so it's quite old.[/quote]

well, that's cool, so we have another version for the origin of mafia...  :D

[quote="ondrob"]We had a version called multifunctional mafia. There's about 30 functions (and every time playing it some new appears) and 6-12 players. It's really cool when having 4 or more functions :)[/quote]

hmm, this looks quite similar to what we in germany call schizophrenic mafia... i have never played it and i don't really want to play it anyway since it seems to me like a big mess, but some people here find it cool...

[quote="ondrob"]but it's hard for the Destiny (we call Destiny a player that organize the game, announce death, ...).[/quote]

that's what we call a moderator ("spielleiter" ;) ). sometimes this turns out to be the most difficult role to play in the game (peter may tell us his story :D ).

[quote="ondrob"]Do you have function mage? He can every 2nd round "change" him and someone else in such way that every attack on mage is going to that person and attacks to that person are going to the mage.[/quote]

that's what we call a "bus driver" (arising from the imagination that two players' seats in a bus are being shuffled - a reason why some active mafia players develop a paranoid fear of buses)... i personally have never played a game with this figure, but i believe some others did, and it is quite a popular figure in onlinemafia.

[quote="ondrob"]If youre interested I can find some time and write funtions that we use :)[/quote]

indeed, this would be very interesting. maybe even a few of them are not known to us here yet :D .

in fact, we have A LOT of special functions here in germany (more precisely, bavaria), but seriously we would always be glad to meet new ones. here is a short selection from what we have:

- the slanderer ("rufmörder"): each night he chooses a person who, if he gets inquired by the detective in that night, is demonstrated as a mafioso (no matter whether he actually is mafioso or citizen).
- the bomb planter ("minenleger"): each night he "lays a bomb" under the house of some player. if this player gets attacked by the mafia, then the bomb detonates and both this player and the mafioso attacking him are dead (this is usually a random mafioso). so far, so nice, but if the doctor or the detective refers to this player in that night, then the bomb also gets blown up, so we have the doctor or the detective and the player both dead... all told, quite a difficult role to play. at the rostock imo training camp this february, we played a round with the bomb planter. result of first night: 3 dead civilians. :D
- the freelancer or the alien ("anderer"): a "third party". his aim is to get killed by the mafia during a night and lynched by the citizens afterwards (variation: lynched by the citizens [i]within the next day[/i]). only in this case he has won. you see, not an easy job, but a very interesting figure. i don't remember any game where the freelancer had really won, except of one, which was played under rather minimalistic conditions. in fact, we had three players: michael müller, christian sattler and me were in a train, going back from an imo training seminar, and it we were so bored that we started to play mafia. i believe we had a mafioso, a detective and a freelancer. the freelancer had won (it was christian)...
- the mathematician: he can choose a group of persons (strictly speaking, a subset of the set of all players), and he is said how many of them are good and how many are evil. usually, the mathematician can do such an action only once in the game.
- the matchmaker: each night he can point on any two players (different from himself) and comes to know whether they are in the same party or not (i. e., he is told the number of mafiosi among them mod 2). a matchmaker playing its role right can often be more useful than a standard detective, and in any case he contributes to the nontriviality of the game.

and many many more... probably one important thing i forgot the "scharping" but i don't manage to define him properly, so i leave that task to others (just to note that, as a consequence of this figure, the mafia players among us have already a number of neologisms like "scharpen", "gescharpt", "entscharpt" which all originate from this figure)...

additionally, it would be nice if somebody explains pentacle mafia ("pentagrammmafia") here... it's a very counterintuitive game but it can turn out quite interesting...

  darij

One time, we had a role called Pinguin, which could swap two people in their roles. Every morning everyone should check his card again.
I'm not interested in mafia a lot. In Greece, it was very funny, and nobody played seriosly -- we broke all rules that one can remember :)

[quote="Feuerbach"]One time, we had a role called Pinguin, which could swap two people in their roles. Every morning everyone should check his card again.[/quote]

we call this role "till eulenspiegel", but i don't recall anybody really having played it...

yes, mafia in athens was funny... but i must say that i was a bit depressed later when i was told that the idea of using a random numbers generator has been known before me...

  dg

Well, as far as nightlife is conserned...
In IMO 2004, I remember that the first day after the exams, we went in a disco...
Some guys from greece,guiders or contestants, zoh, Valentin and me... I can say for sure that Valentine was the best dancer among us.
(I mention zoh seperately, cause i remember that Valentin was flerting her  ;) )
After drinking lots of beers, we went to the pool of the hotel for a midnight swimming, but Valentin didn't come with us cause he had lost his swimming suit.

(Valentine, I was the guy with the long hair)

As far as games is conserned, I remember lots of people playing mafia the first day we arrived in the hotel...
But i think most of the contestants were having conversations rather than playing games.

[quote="ondrob"]Mafia was known in Slovakia 25 years ago (I know from my older friends), so it's quite old. We had a version called multifunctional mafia. There's about 30 functions (and every time playing it some new appears) and 6-12 players. It's really cool when having 4 or more functions :) but it's hard for the Destiny (we call Destiny a player that organize the game, announce death, ...). Do you have function mage? He can every 2nd round "change" him and someone else in such way that every attack on mage is going to that person and attacks to that person are going to the mage. If youre interested I can find some time and write funtions that we use :)[/quote]

Well ... concerning mafia, I guess it's a much older game than a few years, especially because it is so widespread.

This schyzofrenic mafia is cool, I admit, but there's nothing better than the REAL mafia - only a few mafiosos, about twice as many "normal people" and one Cattani (a person who can ask about others' identities, I guess you call it the detective).  It's real psycho, if you play it with right people, one round (= one day!) can last for about two hours, you don't believe it until you play it...

In Germany there are lots of mathematicians and some others, who often played online-mafia-games. At such a game people communicate private by forumpost, eMail, icq....  The night session is about 4, the day session 10 real days long. Its played with a moderator and a special forum for the mafia.

Now a friend of me, Ching, has created a english-languaged mafia-board, where people from all over the world can discuss about the rules/characters/variants of game, so that we can start a international online-mafia-game soon, I hope so.  If you want to take part in that game or if you only want to talk about mafia and meet new people, than follow the link below.

http://www.english.mafiaspiel.de

Hi folks!  :lol:

Julia, are you Julia E. from the QED?

[quote="darij grinberg"]additionally, it would be nice if somebody explains pentacle mafia ("pentagrammmafia") here... it's a very counterintuitive game but it can turn out quite interesting...[/quote]
Well, pentacle mafia doesn't have much to do with the original mafia game anymore, but it is an alternative for 5 (+1 moderator) people who want to play mafia but don't have enough players.

[u]How the game roughly works:[/u]
There are, as in original mafia
- the night, when all the players close their eyes and players with special functions can do certain things
- the day, when the players democratically decide which player they lynch.

[u]There are five roles for the five players (who draw cards etc. to get their role):[/u]
1. Mafioso
2. Soul Saviour
3. Terrorist
4. Freelancer
5. The Pentacle Master

[u]They have the following abilities and features:[/u]

1. Mafioso:
He wakes up and kills another person. He may not kill any person a second time if the person didn't die (and people practically never die at night).

2. Soul Saviour:
Wakes up at night and points at a person who doesn't die at night then. It only makes sense for him to point to himself or to the person that he believes to be the Freelancer (who doesn't become -->[i]sharp[/i] then), because all the others are immortal at night anyway.

3. Terrorist:
He's immortal at night. During the day he can blow himself up, taking another player into the grave with him. He and this other person are immediately dead from then on.

4. Freelancer:
He's immortal at night: When the mafioso tries to kill him at night, the Freelancer becomes [i]"sharp"[/i] unless he was "saved" by the Soul Saviour (and the Mafioso isn't allowed to kill the Freelancer again which kills him in original mafia). He stays sharp until the end of the game (and not for only one day or something like that). The Freelancer wakes up at night and the moderator signalizes him whether he was sharpened or not.

5. Pentacle Master:
Immortal at night.

[u]Only one person can win. The characters win in the following cases:[/u]

1. Mafioso:
He wins if the [i]two next people sitting to his right are dead[/i] (independently from whether he's dead or alive himself).

2. Soul Saviour:
He wins if [i]both his direct neighbors (i. e. the person to his left and the person to his right) are dead[/i] (independently from whether he's dead or alive himself).

3. Terrorist:
He wins if he manages [i]to blow up the [b]sharp[/b] freelancer[/i] or if [i]three people except himself are dead [/i](this is independent from whether he is alive or not then).

4. Freelancer:
He wins if he is [i]made sharp by the Mafioso and afterwards killed by the other players during the day [/i](he does not win if he is killed unsharp or if he is killed by the Terrorist)

5. Pentacle Master:
He wins if [i]anything paradox happens[/i]. This means if two people are winning at the same time (for example if the Terrorist and the sharp Freelancer are the direct neighbors of the Soul Saviour and the Terrorist blows up the sharp Freelancer -->both the Terrorist and the Soul Saviour theoretically win). This rule makes sure that only one player wins and prevents coalitions between several players which would become boring by the time.

I know it sounds complicated but it's actually quite easy if you've played it 2 or 3 times.
BTW: In original mafia everyone claims to be a citizen. In Pentacle Mafia normally everyone claims to be the Pentacle Master.

CU
Annika

@Annika Yes, I´m Julia E. from QED, as you can realize easily at the picture.... :D

Anika actually this a very nice variant of Mafia! I just love the rules ... question is if you can find 5 people that both know the rules and want to play at the same time :D :D

Yes, this might become a problem outside our German (in fact mostly Bavarian) mafia community, and even there pentacle mafia isn't played very often, as there are rarely exactly five people there. Some people try to teach all the others benzene mafia (which is for exactly six players as benzene has a hexagonal structure or whatsoever), but nobody manages to learn all the rules.  :lol:

if the students make it much , they 'll die another days.

[quote="Valentin Vornicu"]well I played basketball with the luxemburg guys (one of them got a silver with 24 points :)) ... any other of you guys?[/quote]

Yeah I remember ;-) I was one of the 3 guys from Luxembourg ;-)
It was quite a funny time, and so I was a bit disappointed that my teamleader, didn´t take me with to mexico and slowenia :-(


assume thatg we have $n_1<n_2<...<n_k\dots $ and the following statement satisfies $Lim_{n\rightarrow \infty}\frac{n_k}{n_1.n_2\dots n_{k-1}}=\infty$ now prove that $\sum \frac{1}{n_k}$ is irrational.

This calls for a slight variation on the ideas behind the usual proof that $e$ is irrational. Assume that the sum in question is rational and can be written as $\frac pq$ in lowest terms. Now choose $m$ such that
$\frac{n_m}{n_1n_2\cdots n_{m-1}} > 2q$.
Note also that for $k>m, n_{k+1}>2n_k$

Now take the whole sum and multiply it by $qn_1n_2\cdots n_{m-1}$. By our assumtion, the result must be an integer. This integer can be written as
$\sum_{k=1}^{m-1} \frac{qn_1\cdots n_{m-1}}{n_k} + \sum_{k=m}^{\infty}\frac{qn_1\cdots n_{m-1}}{n_k}$
The first sum is, by construction, an integer. Our choice of $m$ guarantees that the first term of the second sum is less than $\frac12$; after that, each term is less than half of the preceding term. Hence the second sum is less than $\sum_{j=1}^{\infty}\frac1{2^j}=1$ and thus the second sum cannot be an integer.

This contradicts our original assumption that the number was rational.


group theory
abstract algebra
symmetry
topology
advanced fields
advanced fields unsolved

Let $ G$ be connected and locally compact topological group. Prove that $ G$ is $ \sigma$-compact i.e. countable union of compact sets.

We need a lemma:
[b]Lemma.[/b] Every subgroup $ H$ of $ G$ which is locally compact is closed in $ G$.
[b]Proof.[/b] Because $ H$ is locally compact so there must exist an open neighborhood of 1 in G such that $ \overline{U}$ intersect $ H$ is compact in $ H$ so $ \overline{U}\cap H$ is compact in $ G$ amd so  $ \overline{U}\cap H$ is closed in $ G$. This implies $ H$ is closed in $ G$.

Back to our question, Just take any symmetry and relatively compact $ U$ which is a open neighborhood of 1 (the unity of group $ G$). Now $ V \equal{}\bigcup_{n\geq1}U^{n}$ is an open subgroup of $ G$. Note that $ V$ is locally compact subgroup of $ G$ so $ V$ must be closed, and we're done.

[quote="pluricomplex"]Just take any symmetry and relatively compact $ U$[/quote]
Is relatively compact=locally compact? Or compact with respect to relative topology? I have never seen that term.

[quote="puuhikki"]
Is relatively compact=locally compact? Or compact with respect to relative topology? I have never seen that term.[/quote]

Oh I mean $ U$ is relative compact in $ G$, ie.. $ \overline{U}$ is a compact set in $ G$.


calculus
integration
number theory unsolved
number theory

Find all integral solutions of the equations(independently):--
	(1) $x^4 + x^3 + x^2 + x + 1 = y^2$
                (2) $x^4 + x^3 + x^2 + x = y + y^2$

The 2nd equation is equivalent to the next:
$4x^4+4x^3+4x^2+4x+1=(2y+1)^2$.
And in either 1st or 2nd equations follow my advice, which I wrote:
If you have a polynom of degree $2n$, beginning with $a^2x^{2n}$, which must be perfect square, try to estimate it by squares of polynoms of degree $n$

Pavel can you send any document about this fact ... i can't get this  :(

Davron


hello.i am planning to write the ZIO and want to know how i am supposed to prepare for it.does anyone know some good books that might be useful.also please tell me how i must apply for it and when it is likely to be held?

ZIO is based mostly on reasoning. No programming. See previous year's (7-8 years) ZIO papers in IARCS website.
http://www.iarcs.org.in/inoi/current.php#ioi2009.

The tentative date for ZIO is November 28th.
Registration is through school only. It's not yet begun.

From last year ZCO has started (another way to get qualified for INOI). It is based on programming. An online exam written from the house. 2 problems in 3 hours. I guess ZCO is easier to clear than ZIO, as I feel that there is an element of luck in ZIO- everyone scores around the same.

ZCO registration is also through IARCS website. Tentative date for ZCO- 5ht December.
For, ZCO see problems (Advanced division) and the solutions in IARCS :
http://www.iarcs.org.in/inoi/contests/allproblems.php
You can also submit your codes for instant feedback in the Online Judge:
http://opc.iarcs.org.in/

I guess the 16 solutions given IARCS is good for ZCO.

For books:
An Introduction to Algorithms by Cormen, Leiserson, Rivest, Stein (CLRS).

9th and above can write ZIO and ZCO. But I think they'll allow even lower class students if they ask.

Where can I get the details of the registration process? The latest info I have is [url]http://www.cbse.nic.in/nio1.doc[/url] for 2007.

For those of you who where looking out for the ZIO/ZCO details, They have been put up on the website:
[url=http://www.iarcs.org.in/inoi/current.php#zio2010]IARCS WEBSITE[/url]
[url=http://www.iarcs.org.in/inoi/2010/zio2010/CBSE-Announcement-2010.pdf]CBSE ANNOUNCEMENT[/url]
[quote]The Zonal Informatics Olympiad, 2010 (ZIO-2010) will be held from 10:00 am to 1:00 pm on Saturday, 28 November, 2009.[/quote]
[quote]the Zonal Computing Olympiad 2010, will be held on Saturday, December 5, 2010[/quote]

i wonder if my school will allow 10th standards to write

my teachers,xcept a couple of them,haven't evn heard of olympiad :rotfl:

Considering that the IOI 2009 champion was from class 9 or 10, it would be stupid if Indians don't even allow students from class 10.

The registration for Zonal Computing Olympiad is open:

http://www.iarcs.org.in/inoi/current.php#zco2010

[quote="aadil"]is there a specific time allotted for writing ZCO?[/quote]

Yes.
Last year we had 2 slots. We had asked for the second one, as some of us had school on that day :(
You can submit your solutions in the server only during the stipulated time period.

What are likely cutoffs for ZIO based on past years?

Can someone post the questions ? I thought the exam was from 1:00 to 4:00 pm and missed the exam.  :(

i think i might get 2 and 4 correct[hope so]
my answers:
2.
a.9
b.13
c.19
4.
a.20
b.33
c.63
i'm in 10th.do you think i will qualify?

@aadil: Your 2nd is fully wrong. And your 4th is correct. I dont think you will qualify.

I'm REALLY pissed at the moment. WTH was the second question. Is move =(picking up a number and then moving) or (moving and then picking a number). I interpreted it as the first. And the question said "the game ends when the next move takes you beyond the sequence of tiles" So if that way we can never pick up the number on the last tile as it is a part of the 'move' which takes us beyond the last tile. :(
Did anyone else have this problem? The answers with my interpretation were 4/23/24. Do you think I should contact IARCS about the confusion?
4th one is correct for me too. Third one I did in a real hurry coz I was running out of time- 4/1/8(or maybe 4/8/1?)
And 1st part of first question- 832(again doubtful coz it had soooooo many calculations). I was doing the second part but I realized midway that I make too many careless mistakes to last this long so I should just recheck my other answers :P
Please suggest what I should do about question 2.
And now I'll also be forced to register for ZCO and study C++. Class 9/10 books are enough for syntax prep right?

[quote="mihir0492"]here are my answers:
1.
a: 691
b: 1233
c: 1405
2.
a:14
b:15
c:17
3. not attempted
4.
a:20
b:33
c:63[/quote]

what abt these answers?...u ppl think they're correct?

Can some one give the questions please?
Atleast a general idea.

[b]1.[/b] Given a tree sort of structure, we were asked to find the number of ways in which we can delete nodes (no. of combinations), with the restriction that if a parent node is deleted, all the nodes under it must also be deleted.

[b]2.[/b] Given $ n$ tiles, each with two distinct numbers, we were asked to maximize our "score", wherein "score" is defined as follows - pick a no. from the first tile. If it is the smaller one, got to the next tile. If it's the larger one, skip the next tile and go to the tile following. Again pick a no. and so on... Score = Sum of the nos. picked.

[b]3.[/b] Given four functions with recursive sort of definitions, find $ g(4)$ ... Blah. [i]Verrry[/i] Brutey Forcey

[b]4.[/b] Given $ n$ buildings on a street with integer co-ordinates (only x-axis  :P ), we had to find $ k$ pairs (no building belongs to 2 pairs) such that the sum of the pair-distances is minimum. The question asked for the minimum sum. $ (2k < n)$

Can those people who have taken part in ZCO, INOI etc. recommend a good C++ programming enviroment (editor, debugger) to use?
Thanks.

@Ignormus
Do you use windows? If so, then use Dev-cpp.
DO NOT use turbo C++ while practising for informatics olympiads

[quote="Poincare"][quote="abhitayal"]what's the syllabus for both the exams
and which application/language is used for making programs in ZCO C++ ,VB or SQL[/quote]

I'm sorry but, but who the heck writes in VB or SQL? VB?! I mean, come ON![/quote]

Please excuse me :blush:
I actually don't know anything about ZIO/ZCO and VB,C++,SQL are the only (computer)languages I know.

ZIO and ZCO results announced. Congrats to all selected.
http://www.iarcs.org.in/

Congratulations for keerthan (keyree 10) for making it through both ZIO and ZCO.
Congratulations to Akashnil too for making it through ZIO.

Congrats to all others selected.

(Me not selected again ).


topology
real analysis
real analysis unsolved

Find a one-to-one mapping from the interval (0,1) onto $ \mathbb{R}$ that is a homeomorphism and is a rational function.

You should have some idea of what the denominator should be.


AMC
USA(J)MO
USAMO
AIME
calculus
search
number theory

I'm currently in my freshman year and I really want to qualify for the USAMO next year. How do I [b]get started [/b](I have very little problem solving practice)? What books should I purchase? Should I just jump into AMC and AIME problems? Should I take one of these AoPS classes (Inter. Number Theory looks really interesting)? (Taking Calculus BC next year)

(I got a 5 on the AIME this year) [please excuse the uncapitalized "I" in the title, it won't let me edit  :ninja:)

just do lots of problems, amc, problems from these boards, it does not matter, the aops books are also very good for starting out, the best advice that i can give you is that every time you do a problem, make sure you can pull some strategy from that problem that can be applied generally to other problems in the same topic, ex. when i see solve for integers, i think factoring, mods, etc

Also, don't beat yourself up if you don't make it. A 5 on the AIME is already very good, and the process of preparing has rewards even if you don't "win". Problem solving is a worthwhile skill.

I think that uncapitalized I in the title is a side effect of a feature to prevent people from writing in ALL CAPS or similar annoyances. You only get one capital letter at the beginning of the topic.

Although solving random problems is great practice, I get much more from reading. And actually, when it comes to random problems, I gain most from solutions posted by others. It helps me think in other ways, attack problems differently, and grow as a problem solver.

so for me
understanding required to problem solve = lots of books
new approaches I may not find in books = other people, teachers, etc.

when printing out transcrips for intermediate algebra lessons 1,9 i've been
printing to the "pdf printer" and some of the right side is clipped off. same happens
when i do a "right click" "select all" "print"

also occurs for lessons 4,5,7,8,9 of the introduction to counting class.

all other lessons for both classes print correctly.

i apologize for the lame problem.

Hey surge,

lol I just joined. I really am in the same position as him are except that I got a 2 on the AIME.  :?

Does anyone else have some advice?

[quote="Xantos C. Guin"][quote="LordoftheMorons"]I'm also a freshman. I got a 4 on the AIME II(should have been a 5, but I added incorrectly while brute forcing #11), and will most likely be taking BC calc next year (asuming I don't screw up and miss the 97% in Algebra II required to skip precalc).[/quote]

Your school lets you skip pre-calc with a 97? That's cool.
I have a 101 but our school doesn't allow us to skip. Oh well, easy A.

To "get started" go to this [url=http://www.kalva.demon.co.uk/aime.html]site[/url].

It has AIME problems from 1983 to 2004.

Also, search around the web for practice.

But most importantly, don't get discouraged by what insanely good people on this forum who get 10+ on the AIME say. A 5 is a very good score. Qualifying for the AIME is also good. 40 people from the magnet school I go to took the AMC. Only 11 qualified for the AIME.[/quote]

It's a 97 to be able to take the test required to skip out... but I'm sure that I can pass that.

[quote="keeyounglee"]wow first of all it is really annoying when people quote sombody and somebody else quotes someone who quoted someone.

doing some past AIMEs are excellent practice, and when you're ready, you may consider ordering past olympiad problems from the AMC website as well.

/*
Chris, if you dont get a 97% with Williams I will laugh
*/[/quote]
>_>
<_<
101.42 third quarter.

What's the best books I should buy to 'get started'?

I got a 5 on the AIME I. I'm only in 8th grade, but I find just looking at the high school forums (Getting Started, Intermediate, and Pre-Olympiad) very useful. You can see all the different ways to solve the same problem. It kind of expands your mathematical arsenal of problem solving weapons. :D Wow, that's a wierd metaphor.


Show that 111......1(1997 times)222......2(1998 times)5 is a perfect square.

[quote="lordWings"]By induction,
$(\underbrace{3\dots 3}_{n}5)^{2}=\underbrace{1\dots 1}_{n}\underbrace{2\dots 2}_{n}25$.
Your case is $n=1997$.[/quote]
But there are 1998 2s...

If you look at it carefully, the $(n+1)$-th $2$ is outside the underbrace.
I could've written
$(\underbrace{3\dots 3}_{n}5)^{2}=\underbrace{1\dots 1}_{n}\underbrace{2\dots 2}_{n+1}5$
[hide="Full proof without induction"]Remember
$\underbrace{1\dots 1}_{n}=\frac{10^{n}-1}{9}$.
Therefore
$(\underbrace{3\dots 3}_{n}5)^{2}= (\underbrace{3\dots 3}_{n+1}+2)^{2}=$
$= \left(3\times\frac{10^{n+1}-1}{9}+2\right)^{2}= \left(\frac{10^{n+1}+5}{3}\right)^{2}=$
$=\frac{10^{2n+2}+10^{n+2}+25}{9}= \frac{10^{2n+2}-1}{9}+\frac{10^{n+2}-1}{9}+\frac{27}{9}=$
$=\underbrace{1\dots 1}_{2n+2}+\underbrace{1\dots 1}_{n+2}+3= \underbrace{1\dots 1}_{n}\underbrace{2\dots 2}_{n+1}5$[/hide]


geometry
perimeter
combinatorics unsolved
combinatorics

Let a regular polygon with $p$ vertices be given, where $p$ is an odd prime number. At every vertex there is one monkey. An owner of monkeys takes $p$ peanuts, goes along the perimeter of polygon clockwise and delivers to the monkeys by the following rule: Gives the first peanut for the leader, skips the two next vertices and gives the second peanut to the monkey at the next vertex; skip four next vertices gives the second peanut for the monkey at the next vertex ... after giving the $k$-th peanut, he skips the $2 \cdot k$ next vertices and gives $k+1$-th for the monkey at the next vertex. He does so until all $p$ peanuts are delivered.

[b]I.[/b] How many monkeys are there which does not receive peanuts?
[b]II.[/b] How many edges of polygon are there which satisfying condition: both two monkey at its vertex received peanut(s)?

this question seems rather nice(the second part).
the first part is straightforward. if $a_k$ denotes the position of the monkey receiving the $k^{th}$ peanut then we have, $a_{k+1}=2k+1+a_k(\mod p)$.so we have $a_{k+1}-a_k=2k+1$ for all $k\  \Rightarrow
 a_{k+1}=1+3+\cdots+(2k+1)(\mod p)\Rightarrow a_k=k^2(\mod p)$.
hence the positions which correspond to squares in $\mathbb{F}_p$ get the peanuts and so the number of monkeys which don't receive peanuts is $\frac{p-1}{2}$.(note that $0$ is also a square).

the latter part needs some more work.to count the number of edges which correspond to neighboring monkeys with peanuts, we need to count $\{a\in \mathbb{F}_p|a$ and $a-1$ are both squares in $\mathbb{F}_p\}$.if we denote by $N(x^2-y^2=1)$ the number of solutions to this, then we have
$N(x^2-y^2=1)=\sum_{(a,b),a-b=1} N(x^2=a)\cdot N(y^2=b)$.
Now $N(x^2=a) = 1$ if $a$ is a square and $0$ otherwise so we can write $N(x^2=a)=\frac{1+\left(\frac{a}{p}\right)}{2}$ where $\left(\frac{a}{p}\right)$ is the Legendre symbol.
 now for $u\in \mathbb{F}_p^*$,
\[S_u=\sum_{(a,b),a-b=u} \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\sum_{(a,b),a-b=1}\left(\frac{au}{p}\right)\left(\frac{bu}{p}\right)=S_1\].

so number of solutions to the reqd problem is
\[N=\sum_{a-b=1} N(x^2=a)N(y^2=b)=\sum_{a-b=1} \frac{(1+\left(\frac{a}{p}\right))(1+\left(\frac{b}{p}\right))}{4}=\frac{1}{4}\left[ \sum_{a-b=1} 1 + \left(\frac{a}{p}\right)+\left(\frac{b}{p}\right)+\left(\frac{ab}{p}\right)\right]\].
we can make the following calculations:
 $\sum_{a} \left(\frac{a}{p}\right) = 0$(obvious), $\sum_{a} \left(\frac{a(1-a)}{p}\right)=-\left(\frac{-1}{p}\right)$(for the second sum we use the fact that $S_u=S_1\forall \in \mathbb{F}_p^*$, then calculate $\sum_{u} S_u$ and then simplify-it is easy but needs some writing down which i shall not do here).

if $-1$ is a square $\mod p$ then $0$ occurs in the sums above and doesn't occur if $-1$ is not a square. so if $p\equiv 1(\mod 4)$ then the number of solutions = $1+\frac{p-\left(\frac{-1}{p}\right)}{4}$ and if $p\equiv -1(\mod 4)$ then the number of solutions is $ \frac{p-\left(\frac{-1}{p}\right)}{4}$.

(obviously i have skipped a lot of steps in the calculations,but i think that once the roadmap for the solution is available the calculations can be done individually.i'll elaborate if need be).


Έστω $ f: A \subseteq \mathbb R \to \mathbb R$ τυχούσα συνάρτηση.Υπάρχει ακολουθία πολυωνύμων που να τείνει [b]κατά σημείο[/b] στην $ f$?
Ευχαριστώ εκ των προτέρων για την όποια βοήθεια :) .

Αγαπητέ Γιάννη, η απάντηση είναι όχι τουλάχιστο αν το σύνολο Α έχει μη κενό εσωτερικό. Ας δούμε για παράδειγμα τι γίνεται στην περίπτωση του $ \mathbb{R}$: Τα κατά σημείο όρια συνεχών συναρτήσεων ονομάζονται Baire-1 συναρτήσεις και αποδεικνύεται ότι έχουν σημεία συνέχειας. Αν λοιπόν η συνάρτησή σου είναι ασυνεχής παντού τότε σίγουρα δεν είναι κατά σημείο όριο συνεχών συναρτήσεων. Υπάρχουν βέβαια και απλούστερα παραδείγματα συναρτήσεων που δεν είναι Baire-1. Ένα απλό τέτοιο παράδειγμα είναι η συνάρτηση του Dirichlet (1 στους ρητούς, 0 στους άρρητους).

Η βαλίτσα πάει πολύ μακρύτερα. Τα κατά σημείο όρια Baire-1 ονομάζονται Baire-2, τα κατά σημείο όρια Baire-2 ονομάζονται Baire-3 και γενικά με υπερπεπεπερασμένη επαγωγή μπορούμε να ορίσουμε Baire-α για κάθε αριθμήσιμο διατακτικό α. Η ένωση όλων αυτών των συνόλων των συναρτήσεων συμπίπτει με το σύνολο των Borel μετρησίμων συναρτήσεων.


superior algebra
superior algebra unsolved

Let M be a $ A\minus{}$modul with property:
If $ Q$ is proper submodul of$ M$ then there is a simple modul $ N$ of $ M$ such that $ N \cap Q \equal{} 0.$

Why this implies that $ M \equal{} \Sigma \lbrace N; N$ is simple submodul of $ M \rbrace$?

Use Zorn's Lemma to get a maximal submodule of the form $ \sum_i N_i$ with $ N_i\leq M$ simple. Use the given property to show that this submodule is $ M$.


Determine the positive reals $ x,y$ so that $ 2xy\minus{}x\minus{}y\equal{}0$ and $ x\plus{}y \le 2.$

[hide="hint only"]let a=x-1/2 etc, use am-gm..[/hide]

we have $ 2xy\equal{}x\plus{}y$ adn hence $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}2 \ge \frac{4}{x\plus{}y}$ so $ x\plus{}y \ge 2$ but $ x\plus{}y \le 2$.

AM–GM directly gives $ (x\plus{}y)^2\geqslant4xy\equal{}2(x\plus{}y)\ \Rightarrow\ x\plus{}y\geqslant2.$


How many subsets of the set $ \{1,2,3,4,5\}$ contain the number 5?

5 must be in the subset. Each of the other four elements can either be in the subset or not, so the answer is 2^4=[b]16[/b].


algebra
polynomial
function
Support
integration
real analysis
real analysis unsolved

Let $ \delta$ be Dirac distribution, where $ (\delta,\varphi)\equal{}\varphi(0)$. Find all polynomial $ P(x)$ so that  $ P\cdot \delta\equal{}1$.

All I can say to this is  :what?: (you'd better check the conditions of the problem: you can multiply generalized functions by arbitrary $ C^\infty$ functions, all right, but you have no chance to increase their support this way or to do any of other things required to turn the $ \delta$-function into $ 1$).

[quote="pluricomplex"]Let $ \delta$ be Dirac distribution, where $ (\delta,\varphi) \equal{} \varphi(0)$. Find all polynomial $ P(x)$ so that  $ P\cdot \delta \equal{} 1$.[/quote]

Maybe that dot should be a convolution.

Given that $ g*\delta\equal{}g$ for every generalized function $ g$, this interpretation would hardly be any less strange :?.

you are right fedja, here I need to know how to chose  $ \phi\in C^\infty_c$ so that $ \int \phi dx \equal{}P(0)\phi(0)$ could not hold.


calculus
derivative
symmetry
inequalities unsolved
inequalities

Let $ a,b,c$ be three pairwise distinct real numbers such that $ a\plus{}b\plus{}c\equal{}6\equal{}ab\plus{}bc\plus{}ca\minus{}3$. Prove that $ 0<abc<4$.

The condition for which $ y \equal{} x(x \minus{} 3)^2$ has 3 distinct intersection of points with $ y \equal{} abc$ gives $ 0 < abc < 4$.

In detail [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=207499[/url]

[hide="Not quite sure what Kunny meant..."]
Suppose we have three real $ a,b,c$ that solve the equation. That means that $ x^3\minus{}6x^2\plus{}9x\minus{}abc\equal{}0$ has the three roots $ a,b,c$ which are all real.
Thus the line $ y\equal{}abc$ meets $ x(x\minus{}3)^2$ in three points. Thus $ abc$ lies (vertically) between the two zeros of the derivative of the cubic.

\[ \frac{d}{dx}x(x\minus{}3)^2 \equal{} (x\minus{}3)^2\plus{}2x(x\minus{}3) \equal{} 3(x\minus{}3)(x\minus{}1) \equal{} 0\]
Thus $ abc\in( f(1),f(3) )$, or $ (0,4)$
[/hide]

me@home, have you seen the link that I gave? :wink:

Some Japanese high school student could solve the problem in 30 seconds. :lol:

[quote="kunny"]More difficult problem.

What's the range of $ a^3 \plus{} b^3 \plus{} c^3?$[/quote]

[hide]$ a^3 \plus{} b^3 \plus{} c^3 \equal{} (a \plus{} b \plus{} c)^3 \minus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)^3 \minus{} 3((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc) \equal{} ... \equal{} 54 \plus{} 3abc$

Hence $ 54 < a^3 \plus{} b^3 \plus{} c^3 < 66$ :wink: [/hide]

i have simple solution by delta. $ c=6-(a+b)$ , $ c(a+b)+ab=9$ , $ a^2+a(b-6)+9-6b=0$
now we get delta: $ \Delta = (b-6)^2-4(b-3)^2>0$ so we get  $ \abs (b-6)>abs(2b-6)$

if $ b>3$ then $ 6-b>6-2b$ , so $ b>0$ then by symmetry $ a,b,c>0$ then $ abc>0$
between $ ab,ac,bc$ we know on of them is must greater than 1 like $ ab>1$
we consider $ ab=y>1$ and $ x=a+b$  by $ AM-GM$ ineq $ x>=2\sqrt{y}$ and we knew $ (6-x)(x)+y=9$
$ x^2-6x+9-y=0$  so the solutions are $ x_1,x_2=3\pm \sqrt{y}$
if $ x=3-\sqrt{y}$ then at before we knew $ x>=2\sqrt{y}$ then $ y<1$ contradiction.

thus $ x=3+\sqrt{y}$ now we must prove that $ abc<4$ it means that $ y(6-x)<4$
$ y(3-\sqrt{y})<4$ we know $ y>0$ exist $ s=\sqrt{y}>1$ now our purpose is show that $ 3s^2<4+s^3$ we know $ s>1$ then we consider $ s=t+1$

$ 3(t+1)^2<4+(t+1)^3$ it means that $ t^3+2>3t$ this is obviouse by $ AM-GM$ ineq. then we are done. :wink:

Here is the solution I sent for the contest

[hide]
We have:
$ ab \plus{} bc \plus{} ca \equal{} 9 \Leftrightarrow ab \equal{} 9 \minus{} c(a \plus{} b) \equal{} 9 \minus{} c(6 \minus{} c) \equal{} c^2 \plus{} 9 \minus{} 6c \equal{} (c \minus{} 3)^2 \Leftrightarrow abc \equal{} c(c \minus{} 3)^2$

Similarly we get: $ abc \equal{} b(b \minus{} 3)^2$, and $ abc \equal{} a(a \minus{} 3)^2$

Hence $ a(a \minus{} 3)^2 \equal{} b(b \minus{} 3)^2 \equal{} c(c \minus{} 3)^2 \equal{} abc$

Suppose that $ abc \equal{} 0$, then two of the three numbers must be equal with $ 0$ or $ 3$, but this is a contradiction because the three numbers are distinct, hence we have:
$ abc \equal{} a(a \minus{} 3)^2 \equal{} b(b \minus{} 3)^2 \equal{} c(c \minus{} 3)^2 \neq 0$.
Now obviously the three numbers are not equal with $ 0$ or $ 3$ and all three have the same sign, and because $ a \plus{} b \plus{} c \equal{} 6 > 0$
we have that all three numbers are positive, hence $ abc > 0$

Suppose now that:
$ abc > 4 \Leftrightarrow abc \minus{} 4 \equal{} a(a \minus{} 3)^2 \minus{} 4 \equal{} b(b \minus{} 3)^2 \minus{} 4 \equal{} c(c \minus{} 3)^2 \minus{} 4 > 0 \Leftrightarrow c^3 \minus{} 6c^2 \plus{} 9c \minus{} 4 \equal{} a^3 \minus{} 6a^2 \plus{} 9a \minus{} 4 \equal{} b^3 \minus{} 6b^2 \plus{} 9b \minus{} 4 > 0 \Leftrightarrow (c \minus{} 1)^2(c \minus{} 4) \equal{} (b \minus{} 1)^2(b \minus{} 4) \equal{} (c \minus{} 1)^2(c \minus{} 4) > 0 \Leftrightarrow a,b,c > 4 \Rightarrow a \plus{} b \plus{} c \equal{} 6 > 12$,
which is a contradiction. Hence we have $ abc \leq 4$

I shall prove now that $ abc \neq 4$. Suppose that:
$ abc \equal{} 4 \Leftrightarrow abc \minus{} 4 \equal{} a(a \minus{} 3)^2 \minus{} 4 \equal{} b(b \minus{} 3)^2 \minus{} 4 \equal{} c(c \minus{} 3)^2 \minus{} 4 \equal{} 0 \Leftrightarrow c^3 \minus{} 6c^2 \plus{} 9c \minus{} 4 \equal{} a^3 \minus{} 6a^2 \plus{} 9a \minus{} 4 \equal{} b^3 \minus{} 6b^2 \plus{} 9b \minus{} 4 \equal{} 0 \Leftrightarrow (c \minus{} 1)^2(c \minus{} 4) \equal{} (b \minus{} 1)^2(b \minus{} 4) \equal{} (c \minus{} 1)^2(c \minus{} 4) \equal{} 0$,
which is true only if two of the three numbers are equal with $ 1$ or $ 4$, but this is again a contradiction because the three numbers are distinct. Hence $ abc \neq 4$

Hence we have $ 0 < abc < 4$ :wink:
[/hide]


[quote="soumik"]Prove no integer of the form (a-b)(ab)^(1/2) can be a square of an integer, where a,b are positive integers. (Note root over is only over ab)[/quote]
Let $ N \equal{} (a\minus{}b)\sqrt{ab}$ Assume $ N$ is a square.
Let $ a \equal{} a_1 \cdot d$, and $ b \equal{} b_1 \cdot d$ such that $ (a_1,b_1) \equal{} 1$. Then we have:$ N \equal{} d^2(a_1\minus{}b_1)\sqrt{a_1b_1}$, hence both $ a_1$ and $ b_1$ are squares. So $ a_1 \equal{} k^2$, $ b_1 \equal{} j^2$ (Remember $ (k,j)\equal{}1$). Thus we have. $ N \equal{} d^2(k\minus{}j)(k\plus{}j)kj$ We see that $ (k,j)\equal{}(k\minus{}j,k)\equal{}(k\plus{}j,k)\equal{}1$ and $ (j,k)\equal{}(j,k\minus{}j)\equal{}(j,k\plus{}j)\equal{}1$ so $ j \equal{} j_1^2$ and $ k \equal{} k_1^2$. Alltogether we have $ N \equal{} (dk_1j_1)^2(k_1^2\minus{}j_1^2)(k_1^2\plus{}j_1^2)$ $ \iff$ $ N \equal{} (dk_1j_1)^2(k_1^4\minus{}j_1^4)$. So $ k_1^4\minus{}j_1^4$ is a square, but that's clearly impossible for $ k_1 \neq j_1$.

Any questions are appreciated.

if you have Number Theory in the subject line then post it in the NT forum! :D

[quote="t0rajir0u"]It was known to Fermat and implies the $ n \equal{} 4$ case of Fermat's Last Theorem.  The usual proof is by descent.[/quote]
Yes, I remember that equations $ x^4\plus{}y^4\equal{}z^4$ and $ x^4\plus{}y^4\equal{}z^2$ can be proven by not very hard, but technic methods, but I have forgotten how to prove it. So, I don't know, can it be updated to $ x^2\plus{}y^4\equal{}z^4$ or not.

[quote="Kirill"][quote="Mathias_DK"]So $ k_1^4 \minus{} j_1^4$ is a square, but that's clearly impossible for $ k_1 \neq j_1$.
[/quote]
Thank you for a nice solution, but are you sure that it's well-known and can be easily proved?[/quote]
Not at all...... I'll try something anyways:P

Assume there is a solution, then let $ k_1, j_1$ be the one with smallest $ k_1$ - if more exists take an abritary.

Let $ (k_1^2)^2 \equal{} (j_1^2)^2 \plus{} a^2$. Since $ (k_1,j_1) \equal{} 1$ we know that all the numbers are relatively prime. So we set $ k_1^2 \equal{} m^2 \plus{} n^2$, and deal with the two cases: (But before we go into this we notice that $ k_1 \equal{} p^2 \plus{} q^2$, $ m \equal{} 2pq$ and $ n \equal{} p^2 \minus{} q^2$, for some $ (p,q) \equal{} 1$ and $ p \neq q \bmod 2$)
Case $ j_1^2 \equal{} 2mn$, $ a \equal{} m^2 \minus{} n^2$:
We have $ k_1 \equal{} p^2 \plus{} q^2$, $ j_1^2 \equal{} 2mn \equal{} 4pq(p^2 \minus{} q^2)$, $ a \equal{} (2pq)^2 \minus{} (p^2 \minus{} q^2)^2$
Thus $ 4pq(p^2 \minus{} q^2)$ is a square and $ p \equal{} p_1^2$, $ q \equal{} q_1^2$ and thus $ j_1^2 \equal{} (4p_1q_1)^2(p_1^4 \minus{} q_1^4)$, but $ q_1 < j_1 < k_1$ contradiction the assumtion of the smallest solution.
Case $ j_1^2 \equal{} m^2 \minus{} n^2$, $ a \equal{} 2mn$.
Then $ j_1^2 \equal{} (2pq)^2 \minus{} (p^2 \minus{} q^2)^2$. Now let $ j_1^2 \plus{} (p^2 \minus{} q^2)^2 \equal{} (2pq)^2$. But this can't be right cause $ j_1 \equiv (p^2 \minus{} q^2) \equiv 1 \bmod 2$ and so $ 1 \plus{} 1 \equiv 0 \bmod 4$ Contradiction.

Yes... That is very well known......

Edit: Ok. I should have memorized the proof for fermats last theorem (case $ n\equal{}4$ that is), cause I just used a lot of time to solve this :P:P:P

[color=green]$ A^4\equal{}B^4\plus{}c^2$
pythagorean triple give us

if A and B are both odd then we have

$ A^2\equal{}u^2\plus{}v^2$
$ B^2\equal{}u^2\minus{}v^2$
$ c\equal{}2.u.v$

$ gcd(u;v)\equal{}1$ and $ u\minus{}v$ odd

$ B^2\plus{}v^2\equal{}u^2$

give us that $ v$ is even so

$ B\equal{}k^2\minus{}j^2$
$ v\equal{}2.k.j$
$ u\equal{}k^2\plus{}j^2$

$ gcd(k;j)\equal{}1$ and $ k\minus{}j$ odd

$ A^2\equal{}u^2\plus{}v^2$ give us

$ A\equal{}k`^2\plus{}j`^2$
$ u\equal{}k`^2\minus{}j`^2$
$ v\equal{}2.k`*j`$

$ gcd(k`;j`)\equal{}1$ and $ k`\minus{}j`$ odd

$ k`.j`\equal{}k.j$

$ k`^2\minus{}j`^2\equal{}k^2\plus{}j^2$

remember

$ gcd(k`;j`)\equal{}1$ and $ k`\minus{}j`$ odd
$ gcd(k;j)\equal{}1$ and $ k\minus{}j$ odd

$ (k\plus{}j)^2\equal{}(k`\plus{}j`)^2\minus{}2.j`^2$

$ (k\plus{}j)^2\plus{}2.j`^2\equal{}(k`\plus{}j`)^2$

so by $ mod 4$, $ j`$ is even

TaUrUs*

[/color]

[quote="TaUrUs*"][color=green]TaUrUs*

[/color][/quote]

Sorry, but I do not understand what you are proving. Could you please explain?

Also, Taurus, how do you make the writing in different colors?

You have posted it here http://www.mathlinks.ro/viewtopic.php?t=223230


Let $ a_1,a_2,...,a_n$ be real numbers such that :
       $ I_{n} \plus{} a_1CosA \plus{} a_2Cos(2A) \plus{} .... \plus{} a_nCos(nA) \in S^{ \plus{} }_{n}(R)$ ,$ \forall A \in S_{n}(R)$
Prove that : $ \sum_{k \equal{} 1}^{n}{a_k} \leq n$
 Here  $ CosA \equal{} \sum_{k \equal{} 0}^{ \plus{} \infty}{\frac {( \minus{} 1)^{k}A^{2k}}{(2k)!}}$
   $ I_{n}$ is identity matrix .
   $ S_{n}(R)$ are set real symmetric nxn matrices .
   $ S^{ \plus{} }_{n}(R)$ are set real symmetric nxn matrices and positive semidefinite .
 :wink: :wink:  :wink:

Again reduce to the case that $ A$ is diagonal.

You can give  ?
   Show  the exist : $ (a_1,a_2,...,a_n)\equal{}(\plus{}(\minus{})1,0,...,0)$ then
  $ I_n\plus{}a_1Cos(A)\plus{}...\plus{}a_nCos(nA) \equal{}I_n \plus{}(\minus{})CosA \in S^{\plus{}}_n(R)$ ,$ \forall A \in S_n(R)$
   Now $ a_1,a_2,...,a_n$ such that  $ I_n\plus{}a_1Cos(A)\plus{}...\plus{}a_nCos(nA) \in S^{\plus{}}_n(R)$,$ \forall A \in S_n(R)$
 take $ A\equal{}\omega(\frac {2 \pi}{n\plus{}1},...,\frac {2n}{n\plus{}1}) \omega^{\minus{}1}$ ($ \omega \in O_n(R)$
from $ Tr(I_n\plus{}a_1Cos(A)\plus{}...\plus{}a_nCos(nA) ) \geq 0$ You need !


conics
ellipse
geometry unsolved
geometry

Draw circles $ (F'), (F'')$ with radii equal to $ V_1V_2, L_1L_2,$ respectively. The common focus $ F$ is inside of both $ (F'), (F'')$ (otherwise the 2 ellipses would not intersect). Construct 2 circles $ (A), (B)$ passing through $ F$ and (internally) tangent to $ (F'), (F'').$ This is degenerated Apollonius problem. Then $ AB$ is the common chord of the 2 ellipses $ \mathcal E', \mathcal E''$ with one common focus $ F$ and the other foci $ F', F'',$ respectively.
[hide="Proof"] Suppose $ A', A''$ are the tangency points of $ (A)$ with $ (F'), (F''),$ respectively. Then

$ V_1V_2 \equal{} F'A' \equal{} F'A \plus{} AA' \equal{} F'A \plus{} AF$ $ \Longrightarrow$ $ A \in \mathcal E'$

$ L_1L_2 \equal{} F''A'' \equal{} F''A \plus{} AA'' \equal{} F''A \plus{} AF$ $ \Longrightarrow$ $ A \in \mathcal E''$

In exactly the same way, $ B \in \mathcal E'$ and $ B \in \mathcal E''.$[/hide]
Can you do this degenerated Apollonius problem ?


algebra
polynomial
Vieta
algebra proposed

Solve for all real numbers:

                           $ \&x+y+z=4\\
                             \&x^2+y^2+z^2=14\\
                             \&x^3+y^3+z^3=34$

Big hint below:
Hint#1 [hide]Let $ P(a)$ be a cubic polinomial with roots x,y,and z [/hide]

Hint#2 [hide]Vieta's formula[/hide]

hello, the solutions are
$ x\equal{}3,y\equal{}2,z\equal{}\minus{}1$
$ x\equal{}3,y\equal{}\minus{}1,z\equal{}2$
$ x\equal{}2,y\equal{}3,z\equal{}\minus{}1$
$ x\equal{}2,y\equal{}\minus{}1,z\equal{}3$
$ x\equal{}\minus{}1,y\equal{}3,z\equal{}2$
$ x\equal{}\minus{}1,y\equal{}2,z\equal{}3$
Sonnhard.

[quote="modularmarc101"]Solve for all real numbers:

                           $ \& x + y + z = 4 \\
\& x^2 + y^2 + z^2 = 14 \\
\& x^3 + y^3 + z^3 = 34$

Big hint below:
Hint#1 [hide]Let $ P(a)$ be a cubic polinomial with roots x,y,and z [/hide]

Hint#2 [hide]Vieta's formula[/hide][/quote]

$ x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz=4^2-2(xy+yz+xz)$
$ \implies 16-2(xy+yz+xz)=14$
$ \implies \boxed{xy+yz+xz=1}$

$ x^3+y^3+z^3=(x+y+z)^3-3x^2(y+z)-3y^2(x+z)-3z^2(x+y)-6xyz=$
$ =4^3-3x^2(4-x)-3y^2(4-y)-3z^2(4-z)-6xyz=$
$ =4^3-12(x^2+y^2+z^2)+3(x^3+y^3+z^3)-6xyz=$
$ =4^3-12\cdot 14+3\cdot 34-6xyz= 34 \implies \boxed{xyz=-6}$

The original system may be written

$ \& x+y+z=4\\
\& xy+yz+xz=1\\
\& xyz=-6$

So $ x,y,z$ are the real solutions of $ w^3-4w^2+w+6=0$
The roots of the above polynomial are $ -1,2.3$
So the solution of the original system are
$ (x,y,z)=(-1,2,3)$
$ (x,y,z)=(-1,3,2)$
$ (x,y,z)=(2,-1,3)$
$ (x,y,z)=(2,3,-1)$
$ (x,y,z)=(3,-1,2)$
$ (x,y,z)=(3,2,-1)$


superior algebra
superior algebra solved

Find all homomorphic images of $\mathbb{Z}[x]$.

[quote="grobber"]Let $f:Z[x]\to A$ be an epimorphism. Then $A=f(Z[x])\cong Z[x]/\ker f$. On the other hand, if $I$ is an ideal of $Z[x]$, the canonical projection $\pi:Z[x]\to Z[x]/I$ is surjective.[/quote]
I don't follow it... Go on, please?

I have nothing to add. Look at [url=http://mathworld.wolfram.com/FirstRingIsomorphismTheorem.html]this[/url].

[quote="grobber"]All the factors of $Z[x]$?.. :? What am I missing here?[/quote]
Maybe I should ask,
(because I'm not familiar with English terms)
what does a factor mean here? Quotient Ring ?

I meant factor rings, i/e/ rings of the form $Z[x]/I$, where $I$ is an ideal of $Z[x]$. Is everything Ok now? Any more questions? :)


We call any eight squares in a diagonal of a chessboard as a fence. The rook is moved on the chessboard in such way that he stands neither on each square over one time nor on the squares of the fences (the squares which the rook passes is not considered ones it has stood on). Then what is the maximum number of times which the rook jumped over the fence?

Let the fence to consist of the fileds $ a1,b2,c3,d4,e5,f6,g7,h8$ (like chess-board). It is clear that the rook can jump over $ a1$ zero times, over $ b2$ at most $ 4$ times, over $ c3$ at most $ 8$, over $ d4$ at most $ 12$, etc. The total number of jumping is at most $ 48$.

Consider four squares $ 4X4$: square "x" (a-d)(1-4), square "y" (a-d)(5-8), square "z" (e-h)(1-4), square "w" (e-h)(5-8). If we are going to jump $ 48$ times then we must start from square "y", jump to square "x" or "w", return to square "y", and so on. But in this case we cannot reach square "z". Thus, we cannot jump $ 48$ times.

The example of $ 47$ jumps is not hard to construct.


On a table there are two bags. The first contains one white marble. The second bag contains one red and one white marble. Blindfolded, Billy takes one marble out of the second bag and puts it in the first bag. He then takes off his blindfold and takes a marble out of bag one. It is a white marble.
What is the probability that the other marble in the first bag is also a white marble?

if he puts a red marble into the first bag, he has to take the white one out and the other is a red.
if he puts a white marble into the first bag, he takes out a white one a gets a white next.

1/2 right?

We have this:
W--------WR

If he takes one out of the second one there are two possibilities
WW-------R
WR--------W

I would think 1/2, but it seems like theres more and shouldn't be that easy, maybe like 2/3 or 3/4 or something, idk, my brains dead right now

Yeah its 2/3 because if we look at the probability of just drawing a white ball, it is
1/2*1 when a white ball goes across
+1/2*1/2 when a red ball goes across
So 3/4.
2 of those 3 fourths come when the other ball is white (1/2=2/4)
so the answer is 2/3

i'm not sure what the answer is, but i got 2/3 and i didn't know if that was right so i asked you guys.

I'm sure thats the answer, look at my solution, and tell me whats wrong

[color=indigo]If he already picked out one white ball the diferent possibilities are:
wr w (picks white)
ww r (picks the first white ball)
ww r (picks the second white ball)
there are 2 out of 3 possibilities for picking a white ball next. So the probability is 2/3.[/color]

i think this problem is harder than it seems, you need to know something about conditional probability  :?:   ?_?

Probability if he chose the red marble:

1/2 * 1/2 = 1/4

Probability if he chose the white marble:

1/2 * 2 = 1

Thus, the answer is:

(1/2 * 1/4) + (1/2 *1) =
1/8 + 1/2 =
5/8

Answer: 5/8

lots of different answers posted.  can someone give a definitive answer :) im not really getting any of the posted solutions   :blush:

well the first time i tried it i got 1/2, but the second time i tried it i got 2/3.  I'm pretty sure the answer is 2/3

yeah 2/3 is right. ladykn1ght's solution clarified it.  nice job

[quote="nat mc"]Yeah its 2/3 because if we look at the probability of just drawing a white ball...[/quote]
I think what you guys arent getting is that it is already given that the first marble from the first bag is white.

If the first draw put a white marble in the bag, the probability is 1.
If the first draw put a red marble in the bag, the probability is 0.

Those are the only two possibilities. so the answer is 1/2

ruofanisonfire,

You're missing the fact that it is NOT 1/2 because consider the two cases for what's actually in the bag:

(Type 1) white-white
(Type 2) white-red

there are 2 types of ways to pick a white if it is Type 1 (and thus 2 cases where you get a white as the second marble) and only one way to pick a white if it is Type 2 (and thus 1 case where you get a red as a second marble).  Thus the probability is 2/3.

you guys are good, my dad couldn't get this!

oh i get it!
well, after consulting a numberhappy friend i did.
so basically the white-white situation has two possible outcomes, because either of the white marbles could be drawn, and in both cases the remaining marble is white.
thanks barnacle!

[quote="mathbomberII"]wow so some people are smarter than your dad?[/quote]
yep, that's weird

Hey Barnacle, look had the correct answer and correct solution first, not LadyKn1ght

oops lol.  nat mc had it first!  even still, something about lady knight's wording makes it click in my head better  :D


modular arithmetic
number theory unsolved
number theory

Let $ g$ be a primitive root for $ \mathbb{F}_{p}$.
Suppose that $ x\equal{}a$ and $ x\equal{}b$ are both integer solutions to the congruence $ g^{x}\equiv h\pmod{p}$.

Prove that $ a\equiv b\pmod{p\minus{}1}$.

Given  x=a and x=b are solutions to g^x = h mod p.
=> g^a = g^b mod p
=> g^a-b = 1 mod p
Order of g modulo p = p-1 (By definition of primitive root)
So, a-b is a multiple of Order, i.e., (p-1).
=> a-b= k*(p-1) where k is an integer.
=> a = b mod (p-1)

It is an elementary problem in primitive roots


Hi, I know there is one more week of chapter meets for Mathcounts, so we don't know the full story yet, but I was wondering how many sixth graders are going to their state Mathcounts meets. There are only a few sixth graders posting here on the AoPS Web site, but probably a lot of you know the grade level of someone else who advanced to state from your chapter.

I know sixth graders have gone to state lots of times before, because last year at Nats there were three or four mathletes who were third-time participants at Nats. I also know that this year, much to the disappointment of the sixth graders present, only seventh and (mostly) eighth graders advanced from our chapter to state. (Our chapter has someone go on to Nats pretty much every year, because it includes a city with lots of high tech industry and thus lots of parents who encourage their children's interest in math.)

What's it like where you live? Do lots of sixth graders show up for chapter, or is that mostly for seventh and eighth graders? What grade are most of the mathletes who go on to state from your chapter?

Yes, we had a kid place 5th and make the "allstar" team, His speed is quite impressive on countdown.

well... from Tulsa...

this year, i don't think there were any 6th graders going to state.

last last year, i went to state as a 6th grader, and got 8th there

last year, a 6th grader beat me at countdown to win chapter and go on to state got 3rd and on to nationals

the thing in utah is that junior high is 7-9 and high school is 10-12 so you won't know anything aboiut mathcounts untill in junior high and my school is like totalty no-math so i didn't learn about it in 8tgh grade dang it i really want a ti-83 se.

We had quite a few sixth graders in our chapter competition because the 6th grades are in a separate school from the junior highs (public) and can field there own teams.  I used to be in a chapter where they only had middle schools (6-8) and there rarely are 6th graders at chapter there.


[quote="british8985"]I mean......is there problems from past AMC or AIME
or sth else
coz I've bought one in tw and I'm afraid if there are same Qs[/quote]

There are tons of problems from past AMCs and AIMEs (which I suffer through). There's also questions from competions like MathCounts, ARML, and Mandelbrot. If you wonder whether the questions are the same in vol 1 and 2, they're different. 2's questions are MUCH harder.

I don't think there are problems directly from the AIME, but there's a lot of "AIME-type" problems... and yeah, the books are definitely worth it if you haven't heard that enough on the forums already  ;)

Like the others have said, the AoPS books are amazing.

In my opinion and many others I believe, they are the best books to improve your problem solving skills.  It covers all the necessary materials to do well in the AMCs, AIMEs, etc...


function
limit
analytic geometry
algebra
domain
real analysis

[quote="mathking123"]if one defines a function f such that
------{1, if x is rational}
f(x)={0, otherwise}

the graph of this function is two lines at y=1, and y=0.
clearly each x value has exactly 1 y-value.

my question is, why does the vertical line test fail?
(it fails in the sense that a vertical line intercepts twice and would lead us to beleive this is not a function.)[/quote]

It does not fail, your grapher cannot graph the actual function down to the accuracy needed to determine it is NOT two lines.

actually, i'm pretty sure that they are lines.

focus on the "line" at y=1. even though one may argue that there are "holes" in the line, what is the thinkness of these holes. the thickness would then be infinitessimal, and thus be equal to 0/nonexistent.

this is very similar to saying that .999... =1. if they differ, how much do they differ by?
.00000000000.......1? well, that difference is infinitessimal, and thus is 0/nonexistent.

This type of function is called the Dirichlet function. You cannot graph it with any accuracy at all, since it is not continuous anywhere. See [url]http://en.wikipedia.org/wiki/Nowhere_continuous_function[/url].

[quote="mathking123"]this is very similar to saying that .999... =1. [/quote]

It's actually even more subtle than that.

Consider a vertical line $ x \equal{} r$ with $ r$ rational.  It intersects the graph at $ (r, 1)$ and nowhere else.  However, there are irrational numbers $ r \pm \epsilon$ arbitrarily close to $ r$ such that $ f(r \pm \epsilon) \equal{} 0$.  In other words, it would even be possible to find a sequence of irrational numbers $ r_i$ such that $ \lim_{i \to \infty} r_i \equal{} r$.

The subtlety here is that we [b]cannot[/b] conclude that $ \lim_{i \to \infty} f(r_i) \equal{} f(r)$.  The LHS is equal to $ 0$ but the RHS is equal to $ 1$.  The property that $ \lim_{i \to \infty} r_i \equal{} L \implies \lim_{i \to \infty} f(r_i) \equal{} f(L)$ is true only for [url=http://en.wikipedia.org/wiki/Uniform_continuity]uniformly continuous[/url] functions (correct me if I'm wrong), which this function clearly isn't.

@distracted

i knew it was called a dirichlet function but i thought the name wasnt important( :lol: )

@t0jarou

is there any way to stay away from limits or is this a "limits topic"?

The Dirichlet function is not a very simple object.

First of all, let's consider the nature of the two "lines" that make up the graph.  Neither are lines in the usual sense; they both have an infinite number of gaps.  One line consists only of points with rational $ x$-coordinate, whereas the other line consists only of points with irrational $ x$-coordinate.  At this point, it would be helpful to know two things about the irrational and rational numbers.

[b]Proposition:[/b]  Between every two rational numbers there is an irrational number.  Also, between every two irrational numbers there is a rational number.

So the situation is infinitely bad.  Given any two points arbitrarily close on either line, there is a gap between them (which actually shows that there are an infinite number of gaps between them).

Are the two lines "the same"?  If we put them together, we get the real line.  Separately, though, the $ y \equal{} 0$ line (which contains the irrationals) turns out to have the same [url=http://en.wikipedia.org/wiki/Cardinality]size[/url] (or [url=http://en.wikipedia.org/wiki/Lebesgue_measure]length[/url], in some sense) as the real line, whereas the $ y \equal{} 1$ line is "smaller."  (It actually has "length" $ 0$, but to explain what this really means is to go even further beyond this forum, and we've already gone quite far enough  :| )

[quote="mathking123"]focus on the "line" at y=1. even though one may argue that there are "holes" in the line, what is the thinkness of these holes. the thickness would then be infinitessimal, and thus be equal to 0/nonexistent.[/quote]

It doesn't matter how big the "holes" are
-- While these "holes" are tiny, that doesn't mean you can't fit a line through them. After all, the width of a line is zero.

The meaning of the vertical line test
-- The vertical line test is a visual device to explain a fundamental principle of a function: given a number in the domain, it produces exactly one output. Visually, if we draw an input at this x value, there can be only one y value corresponding to that x value. Here, this is clearly the case -- if x is rational, then y=1, and if x is irrational, then y = 0.

[quote="t0rajir0u"]Neither are lines in the usual sense; they both have an infinite number of gaps.
[/quote]
Also note that each of these gaps are infinitely small. Loosely speaking, their "width" is the width of one point, i.e., 0. Think of the line $ y \equal{} 1$ with the point $ (1, 1)$ removed. If you were to draw that "line", you would still only be able to draw the straight, continuous line; the gap is infinitely tiny. However, this "line" does not intersect the line $ x \equal{} 1$. If you try to graph Dirichlet's function, by a similar argument, you [i]see[/i] two smooth lines with no gaps. But as was said, there are gaps everywhere.

The vertical line test kind of fails when you're dealing with these functions. Well, they still work in a formal test, but as a basic tool that tells you whether a graph is a function or not on first glance, the vertical line test doesn't necessarily give you the correct results.

[quote="t0rajir0u"]The property that $ \lim_{i \to \infty} r_i \equal{} L \implies \lim_{i \to \infty} f(r_i) \equal{} f(L)$ is true only for [url=http://en.wikipedia.org/wiki/Uniform_continuity]uniformly continuous[/url] functions[/quote]

Just continuity is enough (provided the $ r_i$ and $ L$ are in the domain of $ f$, of course).

Uniform continuity is a slightly different concept...the basic idea is that we can pick any point on a graph, and take a range of heights, and the restriction on the inputs is the same for all points on the graph. As an example, $ f(x) \equal{} x^2$ is *not* uniformly continuous on $ ( \minus{} \infty,\infty)$. Since the function is increasing and increasing, the larger positive $ x$ we have, the more we need to restrict the inputs to get the necessary range.

So basically, saying that a function is uniformly continuous is a stronger condition than saying that the function is continuous.


[quote="pkothari13"]I was looking at some AIME problems (21st, 2003 AIME B), and I found this one, but I didn't really understand the solution given by Kalva. The answer is 048, btw.

Find the largest possible integer n such that √n + √(n+60) = √m for some non-square integer m.[/quote]

square both. n + n + 60 + 2√(n+60n) = m
√(n+60n) = m/2 - n - 30
square both. n+60n = m/4 + n + 900 - mn - 30m + 60n
m/4 + 900 - mn - 30m =0
n = (m+3600-120m)/4m = m/4 + 30 + 900/m

so m is a multiple of 4 AND 900/m is integer; 900 = 2.2.3.3.5.5

n=48 gives us m=300=2.2.3.5.5 ; the only bigger options are 450=2.3.3.5.5 and 900=2.2.3.3.5.5.

But 450 is no multiple of 4 and 900 is a square so m=300 is optimal, and since the function (m-60)/4m is ascending for m>=60, this implies that optimal m has optimal n. And the largest n for which m=300 is n=48. :)

Hey Peter VDD, thanks for that explanation! :)

I understand what you did with the problem, but I don't understand what you mean by the XXX case and how it ties into the problem.

The XXX case is... aww a mistake of mine. Just leave it away. (edited my message)

Well theoretically, I just said m/4 + 30 + 900/m had to be an integer. It's most logical that m is a divsor of 900, but not sure. a+b integer doesn't mean a and b integer.

So I thought there would be a problem is a multiple of 900 but not a multiple of 4, too bad that's impossible :D silly me.


Find every function $ f: \mathbb{R}^ \plus{} \mapsto \mathbb{R}^ \plus{}$ satisfy:
$ (x \plus{} y)f(yf(x)) \equal{} x^2f(f(x) \plus{} f(y)),\forall x,y\in\mathbb{R}^ \plus{}$


Medians divide a triangle into 6 smaller ones. 6 circles are inscribed in the smaller triangles, 4 of which are equal. Prove that the triangle is equilateral.

[quote="gotztahbeazn"]wait aren't all six circles equal in an equilateral triangle... since the triangles formed by the medians are identical...[/quote]
[quote="SplashD"][quote="gotztahbeazn"]wait aren't all six circles equal in an equilateral triangle... since the triangles formed by the medians are identical...[/quote]
Yeah, I think he means if [i]at least[/i] 4 are equal.[/quote]
You seem to think i mean a lot of things splashD. :P
The problem tells you 4 circles are equal; that doesn't mean the others aren't equal. In order to prove the triangle is equilateral, you obviously can't use circular reasoning.

Let the triangle be $\triangle{ABC}$ with medians $AR, CQ, BS$ which are concurrent at $P$. We start by examining $\triangle{APQ}$ and $\triangle{PBQ}$ which have equal areas since their bases are equal ($AQ=QB$) and they share the same altitude. We are given that atleast four circles are equal, so we choose two of them by letting the incircles of $\triangle{APQ}$ and $\triangle{BPQ}$ be equal. Using the formula $rs=A$, because the radii of the incircles are the same and the two triangles have the same area, their perimeters must be the same, implying that $AP=PB$. This implies that $PQ$ is the altitude of both triangles, and thus $CQ$ is the altitude of $\triangle{ABC}$, making $\triangle{ABC}$ iscoceles with $CB=CA$. Thus the medians $AR$ and $BS$ are equal. We now use the two "remaining" circles on $\triangle{PCS}$ and $\triangle{PSA}$ and find that $PS$ is the altitude of $\triangle{CPA}$ and thus $BS$ is the altitude of $\triangle{ABC}$ as well. Thus, $BS, CQ, AR$ are all three altitudes and medians to $\triangle{ABC}$ implying that $\triangle{ABC}$ is equilateral. $\boxed{\mathbb{Q}\mathfrak{ED}}$

hm, i think this will hold true if it is given that only three circles are equal as well :wink:

oh wow, this is the last problem on a usamts :o I would say that #3 looks the most challenging on that page :wink:

Actually, number three was my favorite problem that year.

[hide="Hint"]$x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz)$[/hide]

[quote="Jimmy"]You're right, it also holds for just three circles:

[url]http://www.usamts.org/Tests/USAMTSProblems_16_4.pdf[/url][/quote]

Well, in that problem the three congruent circles are given, with two tangent to the same side of the large triangle.  Is the triangle necessarily equilateral if any three of the inscribed circles are congruent?  For example, what if the inscribed circles are each tangent to different sides of the large triangle?

ah, now that makes a good proof.

Based on that diagram.

Prove (or disprove) that $\triangle{ABC}$ is equilateral if three equal circles are inscribed in $\triangle{FBG}, \triangle{GDC}, \triangle{AGE}$.


geometry
3D geometry
number theory unsolved
number theory

prove that every positive rational number can be wriiten as the sum of three cubes of positive rational numbers


What is the significance of a triangle whose angles are in an arithmetic progression?

Take angle A = x. Then angle B = A +(60 - x) and angle C =  A + 2(60 - x) with( x< 60).
 For example A = 20 , B = 20+40 = 60 , C = 20 + 2(60 - 20) = 100.
 This problem must give you more information.

I'll move this to intermediate. I think it involves that law of sines.

Well, you can call the angles $x-y$, $x$, and $x+y$.  Thus $3x=180\Rightarrow x=60.$   So you have $60-y$, $60$, and $60+y$.  What else do you want to know?

Never mind. I was trying to apply it into the Law of Sines somehow to get a result.


factorial
number theory
prime factorization

Find the largest integer $ n $ such that $ n $ has exactly 4 positive divisors and $ n $ divides $ 100! $.

[hide]Since n has two factors besides 1 and itself, it is a multiple of two other distinct prime factors. The biggest primes under 100 are 97 and 89, which multiply to 8633.[/hide]

yif man12 wrote:LynnelleYe wrote:[hide]Since n has two factors besides 1 and itself, it is a multiple of two other distinct prime factors. The biggest primes under 100 are 97 and 93, which multiply to 9021.[/hide]
93=31*3

Edited...I never pay attention to what I write...

[hide]Actually, I think it's 29791=31^3.....[/hide]

Kent Merryfield wrote:Find the largest integer  such that  has exactly 4 positive divisors and  divides .

[hide]n has four positive divisors if its prime factorization is of the form a^3 or ab for primes a, b.  (This is because in the first case, the divisors will be 1, a, a^2, and a^3; in the latter case, the divisors will be 1, a, b, and ab.)

For the first case, the largest number of the form a^3 is 31^3 = 29791.  (If a > 33, then there will be at most two multiples of it among the first 100 positive integers, but for primes under 33, there are at least three.)  For the latter case, the largest number is 89*97 = 8633(these are the largest two primes under 100).  Thus the largest  is 31^3 = 29791.[/hide]

[hide]

a number with 4 divisors will be in the form a^3 or ab, where a and b are prime.

The largest ab is clearly 89*97=8633

To find the largest a^3 divisible by 100!, we find the largest prime number with only three occurences in the factorization of 100!.

This occurs if:

a<100, 2a<100, 3a<100

a<33

a must be 31

31^3 = 27291

[/hide]

i wonder if that makes sense

Scary; we used the same variables and everything.

Thanks, guys, you've got it settled. There's a tempting and reasonable wrong answer (which you found) and then there's the right answer, which you eventually found. Replace 100! with something much larger, and the answer will still be the cube.

[quote="yif man12"]Hehe, at least my answer was reasonable.  :D

So if it were like $n!$, then $a$ would be the largest prime smaller than $n/3$  right?[/quote]

For smaller n it is not  necessary true. eg  for 7! you  have 35vwhich is greater that 2^3.

For very large n', it is most likely true but I am not sure if a proof would be that easy to prove that that is always true.

I should've known it couldn't be that easy...if it were it would definitely not be in the Intermediate forum...better keep that in mind next time...

What does[quote]$ n $ divides $ 100! $[/quote]mean? Does it mean that n/100! = 0, or 100!/n = 0, or what?

n divides 100! means  100!/n=k where k is some integer.  I ll give you some very basic examples to clarify it...  3 divides 6 because 6/3=2 and 2 is an integer.  4 does not divide 3 because 3/4=.75 and that is not an integer.

The phrase "$a$ divides $b$" means, as joml88 said, that there exists an integer $k$ such that $ka=b$. Another way to say it would be to say "$a$ is a factor of $b$". The only problem with the latter wording is that it may fool you into overlooking the fact that 3 divides 0, which is true.

There's a conventional abbreviation for "$a$ divides $b$", namely $a|b$.

Here are a couple of exercises to help cement the notion:

1. Prove that if $a|b$ and $b|c$, then $a|c$.

2. Fill in the blank: For a prime $p$, $p|n!$ if and only if ____________ .


[quote="Interval"]How does one derive the quadratic formula?[/quote]
[hide]Ok so you have a binomoil in the form of $ax^{2}+bx+c=0$. We want to get rid of a so that $x^{2}$ is alone. So divide by a. Now we have $x^{2}+\frac{b}{a}x+\frac{c}{a}=0$. Now we want to complete the square so subtract to have $x^{2}+\frac{b}{a}x=-\frac{c}{a}$. So now we can complete the square by adding $\frac{b^{2}}{4a^{2}}$. So we have $x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}=-\frac{c}{a}+\frac{b^{2}}{4a^{2}}$. So factor the left side since it is a perfect square. $(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}$. Taking the square root of both sides, $x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}$. Finally subtract  to have the quadratic equation! $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.[/hide]

[hide]The general form of a quadratic is $ax^{2}+bx+c=0$.
We complete the square:

$a(x^{2}+\frac{b}{a}x)+c=0$
$a(x^{2}+\frac{b}{a}+(\frac{b}{2a})^{2})-\frac{b^{2}}{4a}+c=0$
$a(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a}$
$(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}$
$x+\frac{b}{2a}=\frac{\pm\sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$[/hide]

EDIT: I was beaten by a few seconds...


Let B1 and B2(retrograd) - two Brocard points of a triangle ABC
and O is it's circumcenter. Prove OB1 = OB2.

Even more can be said: if $K$ is the symmedian point, then $B_1,B_2$ are symmetric wrt $OK$. I think it was proved on the forum.

Sorry, for giving well-known problem, but I can show an interesting solution.
Let T1 and T2 be Brocard points and  < T1AB = < T1BC = < T1CA = <p.
 Then S(BT1C)/S(ABC) = (sin p/sin b)^2, analogously
S(CT1A)/S(ABC) = (sin p/sin c)^2, S(AT1B)/S(ABC) = (sin p/sin a)^2.
Summing we obtain (1/sin p)^2=(1/sin a)^2+(1/sin b)^2+(1/sin c)^2.
Let A’, B’, C’ be projections of T1 on BC, AC, AB respectively.
Then (A’C’/sin b) = BT1 = c*(sin p/sin b) => A’C’ = c*(sin p).
Analogously we can prove that triangle A’B’C’ is similar to ABC with a coefficient (sin p). As a result using podar triangle aria formula.
  S(A’B’C’)/S(ABC)= (sin p)^2= (R^2 – OT1^2)/4R^2
but  as a result  S(A’B’C’) = S(A”B”C”) => OT1 =OT2.

In a triangle ABC,
the circle which goes through A and tangent in B to the side BC,
the circle which goes through B and tangent in C to the side AC
and the circle which goes through C and tangent in A to side AB
are concurent in Ω - Brocard point.
     As you can see there two such points - Ω1 and Ω2.
The interesting properties which they have :
   OΩ1 = OΩ2.
   KΩ1 = KΩ2.  (K - Lemoine point)
   OΩ1KΩ2 lie on a circle and  OΩ1 ┴ Ω1K.
   For demonstration you wiil better ask to Darij Grinberg or Grobber
   who are the Moderators and know geometry very well.

[quote="prowler"]Sorry, for giving well-known problem, but I can show an interesting solution.
Let T1 and T2 be Brocard points and  < T1AB = < T1BC = < T1CA = <p.
 Then S(BT1C)/S(ABC) = (sin p/sin b)^2, analogously
S(CT1A)/S(ABC) = (sin p/sin c)^2, S(AT1B)/S(ABC) = (sin p/sin a)^2.
Summing we obtain (1/sin p)^2=(1/sin a)^2+(1/sin b)^2+(1/sin c)^2.
Let A’, B’, C’ be projections of T1 on BC, AC, AB respectively.
Then (A’C’/sin b) =[b] BT1 = c*(sin p/sin b) [/b]=> A’C’ = c*(sin p).
Analogously we can prove that triangle A’B’C’ is similar to ABC with a coefficient (sin p). As a result using podar triangle aria formula.
  S(A’B’C’)/S(ABC)= (sin p)^2= (R^2 – OT1^2)/4R^2
but  as a result  S(A’B’C’) = S(A”B”C”) => OT1 =OT2.[/quote]
Any one can explain this?  :|

Let $\Omega_{1}$ and $\Omega_{2}$ be two Brocard points.
They satisfy
$\angle \Omega_{1}AB = \Omega_{1}BC = \Omega_{1}CA = w_{1}$.
$\angle \Omega_{2}BA = \Omega_{2}AC = \Omega_{2}CB = w_{2}$.

Then $\frac{S_{B\Omega_{1}C}}{S_{A}BC}= \frac{B\Omega_{1}\cdot BC\sin w_{1}}{AB\cdot BC \sin b}=\frac{\sin^{2}w_{1}}{\sin^{2}b}$, analogously
$\frac{S_{C\Omega_{1}A}}{S_{ABC}}= \frac{\sin^{2}w_{1}}{\sin^{2}c}$,
$\frac{S_{A\Omega_{1}B}}{ S_{ABC}}= \frac{sin^{2}w_{1}}{\sin^{2}a}$.

Summing we obtain $\frac{1}{\sin w_{1}^{2}}=\frac{1}{\sin^{2}a}+\frac{1}{\sin^{2}b}+\frac{1}{\sin^{2}c}$.
Same thing we get for $\Omega_{2}$:
$\frac{1}{\sin w_{2}^{2}}=\frac{1}{\sin^{2}a}+\frac{1}{\sin^{2}b}+\frac{1}{\sin^{2}c}$.

Thus $w_{1}=w_{2}=w$.

Let $A_{1}, B_{1}, C_{1}$ be projections of $\Omega_{1}$ on BC, AC, AB respectively.
Then $\frac{A_{1}C_{1}}{\sin b}= B\Omega_{1}$, because $B\Omega_{1}$ is diameter in the cyclic $BA_{1}\Omega_{1}C_{1}$.
Also form the Sine Theorem in the triangle $AB\Omega_{1}$ we get
$\frac{B\Omega_{1}}{\sin w}=\frac{c}{\sin b}$. It follows that

$A_{1}C_{1}=B\Omega_{1}\sin b= c \sin w$.

It is not difficult to see that triangle $A_{1}B_{1}C_{1}$ is similar to ABC with a coefficient $sin w$. As a result using podar triangle aria formula.
${\frac{S_{A_{1}B_{1}C_{1}}}{S_{ABC}}= \sin^{2}w}= \frac{R^{2}-O\Omega_{1}^{2}}{4R^{2}}$
Analogously for $\Omega_{2}$:
$\frac{S_{A_{2}B_{2}C_{2}}}{S_{ABC}}= \sin^{2}w= \frac{R^{2}-O\Omega_{2}^{2}}{4R^{2}}$
Thus we get $O\Omega_{1}=O\Omega_{2}$ and we are done.

PS. You can prove $A_{1},B_{1},C_{1},A_{2},B_{2},C_{2}$ are concylic and
$O,\Omega_{1}, K, \Omega_{2}$ are concyclic, where $K$ is Lemoine point (intersection of symmedians)

Thanks for your explaination  :)
We can also prove $\omega_{1}=\omega_{2}$ by prove $cot\omega=cotA+cotB+cotC$  :D


42. Which is the largest?

A) $\sum_{i=1}^{48}{64\choose{i}}{64\choose{48-i}}$

B) ${128\choose{64}}$

C) $\frac{2^{128}}{129}$

D) $\frac{(64!)^{2}}{(32!)^{4}}$

E) Two of the preceeding choices are equal.

Correct Answer:
[hide]
B) ${128\choose{64}}$

...but I do not know how to do the eliminations.
[/hide]

Well, you know that a) is clearly $\binom{128}{48}$ by Vandermonde's Convolution, and it's *much* lesser than #2. You also know that it's not e), since c's not an integer and D is just simply $\binom{64}{32}^{2},$ which is lesser than a), which isn't that hard to prove. So it's b)?


function
number theory proposed
number theory

Find all functions $f: \mathbb{N}\to\mathbb{N}$ such that for each $n$ the following holds:

1) $f(n+1)\mid n$;

2) $f\left(f\left(\frac{n}{f(n+1)}\right)+f(n+1)\right)=n$.

(Note: $0\not\in\mathbb{N}$)

EDIT: Corrected mistakes.

[b]If $f ( n ) = 1$ then $n = 1$ or $n = 2$[/b]
Suppose that $f ( n + 1 ) = 1$. Then
\[ n = f ( f ( n ) + 1 ) | f ( n ) + 1 - 1 | n - 1  \]
Therefore $n = 1$ or $n = 2$.

[b]Prove that $f ( 1 ) = 1$[/b]
Consider $n = p > 2$ a prime greater than $2$. Then $f ( p + 1 ) |p$ and $f ( p + 1 ) \neq 1$.
Therefore $f ( p + 1 ) = p$, hence
\begin{eqnarray*} \left. f ( f ( 1 ) + p \right) = p | p + f ( 1 ) - 1 & \Longrightarrow & p|f ( 1 ) - 1 \end{eqnarray*}
Since we have $p|f ( 1 ) - 1$ for all prime $p$, we must have $f ( 1 ) - 1 = 0$.
Finally we have $f ( 2 ) |1 \Longrightarrow f ( 2 ) = 1$. We conclude that $f( n ) = 1 \Longleftrightarrow n = 1 \mbox{ or } n = 2$.

[b]Prove that for all $n$ we have $f ( n + 1 ) = n$[/b]
Suppose that $f ( n + 1 ) \neq n$, then $d = \frac{n}{f ( n + 1 )} \neq 1$.
Hence $f ( d ) |d - 1$ implies that $f ( d ) \leq d - 1$. And since $d|n$ but $d < n$ we have $d \leq \frac{n}{2}$.
Moreover if $f ( n + 1 ) \neq \frac{n}{2}$, then $d \neq 1$ and $d \neq 2$ hence $f ( d ) \neq 1$ and
\begin{eqnarray} \left. f ( f \left( d \right) + f ( n + 1 ) \right) = n & | & f ( d ) + f ( n + 1 ) - 1 \\ f ( n + 1 ) | n & | & f ( d ) + f ( n + 1 ) - 1 \nonumber\\ f ( n + 1 ) & | & f ( d ) - 1 \nonumber\\ f ( d ) & \geq & f ( n + 1 ) + 1 \nonumber \end{eqnarray}
We therefore obtain
\[ \frac{n}{f ( n + 1 )} = d \geq d - 1 \geq f ( d ) \geq f ( n + 1 ) + 1 \geq f ( n + 1 )  \]
\begin{eqnarray*} \sqrt{n} & \geq & f ( n + 1 ) \end{eqnarray*}
Since $d$ is a divisor of $n$ we have $d \leq \frac{n}{2}$. Concluding from (1) we obtain
\begin{eqnarray*} n & \leq & f ( d ) + f ( n + 1 ) - 1\\ n & \leq & \frac{n}{2} + \sqrt{n} - 1 \end{eqnarray*}
A contradiction for $n \geq 4$. (We already proved that $f ( 1 ) = 1, f ( 2 ) = 1, f ( 3 ) = 2$)
Therefore for a given $n \geq 4$ we have
\begin{eqnarray*} f ( n + 1 ) = \frac{n}{2} & \tmop{or} & f ( n + 1 ) = n \end{eqnarray*}
Suppose that $n$ is even and $f ( n + 1 ) = \frac{n}{2}$. Then
\begin{eqnarray*} f \left( f ( 2 ) + f ( n + 1 ) ) \right. & = & n\\ n & | & f ( n + 1 ) + 1 - 1\\ n & \leq & \frac{n}{2}\\ n & \leq & 0 \end{eqnarray*}
A contradiction since $n \geq 1$ is a positive integer.
Therefore for all $n \geq 1$ we have
\begin{eqnarray*} f ( n + 1 ) & = & n\\ f ( 1 ) & = & 1 \end{eqnarray*}

I noticed few mistakes:

1)
[quote="{x}"]We observe that $f ( n ) = 1 \Longleftrightarrow n = 1 \mbox{ or } n = 2$.[/quote]
You have never proved that $f(2)=1$ (which, in fact, is trivial).

2)
[quote="{x}"]\begin{eqnarray*} f \left( f ( 2 ) + \frac{n}{2} \right) = n + 1 & | & 1 + \frac{n}{2} - 1\\ n + 1 & \leq & \frac{n}{2}\\ 2 n + 2 & \leq & n \end{eqnarray*}[/quote]
There is, actually:
\[ f\left(f(2)+\frac n 2\right)=n\mid 1+\frac n 2-1  \]

3) This is the biggest mistake:
[quote="{x}"]\begin{eqnarray*} \left. f ( f \left( d \right) + f ( n + 1 ) \right) = n & | & f ( d ) + f ( n + 1 ) - 1\\ n & | & f ( d ) - 1\\ f ( d ) & \geq & n + 1 \end{eqnarray*}[/quote]
How did you get the second line from the first one?

1) I don't get you. I proved that $f(1)=1$ just above !

[quote]Prove that $f ( 1 ) = 1$
Consider $n = p > 2$ a prime greater than $2$. Then $f ( p + 1 ) |p$ and $f ( p + 1 ) \neq 1$.
Therefore $f ( p + 1 ) = p$, hence

\begin{eqnarray*} \left. f ( f ( 1 ) + p \right) = p | p + f ( 1 ) - 1 & \Longrightarrow & p|f ( 1 ) - 1 \end{eqnarray*}

Since we have $p|f ( 1 ) - 1$ for all prime $p$, we must have $f ( 1 ) - 1 = 0$. [/quote]

3) This follows from the fact that $f(n+1) | n$.

[quote="{x}"]1) I don't get you. I proved that $f(1)=1$ just above ![/quote]
Typo mistake. I corrected my last post, see the objection again.

[quote="{x}"]3) This follows from the fact that $f(n+1) | n$.[/quote]
It does not. It would follow if we knew $n\mid f(n+1)$, but not the other way round.


geometry circumcircle algebra polynomial trigonometry AMC USA(J)MO

So, how did everybody like the competition?  Here are the problems, for reference :
[hide]1. Prove that $x^{2}-y^{2}= A^{3}$ always has integer solutions $(x,y)$ whenever $A$ is a positive integer.

2. In a non-cyclic quadrilateral $ABCD$, points $E$, $F$, $G$, $H$, are the circumcenters of triangles $ABD$, $BCD$, $CDA$, and $CAB$, respectively.  The diagonals of quadrilateral $EFGH$ meet at point $P$.  Prove that $\overline{AP}\cong \overline{CP}$ and $\overline{BP}\cong \overline{DP}$.

3. Let $P(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$ be a polynomial with integer coefficients.  Suppose there exist distinct integers $a,b,c,d$ such that $P(a) = P(b) = P(c) = P(d) = 4$.  Prove that there exists [u]no[/u] integer $m$ such that $P(m) = 7$.

4. The lengths of two sides of an equilateral triangle are doubled, creating an isosceles triangle, as shown in the diagram at the right [which I omit].  The two longer sides are doubled again creating a third isosceles triangle (all three triangles having the same base). This process is continued indefinitely.  If the measure of the vertex angle of each triangle is represented by $A_{1}, A_{2}, A_{3}, \ldots$, determine, with proof, the value of $(1-\cos A_{1})+(1-\cos A_{2})+(1-\cos A_{3})+\ldots$.

5. Let $S$ be a sequence of consecutive positive integers whose terms have a sum of 2007.  Find all such sequences $S$, and prove that you have found them all.  (Do [u]not[/u] include the trivial sequence consisting of 2007 alone.)[/hide]

I'm really glad Mr. Koppelman started this competition—it gives many more people experience with proof competitions than the USAMO could (in Georgia, anyway).

2 I might lose some on and 3 I might lose a little on, I think that I'll get full on the others. Of course, this is *my* guess, based on USAMTS grading; if it's WOOTlike grading, I'll probably get 5 1s :D. How do ya'll think ya'll did?

[quote="Karth"]I thought all of them were pretty straightforward, apart for the second one. Unfortunately, I didn't bring my compass that day, and I could draw an accurate enough diagram, and my teacher was all like *oh my gods, you can't use a straight edge* even though on the IMO you can use pretty much any geometry construction tool, and I got screwed over. Ah, why must life be so painful? :(

But yea, apart from that, it was pretty cool, I guess.[/quote]

#2 was nifty. I started by constructing EFGH and finding from there ABCD and so forth.

It is possible to be interesting and straightforward simultaneously.

[quote="Boy Soprano II"]2 was pretty simple once you realized you should be considering the perpendicular bisectors of the [i]diagonals[/i] of $ABCD$.  I thought 2 and 3 were the most interesting problems.

Anyhow, I this round counts more heavily than the first one. :D

I mostly draw fairly accurate freehand diagrams.  Couldn't you have used the edge of your paper, Karthik? :P[/quote]

No, see I actually tried that. But my paper kept slipping and I got so frustrated, that, at one point, I took four pencils that my teacher gave me, and I took the points and slammed them so hard against my paper that they broke. But then I had nothing to write with so I had to resharpen everything.

*sigh*

But yea, unless it's like the AMC or AIME or easier problems, I can't draw that well without a straight edge or a compass.

EDIT: This was what I thought while I solved #3 in ~ 15 min. or so...

[hide]So we can kinda let $P(x) = a_{n}(x-a)(x-b)(x-c)(x-d)\prod_{i=1}^{k}(x-e_{i})+4.$ Assume that there is, in fact, an integer $m$ such that $P(m) = 7.$ But, since $a_{n}$ can't change and all the coefficients of the polynomials are integers, all $e_{i}$ are integers. So this implies that:

$P(m) = a_{n}(m-a)(m-b)(m-c)(m-d)\prod_{i=1}^{k}(x-e_{i}) = 3$

But oh no!s... $a_{n}\prod_{i=1}^{k}(x-e_{i})= k$ for some integer $k$ since both are integers. So now,

$P(m) = k(m-a)(m-b)(m-c)(m-d) = 3$

Oh my gods! This can't happen, though, because, the maximum possible terms that $3$ can be expressed as a product as of integers is $3 =-1 \cdot 1 \cdot 3,$ or any permutation or such. Because of this, by the Box Principle, two of $m-a, m-b, m-c, m-d$ must be equal. But oh no!s... this implies that two of $a, b, c, d$ are equal, and this is a contradiction!

Therefore, no integer $m$ exists such that $P(m) = 7.$[/hide]

The answers are up at [url]http://math.kennesaw.edu/[/url] under the "mathematics competition" link in the left margin link thingy. I got three of them right. :)

I got them all, assuming that they're going to accept my terribly pauperly trashlike analytic geometry solution for the geometry one. For like the 238th time I could've done it with my compass  :P But apart from that one, I got the rest, I think, for sure.

Not sure if it's enough to place, though, cuz I sucked at the MC. I had a terrible cold that day and half my time was spent changing layers since my math sponsor kept adjusting the air. Another one fourth of my time was spent frantically searching for and wiping my nose with Kleenex's (TM). The rest of the time was spent doing the problems. So yea... I have no idea as to how I qualified, but whatever.

[quote="Karth"]I had a terrible cold that day and half my time was spent changing layers since my math sponsor kept adjusting the air. Another one fourth of my time was spent frantically searching for and wiping my nose with Kleenex's (TM). The rest of the time was spent doing the problems. So yea... I have no idea as to how I qualified, but whatever.[/quote]
Haha, that sounds like my experience at GCTM last year.

I think I got all of them.  I don't remember how well exactly I did in the first round, but I have a cloudy recollection of a rumor that I was second in state to Phillip, so I might have a vague shot at first...


Find the greatest $C$ such that for all $a,b,c,d$ positive numbers the inequality holds

$6(a^2+b^2+c^2+d^2)+(a+b+c+d)^2 \ge C(ab+bc+cd)$.

C=12. a=d=1, b=c=2 yields equality.

I don't have a proof that it works yet though.

Assume that we had some number C that satisfied the requirements. Then replacing (a,b,c,d) with (c,a,d,b) keeps the inequality true because the inequality holds for all values of a, b, c, and d. Adding this inequality to the original gives
$12(a^2+b^2+c^2+d^2)+2(a+b+c+d)^2\ge C(ab+ac+ad+bc+bd+cd)$. From here, since $(a^2+b^2+c^2+d^2)\ge\tfrac23(ab+\cdots+cd)$ it isn't difficult to deduce that $C\le \tfrac{40}{3}$. Now prove that it holds for $C=\tfrac{40}{3}$ (I'm pretty sure it does).

[quote="zabelman"]Now prove that it holds for $C=\tfrac{40}{3}$ (I'm pretty sure it does).[/quote]
This is wrong: $12<\tfrac{40}{3}$ (read my above post) The main flaw is although they are cyclic of each other, the maximums do -not- occur at the same values, so you can't simply add them.

Let a=d=1 ,b=c=2 we have C  <= 12 .
Now we will prove that
 6( a^2 +... +d^2) + (a+...+d) ^2  >= 12(ab +bc+cd) (*)
Put a=d+x ,c=d+x+y ,b=d+x+y+z ,

(*) is equivalent to :
 3(x-z)^2 + 3z^2 +( 2d-2y-z)^2 >=0

The equality hold if and only if a=d ,b=c = 2a  :)

Nice inequality .
Vasc, will you publish a book of all your interesting( and also ugly :D ) inequalities ?
 :)

My method is similar.
Let $a=b+x$ and $d=c+y$. Ineq becomes
$6(x^2+y^2)+(x+y)^2+4(x+y)(b+c)+4(b+c)^2-12bc \ge 0$,
$3(x-y)^2+3(b-c)^2+(2x+2y+b+c)^2 \ge 0$.

I will seriously think to write an inequality book.  ;)

or without substituition: $(2a+2d-b-c)^2 + 3(a+c-b-d)^2 + 3(b-c)^2 \geq 0$... I didn't actually find this although I was looking for it... I need to look harder!

Vasc, are you going to write a new book after old and new inequalities?

[quote="Vasc"]Now I am very busy. I hope to find time for a new book in 2006. :([/quote]

I see, I'll wait for it.  :D