There are 2000 components in a circuit, every two of which were initially
joined by a wire. The hooligans Vasya and Petya cut the wires one after
another. Vasya, who starts, cuts one wire on his turn, while Petya cuts
one or three. The hooligan who cuts the last wire from some component,
loses. Who of them has a winning strategy?

In the beginning , 1 999 000 wires to cut ( even ) . Vasya always have an even position and Petya an odd position , the final position is even so Vasya wons .

If it is Vasya's turn, the number of wires is even and if it is Petya's turn, it is odd. Eventually there will be $ 1000$ wires (even) in total and Vasya will be forced to cut some of them, therefore losing. Petya wins, but he must be careful not to cut the last wire from any of the components. But he's never forced to do it, while Vasya is.

I think both solutions above are wrong. I didn't see the condition that the 2nd hooligan can cut "1 or 3" wires being used. I also didn't see how the 2nd hooligan can avoid the last wire from SOME COMPONENT.

There's another version (harder) for this problem: the only change is that the 2nd hooligan can now cut 2 or 3 wires at a time now.

[quote="sholly"][b]can anybody translate the following russian solutions into english?[/b]
 :blush: [b](solution of 9.8)[/b]
Мысленно разобьем контакты на четыре одинаковых группы: A, B, C
и D. В каждой группе пронумеруем контакты числами от 1 до 500. Петя
будет отвечатьна любой ход Васи так, чтобы для каждого номера k от
контактов Ak, Bk, Ck и Dk отходило поровну проводов. До начала игры
это условие, очевидно, выполняется. Именно благодаря этому условию
проигрышная ситуация впервые случится после Васиного хода.
Опишем подробно Петину стратегию. Если Вася перерезает провод
между контактами одной группы, например, провод AiAj , то Петя пере-
режет провода BiBj , CiCj и DiDj . Если Вася перерезает провод меж-
ду проводами из разных групп и с разными номерами, например, провод
AiBj , то Петя в ответ перережет провода AjBi, CiDj и CjDi. Еслиже
Вася перерезал провод между контактами из разных групп с одинаковы-
ми номерами, например, провод AkBk, то Петя перережет провод CkDk.
Заметим, что из описанной стратегии Пети следует, что провода, которые
он собирается резать, не будут отрезаны до его хода. ПоэтомуПетя всегда
сможет разрезатьт ребуемые провода.
Отметим, что каждый раз после хода Пети от контактов Ak, Bk, Ck
и Dk отходит поровну проводов; при этом от одного из них столько же
проводов отходило уже после Васиного хода. Поэтому ситуация, когда от
одного контакта отрезан последний провод, случится впервые после Ва-
синого хода. Так как количество проводов конечно, проиграет Вася.

[b](solution of 11.8)[/b]Разобьем контакты на четыре одинаковых группы A, B, C иD. В каж-
дой группе пронумеруем контакты числами от 1 до 500. Мысленно покра-
сим в черный цвет провода между контактами с разными номерами, и в
белый цвет—между контактами с одинаковыми номерами.
Петя будет отвечатьна любой ход Васи так, чтобы для каждого номе-
ра k от контактов Ak, Bk, Ck и Dk отходило поровну черных проводов, и
если у одного из контактов больше нет белых проводов, то их не было бы
и у других контактов с таким же номером. До начала игры это условие,
очевидно, выполняется. Именно благодаря этому условию проигрышная
ситуация впервые случится после Васиного хода.
Опишем подробно Петину стратегию. Сначала рассмотрим случай,
когда Вася режет черный провод. Если Вася перерезает провод меж-
ду контактами одной группы, например, провод AiAj , то Петя перережет
провода BiBj, CiCj и DiDj. Если Вася перерезает провод между прово-
дами из разных групп и с разными номерами, например, провод AiBj, то
Петя в ответ перережет провода AjBi, CiDj и CjDi. Такие ходыПетямо-
жет сделать, так как из возможности отрезать один провод от некоторого
контакта следует возможностьот резатьпо одному проводу от вершин с
таким же номером.
Остается рассмотретьслуч ай, когда Вася перерезал белый провод,
т. е., провод между контактами из разных групп, но с одинаковыми но-
мерами. Рассмотрим четыре контакта Ak, Bk, Ck и Dk. Первоначально
любые два из них соединены белым проводом. После того, как Вася пере-
резал первый из этих проводов, например, провод AkBk, Петя перережет
два провода так, чтобы между этих контактов осталосьт ри провода, име-
ющие один общий конец (например,Петя может перерезатьпров одаBkCk
и CkAk, после чего останутся провода AkDk, BkDk и CkDk, что подтвер-
ждает возможностьтакого хода). Если же Вася когда-нибудьперережет
один из этих трех проводов, то от одного из контактов Ak, Bk или Ck он от-
режет последний провод к контактам с этим же номером k, следовательно,
от этого контакта будет отходитьеще какой-то черный провод. Значит, и
от трех других контактов с номером k будут отходитьч ерные провода, сле-
довательно,Петя может перерезать два оставшихся белых провода между
контактами с номером k, что он и сделает.
Отметим, что каждый раз после хода Пети описанное выше условие
выполняется. Следовательно, Петя всегда сможет сделатьход, и, так как
количество проводов конечно, проиграет Вася.[/quote]

Mentally, we divide the contacts on the four identical groups: A, B, C
and D. In each group number the contact numbers from 1 to 500. Petya
will otvechatna any course Wasi, so that for each number k of
Contact Ak, Bk, Ck and Dk departed tie wires. Prior to the game
this condition is obviously satisfied. Precisely because of this condition
a losing situation for the first time will happen after Vasina turn.
Describe Petya strategy. If John cuts the wire
between the contacts of one group, for example, wire AiAj, then Peter re -
cut the wires BiBj, CiCj and DiDj. If Vasya cuts the wire cross -
Do the wires from different groups and with different numbers, for example, wire
AiBj, then Peter in response to sever the wire AjBi, CiDj and CjDi. Eslizhe
Vasya cut the wire between the contacts of different groups with similar -
mi numbers, for example, wire AkBk, then Peter will cut the wire CkDk.
Note that the described strategy Petit follows that the wires that
he is going to cut, will not be cut off before his turn. PoetomuPetya always
able razrezatt rebuemye wires.
Note that after each of the Petit-contact Ak, Bk, Ck
and Dk departs equally wires, while one of them the same
wires departed after Vasina turn. Therefore the situation where from
single contact cut the last wire to happen the first time since the Wa -
sinogo turn. Since the number of wires of course, lose Vasya.

(solution of 11.8) Divide the contacts into four equal groups A, B, C iD. In each
doy group numbered contact numbers from 1 to 500. Mentally, the coating
sim black wire between contacts with different numbers, and
white-between contact with the same numbers.
Peter will otvechatna any course Wasi so that each number be -
ra k from contact Ak, Bk, Ck and Dk departed equally black wires, and
if one of the contacts are no more white wires, then they would not
and other contacts with the same number. Prior to the game this condition,
obviously satisfied. It is because of this condition losing
situation happens for the first time after Vasina turn.
Describe Petya strategy. First consider the case
When Vasya cut the black wire. If Vasya cuts the wire cross -
Do contact one group, for example, wire AiAj, then Peter will cut
wire BiBj, CiCj and DiDj. If Vasya cuts the wire between the conductivity
dami from different groups and with different numbers, for example, wire AiBj, then
Peter in response to sever the wire AjBi, CiDj and CjDi. Such hodyPetyamo -
Jette do so because of the possibility of cutting one wire from some
contact should vozmozhnostot rezatpo one wire from the tops of
the same number.
Remains rassmotretsluch ah, when Vasya cut the white wire,
ie, the wire between the contacts of different groups, but with the same no -
measures. Consider four contact Ak, Bk, Ck and Dk. Initially
any two of them connected to the white wire. After the re-Vasya
cut the first of these wires, for example, wire AkBk, Peter sever
two wires so that these contacts between ostalost ri wires, IME -
yuschie one common end (for example, Peter may pererezatprov odaBkCk
and CkAk, after which the wires will AkDk, BkDk and CkDk, which confirmed
nied by vozmozhnosttakogo move). If Vasya once nibudpererezhet
of these three wires, then one of the contacts Ak, Bk and Ck it from -
cut the last wire to the contacts with the same number k, hence
from this contact will othoditesche a black wire. Hence, the
three other contacts with the number k will othoditch ernye wire, fol -
Consequently, Petya could cut the two remaining white wires between
contacts with a number k, which he did.
Note that after each of the above conditions Petit
performed. Therefore, Peter will always be able sdelathod, and, as
number of wires of course, lose Vasya.

-From Google Translate


analytic geometry
vector
geometry
parallelogram
similar triangles
geometry proposed

Along the sides of triangle $ ABC$ built outside ,equilateral triangles $ ABD,BCE$ and $ ACF$.Show that triangles $ ABC$ and $ DEF$ have the same center of gravity.

According to the Kiepert's theorem, the result can be generalized. Construct outwardly on $ BC,CA,AB$ three similar isosceles triangles $ BDC,CEA$ and $ AFB.$ Then $ \triangle ABC$ and $ \triangle DEF$ share the same centroid.

Let $ \angle BCD \equal{} \angle CAE \equal{} \angle ABF \equal{} \theta.$ Using Conway's notation, the barycentric coordinates of $ D,E,F$ with respect to $ \triangle ABC$ are given by

$ D \ ( \minus{} a^2: S_C \plus{} S_{\theta}: S_B \plus{} S_{\theta})$

$ E \ (S_C \plus{} S_{\theta}: \minus{} b^2: S_A \plus{} S_{\theta})$

$ F \ (S_B \plus{} S_{\theta}: S_A \plus{} S_{\theta}: \minus{} c^2)$

Since the sum of the homogeneous barycentric coordinates of $ D,E,F$ equals $ 2S_{\theta},$ due to the identities $ S_B \plus{} S_C \equal{} a^2,$  $S_A \plus{} S_C \equal{} b^2$ and $S_A \plus{} S_B \equal{} c^2,$ it follows that $ \triangle ABC$ and $ \triangle DEF$ have the same centroid.

I found a elementary solution with vectors but I have a doubt!
Let be $ G$ the center of gravity of the triangle $ ABC$. We have $ \vec{GD}\plus{}\vec{GE}\plus{}\vec{GF}\equal{}\vec{GA}\plus{}\vec{AD}\plus{}\vec{GB}\plus{}\vec{BE}\plus{}\vec{GC}\plus{}\vec{CF}$.Let be $ M$ symmetric of point E in relation to $ BC$.
[b]How can I show that [/b]$ \vec{FM}\equal{}\vec{AD}$.
It looks easy that $ \vec{MC}\equal{}\vec{BE}$.Results that$ \vec{AD}\plus{}\vec{BE}\plus{}\vec{CF}\equal{}\vec{FM}\plus{} \vec{MC}\plus{}\vec{CF}\equal{}\vec{0}$, so $ \vec{GD}\plus{}\vec{GE}\plus{}\vec{GF}\equal{}\vec{0}.$
So how can I get that $ \vec{FM}\equal{}\vec{AD}$.

As said me a friend of mine [b]Nikos Kyriazis[/b], a related more general problem, is coming from the past.

[b][size=100][color=DarkBlue]GENERAL  PROBLEM. - A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup A'BC,\ \bigtriangleup B'AC,\ \bigtriangleup C'AB$ be, three similar triangles erected on its side-segments $ BC,\ AC,\ AB$ respectively, outwardly or inwardly to it, with $ \angle A' \equal{} \angle B' \equal{} \angle C'$ and $ \angle A'BC \equal{} \angle B'CA \equal{} \angle C'AB$. Prove that the triangles $ \bigtriangleup ABC,\ \bigtriangleup A'B'C',$ have the same centroid[/color][/size][/b].

This general problem, is appeared as [b]Theorem 352 - III[/b] in F.G.M book of Geometry.

The proof there, is very elementary and I post here now only the schema, because I think it is enough.

[hide]( We construct the parallelogram $ A'BPC$ and it is enough to prove that $ AB'PC'$ is also a parallelogram. Consider the similar triangles $ \bigtriangleup ABC$ and $ \bigtriangleup C'BP$, to prove that $ B'P\parallel AB'$ ).[/hide]
Kostas Vittas.


inequalities
quadratics
inequalities proposed

Hello there!
Could I ask you for a little favour? I've been trying to do this inequality for quite a long time. I cannot sleep because of it, he-he.

[img]http://img151.imageshack.us/img151/5899/b70d9f7631085e97a0e673dds5.jpg[/img]

It's easy to solve for certain cases. I've proved this for "a"=0 or "b"=0. I also proved this for a=b. But the directing says "Prove it for a non-negative a,b,c". Any hints? I'd really appreciate even a bit of help. Thanks in advance.
Greetings!

It's a quadratic inequality of $ \sqrt{c^3}.$ :wink:


I searched this and found 1 topic, but the information was not adequate.

I have gotten AoPS Vol. 1 from the library. I want to know how much to pace myself. In sixth grade, I got 7th in Texas chapter and 52nd in state (please don't laugh). Now, I have forgotten some formulas, but I have improved a little in Counting and Probability due to Alcumus (thanks admins!). Therefore, I'd still say I'm around the same level. Also, in addtion to how much I should pace myself, should I do the book twice? So as to get a better understanding? Or is it not necessary?

Thanks

Actually I would advise you to get the introductory set, like I did.
I am also in 6th grade and I got 5th Chapter, and 15th State.
I recently bought the introductory set, and it helps A LOT.

I am in 7th grade right now, and I got this from the library; I can't buy.

Besides, you didn't even answer my question.

I did the Aops Vol. 1 Book twice.
The first time I read through it, answered the questions, looked at the solutions if I needed helped.
The second time I tried to get everything correct without looking at the solutions.

Try asking your librarian if he/she can buy some of the AoPS Subject books.

As for going through the book once or twice, just go through it once for now.  If you go through it twice in quick succession, you'll find that you're remembering the exact process for some problems - it's better to wait awhile (say, a month or two) before returning to it.


geometry
perimeter
search
pigeonhole principle
circumcircle
floor function
geometry proposed

Let $ A$ be a set of $ n\geq 2$ points on a plane. Prove that there exists a circle which contains at least $ \left[\frac{n}{3}\right]$ points of $ A$ among circles (involving perimeter) with some end points taken from $ A$ as the diameter, where $ [x]$ is the greatest integer which is less than or equal to $ x$.

isnt this a combinatorial probelm?  :maybe:

Nice problem.

You too can prove that exist a circle with at leat 2[n+2/3] points of A ou from it.

Answer of Reynaldo Gil Pons:
Let $ A_1, A_2 ... A_n$ be the points that belong to the set A.
Let`s suppose that $ A_1 and A_2$ are the farthest points of the set A.
Then all the points of the set A are in the intersection of the circles with centers $ A_1 and A_2$ and radius $ A_1A_2$.

Lemma 1: If $ X_1X_2$ is a string of a circle and is not a diameter and $ X_3$ is a point at the shorter arc $ X_1X_2$, then all points of the area that  determinate the string $ X_1X_2$ and the arc $ X_1X_3X_2$ are inside the circle of diameter $ X_1X_2$. It is easy to prove that.

We are going to prove that exist three circles with end points taken from A as diameter such that all points of the set A belong to someone of these circles. Because if it is true, by the Pigeonhole Principle at least $ \frac {n}{3}$ points of the set A belong to one of these three circles, and the premise is proven.

Let $ A_j$ the point that the angle $ A_1A_jA_2$ is minimum.
Let $ U_1$ the circle that its respective circumference pass through $ A_1, A_2 and A_j$.
Then all points of the set A that are in the same side of the line $ A_1A_2$ with $ A_j$, are inside the circle $ U_1$.
It is easy to see that if the angle $ A_1A_jA_2$ is less than $ 90^o$ then all points of the set A belong to the circle with diameter $ A_1A_2$.
Then we suppose that the angle $ A_1A_jA_2$ is more than $ 90^o$.

Let $ A_k$ the point that the angle $ A_1A_jA_2$ is minimum and $ A_k$ be in the other side of the line $ A_1A_2$ respect to $ A_j$.
Let $ U_2$ the circle that its respective circumference pass through $ A_1, A_2 and A_k$.
Then all points of the set A that are in the same side of the line $ A_1A_2$ with $ A_k$, are inside the circle $ U_2$.

It is easy to see that if the angle $ A_1A_kA_2$ is less than $ 90^o$ then all points of the set A, that are in the same side respect o th line $ A_1A_2$, belong to the circle with diameter $ A_1A_2$. Well all points of the set A belong to the circles with diameters $ A_1A_2$, $ A_1A_j$, $ A_jA_2$, and the problem is solved.
Then we suppose that the angle $ A_1A_kA_2$ is more than $ 90^o$.

Like the angles $ A_1A_jA_2$ and $ A_1A_kA_2$ are more than $ 90^o$, then $ A_j$ and $ A_k$ not belong to the circles $ U_2$ and $ U_1$ respectively.

Let $ B_1$ be the center of the circle $ U_1$. Let $ B_2$ be the center of the circle $ U_2$.

Let divide the line $ B_1B_2$ in three parts: (1) the semiline that begin at $ B_1$ and not contain $ B_2$,(2) the segment $ B_1B_2$ and (3) the semiline that begin at $ B_2$ and not contain $ B_1$.

Let $ C_j$ and $ C_k$ the intersections of the line $ A_jA_k$ with the circles $ U_1$ and $ U_2$ respectively.

Suppose without loose generality that $ A_1$ is nearer to $ A_j$ than $ A_2$.

If the line $ A_jA_k$ intersect the line $ B_1B_2$ at the part (1).
Then the angles  $ A_jA_2C_j$ and $ A_kA_2C_k$ are obtuse.
Else the circle with diameter $ A_jA_k$ contain the circles with diameters $ A_jC_j$ and $ A_kC_k$

Then by the lemma 1 all points of the set A belong to the circles with diameters $ A_jA_k$, $ A_jA_2$ and $ A_2A_k$. And the problem is solved.

The other cases (if the line $ A_jA_k$ intersect the line $ B_1B_2$ at the part (2) or (3)) can be solving similarly.

[quote="ZzZzZzZzZzZz"]I think there might be an easier solution:

Draw the largest circumcircle possible that contains all n points (the circumcircle of ABC).  It is easy to prove that this circle exists.  Then draw the circles with diameters AB, BC, and CA.  These circles intersect at the orthocenter of ABC, so they cover the entire triangle.  Furthermore, they obviously cover the entire circumcircle of ABC.  Thus, one of the circles must contain at least $ \lfloor \frac{n}{3} \rfloor$ of the points.[/quote]
you are wrong for two reasons:
1. these circles don't intersect in the triangle orthocenter, they don't even intersect in one point.
2. if triangle $ABC$ has one obtuse angle, then circles you mentioned don't cover all of it's circumcircle.


[hide="hint#2"]
[quote="Altheman"]test some values and see if you can find a pattern [/quote]if you're good at factoring, try it :)
[/hide]

Hmmm....
[hide] I dunno, at first I was thinking it was factoring by grouping to get two binomials, and could then solve each for -1.  But instead I got stuck with this...  :?

$\displaystyle{6xy(4z+1)+4z(3y+2x)=-1}$

I think I am missing something.
[/hide]

[quote="Altheman"]Find all solutions:

$24xyz+6xy+12yz+8zx=-1$

[hide="hint"]test some values and see if you can find a pattern (this generally is a good tactic for problems where the solution is not immediately obvious)[/hide][/quote]

Are you sure the problem statement is correct?  And are you asking for integer solutions?

i want all triples of numbers that satisfy that...i assure you there is a good solution

Is there a better way to factor this out into 3 binomials other than "guess and check?"

There probably is....lets see... :?

[quote="hoshattack"]Is there a better way to factor this out into 3 binomials other than "guess and check?"
[/quote]
I think that the three binomial have to contain factors $2x$ $3y$ and $4z$, but I dunno if it's possible

I don't know what else to say!  It must not have all 3 of them as factors, but it does... Help?  :(

x, y, and z are not all integers, because then the left side is even, but it has to equal $-1$.  i'm not sure how to factor it...

Hmm... not sure to go past here
$a=2x,b=3y,c=4z$
$(a+1)(b+1)(c+1)=a + b + c$
This is just a factored form, so expanding and simplifying merely gets the orignal expression.  Perhaps there is some trick here...

ok could someone post the solution to this problem?

[quote="Gyan"]What is the point here?
It's not even April 1st!

[hide]There are infinite  number of solutions here,
For example take x=0, y=1, z=-1/12 will work  (and so would infinite number of others..., (x=0, y=2, z=-1/24;or x=0, y=3, z=-1/36 .. you get the idea)

Except taking both x, y to be zero ...
 take x= anything you want...
Y = also anything you want ...

and you can find the value of z which will fit. ..[/hide][/quote]

The point is to characterize all the solutions, I think.  Of course there are infinite ones.

[quote]The point is to characterize all the solutions[/quote]
And that point is ? ??

Mathmatically speaking, or at least  for most of the students I know who like math,  problems like this  make   NO sense.

There are infinite  number of solutions, and to find them, all it is required is  to   randomly pick any two values, and find the third..   Not  a least bit interesting. Isn't it?

( Do not want to be unkind,  but  unless I am missing something this is neither "clever" or even an "equation" which would be asked  to be "solved" in any real  test)

Just some thoughts
Ok Altheman: What have I missed?

[quote="Altheman"]wow, sorry, i messed up the problem   :blush: , i realized this when i was typing this problem up as an example for an arml thing...i was intending it to be like

$(2x+1)(3y+1)(4z+1)=0$, but i forgot the $2x+3y+4z$, but you guys got the idea of the problem so no matter fixing it[/quote]

I knew it.  I kneeeeew it.   :D

lol, no wonder I was having trouble factoring, when there were infinite solutions!

[quote="Altheman"]wow, sorry, i messed up the problem   :blush: , i realized this when i was typing this problem up as an example for an arml thing...i was intending it to be like

$(2x+1)(3y+1)(4z+1)=0$, but i forgot the $2x+3y+4z$, but you guys got the idea of the problem so no matter fixing it[/quote]

I was wondering if it was something like that.  I saw that you could get that by multiplying those factors, but without the $2x+3y+4z$ I was pretty well stumped...


quadratics
modular arithmetic
number theory unsolved
number theory

for some positive integer n, if we can make a square by adding one number on the right side, and adding some numbers on the left side

Let's call it "we can make n a square."

For example, since $ 419904\equal{}648^2$, we can make 1990 a square.

Show that, we cannot make a square with 1991.

Thanks.

[hide="solution"] Assume that we can. We will now find the last digit.

Firstly, all squares are congruent to 0, 1, or 4 (mod 5). Thus, the last digit must be 0, 1, 4, 5, 6, or 9.
Next, all squares are congruent to 0 or 1 (mod 4). From the divisibility rule for 4, we know that the last digit must be 2, 3, 6, or 7. Only the digit 6 is a quadratic residue in both modulos, so the last digit has to be 6.

$ 9916\equiv12\pmod{16}$, so our number is congruent to 12 (mod 16). The quadratic residues (mod 16) are 0, 1, 4, or 9. Thus, our number cannot be a perfect square.[/hide]

What happens if you're allowed to add an arbitrary number of numbers to each side?  More generally, is it true that every sequence of digits occurs in $ k^2$ for some $ k$?


find all x,such that $x^{2}\equiv1(modn)$,$0\leq x<n$,and x is a integer number.

Ummm, do you mean for ALL [tex]n[/tex] because that's kind of difficult...

??
all positive integer x?
maybe i dont get this question..

maybe my expression is not so good.
and you can look at the original problem here:[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21460]www.mathlinks.ro/Forum/viewtopic.php?t=21460[/url]

[quote="zhaobin"][quote="kueh"]They want the number of solutions.... for $p$ prime, it is well known that would be half of it - $\frac{p-1}{2}$. But I am not sure about non-prime...[/quote]
yes.you are right.I don't get the problem.
but you have also a mistake. for p is a prime the number is 2.[/quote]

Argh you are right! I mixed it up with quadratic residues  :blush:  :blush:  :blush:  :blush:

I'll give a hint in spoiler

[hide]think chinese remainder theorem, and solutions for $n = p^k$[/hide]

[quote="Arne"]It is not that hard... Write $n = 2^ep_1^{e_1}p_2^{e_2}...p_k^{e_k}$
You can show that
*for $e = 0$ and $e = 1$, the number of solutions equals $2^k$.
*for $e = 2$, the number of solutions equals $2^{k - 1}$.
*for $e \geq 3$, the number of solutions equals $2^{k - 2}$.
I agree that it is not a good question...[/quote]
I agree with you now.
If we know chinese remainder theorem the problem is simple. :)


Results just came in. I did 2 pure maths, metric spaces 51% & algebra 59%, Partial differential equations 79% and 2 advanced theoretical physics, QM 71% & Stat mech 74% all at the third year level. The scores are exam scores.

I like pure maths but find them hard. I usually have trouble doing the problems in pure maths and it takes me a lot of time to do the assignments. Physics and applied maths on the other hand are more natural to me and fittingly I do find them easier.

There is one more semester left this year and then I would have to specialise next year in physics or pure maths and hopefully go to grad school with one of them. I have a feeling the results will be similar, namely for pure maths subjects to be much poorer than the applied ones. Which should I choose? It would seem more natural to do it in physics because of my better results. But I love pure maths deep down. I love physics as well but pure maths is on another level for me. I am a purist and everything about pure maths seem to suit me except that I am bad at it and obviously don't have any talents for it. So it's quite a big problem. I also never did well in maths competitions in high school.

Whichever I choose, I like to make a career out of it as well so it's a big decision. Advice Appreciated.

This is an interesting question and I hope lots of readers will jump in here.

I would recommend specializing in theoretical physics.  You would probably find it a lot easier to pursue a career path in this area, and could certainly learn and practice mathematics on your own time.

These are just my thoughts.  You should still wait and see what other people have to say.

I made a mistake with regards to the options in the poll. Please use the first 2 options.


Let ABC be a triangle and O its circumcircle. A circle O_A is drawn inside O, tangent to O at A' and tangent to AB and AC. Denote A" the middle of the arc BC of circle O which includes A. Symilarly find B' and B", respectively C' and C".
Then A'A", B'B" and C'C" are concurrent.

Best regards,
Sunken rock

a blind man sees better the details

Hi,

Pls, disregard the subject problem, it was a Turkish NMO 2003 problem (see http://www.mathlinks.ro/viewtopic.php?t=22925 ). Sorry, I was not aware of it be4.
However, I have a nice solution for it.

Best regards,
sunken rock

A Blind man sees better the details


I don't know if this is computers, but I just thought that you guys were really good at programming.

Can anyone post like a program that I could type into my calculator (Not download)? I really want to learn more about how you program. I already have the basics down like Pythag Theorem and such.  :D   Thanks..

I assume you have a TI-83+ and not some other kind of programmable calculator.

I'm also assuming you are learning TI-BASIC.

If so, check this out:
http://www.ticalc.org/programming/

(And for a simple program:

prgmFACTOR
:If X=1
:Input X //Gets user input for an integer
:int(X) -> X //Force-casts input to an integer to prevent infinite loops
:ClrHome //Clears the "home screen"
:
:0 -> Z //List index
:2 -> A //Factor to check next
:
:0 -> dim(L1) //Resets list 1
:0 -> dim(L2) //Resets list 2
:
:While A <= sqrt(x) + 1 //While A is less than the square root of X, keep factoring
:Output(1, 1, A) //Displays the factor currently being checked at the top-left corner, kind of like a progress bar
:If (X/A) = int(X/A) //If A evenly divides X
:Then
:Z + 1 -> Z //Increments list counter
:A -> L1(Z) //Stores the current factor in the next spot in list 1
:0 -> L2(Z) //Sets the exponent associated with that factor to 0
:End //Ends If statement
:
:While (X/A) = int(X/A) //While X still has a factor of A
:X/A -> X //Set X to X/A
:Output(2, 1, "                ") //Clears Row 2 of the screen
:Output(2, 1, X) //Displays the current X
:L2(Z) + 1 -> L2(Z) //Adds 1 to exponent of current factor
:End //End loop
:
:A+1 -> A //Checks next number to see if it factors X
:End //End factoring
:
:If X != 1 //Use the "Does not equal" symbol, not "!="; if X is not 1
:Then
:Z + 1 -> Z
:X -> L1(Z)
:1 -> L2(Z)
:1 -> X
:End

OMG! Thanks so much for that web and the program!


calculus
integration
calculus computations

If $ C$ is any curve from from $ (0,0,0)$ to $ (1,1,1)$ and $ F(x,y,z)\equal{}(z\plus{}4y){i}\plus{}(2z\plus{}4x){j}\plus{}(2y\plus{}x){k}$

What is the line integral $ \displaystyle\int_C  F\,dr$

Parametrize $ C$, which is simple since it's just a line, stick everything in and go from there.


It there anything online that uses Latex code to generate pictures that can be copy and pasted onto e-mails and blogs?

Thanks!

[quote="Stefan_K"]Several more can be found here:

[list][*] [url=http://texblog.net/latex-link-archive/online-compiler/]Online LaTeX Compiler[/url][/list]
Stefan[/quote]
Are there any that I've missed in the list at [url=http://sixthform.info/steve/wordpress/?p=59]Online LaTeX[/url] which I would like to keep as complete as possible?

It seems to me: MonkeyTeX and LaTeXLab. At first I didn't see one site with the title "LaTeX to PDF" but I knew it with the name "Web LaTeX" because that's used as heading (H1), while I'm reading the headings in the content I don't check every time if the browser header has changed.

Your list is very extensive and useful. I took just a brief glimpse but will come back to it for sure.

Stefan

[quote="stevem"][list][*][url=http://sixthform.info/steve/wordpress/?p=59]Online LaTeX[/url]
[*][url=http://sixthform.info/steve/wordpress/?p=65]mathTeX helper[/url]
[*][url=http://sixthform.info/steve/wordpress/?p=67]CodeCogs Equation Editor[/url]
and last, but definitely not least, AoPS's very own [*][url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url][/list][/quote]

[color=darkblue]Thank stevem very much. I need it.[/color]

The way I use is to just choose a forum and use the preview option. When you're done copying and pasting the images, click "cancel".

The only problem with that is that every year or so, I believe they clean the preview latex and other images, so the old url wont work causing your image to not work.

Actually, at least for Google Sites, you can copy the image, not the image location. Besides, on Windows, you can click "Save As".


geometry
3D geometry
icosahedron
combinatorics proposed
combinatorics

Find the greatest possible number of lines in space that all pass through a single point and the angle between any two of them is the same.

It is very well known that the number of equiangular lines in $ \mathbb{R}^d$ is at most $ \binom{d\plus{}1}{2}$.
For $ d\equal{}3$ we can not have more than 6 lines. and this is reached by the long diagonals of the icosahedron  :wink:


This was made to public just today (even though we knew that we made it 3 weeks ago, we were adviced not to make it public knowledge).

[list]
[*]Oleg Ivrii, 11
[*]Jnos Kramr, 11
[*]Dong Uk (David) Rhee, 10
[*]Peng Shi, 10
[*]Jacob Tsimerman, 10
[*]Yufei Zhao, 10[/list]

Leader: Dr. Christopher Small (University of Waterloo)
Deputy Leader: Dr. Edward Wang (Wilfrid Laurier University)
Observer with Leader: Prof. Felix Recio (University of Toronto)

We definitely have a very young team this year.

Congratulations and good luck at the IMO!

Wow, that is a very young team. Four sophomores!!!
Btw congratulations.

obviously this advice was not followed very well ;)  Peng and Yufei was quick to announce it to me immediately after they got a phone call.  I even thought of emailing every canadidate asking them to tell me if they are not on the IMO team, that way using the principle of inclusion/exclusion i could figure out the rest.  :D

A very young team, and what should be a very good team, especially two years from now.  You guys will do Canada proud.


Click the image below to get a zoomed-in version:

[url=http://img81.imageshack.us/my.php?image=triangleproblemfx1.png][img]http://img81.imageshack.us/img81/2026/triangleproblemfx1.th.png[/img][/url]

$\bold{ABC}$ is an equilateral triangle, region $A$ is an equilateral triangle, and so are regions $B$ and $C$. $P$ is some arbitrary point inside triangle $\bold{ABC}$. Regions $A$, $B$, and $C$ are formed by making lines parallel to the three sides passing through point $P$.

Given that the area of region $A$ is $\frac{a^{2}\sqrt{3}}{4}$, the area of region $B$ (the equilateral triangle) is $\frac{b^{2}\sqrt{3}}{4}$, the area of region $C$ (the equilateral triangle) is $\frac{c^{2}\sqrt{3}}{4}$, find the area of the hexagon with bold sides.

As usual, prove your results.

[hide="Hint"]This problem is time-consuming, but there is an easy solution using similar triangles. [/hide]

[hide] Well, for one thing, we know that the sides of the triangles are a b and c respectively.  Then we know that the three quadrilaterals are parellelograms (by angles we can prove them to be parellell).  So the side of the big triangle is a+b+c.
And we know that parrellelograms are split into two congruent triangles by their diagonals.  So from this we know that two times the area of the hexagon minus the areas of the triangles equals the area of the big triangle.  so from this, a few easy algebra steps follow and i got  h= (a^2+b^2+c^2+ab+ac+bc)sqrt(3)/4.  where h is the area of the hexagon.[/hide]

[hide="Alternative"]Or just note that the three scalene triangles which are included in the hexagon have an angle of $60^\circ$ between the pairs of sides $(a,b), (b,c),(c,a)$ - and from that you directly get $\frac{(a^{2}+b^{2}+c^{2}+ab+bc+ca)\sqrt{3}}{4}$[/hide]


modular arithmetic
number theory proposed
number theory

[quote="Mathias_DK"]Prove that: $ 15^3 \mid 4^{n \minus{} 1}(15n \minus{} 32) \plus{} 15n \plus{} 2$ for all nonnegative integers $ n$.[/quote]

Wrong :
$ n\equal{}0$ $ \implies$ $ 4^{n \minus{} 1}(15n \minus{} 32) \plus{} 15n \plus{} 2\equal{}\minus{}6$ not divisible by $ 15^3$
$ n\equal{}2$ $ \implies$ $ 4^{n \minus{} 1}(15n \minus{} 32) \plus{} 15n \plus{} 2\equal{}24$ not divisible by $ 15^3$
...

[quote="pco"]
Wrong :
$ n \equal{} 0$ $ \implies$ $ 4^{n \minus{} 1}(15n \minus{} 32) \plus{} 15n \plus{} 2 \equal{} \minus{} 6$ not divisible by $ 15^3$
$ n \equal{} 2$ $ \implies$ $ 4^{n \minus{} 1}(15n \minus{} 32) \plus{} 15n \plus{} 2 \equal{} 24$ not divisible by $ 15^3$
...[/quote]
Sorry :blush: I have edited the first post now. It should've been $ 2^{4(n\minus{}1)}$ instead of $ 4^{n\minus{}1}$

[quote="Mathias_DK"]Prove that: $ 15^3 \mid 2^{4(n \minus{} 1)}(15n \minus{} 32) \plus{} 15n \plus{} 2$ for all nonnegative integers $ n$.[/quote]
Then :

$ n\equal{}0$ $ \implies$ $ 16^{n\minus{}1}(15n\minus{}32)\plus{}15n\plus{}2\equal{}0$ divisible by $ 15^3$
$ n\equal{}1$ $ \implies$ $ 16^{n\minus{}1}(15n\minus{}32)\plus{}15n\plus{}2\equal{}0$ divisible by $ 15^3$
$ n\equal{}2$ $ \implies$ $ 16^{n\minus{}1}(15n\minus{}32)\plus{}15n\plus{}2\equal{}0$ divisible by $ 15^3$
$ n\ge 3$ :
$ 16^{n\minus{}1}\equal{}(15\plus{}1)^{n\minus{}1}\equiv 1\plus{}15(n\minus{}1)\plus{}15^2\frac{(n\minus{}1)(n\minus{}2)}2\pmod{15^3}$ and so the expression becomes :

$ (1\plus{}15(n\minus{}1)\plus{}15^2\frac{(n\minus{}1)(n\minus{}2)}2)(15n\minus{}32)\plus{}15n\plus{}2\pmod{15^3}$

$ \equal{}(15n\minus{}14)(15n\minus{}32)\minus{}16\cdot 15^2(n^2\minus{}3n\plus{}2)\plus{}15n\plus{}2\pmod{15^3}$

$ \equal{}225n^2\minus{}690n\plus{}448\minus{}(225n^2\minus{}675n\plus{}450)\plus{}15n\plus{}2\pmod{15^3}$

$ \equal{}0n^2\plus{}0n\plus{}0\equal{}0\pmod{15^3}$

Hence the result.

Let $ a_n \equal{} 2^{4(n\minus{}1)}(15n\minus{}32)\plus{}15n\plus{}2$. We can verify that both $ a_0,a_1$ are zero.
Then: $ n \in \mathbb{N} \implies a_{n\plus{}2} \equal{} 32a_{n\plus{}1}\minus{}256a_n\plus{}15^3n$. Now use induction.

[quote="r3d30m3j"]Let $ a_n \equal{} 2^{4(n \minus{} 1)}(15n \minus{} 32) \plus{} 15n \plus{} 2$. We can verify that both $ a_0,a_1$ are zero.
Then: $ n \in \mathbb{N} \implies a_{n \plus{} 2} \equal{} 32a_{n \plus{} 1} \minus{} 256a_n \plus{} 15^3n$. Now use induction.[/quote]

Yesss, quite short and nice! Congrats :)


inequalities
calculus
calculus computations

I want to show directly these sequences are not Cauchy:

1. $ \displaystyle((\minus{}1)^{n})$

So i started with the definition, for $ n>m$

$ |x_{n}\minus{}x_{m}|\equal{}|\displaystyle((\minus{}1)^{n}\minus{}(\minus{}1)^{m})|\le 2$

So how do I use an $ \epsilon$ solution to say that this is not cauchy?

2.$ \displaystyle(n \plus{} \frac{(\minus{}1)^{n}}{n})$

Again, I started with the definition, for $ n>m$:

$ |n \plus{} \frac{(\minus{}1)^{n}}{n}\minus{} \displaystyle(m \plus{} \frac{(\minus{}1)^{m}}{m})|\equal{}|(n\minus{}m)\plus{}\frac{(\minus{}1)^{n}m\minus{}(\minus{}1)^{m}n}{nm}|\le|n\minus{}m|\plus{}|\frac{m\minus{}n}{nm}|< n \plus{}\frac{1}{n}$

should I simply say for the last step that as $ n$ gets large, $ n \plus{}\frac{1}{n} \rightarrow \infty$? or how could i say this using an $ \epsilon$ statement?

as you can tell, I m new to this, so help would be appreciated, especially if I m doing anything wrong.

[quote="mstrip"]I want to show directly these sequences are not Cauchy:

1. $ \displaystyle(( \minus{} 1)^{n})$

So i started with the definition, for $ n > m$

$ |x_{n} \minus{} x_{m}| \equal{} |\displaystyle(( \minus{} 1)^{n} \minus{} ( \minus{} 1)^{m})|\le 2$

So how do I use an $ \epsilon$ solution to say that this is not cauchy?
[/quote]
Choose $ \epsilon \equal{} 1$. How far along in the sequence do you have to go until any two arbitrary elements past that point in the sequence will be within 1 of each other? You've already answered that - never. Consecutive elements are always 2 apart, whether they're the first two consecutive elements, or the billion-trillionth. So it's not Cauchy.

thanks

can anyone explain the second sequence though?

For #2, the inequality you proved isn't in the useful direction. You want to show that some differences are larger than a fixed constant for large $ n$ and $ m$.

This is also an issue in #1: your inequality doesn't prove anything, and indeed some of those differences are zero.

This is the problem. You're used to proving that things are cauchy sequences, but in this case you need to prove that they are not.

In the first case you need the original expression to be less than any fixed constant greater than zero, epsilon in otherwords. So all your inequalities need to have the original expression less than other stuff.

In this case you are trying to do the opposite, so your inequalities will make the original expression greater than other stuff.

thanks, I really appreciate the explanation


vector
advanced fields
advanced fields solved

Hy this is an unsolved problem from a test, i hope someone can solve it because i dont know how..

Planes:
             P1.      A=(0,3,3) B=(0,-1,0) C=(1,-1,0)
             P2.      3x-4z=-9
             P3.      x=5

Find a line L, that for any point R on L, distance is:

d(R,P1)=d(R,P2)=d(R,P3)

The first plane, after doing the math, gives me P1: -3y+4z=3

Then i take this 3 linear equations and construct a system, which has only one solution, the point (x=5,y=7,z=6)

I dont know how to go on to finish, i suspect that point (5,7,6) is the passing point of the line L, and i think that the line L must be like "symmetric" to the 3 planes, no w i need the direction vector of the line L.

Any suggestions please? Sorry for my bad english, i am from Argentina.

I think you may use the formula of the distance between a point and a plane.

hi

this is just a guess but i think you should find four lines

when you have a point (x,y,z) and a plane a*x+b*y+c*z=d
the distance of the point to the plane equals

abs(a*x+b*y+c*z-d)/sqrt(a*a+b*b+c*c)

express that the distance of the points to the first two planes must be equal
and do the same for the first and third

for the first you will find two planes, and for the last also
this gives you four combinations of planes you can cut
resulting into four lines

Thanks to both of you, i have been thinking about the distance formula, and the thing is, when i get the other 4 planes what can i do with them? Should o calculate the intersection of those?

Using the formula of the distance $d$ of a point $P(x_0,y_0,z_0)$ from a plane $ax+by+cz+d=0$ is as follows.

\[d=\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2}}\]

Using this, let the point $(s,t,u)$ on the desired line, we have $\frac{|3t-4u+3|}{\sqrt{3^2+4^2}}=\frac{|3s-4u|}{\sqrt{3^2+4^2}}=|s-5|$,yielding fredbe16's craim.


function
group theory
number theory
number theory solved

Hi, i'm stuck with this. I've learned basic group theory, no sylow theorems yet. How a can prove this? Thanks in advance.
Given $ a$ and $ b$ positive integers with $ (a,b) \equal{} 1$ ( $ a$ and $ b$ only have $ 1$ as common divisor). Show that
\[ a^{\phi{(b)}} \plus{} b^{\phi{(a)}}\equiv 1 (\text{mod}\, ab)
\]
$ \phi$ is the totient function.

hint1: [hide]chinese remainder theorem[/hide]
hint2: [hide]fermat's little theorem on $\mathbb{Z}/a\mathbb{Z}$ and $\mathbb{Z}/b\mathbb{Z}$[/hide].

remark about ma_go's hint1 ;-)
[hide]you don't need the full chinese remainder theorem; just the property $a|z, b|z \Rightarrow ab | z$ if $a,b$ are coprime.[/hide]

$a^{\phi (b)}=1$ mod b
and $b^{\phi(a)}=1$ mod a
 therefor QED


i agree, but i am certainly more amazing

Wait what is the point of this thread? Oh wait, this is the Missouri forum.


i am completly unfamalier to the IMO or internation mathematical olympiad. i want to participate in that but dont know how when and where to do that. can anyone do it ?can i do it ? i am currently in college but finished A-level maths with Grade A. can anyone shed some light on this topic please.

You cannot participate in IMO after you enter a university...  :maybe:

did i say i am in university ?I am not in university i am still in college

In my use of English, term college mostly refers to undergraduate study, so we misunderstood, probably (if you're a highschool student). Where are you from? I'm asking you this because it is important for you to know the procedure of qualifying for IMO in your country, as it may differ to procedures in other countries.

i am in England. dont know i have never heard of the IMO thing in here from anyone , but i saw it on Aops i want to participate.currently i am doing my GCSE and A-levels.

If you're elligible for IMO, only way for you to qualify is:

1. enter BMO1
2. qualify for BMO2
3. be among top 20
4. be among top 6 in the IMO UK team selection

I am 18, and i dont know what grades you mean ? you mean the A-level grades  or ? if you mean that then i got a grade A in A-level maths.

When do you study calculus, say, Integral in your country?

and do you have a university entrance exam in your country?

I have just examined the educational system of England.

[url]http://www.sa.il24.net/~ace-nice/newseido.html[/url]

From top, Japan, U.S.A., Canada, Australia, Newzealand, England

Are you in 6th grade?

if you mean Integration and defrentiation, we did integration and defrentiation and some applications a years ago.  let me explain abit clearly. In england after doing GCSE maths for one year and passing it with a a grade C or higher, you study AS pure core maths for the first year (in As you study pure core 1 and core 2) then the second year, you study A2 (which contains pure core 3 and core 4) after finishing that two years with grades higher than C, then that is called a full A-level. and overal for A-level (As+A2) you get a grade after 2 years. which i got an A.
and ya i did integration and defrentiation a year ago and still i am doing , because i am doing further maths this year.

so any chance of taking part in IMO ?or any other international competation for me ?

Judging from the table I linked, you are a university student in Japan, rather than other countries we may say except England

lol i cant read that  unless u teach me how to.

binomial_4eva, wait a moment, please.

kunny

Based on your posts and the general level of knowledge you have shown, you will not qualify for the British IMO team. You are welcome to participate in whatever local open competitions you can find, and the national system used to select the IMO team is described [url=http://www.artofproblemsolving.com/Wiki/index.php/United_Kingdom_mathematics_competitions]here[/url].


This is something of a subsection of the favorite music/bands thread.

I am an oldies fan, a group rapidly shrinking in number in the teenage demographic.  I am curious to find out who else here enjoys this particular genre of music, and what particular nuances of said genre are considered most enjoyable.

I like the Beach Boys, Stevie Wonder, Elton John, the Righteous Brothers, the Jackson Five, and nearly any oldie to which one can dance.  I do not personally like rock (as opposed to rock'n'roll), and am no particular fan of the British Invasion period (with a few exceptions, of course).

I will most likely think of more to add here, but until then, the floor is open.

I don't know if you are considering it part of the invasion but I think the Beatles are great.

Part of the Invasion, but also some of their songs are constituents of the exceptions.  For the most part, however, I consider the Beatles merely so-so.

The Beatles ROCK!  lol.  I love the oldies.  Beach boys, Beatles, 5 dog night, and others which I don't have time to name right now.

When I was little, my parents would listen to the local oldies station (Oldies 103.7 - which later became The River); as a result, I enjoy oldies today.  It must be the nostalgia factor.

My favorite oldies band is (of course) the Beatles.  Aside from that, I like a few songs by a variety of artists, but have no particular faithfulness to any one of them.  The Righteous Brothers (You've Lost That Lovin' Feeling), The Turtles (Happy Together), The Lovin' Spoonful (Summer in the City), Van Morrison (Brown-Eyed Girl), CCR, and the Temptations (My Girl, etc.) are a few such examples.  If you could consider music from the 70's, this list increases significantly: Paul Simon, Jim Croce, James Taylor, etc.

In any event, I think you could call me an oldies fan.

Is anyone else amused at the prospect of liking music that's 100 years old by the time we're 50 or 60?

Yes, the Temptations are excellent, athough my favorite Righteous Brothers song is probably Little Latin Lupe Lu.

Also Sam Cooke, as well as the Drifters and Jay and the Americans.

And for the most part, I think that the term 'oldies' can definitely be expanded to encompass the 70s.

[quote="BHorseMath"] 5 dog night[/quote]
Three Dog Night?
They rock, but
SIMON AND GARFUNKEL RULE!

Also, I happen to be playing a beatles record (that's right, record). They rock too.

The Beatles, The Who, ELO, and Bob Marley and the Wailers are my favorite oldies bands. I happen to like both classic rock and more modern rock (plus Nirvana, which category they belong in IDK)

The Who is Oldies?
I'd say they're more Classic rock...


Given a circle of circumference 1 and a ditch of size $\epsilon > 0$, you can travel in steps of an irrational number $\alpha$. Prove that eventually you will fall into the ditch.

You can solve this problem using Dirichlet's Box Principle, but the following is a possible solution using a   strong theorem.  The solutions assume that the person have feet that are of thickness less than or equal to $\epsilon$.

[b]Solution.[/b] Here, modulo $1$ signifies the fractional part of a number.

[b]Weyl's Theorem.[/b] [i]If $\alpha$ is an irrational number, then the numbers in the set $S=\{n\alpha\}_{n=1}^{\infty}$ are equidistributed in the interval $(0,1)$.[/i]

The person who takes $n$ steps of irrational length $\alpha$ is a distance of $n\alpha\bmod 1$ from the starting point.  Let the ditch of size $\epsilon$ be located a length of $d$ from the starting point.  Since $d\in (0,1)$ and $S$ is equidistributed in the interval $(0,1)$, $d\in S$.

The desired answer is thus yes; you will always fall into the ditch of any size $\epsilon$ in a finite number of steps. $\Box$

[quote="boxedexe"]

[b]Weyl's Theorem.[/b] [i]If $ \alpha$ is an irrational number, then the numbers in the set $ S \equal{}\{n\alpha\}_{n \equal{} 1}^{\infty}$ are equidistributed in the interval $ (0,1)$.[/i]
[/quote]

Uhhh...what is the point of citing a much stronger theorem to prove something?

I think that a first step would be to prove:

For $ \alpha$ irrational and all $ \epsilon>0$, there exists $ n\in\mathbb{Z}$ such that $ \{ n\alpha\}<\epsilon$.

[hide="For those who are stalking my posts"]yes i am looking through engel...[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101656 (and many others)

So maybe it's been solved a billion times already, but I'd like to make sure this solution is valid.

Consider the minimum positive integer $ n$ such that $ \{n\alpha\}<\{\alpha\}$. We have a few cases:
- $ \{n\alpha\}\equal{} 0$ or $ \{n\alpha\}\equal{}\frac{\{\alpha\}}{2}$ lead to quick contradictions since $ \alpha$ must be rational in either case.
- $ 0 <\{n\alpha\}<\frac{\{\alpha\}}{2}$: Consider only steps of length $ n\alpha$. These are less than half the magnitude of $ \alpha$.
- $ \frac{\{\alpha\}}{2}<\{n\alpha\}<\{\alpha\}$: Consider only steps of length $ (n\minus{}1)\alpha$. These are also less than half the magnitude of $ \alpha$, though in the other direction.

Continually generate smaller $ \alpha$ from the last two (and only possible) cases, and we can get any $ \epsilon$, since $ 2^{\minus{}n}$ goes to 0 as $ n$ goes to infinity.

Yes that looks right.

[hide="For those not stalking my posts"] Yes I am looking through Engel.[/hide]


MIT college Duke trigonometry ARML Columbia

I'm invited to the University of South Carolina-Upstate Math Contest. I got picked for it even though I'm in middle school. It will be held this saturday November 11th. Is any one else going? What is the test like? Is it hard?

Our school is not going this year, but we went in the past.  I have never personally gone, but I hear that the contest is decent.  From what I understand, the top prize is a scholarship to USC Upstate, which I'm sure the winner will not take (MIT or something for the winner  :wink: ).  But there will be good competition against you guys if Spring Valley HS goes, and it will be fun.  We're going to Duke instead this week, but let me know how USC Upstate goes.  Oh, by the way, what school are you going with?  I'll ask Asif if he saw you guys.

Oh, and I would definitely use [url=http://faculty.uscupstate.edu/kwaters/MathContest/Results_2005.html]THIS[/url] to prepare for USC Upstate!

[quote="not_trig"]Just wondering... as a North Carolinian, WHY AREN'T YOU GOING TO DUKE MATH MEET????????

(Which, btw, is for all the good people)

OK, not really....  :rotfl:[/quote]

I don't know. I'm in middle school.

Any school has a 10x better chance to win USC Upstate, because all the good teams are at Duke (except for SV, who might not be going to either competition).

[quote="not_trig"]Yeah, that's true

Who's SV?

Southern Virginia?  :ninja:[/quote]

Spring Valley, which is (arkmastermind) Asif's team.

^^^ lmao it's funny how you deleted not_trig and joe's posts ;) ;)

anyways, the results went ok i guess, but asif, phillip, and junda (perfect score) beat me :(...i suck

compared to last year, the test had harder questions, but a lot fewer stupid questions (like mixture problems, etc). The average score was about a 60 and a about a 140 ppl participated (some from NC and GA schools). But the cool part though was that there was a lot more competition this year than last year (judging by the scores). The scores went 150, 130, 130, 120, 110, (the scores which all the ARML A teamers got) and then to like the 90s... , as opposed to last year's 130, 120, 93, and then to the 80s and 70s...

But surprisingly GSSM sweeped everything, individual, team, and relay, Dreher got 2nd, and SV got 3rd

I can say one thing though, the competition for the top 3 spots at USC columbia will be fierce ;) especially since Fangda and Brian didn't go

[quote="joml88"]And Harry.[/quote]

whoops, forgot about him, but yea obviously

Did you mean Junda had a perfect score, or all three had a perfect score?  Also, can you post the Top 10 individuals, in order?  Where did you place?

Team:

1) GSSM
2) Dreher
3) Spring Valley

Where was Summerville? Hammond?

dude no, junda got a 150, asif 130, phillip 130, and noah 110, so you can figure out how i did and placed based on what i said on my previous post ;)

and hammond got 5th :)

and I don't know the EXACT top 10, but that was the top 5, and unfortunately they don't post this stuff till like three months later on their website...i do know how other ARMLers did and last year's MC ppl. (but i don't want to disclose this information in public ;))

[quote="mysmartmouth"]
Where was Summerville? Hammond?[/quote]

Uh, S'ville lost their #1 and #(2 or 3) men from last year.  They certainly won't be able to fill in those spots to where they were last year.  Maybe some people are stepping it up though to make the team decent.  There are certainly some very capable people on the team...whether or not they've practiced enough we'll have to see.

Additionally, Spartanburg is 4 hours from S'ville...

junda's not going to usc columbia btw (exams or something)

so i guess it won't be AS competitive as originally thought

[quote="mysmartmouth"]It will still be extremely competitive, with 3 out of last year's top 4 returning, plus you and a few other good people.[/quote]

make that four out of the top 5 ;)

I didn't do so good. I got about 35th on the competition. Oh well. Even if I did finish in the top 3, I wouldn't go to USC-Upstate. Who would? :D

Did that overzealous admissions lady give her pitch again this year?  That was funny :D

[quote="dark_sunburst"]Um, fifth place was 115...

I happen to know this because I was fifth.

And indeed, she gave her pitch, and it was funny.[/quote]

hey i thought u died off aops? oh ok sorry about that, ok so noah's 110 was 6th...anyways usc columbia is comming up....maybe we should discuss that....er, i'm just glad gssm isn't going ok

[quote]hey i thought u died off aops?[/quote]
More or less, because of GSSM and all, but it's Thanksgiving Break, so I thought I'd spend a little while on here.

No, we're not going...curse you, math exams.


Let ABC a triangle with the fixed vertices B,C  . Let G and H be the centroid and orthocenter of the triangle. A moves on a fixed parallel line to BC

Prove that the locus of G and H are a line and a parabola respectively

[geogebra]860f25d861aaf1bb129896b5a95673cb7a4b5dca[/geogebra]

About G:[/hide]It lies on a line parallel to BC between BC and the line where A moves. If $ d1$,$ d2$ are the distances of that line from BC and the line where A moves respectively we have $ d2$=$ 2$$ d1$


hmm... try constructing a hexagon and go from there.

[quote="mathnerd314"]It's quite easy to construct a dodecagon--simply draw any 12 line segments that form a convex 12-sided figure.

I think you meant a [b]regular[/b] dodecagon :P...

I would construct a regular hexagon inscribed in a circle, and then bisect each of the angles.[/quote]

Way to find a flaw in what everyone says.

As for construction.  You have to draw a hexagon, bisect each of the six central angles and connect the dots, I think.  It may not be regular, because you can never construct a perfect figure, but it will be pretty close.

I would just construct 12 congruent sides that form 150-degree angles.


Given two triangles with vertices A1, B1, C1 and A2, B2, C2 respectively. A1A2, B1B2, C1C2 are extended to meet at a point  say. Now, B1C1 and B2C2 are extended to meet at L, A1B1 and A2B2 meet at N and A1C1 and A2C2 meet at M. Prove that L, M and N are concurrent.

Proof (as given in text):

[i]Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).

Now, the line B2C2 is

det( x  y    z
      1  1-q  1         = 0.
      1  1     1-r)

The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0

i.e. x=0,y/q + z/r  = 0

Therefore, L lies on the line  x/p + y/q + z/r = 0. By symmetry, so do M and N.

Hence proved[/i]

[img]http://www.physicsforums.com/attachment.php?attachmentid=21094&d=1255422239[/img]

The above is the statement of the sum and it's proof. I don't know what a reference triangle is. Nor do I know how he sets the co-ordinates of A2 to be what they are? Could someone please explain this proof to me?

Anyone know how to explain this one? Mr. Kent Merryfield?

Use the barycentric coordinates. See
http://mathworld.wolfram.com/BarycentricCoordinates.html
A curious detail: $ V$ is the centroid of $ A_1,A_2,A_3$; I have not seen this hypothesis.

Could you explain a little bit more, sir? This is quite new to me, and am having trouble understanding it.


LaTeX
inequalities unsolved
inequalities

a,b>0 and and a is not equal to b. we know that A=a+b/2 and B=(ab)^1/2.
prove that:B<(a-b)^2/8(A-B)<A

hello, do you meant $ A\equal{}\frac{a\plus{}b}{2}$,$ B\equal{}\sqrt{ab}$ and $ B<\frac{(a\minus{}b)^2}{8(A\minus{}B)}<A$ for $ a,b>0$ and $ a \ne b$?
Sonnhard.

yes.I am sorry that i did not write it in latex format.

\[ \frac {(a \minus{} b)^2}{8(A \minus{} B)} \equal{} \frac {(a \minus{} b)^2}{8(\frac {a \plus{} b}{2} \minus{} \sqrt {ab})} \equal{} \frac {(\sqrt {a} \plus{} \sqrt {b})^2(\sqrt {a} \minus{} \sqrt {b})^2}{4(\sqrt {a} \minus{} \sqrt {b})^2} \equal{} \frac {(\sqrt {a} \plus{} \sqrt {b})^2}{4}
\]

\[ \text{By AM \minus{} GM : }\frac {(\sqrt {a} \plus{} \sqrt {b})^2}{4} > \frac {(2\sqrt [4]{ab})^2}{4} \equal{} \sqrt{ab}\equal{}B
\]

\[ \text{By Cauchy Schwartz : }A \equal{} \frac {a \plus{} b}{2} \equal{} \frac {(1^2 \plus{} 1^2)((\sqrt {a})^2 \plus{} (\sqrt {b})^2)}{4} > \frac {(\sqrt {a} \plus{} \sqrt {b})^2}{4}
\]


probability
expected value
combinatorics unsolved
combinatorics

In a permutation a1. . . .an of n distinct integers, an inversion is a pair (ai , aj) such that i < j and ai > aj
If all permutations are equally likely, what is the expected number of inversions in a randomly chosen permutation of 1. . . . . n ?


quadratics
search
LaTeX
Diophantine equation
algebra unsolved
algebra

[quote="KapioPulsar"]find all Z solutions of $ x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xyz \equal{} 1.$[/quote]

Considering this as a quadratic in $ x$, for example, we get the discriminant equation $ (y^2\minus{}1)(z^2\minus{}1)\equal{}d^2$ and so a general solution :

Choose any integer $ k\ge 0$ and consider the solutions of Pell equation $ u^2\minus{}kv^2\equal{}1$

Take any two solutions $ (a,b)$ and $ (c,d)$ of this equation and you get a solution of the original problem : $ (x,y,z)\equal{}(ac\pm kbd,a,c)$

Example : $ k\equal{}2$
Set of solutions of Pell equation $ u^2\minus{}2v^2\equal{}1$ : $ \{(3,2),(17,12),(99,70),(577,408), ...\}$
Solutions of $ x^2\plus{}y^2\plus{}z^2\minus{}2xyz\equal{}1$ :
$ (3,2)$ and $ (3,2)$ give the two solutions $ (17,3,3)$ and $ (1,3,3)$
$ (3,2)$ and $ (17,12)$ give the two solutions $ (99,3,17)$ and $ (3,3,17)$
$ (3,2)$ and $ (99,70)$ give the two solutions $ (577,3,99)$ and $ (17,3,99)$
$ (3,2)$ and $ (577,408)$ give the two solutions $ (3363,3,577)$ and $ (99,3,577)$
$ (17,12)$ and $ (17,12)$ give the two solutions $ (577,17,17)$ and $ (1,17,17)$
$ (17,12)$ and $ (99,70)$ give the two solutions $ (3363,17,99)$ and $ (3,17,99)$
$ (17,12)$ and $ (577,408)$ give the two solutions $ (19601,17,577)$ and $ (17,17,577)$
$ (17,12)$ and $ (3363,2378)$ give the two solutions $ (114243,17,3363)$ and $ (99,17,3363)$
$ (17,12)$ and $ (19601,13860)$ give the two solutions $ (665857,17,19601)$ and $ (577,17,19601)$
...

:maybe: ... somehow like [url=http://www.mathlinks.ro/viewtopic.php?search_id=1850612055&t=305542]this[/url]

[quote="petros r"]WE KNOW BY AM-GM INENGUALITY THAT: x^2+y^2+z^2-2xyz>=τριτηριζαx^2+y^2+z^2 -2ριζαx^2+y^2+z^2 let εκτηριζαx^2+y^2+z^2=u then we have 1>=u^2-2u^3 :arrow: u^2-2u^3=0 or u^2-2u^3=1 from the first one we have u=o or u=1/2 if u=o then x=o or y=o or z=o. without the loss of generality we have x=o and then y^2+z^2=1 so y=o and z=1 or y=1and z=0 SO WE HAVE THE TRIPLETS (x,y,z)=(0,0,1) or (0,1,0) or(1,0,0). if u=1/2 then we have no integers solutions. so the only case which left is: u^2+u^3=1 then we have that u=1 or u=0.we have done the case where u=0 so if u=1 then:x^2y^2z^2=1 so one other triplet is (x,y,z)=(1,1,1)  :wink:

kapio pulsar please translate to the others my solution because i am a new member :wink:  :maybe:[/quote]

If I well understood, you claim that there are only $ 4$ solutions $ (1,0,0),(0,1,0),(0,0,1),(1,1,1)$

Have you read my post?
Are you claiming the $ 18$ solutions (from infinitely many) I gave as examples are all wrong ?

[quote="litongyang"][quote="shoki"]we can assume $ x,y,z \in N^3$
Suppose that $ z \equal{} \min \{x,y,z\}$ then if $ z \equal{} 0$ we have $ (x,y) \sim (1,0)$ and if $ z \equal{} 1$ then $ x \equal{} y$.now suppose that $ z > 1$.
then use the same method which is used for finding all solutions of $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3xyz$.[/quote]
Why can assume $ x,y,z \in N^3$?[/quote]

If $ xyz \equal{} 0$, solutions are $ (\pm 1,0,0),(0,\pm 1,0),(0,0,\pm 1)$
If $ xyz < 0$, no solution since $ LHS\ge 2$
If $ xyz > 0$, either $ x,y,z\in \mathbb N^3$,  either two of them are negative, wlog $ x,y$ and then $ ( \minus{} x, \minus{} y,z)$ is a solution in $ \mathbb N^3$

So it's enough to look for solutions only in $ \mathbb N^3$

TO PCO: NO I READ YOUR SOLUTION I JUST ADD SOME SOLUTIONS WHICH YOU IGNORE :(  I AM NOT CLAIMING THAT THIS EQUATION HAS ONLY 4 SOLUTIONS :wink: I WANT TO LEARN TO WRITE IN LATEX FORM CAN YOU HELP ME PLEASE??????
????? :blush:

here is a pdf file which can help u a lot.  :)


for my question see   http://www.picturerack.com/ims/guest/8476.gif

These are known as the Malfatti circles:
[url=http://mathworld.wolfram.com/MalfattiCircles.html]mathworld.wolfram.com/MalfattiCircles.html[/url]

I know what is the formula but I am looking for proof of the formula.


logarithms
calculus
calculus computations

Find all $ b$ such:
$ \displaystyle\sum_{n\equal{}1}^{\infty} (e^{\ln b})^{\ln n}$ converges

Hint: For what value of $ p$ does $ \sum_{n\equal{}1}^{\infty} n^p$ converge?

[quote="sos440"]Hint: For what value of $ p$ does $ \sum_{n \equal{} 1}^{\infty} n^p$ converge?[/quote]
I'm so stupid... thanks a lot!  :blush:


rotation
geometry
geometric transformation
parameterization
advanced fields
advanced fields unsolved

Today in class we were attempting to prove the Fundamental Theorem of Space Curves (at least the uniqueness part). If $C$ and $C^{*}$ are two curves in space such that $\kappa(s)=\kappa^{*}(s)$ and $\tau(s)=\tau^{*}(s)$ for all $s$, then $C$ and $C^{*}$ are the same except for their position in space.

So my prof right at the start made the (not so obvious, but seemingly so at the time) claim "Let $C^{*}$ be translated so that for some $s=s_{0}$ the corresponding points on $C^{*}$ and $C$ are made to correspond.

To see where this is going, we will then rotate about $s_{0}$ to make $(\textbf{t}_{0}^{*}, \textbf{n}_{0}^{*}, \textbf{b}_{0}^{*})$ and $(\textbf{t}_{0}, \textbf{n}_{0}, \textbf{b}_{0})$ also coincide. Etc...

We argued (as a class against the teacher) about that first step for the rest of the class and then for about 40 minutes after class. We fairly convinced him that we can't just state it, but he couldn't come up with how to show it. How do we know that we can translate so that the next rotation is possible? Any ideas? I'll post with an example we used to dissuade him if you would like that.

If you admit that the translation is without a problem, then the rotation is independent of that translation.  I am kinda confused what's wrong you found.

We admit without a problem that it is possible to do, but the problem comes that we can't assume they are the same curve (that is what we are trying to show). How do we choose a $P$ on $C$ and a point $P^{*}$ on $C^{*}$ to translate on to eachother that is going to work.

You can't just choose two random points, or else there has to be some more stuff in there to get the curves to line up.

Does anyone know of a book that has a good treatment of this Theorem, then, if no one knows how to handle this situation.

But you have the correspondence between the curvature and torsion already, we ought to choose the point correspond to the parameter 0 for each.

Do Carmo's book will be fine.


vector
superior algebra
superior algebra solved

$n>2$
Let $A \M_n(C)$, $rank(A)=2$ Prove that there exist $p_i,q_i,r_i,s_i \in \mathbb C$ such that $a_{ij}=p_iq_j+r_is_j$ for any $i,j$ in {1,2,...,n}

From Mircea Becheanu

[quote="Diogene"]Trivial : rank(A)=2  ===> the line(or column) vectors are generated by a 2 vectors family .
:cool:  Frankly man , is it a  " Superior Algebra Problem"  ??!!  - :cool:[/quote]

Yes it is, just not a superior problem in algebra. Still it is quite useful.


7 is hawt, i think  (heh that's how I first read yours).  Only cuz I got a 7 and i wanna make it  :)  :(  but i didn't

yeah, but nothing is ever done by the book...  :P  heh

[quote="idupda"]I hope the floor is 7...
When is it announced? According to the handbook, it says the results are out by today, the 28th...[/quote]

No, according to the handbook it says the decision will be made on the 28th.  Not that they'll post it by then.

Is a $5$ even possible for a floor score?  Has it ever gone that low?

I think it's gotten that low, but not in recent memory. Most likely, it won't happen this year.

No, the floor was never that low, but in 2002, the floor was 6, which would probably be 5 with the expansion.

Hm, I guess the floor will be an 8. Lol a little late because it's already posted and official.

[quote="JSteinhardt"][quote="mathdool"]The majority wins!
That is, if not too many people voted after the results were out.[/quote]

I predict an 8![/quote]

I predict 8.5!

Floor score = 0 with a tiebreaker honoring people with less than 100 on the AMCs


Prove that sequence of points $(x_{n})=(\xi_{k,n})\in l$ $(n=1,2,...)$ is convergent to the point  $x=(\xi_{k})\in l$ if and only if
    1. $\xi_{n,k}\rightarrow\xi_{k}$
  and
    2. $\sum_{k=1}^{\infty}|\xi_{n,k}|$ is uniformly convergent, that is $\forall \epsilon>0 \exists k_{0}\sum_{k=k_{0}}^{\infty}|\xi_{n,k}|<\epsilon, n=1,2,...$.
$l$ is the space of all sequences satysfying the condition:
$\sum_{n=1}^{\infty}|x_{n}|<\infty$.

Proving the first implication (that is. proof that from convergence we can conclude that 2 conditions are satisfied is not difficult). I have a problem with showing that from these 2 points we can conclude the first part.

Thanks all for help

Fix some $N_{0}$ large enough such that condition $2$ is satisfied with some small $\epsilon$ [b]and[/b] such that the similar condition holds for the series in $\xi_{k}$. Now, for the $N_{0}-1$ first terms, use pointwise convergence (condition $1$) to find a $N_{1}$ large enough such that ... sorry I just give the beginning I'm kind of tired  :D (end of the day)


calculus
derivative
calculus computations

[hide=" I hope you learn derivative rules"]

$ v(t) \equal{} (2\minus{} 9t^{2}) \vec{i} \plus{} 8\sec^{2}(2t) \vec{j} \plus{} 4e^{4t} \vec{k}$
$ a(t) \equal{} \minus{}18t \vec{i} \plus{}32\sec^{2}(2t)\tan(2t) \vec{j} \plus{} 16e^{4t} \vec{k}$
$ a\left(\frac{\pi}{8}\right) \equal{} \minus{}\frac{9\pi}{4} \vec{i} \plus{} 64 \vec{j} \plus{}16e^{\frac{\pi}{2}}$

So the answer's D.

[/hide]

this is solved..moderator can close my topic


When we divide the numbers 272 758 and 273 437 by a two digit number $N$, we get remainders of 13 and 17 respectively. Find the sum of the digits of $N$.

Source: Australian Math Competition, S21, 2002

[hide]$272758 \equiv 13 \bmod N$, $273437 \equiv 17 \bmod N$. Hence $N \mid 272745$ and $N \mid 273420$. So $N \mid 675$.
The possible values of $N$ are 15, 25, 27, 45, 75. But only $45$ is solution. So the sum is $9$.[/hide]

Err, if $N = 15$, we can't have a remainder of 17.

aidan, the lyrics in your signature are wrong.

[hide]
gcd(272745,273420) = 45
4+5 = 9
lucky the gcd was already a two digit number, no fiddling :p
[/hide]


Given $\sec{A} = \frac{41}{40}$ and $\sin{A} \leq 0$, express $\cot{(A - 90)}$ as a fraction in lowest terms.

[quote="L_Li"][hide]cot (A-90) = - tan (A)

Since cos A is positive and sin A is negative, A is in the 4th quadrant, and tan A is negative.
Also, A-90 is in the third quadrant, so cot (A-90) is positive.

either draw a unit circle or use the sin and cos angle addition formulae to get:
cot (A-90) = - tan (A)

sec A = 41/40
cos A = 40/41
1 - cos^2 A = sin^2 A       <--- solve for sin A
-9/41 = sin A

tan A = (-9/41) / (40/41) = -9/41
cot (A-90) = - tan A = 9/41[/hide][/quote]

Close, but the final answer is not correct...

[hide]Your steps are all correct, except for one minor error:

$\tan{A} =\left(\frac{-\frac{9}{41}}{\frac{40}{41}}\right) = -\frac{9}{40}$

This means that cot (A-90) = - tan A = 9/40, and in lowest terms the answer would be $\frac{3}{20}$[/hide]

[quote="Mario_KaRt"] [quote]
$\tan{A} =\left(\frac{-\frac{9}{41}}{\frac{40}{41}}\right) = -\frac{9}{40}$

This means that cot (A-90) = - tan A = 9/40, and in lowest terms the answer would be $\frac{3}{20}$[/quote][/quote]

Lol, 9/40 doesn't simplify to 3/20 :-)


geometry
3D geometry
Support
number theory
prime numbers

[quote="Gza"]92931259 --> it is an 8 digit prime number (it's reverse 95213929 is also prime!)

9293 --> it's first four digits form a prime number

1259 --> it's last four digits form a prime number

3929 --> it's first four digits in reverse form a prime number

9521 --> it's last four digits in reverse form a prime number

39299521 --> concatenating the two reverse primes forms an 8 digit prime number

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

Just thought I'd share  :P[/quote]

Thats so cool!!! (I'm a nerd) my house number is the xth prime number, where x is a perfect cube, which is really cool. My screen name number is also a cube number :P!!!

Edit: Welcome to AoPS (if you're new) and congrats on your first post.

Seems like a fairly unique number, but upon closer inspection I discovered that exactly 112649 8 digit prime numbers are the result of concatenating two 4 digit primes.  So if you find two 4 digit prime numbers, concatenate them, and make sure the sum of the digits isn't divisible by 3, there is almost exactly a 20% chance the 8 digit number is prime (there are 1061 four digit primes, so 1125721 unique numbers created by concatenating them).

I do not know, however, how many 8 digit primes are prime in reverse, have two 4 digit prime components which are primes in reverse and can be concatenated to form a new 8 digit prime.  My intuition tells me there is at least one other 8 digit prime with all of these properties.

So unless I made some error somewhere, if you were to make a guess of a four digit prime in your head by:

(1) number ends in 1, 3, 7, or 9
(2) number's digits do not add to a number divisible by 3

you are knocking off all numbers divisible by 2,3,5 so lowest number it can be divisible by is 7

any 4 digit number with these 2 properties (which are easy to check in your head) has a 44% chance of being prime.

So say you chose 2 of these and concatenate them, there is a 20% chance they are both prime and a 4% chance the concatenated number is the construction of two primes and also prime itself.

But perhaps just simply these two properties leaves you with a 20% chance of the 8 digit number being prime regardless if the two 4 digit components are actually prime or not.  The prime number theorem seems to support this.

So hey... if you pick 5 8 digit numbers this way, one of them is probably prime.  Time to go pretend you are an idiot savant?

Wow...that sounds really cool! How'd you find out about this?

I found out that all prime numbers are in the form: $p=\frac{2\times3\times\times\cdots\times q}{n_1\times\cdots\times n_m}\pm r$ where q is a prime number,and $n_1\times\cdots\times n_m$ are the products of all distinct prime factors of r.


Let k be a integer number.Find all f:Z->Z such that:
f(f(f(n)))=f(n+1)+k

[quote="dhthstn"]Let $k$ be an integer. Find all $f: \mathbb{Z} \to \mathbb{Z}$ such that :
$f(f(f(n)))=f(n+1)+k$[/quote]

 :)

EDIT : Sorry.  :oops: See : http://www.mathlinks.ro/Forum/viewtopic.php?t=14027


algorithm
number theory
Euclidean algorithm

Any short cut for a problem like this? I used brute force.

$ 34x \equiv 60 (mod 98)$

Not sure what you mean by brute force. I think the standard way is to use the Euclidean algorithm to solve $ 34x \plus{} 98y \equal{} 60$.


[quote="thanhnam2902"]Find sum: $ S \equal{} 1.2 \plus{} 2.3.x \plus{} 3.4.x^2 \plus{} ... \plus{} n(n \plus{} 1).x^{n \minus{} 1}$[/quote]

consider the series,
x^2 + x^3 + x^4 + ............. + x^(n+1) = x^2(x^(n) - 1)/(x-1)

so differentiate this equation twice wrt x,
LHS contains all the terms of S and the sum is equal to RHS.

hello, i have found $ \sum_{i\equal{}1}^n i(i\plus{}1)x^{i\minus{}1}\equal{}\frac{\minus{}2 n^2 x^{n\plus{}1}\plus{}n^2 x^{n\plus{}2}\plus{}n^2 x^n\minus{}4 n x^{n\plus{}1}\plus{}n x^{n\plus{}2}\plus{}3 n x^n\plus{}2 x^n\minus{}2}{(x\minus{}1)^3}$
Sonnhard.


calculus
integration
inequalities
LaTeX
binomial coefficients
Pascal\u0027s Triangle

How many integrals solutions have

$ a.) x+y+z+r+t < 2005$
$ b.)1900 \leq  x+y+z+r+t < 2005$

[quote="i_like_pie"]I hope you mean positive integers...[/quote]
oops, sorry, its non-negative integers

I can try this:
[hide="1"]
Add an extra variable $ n$ so that it becomes an equality.
$ x+y+z+r+t+n = 2005$
Using balls and urns, there are $ \binom{2010}{5}= 272,042,500,325,652$ solutions (I hope I typed all the digits correctly..)

Subtracting the case where $ n=0$, which does not satisfy the inequailty, we have $ \boxed{272,042,500,325,651}$ solutions
[/hide]
[hide="2"]
This is the same as #1, only we restrict that $ x+y+z+r+t \ge 1900$.
To do this we find the solutions of $ x+y+z+r+t < 1900$, then subtract that from the answer to #1.
$ x+y+z+r+t+n = 1900$
Solutions = $ \binom{1905}{5}-1= 207,975,461,064,756-1 = 207,975,461,064,755$
$ 272,042,500,325,651-207,975,461,064,755 = \boxed{64,067,039,260,896}$
(again, I hoped I typed everything correctly..)
[/hide]

[quote="p4fn2w"]I can try this:
[hide="1"]
Add an extra variable $ n$ so that it becomes an equality.
$ x+y+z+r+t+n = 2005$
Using balls and urns, there are $ \binom{2010}{5}= \boxed{272,042,500,325,652}$ solutions (I hope I typed all the digits correctly..)
[/hide][/quote]

Im just wondering why do you have to add a single variable.  I've seen this in books, but i don't know why. Im using a different method. Can you explain?

I can explain, but then Im assuming my answer is correct?

Basically, the LHS is less than the RHS, so adding one single variable that accounts for the difference of the LHS and the RHS (which we know is always nonnegative, since the LHS < RHS) would make the LHS = RHS.

So if $ 3 < 5$, adding a single variable that varies by case, and in this case the variable should be $ 2$, would make $ 3+2 = 5$ which is true. Similarly $ 1 < 5$, so in this case the variable would be $ 4$.

We add one single variable because that way we know every case is unique, so we avoid over and under counting. If we add two variables we woud have to consider the case where the two added variables are symmetric while keeping the original variables the same value, which is a case of over-counting (since we are only looking at the different values of the original variables, not the added ones). Adding more variables would make the solution even uglier...

I think i got the point. However,

if $ x+y+z+r+t+n = 2005$, if $ n = 0$, then $ x+y+z+r+t = 2005$ which is does not satisfy the given equation if $ x+y+z+r+t < 2005$. Aren't you going to subtract $ 1$, that is the case of $ n = 0$? Im not really sure if i understood it correctly.

[quote="p4fn2w"]Yes, you are right.
I'm sorry... I shall change my solution[/quote]

Please check if my analysis is correct. If we want to get for the number of non-negatve solutions of x + y + z + r + t < 2005, then we should get the number of solutions for the following:

Assuming that $ N_{i}$ is the number of solutions with the sum from 1 to 2004

N1: x + y + z + r + t = 0
N2: x + y + z + r + t = 1
N3: x + y + z + r + t = 2
...
N2003: x + y + z + r + t = 2004

Right?

Now, we know that for the equation
$ n_{1}+n_{2}+...+n_{k}= r$, the number of solutions is $ \binom{k+r-1}{r}$
which means that we have to sum $ N1$ up to $ N2003$. That is the summation of $ \binom{5+r-1}{r}$ where $ r = 1, 2,..., 2004$

Of course, we can solve that using only a computer or a calculator.

Please check if my explanation is correct.

I might not be the most qualified person to answer this, since my combinatorics skill isnt olympiad level yet, but...

Yes, that seems very reasonably correct (however, I think you made a small miscalculation. $ N_{i}$ should go up to $ N_{2005}$, not $ N_{2003}$. its $ N_{2005}: x+y+z+r+t = 2004$ and $ N_{2003}: x+y+z+r+t = 2002$).

Also, since $ N_{1}$ is the number of solutions when the variables sum up to $ 0$ and not $ 1$, you should find the summation of $ \binom{5+r-1}{r}$ starting from when $ r=0, 1, 2, ..., 2004$, not when $ r = 1, 2 , ..., 2004$.

And again, method this is more rigorous than adding an extra variable to make the inequality into an equality. Although both methods are difficult to calculate,
finding $ \binom{2010}{5}-1$ is much more easier than finding $ \sum_{r=0}^{2004}\binom{5+r-1}{r}$ (unless there is some binomial summation property I'm not aware of).

Just remember though, I'm not that great at combinatorics, so I might make incorrect assumptions.

[quote="p4fn2w"]Although both methods are difficult to calculate,
finding $ \binom{2010}{5}-1$ is much more easier than finding $ \sum_{r=0}^{2004}\binom{5+r-1}{r}$ (unless there is some binomial summation property I'm not aware of).[/quote]
Sorry but $ \binom{2010}{5}-1\ne\sum_{r=0}^{2004}\binom{r+4}{r}$. Off by 2% (700 billion).

However, $ \binom{2010}{5}-1 = \sum_{r=1}^{2005}\binom{r+4}{r}$.  This is one case of a standard identity on binomial coefficients, sometimes known as the Hockey Stick Identity.  Your differing solutions can be modified slightly to provide a proof of it in general.

[quote="p4fn2w"]

Yes, that seems very reasonably correct (however, I think you made a small miscalculation. $ N_{i}$ should go up to $ N_{2005}$, not $ N_{2003}$. its $ N_{2005}: x+y+z+r+t = 2004$ and $ N_{2003}: x+y+z+r+t = 2002$).

Also, since $ N_{1}$ is the number of solutions when the variables sum up to $ 0$ and not $ 1$, you should find the summation of $ \binom{5+r-1}{r}$ starting from when $ r=0, 1, 2, ..., 2004$, not when $ r = 1, 2 , ..., 2004$.

[/quote]

You're right.

Now simplifying, $ \binom{5+r-1}{r}= \binom{4+r}{r}$ where $ r=0, 1, 2, ..., 2004$, not when $ r = 1, 2 , ..., 2004$.

Sorry, i don't know the latex for summation.

Now if r = 0, 1, 2,..., 2004

The summation of $ \binom{4+r}{r}$ = $ \binom{4}{0}+\binom{5}{1}+\binom{6}{2}+...+\binom{2008}{2004}$

Since $ \binom{4}{0}$  = $ \binom{5}{0}$
$ \binom{4+r}{r}= [\binom{5}{0}+\binom{5}{1}]+\binom{6}{2}+...+\binom{2008}{2004}$
$ = [\binom{6}{1}+\binom{6}{2}]+\binom{7}{3}+...+\binom{2008}{2004}$
$ = [\binom{7}{2}+\binom{7}{3}]+...+\binom{2008}{2004}=...= \binom{2008}{2003}+\binom{2008}{2004}$
$ = \binom{2009}{2004}$

Is that correct?

The latex code for summation is /sum_{bottom text}^{top text}.

To be honest, I didnt know there existed a property such that $ \binom{x}{y}+\binom{x}{y+1}= \binom{x+1}{y+1}$ (thats how bad at combinatorics I am!)

$ \binom{2009}{2004}= \binom{2009}{5}$ is equivalent to the number of solutions for $ x+y+z+r+t+n = 2004$ using the balls and urns formula, which, I believe, should [b]not[/b] be equal to the number of solutions for $ x+y+z+r+t < 2005$.

I think we made an error somewhere, but I cant put my finger on it... maybe as JBL and i_like_pie have pointed out, we should have summed up $ r$ to 2005 and not 2004, but Im not sure why

[quote="p4fn2w"]
To be honest, I didnt know there existed a property such that $ \binom{x}{y}+\binom{x}{y+1}= \binom{x+1}{y+1}$ (thats how bad at combinatorics I am!)
[/quote]

Its called Chu-Shih Chiek's Identity

It's alright, I am also a beginner in combinatorics, im taking the subject now for the first time. Maybe i just have a good book and a good teacher.  :)

[quote="p4fn2w"]
$ \binom{2009}{2004}= \binom{2009}{5}$ is equivalent to the number of solutions for $ x+y+z+r+t+n = 2004$ using the balls and urns formula, which, I believe, should [b]not[/b] be equal to the number of solutions for $ x+y+z+r+t < 2005$ which does not satisfy the inequality...

Maybe we just dont know the answer but I have a feeling that our answer is correct.
[/quote]

I think I got it, I forgot to change $ N_{i}$, you reminded me that it should go up to 2005 which makes our last term on the summation $ \binom{2009}{2005}$.

Therefore, your answer is correct. That is
$ \binom{2010}{2005}=\binom{2010}{5}$ Whew!

What do you think? Why aren't other good people at this topic helping us.
I have a feeling that your first answer is correct, we can't just find an explanation for my question that if $ n = 0$, then $ x+y+z+r+t$ can equal $ 2005$.

[quote="guile"]Sorry, i don't know the latex for summation.[/quote]
If you hover your mouse over a latex image, it should show you what the actual code is.  (p4fn2w had a slash going the wrong direction.)

[quote="guile"]What do you think? Why aren't other good people at this topic helping us. [/quote]
Looks good to me.  Other people probably aren't helping you because (1) you're doing fine on your own and (2) they have better things to be doing at this particular moment.  (In my case, I was sound asleep when you wrote the question -- it was nearly 4AM where I live.)

[quote="guile"][quote="p4fn2w"]
To be honest, I didnt know there existed a property such that $ \binom{x}{y}+\binom{x}{y+1}= \binom{x+1}{y+1}$ (thats how bad at combinatorics I am!)
[/quote]

Its called Chu-Shih Chiek's Identity[/quote]

Really?  I usually see it called Pascal's Formula or Pascal's Rule.  Then again, mathematical canon tends to attribute ideas poorly.

In the West, we attribute it to Pascal.  It would be totally unsurprising if in China (I assume that's where guile is from) it's attributed to a Chinese mathematician.  [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Zhu_Shijie.html]Here[/url] is a biography.

And of course, it was  also published  in Bhagabati Sutra -- (About 300 BC)  and quite  popularly used  in text books  by, say, Mahavira (in 900AD) or Bhaskara (1100 AD)  ...Centuries before Pascal.

(google bhagabati sutra and pascal triangle - and one may find some interesting material ....)
)BTW in Italy it is known as " Tartaglia's triangle")


I haven't gotten the September segment... just wondering if anyone else got it yet?

i emailed titu about this; he said it will be coming near the end of september

@phelpedo: [url]http://www.awesomemath.org/AMY2007Brochure.pdf[/url]  (scroll down)

Interesting...these seem to be pretty narrow topics for the most part.

Well, if they use topiocs that are too general, they will run out of topics much more quickly than if the topics are narrower.


geometry
trapezoid
trigonometry
quadratics
perpendicular bisector
trig identities
Law of Sines

ABCD is a trapezoid (AD || BC); angle ACB is twice angle ADB; AB=BC=13/2, AD=12.  Find the area of the trapezoid.[/tex]

[hide]angleADB=angleDBC=x=angleABD,

so angleABC=2x=angleACB, thus AB=AC=13/2.

Let AH be the altitude of ABC(which is altitude of ABCD).

AHC is right angle with hypotenuse 13/2 and 12/2.

so AH=5/2

Area is (12+13/2)(5/2)/2=185/8[/hide]

Can you explain how you found AH again?  Thanks.

Beta -- where on earth did the 12/2 come from?  And could you be more detailed in your explanation in general, like explaining what triangles you're using and so on?  (Also, I think you have a typo: you wrote ABC and I think you meant BAC.)

[hide]Triangle ABC is isoseles triangle. Point A must lies on the perpendicular bisector of BC. Therefore H is the midpoint of BC. HC=6, AC=13/2, and AHC is rt. triangle. Using Pythagorian theorem we can find AH.[/hide]

beta wrote:[hide]Triangle ABC is isoseles triangle. Point A must lies on the perpendicular bisector of BC.[/hide]

This is false -- [hide]point B must lie on the perpendicular bisector of AC.[/hide]

In fact, it appears that all of your point labeling is not the same as given in the problem.

Hmm.. is angle ACB the same as angle ABC (and ADB the same as DAB)?  If so, my answer is different from beta's; if not, ignore this, and please explain where angle ACB is (I probably drew my diagram incorrectly).

[hide]Let angle DAB=x, so angle ABC=2x.  Since they are same-side interior angles, they sum to 180, so x=60.  Draw the altitude from B to AD, and call the point it meets H.  Then angle CBH=90, and angle ABH=120-90=30.  We have triangle ABH as a 30-60-90 triangle, and we have the hypotenuse as 13/2.  Then AH=13:rt3:/4 by the properties of the 30-60-90, and this is also the altitude of the trapezoid.  We have AD=12 and BC=13/2, so [ABCD]=((13/2)+12)/2 * 13:rt3:/4 = 481:rt3:/16[/hide]

This problem has a really ugly answer, that I can only get with fairly complicated trigonometry, which leads me to the conclusion that either (1) it's incorrectly stated or (2) It's just really hard and ugly.

Here's a diagram, from which all of you can check if what you're saying in fact pertains to the question or just sounds kind of like it:

Yup, that's very different from what I did, but it fits the problem better and actually makes sense to be at the intermediate level (as opposed to beginning level).

So this problem only has an ugly solution?  That's the only one I found, but this problem was on a Russian mail-in challenge (similar to the camp ones) so I thought there would be a simpler solution... all i've found is something involving solving a quartic for the cosine of an angle lol

Thank you!

Well, that looked uglier than mine, but I was using Mathematica for the arithmetic.  I reduced the givens to the equation [tex]\displaystyle{24\sin(x) = 13\sin(5x)}[/tex], while the area is represented by [tex]\displaystyle{481\sin(4x)/8}[/tex], and you can solve this uglily -- the first equation is solvable for [tex]\displaystyle{\sin^2(x)[/tex] without too much trouble, and then substituting.  Ain't pretty, though.

[quote="JBL"]I was using Mathematica for the arithmetic.  I reduced the givens to the equation [tex]\displaystyle{24\sin(x) = 13\sin(5x)}[/tex].[/quote]

I don't have Mathematica, but I got comparably ugly results.  Chasing angles and then using law of sines, I get your equation.  Then substituting a 5x-angle formula:
[tex]\displaystyle{13(5\sin(x)-20\sin^3(x)+16\sin^5(x))=24\sin(x)

41\sin(x)-260\sin^3(x)+208\sin^5(x)=0

(sin(x))(208\sin^4(x)-260\sin^2(x)+41)=0
[/tex]

[tex]\sin(x)=0[/tex] is a trivial solution.

Use quadratic to solve for [tex]\sin^2(x)[/tex]:

[tex]\displaystyle\sin^2(x)=\frac{260\pm \sqrt{260^2-4\cdot 208\cdot 41}}{2\cdot 208}

\sin^2(x)\approx 0.18510272178085099611

\sin(x)\approx 0.43023565842553194409

x\approx 25.48251654554402141157[/tex]

I won't slog through the rest, but I get an area of about 58.8263.  This differs by about 0.7% from JBL's formula for the area.  I don't know who is correct.


[quote="Fanatic"]After a price reduction of x% an item has its price increased to its original value. What was the precent of increase/ Express your answer as a common fraction in terms of x.

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

I keep on getting it wrong by a hair and I don't know how to fix it  :wacko:  :wallbash_red:  :diablo:[/quote]

[hide]ARgh...
I keep getting $\frac{x}{100-x}$...but i saw the answer key and it's like $\frac{100x}{100-x}$
I thought that the base is 100 so then it's decreased by x. so then the denominator would be the total that is left after subtracting x so it is 100-x... so it's $\frac{x}{100-x}$...[/hide]

[hide]
I think I got it.
let's say the original price is m
then the discounted price is ((100-x)/100)m
let y be the percent increase
so we multiply the previous equation by ((100+y)/100)
and get ((100-x)(100+y)/10000)m=m because we get back to the original amount of money.
so dividing by m, multiplying by 10000, and distributing...
10000=10000-100x+100y-xy
100x=100y-xy
(100x)/(100-x)
this seems pretty straightforward, if a little variable-heavy...[/hide]

[hide]
Say that it was originally 1.

Then it decreased $x\%$

Now it is
$\frac{x}{100}$

We need to multiply by something to get 1.

$1 / \frac{x}{100}=1\cdot \frac{100}{x} = 100/x$
That is the amount of increase that we need.
To fund percent, multiply by 100.
Get $\frac{10000}{x}$[/hide]

K8, you've got it.

ch1n353ch3s54a1, I keep getting what you get except I have the 100 part and NOT the x part, lol.

Thanks k8.


Anyone done any CS Research?

If so, what was your topic and how did you come up with it?

I'm trying to come up with something, but really getting no where.  I was thinking about trying something with Data Compression, but there's not really much more that can be done.

Umm I'm pretty sure that there still is a lot more research to be done in the area of data compression... People will never be happy with the size of compression, and will always want to further compress.

I suggest you start by analyzing basic compression algorithems that are around right now. I'd say a classic one is LZW.

Another topic that is closely linked to data compression is error correction, might want to take a look at that as well.

If I would have to choose a topic, I will choose character recognition or cryptography. You can try them too.


Hi all,

I've tried all the Asy packages for the Macintosh I could find: and it still won't work.  (I mean other than the message board)

See this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223490

my brother tried too rofl

I made the link above into an AoPSWiki page: [url]https://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Macintosh[/url]

Let me know of any problems and I'll add solutions (hopefully) to the troubleshooting section.


modular arithmetic
number theory unsolved
number theory

Suppose that x^2 + y^2 = 3...31 (a number with a series of 3 then 1, like 31, 331, 33331, 333333331, etc), with x, y integers. Prove that either x or y (but not both) is divisible by 3.

i only know how to show that if both x and y are divisible by 3 the equality is impossible to hold. but how to show the equality works if one of x or y is divisible by 3?

Just show that if both are not divisible by three then equality also does not hold.

How do you show that? Please post your proof :)

[quote="zgorkster"]If $3$ does not divide $x$, then $x^{2}\equiv 1\pmod{3}$.  But then $x^{2}+y^{2}\equiv 2\pmod{3}$ which is a contradiction.[/quote]
Can be 3|y, $3\not |x$.
Consider mod 4 $...31=3(mod 4)$, but $x^{2}+y^{2}\not =3(mod 4)$.

Right, that shows there are no solutions at all, I just wanted to show what the original problem was: if there is a solution, only one can be divisible by $3$.


function
inequalities
integration
calculus
logarithms
floor function
real analysis

[quote="zgorkster"]Let $g=\sum_{i=1}^{2006}\frac{1}{i+2006^{2}}$.  Then if we consider this sum as a partition of $[0,2006]$ as a Riemann sum of the function $f(x)=\frac{1}{x+2006^{2}}$, we have the inequality

$g<\int_{0}^{2006}\frac{dx}{x+2006^{2}}$

After integration and multiplying both sides by $2006$, we have

$2006g<2006\ln\frac{2007}{2006}=\ln\left (1+\frac{1}{2006}\right )^{2006}$

But the function $g(n)=\left (1+\frac{1}{n}\right )^{n}$ is monotone increasing on $\mathbb{N}$ and has a limit of $e$.  Therefore the result above gives

$2006g<\ln e=1$, and so $\lfloor 2006g\rfloor=0$.[/quote]

Thanks for your inspiration.

Then this problem becomes simple ----
g=1/(1+2006^2)+1/(2+2006^2)+...+1/(2006+2006^2)
<2006*1/(1+2006^2)
2006*g=2006^2/(1+2006^2)
0<2006*g<1
so [2006*g]=0

Am I right?
thanks again.

[quote="ice_age"]g=1/(1+2006^2)+1/(2+2006^2)+...+1/(2006+2006^2)
find [2006*g], where [x] is the maximum integer <=x.

I forget the sqrt.
The actual problem should be
g=1/sqrt(1+2006^2)+1/sqrt(2+2006^2)+...+1/sqrt(2006+2006^2)
find [2006*g], where [x] is the greatest integer <=x.[/quote]

The actual problem should be
g=1/sqrt(1+2006^2)+1/sqrt(2+2006^2)+...+1/sqrt(2006+2006^2)
find [2006*g], where [x] is the greatest integer <=x.


induction
Inequality
number theory
algebra
IMO
IMO 1994
IMO Shortlist

Let $ m$ and $ n$ be two positive integers. Let $ a_1$, $ a_2$, $ \ldots$, $ a_m$ be $ m$ different numbers from the set $ \{1, 2,\ldots, n\}$ such that for any two indices $ i$ and $ j$ with $ 1\leq i \leq j \leq m$ and $ a_i \plus{} a_j \leq n$, there exists an index $ k$ such that $ a_i \plus{} a_j \equal{} a_k$. Show that
\[ \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_m}{m} \geq \frac {n \plus{} 1}{2}.
\]

Take $a_1 < a_2 < ... < a_m$. Take $k \leq (m+1)/2$. We show that $a_k + a_{m-k+1} \ge n + 1$.
If not, then the $k$ distinct numbers $a_1 + a_{m-k+1}, a_2 + a_{m-k+1}, ... , a_k + a_{m-k+1}$ are all $\leq n$ and hence equal to some $a_i$.

But they are all greater than $a_{m-k+1}$, so each $i$ satisfies $m-k+2 \leq i \leq m$, which is impossible since there are only $k-1$ available numbers in the range.

Re-order the set $A$ such that $a_1 < a_2 < \dots < a_m$. Now $a_i < a_j$ if and only if $i<j$.

[b]Lemma:[/b] $a_i + a_{m+1-i} \not\in A$.

Assume for contradiction that it is in $A$. Then $a_i + a_{m+1-i} \le n$ which implies that for all $q$ less than or equal to $i$, it follows that $n \ge a_{m+1-i} +a_q = a_p > a_{m+1-i}$. This implies that $m \ge p > m+1-i$ and hence that there are $i-1$ possible values of $p$. However there are $i$ possible values of $q$, each of which determines a unique $p$. This is a contradiction. $\Box$

Since $a_i + a_{m+1-i} \not\in A$, it follows that $a_i + a_{m+1-i} > n$ or that $a_i + a_{m+1-i} \ge n+1$. Therefore

\[(a_1 + a_m) + (a_2 + a_{m-1}) + \dots \ge \frac{m}{2}(n+1)\]
\[\frac{a_1 + a_2 + \dots + a_m}{m} \ge \frac{n+1}{2} \quad \blacksquare\]

[quote="stsamster"]fix $n$ and use induction on $m$.

If $m=1$, then $a_{1}+a_{1}$ doesn't contained in $A$ since $A$ is singalton. Thus $a_{1}+a_{1}>n$. So we get $a_{1}\geq(n+1)/2$.

Assume result is true for $m-1$.
Let $s=a_{1}+a_{2}+...+a_{m}$. So $s-a_{i}=a_{1}+a_{2}+...+a_{i-1}+a_{i+1}+...+a_{m}\geq\frac{n+1}{2}(m-1)$ for all $i$ by the assumption for $m-1$.

$s-a_{1}\geq\frac{n+1}{2}(m-1)$
$s-a_{2}\geq\frac{n+1}{2}(m-1)$
.
.
.
$s-a_{m}\geq\frac{n+1}{2}(m-1)$

Thus $ms-(a_{1}+a_{2}+...+a_{m})\geq\frac{n+1}{2}(m-1)(m)$. So we get $s\geq\frac{n+1}{2}(m)$.

So we are done...[/quote]

I'm not sure this is correct ... when we remove $a_i$, the condition of the problem where there exists an $a_k$ does not necessarily hold anymore for the remaining $m$ values, and therefore we cannot apply the induction hypothesis. Am I correct?

yes you are right. Thanx for pointing it out.

I have been trying to work on proof writing skills, as I am new to olympiad math, and I would be really thankful if someone could verify the mathematical correctness of my proof and give me pointers on style.
[hide="Solution"]
Let $a_1, a_2, \dots a_m$ satisfy the given conditions. We will prove that for all $j, 1 \le j \le m,$

$$a_j+a_{m-j+1} \ge n+1$$

WLOG, let $a_1 < a_2 < \dots < a_m$. Assume that for some $j, 1 \le j \le m,$

$$ a_j + a_{m-j+1} \le n$$

This implies, for each $i, 1 \le i \le j,$
$$ a_i + a_{m-j+1} \le n $$
because $a_i \le a_j$

For each of these values of i, we must have $a_i + a_{m-j+1} = a_{k_i}$ such that $a_{k_i}$ is a member of the sequence for each $i$. Because $a_i > 0, a_{k_i} > a_{m-j+1}$.
Combining all of our conditions we have that each of $k_i$ must be distinct integers such that
$$m-j+1 < k_i \le m$$

However, there are $j$ distinct $k_i$, but only $j-1$ integers satisfying the above inequality, so we have a contradiction. Our assumption that $a_j + a_{m-j+1} \le n$ was false, so $a_j + a_{m-j+1} \ge n+1$ for all $j$ such that $1 \le j \le m$ Summing these inequalities together for $1 \le j \le m$ gives
$$ 2(a_1+a_2+ \dots a_m) \ge m(n+1) $$
which rearranges to
$$ \frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2} $$
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[hide = Solution]
Let $a$ be the smallest element of $S = \{a_1, \ldots, a_m\}$. Then for any $s \in S$, $s + a$ is either in $S$, or greater than $n$. This means we can partition $S$ into chains \[s \to s + a \to s + 2a \to \cdots \to s + ka,\] where $s + (k + 1)a \geq n + 1$ and $s \geq a$. The average of the elements in this chain is \[\frac{2s + ka}{2} \geq \frac{s + (k + 1)a}{2} \geq \frac{n + 1}{2},\] so the average over all elements is also at least $\tfrac{n + 1}{2}$.
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topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Let $S_{n}$ be the $n$-dimensional sphere(not a subset of $\mathbb R^{n+1}$). Define $\alpha: S_{n}\to S_{n}$ to be the antipodal map $x\to-x$, where $-x$ denotes the point of $S_{n}$ opposite to $x$.

 a) For the case $n=2$ show that $\alpha$ is a diffeomorphism;
 b) Let $G$ be the group of diffeomorphisms $S_{2}\to S_{2}$ spanned
     by $\alpha$. Show that $S_{2}/G$ is aidfferential manifold called the
     projective sphere;
 c) Is it possible to represent $S_{2}/G$ as a topological space by identifying the edges of a polygon?
 d) Repeat by a) and b) for arbitrary $n$.
 e) Show that $\alpha$ is a spherical isometry and consequently $S_{2}/G$ is a spherical manifold.

You say that $S_{n}$ is not a subset of $\mathbb R^{n+1}$. Then how is it defined: as a metric space, a topological manifold, a $C^{k}$-manifold (which $k$?), or something else? And how is this structure introduced on $S_{n}$, if not from its embedding into $\mathbb R^{n+1}$?