calculus derivative inequalities LaTeX function

What is the maximum value of $ x \minus{} x^3$ for x being between 0 and 1?

hello, let $ f(x)=x-x^3$, then we get as the first derivative $ f^'(x)=1-3x^2$, solving the equation $ f^'(x)=0$ we have as the solutions $ x_1=\frac{1}{\sqrt{3}}$ or $ x_2=-\frac{1}{\sqrt{3}}$.You must plug $ x_1$ in the second derivative.
Sonnhard.

We can solve this without calculus, too. Consider the square:

$ (x \minus{} x^3)^2 \equal{} [x (1 \minus{} x^2)]^2$

$ \equal{} 4 (x^2) \left( \frac {1\minus{}x^2}{2} \right) \left( \frac {1\minus{}x^2}{2} \right)$

$ \equal{} 4 \textbf{GM} [x^2, \; (1\minus{}x^2)/2, \; (1 \minus{} x^2)/2] ^3$

$ \leq 4 \textbf{AM} [x^2, \; (1\minus{}x^2)/2, \; (1 \minus{} x^2)/2] ^3$

$ \equal{} 4 \left( \frac {1}{3} \right)^3 \equal{} \frac {4}{27}$

Where $ \textbf{AM}[ \ldots ]$ and $ \textbf{GM} [ \ldots ]$ denote arithmetic and geometric means.

Because $ x \minus{} x^3$ is positive over the desired interval, we can just take the squareroot: $ x \minus{} x^3 \leq \frac {4}{9} \sqrt {3}$

[b]Edit:[/b] Right, thanks

The problem with using AM-GM is that you need to guarantee that the equality case is attainable, which is trickier than it sounds.  Since two of the terms you've got are equal, it's possible in this case, but if you don't set it up correctly you end up getting a non-optimal bound.  By the way, that should be $ 4x^2 \left( \frac{1 \minus{} x^2}{2} \right) \left( \frac{1 \minus{} x^2}{2} \right)$.

[quote="t0rajir0u"]The problem with using AM-GM is that you need to guarantee that the equality case is attainable, which is trickier than it sounds.  Since two of the terms you've got are equal, it's possible in this case, but if you don't set it up correctly you end up getting a non-optimal bound.  By the way, that should be $ 4x^2 \left( \frac {1 \minus{} x^2}{2} \right) \left( \frac {1 \minus{} x^2}{2} \right)$.[/quote]
You're absolutely right, t0rajir0u :) . In fact, the "weighted AM-GM" is alternative way to solve most of "maximum and minimum 1-variable problems", but it is derived from derivative. We always, always have to make sure that the inequality we use in every steps has the same equality. And not everybody can predict exactly where the equality is without using derivative first. Basically, we all use derivative to find the equality first, then present it in another way(in this case, weighted AM-GM) to make our solution look nicer and more elementary. And another problems when using weighted AM-GM straightforwardly is that the equality can happen somewhere out of the interval restriction(e.g: let 0<x=<1/square root6, find the maximum for: x-x^3). In this case it attains at the upper bound of x(=1/square root6), which I believe most of people know this property!

When I solved it, I did not use calculus. So the idea is to find a way to solve it without calculus. After all, the admin said not to post calculus problems.

[quote="AIME15"]Well how'd you solve it mathelete?[/quote]
The only way I can explain is if we were actually talking to each other. But I can say that one way is to form an inequality to solve for the answer in terms of a guess.

[quote="t0rajir0u"]Do you get the right answer that way?  If you can't explain your solution, it is probably too vague to be completely correct.[/quote]
I can explain it, but you would have to be here for it to happen. Truth is, it would require brutal latex.

Well then....show us with brutal latex.......

$ \text{BRUTAL}^\text{\LaTeX}$

[quote="t0rajir0u"]If it is more complicated than either TZF's solution or the calculus solution, it is probably not correct.  Please share it.[/quote]I am not very good with latex.

If the $ \text{\LaTeX}$ is really such a problem, then write it out on paper, scan it, and post. Or if no scanner is available, most text editors have a built in what-you-see-is-what-you-get math typesetting function (although with a less nice rendering than $ \text{\LaTeX}$).

[quote="DiscreetFourierTransform"]Not to sound too cynical, but it seems as if mathlete doesn't really have a solution, and he's just fishing for answers for a homework problem... He's thwarted several different attempts to coax out a solution and it's already been explained to him that the elegant solutions are primarily motivated by the underlying calculus.

Mathlete, if in fact you DO have a nice solution, I apologize for this post.

Otherwise, I'm going to remain cynical until you prove me otherwise.[/quote]
I will write it on a piece of paper, scan it, then send it in a pdf file. Otherwise I will have to tell you face-to-face. I have a nice solution. This will take some time, because I do not actually own a scanner.

Okay, I will do latex.

We are looking for the maximum of $ f(x) \equal{} x \minus{} x^3, 0 < \equal{} x < \equal{} 1$. If our guess for x is b, and the correct answer is $ a$, we must make sure that $ a \minus{} b > \equal{} a^3 \minus{} b^3$. By factoring, $ a \minus{} b > \equal{} (a \minus{} b)(a^2 \plus{} ab \plus{} b^2)$. If $ b < \equal{} a$, then $ a^2 \plus{} ab \plus{} b^2 < \equal{} 1$. If $ b > \equal{} a$, then $ a^2 \plus{} ab \plus{} b^2 > \equal{} 1$. Solving, we have that if $ b < \equal{} a$, then $ a < \equal{} \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. If $ b > \equal{} a$, then $ a > \equal{} \frac { \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}}{2}$. So we must set $ a$ and b equal. Now we have $ 2b \equal{} \minus{} b \plus{} \sqrt {4 \minus{} 3b^2}$. Add each side by b and square it. $ 9b^2 \equal{} 4 \minus{} 3b^2$. Add $ 3b^2$ to each side and take the square root. $ b \equal{} \frac {\sqrt {3}}{3}$. Plug it in to f(b) and we get that the maximum is $ \frac {2\sqrt {3}}{9}$.

[quote="people"]Is calculus required?

Yes(which means I did not do it correctly)
  33%  [ 2 ]
No(which means I probably did it right)
  67%  [ 4 ]
Calculus cannot be used(I am curious)
  0%  [ 0 ]

Total Votes : 6
[/quote]
The poll originally said 66%. This is the fixed version of the poll.

As long as we are doing stupid proofs,

$ \frac {4}{27} \minus{} (x \minus{} x^3)^2 \equal{} \frac {1}{27}\cdot (4 \minus{} 3 x^2 ) ( \minus{} 1 \plus{} 3 x^2)^2\ge 0$

since $ 4 \minus{} 3x^2 \ge 4 \minus{} 3 \equal{} 1 > 0$. Equality for $ x \equal{} \frac {1}{\sqrt {3}}$.

(Note $ x \minus{} x^3\ge 0$ so we can square root to come up with the original inequality).

[quote="people"]Is calculus required?

Yes(which means I did not do it correctly)
  17%  [ 2 ]
No(which means I probably did it right)
  66%  [ 8 ]
Calculus cannot be used(I am curious)
  17%  [ 2 ]

Total Votes : 12[/quote]
This poll is bad at calculating. Come on, 16%? It is 17%!

[quote="DiscreetFourierTransform"]Not to sound too cynical, but it seems as if mathlete doesn't really have a solution, and he's just fishing for answers for a homework problem... He's thwarted several different attempts to coax out a solution and it's already been explained to him that the elegant solutions are primarily motivated by the underlying calculus.

Mathlete, if in fact you DO have a nice solution, I apologize for this post.

Otherwise, I'm going to remain cynical until you prove me otherwise.[/quote]

I proved you otherwise on 8/20.

[quote="DiscreetFourierTransform"]Wait... where? I've been checking my topics.. I haven't seen any new post.. could you possibly repost what you had said?[/quote]
Post #17

"Okay, I will do latex."

What are you talking about?
Either you're playing some sort of game, or I'm looking at the wrong post #17?

[quote="azjps"]Emoticon hides ...  :huh: I hope you aren't trying to prove some point, mathelete.[/quote]
You caught me, this was meant only for a specific user.

[b]@ OP:[/b]

Here are some things that happened in this thread that should not happen in subsequent threads -- please read all of them:

[b]1.[/b] [color=blue]Never use an emoticon hide[/color] unless you specifically point it out. Even then, it's undesirable.
[b]2.[/b] If you don't know how to post your solution, you could [color=blue]try explaining the general steps[/color] or concepts you used.
[b]3.[/b] The poll really didn't need to be there. [color=blue]Mathematics is not a democracy[/color]. Ask it in your post and you will likely get a reasonably good response, at the least.
[b]4.[/b] We don't care to hear complaints about the poll rounding up vs rounding down. Or anything about the poll. See above.
[b]5.[/b] Repeatedly saying that you don't know how to post your solution just makes you look a bit lazy. I don't see any LaTeX element in your solution that wasn't already used in the thread (by the way, [color=blue]clicking on a LaTeX formula will reveal the source code in a pop-up window[/color] -- this is the fastest way to learn how to use it).
[b]6.[/b] Although [color=blue]October[/color] begins with "Oct-", it [color=blue]is the 10th month[/color] of the year, not the 8th

Addendum: don't post ANYTHING hidden below an emoticon that you want people to read. It's confusing, and unless you WANT the information to be obscured from everyone, just don't do it.

Alright, so now I've taken a look at your solution. It's a bit confusing the way you've worded it (the LaTeX is fine), but your gut was taking you in the right direction.

You had $ (a \minus{} b) \geq (a \minus{} b)(a^2 \plus{} ab \plus{} b^2)$. Then you divided by $ a \minus{} b$. Now you set $ a \equal{} b$, which is essentially setting $ a \minus{} b \equal{} 0$. You need to justify your division by zero there. Rather than straight-up dividing by zero (which is illegal), you need to use limits to accomplish this, which would be calculus.

However, it is a decent heuristic approach.

[hide="Another popular approach that does not use calculus (though it is calculus motivated)"]
Invoking the fact that if $ M$ is the maximum value you want, then $ (x \minus{} x^3) \minus{} M$ would have a double root* at the desired value of $ x$.

So then, we can write $ x^3 \minus{} x \plus{} M \equal{} (x \minus{} r)^2 (x \minus{} p)$, where $ r$ is the desired value of $ x$.

Equating coefficients, we find that $ 2r \plus{} p \equal{} 0$ and $ 2rp \plus{} r^2 \equal{} \minus{} 1$.

The first equation gives $ p \equal{} \minus{} 2r$. Subbing into the second equation gives $ \minus{} 4r^2 \plus{} r^2 \equal{} \minus{} 1 \implies r^2 \equal{} \frac {1}{3} \implies r \equal{} \frac {1}{\sqrt {3}}$

* - [i]Consider the intersections between $ y \equal{} x \minus{} x^3$ and lines of the form $ y \equal{} k$ for values of $ k < M$. What happens to the larger two roots as $ k$ approaches $ M$?[/i][/hide]

[b]Note:[/b] [i]There is a typo at the end of my AM-GM solution in post 3 -- the answer should read $ \frac {2}{9} \sqrt {3}$ rather than $ \frac {4}{9} \sqrt {3}$. Everything else in that post is correct.[/i]


1. How many numbers are there between 9 and 101 that are multiples of 2 and 5?

2. How many numbers are there between 7 and 101 that are multiples of 2, 5, and 3?

[hide="1"]Since 2 and 5 are relatively prime, we look for numbers that are divisible by 10.

Between 9 and 101, the desired numbers are 1*10, 2*10, ..., 10*10, or 10 numbers. [/hide]

[hide="2"]We look for numbers that are divisible by 2*3*5=30.

There is 30, 60, and 90, or three numbers. [/hide]

I like 236factorial's way much better.   :D

here my answers

1.  [hide]10[/hide]
2.[hide]3[/hide]


$ ABCDE$ is a convex pentagon.$ AE$ and $ BC$ are parallel and $ <ADE$ = $ <BDC$.$ AC$ and $ BE$ intersect each other at $ P$.prove that $ <EAD$=$ <BDP$ and $ <CBD\equal{}<ADP$

Let's keep to one thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=285686


With $ a,b,c$ are three lenghs of sides of a triangle such that  $ a\plus{}b\plus{}c\equal{}2$ Prove that:
  $ x^3\plus{}y^3\plus{}z^3\plus{}4xyz \geq 2 \geq x^3\plus{}y^3\plus{}z^3\plus{}3xyz$

I think you made a typo my friend: [b]x+y+z[/b]=2

The left part:
The given inequality is equivalent with:
$ 4 \sum x^3\plus{}16xyz\ge \sum x^3\plus{}3\sum_{sym}x^2y\plus{}6xyz \Longleftrightarrow
3 \sum x^3\plus{}10xyz\ge 3\sum_{sym}x^2y$
Which is straight schur....

[quote="2424"]With $ x,y,z$ are three lenghs of sides of a triangle such that  $ x\plus{}y\plus{}z \equal{} 2$ Prove that:
  $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 4xyz \geq 2 \geq x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz$[/quote]

We have : $ 2p\equal{}a\plus{}b\plus{}c\equal{}2;abc\equal{}4Rr;a^3\plus{}b^3\plus{}c^3\equal{}2(1\minus{}3r^2\minus{}6Rr)$

The left part is equivalent to : $ r(2R \minus{}3r) \geq 0$ ( right because $ R \geq 2r$ )

The right is equivalent to : $ 3r^2 \geq 0$

Done :)

[quote="earldbest"][quote="Materazzi"]

abc = 4Rr

[/quote]

how? sorry.  :([/quote]

I think:
$ abc \equal{} 4SR\equal{}4Rr$  :)

[quote="caubetoanhoc94"][quote="earldbest"][quote="Materazzi"]

abc = 4Rr

[/quote]

how? sorry.  :([/quote]

I think:
$ abc \equal{} 4SR \equal{} 4Rr$  :)[/quote]
ặc, Nghĩ gì vậy em.
We have: $ abc\equal{}4SR\equal{}4pSr$

[quote="math10"][/quote][quote="caubetoanhoc94"][quote="earldbest"][quote="Materazzi"]

abc = 4Rr

[/quote]

how? sorry.  :([/quote]

I think:
$ abc = 4SR = 4Rr$  :)[/quote][quote="math10"]
ặc, Nghĩ gì vậy em.
We have: $ abc = 4SR = 4pSr$[/quote]
Nhầm à anh :)
$ abc=4SR=4prR$ chứ anh :P
But
$ p = \frac {x + y + z}{2} = 1$  :P
So:
$ abc = 4SR = 4pRr = 4Rr$

[quote="caubetoanhoc94"][/quote][quote="math10"][/quote][quote="caubetoanhoc94"][quote="earldbest"][quote="Materazzi"]

abc = 4Rr

[/quote]

how? sorry.  :([/quote]

I think:
$ abc = 4SR = 4Rr$  :)[/quote][quote="math10"]
ặc, Nghĩ gì vậy em.
We have: $ abc = 4SR = 4pSr$[/quote][quote="caubetoanhoc94"]
But
$ p = \frac {x + y + z}{2} = 1$  :P[/quote]

he he,quên mất,chưa đọc đề đã post bài  :D .

Dear 2424,you can say:"can you prove it?"
For the right part:
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \le 2$
<=>$ (x\plus{}y\plus{}z)^3\minus{}3(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\plus{}6xyz \le 2$
<=>$ 8\minus{}6(xy\plus{}yz\plus{}zx)\plus{}6xyz \le 2$
<=>$ xy\plus{}yz\plus{}zx\minus{}xyz \ge 1$
because a,b,c is side length in triangle,we can setting: $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y$.From a+b+c=2,we get x+y+z=1
$ xy\plus{}yz\plus{}zx\minus{}xyz\equal{}(a\plus{}b\plus{}c)^2\plus{}ab\plus{}bc\plus{}ca\minus{}ab\minus{}bc\minus{}ca\plus{}abc\equal{}1\plus{}abc \ge 1$
Our proof are completed,equality occur if and only x=y=1,z=0 or any cyclic permution.:)

[quote="nguoivn"][quote="2424"][quote="Dimitris X"]I think you made a typo my friend: [b]x+y+z[/b]=2

The left part:
The given inequality is equivalent with:
$ 4 \sum x^3 \plus{} 16xyz\ge \sum x^3 \plus{} 3\sum_{sym}x^2y \plus{} 6xyz \Longleftrightarrow 3 \sum x^3 \plus{} 10xyz\ge 3\sum_{sym}x^2y$
Which is straight schur....[/quote]
yes,the right part?You can't prove?It is easy! :P[/quote]
@2424: [hide]Anh nghi~ em ko nen dung` n~ cau kieu nhu "you can't prove it?", nhat la` voi n~ ng` nuoc ngoai` vi` nhu the la` ko ton trong. ho. va` thieu lich su. Hi vong. em se~ chu' y' hon. Du` sao anh cung~ thay' em hoc. bdt rat tot', neu sua? dc 1 vai` nhuoc diem? nho? chac chan se~ rat tuyet voi`  :) [/hide][/quote]
@nguoivn:[hide]E cam on anh de chi bao.E se rut kinh nghiem.Hi vong dc anh chi bao them a!E moi len mathlinks len chua hiu luat lem.E cung chi noi cau do cho vui,khong ngo lai khong tot nhu vay.
E co the lam quen vs anh duoc khong a?[hide]

[quote="secrets"]2424[hide]Em dùng lệnh hide và /hide nhé,à,anh hỏi,những bài này đều do em creat à,ghi chữ own ở dưới sẽ tốt hơn [/hide][/quote]
@secrets:[hide]Vang ah,nhung bai nay deu do e sang tac.E cam on anh nhieu nhe!Hic,hinh nhu anh cung len mathscope ah?anh sinh nam bao nhieu vay?[/hide]

the left part:
$ 4\sum x^3 \plus{} 16xyz \geq \sum x^3 \plus{} 3\sum x^2y \plus{} 3\sum xy^2 \plus{} 6xyz$ so
$ 3\sum x^3 \plus{} 10xyz \geq 3\sum x^2y \plus{} 3\sum xy^2$
with schur inequality we have $ \sum x^3 \plus{} 3xyz \geq \sum x^2y \plus{} \sum xy^2$ so
$ 3\sum x^3 \plus{} 9xyz \geq 3\sum x^2y \plus{} 3\sum xy^2$ so
$ 3\sum x^3 \plus{} 10xyz > 3\sum x^2y \plus{} 3\sum xy^2$.

the right part:
let $ x\equal{}a\plus{}b,y\equal{}b\plus{}c,z\equal{}c\plus{}a$ then $ a,b,c$ is positive (because of $ x,y,z$ be lenghs of sides of a triangle ) so
$ a\plus{}b\plus{}c\equal{}1$ and i prove $ 2(a\plus{}b\plus{}c)^3 \geq (a\plus{}b)^3\plus{}(b\plus{}c)^3\plus{}(c\plus{}a)^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$.
$ 2\sum a^3\plus{}6\sum a^2b \plus{}6\sum ab^2\plus{}12abc \geq 2\sum a^3\plus{}6\sum a^2b\plus{}6\sum ab^2\plus{}6abc$ so
$ 6abc \geq 0$.
so inequality true.


inequalities
geometry
search
geometry unsolved

Let $ABC$ be a triangle and $P$ a point inside the triangle. If $BC$ is the smallest side of triangle $ABC$, show that :
\[PA+PB+PC < AB+AC \]

[i]Alternative formulation.[/i] The point $P$ is in the interior of a triangle $ABC$, whose two longer sides are $b$ an $c$.  Let $l$, $m$, $n$ be the distances from  $P$ to the vertices. Then we have $l+m+n < b+c$.

If the point P inside triangle ABC then we have following property:
$PA+PB+PC \leq\max\left\{ AB+AC, BA+BC, CA+CB \right\}$
thus BC is the shorstlestside, we desire easyly

Ok but how do you prove your result?  :?

[quote="hien"]If the point P inside triangle ABC then we have following property:
${PA+PB+PC \leq\max\{AB+AC,BA+BC,CA+CB}\}$
thus BC is the shorstlestside, we desire easyly[/quote]

Ok, wait me for two days. Now I'm very busy. I'll post completed solution on thursday

Now it seems to me that the problem is difficult. The friend who tell me about the problem says that the source is a list of problems to be admitted at Mine Engineering studies in Spain at 50s

Any idea about a solution?

[quote="Labiliau"][i]Alternative formulation.[/i] The point $P$ is in the interior of a triangle $ABC$, whose two longer sides are $b$ an $c$.  Let $l$, $m$, $n$ be the distances from  $P$ to the vertices. Then we have $l+m+n < b+c$.[/quote]

by Erdos Morden
$l+m+n<MA+MB+MC<max\{b+c,c+a,a+b\}=b+c$:D

I'm sorry, I don't understand who is $M$; if you are using the Erdos-Mordell inequality, and $DEF$ is the pedal triangle of $P$ we have

$l+m+n = PA+PB+PC \geq 2 (PD+PE+PF)$.

?

oh sorry I write not clearly :D
by Erdos Morden $PA+PB+PC>2(l+m+n)$
we can prove $PA+PB+PC<mac\{b+c,c+a,a+b\}$
indeed:we assume
$x\vec{PA}+y\vec{PB}+z\vec{PC}=\vec{0}(x,y,z>0)$
$\Rightarrow y\vec{AB}+z\vec{AC}=(x+y+z)\vec{AP}\Rightarrow PA=\frac{|y\vec{AB}+z\vec{AC}|}{x+y+z}\le \frac{yc+zb}{x+y+z}$
similar $PB\le \frac{za+xc}{x+y+z},PC\ge \frac{xb+ya}{x+y+z}$
therefore
$PA+PB+PC\le \frac{yc+zb}{x+y+z}+\frac{za+xc}{x+y+z}+\frac{xb+ya}{x+y+z}=\frac{x(b+c)+y(c+a)+z(a+b)}{x+y+z}\le max\{b+c,c+a,a+b\}$
thus we have the stronger inequality
$2(l+m+n)\le max\{b+c,c+a,a+b\}=b+c$ (if two longer sides are $b$ and $c$)  :D

gemath, thank you for your attention,

you still write that $PA+PB+PC > 2(l+m+n)$ while we define $l=PA$, $m=PB$, $n=PC$.

Also, I think that the identity
$\frac{yb+zc}{x+y+z}+\frac{zc+xa}{x+y+z}+\frac{xa+yb}{x+y+z}=\frac{x(b+c)+y(c+a)+z(a+b)}{x+y+z}$

is wrong...

Is there anything I can't see?

ok thanks pacoga for your remark I have edit it, sorry for some stupid mistake, I only proved more an inequality $d_{a}+d_{b}+d_{c}<b+c$ but your inequality $PA+PB+PC<b+c$ I am also done too!
There is a nice following problem that I proved it:
$\alpha MA+\beta MB+\gamma MC \le \max\{\beta c+\gamma b,\gamma a+\alpha c,\alpha b+\beta a\},\forall M$ inside triangle
your inequality when $\alpha=\beta=\gamma=1$  hope you are interested in :)

Let ABC be the given triangle and we suppose that BC < AB < AC. For every point P inwardly to ABC, we will prove that
 PA + PB + PC < AB + AC ( l + m + n < b + c ).

Because of AC > AB, => angleABC > angleACB, (1).

We draw a line through P and parallel to BC, which intersects the sidelines AB, AC, at points D, E, respectively.
So we have that, angleADE > angleAED, (2).

From triangle APD, => angleAPE > angleADE, (3).

From (2), (3), => angleAPE > angleAED, => PA < AE, (4).

From triangles PBD, PCE, => PB < BD + DP, (5) and PC < PE + EC, (6).

From (4), (5), (6), => PA + PB + PC < AE + EC + BD + DP + PE, (7).

From (7), => PA + PB + PC < AC + BD + DE, (8).

Because of angleBAC < angleACB = angleAED, from triangle ADE we have that

DE < AD, => DE + BD < AD + BD, => BD + DE < AB, (9).

From 98), (9), => PA + PB + PC < AB + AC and the proof is completed.

Kostas Vittas.

PS. I have found this solution in Hellenic textbook by J. Panakis – Triangle’s geometry – volume A – page 125.
''Guttenberg'' publications – Athens – 1969.

In another Hellenic textbook by D. Kontogiannis – Iqualities and inequalities in a triangle – page129, there is the reference
''Visschers, 1902''. Somebody else more experienced than me, probably may give you more details about this proposition.

Dear Kostas,

thank you for your solution and references. A search with Google from +Visschers +1902 gives these solutions of CRUX problems

http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv24n2/body/PDF/page110-128.pdf

In turn, the pdf gives the reference:

Visschers (1902)
FGM Exercises de Geometrie,. p. 228, Th. 25 IV.

(In http://garciacapitan.auna.com/problemas/distancias I have put in Spanish these solutions).

Francisco Javier García Capitán
http://garciacapitan.auna.com

Let $P$ be a point inside or on the boundary of a convex polygon $A_{1}A_{2}\cdots A_{n}$, then the maximum of $\sum_{i=1}^{n}PA_{i}$ be achieve only if $P$ is a vertex of $A_{1}A_{2}\cdots A_{n}$.

This is a theorem of my essay be writen 23 year ago:

Ji Chen, A generalization of the reverse Fermat's problem (Chinese), [url=http://www.shuxuetongxun.com/]Bulletin of Mathematics (Wuhan)[/url], 1984, No. 5, p. 26.

lemma. Let $A,B,M$ be points such that $M\in\ [AB]$ then exists $0\leq k\leq 1$ such that $OM\leq kOA+(1-k)OB$ for all points $O$.

Now problem original is trivial.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $G$ is a finitely generated group and $H$ is a subgroup of $G$ with finite index. Show that $H$ is finitely generated.

I first saw this problem and its solution this spring, as a minor theorem with computational uses . The key to the usefulness is its constructive nature; we can easily find a generating set of reasonable size from a list of generators and coset representatives.
The applications dealt mainly with modular groups, and mere finiteness is not enough; I stated the problem that way for simplicity and to hide the constructive solution.


geometry
rectangle
number theory
relatively prime

The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares.  The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Solution"]First we calculate $r.$  There are $\binom{9}{2}$ ways to choose 2 vertices from 9 of them that make up a row or column.  Thus, we have $\binom{9}{2}$ ways to choose a length, $\binom{9}{2}$ ways to choose a height, and in counting this way we are left with only one positive for this rectangle to occupy.  So $r=\binom{9}{2}^2.$

The number of $1\times1$ squares is $8^2=64.$  The number of $2\times 2$ squares is $7^2=49.$  In general, the number of $k\times k$ squares is $(9-k)^2$ for $k=1,2,\ldots,8.$  So $s=1^2+2^2+\cdots+8^2=\frac{8\cdot9\cdot17}{6}=204.$

So we have that $s/r=17/108$ and our answer is 125.[/hide]

[hide=Solution]
We give the checkerboard coordinates with one vertex being the origin and the lines parallel to the coordinate axes. There are $\binom{9}{2}^2=36^2$ rectangles. Pick the x coordinate of the vertical sides and the y coordinate of the horizontal sides. Of them, $8^2+7^2+ \cdots+1^2=\frac{8*9*17}{6}=4*3*17$ squares. Pick the side length, then pick the corner closest to the origin. Hence $\frac{m}{n}=\frac{4*3*17}{36^2}=\frac{4}{16}\frac{3}{81}\frac{17}{1}=\frac{17}{108}, the answer is $\boxed{125}$.[/hide]

Combo AIME grind: There are $\binom{8+1}{2}^2 = 36^2 = 1296$ ways to select a rectangle. There are $1^2+2^2+\ldots+8^2 = 204$ ways to select a square, so the answer is $\frac{17}{108} \implies \boxed{125}$.


topology
real analysis
real analysis unsolved

Maybe this problem is too simple and should be a computations and tutorial section, but anyway:

A collection $ {V_{a}}$ of open subsets of $ X$ is said to be a base for $ X$ if: for every $ x \in X$ and every open set $ G \subset X$ such that $ x \in G$, we have $ x \in V_{a} \subset G$ for some $ a$. In other words, every open set in $ X$ is the union of a subcollection of $ {V_{a}}$.

Prove that every separable metric space has a countable base.
(A metric space is separable if it contains a countable dense subset).

[hide] Hint: Take all neighbourhoods with rational radius and center in some countable dense subset of $ X$. [/hide]

Since this question has remained unanswered, I will post a solution.  I hope that I have not made any mistakes. :)

Let $ S$ be a countable dense subset of metric space $ X$.  For each point $ x\in S$, construct the set $ \{\{q|d(x,q)<r\}|r\in\mathbb{Q}^\plus{}\}\equal{}N(x)$.  Since $ \mathbb{Q}^\plus{}$ is countable, $ N(x)$ is countable, and so $ N\equal{}\bigcup_{x\in S}N(x)$ is countable.  All that remains to be proved is that $ N$ is a base of $ X$.  Let $ p\in X$ be a point contained in the open subset $ G\subseteq X$.  Since $ G$ is open, there is a neighborhood $ N_{\delta}(p)\subseteq G$.  Take a rational number $ r\leq\frac{\delta}{2}$.  Since $ S$ is a countable base, there is a point $ q\in S$ that is also in $ N_r(p)$.  Then $ p\in N_r(q)\subset N_{\delta}(p)\subseteq G$.  Since $ N_r(q)\in N$, we are done. $ \Box$


How can you prevent yourself from getting bored.
I think most of you agree you can't do math all day long, so what else can be done?

[quote="amfulger"]what else can be done?[/quote]

Trying to find a nice woman mathematician .....

I'll work on that. I see that you have a keen a sense of humor.
Of course the question was about what [b]YOU[/b] do in order to prevent [b]yourself[/b] from getting bored.

I would like you to post your hobbies and the reason you like them.

I'll go first.

I like music. I don't like it because everybody does. I like it because I find it quite relaxing and useful. The reason is in one of my flaws: I just can't stay focused for too long unless I have a realy strong motivation of course. That's why if I think too much at something I don't find quite interesting, my brains start boiling and then, music is the only thing that relaxes me.

I like computer games (not all of them), but I don't like me liking this hobby so much because it's more like a drug to me.

I like some sports: football (or soccer) and tenis. Sometimes is good to move.

I like Formula 1 but I woun't tell you why. :D

I like to read a sports magazine. This hobby is kind of accidental. My mother bought me an issue 7 years ago while I was sick, and I've been reading that magazine ever since (not the same issue for some of you with a keen sense of humor). Guess is one of my links to the world.

What do I like ??

See, I like people who say what they like.   :D

So you are like a philanthropist?
Or are you Santa Claus?

Unfortunately I can't relax myself with math
I'm major in math and there're tons of work to do

indeed I must concur with Charlie. Math was much more enjoyable on IMO participating times :D. But I'm sure that's gonna change at some time in the future :D

YOU CANNOT REST VALENTIN, VECTOR HAS CHALLENGED YOU IN SOME OF THE OTHER POSTS IN THE ADVANCED SECTION GEOMETRY AND ALGEBRA I THINK. GOOD LUCK

hmm. it is strange... looking up the ip for our first spammer :D on mathlinks I was very intrigued to quickly find out who it is.

I think that the question you asked above is pointless, because the guest is you 3x.lich :? ...

:D  :D  :D This really explains you can never be bored at any time.  :D  :D  :D

BTW, seriously a user named 'vector', he proposed you some problems and he asked if you can explain him the concept of infinity.

he's he did. unfortunately I am very busy with other projects as well, and also with college.

I thought about the problems and came up with some partial answers. about the infinity I have to explanation to give him, you already gave one. If he is concerned with such things, he might choose to open a "fundamentals in math" book :D

I'm also in college Valentin. And yeah it is extremely busy down here. At least I didn't get bashed by the amount of tough courses that I'm having. Hmm, I can't login, something is wrong with my browser and security settings... :?

[quote="amfulger"]How can you prevent yourself from getting bored.
[/quote]

Actually, I try to taste when I have possibility french traditional food.
Help the best I can my high school students for their exams.
Follow the news.

I like computer games, and I've lost days trying to complete certain quest/adventure type of games. That doesn't really sound like relaxing, does it? :).

However, I'm currently in a situation where maths is a very good relaxation source: I have to study from my final high school exams, so I became unusually interested in other subjects. When I get bored with those, solving a maths problem sems like bliss to me :D:D.

as for me,i like to unwind myself by listening metallica,megadeth,anthrax,guns and roses,dream theatre etc
hey btw anyone here who really loves hard rock,heavy metal and stuff

and since my final xams are coming up (ist march that is) i'm having to study the boring subject that is biology (well my teacher says i'll fail if i dont study particularly hard,of course that's wot one xpects when u get 30/70 :D )
but sumthing that really gets me pumped is PHYSICS and ORGANIC CHEMISTRY.
i play lawn tennis 1.5 hrs daily
that's it

I juggle :-)

1) It passes time
2) It's not overly exerting, unless you want it to be
3) It impresses all the girls (helps with that "get yourself a girl" suggestion)
4) It makes you smarter ( http://www.cnn.com/2004/HEALTH/01/22/offbeat.juggling.brain.reut/ )
5) It lets you think about other things at the same time.

I don't know how many of you have been to IMO 2002 in Scotland, but I can tell you that in one of the opening shows, there was this juggler who was just fantastic. He could talk with use, juggle 7 balls in the same time, with different juggling schemes.

In fact the researched mathematical representations for juggling schemes, and it seems that he has gone pretty far with that thing. Anyway it was the coolest presentation of them all. And after that I myself bought a set of jugglin' balls which I learned how to use, but still I can only juggle with 3 balls :( ... :D

3 balls isn't bad. I'm still working on five, and I've been juggling for 3 or 4 years now. There are lots and lots of things you can do with three balls...

And yes, there is also some very interesting stuff with mathematics and juggling. I've just started to get into it a little--I bought a book (The Mathematics of Juggling, Burkard Polster) but haven't had much of a chance to read it, so I can't tell you if it's any good.

.dzhonatan

yeah... listening the music is very relaxing.. actually i listen music most of the time(excepts when i'm learning at german)...
i like rock to...most i like metallica, ramms+ein, evanescence, Iris, Phoenix, Iron Maide,..

i hate most of computer games... i can't play a game more than 30 min... i'm getting bored..:(...BUT recently i found an "educative" game.. BLUE SKY.. it's simulator for cracking systems and steeling money :D... that one is nice... but pretty hard[at least for me]...

i like to play table tennis too... but i haven't play for some while :(...

I'd love to know how to juggle, but I guess just fooling around with a couple of oranges doesn't really do the trick :D.

Well, most people think about juggling with 3 objects, though 2 ("a couple") is also perfectly acceptable.

Honestly, it's not really as hard as most people think. Most of the obstacles are purely psychological. I know there are some sites that are supposed to teach you how to juggle, though I can't cite any references off the top of my head, and I don't know how good they really are.

yeah. i learned to juggle in about 15 mins with 3 balls. can't get pass that tough :D.

Well, to be honest, I'm not even sure I can juggle 2 balls! I don't have those oranges with me right now..:D:D

Actually, my personal opinion is that 2 balls is harder than 3. The reason being, you have to do just as many motions but you only have 1 hand to do them with, whereas with 3 balls you are using both hands.

I know that it's a lot harder to juggle 2 clubs in one hand than 3 clubs in 2 hands!

:? maths IS what I do as a hobby, with the time I got left from school work :(

I like people who revive a 14 year old dead thread

[quote=NoDealsHere]Idk what to do[/quote]

these posts are older than you. Go to bed or get a hobby for christ's sake.

Yuppers, I just was browsing, as I was a bit bored.


Find all positive integers m,n satisfying:

(m-1)! + 1= m^n

notice that if $ m$ is not prime, then $ (m\minus{}1)!$ and $ m^n$ have a factor $ >1$ in common, so the equation can't have a solution. so $ m$ must be prime. by wilson's theorem, $ (m,n)\equal{}(p,1)$ is a solution for all primes $ p$. for $ n>1$, $ (5,2)$ is a solution... hmm... we know that $ p|(p\minus{}1)!\plus{}1$, but what about higher powers of $ p$? let's think about it


A plus is a polygon made of 5 congruent squares (one in the center, and 4 built on each side) for 12 total sides, all the same length.
The 8 outer points are inscribed in a circle. If the length of one side of a plus = p, what is the ratio of the circle area to the plus area?

[hide]$\frac {\sqrt {(3p)^2 + p^2 }}{2} = \frac {p \sqrt {10}}{2}$ is the radius of circle so the ratio of the areas is $\pi *\frac {\left(\frac {p\sqrt {10}}{2}\right)^2}{5p^2} = \frac {\pi}{2}$.[/hide]


function
inequalities
algebra proposed
algebra

Find all function $ f: R\minus{}>R$ such that $ 2f(x\plus{}y)\equal{}f(2x)\plus{}f(2y)$

Put $ y\equal{}x\plus{}2$. This gives you

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

Now, if $ g(x)\equal{}f(2x)$ we have

$ 2g(x\plus{}1)\equal{}g(x)\plus{}g(x\plus{}2)$

Using basic recurrence solving techniques we get

$ f(x)\equal{} \frac{A}{2}x\plus{} B$

and this works for any $ A$ and  $ B$.

[quote="milin"]Put $ y \equal{} x \plus{} 2$. This gives you

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

Now, if $ g(x) \equal{} f(2x)$ we have

$ 2g(x \plus{} 1) \equal{} g(x) \plus{} g(x \plus{} 2)$

Using basic recurrence solving techniques we get

$ f(x) \equal{} \frac {A}{2}x \plus{} B$

and this works for any $ A$ and  $ B$.[/quote]

I think u are wrong, please post ur full solution..

That solution is wrong, as any solution of the Cauchy equation $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$ is a solution of the equation given in the problem (and there are others). In particular we have solutions that are not continuous.

I never said it was full! :) But it is a part of one..

What milin's argument shows is that $ f(x) \equal{} \frac{A}{2} x \plus{} B$ for numbers of the form $ \alpha \plus{} \mathbb{Z}$ for a particular $ \alpha$ (which you can also conclude from the Cauchy equation).  The problem, of course, is that $ A, B$ depend on $ \alpha$.

hix,$ A\equal{}?,B\equal{}?$.I don't know
Now,who can give me a solution for that problems?

For any additive function g(x): $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ we get $ f(x)\equal{}g(x)\plus{}Const$ is solution. There are many non continiosly (non linear) additive functions.

[quote="every"]Find all function $ f: R \minus{} > R$ such that $ 2f(x \plus{} y) \equal{} f(2x) \plus{} f(2y)$[/quote]
$ \{x\}\equal{}x\minus{}[x]$

for $ r\in[0,1[$
we take $ U_n\equal{}f(2(nx\plus{}r))$, obviosly $ U_{n\plus{}1}\equal{}\frac{U_n\plus{}U_{n\plus{}2}}{2}$
then $ f(2(nx\plus{}r))\equal{}U_n\equal{}2A(r)(nx\plus{}r)\plus{}B(r)$ for $ A,B: [0,1[\to R$ choosen arbaitrary

then $ \boxed{f(x)\equal{}2A(\{\frac{x}{2}\})x\plus{}B(\frac{x}{2})}$

for exemple if $ \forall r\in[0,1[: \ A(r)\equal{}B(r)\equal{}r$: $ f(x)\equal{}2\{\frac{x}{2}\}x\plus{}\frac{x}{2}$ is solution

[quote="BG Yoda"]this is jensen equation .[/quote]
No,we need $ f$ is continuous  function to have Jensen equation


Harvard college Princeton AMC USA(J)MO USAMO MIT

Hello, everyone!

I am a sophomore this year, and my parents always talk about how I should apply to college next year instead of senior year, which is a "waste of a year" to them.

How easy is it to get in for people who apply early?  I know someone from my school who applied and got into Caltech in his junior year, but it seems it's pretty rare becausehe didn't have that many qualifications...

I don't have that many either...
I am 3rd in my class right now (though I'll probably be 2nd after this semester).  I took the SAT once and got 1580 (should I retake it before applying?)  I am also going to take  3 SAT II tests - math, physics, and writing.  I got into USAMO twice, and into MOP this year (not by being one of the 5 invited sophomores).  I am also president of a club (Mu Alpha Theta).  I play the violin in two orchestras as well as learning it for myself.

Also, All my classes will be AP next year, but so far, I only took 5.

What do I think? Can I get in a good college like Caltech or MIT next year or should I wait a year?

Thank you!

-OR

That's a pretty impressive resume.  I don't know for sure if you can get in but if you really want to it won't hurt you to try(at least I don't think it will).  Just give it a shot if you get in you get in if you don't you don't.  Your resume is probably better than the average person who goes to Caltech or MIT but no one knows if they ll get in for sure at those kinds of schools.

Do you mean just applying to Universities a year early? Or actually going a year early?

If you're talking about applying a year early in order to sit on the admission, that's not a good idea. There's no advantage in doing that, and I don't think very many, if any, Universities will just give you an early admission to put in the bank and wait for a year. There is deferment, but they're not going to let a student defer for a year to finish high school.

If you mean actually going to college early, then that depends on whether you will have fulfilled all of the requirements for your high school diploma first. While I'm not sure if its a set-in-stone requirement to have one before entering a University, it would be a bad idea to go without one.

If you'll be done with the requirements for your high school diploma next year and you're talking about graduating early, then this is a question better asked of each specific University. I know that some schools, like Harvard and Princeton, tend to frown on this unless you can show that you've truly maxed out all of your high school academic options.

But people skip grades or complete high school early all the time and go on to great Universities - so its certainly possible. I'm sure that individual schools could give you better advice about this than we could though. So you should contact them and ask what they think about it.

I was wondering the same as gauss i.e. if he was asking about just applying and waiting or going.  I assumed he meant going since he made the comment about his parents thinking his senior year would be a waste.  But I could be wrong.  I can't offer any help in terms of experience but I think that this is a decision that you ll have to make mostly on your own.  Also I don't think it isn't that big of a deal.  You ll learn more if you go to college but I'm sure that you could still find something to do for a year.  You should also consider if you want to do the high school activities one more year.  I am talking about stuff like the AMC's, various other math competitions, sports(if you do any),  etc.  Keep in mind that you only have one shot at high school but college can last a lot longer.

[quote]You ll learn more if you go to college but I'm sure that you could still find something to do for a year. You should also consider if you want to do the high school activities one more year. I am talking about stuff like the AMC's, various other math competitions, sports(if you do any), etc. Keep in mind that you only have one shot at high school but college can last a lot longer.[/quote]

I highly agree. If you choose to go to college a year early, you will be a year younger and a year less mature than everyone else for the rest of your college years. You will also miss out on all of the fun parts of senior year and high school: fooling around with friends, partying with your classmates, finding young romances, etc... You will never again be able to experience the joy of high school ever again in your life. So what if you get a year ahead in your schooling? Big deal. There is more to your life than that.

Unless you are some super-genius weirdo, I would definitely not encourage you to skip your senior year.

Hi Celeborn, this is my advice/comments:

(1) Senior year is a waste compared to freshman year in college.  I mean you can take the hardest classes offered in your school, but most people start to slack off really early and teachers can't help but go along with it, so the quality of the classes is lower (it's called "senioritis")

(2) It is very hard to get into MIT or Caltech as a Junior.  For some reason they don't really like to admit kids that are not seniors unless they are a super-genius or have maxed out what their HS offers.  There are a lot of schools similar to MIT and Caltech that would accept you.  If you could make it into Caltech or MIT a year later, I would not settle for a place that is not your first choice just to get out of HS.  BUT, there is no harm in applying.  If you make it in, then great!  If you don't, then know that it is just because of your status as an 11th grader, not because you are not qualified, and apply again next year.  Whatever you do, do not choose a lesser college since it's better than HS, but plan to transfer to MIT/Caltech later.  Tranferring to MIT/Caltech is VERY difficult, so you have a much better chance trying to get in as a freshman.

(3) If you are staying in HS for another year, why not take some math/science classes at your state school?  They are cheaper and usually good.

(4) As gauss said, you have to graduate from HS to go to college.  At my HS I could have graduated from HS Junior year if I had taken 2 english classes, 2 PE, and 2 Health classes.  Usually this is not too hard.

Good Luck!

I think Fierytycoon and I have the same basic viewpoint here -- it is definitely within the realm of possibility to skip senior year and go to college (although, as JS was just writing, I'm not so sure you could as easily go to the college you wanted to), but I'm not so sure you'd really want to.  I mean, there is no cost in going to high school, unless you are really miserable there.  Relaxation, taking your time -- there's nothing wrong with that.  In fact, it seems to me that many extremely successful people are unhappy just because they've never learned how to relax.  Perhaps senior year wouldn't be as academically rewarding as freshman year of college -- that, by itself, does not seem to me to be a compelling reason to skip it.

If he went to college early, would he really be "skipping" his senior year or would next year just be his "senior" year?

What's the difference between skipping your "senior" year in high school, and skipping a grade sometime earlier in school?

Well, the major difference is that American high schools teach more than American elementary schools.  In general, though, if someone is at a good school, unless they are (very, very) exceptionally talented, the reasons to skip a grade are very insignificant.  Now, I don't know anything about Celeborn's school -- I don't know if it's good, I don't know if celeborn likes it or not, etc.  And I don't know anything about Celeborn.  But if we assume that it is somewhere in the closed interval between not bad and very good, and that he is somewhere in the open interval between average and unusually talented, then I don't see a convincing case for skipping a grade rather than taking his time.  Besides, if he really wanted, he could do work on his own or take classes at a local college instead of doing away with a year of school entirely.

Or... he could apply to college early. And use the "extra time" to travel abroad, do a service project, or enter a 5 year masters program.

I think there are lots of very good reasons to graduate early from high school. I don't see it as a rush th rough life at all. It all comes down to what his options are if he does graduate early, whether he feels he is prepared, and where he would be most happy. If he has a substantial love for learning (which it appears that he does), then its more than likely that he would enjoy his time in college a lot more than he would enjoy his time in high school.

If he can get into a school that he really wants to go to, and he feels that he is academically, emotionally, and socially prepared - then I say go for it.

If someone's parents are pushing him to skip his senior year of high school, I would give very good odds that (1) he is of recent Asian descent and (2) they are not pushing him to do anything other than go to a high-pressure school such as CIT.  Now, I might be wrong there, but I don't think he's discussing skipping senior year to spend time in Europe or to work for Habitat for Humanity.  I think those are perfectly reasonable things to do -- in fact, I know several people who took off the year between high school and college in order to do a variety of things: a girl downstairs trained with the British Olympic swim team, for example -- but I don't think that's the kind of thing his parents are pushing for.

I do wonder -- Celeborn, what do you think of high school?

That is very much not the same thing as saying that anyone who reached USAMO could get into Caltech, though.

But I would think that most people who did qualify for USAMO would indeed apply to Caltech, right?  They certainly have a decent chance of getting in in most cases.

Personally I don't understand how anyone who is qualified and would probably get in would decide not to go or not to apply to MIT/Caltech.  I didn't apply to them because I know I would not be accepted by a long shot, so I decided I would save that ticket for grad school.

Financial reasons.  If I got accepted to MIT I don't know if I would go since I am middle class.  I am not poor enough to get that kind of financial aid but my family probably wont have enough money for me to go.  It s hard to get scholarships from these kinds of schools.
Now if the person can afford the school I dont know why they wont apply unless they applied to an equally good(in their mind at least) school.

I did not apply to either MIT or Caltech.  I would probably count in the group of qualified people you mentioned.

Well, as long as we're massively digressing anyhow :D

I didn't apply to any schools further west than Chicago, so I wouldn't have applied to Caltech for that reason.  I probably wouldn't have applied anyways, though -- the idea of the culture there was very unappealing to me.  I mean, I really do enjoy relaxing.  I'm not sure anyone at Caltech relaxes, ever :).  A similar principle applied to MIT, although as a bigger and more diverse (in a variety of ways, but mostly just as a consequence of being bigger and slightly less hardcore) school it was less the case.  I visited MIT when I went to USAMO in 2002 and I just didn't like the feel of it at all.  That's really not a very good explanation, but it's the only one I have.

The schools I was really interested in were (initially) Harvard, Princeton and the University of Chicago, and later (by the time I had started getting admissions decisions) just Harvard and Chicago.

[quote="JS1527"](4) As gauss said, you have to graduate from HS to go to college.[/quote]

Not so. It is almost never a requirement to graduate from high school before going to college. Ironically, the colleges most likely to insist that their applicants be high school graduates are the lowest-echelon colleges. The top colleges have been admitting applicants who never completed high school for generations.

This is something homeschoolers have to know, of course. In most states, homeschoolers don't have a channel to obtain a "high school diploma," yet [url=http://learninfreedom.org/colleges_4_hmsc.html]homeschoolers are readily admitted by hundreds of top colleges[/url]. And once a person has completed a college degree, few people care whether or not that person previously graduated from high school or not.

Tokenadult, my bad.

JBL -- Do you take "Math 55" at Harvard?  I am not sure if this exists or too much about it, but I have heard that even people who do USAMO and such have trouble in it.

[quote="JS1527"]JBL -- Do you take "Math 55" at Harvard?[/quote]

I was wondering the same thing. The Harvard Mathematics Web site has [url=http://www.math.harvard.edu/undergrad/Pamphlets/freshmenguide.html]an interesting page[/url] about the different beginning math courses at Harvard that makes Math 55 sound quite interesting.

You certainly have the resume to get into one of those schools.  It's of course not guaranteed, but it won't be senior year either.  I would advise graduating a year early, unless you are particularly attached to the friends you have there.  The only reason to stay for another year would be for your friends.  I at this point, about to go into senior year, wish that I had graduated early.  I've run out of interesting things to do at my high schools, unfortunately.

[quote]I was wondering the same thing. The Harvard Mathematics Web site has an interesting page about the different beginning math courses at Harvard that makes Math 55 sound quite interesting.[/quote]

Yes, that does sound very interesting. I wonder whether they have a copy of that Math 55 diagnostic exam online for people to check out.

[b]Celeborn[/b], I think you have a good shot. Whether you want to do it or not - that's completely different question...

Our daughter  [b]wanted[/b] to graduate early. We didn't want her to leave us at 17, and besides, MIT website said something like "We do accept younger students, but..."  So we offered her either graduate after 11 grade and go to a (top-rated) public university in our city, or graduate in time and try to apply to MIT... She really didn't like school, and honestly, there was not much to do there. So she decided to go with the public university (seeing how it outranked Ivy League in her chosen major). I still don't know if it was the right decision, but it was how she wanted it...

In your case - if you'll have enough credits to graduate after 11 grade, you will be no different from all those graduating 12-graders. Our daughter was called "graduating junior", and she had "senioritis" her junior year, and she went to Prom and to Senior Trip to Florida. The universities will treat you the same (or almost the same, like some posters say here).

If you just will apply in 11 grade, you still have a good chance of being admitted. I know Princeton doesn't require HS diploma, and currently there is a student accepted after 11 grade; she is enrolled both to Princeton and to 12 grade of her High School. I don't know exactly about other universities, but I'm pretty sure most of top-notch places will not demand your High School diploma, just like tokenadult said.

Think of it this way: it is actually an [b]extra[/b] chance! You can try it this year, and if you are not accepted, you will try it agani next year! :-)

That is, if [b]you[/b] want it yourself.  And you may want to check with the universities if they will allow you to take that year off... To travel, to do Math, to prepare for the competitions and maybe make it to IMO or IPhMO as a homeschooler... instead of spending time in the "required for graduation" classes, like PE or Consumer Education.

On the other hand, maybe there is no required PE at your school, and you actually like History and English, and your school has a strong Math club, and you have lots of friends there, and you are looking forward to all that senior year social life...  Then better try to explain all that to your parents. Otherwise you will be tempted to blame them for this pushing, if something will go wrong. This should be your decision.

Celeborn,

I don't understand why you are worried about getting in. Most people who get into the top colleges don't necessarily have the most qualifications. It's really about how you present yourself, and the level of confidence you present yourself with. Be proud ~ and let it reflect in your application and your interviews. I am certain you will get in anywhere you wish to go. Go ahead to MIT and Caltech -- I wish you the best of luck...

And if you would like the application essays from that junior who got into Caltech... I saved it just for you...

Don't forget to apply for scholarships ~ that's all you should worry about..

Once again... good luck at MIT and Caltech...

[quote="JS1527"]Now that I think about it, I have heard before that Caltech hardly ever rejects someone who qualified for USAMO.  I don't have any facts to base this on though.[/quote]

Absolutely not true for either MIT or CIT at all! I don't know much about CIT, but I know MIT frequently rejects MOPers and they have in the past rejected members of the IMO team.

Too many young kids value the USAMO as a measure of ones abilities, when sadly... it is not. It is one set of tests that you take during the year that measure some math concepts and how well you don't screw up. You also have to realize that the opportunities are not always available to everyone. Many people who don't qualify for AIME even in one high school could probably easily qualify for USAMO if they went elsewhere. The USAMO and other math contests represent the opportunities the individual has had and how they chose to make the best of them. I didn't even know what the USAMO or AIME was until the summer after 11th grade... I come from a high school where there are 10 students in calc AB, and none of the teachers are qualified to team the basis skills of geometry and algebra, much less teach or even know about math on the AIME/USAMO level. Is that fair to someone like me? Of course not... is it my fault? No... Colleges look at these types of situations for each person, their final decision comes down to "Will this person make MIT a better place". His list of qualifications are quite impressive but I don't think it's enough information to draw a conclusion to whether or not you would be accepted.

I think you should go for it though, what the hell. Have a good interview, call admissions and have some chats... Show them you're a competant and nice person who loves there school and wants nothing more than to be there. You're in for sure if you do that.

*goes to read then to bed*

As for math 55 --

Harvard has lots of websites and information about their tough multi/lin-alg classes, only some of which actually reflects reality.  So I'll just tell you everything.

At Harvard, math 21 is your standard multivariable calculus and linear algebra course.  Above math 21, there are 3 courses: math 23, 25 and 55.

Math 23, which had more than 100 students first semester and more than 80 second semester this year, is a theoretical multi/lin-alg course.  It is proof based, which is what gives so many people trouble.

Math 25, which I took, had 25 students fall semester and 10 spring semester.  It is similar to 23, but there is MUCH more work.  More mathematical experience was expected, as well as some experience with proofs.  It also covered a somewhat broader range of topics.

Math 55, which a lot of IMO people took, had 20 students first semester and 19 second semester.  It has somewhat more work than 25 and covered a somewhat broader range of topics.  It expects more mathematical/proof-writing experience than 25.

Math 25 and 55 are not extremely different courses -- the gap between 25 and 23, I would say, larger.  Normally, one has to choose his courses within a week or two after classes have begun.  However, students choosing between 25 and 55 are given 2 additional weeks to make the decision.  I personally sat in on both courses for a week past the required time.  At that time, there were still a lot (maybe 30?) kids in 55 and the professor gave a "quiz."  I had already decided that I was going to take 25, but I took the test in 55 just for fun.  Students with scores below a certain level were I believe "talked to" by the professor, although not explicitly asked to leave.  I would imagine that all such students had figured out for themselves that they didn't belong there, however.  I don't actually know for a fact that I was above the cutoff, but it worked out to something like half a question out of 5.  Anyhow, the main reasons that I took 25 and not 55 were that I did not much like the 55 professor and I also realized that, if I had taken 55, I would not have done the work necessary to make it a worthwhile experience.  (I place a high value on not being over-worked and over-stressed.)  There is no difference in terms of future coursework between 23, 25 or 55 students -- there is no course for which 55 is a prerequisite that 23 or 25 is not.

By the way, if you want a sense of how difficult these classes were, on our final from first semester, the top 3 scores were 80, 77, 65.  No other scores were above 55, I think.  Then there were 5 or more scores below 30, including a 10 and a 2.  (All out of 100.)  People getting an average of 2/3rds of everything were at the top of the class.

Hmm, that's all I can think of to say -- if anyone has questions, I'd be happy to answer them.

[quote="JBL"]
By the way, if you want a sense of how difficult these classes were, on our final from first semester, the top 3 scores were 80, 77, 65.  No other scores were above 55, I think.  Then there were 5 or more scores below 30, including a 10 and a 2.  (All out of 100.)  People getting an average of 2/3rds of everything were at the top of the class.

Hmm, that's all I can think of to say -- if anyone has questions, I'd be happy to answer them.[/quote]

What do you have to get to pass the class?  I mean if ppl who get 2/3 right are the top of the class then that's just crazy.

Btw, what did you get on the final?

Would you care to post a sample problem (for any one or more of the courses) so we can get an idea?

Well, we had as a homework problem to prove Holder's and Minkowski's inequalities fairly recently.  The course website is at
http://abel.math.harvard.edu/~laurent/math25/math25.html
if you would like to take a look at our problem sets, midterm and final from this semester.
For the math 55 website, try
http://www.math.harvard.edu/~siu/math55b/index.html

First semester, I was the 65.  There may actually have been another score on the final a bit better than mine, like a 67 or something.  I don't commit all of this to memory exactly :).  Second semester, I had a 102/140 on the final, which was the highest in the class.

I don't *think* anyone failed the class.   There isn't an expectation that people will be getting in the 80's and 90's, so people who got in the 30's on the exams were still probably getting B's or B+'s in the class.  (The homework was also significant in grading, and people did relatively better and were less spread out on those grades.)

Its not unusual for some college courses to have top scores of 2/3rds of the total number of possible points. In college courses, grades are usually curved in some way. Different professors have different ways of doing this - some use a formula of some kind, and some just do it subjectively.

You can read the course descriptions and see the homework and exams for some of the Harvard courses at http://www.registrar.fas.harvard.edu/Courses/Mathematics.html


What is the greatest positive integer that must divide the sum of the first ten terms of any arithmetic sequence whose terms are positive integers?

We can write this as:

$ n\plus{}(n\plus{}1)\plus{}(n\plus{}2)\plus{}\ldots\plus{}(n\plus{}9)\equal{}10n\plus{}45$, so the answer is clearly $ \boxed{5}$.


There is a Favorites button when you are logged in, but I haven't figured out how to add a Favorite. How do you use this feature?

The attached picture of the lower right corner of a post screen should explain it.  Note that what you see will be slightly different if you use the AoPS skin.

Thanks.

Does this only work for topics, or can forums also be added as favorites?

You can "watch this forum" to a forum, but only topics can be favorites.

It would not make sense to have forums in the favorites list. However, if you only visit certain forums and you are totally uninterested by the rest, you can always "clean" your view in the Profile >> Hidden Forums option.


Let $ v$, $ w$, $ x$, $ y$, and $ z$ be the degree measures of the five angles of a pentagon. Suppose $ v < w < x < y < z$ and $ v$, $ w$, $ x$, $ y$, and $ z$ form an arithmetic sequence. Find the value of $ x$.

$ \textbf{(A)}\ 72 \qquad \textbf{(B)}\ 84 \qquad \textbf{(C)}\ 90 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 120$

[hide] The angles in a pentagon add up to (5-2)(180)=540 degrees. We let the common difference be a so $ (x\minus{}2a)\plus{}(x\minus{}a)\plus{}x\plus{}(x\plus{}a)\plus{}(x\plus{}2a)\equal{}540$ and we find $ x\equal{}108$ degrees. Our answer is $ \text{(D) 108}$. This also proves that in an arithmetic progression with an odd number of terms, the middle term is the average of all the terms.[/hide]

lets say that the common difference is a.

then v, w, x, y, z is $ x \minus{} 2a, x \minus{} a, x, x \plus{} a, x \plus{} 2a$.

the sum of the angles in pentagon is 540 degrees.

$ 5x \equal{} 540 \implies x \equal{} 108$

answer: D

blah beaten again  :(


abstract algebra
Princeton
college
Harvard
geometry
topology
MIT

I finished my undergrad degree this last June, I had applied to grad schools in December of my senior year like you're supposed to, but didn't get in anywhere.

I've been trying to find a job as an actuary, passed an actuary exam, looking to take another one, but finding a job is still hard.

Anyway, that doesn't matter since I don't want to be an actuary, I want to study math. Please help me figure out where I have a reasonable shot of getting accepted in my current situation.

school: UCSC
GPA: 3.0
major GPA: 3.3
major: Math (pure concentration)
GRE: 500 verb, 800 quant
math GRE: 620 (should have signed up to retake this a couple months ago)

I took as many math classes as I possibly could, once I figured out I wanted to do math. I ended with 20 undergrad math classes (~100 credits) and 5 graduate classes (25 credits), graduated in 4 years and did relatively well in the grad classes(B/B+'s with one A-), also wrote a thesis.

grad classes covered: rings & modules, fields & galois theory, commutative algebra (ch1-9 of atiyah), measure theory, Lp spaces, functional analysis, some spectral theory

I know I'm not going to Princeton or Harvard, but I want to find the place where I can do my best possible phd work under the circumstances.

My main interests are in the direction of Commutative algebra, Algebraic Geometry, and soft analysis. Though I haven't learned any algebraic number theory or much algebraic topology, I am fairly confident if I continue studying homological algebra (as I currently am), that I will end up set on Algebraic Geometry.

any personal experiences or lesser known places with great faculty are also appreciated

[quote="shallow"]graduated in 4 years and did relatively well in the grad classes(B/B+'s with one A-)[/quote]

At most graduate programs that I am familiar with, anything less than an A is a poor grade, and any type of B is a horrible grade.  It may be different at your particular program but this is very unlikely to impress prospective graduate programs.  As is, I don't think you have a chance at any top 30 graduate Ph.D programs in math.  It's possible that you could get in somewhere for a master's, do great, and then reapply to better Ph.D programs though.

I wouldn't go so far as to say a B in a graduate course is a horrible grade. There are certainly people with similar GPAs to yours who get into Math grad schools. It's all in the context of what type of school you're hoping to get into. If you don't mind saying, what grad schools did you apply to in your senior year? (And you should think about who you had to write your recommendations for you and if they were the best people to have asked.)

[quote] If you don't mind saying, what grad schools did you apply to in your senior year?  [/quote]

I applied to several schools (some I'd rather not mention out of embarrassment) including a few of the UC's (santa barbara, san diego, and la), the last two were long shots for sure, but I thought it was worth a shot considering the size of their departments.

In hindsight, I should have definitely applied to my undergrad school, but I had heard a change of scenery and department was recommended in the transition from undergrad to grad school and figured santa barbara was about the same size and type of department anyway.

I realize it may be hard to (or some may not want to) suggest realistic schools for me at a forum where people are deciding between mit and harvard, but could you try looking at this list of my current planned apps and make any additions or removals as you see fit.

PhD:
UCSC
UCSB
Notre Dame (no masters)

Masters:
U Maryland
U Oregon
Texas A&M
U Arizona

I'm not planning on applying to all of these necessarily. Feel free to slap me in the face if I'm being too ambitions with my selections(again).

[quote="shallow"]I finished my undergrad degree this last June, I had applied to grad schools in December of my senior year like you're supposed to, but didn't get in anywhere. I've been trying to find a job as an actuary, passed an actuary exam, looking to take another one, but finding a job is still hard.

Anyway, that doesn't matter since I don't want to be an actuary, I want to
study math. [/quote]

Well, If you didn't want to be an actuary, why did you attempt those actuary examinations? (I do understand that the job market isn't great). You could have invested the time which you had spent in those actuary examinations to prepare well for the Math subject GRE. Probably, you might have got a 'stellar' score in the first attempt itself.

[quote="shallow"]I'm not planning on applying to all of these necessarily. Feel free to slap me in the face if I'm being too ambitions with my selections(again).[/quote]

Based upon your posts I had observed that you're displaying a low degree of self-confidence. Just because, [b] someone [/b] remarked rather implicitly that your grades are horrible doesn't mean that you're completely worthless. Actually, it would have been far wise had you -- instead of investing your time in those actuary examinations-- invested in preparing well for the Math subject GRE. Also, as an addendum to [b] blahblahblah [/b]sir's advice-- I would advise or rather suggest you to re-take the Math subject GRE and perform really well on it. (Say, getting in the 98 or 99 percentile range). It might shore up your application to some extent.

(In the given circumstances I think it would be better or wise if you could follow [b] blahblahblah [/b] sir's advice-- along with the addendum that I had suggested).

[quote="shallow"]I realize it may be hard to (or some may not want to) suggest realistic schools for me at a forum where people are deciding between mit and harvard, but could you try looking at this list of my current planned apps and make any additions or removals as you see fit. [/quote]

Not everyone on this forum is from MIT or Harvard. This forum constitutes a high percentage of middle-schoolers. :D

[quote]Well, If you didn't want to be an actuary, why did you attempt those actuary examinations? [/quote]

As much as I'd like to, I can't exactly to sit around not working for a year until I can start grad school. So I wanted to find a decent paying job that was at least somewhat mathematically intensive in order to have some real world experience, a savings account, and a possible backup plan.

In my "zeal" of pursuing the actuarial track, I neglected to sign up for the October and November GRE subject test dates (next available one is in April). I know this was a HUGE mistake.

[quote]Based upon your posts I had observed that you're displaying a low degree of self-confidence[/quote]

I hope understandably so, after being rejected from [i]every[/i] graduate school I applied to.

[quote]In the given circumstances I think it would be better or wise if you could follow  blahblahblah  sir's advice-- along with the addendum that I had suggested[/quote]

Thank you both for the advice. I'll try mostly masters programs this year, and if that doesn't work I'll be sure to retake the GRE subject test and reapply the year after that.

If anybody has any comments or opinions on the programs I listed earlier, particularly the master's programs, I'd greatly appreciate it.

shallow, one question that you should ask yourself and answer honestly is:  Why is your application so weak?  Specifically, does it reflect your actual mathematical abilities or not?  And assuming that the answer is no, why is that?  Have you changed in some way?

These are tough questions, but they are important.  While I sincerely believe that anyone who wants to major in math in college should do so, I don't believe that anyone who wants a PhD in math should pursue one.  The reason is that embarking on a PhD is embarking on a specific career path, so you should have some idea of what you intend to *do* with that PhD and what your chances of success are.

While I agree with the idea that it's nice to go somewhere different for graduate school than undergrad, it is sometimes the case that the best school that will accept you is your alma mater.

Finally, the number of people who regularly post on AoPS who could have chosen between Harvard and MIT for a math PhD can surely be counted on two hands.  Even the number of AoPSers who will face such a choice in the future is probably much smaller than you think.  You're talking about a population of at most a couple dozen people in the world per year.

[quote]shallow, one question that you should ask yourself and answer honestly is: Why is your application so weak? Specifically, does it reflect your actual mathematical abilities or not? And assuming that the answer is no, why is that? Have you changed in some way? [/quote]

[quote]The reason is that embarking on a PhD is embarking on a specific career path, so you should have some idea of what you intend to *do* with that PhD and what your chances of success are.[/quote]

I appreciate the questions and I have been thinking about these things ever since I started thinking about graduate school. I don't know if you intended me to answer these, but my intent to pursue a PhD is not half-hearted or nonchalant in any way, I have very clear career and life goals, and a PhD in math plays an integral role in that.


In your chapter, how many teams are allowed to go to state? Nationals? Top 2 in the chapter or state.....top3...top team?

How many should a team expect to get right in order to go to state? Nationals?

top 3 teams plus top 10% not on a team go to states
top 4 go to nationals

Top four teams in each chapter go to state, and top 2 individuals not on a team go to state.  Then, top 4 individuals and coach of the top team go to nationals.  As for the score, if your in california or someplace smart, your chapter team score should be like [color=red]XX-XX[/color].  If your in Arkansas(where i'm from) or someplace stupid, you can make it to state with a [color=red]XX-XX[/color](which our team did).

[color=red]Moderator rcv's note:  No Chapter scores until the chapter competition window is over.[/color]

[quote="xxm0rg07hxx"]  As for the score, if your in california or someplace smart, your chapter team score should be like [color=red]XX-XX[/color].  [/quote]
I live in like the weakest chapter in California, and last year the teams that made states had a 59.5+.
As for how many people make states, it depends on how many schools are registered, whether the winning school is a magnet, and so forth. There might be something on the mathcounts website about that.

[quote="easyas3.14159..."]
I live in like the weakest chapter in California, and last year the teams that made states had a 59.5+.
[/quote]
WOW, I our team had like a 32 last year(in Arkansas), but I was the only one that was good.  Everyone else got less than a 15.

Aww, lucky, I bet it'd be easy for you to make nats in Arkansas. As for me, I'd be thankful to make state countdown.

[quote="xxm0rg07hxx"]I did make nats last year.  I got 2nd.  To garentee a place at nats was third and they got a 28, and in your state, a 4_ something.  So I guess I am lucky.[/quote]

NO FAIR!!!I'm moving to Arkansas. What grade are you in?

To the moderator,
[hide]I was talking about last year's scores, not this years scores, so I thought we were aloud to talk about last year's score.  Because I haven't even been to this year's chapter yet.[/hide]

Last year I was in 7th, i'm my second year of MATHCOUNTS now, 8th grade.

Clarification:

At Chapter, for every 10 teams that participate, one goes on to state.  So for 9 teams, only 1 goes on to state.  For 11 teams, 2 go on to state, and for 21 teams, 3 go on to state, etc.

Also, the top 2 individuals not already going to state (on a team) will be invited to attend as individuals.

im n individual...:furious:
didnt turn in those stupid pkts.. :mad:
for CA state individual, how many individuals from each chapter go to states?

The rulebook officially says that it is REQUIRED that a chapter sends one team and two individuals.

HOWEVER: Sending more than the requirement is at the discretion of chapter coordinators.

In most cases, they send more. My chapter sends 4 teams, 4 indiv.

is Neal Wu the 1st place nat last year??

My school gets 1st in Chapter every year since...[i]forever[/i]
But we get slautered in states. Detroit CountryDay takes 1st team with 2-3 team mates going to nationals. Troy Smith takes 2nd team with 1-2 team mates going to nationals. My school used to get National top 5 (team and individuals) 16 and before years go. *sigh* my school was beast 1 and before years ago.

[quote="tz"]is Neal Wu the 1st place nat last year??[/quote]

yeah and that other guy who went to nats last year too

[quote="biffanddoc"][quote="sferics"]I live in the weakest chapter of Louisiana and we make it to states almost every time. Glasgow really dominates there though...[/quote]
Might have to do with Neal Wu a little...[/quote]

Heh, that's probably the only disadvantage from living in one of the weakest states. Although, If I do make nationals on the Louisiana state team, I'll ask for some advice. :)

[quote="kyyuanmathcount"]You're Michigan? What school?[/quote]

He goes to Clague in my chapter.  They beat us by [color=red]XX[/color] points.  :mad:

tz: next year, we will beat you  :rotfl: just kidding.

kinda like indiana's team last year. they lost by $\frac{0}{\ln{\pi^4}}$

For my chapter, the top 5-6 teams go to state (out of 25-30 teams). Kansas only has 4 chapters.

[quote="Klebian"][quote="xxm0rg07hxx"]I did make nats last year.  I got 2nd.  To garentee a place at nats was third and they got a 28, and in your state, a 4_ something.  So I guess I am lucky.[/quote]

Man...
A 28 in Texas last year wouldn't get you in the top 25... probably even not top 40 or something.

It took 32 or 33 to get into top 10.[/quote]

How on earth did Texas beat us last year?

Robin He, Clague is [i]ALWAYS[/i] better than ur school Greenhills.
We get 1st in Chapter since...[i]forever[/i]
Greenhills comes 2nd in Chapter since....[i]forever[/i]
Yeah, we beat them by [color=red]XX[/color] points.
 :D :D

[color=red]Edited by moderator rcv[/color]

If you had a 28 you would've barely made top 25 Indiana. I believe #24 had a score of 27 in Indiana. Maybe it was 29?

Wow, i'm amazed we even had 1 nats champion like 16 years back or something.

Hey.....How much should you get on the team round? Individual score? to get to state or nationals?

in one of the most competitive states, california, and in one of the most competitive chapters, orange county, to 2 teams go to nationals, top 2 or 3 not on a winning team go on to state

in state, last year, the top four people had 44,45,46,46 on the state test

[quote="kyyuanmathcount"]The rulebook officially says that it is REQUIRED that a chapter sends one team and two individuals.

HOWEVER: Sending more than the requirement is at the discretion of chapter coordinators.

[/quote]

Thats true.  We have the strongest chapter in Northern California (I think something like 3/4 ths of the team last year was from our chapter), and had 30+ schools this year.  This year 4 teams went, and last year I think 2 were supposed to go, but there was a tie and 4 went, or something like that.

[quote="tz"]Robin He, Clague is [i]ALWAYS[/i] better than ur school Greenhills.
We get 1st in Chapter since...[i]forever[/i]
Greenhills comes 2nd in Chapter since....[i]forever[/i]
Yeah, we beat them by [color=red]XX[/color] points.
 :D :D[/quote]

No, we lost to you by [color=red]XX[/color] points.

Your score was [color=red]XX[/color]
Our score was [color=red]XX[/color].

[color=red]Edited by moderator rcv[/color]

In our chapter they take the top 7 teams and the top 5 individuals not on a winning team.  we have a very large and competitive chapter.  It varies from chapter to chapter.  In Texas the state mathcounts committee decides how many teams and individuals each chapter gets to send to state.

[quote="Klebian"][quote="xxm0rg07hxx"]I did make nats last year.  I got 2nd.  To garentee a place at nats was third and they got a 28, and in your state, a 4_ something.  So I guess I am lucky.[/quote]

Man...
A 28 in Texas last year wouldn't get you in the top 25... probably even not top 40 or something.

It took 32 or 33 to get into top 10.[/quote]

in oklahoma last year a got a 21 or so and made 15th...at state.
oklahoma is an easy state  :lol:

[quote="apusey"]
in state, last year, the top four people had 44,45,46,46 on the state test[/quote]
That's crazy, 44, 45, 46, and 46.
I wish I had that score in school or chapter

i think the top person at oklahoma state competition got a 36 or 37.

The posting of scores and score comparisons has gotten out of control.  If you are posting to this Forum, you are expected to know the rules of this Forum.  There will be no discussion of the Chapter Competitions until the Chapter Competition window is over.

Those who are discussing current-year scores are jeopardizing your participation in this Forum.  Depending on MATHCOUNTS' rules, you may also be at risk of disqualifying your own and/or your team's participation in your State competitions.

Since a couple of you have been unable to keep the discussion and comparisons to prior years, I am locking this thread.  If you read the posts from last year, you can get some idea of what was necessary to get to the State and National Competitions.


Hi, I 'm using a Mac machine with the Mac version of Microsoft Office. Which version of Latex should I download? By version, I mean that there's a windows machine version and then there's TexShop on the FAQ page. Any suggestions would be much appreciated.

There are some misconceptions in what you've said:
1. LaTeX has nothing to do with Microsoft Office;
2. You can't run Windows versions on a Mac (without using virtual machines - or is that what you meant by 'Windows desktop'?);
3. TexShop is an editor for LaTeX files, you also need LaTeX itself.

For a guide to installing LaTeX on a Mac have a look at [url=http://guides.macrumors.com/Installing_LaTeX_on_a_Mac]Installing LaTeX on a Mac[/url]

Thanks for the link and for fixing up the confusions.

By Mac version of MS Office, I meant the MS office for Mac. Sorry for the switchup.

New Macs have dual-boot capabilities; if you have Windows, you can boot into it or into Mac OS X as desired.

You should get the version of $ \text{\LaTeX}$ that matches the OS you intend to use it in. If you are likely to use both, get both versions.


Please solve the following problem.
For each +ve $ x \in \mathbb{R}$, define

$ E(x)=\{[nx]: n\in \mathbb{N}\}$
Find all irrational $ \alpha >1$ with the following property:
 If a positive real $ \beta$ satisfies $ E(\beta)$ is a proper subset of $ E(\alpha)$ then $ \frac{\beta}{\alpha}$ is a Natural number.

Anybody please help. :( (It might be a number theory problem :maybe: , idon't know):help:

the question was more like find the irrationals $ \alpha$ for which...
i will give a solution for $ \alpha > 3$ (i have been able to show that it is rational for $ \alpha > 2$)
i don't seem to have anything for $ 1 < \alpha < 2$
for $ \alpha > 2$
let $ {k_{n}}$ for which $ [n \beta ] \equal{} [k_{n} \alpha ]$
this implies that $ n \beta$ and $ k_{n} \alpha$ lie in the same interval of consecative integers
$ \Longrightarrow \mid n \beta \minus{} k_{n} \alpha \mid$
$ \Longrightarrow \mid n \frac {\beta}{\alpha} \minus{} k_{n} \mid < \frac { 1}{\alpha}$
  now as $ \alpha > 2$ we can also find $ r \in \mathbb{Z^{ \plus{} }}$ such that $ \frac { 1}{\alpha} < \frac {1}{2} \minus{} \frac {1}{2r}$
$ \Longrightarrow \mid n \frac {\beta}{\alpha} \minus{} k_{n} \mid < \frac {1}{2} \minus{} \frac {1}{2r}$

now we define some new notations and definations
let numbers like $ \frac {1}{2},\frac {3}{2},\frac {5}{2},\frac {7}{2},...$ be called $ spdp's$
now let $ bhel(x)$ be defined as largest $ spdp$ smaller than $ x$
and let $ dahi(x) \equal{} x \minus{} bhel(x)$ (quite analogous functions to integer part and fractional part of a real)

now case 1
i.e.$ \frac {\beta}{\alpha}$ is irrational
then as $ \Longrightarrow \mid n \frac {\beta}{\alpha} \minus{} k_{n} \mid < \frac {1}{2} \minus{} \frac {1}{2r}$
we have the mod of (some integer multiplied by an irrational greater than $ 1$ minus an integer strictly less than half)
$ \Longrightarrow$ fractional part of $ n \frac {\beta}{\alpha} < \frac {1}{2} \minus{} \frac {1}{2r}$ or $ > \frac {1}{2} \plus{} \frac {1}{2r}$
i.e. $ dahi(n \frac {\beta}{\alpha}) \not < \frac {1}{2r}$ and $ \not > 1 \minus{} \frac {1}{2r}$ for each $ n \in \mathbb{Z^{ \plus{} }}$
now consider the following numbers
$ dahi(n \frac {\beta}{\alpha})$ for $ 1 \le n \le 2r$
we divide the interval $ [0,1)$ into intervals as $ [0, \frac {1}{2r}) ; [\frac {1}{2r},\frac {2}{2r}),[\frac {3}{2r},\frac {4}{2r}),[\frac {5}{2r},\frac {6}{2r}),..,[\frac {2r \minus{} 1}{2r},1)$
now if any of them lie in the first or the last interval then we are done (i.e. we have a contradiction to the condition mentioned above)
else some $ 2$ say $ dahi(a \frac {\beta}{\alpha}),dahi(b \frac {\beta}{\alpha})$  with $ a > b$
as a result we consider $ dahi((a \minus{} b)\frac {\beta}{\alpha})$ this is either in the first or the last interval.
again a contradiction
hence
$ \frac {\beta}{\alpha}$ is rational and the denominator in its reduced form is odd and $ < r$
for $ \alpha >3$
$ r\equal{}3$ satisfies hence the denominator has to be $ \le 1$ and hence $ \equal{}1$
therefore it is natural


function
inequalities
geometry
cyclic quadrilateral
national olympiad

The morrocan TST's for IMO 2005 are over. Here are the problems :
[b][size=150]First day[/size][/b]
27 May 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39108]Problem 1 :[/url]

Find all the nonegative integers satisfying $3y^{2}=x^{4}+x$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39110]Problem 2 :[/url]

Let $A$ be a set of positive integers such that
a) if $a \in A$, the all the positive divisors of $a$are also in $A$;
b) if $a,b \in A$, with $1<a<b$, then $1+ab \in A$.
Prove that if $A$ has at least 3 elements, then $A$ is the set of all positive integers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39109]Problem 3 :[/url]

The real numbers $a_{1},a_{2},...,a_{100}$ satisfy the relationship :
$a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}+( a_{1}+a_{2}+\cdots+a_{100})^{2}= 101$
Prove that $|a_{k}| \leq 10$ for all $k \in \{1,2,...,100\}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39107]Problem 4 :[/url]

Consider a cyclic quadrilateral $ABCD$ and let $S$ be the intersection of $AC$ and $BD$
Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively.
Prove that the bisector of segment $EF$ meets $AD$ and $BC$ at their midpoints.

[b][size=150]Second day[/size][/b]
28 May 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39207]Problem 5 :[/url]

Find all the functions $f: \mathbb R \rightarrow \mathbb R$ satisfying :
$(x+y)(f(x)-f(y))=(x-y)f(x+y)$ for all $x,y \in \mathbb R$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35937]Problem 6 :[/url]

Let $a,b,c$  be positive real numbers. Prove the inequality:
$ \frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a\geq a+b+c+\frac{4(a-b)^{2}}{a+b+c}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=35935]Problem 7 :[/url]

Find all primes $p$ such that $p^{2}-p+1$ is a perfect cube.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39209]Problem8 :[/url]

Consider in the interior of a convex quadrilateral four circles such that each one of them is tangent to two sides of the quadrilateral and tangent to two of the other circles.
suppose that the qudrilateral has in inscribed circle.
Prove that two of the four circles have the same radius.

[b][size=150]Third day[/size][/b]
28 May 2005
Time allowed : 4h

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39312]Problem 9 :[/url]

Find all the positive primes $p$ for which there exist integers $m,n$ satisfying :
$p=m^{2}+n^{2}$ and $m^{3}+n^{3}-4$ is divisible by $p$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39314]Problem 10 :[/url]

Consider the set $A=\{1,2,...,49\}$. We partitionate $A$ into three subsets. Prove that there exist a set from these subsets containing three distincts elements $a,b,c$ such that $a+b=c$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39315]Problem 11 :[/url]

Consider a sequence $\{a_{n}\}_{1}^{\infty}$ of positive reals verifying :
a) there exist $c \in \mathbb R$ such that $a_{i}\leq c$ for $i=1,2,...$
b) $|a_{i}-a_{j}| \geq \frac{1}{i+j}$ for all $i \neq j$
Prove that $c \geq 1$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39316]Problem 12 :[/url]

Let $ABCD$ be a cyclic qudrilaterlal such that $AB.BC=2.CD.DA$
Prove that $8.BD^{2}\leq 9.AC^{2}$

I see that my problem (no. 2) was proposed :D What did you do Erdos? :)

The second test has been posted .
To [b]valentin[/b], I didn't do well in the first TST  :( , but I did better in the second  :lol:
I'll be glad if you could make a links to the morrocan tst's in the ML home page  :lol:

;)  :mad:  you cant call this Moroccan olympiad since the problems are just recycled from the true olympiads as their precedents
 :lol: anyway, we were able to solve only few problems

[quote="erdos"]I'll be glad if you could make a links to the morrocan tst's in the ML home page  :lol:[/quote]Ok, you can be glad now :)

I have a question:
Why do two of the problems (Day 2) coincide with the ones from the Balkan MO 2005? ;)

Well, none of the problems are actually original I think :) They either presumed that students don't use internet, or they want to select exactly those who do use internet :D :D :D

many problems are duplicated in a lot of contries...  ;)

[quote="math_sipo"]i'll tell you what!
i know these guys who prepare the Moroccan TST!
they are lazy people!
in the IMO XYZT they meet their mates from other countries
they give them samples from the TSTs  :mad:
so morocan TST XYZ(T+1) or XY(Z+1)0 or etc ... :D
is a kind of "combinaison" of foreign's TSTs :)
do you see what i mean ;) ;)[/quote]

this is quite normal for a country such as Morocco, I explain: the people concerned with the IMO team selections there are not really involved in mathematics and do not have a very big mathematical knowledge so that they can create, the moroccan mathematicians are almost all living in Europe and in north America. So it is somewhat "natural" that the problems are not original. However, this is the case for the most countries except those with a very long mathematical history.

[quote="Michael"]this is quite normal for a country such as Morocco, I explain: the people concerned with the IMO team selections there are not really involved in mathematics and do not have a very big mathematical knowledge so that they can create, the moroccan mathematicians are almost all living in Europe and in north America. So it is somewhat "natural" that the problems are not original. However, this is the case for the most countries except those with a very long mathematical history.[/quote]Not quite. Even those countries are beginning to do the same (ie copy problems). And some other countries (for example France) are replacing their old ways of making tst-s (of copying problems from other countries) with new ways (ie selecting original problems).

[quote="erdos"]The last test has been posted. And finally the Morrocan team for IMO 2005 has been announced, I am sorry I don't have the results. here is the team members :

[b]Mehdi Kamill [/b] (IMO 2004)
[b]Sif maach[/b] (IMO 2004)
[b]Mohammed karim moustaid[/b]
[b]Bellahsayn Ibrahim[/b]
[b]yahyaoui fayçal[/b]
[b]Badr Sbai[/b] (Erdos  :lol: )[/quote]

the best result for the Moroccan IMO team was in 1995 (one silver and four bronze medals), and the best individual result was achieved by Youssfi Chrif in 1991 (30 points out of 42).

I hope that the team of this year will do better individually and collectivily :P  :P  :P

i hope so!
and for u erdos (Badr) get ready boy!
we want Gold this time allright!???
 ;)
good luck

It's true that the problems are not original however no one talked about the difficulty of the tests !? Some said that it was the most hard ever given morrocan test!!! So i want to know different opinions if we consider that the problems were new for students and the tiring 3 consecutive days??

i admit that it was difficult comparing with other TSTs :)

it's high time we forgot the imo :) the bac exam won't wait us , and after we can restart . i hope we honour our country like in 1995 , but nowadays it's getting harder...anyway we will do our best  :blush:

i hope so wisemaniac!
good luck for the bac also ;)


Find the value of $ xy$, given that $ x\minus{} \frac{1}{x} \equal{} y\minus{}\frac{1}{y}$ and $ x \neq y$.

$ x\minus{}\frac{1}{x}\equal{}y\minus{}\frac{1}{y}$

$ x\minus{}y\equal{}\frac{y\minus{}x}{xy}$

$ xy\equal{}\minus{}1$


I have problem with this:
We have:
$ a,b,c > 0$ and $ a \plus{} b \plus{} c \geq abc$
prove, that:
$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {3}abc$
Any ideas?  :huh:

Assume $ a^2\plus{}b^2\plus{}c^2<abc\sqrt{3}$

By AM-GM, $ abc\sqrt{3}>a^2\plus{}b^2\plus{}c^2\ge 3(abc)^{\frac{2}{3}}\Rightarrow abc>3\sqrt{3}$.
By Cauchy, $ 3abc\sqrt{3}>3(a^2\plus{}b^2\plus{}c^2)\ge (a\plus{}b\plus{}c)^2\ge (abc)^2\Rightarrow abc<3\sqrt{3}$.

Contradiction.

See here: http://www.mathlinks.ro/viewtopic.php?p=495326#495326  :wink:


What is the missing term of the sequence $1, 4, 15, 40, \_\_\_{,}156, 259$?

[hide]http://www.research.att.com/~njas/sequences/?q=1%2C4%2C15%2C40&language=english&go=Search

So it is just 1111 in base 4 converted to base 10 which is 64+16+4+1=85. I would never have thought of that =s. Nice guess though hahahah [/hide]

:mad:

Ugh, what a dumb question to put on a contest.  (Good thing we guessed it right anyway!   :rotfl: )

Wow. My guessing pro-ness is overflowing.

[quote="Phoenix257"]You can use the method of finite differences, which does get you the answer 85.[/quote]

What do you mean by this?

Given a polynomial $f_{0}(n)$, we define $f_{k+1}(n) = f_{k}(n+1)-f_{k}(n)$.  If $f_{k}(n)$ is constant for finite $k$ then $f_{0}(n)$ is a polynomial by induction.  $f_{i}(n)$ are called the "finite differences."

Well, it's a cubic polynomial, so the 3rd difference between terms should be constant.

I thought of using it and then I sorta gave up since time was running out  :oops:

[quote="Phoenix257"]You can use the method of finite differences, which does get you the answer 85.[/quote]

True, but that's a lot of finite differencing. And of course you wouldn't call it random after you figured it out... so the topic title is still appropriate.

Really??  Maybe you guys are thinking of a more complicated version of the method of finite differences that I don't know about, which is probably the case since you guys know a lot more math than I do.

This is what I used to solve the problem: http://jeff.ct.wong.googlepages.com/MethodofFiniteDifferences.pdf

It should take only 30 seconds to solve?

No, we're thinking of the same thing, but this problem appeared during the MathIsCool Pressure Round, and we were lazy.


geometry
conics
hyperbola
Asymptote
angle bisector
geometry unsolved

It is fairly easy to draw a bisector which passes a summit and also divides the area of the triangle ABC.
It is also not that hard to draw another bisector that goes through an unidentified side P.
then, how can we draw a bisector which passes a point P located at the outside of the triangle ABC?

It was posted before, but I would do it simpler now. Let $ A', B', C'$ be midpoints of the triangle sides $ |BC|, |CA|, |AB|.$ Medians $ AA', BB', CC'$ concur at the centroid $ G.$ WLOG, assume the point $ P$ to be inside the angle $ \angle BGC'$ or $ \angle CGB',$ between the B-, C-medians $ BB', CC'$. Any line $ l$ cutting off half the triangle area on the rays $ (AB, (AC$ is tangent to the same hyperbola branch with asymptotes $ AB, AC.$ If $ l$ cuts $ AB, AC$ at $ X, Y,$ so that $ |\triangle AXY| \equal{} \frac{_1}{^2}|\triangle ABC|,$ the midpoint of the segment $ |XY|$ is the tangency point. The B-, C-medians $ BB', CC'$ are tangent to this hyperbola at the midpoints of $ |BB'|, |CC'|.$

To find the hyperbola vertex and focus, construct segments $ |AD| \equal{} |AE| \equal{} \sqrt{|AB| \cdot |AB'|}$ on the rays $ (AB, (AC.$ Then $ |\triangle ADE| \equal{}|\triangle ABB'| \equal{} \frac{_1}{^2}|\triangle ABC|.$  Line $ DE$ cuts the internal bisector of the angle $ \angle CAB$ at the hyperbola vertex $ V.$ Circle with center $ A$ and radius $ |AD|$ cuts this angle bisector at the focus $ F.$ Circle $ \mathcal A$ with center $ A$ and radius $ |AV|$ is the hyperbola pedal circle, locus of the feet of perpendiculars from the focus $ F$ to the hyperbola tangents. Circle $ \mathcal Q$ with diameter $ PF$ cuts the pedal circle $ \mathcal A$ at $ M, N.$ The angles $ \angle PMF \equal{} \angle PNF$ are right, $ PM, PN$ are hyperbola tangents through $ P.$ Because of the assumed position of $ P$ between the B-, C-medians, one tangent cuts the segments $ |AB|, |AC|$ and the triangle area in half. The other tangent belongs to the other hyperbola branch and misses the triangle.


Solve the sysstem
$ \left\{\begin{array}{l}
[x]+\{\ y\}=1,2
\\\{ \ x\}+[y]=3,3
\end{array}\right.$ :?:

________________________________
$ \begin{array}{c}
\mathfrak{Azerbaijan } \\
\mathfrak{Land } \\
\mathfrak{of} \\
\mathfrak{Fire} \end{array}$

this should go to  basics :P

                                            {y}<1 and [x] its an integer rezults that [x]=1,{y}=0,2
                                            {x}<1 and [y] its an itneger rezults that [y]=3,{x}=0,3
 x=[x]+{x}=1,3
 y=[y]+{y}=3,2


solve $ x^2\plus{}y^2\equal{}2z^2$ where $ x,y,z$ are pairwise relatively prime.


show that  21x^3-3x^2+2x+8 is irreducible over Q?

Let y=21x, then if polinom is not irreducible  $p(y)=y^{3}-3y^{2}+42y+8*21^{2}$ have integer root. Reducing by modul 3 and 7 give $y_{0}=3 \ or \ 21 (mod \ 63)$. But if $y\ge 3$, then $P(y)>0$. We have $p'(y)=3(y^{2}-2y+14)>0$, therefore $y_{0}>-20$. It proved, that p(y) have not integer roots, therefore p(y) irredicible.


linear algebra
matrix
linear algebra unsolved

for each of the following, define, if possible, an endomorphism $ \phi$ of $ \mathbb{R}^3$ that suits it:
a) $ ker\phi\equal{}im\phi$
b) $ im\phi \subseteq ker\phi$
c) $ ker\phi \subseteq im\phi$ and $ ker\phi \ne 0$
d) ker and im are complementary
e) $ ker\phi \ne im\phi$ and their sum is not direct

a) Not possible. $ \dim(\text{ker}\phi) \plus{} \dim(\text{Im}\phi) \equal{} 3.$ Hence, the two dimensions can't be equal.

b) $ \begin{bmatrix}0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0\end{bmatrix}.$

c) $ \begin{bmatrix}0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0\end{bmatrix}.$

d) $ \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}.$

e) Are you sure that's the question? As stated, any answer to (b) or (c) would work - nontrivial containment is an extreme way of not having a sum be direct.

I forgot that the endomorphism $ \phi: \mathbb{R}^3\minus{}>\mathbb{R}^3$ must be different from zero;
for e), if you give an example of nontrivial containment, it's ok; and the bonus question was to generalize to $ \mathbb{R}^4$

Zero is an endomorphism, and it's an example of (b).

In $ \mathbb{R}^4$ and in every even dimension, (a) is possible. Here's an example:

$ \begin{bmatrix}0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}$

Note that $ \text{Im}\phi\subseteq\text{ker}\phi$ implies that $ \phi^2\equal{}0.$

For the cheapest examples of a), b), c), and e), I'll look at nilpotent matrices; for the cheapest examples of d), I'll look at idempotent matrices.


inequalities
ratio
geometry
geometry proposed

Let m_a, ,m_b and m_c be the lengths of three medians in triangle ABC, and p, q and r are nonnegative integers that at least one of them is larger than zero. Prove that:
(1) pm_a + qm_b + rm_c, pm_b + qm_c + rm_a and pm_c + qm_a + rm_b must be three side lengths of a triangle(namely DEF).
(2) The ratio of the area of triangle DEF to the area of triangle ABC is no less than (3/4)*(p + q + r)^2.

The first question becomes obvious when $m_a,m_b,m_c$ are sides of a triangle.
  We notice that triangle 's area is equal to 3/4 the area of ABC.
  By denoting $x=m_b+m_c - m_a, y=m_a+m_c-m_b, z= m_a+m_b-m_c $, your inequality is equivalant to
     $(px+qy+rz)(py+qz+rx)(pz+qx+ry)\geq xyz(p+q+r)^3$
  By expanding the LHS and applying Cauchy 's inequality, we finally obtain what we want. :lol:


If I made no mistakes, then the answer is 300 cents.

yea its three hundred.  I did it in my head at first, now i wrote it out and i got 300.

[code]Anjali has 230 quarters, 300 dimes and 165 nickels. Niki has 210 quarters, 316 dimes and 173 nickels. How many more cents does Anjali have than Niki?
[/code]

[hide="solution"]
 Anjali has (230 X 25 + 300 X 10 + 165 X 5)cents = 9575 cents
 Niki has (210 X 25 + 316 X 10 + 173 X 5)cents = 9275 cents

 so Anjali has (9575 - 9275)cents or 300 cents more than Niki[/hide]