linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ R$ be a ring. For m,n $ \in \mathbb{N}$ , denote by $ M_{m,n}(R)$ the set of (mxn)-matrices over $ R$, and let $ IP_n(R)={\{p \in M_n(R) : p^2=p\}}$, be the collection of all idempotent (nxn)-matrices over $ R$.Further, set $ IP(R)=\displaystyle\bigcup_{n=1}^{\infty}{IP_n(R)}$, so that $ IP(R)$ is the collection of all the idempotent matrics over R of arbitrary size, and define a relation $ \sim$ on $ IP(R)$, as follows. Suppose that $ p \in IP_m(R)$, and $ q \in IP_n(R)$, where $ m,n \in \mathbb{N}$. Then, we stipulate that $ p \sim q$, if there are $ a \in M_{m,n}(R)$ and $ b \in M_{n,m}(R)$, such that $ p=ab$ and $ q=ba$.

a) Suppose that $ p \sim q$, where $ p \in IP_m(R)$, and $ q \in IP_n(R)$. Show that there are $ c \in M_{m,n}(R)$ and $ d \in M_{n,m}(R)$, such that $ p=cd$, $ q=dc$, $ cdc=c$, and $ dcd=d$.
Hint : Take $ c=aba$ and $ d=bab$, where $ a$ and $ b$ are as in the definition of $ \sim$ above.

b)Show that $ \sim$ is an equivalence relation on $ IP(R)$.

I think that this is a very difficult problem and unfortunately i don't have any solution. It is bothering me for a long time..
Thanks.

The problems are not hard. They are solved by straightforward calculations applying the definitions.

a) Use the hint, i.e. set $ c=aba=pa\in M_{m,n}(R),d=bab=qb\in M_{n,m}(R)$, then $ cd=(ab)^3=p^3=p$ and $ dc=(ba)^3=q^3=q$. Moreover, $ cdc=pc=p^2a=pa=c$ and $ dcd=qd=q^2b=qb=d$.

b) 1. For arbitrary $ p\in IP_n(R)$ set $ a=p,b=I_n\in M_{n,n}(R)$, then $ p=ab=ba$ to prove $ p\sim p$

2. For arbitrary $ p\in IP_n(R),q\in IP_m(R)$ determine $ a\in M_{n,m}(R),b\in M_{m,n}(R)$ with $ p=ab,q=ba$. Set $ a'=b,b'=a$ to verify $ a'b'=q,b'a'=p\Longrightarrow q\sim p$

3. For arbitrary $ p\in IP_n(R),q\in IP_m(R),r\in IP_k(R)$ with $ p\sim q\sim r$ determine $ a\in M_{n,m}(R),b\in M_{m,n}(R),c\in M_{m,k}(R),d\in M_{k,m}(R)$ with $ p=ab,q=ba=cd,r=dc$. Now set ${ \alpha =ac\in M_{n,k}(R}),\beta=db\in M_{k,n}(R)$ and calculate $ \alpha\beta=acdb=aqb=abab=p^2=p$ and similarly, $ \beta\alpha=dbac=dqc=dcdc=r^2=r$, i.e. $ p\sim r$

The question then follows (and this is the hard part) :
c) For $ p,q \in IP(R)$, define
$ {p \oplus q}\equal{}
\left[\begin{array}{cc}
p & 0 \\
0 & q \\
\end{array}\right]$
Show that $ p \oplus q \in IP(R)$, and that $ p \oplus 0_n \sim p$, for each $ n \in \mathbb{N}$, where $ 0_n$ denotes the (nxn)-zero matrix over $ R$.
d) For $ p \in IP(R)$, let $ [p]\equal{}\{q \in IP(R) : p \sim q\}$, denote the equivalence class of $ p$ with respect to $ \sim$, and set $ V(R)\equal{}\{[p] : p \in IP(R)\}$.
Show that if $ p_1 \sim p_2$ and $ q_1 \sim q_2$, where $ p_1,p_2,q_1,q_2 \in IP(R)$, then $ p_1 \oplus q_1 \sim p_2 \oplus q_2$, and conclude that it makes sense to define an operation $ \plus{}$ on $ V(R)$ by $ [p]\plus{}[q] \equal{} [p \oplus q] ,   (p,q \in IP(R))$.
e) Show that $ V(R)$ is a commutative monoid under the operation $ \plus{}$, defined in (d).

Thank you very much for any interst on this problem..

It is not harder at all, again just using the definitions. I am confident about your capabilities doing that.

I strongly feel that I do somebody else's homework (which course do u take ?). However, attached are the straightforward calculations.

c) $ [p\oplus q]^2\equal{}[p^2\oplus q^2]\equal{}[p\oplus q]$ and for $ p\in M_{n,n}(R)$ and arbitrary $ m\in\mathbb{N}$ set $ a\equal{}\left[p\atop 0_{m\times n}\right]\in M_{n\plus{}m,n}(R),b\equal{}\left[I_n\;0_{n\times m}\right]\in M_{n,n\plus{}m}(R)$ to observe that $ ab\equal{}\left[p\oplus 0_{m\times m}\right]\,,\,ba\equal{}p\Longrightarrow \left[p\oplus 0_{m\times m}\right]\sim p$

d) For $ i\equal{}1,2$ let $ p_i\in M_{n_i}(R),q_i\in M_{m_i}(R)$ with $ p_1\sim p_2,q_1\sim q_2$, i.e. we can choose matrices $ a_i\in M_{n_i,m_i}(R),b_i\in M_{m_i,n_i}(R)$ with $ a_ib_i\equal{}p_i,b_ia_i\equal{}q_i$ for $ i\equal{}1,2$.

Now set $ a\equal{}\left[\begin{array}{cc}a_1 & 0_{n_1\times m_2}\\0_{n_2\times m_1} & a_2\end{array}\right]\in M_{n_1\plus{}n_2,m_1\plus{}m_2}(R)$ and similarly,

$ b\equal{}\left[\begin{array}{cc}b_1 & 0_{m_1\times n_2}\\0_{m_2\times n_1} & b_2\end{array}\right]\in M_{m_1\plus{}m_2,n_1\plus{}n_2}(R)$

and compute $ ab\equal{}[p_1\oplus p_2],ba\equal{}[q_1\oplus q_2]\Longrightarrow [p_1\oplus p_2]\sim [q_1\oplus q_2]$

e) The [b]identity element [/b]$ [0]$ and [b]closure[/b] of the monoid follow both from c). [b]Associativity[/b] follows from the definition of the operation, i.e. $ ([p]\plus{}[q])\plus{}[r]\equal{}[p\oplus q\oplus r]\equal{}[p]\plus{}([q]\plus{}[r])$. As for the [b]commutativity[/b] let $ p\in M_n(R),q\in M_m(R)$ and define matrices $ a\equal{}\left[\begin{array}{cc}0_{n\times m} & p\\q & 0_{m\times n}\end{array}\right]\,,\,b\equal{}\left[\begin{array}{cc}0_{m\times n} & I_m\\I_n & 0_{n\times m}\end{array}\right]$ in $ M_{n\plus{}m,n\plus{}m}(R)$. Direct calculation yields $ ab\equal{}[p\oplus q],ba\equal{}[q\oplus p]\Longrightarrow [p\oplus q]\sim [q\oplus p]\Longrightarrow [p]\plus{}[q]\equal{}[q]\plus{}[p]$.

Thanks for your help.
Please allow me to ask something final:
In the last post, in the proof of part d, i don't see how the fact that $ [p_1\oplus p_2]\sim [q_1\oplus q_2]$ implies what the problem is asking, i.e that $ {p_1\oplus q_1}\sim {p_2\oplus q_2}$, and how we conclude that it makes sense to define this + operation.
Thank you very much again for helping me understand this problem..


function
limit
real analysis
real analysis unsolved

Let $ f(x) \equal{} \frac {1}{q^3}$ if $ x \equal{} \frac {p}{q}$ where p and q are integers with no common factor with q>0 and f(x)=0 if x is irrational. At which points is f differentiable?

It's differentiable almost everywhere, but there isn't really a simple description of the set. The criterion for $ f$ to be differentiable at some irrational $ x$ is that $ x$ has at most finitely many rational approximations $ \frac{p}{q}$ with $ \left|x\minus{}\frac pq\right|<K\frac1{q^3}$ for each $ K$. This is very hard to test- it's not known whether this is true for $ \pi$, for example.

so there's no easy way to do it? ok thanks

what about for algebraic numbers only?

The following problem is well-known. Consider the function $ f$ defined on $ \mathbb R$ a follows
\[ f\left( x \right) \equal{} \left\{ \begin{gathered}
  0 \quad {\rm if} \quad x \equal{} 0, \hfill \\
  0\quad {\rm if}\quad x \notin \mathbb{Q}, \hfill \\
  a_q \quad {\rm if}\quad x \equal{} \frac{p}
{q},p \in \mathbb{Z},q \in \mathbb{N},\left( {p,q} \right) \equal{} 1. \hfill \\
\end{gathered}  \right.
\]
where the sequence $ \{a_q\}$ is such that $ \mathop {\lim }\limits_{n \to \infty } n^k a_n  \equal{} 0$ for some $ k \geq 2$. Then $ f$ is differentiable at each irrational which is algebraic of degree at most $ k$.

The proof is based on a Liouville theorem in sense that there exists $ M>0$ such that
\[ \left| {x \minus{} \frac{p}
{q}} \right| > \frac{1}
{{Mq^k }}
\]
for all rationals $ p/q$. Have fun!

so if you had $ e^{\minus{}q}$, then it would be differnetiable for all algebraic irrationals?


Let d(n) be the digital sum of n(an integer n>0). Solve n+d(n)+d(d(n))=1997.

Oh... I feel like a fool now. :blush: Working on it...

well, d(n)=n (mod 9) so n + d(n) + d(d(n))=0 (mod 3) but 1997=2 (mod 3) so there is no such n.

This question has been posted before on the boards very recently.  It's also from Engel's book and thus should probably go into Intermediate (since it's one of the easier Engel problems).


Let be given a set of $ 1996$ equal circles in the plane such that no two of them having common interior points.prove that
there exist a circle touching at most three other circles.

Since there is a finite number of circles, there is a finite number of center of circles, so we may find a line $ d$ which is not parallel to any line which goes through two centers of circles, and such that all the circles are on the same side of $ d$.
Now, let's consider, say $ C$, the nearest circle from $ d$ (the choice of $ d$ ensures that there is only one nearest circle).
Draw the tangent, say $ D$, to $ C$ parallel with $ d$ such that all the circles are on the same side of $ D$. Note that none of the other circles is tangent to $ D$.

Assume that four circles, say $ C_i$ for $ i\equal{}1,2,3,4$ are tangent to $ C$, and let $ c_i$ be the center of $ C_i$, and $ c$ be the center of $ C$.
Assume that the rays $ [c,c_i)$ are in the same order than the subscripts (in the clockwise sense). Then, each angle $ c_icc_{i\plus{}1}$ is at least $ \frac {\pi}{3}$. But the sum of these angles is less than $ \pi$ (since they do not overlap and the circles are in the same side of $ D$, and not tangent to $ D$). A contradiction.

Pierre.


algebra
polynomial
search
superior algebra
superior algebra unsolved

$\mathbb{K}$ is a non algebraically closed field.
Show that there is an integer $n>1$ and a homogeneous polynomial $P$ of degree $n$ over $\mathbb{K}[X_{1},\ldots,X_{n}]$, s.t. $P(x_{1},\ldots,x_{n})=0$ has no solution in $\mathbb{K}^{n}$.
P.S.: I couldn't find it on this forum.

I posted it some time ago and it received solutions, so please search for it.

You are right harazi!
It's http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=closed&t=89068
I couldn't find it in my initial search ("Homogeneous polynomial"), because you spelt homogeneous wrongly in your post :wink:

I will never be able to spell corectly such words...  :wink:


let a,b,c>0 such that a+b+c=$ \sum\frac{1}{a}$  prove that
$ a^2b^2c^2\geq \frac{min({a;b;c})}{max{(a;b;c)}}$

Assume that :$ a\ge b \ge c$
=>$ RHS \equal{}\frac{c}{a}$;$ a^2b^2c^2\equal{} (\frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c})^3.\frac{1}{abc}$
Setting :$ a\equal{}x^3;b\equal{}y^3;c\equal{}z^3$
<=>$ f(x)\equal{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3 \minus{} z^2y(x^3\plus{}y^3\plus{}z^3) \ge 0$
$ f'(x)\equal{}3x^2y^3\plus{}3x^2z^3\minus{}3z^2yx^2 \ge 0$
=>$ f(x) \ge f(y)\equal{} y^6\plus{}2y^3z^3\minus{}z^2y(2y^3\plus{}z^3)\equal{}y(y\minus{}z)(z^4\plus{}z^3y\minus{}z^2y^2\plus{}y^3z\plus{}y^4) \ge 0$
done .
[b]Note:[/b] I'm not clever ;)


I don't know if all of you know about the Canadian ARML, known as McTic.  It's on May 20-22, in London, and it's a team math competition.  It's like ARML - there's team, individual, power, relay, and mental math rounds.  And plus, it's a great time to socialise with math friends.  I'll be there, and so will (I think) Shaun White, Jimmy Shen, Taotao Liu, Francis Chung, Chen Sun, and hopefully even Jenn Park and Elyot Grant.

It's not too late to sign up!  You just need to find a team of nine people to come - e-mail me at gmsiu(at)hotmail.com if you're interested, and I'll put you in touch with the right people.  The cost is quite reasonable.  I really want to see you all there!  Torontonians, talk amongst each other - but do it quickly!

Yeah, we're trying to get something worked out in the "questions about canadian ARML" thread.

I'm just going to interject and speak honestly.

Who wants to bother to travel all the way to london, wasting all that time for a non-interesting informal contest?

I've already asked in the other antiquated thread, but is any non-london/windsor team coming for sure?


Prove that every sequence of $mn+1$ real numbers contains either a nondecreasing subsequence of length $m+1$ or a nonincreasing subsequence of length $n+1$.

[hide="Hint"]
Consider the sign of $\Delta= a_{k+1}-a_{k}$.
It's either nondecreasing (nonnegative), or decreasing (negative); and it's either nonincreasing (nonpositive), or it's increasing (positive).

How many $\Delta$s are there?  [/hide]

[hide="Further Hint"]
Consider $\Delta_{i}= a_{k+i}-a_{k}$

How many $\Delta_{i}$s are there? [/hide]

What do you mean? What about 0,1,0,1,0,1,0,1...0,1

Where is a subsequence of non-decreasing or non-increasing integers?

I saw this in the AMY pigeonhole packet and was confused. I think the problem should specify distinct reals, although you can still find a similar counterexample...

A subsequence does not need to be contiguous.  In Zuton Force's example, the 0's form a large nondecreasing subsequence.

This is a case of Dilworth's lemma.
For some references: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1788371293&t=75555


Consider two solutions, one formed by adding 20 g of glucose to 1 L of water and another formed by adding 10 g of sucrose to 1 L of water. Are the vapor pressures over the two solutions the same? Why or why not?

Are you aware of the colligative properties? [url=http://en.wikipedia.org/wiki/Colligative_properties]Colligative properties[/url]

In the case of a solution of a non-volatile solute, the vapour pressure is due to the force exerted on surface of the solution by the molecules of solvente that managed to escape to the vapour phase. If we have a solution, that's because the solvent was capable of solvating the molecules/ions of solute. So, because of the presence of the solute, there are less solvent molecules available to escape to the vapour - some of them are bounded to the solute. Therefore, the vapour pressure is necessarily smaller. Of course, the more solute we have, less will be the vapour pressure.


[quote="arqady"]Let $ \ x \in\mathbb R.$  Prove that
\[ \frac {x^{2n} \plus{} ... \plus{} x^4 \plus{} x^2 \plus{} 1}{n \plus{} 1} \geq \frac {x^{2n \minus{} 1} \plus{} ... \plus{} x^3 \plus{} x}{n}
\]
[/quote]

We have
Case 1. $ x \leq 0,x \equal{} 1$  then done.

Case 2. $ x>0,x \neq 1$ then

$ \frac {x^{2n} \plus{} ... \plus{} x^4 \plus{} x^2 \plus{} 1}{n \plus{} 1} \geq \frac {x^{2n \minus{} 1} \plus{} ... \plus{} x^3 \plus{} x}{n}$ $ \Longleftrightarrow$ $ \frac {nx^{2n\plus{}2} \minus{} (n\plus{}1)x^{2n\plus{}1}\plus{} (n\plus{}1)x\minus{}n}{x^2\minus{}1} \geq 0$ $ \Longleftrightarrow$
$ \frac {nx^{2n\plus{}1}\plus{}n \minus{} (x^{2n} \plus{} ... \plus{} x^2 \plus{} 1)}{x\plus{}1} \geq 0$ $ \Longleftrightarrow$$ \sum_{k\equal{}1}^{n}(x^{k}\minus{}1)(x^{2n\plus{}1\minus{}k}\minus{}1) \geq 0$
the inequality holds!

[quote="Elio (n)"]Can we use the mathematic induction or.......... this could not solve by mathematic induction or.......I don't know where is mathematic induciton used.....    :maybe:
Can you help me please[/quote]

Mathematical induction can only be used on ordered sets, which the reals are not. There is not induction used in this problem

The reals are very clearly an ordered set -- what they aren't is [i]well-ordered[/i].  You can't induct on $ x$, but it might in principle be possible to induct on $ n$.  (I don't have any thoughts on whether that will be easy or hard -- give it a try!)

I think this problem is solvable by weighted AM-GM, but calculating the weights out is quite hellish: almost makes me want to use some rather powerful linear algebra on it. Anyone can confirm this?

Hrmm...

Let $ E \equal{} x^{2n} \plus{} \cdots \plus{} x^2 \plus{} 1$, and let $ O \equal{} x^{2n \minus{} 1} \plus{} \cdots \plus{} x^3 \plus{} x$.  Let's be lazy and combine all the terms containing $ x$ on one side.  Obviously $ E \equal{} Ox \plus{} 1$.  We thus get $ \frac {E}{O} \ge 1 \plus{} \frac {1}{n}$. But $ \frac {E}{O} \equal{} x \plus{} \frac {1}{O}$.  So we need show that

$ x \plus{} \frac {1}{x^{2n \minus{} 1} \plus{} \cdots \plus{} x^3 \plus{} x} \ge 1 \plus{} \frac {1}{n}$.

Now $ O \equal{} x\cdot\frac {x^{2n} \minus{} 1}{x^2 \minus{} 1}$, so our inequality becomes

$ x \plus{} \frac {x^2 \minus{} 1}{x\cdot(x^{2n} \minus{} 1)} \ge 1 \plus{} \frac {1}{n}$.

The LHS can be rewritten as $ \frac {x^{2n \plus{} 2} \minus{} 1}{x^{2n \plus{} 1} \minus{} x}$.  Thus we get the slightly nicer inequality

$ (x\minus{}1)(x^{2n\plus{}1}\minus{}1) \ge n(x^{2n\plus{}1}\minus{}x)$, which I'm too lazy to prove.


inequalities
geometry
inequalities unsolved

[quote="gemath"]Please help me :
$\forall x,y,x>0$ then
$\frac{8xyz}{(x+y)(y+z)(z+x)}\le \frac{3(x^{2}+y^{2}+z^{2}+3(xy+yz+zx))}{4(x+y+z)^{2}}$
Thank you!  :)[/quote]

$9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx) \iff \sum x^{2}y \geq 6xyz$
$\rightarrow \frac{8xyz}{(x+y)(y+z)(z+x)}\leq \frac{9xyz}{(x+y+z)(xy+yz+zx)}$

$\frac{9xyz}{(x+y+z)(xy+yz+zx)}\leq \frac{3(x^{2}+y^{2}+z^{2}+3(xy+yz+zx))}{4(x+y+z)^{2}}$
$\iff 12\sum x^{2}yz \leq (\sum xy)(\sum x^{2}+3\sum xy)$
$\iff 5\sum x^{2}yz \leq \sum x^{3}y+3\sum x^{2}y^{2}$
$\leftarrow \sum x^{3}y \geq 2\sum x^{2}y^{2}\geq 2\sum x^{2}yz$
which is true by AM-GM.

Ok thank you for very good and cold solution !! :D With this inequality I had a nice problem in geometry! Thank again Sung-yoon Kim!


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

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

$\text{trace}A^{n}= \text{trace}B^{n}$ for all $n \in \mathbb{N}$ then $A$ and $B$ are similar.
I have a solution, but i think there must be a nicer one.

That's wrong: any two matrices having exactly the same collection of eigenvalues will do the job!

Are two matricies having the same eigenvalues not similar?

By similar i mean $A=PBP^{-1}$ for some $P$

Of course there are matrices having the same characteristic polynomial and which are not similar: take the identity and any triangular unipotent matrix different from the identity.


Problem 3:
Let $ m$ and $ n$ be positive integers such that the equation $ x^{2}\minus{}mx\plus{}n \equal{} 0$ has real roots $ \alpha$ and $ \beta$. Prove that $ \alpha$ and $ \beta$ are integers if and only if $ \left[m\alpha\right] \plus{} \left[m\beta\right]$ is the square of an integer. (Here $ [x]$ denotes the largest integer not exceeding $ x$.)


Prove the inequality $a^3b+b^3a+2a^3+2b^3+2\ge 2ab^2+2ab^2+a^2+b^2+2ab$, where $a,b\in \mathbb{R}$.

I guess the right-hand side is $2ab^2+2a^2b+a^2+b^2+2ab$. Moreover I think the inequality is only true when $a,b$ are non-negative, try $a=2, b=-2$.
\[a^3b+ab^3+2a^3+2b^3+2 \ge 2ab^2+2a^2b+a^2+b^2+2ab \Leftrightarrow\]
\[(a+1)(b+1)[(a-1)^2+(b-1)^2] + (a-b)^2(a+b) \ge 0\]
which is obviously true.


A rocket is launched straight up from ground level.  After t seconds, its height, h, in feet is given by the function h(t) = -16t^2 + 376t.

(a) How many seconds after the rocket is launched will it strike the ground?

(b) During what interval of time, to the nearest tenth of a second, will the height of the rocket exceed 1800 feet?

I think the means that :[hide]

domain $D_{istance of rocket > 1800}$ : $\left(\frac{47-\sqrt{47^{2}-450}}{4}, \frac{47+\sqrt{47^{2}-450}}{4}\right)$[/hide]


Hello, everyone:

How do you rate Mathcounts Handbooks' difficulty level?
Are they at School, Chapter, State, or National level?

Handbooks have WARMUPs, WORKouts, ...etc.
What is the difference of them?
Just different difficulty levels?

What is the difference between volume 1 and volume 2? Is it just the handbook in two parts?

Warm-ups are similar to Sprints: you shouldn't use a calculator.
Workouts are similar to Targets: you're allowed to use a calculator.

Not too sure about the difficulty level though

I'm pretty sure that volume I is warmups 1-6 and workouts 1-3, and volume II is warmups 7-18 and workouts 4-9.

the problems in MC handbooks are a REALLY easy... However, some problems(which sometimes have asterisks) are either state or national level.

Actually, many problems in the Handbooks are more difficult than the most difficult problems at Nationals.  The Handbooks are designed to be used for preparing for MATHCOUNTS, and thus are not "REALLY easy".  For example, look at the last two Target round problems of this year's National test, and the last two problems of Workout 9.  At least in my opinion, the last two problems in Workout 9 are much more time-consuming and difficult.

To answer the OP's questions: The easiest problems are at School Round level, but the hardest problems are very difficult.  Warmups are designed to prepare for the Sprint Round, so no calculators are to be used and speed is emphasized.  Workouts are designed to prepare for Target and Team Round problems, so calculators are allowed and problems generally take more time to solve.

While I agree with most of what you said, this year's National Target was ridiculously easy. The last two questions could've easily been Chapter Target questions.

Heh, I suppose most of us AoPSers agree that this year's MATHCOUNTS tests were easier than normal.

However, the point is, [b]the hardest handbook problems are generally harder than the hardest competition problems[/b].


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Let $\mathbb S^n$ be the $n$-dimensional unit sphere, and let $f: \mathbb S^n\to\mathbb R^m,\ g: \mathbb R^m\to\mathbb S^n$ be two continuous functions.

Prove that $g\circ f: \mathbb S^n\to\mathbb S^n$ has a fixed point.

Well, let me clarify it more.

The induced map to the top homology group of $S^n$ should be zero, that is, $f_*: H_n(S^n; \mathbb{Z})\simeq \mathbb{Z} \rightarrow H_n(\mathbb{R}^m ; \mathbb{Z})=0$ which is the zero homomorphism. Hence, $\deg (f)=0$, therefore, by Hopf theorem, it is homotopic to a constant map and $g \circ f$ is homotopic to a constant map, too. Then, it has a fixed point, otherwise, $\deg (g \circ f)=(-1)^{n+1} \not =0.$


analytic geometry
graphing lines
slope
geometry
conics
ellipse
hyperbola

Today I was a math problem which I don’t understand how to solve let alone how it is possible. Here is the question:
Why, how and when do parallel lines meet?
Any help would be much appreciated.
Thank You

Parallel lines only meet when they are equal... which is pretty obvious.

The definition of parallel lines is lines with same slope that never meet.

Parallel lines never meet in Euclidean geometry...the definition is lines which never meet...so i don't know if they can be the same line.  I don't know if parallel lines are defined in other planes, such as the surface of a sphere.

Parrallel lines meet at infinity, that is the same as saying they never meet.

Parallel lines never meet, however, if they have the same slope and [b]y-intercept[/b], they will coincide (overlay) each other.

Masoud Zargar

If you have a 3D thing and 2 lines are parallel then they can appear to be intersecting from some angles.

well appearing to intersect means very little now doesn't it

generally, parallel lines do not intersect, unless you consider 2 lines that are the same, like
$ y=2x +1 $ and
$ 2y = 4x +2 $
they have the same slope, so the are parallel, but they have the same y intercept, so they are the same line, it depends if you consider 2 definitions of the same line being different lines

Riddler - that doesn't work for parallel lines, only for skew lines.

Ok I got a lot of mixed answers and i am still not sure so here is some more detail

We where talking about how you would put 1/3 and (6,4) into standard form and into well i forget the name but it is y=mx+b form and so we did it and then he asked us the graph it and then asked how to get a parrell line of it and so we said y=1/3+5. Then he said for our purposes in this class these lines will never meet but he said they will meet and to trust him on that and then he asked us if we knew how, why and when and of course no one knew.

Thanks for all the responses and I hope that clarifies it a little more.

I don't know how or why they meet, but I know where: in the mind of your math teacher. That's it. Assuming you're using Euclidean geometry (which seems like a quite valid assumption), parallel lines by definition never meet. You could argue that if you let $ x=\infty $ then $ y=\infty $ in both cases, but that is just nonsense, as $ \infty $ is not a number. They both take on infinitely large values of $ y $ when $ x $ becomes infinitely big, but that doesn't mean they meet. So unless these equations are being plotted on an ellipse, a hyperbola, or some other non-planar surface, the lines don't meet. That seems to be the consensus with everyone here.


So assuming i take AP physics B, AP chemistry and AP computer A senior year, which of the olympic will be easier for me?  Physics olympic or chemistry olympic or computer science olympic?  BTW, do you guys know how many kids taking these three exams respectively in the nation every year?

Merely being in an AP-level class is usually not going to help you much for the science olympiads. All three of those olympiads you mentioned require significant, regular practice to do well on.

you could make semifinals for USNCO or USAPhO with that knowledge base, but there's more to it than just taking the AP...

Depends on what floats ur boat really....

For physics, the very fact that F=ma is prolly enough to get through half the mechanics exam, yet people still manage to bomb it either way..... You just have to get comfortable with the problems and to a lot of them

For chemistry, you gotta get good at ur theory and understand how it pertains to the mathematical section.

And ive never even heard about an olympiad for computers, although im thoroughly amused by the idea.....

Whatll they come up with next... an english olympiad??  :rotfl:  :rotfl:

In terms of people, I think participation goes in the order $ \text{USNCO} < \text{USAPHO}\approx\text{USACO}$.  I could be wrong about that, however.  USACO is probably the hardest of these Olympiads to dive into, while I'd say chemistry is the easiest.  @atessler, the computer olympiad is a lot different from your standard, sit down and take a written test, model of an Olympiad.  Rather, they give you some computational mathematical problems and turn you loose with an actual computer.  I agree that a written portion would be kind of odd, to say the least :).

Anyway, I recommend that if you are interested in doing well in an Olympiad, you figure out which one you are most interested in and then try to study for it.  You could also try studying for multiple, as long as you realize that breadth necessarily substitutes for depth at this level.  Or you could not study for any at all, but that won't give you much of a chance of making it past the second stage of selection...

It took just AP Chemistry to get me to Chem Olympiad study camp...

Wrong, you studied out of atkins for 3 weeks (or so you say), which is a beyond-AP book. Not by a lot, but still not AP.

And stop running away from me :rotfl:


Two congruent cones with radius 12cm and height 12cm are enclosed within a cylinder. The base of each cone is a base of the cylinder, and the height of the cylinder is 24cm. What is the number of cubic centimeters in the volume of the cylinder not occupied by the cones? Express your answer in terms of $ \pi$.

The volume of the two cones together is $ 2(12^2 \times 4 \times \pi)\equal{}1152\pi$ and the volume of the cylinder is $ 12^2 \times 24 \times \pi\equal{}3456\pi$ so the desired volume is $ 3456\pi \minus{} 1152\pi \equal{}\boxed{2304\pi}$.


Find all positive integers $ n$ such that $ 3n(n\plus{}1)$ has just 12 positive divisors.

[hide="Solution"]We have $ 2|n(n\plus{}1)$, so $ 6|3n(n\plus{}1)$. We must have $ 3n(n\plus{}1)\equal{}2\cdot3p^2,2^2\cdot3p,2\cdot3^2p,2^2\cdot3^3,2^3\cdot3^2,2\cdot3^5$, or $ 2^5\cdot3$. Dividing through by $ 3$, we have
$ n(n\plus{}1)\equal{}2p^2,4p,6p,36,24,162$, or $ 32$. It is easy to verify that the last 4 do not have integer solutions for $ n$, so we have $ n(n\plus{}1)\equal{}2p^2,4p,6p$. Since $ n, n\plus{}1$ are relatively prime, the first one is clearly impossible, since one factor must have only $ 2$, and the other must have only $ p^2$. Similar reasoning tells us that only $ n\equal{}4$ works for the $ 4p$ case. For $ n(n\plus{}1)\equal{}6p$, we must have the factors split up into $ \{2p,3\},\{3p,2\}\{p,6\}$. We see that the first two are impossible, and the last yields solutions $ n\equal{}5,6$.
So the only $ n$ which work are $ n\equal{}4,5,6$.[/hide]


This is easy but I want a lot of silutions.This is the reason I posted is Olympiad section

 If x, y, z > 0, k > 2 and a = x+ky +kz, b = kx+y+kz, c = kx+ky+z,
show that $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq\frac{3}{2k+1}$

Silouan,this is too easy. This inequality is enough: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\geq\frac{(x+y+z)^2}{ax+by+cz}$ ;)

Why? You can't do the calculations? It's really that hard to write it down. :?

[quote="silouan"]If $x, y, z>0$ and $k\geq1$, show that $\frac{x}{x+ky+kz}+\frac{y}{y+kz+kx}+\frac{z}{z+kx+ky}\geq\frac{3}{1+2k}$.[/quote]
Let we generalize it as follows :

[b]Theorem.[/b] If $x, y, z$ are positive real numbers, then the inequality
\[\frac{x}{x+uy+vz}+\frac{y}{y+uz+vx}+\frac{z}{z+ux+vy}\geq\frac{3}{1+u+v}\]
holds if and only if $u\geq0\wedge v\geq0\wedge u+v\geq2$.
[hide="Proof."]
$\frac{x}{x+uy+vz}+\frac{y}{y+uz+vx}+\frac{z}{z+ux+vy}-\frac{3}{1+u+v}$

$\equiv\frac{F(x,y,z)}{(1+u+v)(x+uy+vz)(y+uz+vx)(z+ux+vy)}$.

Without losing generality, we may assuming $x=\min\{x,y,z\}$.

$F(x,y,z)=F(x,x+s,x+t)$

$=(u+v+1)(u^{2}+v^{2}-u-v)(s^{2}-st+t^{2})x$

$+(us+vt)[v(u+v-2)s^{2}-(3uv-2u-2v+1)st+u(u+v-2)t^{2}]\geq0$,

which is clearly true for

$u^{2}+v^{2}-u-v=\frac{(u-v)^{2}+(u+v)(u+v-2)}{2}\geq0$,

and

$v(u+v-2)s^{2}-(3uv-2u-2v+1)st+u(u+v-2)t^{2}$

$=(u+v-2)(s\sqrt{v}-t\sqrt{u})^{2}+(\sqrt{uv}+1)[2(\sqrt{u}-\sqrt{v})^{2}+\sqrt{uv}-1]st$

$=(u+v-2)(vs^{2}+ut^{2})+[2(u+v-2)+3(1-uv)]st\geq0$.

The necessity is easy.[/hide]
[b]Corollary.[/b] If $x, y, z>0$ and $k\geq0$, then $\frac{x}{ky+z}+\frac{y}{kz+x}+\frac{z}{kx+y}\geq\frac{3}{k+1}$.


superior algebra
superior algebra solved

Prove that $ \textrm{Tor}_1(R/I,R/I) \cong I/I^2$ for every $ 2$-sided ideal $ I$ of $ R$.

the long exact sequence gives $ Tor_1(R,R/I) \to Tor_1(R/I,R/I) \to I \otimes R/I \to R \otimes R/I$, i.e. $ 0 \to Tor_1(R/I,R/I) \to I/I^2 \to R/I$, where $ I/I^2 \to R/I$ is trivial.


Prove for every integer $n>0$, there exists an integer $k>0$ such that $2^nk$ can be written in decimal notation using only digits 1 and 2.

This one is VERY old, given in the Russian competitions in the sixties, I think.


Let $ A$ be a point inside a given circle with center $ O (A \not\equal{} O).$ Find all points $ P$ on the circle for which $ \angle OPA$ is maximum.

Call $ r$ the radius of the circle, $ d\equal{}OA$, and $ x\equal{}AP$.  By the Cosine Law, $ \cos\angle OPA\equal{}\frac{r^2\plus{}x^2\minus{}d^2}{2rx}\equal{}\frac{1}{2r}\left(x\plus{}\frac{r^2\minus{}d^2}{x}\right)$.  The angle will be maximum when the cosine is minimum, and the sum of $ x$, $ \frac{r^2\minus{}d^2}{x}$, whose product is constant, is minimum when both are equal, ie, when $ x^2\equal{}r^2\minus{}d^2$, or when $ AP\perp OA$.  The points $ P$ that maximize the angle are thus the two points where the perpendicular to $ OA$ through $ A$ intersects the circle.


yet another 'wow' .....my post count reduced from 3233 to 1725 :o  :rotfl:  :rotfl:

look at [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=36236]chappli's[/url] post count ...it's 0 :o

the posts u posted in the national forums have been taken off the post count :rotfl:  :rotfl:

I'm @ 58... Hahahaha!!! That really rules.....  :rotfl:  :rotfl:  :rotfl:

Anyway who cares.. That's just a number after all.

I think some of the higher-ups thought that we are over-posting here just to increase our post count. But I least bother.

LOL! Pardesi is this going to be a weakly affair???

no da as u wud have seen there was some problem yesterday writh post counts...so

Atleast I am above 200 so one solace for posting in other forums :rotfl:
Yes I have been wiped of 300 posts too :o

Lol... my post count jus reduced 25 ...  :D edit: later realised it has reduced by about 50

My posts remain MoL the same :). I haven't posted stuff in the India forum that much :).

[quote="shruthibaskaran"]Listen, we are talking about 300 posts here :( You probably had a few posts on you, 10 maybe? Now pardesi, you are a vethu mod ;)[/quote]loss is a loss  doesnt matter by how much

if u miss qualification in an exam by one place u remain at same level as last
dont u remember [i]all or none law [/i]????????????

[quote="rituraj007"][color=darkred]
i ve found a way u can know ur exact post no.
just go to ur profile an find all ur posts

but perhaps u knew this already??
[/color][/quote]

Ofc for ppl like paradesi who have 3335+ posts, its very simple to count??

well its written at top how many search results were found

Meh who cares as long as we can post  :P   :D   :rotfl:

[quote="Chapplikka"]Probably aops couold bear the wt of our posts in the forum.. Or probably our forum is the only forum wit more posts than even the average no of posts per forum... So it was slashed in this forum alone to get it equal to the mean..:P:D:)[/quote]

My dear mathematician. Unless all the quantities are equal, there has to be quantity greater than the mean no?? If you delete the greatest weight, mean also decreases.

My dear logistician,
                         I never claimed they deleted all our posts. They deleted some of them so that the net mean turns out to equal to our no. of posts, jus a diminishingm not an eliminating situation..

Oh yeah. Its happening all over again. And this time, I really dont care

As I said before, I guess they think we just want to inflate our post count by posting crap. :D Who cares, back to business. :D

let the no of posts be 1 or 1000  whats the diff.?
arent o's insignificant

yay, the post count has returned to normal :D :D

what is this poincare conjecture all about????? :rotfl:  :D


function
real analysis
real analysis unsolved

Let $f: A \to A,$ where $(A,d)$ is a compact metric space. If $d$ is an isometry (i.e. $d(x,y)=d(f(x),f(y))$ for each $x,y\in A$), prove that $f$ is a surjective function.

sketch of the proof- first note that every isometry is a continuous map. now suppose f is not surjective, consider   $B_{0}=f(A),B_{1}=f(B_{0}).... etc$  the above form a sequence of nested compact sets ( non empty). can you carry on from here?

Then $\cap B_{i}\ne \emptyset$ and ...

Help, please.

Thanks

My solution:
Let $x_{0}\in A,x_{n}=f_{n}(x_{0})$,since $A$ is compact ,there exists a number $a$ is the limit value of $(x_{k})$.So $a\in f(A)$,put $a=f(a_{1})$,so $a_{1}\in f(A)$,so on...
And $d(x_{0},a{}_{n}{}_{k})=d(x{}_{n}{}_{k},a)\to 0$ so $x_{0}\in f(A)$.
Another solution:
If $f(A)\subset A$,let $B(x_{0},r)\subset A-f(A)$,so consider the sequence:$(f_{n}(x_{0}))$.

In the above solution, how do we know f(f(x_0)) exists?  Inverse of f isn't necessarily surjective, is it?

This problem was discussed here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=117359[/url] .


Show that the following equation does not have a solution if a and b are positive whole numbers:

$\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{b^2} = 1$

[hide]$a^3b^3 - a^3b - a^2b^2 - ab^3 = 0$
$a^2b^2 - ab - a^2 - b^2 = 0$
$a^2 + b^2 = ab(ab - 1)$

$\frac ab + \frac ba = ab - 1$

If $a$ and $b$ are natural numbers than the RHS must be an integer. The only way that $\frac ab + \frac ba$ will be an integer is when $a = b$. The equation, then, would only be true if $a = \sqrt 3$ but $a$ cannot be an irrational number.[/hide]


If you pm me the top scores you know and I will compile them and send back a file to you.
____________________________________________________________
Western Carolina Chapter

Top 4:

Marie Mauhar, 1st place, Valley Springs team
Rusty Mau, 2nd place,  Cullowhee Valley School
Tyler Pressley,  3rd place,  Enka Middle School
Myles Kimble,  4th place, Owen Middle School

NW Piedmont Chapter  (total 134 students)

Top 4:
Bryce (1), Hanes
LiAn (2). Hanes

Eastern Carolina Chapter  (total 230 students)

Jenny Chen, 1st place, Arendell Parrott Academy
Christine Hong, 2nd place, Arendell Parrott Academy
Bailey Hu, 3rd place, Arendell Parrott Academy
Tony Philips, 4th place, Arendell Parrott Academy

Charlotte Chapter

Brendan Fletcher (1st)
David Lucia (2nd)

Could someone give me an estimate of what sort of scores are needed to make state (say in the triangle area)?

No, we can't. Besides, the cutoff would be very different in the triangle area anyway. In the southeast chapter, I know (some of) the top four, but not their scores. I'll pm them to you as soon as I do. I'm looking forward to your spreadsheet, since I coached the top two here :D.

1. Kailey Rose (Wilmington Academy of Arts and Sciences)
2. Zack Shiro (Wilmington Academy of Arts and Sciences)
3. ? (Roland Grise Middle, I think)
4. Kami Liu (Myrtle Grove Middle, I think)

I'd just like a (very rough) ballpark estimate.  I know it is an ignorant question, but bear in mind that I never did well at all on MathCounts myself.  I sort of started competitions in high school...

I meant we are not allowed to, for whatever reason. Please wait until Saturday.

Oh ok... I just meant in general (not as regards this specific chapter round or anything).  It's not terribly important anyways.

South Centra  (Fayetteville)

1. Jordan Strunk	West Pine
2. Victor Gardner	South View
3. Zac Lewie	West Lee
4. Tammy Yu	Seventy-First

North Piedmont  (Greensboro)

(1) Tom Lu	           The Academy at Lincoln
(2) Yimo Chen       The Academy at Lincoln

n^4:  jb's right about the variability in the scores from year to year and between chapters.  For instance, in 2006, in NW Piedmont the countdown cut was 35.  It was probably several points higher in the Triangle.  In some extremely competitive chapters in Texas, cutoffs were around 44 last year.

At the other extreme, in 1999, average chapter scores were approximately 8 with countdown cutoffs in the teens.

How do you know those statistics (for 1999)? If you (or anyone else) has statistics for NC countdown cutoffs for past state tests, could you please post them or send them to me?

The winter 2000 newsletter has some average scores on chapter rounds from the 90's.  An potentially valid argument for dumbing down the contests after 1999 is presented in the article.  Countdown cutoffs are not presented but you can get some idea about relative difficulty of the contests from different years.

[url]http://www.mathcounts.org/webarticles/articlefiles/31-Newsw00.pdf[/url]

Any news from the Raleigh Chapter competition yesterday?

I heard from a friend whose brother took it that Ligon won.

Carnage (spelling?) got 8th, I think (or maybe it was 2nd).

I will post more when I find out more - my friend had forgotten a lot of the stuff but I can find out more later.

Apparently, Allan Yang won written... :maybe:

Allen Yang from West Cary Middle School.

Yeah, and apparently he lost in the first round to #16, who goes to my school... (plus I'm a "coach") :D

shut up not_trig >_> I was nervous

and don't forget how to spell my first name ever again

b-flat, west cary got 2nd


number theory
least common multiple
algebra
polynomial

Square both sides.
Add three.
Divide by four.
Square both sides.
Add three.
Divide by four.
Square both sides.
Add three.
Divide by four.
Multiply by the LCM
You'll have
$x^{8}+12x^{6}+114x^{4}+36x^{4}+684x^{2}+15537 = 16384x$
Subtract the RHS from both sides, getting a polynomial that equals zero.
$x^{8}+12x^{6}+114x^{4}+36x^{4}+684x^{2}-16384x+15537 = 0$
Use substitution, and other methods taught in algebra to deduce that $\boxed{x = 1; x = 3}$

that statement is the same as x = sqrt(-3 + 4x), only x is substituted for itself twice, so the solutions are the same as what i just posted, which is easy to solve.

[hide]We can plug in x into the radical part for infinity, and see that

$x=\sqrt{-3+4\sqrt{-3+4\sqrt{-3+4\sqrt{\cdots}}}}$

We can square, add 3, and divide by 4:

$\frac{x^{2}}{4}+\frac{3}{4}=x$

$\frac{x^{2}}{4}-x+\frac{3}{4}=0$

$x=\frac{1\pm\sqrt{1-\frac{3}{4}}}{\frac{1}{2}}=2\pm1=1,3$[/hide]


vector
combinatorics proposed
combinatorics

A meteorology researcher is recording weather records in $ n$ cities for 10 days. For each day, a city's weather can be "Sunny" or "Rainy."
We are told that for any two different days, there are equally many cities that have same weather on those two days as the ones that have different weathers on those two days.

What are the possible values of $ n$?

Or, to put this in another language, we have $ k$ (in this case $ k \equal{} 10$) vectors in $ {\bf F}_2^n$ such that the sum of any two of them contains the same number of 1s and 0s; for which $ n, k$ is this possible?  (One could also phrase it in terms of set systems and their symmetric differences.)

Certainly for $ k > 1$ we must have $ n$ even, and for $ k \equal{} 2$ this is good enough.  For $ k \geq 3$ we must have $ 4 | n$: otherwise, WLOG take the first vector to be the zero vector.  Then the second and third vectors each have an odd number of 1s, so their sum has an even number of 1s, a contradiction.  $ 4 | n$ is good enough for $ k \equal{} 3, 4$, but no longer for $ k \equal{} 5$.  I'm sure there's a smarter way to go about this ...

I'm sure it has something to do with array. :D

JBL, I think you're on the right track.
My conjecture is that, the biggest power of two that divides $ n$ grows linearly with $ k$.

This is my approach:
For a city $ c$, if you flip the weather for that city for all $ k$ days, the condition still holds, because the condition only says "same" or "different".

So WLOG, we can assume that the first day, everybody has a Sunny day.
For second day, exactly half has a Sunny day. Thus, $ n$ must be even.

For third day, notice that day 2 and day 3 must satisfy that condition, so $ n$ must be a multiple of four.
For fourth day, same.

I still haven't been able to generalize this.

Clearly, $ n \equal{} 2^km$ satisfies the problem. So the set of multiples of $ 2^k$ is the superset of the answer.


A sequence of real numbers $ \{a_{n}\}$ is defineds by $ a_{0}\neq 0,1$, $ a_1\equal{}1\minus{}a_0$,$ a_{n\plus{}1}\equal{}1\minus{}a_n(1\minus{}a_n)$, $ n\equal{}1,2,...$.

Prove that for any positive integer $ n$, we have
$ a_{0}a_{1}...a_{n}(\frac{1}{a_0}\plus{}\frac{1}{a_1}\plus{}...\plus{}\frac{1}{a_n})\equal{}1$

That's way too obvious,isn't it?
We may proceed by induction on $ n$,assume that for all integers $ \leg k-1$,the statement holds.If $ n=k$:
$ a_k\cdot a_0a_1\dots a_{k-1}\left(\frac{1}{a_0}+\dots+\frac{1}{a_{k-1}}\right)+a_0\cdot\dots\cdot a_{k-1}=a_k+a_0\dots a_{k-1}$
$ =1-a_{k-1}(1-a_{k-1})+a_0\cdot\dots\cdot a_{k-1}=1-a_{k-1}(1-a_{k-1}-a_0a_1\dots a_{k-2})=1$

$ 1\minus{}a_1\equal{}a_0$.
Let \[ 1\minus{}a_n\equal{}\prod_{k\equal{}0}^{n\minus{}1}a_k,\] then \[ 1\minus{}a_{n\plus{}1}\equal{}a_n(1\minus{}a_n)\equal{}\prod_{k\equal{}0}^na_n.\]
By induction it is true for all n.
$ a_0a_1(\frac{1}{a_0}\plus{}\frac{1}{a_1})\equal{}1$.
Let it is true for n-1, then
$ a_0...a_n(\frac{1}{a_0}\plus{}...\plus{}\frac{1}{a_n})\equal{}a_n\plus{}a_0...a_{n\minus{}1}\equal{}a_n\plus{}1\minus{}a_n\equal{}1.$

[hide=Solution]
[quote=CWMI 2008 P1]A sequence of real numbers $ \{a_{n}\}$ is defineds by $ a_{0}\neq 0,1$, $ a_1\equal{}1\minus{}a_0$,$ a_{n\plus{}1}\equal{}1\minus{}a_n(1\minus{}a_n)$, $ n\equal{}1,2,...$.

Prove that for any positive integer $ n$, we have
$$ a_{0}a_{1}...a_{n}(\frac{1}{a_0}\plus{}\frac{1}{a_1}\plus{}...\plus{}\frac{1}{a_n})\equal{}1$$[/quote]
[b][color=#000]Solution:[/color][/b] Suppose for some $n \in \mathbb{N}$, we have: $a_{n}=1-\prod_{k=0}^{n-1} a_k$ $\implies$ $a_{n+1}$ $=$ $1-a_n(1-a_n)$ $=$ $1-\prod_{k=0}^n a_k$ $\implies$
$$a_{n+1}=1- \prod_{k=0}^{n} a_k$$
Now when, $n=2$ $\implies$ $\text{ LHS }$ $=$ $a_0a_1+a_1a_2+a_0a_2=a_0a_1+a_2(a_0+a_1)=a_0a_1+1-a_0a_1=1$. Suppose for some $n \in \mathbb{N}$, it holds that:
$$\sum_{k=0}^n \left( \frac{\prod_{i=0}^n a_i }{a_k} \right) =a_1a_2 \ldots a_n + a_0a_2a_3 \ldots a_n + a_0a_1a_3 \ldots a_n +\ldots + a_0a_1a_2\ldots a_{n-1}=1$$
Then, $$\sum_{k=0}^{n+1} \left( \frac{\prod_{i=0}^{n+1} a_i }{a_k} \right) =\left( \prod_{k=0}^{n} a_k \right) +a_{n+1} \left( \sum_{k=0}^n \left( \frac{\prod_{i=0}^n a_i }{a_k} \right) \right) = a_{n+1} +\prod_{k=0}^{n} a_k =1 \qquad \blacksquare$$
[/hide]


[quote="geotrig"]Can someone prove that e^(theta*i)=cis(theta) ?

Sorry...I still don't know how to use LaTeX.[/quote]

From wikipedia ,[url]http://en.wikipedia.org/wiki/Euler%27s_formula[/url]

It seems that the most common way to understand it is using Tayor series definition of $ e^x$  (This is just my feeling  :) ).

$ e^x \equal{} 1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\frac{x^3}{3!} \plus{} \frac{x^4}{4!} \plus{} \cdots$

Plug in x = i * theta and then use the Taylor series for sin and cos, which are

$ \cos x \equal{} 1 \minus{} \frac{x^2}{2!} \plus{} \frac{x^4}{4!} \minus{} \frac{x^6}{6!} \plus{} \cdots$

$ \sin x \equal{} x \minus{} \frac{x^3}{3!} \plus{} \frac{x^5}{5!} \minus{} \frac{x^7}{7!} \plus{} \cdots$


quadratics
calculus
integration
Gauss
algebra
system of equations
quadratic formula

Solve the following system of equations:

A(B+C+D+E+F) = 184
B(A+C+D+E+F) = 225
C(A+B+D+E+F) = 301
D(A+B+C+E+F) = 369
E(A+B+C+D+F) = 400
F(A+B+C+D+E) = 525

where A,B,C,D,E, and F are positive integers.

Very cute problem   :mrgreen:

[hide]My solution starts from the primes decomposition of 301 !? [/hide]

May I give an spoiled hint

[hide]

. prove 1 <= A < B < C < D < E < F

. then A+B+D+E+F > 7

. E follows easily [/hide]

Okay, this is a solution I think, although I have no idea where i/3's second hint would come into it.

[hide]We can write each of the equations as follows, letting

S=A+B+C+D+E+F

AS - A:^2: = 184 = 2:^3: * 23

BS - B:^2: = 225 = 3:^2: * 5:^2:

CS - C:^2: = 301 = 7 * 43

DS - D:^2: = 369 = 3:^2: * 41

ES - E:^2: = 400 = 2:^4: * 5:^2:

FS - F:^2: = 525 = 3 * 5:^2: * 7

Now, each of these makes a quadratic equation in their respective capital letter, which we know must have positive integer solutions for an integer value of S.  So, following after i/3's hint, we focus in on C, since 301 has a relatively simple prime factorization.

C:^2: - SC + 301 = 0

(C - r1)(C - r2) = 0

r1 + r2 = S

r1*r2 = 301

which leads to solutions (C, S) = (1, 302), (7, 50), (43, 50), or (301, 302).  Clearly, (301, 302) doesn't work for positive integer solutions.  Also, we can check quickly that (43, 50) won't work either.  This leaves us with 2 possibities.  We first consider if S = 302.  Then let us look at the equation AS - A:^2: = 184.

A simple check with the quadratic formula shows that this has no integral solutions when S = 302.  Thus, S = 50.  We now just plug into the other equations to find A = 4 or 46, B = 5 or 45, D = 9 or 41, E = 10 or 40, and F = 15 or 35.  This forces us to choose the smaller option for each variable, giving us the final solution

(A,B,C,D,E,F) = (4, 5, 7, 9, 10, 15).

I was thinking as I was doing this that it must have been hard to come up with this problem, but I just realized you need only to pick the six numbers and then find out what the results are and replace the numbers with variables.  Go figure.[/hide]

Since my solution is different from JBL's (but with the same result !) :

. if x(y+z) > y(x+z) then x > y (difference = z(x-y)) so 1 <= A < B < C < D < E < F

. Then  A+B+D+E+F > 1+2+4+5+6 > 7

. C(A+B+D+E+F) = 301 = 7*43 so C = 7.

. Now D >= 8, D | 9 * 41, A+B+C+E+F > 9 so D = 9.

. A+B>=3, F>=11, A+B+C+D+F >= 29, so E >= 10, E <= 13, E divides 400 so E = 10

. now B = 3 or 5, so A+B >= 4 and we obtain 15 <= F < 21, F divdes 3*5*7 so F = 15, A = 4 and B = 5 follows immediatly.

So, gauss202, what's the official solution ?

I basically did what Gauss did.. first you get C=7, and straight away you know the sum of all 6 must then be 50. So you just get each equation in one variable.


Exactly forty-eight non-overlapping square tiles, each 1 inch
by 1 inch, fit within a rectangle. What is the least possible
number of inches in the perimeter of the rectangle?

This means that the area is 48. The smallest possible perimeter occurs when the quadrilateral is as close to a square as possible. So the closest this rectangle can be is $ 6 \times 8$. The perimeter of such a rectangle is $ \boxed{28}$.


Hello. I need some help with this problem. I attached a file with the diagram.
Here's the problem though

10. ABCD is a square. Parallel lines [i]m[/i], [i]n[/i] and [i]p[/i] pass through vertices B, C, and D, respectively. The distance between [i]m[/i] and [i]n[/i] is 7 units, and the distance between [i]n[/i] and [i]p[/i] is 9 units. Find the number of square units in the area of square ABCD

And here's what they say is the answer.
[hide]130 sq units[/hide]

Does anyone have an explanation of how to solve this?[/i]
Thanks

Draw perpendiculars from the B and D to $ n$ and look for congruent right triangles...

Ah...I get it now...

But I wonder, how do you know they're congruent?

No...
They are congruent because, firstly, they have 2 angles congruent: a right angle and the 2 angles that meet at C are complementary.
Then, their hypotenuses are congruent since it's a square, so AAS congruence gives congruent triangles.
Finish it off with Pythagorean Theorem.

eh, yeah, I found the congruent triangles while actually doing the problem

although all congruent triangles are similar to each other...


ratio
combinatorics unsolved
combinatorics

we choose $n+1$ elements from$\{1,2,\dots,2n\}$ prove that there are $a,b$ such that$a|b$

Really well-known Erdös-type problem.

Pierre.

For each number $n$ let $f(n)$ be its odd part (i.e. $n$ divided by the highest power of $2$ that divides it). Then, since each number has odd part equal to one of $1, 3, \ldots, 2n-1$ ($n$ choices), there will be two with equal odd part. Then the ratio between the bigger and the smaller will be a power of $2$ and thus an integer.

i think thisproblem is east but useful and you can learn good ideas from it.


geometry
geometric transformation
reflection
ratio
parallelogram
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14578[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14579[/img]

 :thumbup:

[quote="¬[ƒ(Gabriel)³²¹º]¼"]from the italian WC08 we know that the only point that have that propriety is the Symmedian Point (i.e. the isogonal conjugate of the centroid), it's a nice problem with a nice proof.[/quote]
Please, next time around, could you provide a link or outline the proof ?

$ \triangle DEF$ is a pedal triangle of $ P$ WRT the reference triangle $ \triangle ABC$ with $ D, E, F \in BC, CA, AB$.,

Let $ P$ be centroid of the pedal triangle. Reflect $ P$ in the midpoint $ D'$ of $ EF$ into $ Q.$ $ DP \equal{} PQ,$ because $ P$ divides the median $ DD'$ in the ratio 2 : 1. $ PEQF$ is a parallelogram, $ EP \equal{} QF.$ $ \angle FPQ \equal{} \angle B$ from the cyclic $ BDPF$ and $\angle PQF = \angle EPQ = \angle C$ from cyclic $CEPD$ $ \Longrightarrow$ $ \triangle FPQ \sim \triangle ABC$ $ \Longrightarrow$ $ \frac {PQ}{BC} \equal{} \frac {QF}{CA} \equal{} \frac {FP}{AB}$ $ \Longrightarrow$ $ \frac {DP}{BC} \equal{} \frac {EP}{CA} \equal{} \frac {FP}{AB}$ $ \Longrightarrow$ $ P$ is the symmedian point of $ \triangle ABC.$

Conversely, let $P$ be symmedian point of the reference triangle  $ \Longrightarrow$ $ \frac {DP}{BC} \equal{} \frac {EP}{CA} \equal{} \frac {FP}{AB}$. Reflect $D$ in $P$ into $Q$. $ \angle FPQ \equal{} \angle B$ from the cyclic $ BDPF$ and $ \frac {PQ}{BC} \equal{} \frac {DP}{BC} \equal{} \frac {FP}{AB}$ $ \Longrightarrow$ $\triangle FPQ \sim \triangle ABC$ by SAS. $\angle QPE \equal{} \angle C$ from the cyclic $ CEPD$ and $ \frac {QP}{BC} \equal{} \frac {PD}{BC} \equal{} \frac {PE}{CA}$ $ \Longrightarrow$ $\triangle EQP \sim \triangle ABC$ by SAS. Combining, $\triangle FPQ \cong \triangle QPE$ are congruent  $ \Longrightarrow$ $PEQF$ is parallelogram $\Longrightarrow$ $PQ, EF$ cut each other at their common midpoint $D'$ and $\frac{DP}{PD'} = 2$ $ \Longrightarrow$ $P$ is centroid of $\triangle DEF$.

Beautiful synthetic proof, Vladimir! See also [url=http://forumgeom.fau.edu/FG2008volume8/FG200812index.html]Cosmin Pohoata, A short proof of Lemoine's theorem, Forum Geometricorum, 8 (2008) 97--98[/url] for a proof using a theorem of Daneels and Dergiades on orthologic triangles and for some other references.


function
pigeonhole principle
combinatorics proposed
combinatorics

Let $A$ and $B$ be two subsets of $S = \{1, 2, . . . , 2000\}$ with $|A| \cdot |B| \geq 3999$. For a set $X$ , let $X-X$ denotes the set $\{s-t | s, t \in X, s \not = t\}$. Prove that $(A-A) \cap (B-B)$ is nonempty.

a is an element of A and b is an element of B, there are 3998 values of a+b also there are at least 3999 pair of a,b so there are four numbers a,b,c,d such that a and c are elements of A; b and d are elements of B and a+b=c+d hence a-c=b-d gives the desired result.

Consider the function $F: A \times B \rightarrow \{2,3,....,4000\}$ defined by $F(a,b)= a+b$. Let us consider two cases:

1. $F$ is onto. Then both $A$ nad $B$ contain $1$ and $2000$. Therefore, $1999= 2000-1$ belongs to $(A-A) \cap (B-B)$.

2. $F$ is not onto. Since $|A| \times |B| \geq 3999$, it follows by the pigeonhole prinbiple that $F$ cannot be injective (since $F$ may have a maximum of $3999$ possible values and $F$ does not necessarily assume all of them). So, we must have $a_{1},a_{2}$ in $A$ and $b_{1},b_{2}$ in $B$ such that $a_{1}+b_{1}=a_{2}+b_{2}$. From this we get $a_{1}-a_{2}=b_{2}-b_{1}$, and since $a_{1}-a_{2}\in A-A$ and $b_{2}-b_{1}\in B-B$, it follows that $A-A \cap B-B$ is non-empty, which completes the proof.


I have an idea for the first one.
Let a<=b<=c<=d,then we must prove:
       ax+by+cz+dt>=(a+b)/2
<=>a(x-1/2)+b(y-1/2)+cz+dt>=0
By the Abel formula,we have:
a(x-1/2)+b(y-1/2)+cz+dt=(x-1/2)(a-b)+(x+y-1)( b-c)+(x+y+z-1)(c-d) and this one is bigger than 0 clearly. :)


Let A be any set of twenty integers chosen from the arithmetic progression 1,4, . . . ,100. Prove that there must be two distinct integers in A whose sum is 104.

[hide="Attempt"]
There are only $ 33$ possible terms that can be in the set, even if one were to pick the $ 17$ that did not add up to $ 104$, one is still forced to pick a $ 3$ more, of which would add to the previously chosen to make $ 104$ .
[/hide]

[hide="solution"]There are 34 numbers total and 16 possible ways to make 104 and two numbers that can't be paired with another to make 104.  If we pick one number from each of the 16 pairs and also pick the two leftover numbers, we have only chosen 18.  Therefore we can't choose anymore numbers.[/hide]


Let $ a,b,c\ge 0$. Prove that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\le \sqrt{\frac{a\plus{}2b}{3}}\plus{}\sqrt{\frac{b\plus{}2c}{3}}\plus{}\sqrt{\frac{c\plus{}2a}{3}}$


$ x \equal{} 3 \plus{} \frac{1}{5 \plus{} \frac{1}{3 \plus{} \frac{1}{5 \plus{} \frac{1}{3 \plus{} \frac{1}{5 \plus{} \frac{1}{3 \plus{} \frac{1}{5 \plus{} }...}}}}}}$

What is the value of $ x$?

I hope I got the equation right... basically I meant an infinitely repeated continued fraction...

What he is saying is, by looking at its simplified form, denoted as $ x \equal{} [3,5,3,5...]$ and there is a pattern. The period of 2, with $ [3,5]$ can be replaced by some other variable, namely $ x$. Then substituting this back in the first place you can see it repeat (to make it easier) makes a quadratic after simplifying.

Actually a neat thing comes up.

$ x \equal{} \frac{16x\plus{}3}{5x\plus{}1}$

Can you see what is neat about this?

[hide="Hint"]What form is $ x$ in?
[/hide]

it am clear that they neput not give a precise value of an continued fraction, but they can the evaluer, it has a term it which repete in this fraction, it is for it that they can the evaluer, the equation which I gave he is not going to give you the value precise.


abstract algebra
superior algebra
superior algebra unsolved

May I know if we can move the whole theory of algebraic extension of field to algebraic extension of rings? (result like F in K in W, all fields, implies $[W: F] = [W: K][K: F]$ or a is algebraic over F iff F adjoining to a is a finite extension of F)
I think it is okay, at least on commutative ring with unity. May I know if I am correct?
Thanks.

[quote="ZetaX"]In generally (also when you restrict to commutative rings with units), no.
One problem e.g. is that $[A: B]$ can't be defined as you are used to.

The adjoining part is indeed correct when you define finite extension properly.[/quote]

emm.. may I know how to define finite extension then? Is it possible to define $[R_2: R_1]$ for $R_1$ in $R_2$ (how to type subset symbol?), then define $[R_2: R_1]$ to be the rank of the basis of the module of $R_2$ over $R_1$ whenever the module is free?


Lines L1 and L2 are not parallel,
points A, C, E on L1
points B, D, F on L2
such that AB//ED and BC//FE
prove that CD //FA....  :wacko:

Let L1 and L2 meet at O

AB//ED implies $\frac{OA}{OB}=\frac{OE}{OD}$
BC//FE implies $\frac{OB}{OC}=\frac{OF}{OE}$

Multiplying the two gives $\frac{OA}{OC}=\frac{OF}{OD}$, thus AF//CD


quadratics
number theory unsolved
number theory

find all prime $p$ such that for some $x.y$ in $\mathbb{Z}$, $3x^2-y^2=p$

if $3$ is a quadratic residue $mod$ $p$ (forms of these numbers come from the Gaussian quadratic reciprocity theorem) there exist such $x,y \in Z$ that $0 \leq x,y \leq \sqrt{p}+1$
$3x^2-y^2=2p$ or $3x^2-y^2=p$, if $3$ is not a quadric residue then the equality never holds.

Right Rafal I reached here.
so what will happend if$3y^2-x^2=2p$????

I should wright before $x,y < \sqrt{p}+1$

then $x^2-3y^2 <p$ so it cant be $x^2-3y^2=2p$.

sorry Rafal I took a mistake I edited my post.


Welcome.
I am Donald Trump, believe it or not.

If you sign up now, I will or will not accept you to become a qualifying member...

To sign up, please answer these questions (py pm)

On a scale of 1-5, 5 being the highest:

1. How much leadership do you believe you have?
2. How smart are you?
3. How modest are you?
4. How athletic are you?
5. How musical are you?

Yes/no:

1. Have you ever "managed" a project outside of AoPS?
2. Do you play a musical instrument?
3. Are you absolutely, absolutely sure that you will check this thread, as with any other corresponding quickthreads as often as possible? (Most likely once a day.)

Give me answers to these questions, and I will decide if I should let you in or not.

Here is the run-down on the rules.

I will chose 2 groups of people.
Both groups will decide, in thread, who the project manager of their team is.
The groups will both do their assigned project. I will decide which team does better on each task with a rubric. The winning team will win money and the losing team will... not.

The losing project manager then picks 1-3 people who he believes contributed to the loss. Then those 3 people each say their pleas and he fires at least one of the people. The manager pm's me his reasoning.

I have the right to not allow the project manager to choose 3 people to a final hearing, fire/eliminate any candidate without a final boardroom session if there's enough information to warrant so, and fire multiple people if two or more people are found liable for the weak performance.

[quote="'Donald Trump'"]I am Donald Trump, believe it or not.[/quote]

cough. How believable.

Just to make sure, Dojo did you get my pm?

how many signed up?

......Donald Trump.....?

:lol:

ps, if there are alot, i may sign up......

@dragon96: Why, it is so believable.
@jjx1: yes
@Joe: 6 appplications so far. EDIT: 7, phail.

um.......dojo, can u tell us who signed up?

and did u get my pm??

yeah.

nt(insert number here)
hedgehog12
isabella2296
you
eggylv999
jjx1
BOGTRO

When I get enough people.

I have recieved 2 more.

And so we begin.

Since I have not recieved a plentiful number of applications, I will reduce the number of people the project manager can fire to 2.

So teams are:

TEAM A:
nt811581147
hedgehog12
isabella2296
Joe10112
crazypianist1116

TEAM B:
eggylv999
jjx1
BOGTRO
hongsquare
Alekim

Teams, please pm me your top 3 choices on who should be the project manager to me.

In the meantime. I will assign tasks.

[size=200]Task 1: Architecture[/size]
Each team will be assigned a different task. To kick everything off, I have assigned a realitivly "easy" task.

I will send quicktopics to the project manager once he/she has been decided.

Anyway:

TEAM A: Go to http://www.grc.nasa.gov/WWW/K-12/rocket/rktsim.html and using nomal Earth conditions, show me how to make the Rocket go as far as possible.

TEAM B: Go to http://bridgecontest.usma.edu/download.htm and design the cheapest amd strongest bridge. Take the picture of the bridge and send it to me.

Bleh, when is the deadline

Away at Stanford (6.5 hour flight  :lol: ) from Thursday-Sunday with POSSIBLE LIMITED internet access

You have time.

Supposedly I'm supposed to give 1 week.

Should we discuss with our team memebers for the team manager? Cause I don't really know like 2 people on our team.

well i know you.. you're the corrupt mod xD no harsh feelings, lets pwn

I guess press new and make a new file out of it.

I'll try it out by myself in few moments, as soon as I finish some work.

eh... me as your team leader? are you trying to get me out of the way ASAP? haha

After you extract it, go to the .exe file. I believe it leads to an installation process.

No, Alekim, we dont want 2 get rid of u......

we think u can be the best.......
:)

Can I have an extension? I was away at stanford from Thursday-very late Tuesday because our flight was canceled.

How can we work on a rocket building together?

do we communicate in this thread (i tink this is called a thread, right?), of do we pm each other?

WHEN is the deadline?

I will send your manager a quick topic in hopes that he/she will tell you it.

Managers:

isabella2296
alekim

HELLO??

ANYONE HERE??

DOJO?

HELLO?

DID WE ABANDONE tHIS GAME???

I THINK WE DID!!!!!!!

NOT SAYING I WANT TO ABANDON THIS GAME............

Well apparently noone is participating. :roll:

wait, did u send the topics to izzy and alekim?

and pm everyone, asking if they want to abandon this game.

we'll take a vote


Put $x=\frac{b}{a},y=\frac{c}{b},z=\frac{d}{c},t=\frac{a}{d}$ we have a problem from China TST.


Just curious - Any one here  paritcipated in Virgina Tech Math contest here this year?

Hey i participated in 2004 (sorry im replying so late).  i got 36, kind of irritated though cuz the problems were easy, but I made the leading scorers list :)


Five marbles of various sizes are placed in a conical funnel. Each marble is in contact with the adjacent marble(s). Also, each marble is in contact all around the funnel wall. The smallest marble has a radius of 8, and the largest marble has a radius of 18. What is the radius of the middle marble?

I know it's the geometric mean, but is there any easy but concrete way to see this?

He asked for why this is true, so I gave a possibility.

Take the cross section, so it looks like 2 lines that eventually intersect, and 5 circles inside of it. Take any 3 consecutive circles (call their radii r, q, and w, from smallest to biggest) and draw perpendiculars to one of the lines (call it line L). Now from the center of the smallest circle, draw a line parallel to line L. We now have 2 similar triangles. Taking ratios of the short leg, which is the difference of the radii, and the hypotenuse, which is the sum of the radii. So we have the ratio \[\frac{w-r}{q-r}= \frac{w+2q+r}{q+r}\]\[wq+wr-qr-r^{2}= wq+2q^{2}+qr-wr-2qr-r^{2}\]\[2wr = 2q^{2}\]\[q = \sqrt{wr}\]

This is just like a problem from the 1983 (I think) AHSME.

1982 is the year.

I am just trying to say, how do you remember the years?

I've only done 2 or 3 of the 80's ahsme's, and I've seen that problem (with different wording of course) before. We also did it in the AMC 12 problem series.

and this was on an old guts round...but i was like OMG I THINK THIS IS DIFFERENT>>>LETS TO ARITHMETIC MEAN

yea...that was stupid...

[quote="Recursive Acronym"][quote="mathgeniuse^ln(x)"]WEll, I think one could use similar triangles and get that the ratio of consecutive radii, but I don't know.[/quote]
You don't have to find the ratio, you simply have to know that it is constant, and it is because each pair of adjacent marbles are similar.  Thus, the answer is the geometric mean (12).[/quote]

Just because the triangles are similar doesn't mean that it's the geometric mean. I am just wondering if there is a natural way to see this without chugging through algebra.

Well I noticed that 8*18=144=perfect square.
Otherwise I would've put the AM down.


how difficult are the AIME tests from, say, 1983 and now? where can i find the USAMO cutoff indexes for past years?

and i know that since the ahsme is scored differently (5 pts per question, 2 per blank or something like that), does that mean it's about one the same difficulty level as the AMC?

The AHSME, at least in the 1990s, should be considered comparable to the AMC-12. The main difference is that the AHSME was 30 questions in 90 minutes, while the AMC-12 is 25 questions in 75 minutes; the longer test made perfect scores rarer, but didn't change much else.

USAMO qualification rules were also different in that time, without the "floor". One case I remember: In 1998, my 10th grade year, it took an index of 200 to qualify as a senior, and 190 to qualify as a 10th grader. My 130/7 was good enough. I don't know of any source with old cutoff information.

soo. if i get around a 200-210 index that should be pretty indicative of what i'll get now, even if the test's from quite a long time ago?

No. The difficulty varies too much from year to year.

For example, here are the index scores required to qualify for the USAMO (as an 11th or 12th grader) for the last few years:

2005: 233
2006: 217
2007: 197.5

The score levels could vary by 20 points in either direction, and there's no way to predict these things. There's also the difference between a practice test and the real thing- how do you respond to all of that adrenaline?

You have to remember that in 2005 only ~200 people got to take the USAMO
in 2006 they started expanding it
and in 2007 500+ people got to take the USAMO

if you are trying for 2008,
i think if you can pretty consistently get 200+ for index, you are pretty safe.

[quote="Narcissa"]hmm is the 2002 amc12a easier than usual?[/quote]
It depends on which group of people you are (the group going for a perfect, the group going for a 130+, the group going for 100-130, and the group of people who don't expect to qualify).

You can find statistics on the AIME website.

What index would be required from freshmen?

Under the current system, 9th graders and younger must meet exactly the same standards as 10th graders. This is not a simple index number; for most years since the AMC-10/AMC-12 split, only the AIME score has mattered.

So about what AIME score would a freshman need?

...i got a 138
when all i really wanted to do was, like, pass (100)
on the AMC12A 2002

[quote="jmerry"]The "floor" scores on the AIME since that system started:
2002: 6
2003: 8
2004: 7
2005: 9
2006: 8
2007: 6[/quote]

So I guess we can expect something near 6 this year?

[quote="mountain.dew"][quote="mihail911"][quote="h_s_potter2002"]Well, it depends on the difficulty of the test, you can't really guess anything from previous scores.[/quote]

true, they don't really have a particular recurring pattern. so, if you want to set your mind at rest, just get a 10. you have a good chance of qualifying, not 100% though  :D[/quote]

10 would be overkill to make it since they now take 500 people.[/quote]
sarcasm? :wink:

[quote="mountain.dew"]From me or mihail911?[/quote]
sheesh, people. considering most of us are still on break, id think you're joke brains would be on. i very highly doubt that most of the USAMO qualifiers will score 10s on aimes.

[quote="ZzZzZzZzZzZz"]You can't really conclude that most people are on break.  I know some people had to go to school starting today, but that's beside the point.

It's probably true that a lot of USAMO qualifiers get 10 or higher on the AIME, and it could have easily been taken literally, since 10 isn't especially hard for a lot of people.[/quote]
Yes, but the AIME floor has never been 10, and now they're taking more people than ever before.

[quote="the future"]Hey how hard would you guys say the AHSMEs of the 1980s and early 1990s were?

Would you say they were on par with the current AMC 12s?

(both no calculator, same time for each problem (3 min))[/quote]

Any post-1983 AHSME is comparable to an AMC 12, though difficulty may vary a bit. Pre-1983 AHSME's are harder than AMC's because there was no AIME then; AHSME was used directly as qualification for USAMO.


Prove of disprove:
$ 2^{199}\minus{}1$ is a prime.

[hide]
well no, it has two factors of (2^98+1) and (2^98-1)
difference of two squares. [/hide]

[quote="cipos2"]it has two factors of (2^98+1) and (2^98-1)[/quote] I don't really understand what you want to say. :maybe:
[quote="t0rajir0u"]That product is $ 2^{196} \minus{} 1$.  Anyway, I can tell you that this number is indeed composite because it's not on the list of Mersenne primes.  What property must its prime factors have?[/quote] Yes, I know that  :wink: . But there should be some elementary way to prove that it is composite.

I *attempted* a mathematica bash, and aborted, not finding any prime factors, but I did go all the way up to 303689053, the 16441268th prime.
So yeah, not thinking there's a good mod bash way.

I don't know a good way to do this by hand, but executing FactorInteger[$ 2^{199} \minus{} 1$] with mathematica gives $ 2^{199} \minus{} 1 \equal{} 164504919713 * 4884164093883941177660049098586324302977543600799$.  Hence it is not prime.

Wait.. Brut3Forc3, there's like a single command to do this in Mathematica...

PrimeQ[2^199 - 1] returns a boolean value: True if the argument is a prime, False otherwise.

Then, after seeing that PrimeQ[2^199 - 1] = False, you can try

FactorInteger[2^199 - 1] which IMMEDIATELY yields
$ 164504919713 \cdot 4884164093883941177660049098586324302977543600799$ as the two factors.

Edit: ninja'ed by facis.

[strike]You can take it's $ mod 4$ because every prime has either 1 or 3 as remainder, when divided by 4. You can then, find the smaller values of $ 2$'s remainders, then go, go, go till you got like 196, which will give the same remainder. Then multiply both sides by $ 2^3$ and do proper divisions again, substract 1 and see if the remainder is 1 or 3.[/strike]

Non-primes also have remainders of 1 and 3...

[quote="t0rajir0u"]What property must its prime factors have?[/quote]

Factors of $ 2^{199} \minus{} 1$ are of the form $ 398k \plus{} 1$. Even so, the smallest factor occurs at $ k \equal{} 413328944$...


inequalities
search
inequalities proposed

I think it is a particular case of an inequality in VASC's book. This kind seems to be so familiar.

If you have solution or hint or some related things, post it don't spam, if it is in Vasc's book show me the page?  :D

[quote="skywalker"]I think $\frac{20}{3}$ be infimum, there isn't equality, take
$a\rightarrow 0,b\rightarrow 0,c\rightarrow 0$[/quote]
Eum.....What do you mean?

I meant when take $a\rightarrow 0,b\rightarrow 0,c\rightarrow 0$ then the express $\rightarrow \frac{20}{3}$ and it will be less than $\frac{22}{3}$
$\frac{20}{3}$ is the best const because it is infimum of this express, I think so!

I don't spam of course. The reason why I remember this problem is that: I once tried to create some problems of 3 variables which have the same kind as yours and... happy because of a lot of nice inequalities but when I saw VASC' book, I am so sad because they are involved in this book. Your problem is a particular case of VASC problem (Problem 10, Solution on page 334). If you had this book, please see it. And please never say that I am spamming.

ok that's right I saw it in Vasc book, thus it really has minimum be $\frac{22}{3}$ actually mine is from
$\frac{a^{2}+16bc}{(b+c)^{2}}+\frac{b^{2}+16ca}{(c+a)^{2}}+\frac{c^{2}+16ab}{(a+b)^{2}}\geq 10$
It seemed not in Vasc book!
Note I see a nice and similar problem above in Vasc's book
$\frac{a^{2}+64bc}{(b+c)^{2}}+\frac{b^{2}+64ca}{(c+a)^{2}}+\frac{c^{2}+64ab}{(a+b)^{2}}\geq 18$
Who did think about general?
$f(a,b,c)=\frac{a^{2}+qbc}{(b+c)^{2}}+\frac{b^{2}+qca}{(c+a)^{2}}+\frac{c^{2}+qab}{(a+b)^{2}}$

[quote="GacKiem"]All above are in VASC's book. CHeck it for you.
[/quote]
What's is the page of the general inequality?  :D

[quote="GacKiem"]All above are in VASC's book. CHeck it for you.

To TheSoul ofRock:
[quote]Please check his previous post![/quote]

what do you mean?[/quote]
Because if his's true then Vasc's wrong. Please se his problem. But I think Vacs is true :D

p/s: Please post the solution, not the name of book. It's seem so impolite. May be some member can't have yours books.

I will never post a complete solution in a book. It's normal. Why you say that's impolite? Or you are impolite? I post the origin of a problem (in some books) because that fact shows it is not a new problem, and Mathlinkers can search in this forum, not try to prove it again. It is helpful. Poor you, thesoulofrock.

To Skywalker. The problem you just say is also a particular case of problem 10, page 334. This problem is so general that we shouldn't try to discover more (because if we do that, this problem is still not of our own, it's of VASC  :( ).

Why search? The only way to learn mathematics is do mathematics!

I don't want to make this topic become a chaos of unnecessary quarrel, but I must say that when I read Vietnamese Mathlinkers' posts in Mathlinks, I am annoyed of The Soul of Rock with a lot of bad post. I knew many my friends and other Vietnamese Mathlinkers share my view so please don't annoy us, THE SOUL OF ROCK.

Mod can delete my post. It's not good but I must say.

I'm wrong, I saw the general problem in Vasc book!

the soul of rock is absolutely right . giving the pages of the books which
to some like me are not available never serve the purpose. mathlinks is problem solving forum and not a chat about the number of pages of a book.

If you don't want, please post a new problem, not an old one.  :P

[quote="me@home"]Sorry, but can someone post the general inequality?[/quote]
General problem in Vasc's book in page 334
$a,b,c\ge 0$ and no two of which are zero let $q,r\in R, r>-2$ let
$E(a,b,c)=\sum\frac{a^{2}+qbc}{b^{2}+rbc+c^{2}}$
then
$E(a,b,c)\ge \frac{3(q+1)}{r+2}$ for $q\le\frac{2r+1}{2}$
$E(a,b,c)\ge \frac{q}{r+2}+2$ for $\frac{2r+1}{2}\le q\le 4(r+2)^{2}$
$E(a,b,c)\ge 4kr+12k^{2}-2$ for $q=4k(r+2k)^{2}, k\ge 1$


Prove : If a polynomial $ f(x)$ with real coefficients takes only non-negative values, then there exists a positive integer $ n$ and polynomials $ g_1(x),g_2(x),\dots,g_n(x)$  such that $ f(x)\equal{}g_1(x)^2\plus{}g_2(x)^2\plus{}\dots\plus{}g_n(x)^2$.

Take $ n\equal{}2$:
$ P(x)\equal{}a(x^2\plus{}a_1x\plus{}b_1)*...*(x^2\plus{}a_kx\plus{}b_k)$ and all $ x^2\plus{}a_jx\plus{}b_j$ is a summ of two squares of polynomials. So, the product $ a(x^2\plus{}a_1x\plus{}b_1)*...*(x^2\plus{}a_kx\plus{}b_k)$ is a summ of two squares of polynomials, too.

why can we factor $ f$ as $ a(x^2\plus{}a_1x\plus{}b_1)(x^2\plus{}a_2x\plus{}b_2)\dots (x^2\plus{}a_ix\plus{}b_i)$?

sorry for my silly question, thx
Raja

Every polynomial with real coefficients is the product of two-degree and one-degree polynomials. Use fact, that if $ a\plus{}bi$ is a root of polynomial, than $ a\minus{}bi$ is a root, too. So, $ (x\minus{}a\minus{}bi)(x\minus{}a\plus{}bi)\equal{}x^2\minus{}2ax\plus{}a^2\plus{}b^2$.


Let $a$, $b$, and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$. What is $a^{2}-b^{2}+c^{2}$?

Both solutions are essentially correct but with many sign errors/typoes.


median point of AB
the circumcircle
locus (circle) of orthocenters of triangles with this circumcircle intersect Lh in orthocenter point of the ABC triangle

 :lol:


pigeonhole principle
inequalities
Additive Number Theory

Prove that for each positive integer $K$ there exist infinitely many even positive integers which can be written in more than $K$ ways as the sum of two odd primes.

We will prove the following stronger statement (implying the above with $ S\equal{}\{2\}$, and we don't even need the distinctness of the primes):

Let $ \mathbb P$ be the set of primes and let $ S\subset\mathbb P$ be finite. Then for any $ K$, we find $ \infty$ integers expressible as sum of two distinct primes in $ \mathbb P\backslash S$ in more than $ K$ ways.

Proof:
Call $ \pi(n)$ the number of primes $ \leq n$.
Lets see that we find at least one such integer $ x$ for any $ K$:
Take some integer $ n$. There are $ m: \equal{}\binom{\pi(n)\minus{}|S|}2$ different pairs $ (a,b)$ of primes $ a < b\leq n$ not in $ S$, their sum $ a\plus{}b$ trivially being $ <2n$. If $ m > (K\minus{}1)\cdot 2n$, $ K$ of these sums are the same by Pidgeonhole-Principle, thus such $ x$ exists. Thus lets prove that inequality for some large $ n$:
By Chebyshev, there is a $ c>0$ such that $ \pi(n) > c\cdot\frac {n}{\log(n)}$. As we also have $ m\equal{}\binom{\pi(n)\minus{}|S|}2\geq\frac{1}{2}(\pi(n)\minus{}|S|\minus{}1)^{2}\geq\frac{1}{2}\left( c\cdot\frac {n}{\log(n)}\minus{}|S|\minus{}1\right)^{2}$, we just want that $ \frac{1}{2}\left( c\cdot\frac {n}{\log(n)}\minus{}|S|\minus{}1\right)^{2}> (K\minus{}1)\cdot 2n$, clearly happening for some $ n$ (the LHS grows like $ c^{2}\cdot\frac{n^{2}}{\log(n)^{2}}$, the RHS only linear in $ n$).

Now that we found one $ x_{K}$ expressible as sum of two different primes from $ T: \equal{}\mathbb{P}\backslash S$ in $ K$ different ways for any $ K$, we even get infinitely many of them:
If $ x_{K}$ can be expressed in exactly $ L\geq K$ different ways as sum of two different primes in $ T$, we simply take $ K^{\prime}\equal{} L\plus{}1$, then $ x_{K^{\prime}}\neq x_{K}$ but still fulfills our requirements. Continuing this, we get infinitely many $ x$ with the desired property.

Note: in the proof above, we used Chebyshevs theorem. But it's easy to see that the following suffices (being slightly easier):
$ \frac{\pi(n)}{\sqrt n}$ is unbounded.


John uses the average of the perieters of the circumscribed and incrsibed squares to a circle of radius 1 too estimate its circumfrence. Using this approximation, what is his estimate of pi using the formula for the circumfrence?

I solved the problem, but don't have the answer with me. It would be great if someone could check my answer:

[hide]2+root 2[/hide]

The circumscribing square is going to have side length of 2, so its perimeter will be $ 4\cdot 2 \equal{} 8.$
The inscribed square is going to have diagonal length of 2, which is the hypotenuse of a 45-45-90 triangle, so the side length of the inscribed square must be $ \frac {2}{\sqrt {2}} \equal{} \sqrt {2}$, and its perimeter is $ 4\sqrt{2}$.
Averaging these, we get the circumference of the circle as $ \frac {8 \plus{} 4\sqrt {2}}{2}\equal{}4\plus{}2\sqrt{2}$
We know that $ 4\plus{}2\sqrt{2} \equal{} 2\pi \cdot 1\implies\pi \equal{} \frac {4 \plus{} 2\sqrt {2}}{2}\equal{}2\plus{}\sqrt{2}$.

Edit: Sorry, stupid mistakes...  :oops:

Yes, you're right. (Unless I'm wrong)

Yes, you're right. (grn_trtle use \approx instead of = for estimates.)

The average of the perimeters is $ 4 \plus{} 2\sqrt {2}$
and since the formula for circumference is $ 2\pi r$, and $ r \equal{} 1$
we have $ 2 \pi \approx 4 \plus{} 2\sqrt {2} \rightarrow \pi \approx 2 \plus{} \sqrt {2}$


trigonometry
function
real analysis
real analysis theorems

could anyone prove this
\[ \frac{1}{\cos x}\equal{}\sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n(2n\minus{}1)\pi}{x^2\minus{}\left(n\minus{}\frac{1}{2}\right)^2\pi^2}\]

http://math.stackexchange.com/questions/8319/series-representation-for-1-cos-x

Theory of Functions of a Complex Variable, Volume 1 By A. I. Markushevich, Richard A. Silverman page 60.


I've browsed through this forum for a bit, and I apologize if a similar question has already appeared and been answered.  Anyway, here is the question:

Does anyone have any idea how to choose what part of a string to use for alignment (in Asymptote)?  In particular, I am looking to correct something like the following:

[asy]size(100); draw((0,0)--(40,0)); dot("$Q$",(40,0),N); dot("$R$",(30,0),N); dot("$S$", (10,0),N); dot("$P$",(0,0),N);[/asy]
(code: size(100); draw((0,0)--(40,0)); dot("$ \$Q\$$",(40,0),N); dot("$ \$R\$$",(30,0),N); dot("$ \$S\$$", (10,0),N); dot("$ \$P\$$",(0,0),N); )

It's not hard to see that the $ Q$ is not properly aligned, and this is because it is a character that extends below the line, while Asymptote is measuring from the bottom of the string.  I would like to find a way to have the tops of the strings all be collinear.  Of course, one solution is to send the characters to S rather than N, but I am wondering if there is a better solution than this.  Thanks!

Alternatively, you can use the $ \text{\LaTeX}$ tools like \vphantom command to create a box of proper vertical size. Since $ Q$ is the only "non-standard" size capital letter, the following code will work pretty well:
[asy]size(100); void mylabel(string s,pair A,pair dir=N) { label("$\vphantom{Q}"+s+"$",A,dir); } pair P=(0,0), S=(10,0), R=(30,0), Q=(40,0); draw(P--Q); dot(P); dot(S); dot(R); dot(Q); mylabel("P",P); mylabel("Q",Q); mylabel("R",R); mylabel("S",S);[/asy]

[code]
size(100);

void mylabel(string s,pair A,pair dir=N)
{
label("$\vphantom{Q}"+s+"$",A,dir);
}

pair P=(0,0), S=(10,0), R=(30,0), Q=(40,0);
draw(P--Q);
dot(P);
dot(S);
dot(R);
dot(Q);
mylabel("P",P);
mylabel("Q",Q);
mylabel("R",R);
mylabel("S",S);
[/code]
If you think that the letters are too far from the dots now, replace $ N$ by something like $ (0,0.5)$ everywhere. The advantage of this solution is that you do not need to figure out those values like 0.5 and 1.15 that align the boxes the way you want.


modular arithmetic
real analysis
AMC
AIME
calculus
derivative
geometric series

Let $ S \equal{} \{2^0,2^1,2^2,\ldots,2^{10}\}$. Consider all possible positive differences of pairs of elements of $ S$. Let $ N$ be the sum of all of these differences. Find the remainder when $ N$ is divided by $ 1000$.

[hide="bleh, i feel retarded reponding to my own post"]
Each number in S is included in 10 different differences. We see that $ 2^{10}$ is counted as positive in each of the 10 differences. Thus, the total sum produced from the $ 2^{10}$s is $ 10{}^2{}^10$. $ 2^9$ is positive in 9 differences, and counted as negative in 1.  Thus, the total sum produced from the $ 2^9$s is $ 9*2^9-2*9=8*2^9$. We can continue, and we find N to be $ 10*2^10+8*2^9+6*2^8+...-8{}^2{}^1-10*2^0$. Taking this mod 1000, we find $ N \equiv 240+96+536+512+128-32-32-24-16-10 \equiv \boxed{398} \pmod{1000}$[/hide]

[hide]Split the differences up into differences between terms that are 1 apart, 2 apart... and so on until you reach the last difference between 1024 and 1, which is 10 terms apart.

For the terms that are one apart, you get 1 + 2 + 4 + .... + 512 for a sum of 1023

For the terms that are two apart, you get 3 + 6 + 12 +... 3(256) = 3(1 + 2 + 4 + ... + 256) for a sum of 3(511) = 1533

3 apart: 7 + 14 + 28 + ... + 896 = 7(1 + 2 + 4 +... + 128) = 7(255) = 1785

Continue in this fashion, factoring out a common factor until you get to six apart, when it appears the sums 6-10 apart will mirror your sums from 5-1 apart (which makes sense, considering how you get the common factor to be factored out is by an initial difference from a term and 1, which gives you a sum for a certain number of powers of two). Sum together all the parts to get 16398, and dividing that by 1000 and taking the remainder gives [b]398[/b][/hide]

[hide="solution"] Consider when $ 2^{10}$ is the larger number. There are ten numbers that we may subtract from it, so our sum from that is $ 10(2^{10})\minus{}(2^0\plus{}2^1\plus{}...\plus{}2^9)$. Using the fact that $ 2^0\plus{}2^1\plus{}...\plus{}2^k\equal{}2^{k\plus{}1}\minus{}1$, we compute this sum as $ 9(2^{10})\plus{}1$.

Proceeding in a similar fasion, our complete sum is $ 9(2^{10})\plus{}8(2^9)\plus{}7(2^8)\plus{}6(2^7)\plus{}5(2^6)\plus{}4(2^5)\plus{}3(2^4)\plus{}2(2^3)\plus{}(2^2)\plus{}10$. When we consider the sum (mod 1000), we end up with $ \boxed{398}$.[/hide]

[hide="Solution"]
Alternatively, consider the unique differences $ 2^{a \plus{} n} \minus{} 2^a$. Simple casework yields a sum of $ \sum_{n \equal{} 1}^{10}(2^n \minus{} 1)(2^{11 \minus{} n} \minus{} 1) \equal{} \sum_{n \equal{} 1}^{10}2^{11} \plus{} 1 \minus{} 2^n \minus{} 2^{11 \minus{} n} \equal{} 10\cdot2^{11} \plus{} 10 \minus{} 2(2 \plus{} 2^2 \plus{} \cdots \plus{} 2^{10})$
$ \equal{} 10\cdot2^{11} \plus{} 10 \minus{} 2^2(2^{10} \minus{} 1)\equiv480 \plus{} 10 \minus{} 4\cdot23\equiv\boxed{398}\pmod{1000}$.
[/hide]

So far I haven't seen a really clever solution. I did what basically everybody else did and bruteforced it out with the "pattern". I thought there might be a clever way to solve it by turning the numbers into base 2 numbers, but that didn't give me any success.

Did anybody find a quicker way? This ate up a lot of time multiplying and adding all those numbers. I'm sure most people double checked the arithmetic too because there was so much of it.

My method is not arithmetic-intensive at all, and the pattern is very easily proven...it took less than one minute for sure.

Wow...nice. :P I would get that, give or take a hundred, if I did that...

[quote="alkjash"]I found a slightly better solution using aritmetico-geometric series but it's still pretty messy.[/quote]

Same here, but not as messy. It eased the computation quite a lot.

the calculations weren't that bad...

vorfield:  :rotfl:

I did exactly what you guys did (sort of brute forcey) and got 398.  Then, I tried checking it with another method, basically simplifying with identity $ 2^0\plus{}2^1\plus{}...\plus{}2^n\equal{}2^{n\plus{}1}\minus{}1$, and got 2 more different answers.  Brute forcing again, I got 399 (which is impossible, because it has to be even) and put that down.  Ugh.

Personally, I like math154's solution quite a lot.  Here's my version.

[hide]
The answer is the discrete convolution of the first few Mersenne numbers.

$ \sum_{i \equal{} 1}^{n}\Big((2^i \minus{} 1)(2^{n \plus{} 1 \minus{} i} \minus{} 1)\Big)$

$ \sum_{i \equal{} 1}^{n}\Big(2^{i \plus{} n \plus{} 1 \minus{} i} \minus{} 2^{n \plus{} 1 \minus{} i} \minus{} 2^i \plus{} 1\Big)$

$ \sum_{i \equal{} 1}^{n}2^{n \plus{} 1} \minus{} \sum_{i \equal{} 1}^{n}2^{n \plus{} 1 \minus{} i} \minus{} \sum_{i \equal{} 1}^{n}2^i \plus{} \sum_{i \equal{} 1}^{n}1$

$ n\times 2^{n \plus{} 1} \minus{} \sum_{i \equal{} 1}^{n}2^i \minus{} \sum_{i \equal{} 1}^{n}2^i \plus{} n\times 1$

$ n\times 2^{n \plus{} 1} \minus{} (2^{n \plus{} 1} \minus{} 2) \minus{} (2^{n \plus{} 1} \minus{} 2) \plus{} n$

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

$ (n \minus{} 2)(2^{n \plus{} 1}) \plus{} n \plus{} 4$

That's a pretty simple general formula, and is easily checked for very small values of $ n$.  For the specific case of $ n \equal{} 10$ the arithmetic is quite easy:

$ 2^3\times 2^{11} \plus{} 10 \plus{} 4$

$ 2^{14} \plus{} 14$

$ 16384 \plus{} 14$

$ 16398$[/hide]

[quote="Vorfield"]pffft [i][b]this[/b][/i] is the most elegant solution

[hide][img]http://img.photobucket.com/albums/v139/Samuru/CCF03182009_00000.jpg[/img][/hide][/quote]

I did that hahaha

[quote="Vorfield"]pffft [i][b]this[/b][/i] is the most elegant solution

[hide][img]http://img.photobucket.com/albums/v139/Samuru/CCF03182009_00000.jpg[/img][/hide][/quote]

You are my hero.

i found the expression for the sum differences between 2^k and 2^i (i = 0 to i = k-1)
and added it from k = 1 to 10
so i turned out as

sigma (from k = 1 to 10) of [(k-1)*2^k + 1]

did everything correct with summing the powers of two..and then i confused 1+1+ ... + 1 as 1 + 2 + ... 10
what a silly mistake

meh, i brute forced this one.
brute force got me like 3 extra points on the AIME this year.

With calculus, it's possible to find a formula for $ \sum_{i \equal{} 1}^{10} (i \minus{} 1)*2^i$, greatly simplifying the calculation of the answer. (To find the formula, just take the formula for a geometric series, take the derivative, and multiply by the base squared; I'll include the derivation if people are interested)

Is it possible to find such a beast without using calculus?

Yes...this is known as an [url=http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series]arithmetico-geometric series[/url]. However, it's not too pretty in the end (the case for $ d\equal{}1$ is a bit nicer, though).

Can somebody show the calculus derivation?

Here's a fun derivation:

$ \sum_{n\equal{}0}^{N}n2^n \equal{} \sum_{n\equal{}0}^{N}\sum_{m\equal{}0}^{n\minus{}1}2^n \equal{} \sum_{m\equal{}0}^{N\minus{}1}\sum_{n\equal{}m\plus{}1}^{N}2^n \equal{} \sum_{m\equal{}0}^{N\minus{}1}2^{N\plus{}1}\minus{}2^{m\plus{}1} \\\equal{} N2^{N\plus{}1} \minus{} 2^{N\plus{}1} \plus{} 2 \equal{} (N\minus{}1)2^{N\plus{}1} \plus{} 2$.

Note that each possible difference must be in the form $2^a - 2^b$. Now we take cases on $a$. If $a=1$, there are $1 \cdot 2^1 - (2^1-1)$ ways. If $a=2$, then there are $2 \cdot 2^2 - (2^2-1)$ ways, and so on. The answer is $(1 \cdot 2^1 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + 10 \cdot 2^{10}) - (2^1+2^2+2^3 + \ldots + 2^{10} - 10) \equiv 16398 \equiv \boxed{398} \pmod {1000}$.

[hide=Solution]$2^{0}$ is the larger number of $0$ differences and the smaller number of $10$ of them, so overall it is subtracted $10$ times.

$2^{1}$ is the larger number of $1$ and the smaller number of $9$, so it is subtracted $8$ times.

$2^{2}$ is the larger number of $2$ and the smaller number of $8$, so it is subtracted $6$ times.

And so on, until $2^{10}$ which is the larger of $10$ differences and the smaller number of none, so overall it is added $10$ times.

Thus we have \[N=10\cdot 2^{10}+8\cdot 2^{9}+6\cdot 2^{8}+\cdots-6\cdot 2^{2}-8\cdot 2^{1}-10\cdot 2^{0}=10240+4096+1536+512+128+0-32-32-24-16-10=16398\] so $N\equiv{\textbf{398}}\pmod{1000}$.[/hide]


I just got:
AoPS Volume 1
Introduction to Counting & Probability
Introduction to Geometry

I have a math contest similar to MathCounts on March 10th, and I wanted to know what is the best way to practice from these books (i.e. finish Geometry first, then Volume 1, or do a little of two).

I am pretty confident in my basic Counting and Probability skills, so if there are some quick ways that are described in the C&P book, could you tell me about them?

Thanks for being my SparkNotes  :P !

actually do vol 1 first i think...it covers some of the topics in your other books (i think i just got aops vol 1 today  :lol: ) then if u have any questions the other books might help


[b]Question 1:[/b]
    Find two numbers such that the square of the first plus the second equals $11$, and the square of the second plus the first equals $7$.

[b]Question 2:[/b]
    $a, b, c, d$ represent four different digits such that $addd$, $aaca$, $bcdb$ and $bdac$ are prime numbers. What digits do the letters represent?

[b]Question 3:[/b]
EDIT: This question has been deleted.

[b]Question 4:[/b]
   Using all digits from $1$ to $9$ (each digit to be used once and once only), it is possible to form a fraction equal in value to a half, thus: $6729/13458$. Using these nine digits form fractions equal in value to:
a) $1/3$
b) $1/4$
c) $1/5$
d) $1/6$
e) $1/7$
f)  $1/8$
g) $1/9$

1. [hide]$x^2+y=11$........(1)
$x+y^2=7$.............(2)

(1)-(2)= $(x-y)(x+y-1)$

${(x-y), (x+y-1)}$ can be ${4,1}, {2,2}$ (when they are integers), in which only integer pair $(x,y) = (3,2)$. There are non integral answers, but the question said find one pair... meh.[/hide]

[quote="xxxyyyy"][quote="José"][quote="hydro"][hide="2"]
clearly the only digits we have to consider are 1,3,7,9. If $d=1$ then it must be $a=7$, cause if not $addd$ would be divisible by 3. but $77c7$ is divisible by 3. So $d\neq1$ and $d\neq7$ for the same reason. Then $d=3 \vee 9$ and $a=1 \vee 7$. Now consider $bdac$. Clearly, $b+a\neq d+c$, cause if not the number would be divisible by 11. Try $d=3$. Then $a=7$, $b=9$, $c=1$. but in this case $bcdb=9139$, that is obviously divisible by 13. So the only possibility is $a=1$, $b=3$, $c=7$, $d=9$[/hide]

what's the question 3?[/quote]

Maybe find all pairs[/quote]

Yes, find all such pairs.[/quote]

here is the answer to question-4
[hide]$1/2$: $6729/13458$ ($12$ such fraction)
    $1/3$: $5823/17469$ ($2$ such fractions)
    $1/4$: $3942/15768$ ($4$ such fractions)
    $1/5$: $2697/13485$ ($12$ such fractions)
    $1/6$: $2943/17658$ ($3$ such fractions)
    $1/7$: $2394/16758$ ($7$ such fractions)
    $1/8$: $3187/25496$ ($46$ such fractions)
    $1/9$: $6381/57429$ ($3$ such fractions)[/hide]

[quote="xxxyyyy"][quote="José"][quote="hydro"][hide="2"]
clearly the only digits we have to consider are 1,3,7,9. If $d=1$ then it must be $a=7$, cause if not $addd$ would be divisible by 3. but $77c7$ is divisible by 3. So $d\neq1$ and $d\neq7$ for the same reason. Then $d=3 \vee 9$ and $a=1 \vee 7$. Now consider $bdac$. Clearly, $b+a\neq d+c$, cause if not the number would be divisible by 11. Try $d=3$. Then $a=7$, $b=9$, $c=1$. but in this case $bcdb=9139$, that is obviously divisible by 13. So the only possibility is $a=1$, $b=3$, $c=7$, $d=9$[/hide]

what's the question 3?[/quote]

Maybe find all pairs[/quote]

Yes, find all such pairs.[/quote]

ok, so I try...
[hide="3, a part"]
write the condition as $(10a+b)(10c+d)=(10b+a)(10d+c) \Longleftrightarrow ac=bd$ with $a,b,c,d \in \{1,2,3,4,5,6,7,8,9\}$
Clearly we want that at least 3 among $a,b,c,d$ are different digits, because we don't want to consider number such for example $12*21$
is there now a fast method to find all pairs? cause they are a lot...
[/hide]

ooops... I'm very sorry :blush:  :blush: I didn't read this topic

[quote="xxxyyyy"][quote="JBL"][quote="vulgarfraction"]Funny.. Q3 is the basis for the whole of an ongoing program called [url=http://www.mansw.nsw.edu.au/studentservices/mathsearch.htm]Mathsearch[/url]. I just got the stuff for it today. And it's the main criterion for entry into a camp next January so maybe it shouldn't be discussed here because help can be easily accessed.[/quote]

In light of this fact, could a moderator delete hydro's last response and lock this thread please?[/quote]

Thanks for bringing this to my notice. I am really sorry for the post.

I have edited the post and deleted the question.

Still, if the MODS feel, they can close this thread.

Once again, I sincerely apologize.[/quote]

What have happened????


[quote="chess64"]A positive integer $n$ is said to be $\textit{increasing}$ if, by reversing the digits of $n$, we get an integer larger than $n$. For example, 2003 is increasing because, by reversing the digits of 2003, we get 3002, which is larger than 2003. How many four-digit positive integers are increasing?[/quote]
[hide]
If the fourth digit is larger than the first digit then ... yeah. The middle digits can be anything, $10\cdot 10=100$, and there are 8+7+6+5+4+3+2+1 ways to choose the first and last digit. So $\frac{8\cdot 9}{2}\cdot 100=3600$.
If the first digit is equal to the last digit (9 ways), then there are 9+8+7+6+5+4+3+2+1 ways the third digit can be larger than the second digit (45 ways)

So is the answer 3600+45=$\boxed{4005}$?
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[quote="E^(pi*i)=-1"][hide]

Define a decreasing number to be the opposite of an increasing number: a number which becomes smaller when we reverse its digits, and a palendromic number to be a number which has the same value when its digits are reversed.  All numbers are increasing, decreasing, or palendromic, and the number of increasing numbers is the same as the number of decreasing numbers (since when we reverse the digits of a decreasing number, we get an increasing number, and vice versa).

We specify a four digit palendromic number simply by specifying its first two digits, as the last two must be the reverse of them.  There are 90 possible first two digits, so there are 90 four digit palendromic numbers.

Of the 8910 four digit numbers remaining, half are increasing, so the the total number of four digit increasing numbers is 4455.

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There are numbers with last digit 0...does that matter in this case?  :?

ch1...whatever is right, but he meant to do $\binom{9}{2}\cdot 10^2+\binom{10}{2}\cdot 9=\boxed {4005}$. ;)


What do you thing about... Can you find any better solutions?    :arrow:   [b][color=red][url=http://mathzone.net.tc/]mathzone.net.tc/[/url][/color][/b]

Well , the concept's pretty interesting....As far as better solutions are concerned , I doubt I'll be able to provide them....


Let $k$ be a natural number smaller than or equal to n-1. Let $ x_1,...,x_n$ positive numbers such that the sum of their k-th powers is 1. What is the minimal value of $\prod_{1\leq i\leq n}{(\frac{1}{x_i}-1)}$?

[quote="Myth"]Uph...
I did it via IPT. The minimal value is for equal $x_i$.[/quote]

what's ipt?

Obviously, Myth did not solve the problem!  :D  :D  :D


calculus
integration
calculus computations

$\int \frac x{(x+6)^{2/3}}$

the (2/3) is supposed to be an exponent.

This problem has me stumped.  I think it's supposed to be a u-substitution but nothing there it's not obvious to me how to do it.   :huh: Any hints would be appreciated.  Thanks.

The substitution you'd like to make is $u=x+6.$ The apparent obstacle: what do you do with the $x$ in the numerator? Try solving the substitution "backwards": $x=u-6.$ So replace that $x$ by $u-6$ and $dx$ by $du.$ Now what do you have?

Thank you, I see what I did wrong.  I didn't separate the integral into two integrals before taking the anti-derivative.  Thanks for saving the hours of confusion.

 :rotfl: (me dying)