inequalities
calculus
calculus computations

1) prove that $ \inf_{x\in X} f(x)\plus{}\inf_{x\in X} g(x)\leq \inf_{x\in X}\{ f(x)\plus{}g(x)\}$

2) suppose that $ \frac{f(x)}{x} (x\in \mathbb{R}_\plus{})$ is monotone decreasing,  $ a,b \in \mathbb{R}_\plus{}$,  prove that

   $ f(a\plus{}b)\leq f(a)\plus{}f(b)$

[quote="red_dog"]2)
$ a \plus{} b\geq a\Rightarrow \frac {f(a \plus{} b)}{a \plus{} b}\leq\frac {f(a)}{a}\Rightarrow af(a \plus{} b)\leq(a \plus{} b)f(a)$

$ a \plus{} b\geq b\Rightarrow\frac {f(a \plus{} b)}{a \plus{} b}\leq\frac {f(b)}{b}\Rightarrow bf(a \plus{} b)\leq(a \plus{} b)f(b)$

Add the inequalities:

$ (a \plus{} b)f(a \plus{} b)\leq(a \plus{} b)(f(a) \plus{} f(b))\Rightarrow f(a \plus{} b)\leq f(a) \plus{} f(b)$[/quote]

Thank you... I was being silly then

for a: $ \inf_{x\in X}f(x)\leq{f(a)}$  $ \forall a\in X$

$ \inf_{x\in X}g(x)\leq{g(b)}$  $ \forall b\in X$

adding the inequalities we have

$ \inf_{x\in X}f(x) + \inf_{x\in X}{g(x)}\leq{f(a) + g(b)}\forall a,b\in X$

so also $ \inf_{x\in X}f(x) + \inf_{x\in X}{g(x)}\leq{f(a) + g(a)}\forall a\in X$

cause $ \inf_{x\in X}{(f(x) + g(x))}$ is the maximum minimum bound

we got the desired inequality

$ \inf_{x\in X}f(x) + \inf_{x\in X}{g(x)}\leq{\inf_{x\in X}{(f(x) + g(x))}}$


hello !! if $ x \plus{} y \plus{} z \equal{} 1$ and $ x,y,z>0$ prove that :
[img]http://www.texify.com/img/%5CLARGE%5C%21x%5E3%2By%5E3%2Bz%5E3%2B6xyz%5Cgeq%5Cfrac%7B1%7D%7B4%7D.gif[/img]

This Inequality is incorrect.

If $ (x, y, z)\equal{}\left(\minus{}\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$,
$ x^3\plus{}y^3\plus{}z^3\plus{}6xyz\equal{}\minus{}\frac{1}{3}$.

[quote="Kouichi Nakagawa"]This Inequality is incorrect.

If $ (x, y, z) \equal{} \left( \minus{} \frac {1}{3}, \frac {2}{3}, \frac {2}{3}\right)$,
$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \equal{} \minus{} \frac {1}{3}$.[/quote]
 :blush: sorry, i have forgotten : $ x,y,z > 0$

After homogenising the inequality is equivalent to $ \sum x^3 \plus{} 3xyz \ge \sum (x^2y\plus{}xy^2)$, which is Schur.

I'm assuming that $ x,y,z \ge 0$, since the ineq is wrong otherwise.

Let $ f(x,y,z) \equal{} x^3\plus{}y^3\plus{}z^3\plus{}6xyz$.
Then $ f(x,y,z)\minus{}f(0,x\plus{}y,z) \equal{} 3xy(2z\minus{}x\minus{}y)$.
Because the ineq is symmetric we may wlog assume $ z \ge x,y$, And hence $ f(x,y,z) \ge f(0,x\plus{}y,z)$. Let $ w \equal{} x\plus{}y$. Then we only have to prove that $ w^3 \plus{} z^3 \ge \frac{1}{4}$ when $ w\plus{}z \equal{} 1$. Which is obvious from the power-mean inequality:
$ \frac{w^3\plus{}z^3}{2} \ge \left ( \frac{w\plus{}z}{2} \right ) ^3$

[quote="mahanmath"]hello !! if $ x \plus{} y \plus{} z \equal{} 1$ and $ x,y,z > 0$ prove that :
[img]http://www.texify.com/img/%5CLARGE%5C%21x%5E3%2By%5E3%2Bz%5E3%2B6xyz%5Cgeq%5Cfrac%7B1%7D%7B4%7D.gif[/img][/quote]

Ineq
$ <\equal{}>1\minus{}3q\plus{}9r \ge \frac{1}{4}$
With:$ p\equal{}a\plus{}b\plus{}c\equal{}1;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$
Case 1:$ q \le \frac{1}{4}$
So:$ LHS \ge 1\minus{}3q \ge \frac{1}{4}$
Case 2:$ \frac{1}{4} \le q \le \frac{1}{3}$
We have:$ 9r \ge 4q\minus{}1$
$ LHS \ge 1\minus{}3q\plus{}4q\minus{}1 \ge \frac{1}{4}$

[quote="KrazyFK"][quote="FelixD"]After homogenising the inequality is equivalent to $ \sum x^3 \plus{} 3xyz \ge \sum (x^2y \plus{} xy^2)$, which is Schur.[/quote]

Am I being silly, but when I homogenise I get $ \sum x^3 \plus{} 6xyz \ge \sum (x^2y \plus{} xy^2)$ which is still true, but different to what you got.

I'm quite new to this topic, so I'm probably wrong, but could someone please explain this to me?[/quote]
You are correct and Felix is wrong. That's also why we don't have equality when $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$, which we would have had if it was equivalent to $ \sum x^3 \plus{} 3xyz \ge \sum (x^2y \plus{} xy^2)$ :)

OK, thank you for clarifying that for me  :)


For $ x,y,z \in R_{\plus{}}$ such that $ \sqrt[3]{xyz}\equal{}8$ prove:
$ \sum_{cyc}\sqrt{\frac{1}{x\plus{}y\plus{}2}}\geq \frac{\sqrt{2}}{2}$
:wink:

Yes, but this inequality can turn out to be untrue :wink:

Yes. Indeed. The inequality is untrue.
$ x\equal{}100,y\equal{}\frac{1}{100},z\equal{}8^3$ gives the value $ \approx 0.0133771124311$ (Thanx to mr.Comp. :wink: )
I would suggest $ x,y,z<\equal{}26$. Then it is easy with convex functions. May be this 26 can be increased too.

Well if that's the case...
Sorry it turned out like that :wink:
Then I'd like to change the problem slightly:
Find (if exists) the biggest $ m\in R_{ \plus{} }$ such that:
$ x,y,z \leq m$ and x,y,z must have the previous condition :wink:


superior algebra
superior algebra unsolved

Let G be a finite group of order N.
 Prove that if we have 2N-1 elements, then there are N with sum Zero,

a)if G is solvable
b)General G

(a) is quite easy, but i have no idea for b))

Can we write and add the $N$ elements we choosed in any order or in a given one¿

I think it is not known if this is true if we can only add in the given order, i think it is a conjecture. (probably for the solvable case its true, with the same proof (but i´m not sure)).


algebra
polynomial
complex numbers
algebra unsolved

A complex number $ z$ is called [i]good[/i] if it is a root of a monic polynomial with integer coefficients. Prove that the sum of two good complex numbers is also good.

This is an assertion in Algebraic Number Theory, on algebraic integers.


If there is a function y = x^ 2 - x + 41 can you find a value of x where y is not prime? If x is a natural nuumber.

There is a cool way to do this.  Just think about that equation.

For above, I think the author means x  (and y) are positive integers.
BTW this is same as noticing $  41, (41+2), (41+2+4), (41+2+4+6)  $
or $ 41, 43, 47, 53, 61, 71, 83 ...$ all are prime....

(You add 2, 4, 6, 8 to the  previous number ..)
IOW How far you have to go up, using the above method,  till you no longer get a prime ... Orwill you  always get a prime?

Thanks for catching my error. I fixed it now.

Other famous expressions are:

$x^2-79x+1601$   which gives primes from n=0 to 80

Challange: How do you derive this  if you know that $x^2-41x+41$ gives prime for  40 such numbers....?

Also curious  is
$2x^2 + 29 $  which gives primes when x=0,1,2 ...28

Another very 'fantastic'  one is:

$N = 12150 - 1710 x + 60 x^2 $
 here for x =  1,2,...20  both $N+1$ and $N-1$ are prime !

[hide][quote]If there is a function y = x^ 2 - x + 41 can you find a value of x where y is not prime? If x is a natural nuumber. [/quote]
 y=x 2 -x+41=x 2 -x-2+43=(x-2)(x+1)+43
So that for x=43n+2,x=41n or x=43n-1 there are infinitely many solutions ( n is natural number)  :)   . [/hide]


solve this system in rationals:

$ a^2b\plus{}2ac^2\plus{}2b^2c\equal{} ab^2\plus{}a^2c\plus{}2bc^2\equal{}0$

First assume that $ abc\neq 0$. Dividing by $ abc$ we get $ \frac ac\plus{}2\frac cb\plus{}2\frac ba\equal{}0\equal{}\frac bc\plus{}\frac ab\plus{}2\frac ca$. Letting $ x\equal{}\frac ab,y\equal{}\frac bc,z\equal{}\frac ca$ this becomes
\[ x\plus{}y\plus{}2z\equal{}0\\ \\
\frac 1z\plus{}\frac 2x\plus{}\frac 2y\equal{}0\]
Substituting $ z\equal{}\minus{}\frac{x\plus{}y}{2}$ into the second equation we get $ 2\left(\frac 1x\plus{}\frac 1y\right)\equal{}\frac{1}{2(x\plus{}y)}$.

Or $ (x\plus{}y)\left(\frac 1x\plus{}\frac 1y\right)\equal{}\frac 14$. But $ (x\plus{}y)\left(\frac 1x\plus{}\frac 1y\right)\ge 4\iff (x\minus{}y)^2\ge 0$. Thus the system has no solutions in real numbers when $ abc\neq 0$.

Now we consider $ abc\equal{}0$. Note that if any one of $ a,b,c$ is $ 0$ then one of the remaining two must be zero as well. Thus $ (0,0,0),(0,0,t),(0,t,0),(t,0,0)$ are the only solutions in this case and in all cases, where $ t$ is any rational number.


function
geometry
geometric transformation
reflection

I think you should give only the positive solution. Since $f(x)$ is a [i]function[/i], for each value of x it can give one and only one value of y.
If you consider both positive and negative cases, $f(x)=\sqrt{x}$ is not a function, because to one value of x are given 2 different values of y

$\sqrt{x}$ is the principal square root, which gives only a positive real value. I'm pretty sure $x^{\frac{1}{2}}=\sqrt{x}$ so only the positive answer applies.

[quote="Jhonny Ruiz"]So applying that to the problem we'll have...
$f(9)=8(9)^{\frac 1 2}-(9)^{-1}$
$=24-\frac 1 9$
$=\frac{215}{216}$[/quote]
 :?  :huh: Isn't the answer $23\frac{8}{9}=\frac{215}{9}$?

[quote]Isn't the answer $23\frac{8}{9}=\frac{215}{9}$
[/quote]
Yeah of course I don't even know why i put 216......... sorry there guys.... the answer is $\frac{215}{9}$

[quote="coffeym"]Yes, but then again the function could also be defined for the negative root and not the positive root, which would maintain the 1:1 mapping.[/quote]

But you need to know that that the function is [i]defined[/i] using the positive square root...generally you are overthinking the problem

[quote="coffeym"]I just took an exam there in which a question was
$f(x)=8x^{\frac{1}{2}}-x^{-1}$
$Find\ f(9)$

Now seeing as there is a power of a half, I gave solutions in the cases of both a positive and a negative square root.  Should this have only been the positive square root? Thanks[/quote]

I think you have to put both result, the negative and the positive one. Then, the two results are $\frac{215}{9}$ and $-\frac{217}{9}$.

I [i]think[/i] that you only consider the positive value when taking the square root. You only consider the negative root when you are solving an equation. Because $(-x)^2=x^2$, and the original number could have been $\pm x$. One more thing, on my Ti-89, if I type in $\sqrt{x^2}$ I get $|x|$, not $\pm x$.

[quote="coffeym"]Yes, on reflection it seems that it is only the positive that is required, based on the mathematical assumption that you just deal in positives under the situation.  However I doubt I could be penalised for providing 2 possibilities :) . Thanks[/quote]

Surely no penalisation


We have: $\frac{a^3}{(2a^2+b^2)(2a^2+c^2)}\leq\frac{a}{(a+b+c)^{2}}$ and analogues.
The conlusion is obvious.

[quote="Vasc"]If $a,b,c$ are positive numbers then
$\frac{a^3}{(2a^2+b^2)(2a^2+c^2)}+\frac{b^3}{(2b^2+c^2)(2b^2+a^2)}+\frac{c^3}{(2c^2+a^2)(2c^2+b^2)} \leq\frac{1}{a+b+c}$.[/quote]
My inequalitie is harder :lol:

Actually, my original inequality is the following:
If $a,b,c$ are positive numbers then
$\frac{a^3}{(2a^2+b^2)(2a^2+c^2)}+\frac{b^3}{(2b^2+c^2)(2b^2+a^2)}+\frac{c^3}{(2c^2+a^2)(2c^2+b^2)} \leq\frac{a+b+c}{3(a^2+b^2+c^2)}$. ;)

[quote="Vasc"]This is also my solution.
But, according to Darij's Lemma, it suffices to show that
$\frac{a^3}{(2a^2+b^2)(2a^2+c^2} \geq \frac{b^3}{(2b^2+c^2)(2b^2+a^2)}$.[/quote]
I used the arqady's Lemma ( :lol: ):
Assume $a\geq b\geq c>0,x>0,y>0,z>0.$
If $(a-c)x\geq(b-c)y$ or $(a-c)z\geq(a-b)y$ then
$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)\geq0.$
This is stronger! ;)


Find all 9-digit numbers such that the numbers formed by looking at the first, middle, and last three digits are all squares and the number itself is square.

There are a lot of answers, and the casework is quite messy.  Where is this problem from?

MMPC (Michigan Math Prize Competition) 2006

[quote="shyong"]there are more than that (28 in total if I did not miscalculate ) .

define the 9 digit number as

$S=10^{6}c^{2}+10^{3}b^{2}+a^{2}$

we find that $0\leq a,b\leq 31$ and $10\leq c\leq 31$

Let $S=(10^{3}c+a)^{2}$ we get

$b^{2}=2ac$

the rest is case work[/quote]

Provided, of course, that strings such as "000" are considered legitimate numbers. If we limit all three numbers to $[10,31]$, there are $17$ solutions shown above.

[quote="Hoborobo"]This doesn't explain the solutions that aren't in the form $(1000a+c)^{2}$ where $b^{2}=2ac$, like $12035^{2}$[/quote]

Yes, shyong only considered the case $c < 32$ but the case $c \ge 32$ produces a few more answers.


topology
vector
advanced fields
advanced fields theorems

"Suppose that $M$ is a general manifold and that $\psi$ is a differentiable map of $M$ into a vector space $V_{1}$. Then $\psi_{*}(T_{x}(M))$ is a subspace of $T_{\psi(x)}(V_{1})$. If we regard $\psi_{*}(T_{x}(M))$ as a subspace of $V_{1}$ and consider the corresponding hyperplane through $x$, we get the "plane tangent to $\psi(M)$ at x" in the intuitive sense."

My questions: why doesn't the author assume $\psi$ to be bijective ? Is $V_{1}$ without any extra hypotheses ? What does the word "hyperplane" really mean in this context ?

Looks like a poorly written statement to me.  Is this a reputable text?  I'm not even sure what point the author is trying to make.

Perhaps the author meant to say that \psi_{*}(T_{x}(M)) is the "plane tangent to \psi(M) at \psi(x)" in the intuitive sense."  In fact, as long as \psi(M) is a smooth manifold at \psi(x), then it is true the rigorous sense.

Yeah, I'm struggling to follow his ideas. Maybe I should change the material ...


vector
quadratics
algebra
polynomial
Functional Analysis
real analysis
real analysis unsolved

Suppose that $T$ is a linear application on a Hilbert space such that $\langle x,T(x)\rangle\geq 0$ for all $x$. Then $T$ is continuous. Isn't it cute?

Is the Hilbert space complex or real? I have a proof that works only for complex spaces:

The important thing (if the field of scalars is $\mathbb C$) is that all $(Tx,x)$ are. This implies that for all $x$ in our Hilbert space $\mathcal H$ we have $(Tx,x)=(x,Tx)$.

Now, for any sesquilinear (that is, linear in the first variable and conjugate-linear in the second) form $\langle\cdot,\cdot\rangle$ on a complex vector space, we have $\langle x,y\rangle=\frac{1}{2}[\langle x+y,x+y\rangle-\langle x,x\rangle-\langle y,y\rangle+i(\langle x+iy,x+iy\rangle-\langle x,x\rangle-\langle y,y\rangle)]$. Applying this to the forms $\langle x,y\rangle_{1}=(Tx,y)$ and $\langle x,y\rangle_{2}=(x,Ty)$, and using $(Tx,x)=(x,Tx),\ \forall x$, we conclude that $(Tx,y)=(x,Ty),\ \forall x,y$. Now the problem is a particular case of the following more general situation:

Let $A,B: \mathcal H\to\mathcal H$ be linear transformations such that $(Ax,y)=(x,By),\ \forall x,y$. Then $A$ is continuous.

By the Closed Graph theorem, it suffices to prove that $A$ has a closed graph. Assume that $x_{n}\to x$ and $Ax_{n}\to u$. We then have $\forall y,\ (Ax_{n},y)=(x_{n},By)\to(x,By)=(Ax,y)$ on the one hand, and $\forall y,\ (Ax_{n},y)\to(u,y)$ on the other. This says that $\forall y,\ (Ax,y)=(u,y)$, which implies that $Ax=u$.

Yes it does, apparently :). Thank you. Here's the proof for real spaces:

For all $x,y\in\mathcal H$ and all (real) scalars $\alpha$, we have $0\le (T(x+\alpha y),x+\alpha y)=(Ty,y)\alpha^{2}+[(Tx,y)+(Ty,x)]\alpha+(Tx,x)$. The RHS is a quadratic polynomial in $\alpha$, so its discriminant must be non-positive. In other words, we have found that for all $x,y\in\mathcal H,\ [(Tx,y)+(Ty,x)]^{2}\le 4(Tx,x)(Ty,y)\ (*)$ holds.

Again, we employ the Closed Graph theorem. Suppose $z_{n}\to z$ and $Tz_{n}\to u$. Plug $x=z_{n}-z$ in $(*)$, and make $n\to\infty$. The LHS goes to $(u-Tz,y)^{2}$, while the RHS goes to $0$ (because $(Ty,y)$ stays constant, and $(T(z_{n}-z),z_{n}-z)\le\|T(z_{n}-z)\|\cdot\|z_{n}-z\|$ by Cauchy-Schwartz, and $T(z_{n}-z)\to u-Tz$ so in particular $\|T(z_{n}-z)\|$ stays bounded, while $z_{n}-z\to 0$). This shows that for all $y,\ (u-Tz,y)=0$, i.e. $u=Tz$.

[quote="grobber"]
Let $A,B: \mathcal H\to\mathcal H$ be linear transformations such that $(Ax,y)=(x,By),\ \forall x,y$. Then $A$ is continuous.
[/quote]

This fact is called Hellinger–Toeplitz theorem usually.


[quote="shobber"]The quadratic ineqaulity $ax^2+bx+c \geq 0$ is true for all $x \in R$. If $b>a$, then find the minimum value of $\frac{a+b+c}{b-a}$.[/quote]
What's the matter?
Let $x=1$ then $a+b+c > 0$
and $b>a$...

$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\ge 0$ for all $x\in\mathbb{R}$ iff $c-\frac{b^2}{4a}\ge 0\Rightarrow c\ge\frac{b^2}{4a}$ and $a>0$.

$\frac{a+b+c}{b-a}=\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$

Let $b=a+t$, $t>0$

\[ \frac{(2a+b)^2}{4a(b-a)}=\frac{(3a+t)^2}{4at}=g(t) \]

Keep $a$ fixed, minimum of $g(t)$ occurs when $t=3a$ ($t>0$) and minimum is 3

shobber, what is answer?  :blush:


AMC
USA(J)MO
USAMO
inequalities proposed
inequalities

It can be done in the same manner as USAMO 1997, problem 5: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=948


linear algebra
matrix
inequalities
graph theory
linear algebra unsolved

Let $G$ be a graph with $n$ vertices and $m$ edges, and suppose that $A$ is the adjacency matrix of $G$, with $\lambda_{1}\ge \lambda_{2}\ge \dots \ge \lambda_{n}$ as its eigenvalues.  Prove that $\lambda_{1}\leq \sqrt{2m-n+1}$.

This is trivially false, since we can add vertices without adding edges, and this decreases $\sqrt{2m-n+1}$ without changing any nonzero eigenvalues. If we require connectedness, there's a chance.

You're right jmerry. Assume that $G$ is also connected.
PS. As far as I remember we can even assume that $G$ does not have a single vertex. And note that this is not a simple problem. :wink:

Indeed. Some equality cases: the complete graph, the "hub and spokes" graph with one vertex connected to each other. With that much variety, there's a serious constraint on methods. The basic inequality $\lambda_{1}^{2}\le \text{tr }A^{T}A=2m$ is easy, but the improvement by $n-1$ is not.

we have $\sum_{i=1}^{n}\lambda_{i}^{2}= 2m$, so $\lambda_{1}^{2}= 2m-\sum_{i=2}^{n}\lambda_{i}^{2}$ i think it might help


A truck travels 40 miles from point A to point B in exactly 1 hour. When the truck is halfway between point A and point B, a car starts from point A and travels at 50 miles per hour. How many miles has the car traveled when the truck reaches point B?

[hide="Answer."]We know that the time it takes from halfway to B is 30 minutes, therefore the car travels for 30 minutes. Then the answer would be $50(.5)=\boxed{25}$.[/hide]


A license plate consists of 2 letters followed by 2 digits; for example, MP78. Neither the digits nor the letters may be repeated, and neither the letter "O" nor the digit "0" may be used. When reading from left to right, the letters must be in alphabetical order and the digits must be in increasing order. How many different license plate combinations are possible?

[hide="hint"]Notice you can make 2 numbers alphabetical in exactly $ \binom{9}{2}\minus{}9$ ways. Same idea for letters.[/hide]
EDIT-Forgot to add something.

[hide="Solution"]To find the number of ways to make two distinct letters alphabetical, we first find how many choices we have for the first letter, namely $ 24$.
From here, we can simply find the average number of choices for the second letter, which is $ \frac {1 \plus{} 2 \plus{} ...\plus{} 24} {24} \equal{} \frac {24*25} {2*24} \equal{} \frac {25} {2}$.

So the number of ways to make two distinct letters alphabetical is: $ \frac {25} {2}*24 \equal{} 300$.

Then, the number of ways to find two numbers between $ 0$ and $ 10$ and arrange them so that they are in increasing order is: $ 8*\frac {1 \plus{} 2 \plus{} ... \plus{} 8} {2*8} \equal{} 36$

$ 300*36 \equal{} 10,800$[/hide]

Edit: Sorry.  You guys are so mean to me.  Just because I messed up T.T.

Ermm... the no 'O' was kinda annoying -_-.

[hide]I disagree, there are 300 ways for letters. If we start with A, there are 24 others, if B, then 23...down to 1. The sum is 300. Your idea for numbers was right, so the answer should be 10,800.[/hide]

[hide]The lack of a 0 or O doesn't cause any variations int he formula; it just reduces the number of possible characters by one. There's no gap between N and P that needs accounting for; as far as the problem is concerned, O never existed. The solution that xpmath posted is correct; you get 300 possibilities for the letters and 36 for the numbers, which, when multiplied, yields 10800.[/hide]

For alphabet total possibilities $ 25\times24$. Since they must be in decreasing order half do not count.
So possibilities are $ \frac{25\times24}{2}$

For numbers same concept $ \frac{9\times8}{2}$
Total possibilites $ \frac{25\times24\times9\times8}{2\times2}\equal{}10800$


The sum of two numbers is $ S$. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$ \textbf{(A)} \ 2S \plus{} 3 \qquad \textbf{(B)} \ 3S \plus{} 2 \qquad \textbf{(C)} \ 3S \plus{} 6 \qquad \textbf{(D)} \ 2S \plus{} 6 \qquad \textbf{(E)} \ 2S \plus{} 12$

[hide]Let the 2 numbers be $x$ and $y$. So then $x+y=S$. We are trying to find $2(x+3)+2(y+3)$. This simplifies to $2(x+y)+12$, which is $2S+12$. So the answer is $\boxed{E}$. [/hide]

[hide="Answer"]$3$ is added to each of the two numbers, so we have $+6$, then each number is doubled, so we double the sum and have $2S+16\Rightarrow \boxed{E}$.[/hide]


Could  anyone  prove  the  following :

   $\sum\limits_{n=0}^{2001} [\frac{2^k}{25}] \equiv 0 (\bmod25)$

   where  [x] =  the  integer  part  of  the  $x\epsilon \Re$


inequalities
number theory unsolved
number theory

Suppose that $m$ and $n$ are positive integers, with $m>n$.  If $m^2+mn+n^2|mn(m+n)$, prove that $(m-n)^3>mn$.

Hmmm... I've found a stronger inquality, namely : $(m-n)^3 \geq 3mn$.

Let $m,n$ be two positive integers such that $m^2+mn+n^2$ divides $mn^2+m^2n$ and $m>n$.
Since $mn^2+m^2n = n(m^2+mn+n^2) - n^3$, it follows that $m^2+mn+n^2$ divides $n^3$. (1)
Let $m=da$ and $n=db$ where $a>b$ are coprime positive integers.
Then (1) leads to $a^2+ab+b^2$ divides $db^3$ and $a^2+ab+b^2$ is coprime with $b^3$. Thus $a^2+ab+b^2$ divides $d$, which leads to $a^2+ab+b^2 \leq d$.

Therefore, from AM/GM (for the last inequality), we have :
$(m-n)^3 = d^3(a-b)^3 \geq d^3 \geq d^2(a^2+ab+b^2) \geq d^2\cdot 3ab = 3mn$, as claimed.

Pierre.

You are so quick, Pierre!
I was ending my solution when you posted yours.  :)

I wrote you a pm to slow you down  :D  :D  :D

Pierre.

Nice solution!

I have one question though... since $a>b$, doesn't that mean we can't have equality in AM-GM? i.e. the equality is strict?


topology
advanced fields
advanced fields unsolved

Let $X$ be the subspace of $\mathbb{R}^{2}$ consisting of the horizontal segment $[0,1] \times \{ 0 \}$ together with all the vertical segments $\{ r \} \times [0,1-r]$ for $r$ a rational number in $[0,1]$. Show that $X$ deformation retracts to any point in the segment $[0,1] \times \{ 0 \}$, but not to any other point.

Hint given: Use the following fact:

If $X$ deformation retracts to a point $x \in X$, then for each neighbourhood $U$ of $x$ in $X$ there exists a neighbourhood $V \subset U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.

Take $x \in [0,1] \times \{ 0 \}$. Then there is a deformation retract of $X$ to $[0,1] \times \{ 0 \}$ by $f_t(x,y)=(x,(1-y)(1-t))$ and the rest is easy.

When$x \not\in [0,1] \times \{ 0 \}$:
Take any $U$ that has no point in common with $[0,1] \times \{ 0 \}$. Then assume you have such a neighborhood $V$. But this $V$ can't be nullhomotopic, it's not even path-connected (since the rationals are dense in the reals, there are at least parts of two of these rational segments contained, but not the part $[0,1] \times \{ 0 \}$ connecting them).

Let $ Y$ be the subspace of $ \mathbb{R}^2$ that is the union of an infinite number of copies of $ X$ arranged so that each copy is at alternating $ \pi/2$ angles, and $ Z$ be the subspace of $ Y$ homeomorphic to the real line. Show that $ Y$ deformation retracts in the weak sense onto $ Z$ but no true deformation retraction.

(picture at [url]http://www.math.cornell.edu/~hatcher/AT/ATch0.pdf[/url] p.18)

I do not see why there is no true deformation retraction. Isn't the map simultaeously deformation retracting each copy of $ X$ onto $ [0,1]\times\{0\}$ a deformation retraction?

For a true deformation retract, you have to fix every point of $ Z$. Assume there is such a retraction, then if restricting to $ X$ (where $ X$ means some arbitrary of these $ X$-like subparts), we get a deformation of $ X$ to $ Z \cap X$. Then again, we can deformation retract $ Z\cap X$ to any of it's points $ z$ (it's homeomorphic to $ [0,1]$!). So in total, we can deformation retract $ X$ to the point $ z\equal{}(0,1)$, which we have already shown to be impossible.


Isn't that so cool?

It's going to be exciting to see who that person is.

didn't think the people at myspace could count that high

umm..cool... :?:  myspace isn't as great as nationstates

LOOK HERE -
[url]http://profile.myspace.com/index.cfm?fuseaction=user.viewprofile&friendid=100000002[/url]


geometry
rectangle
factorial
trapezoid
3D geometry
induction
LaTeX

i'll do them later...my dad "DEMANDS ME TO GET OFF THE COMPUTER IMMEDIATELY....."

[quote="LuCy4EvA"]1. A swimming pool whose surface forms a rectangle measures 25m long by 15m wide. THe pool is 2m deep at the shallow end and the depth increases at a constant rate to 4m deep at the other end. How many liters of water will the pool hold?

2. what is the number of square units in the area of the region encosed by the graphs of the lines represented by teh equations y = x, y = -x, and y = 3?

3. Billy Bob is reading a volume of the Britannica Encyclopedia. He notices that he is on page 238. How many digits are in the book up to and including the page he is on if the book starts on page 5?

4. If 4 bums have 5 thumbs, then 2 chums have 3 thumbs. However, the thumbs of bums and chums sometimes get mixed up with gums. If a bum's thrumbs get mistaken for his gums 1/5 of the time, and if a chum's thumbs get mistaken for gums 1/3 of the time, find the sum of the bums' and chums' thumbs that will be mistaken for gums if there are 8 bums and 10 chums.

5. Find the sum of the first 15 perfect squares

6. It takes Adam five hours to solve a certain math problem alone. It takes Vicky four hours to work the same problem alone. Vicky  works on the problem alone for one hour before Adam joins her. How many hours does it take them to finish the problem together?

7. Coach Pratt lined his students up in a row. He said to Sally, "Pick the 12th and 13th kids in the line for they will be today's team captains". sally replied, "Coach, from which end do I start?" "Sally," replied Coach, 'It does not matter." How many kids were in line?

8. If 387,a42 is divisible by 3, then how many possible values of a are there?

9. Give that n factorial ends in exactly 27 zeros, and n is a multiple of three, what is the largest possible value of n?
a. 120    b. 132     c. 118     d. 117

10. If w is increased by 400% of w and the result is decreased by 80% (of the result) then 54 is the final value. What is 80% of w?[/quote]

AAAH! Chatspeak! *burns retinas*

Hints:

[hide="1"]
Think of the average depth.
[/hide]

[hide="2"]
Triangle.
[/hide]

[hide="3"]
Think before you subtract.
[/hide]

[hide="4"]
What the...
It's not too hard, just don't kill yourself trying to get the gist of the problem.
[/hide]

[hide="5"]
Calculator time!
[/hide]

[hide="6"]
Well, you could draw a diagram. Or...

Combined rate formula:

1/x + 1/y = 1/z

x = hours taken by person 1
y = hours taken by person 2
z = total hours
[/hide]

[hide="7"]
Draw a diagram or use simple arithmetic methods.
[/hide]

[hide="8"]
If the sum of the digits is divisible by 3, then the number is divisible by 3.
[/hide]

[hide="9"]
The number of 0s n! ends in is the largest possible value of x, where (n!)/(5^x) is an integer.
[/hide]

[hide="10"]
Just read it through. Remember, 100% is a whole.
[/hide]

[quote="3X.lich"][quote="Treething"][quote="LuCy4EvA"]...
5. Find the sum of the first 15 perfect squares
...[/quote]

AAAH! Chatspeak! *burns retinas*

Hints:...
[hide="5"]
Calculator time!
[/hide]
... [/quote]
There is another way to this problem without resorting to your 'artificial method'. But the difficulty seems a little off for Middle School people.[/quote]

can you tell me what it is and try to explain it?

[quote="LuCy4EvA"]can you tell me what it is and try to explain it?[/quote]

There are nice formulas for the sum of the positive integers, $1 + 2 + 3 + \ldots + n$, for the sum of squares, $1^2 + 2^2 + 3^3 + \ldots + n^2$, and for the cubes, $1^3 + 2^3 + 3^3 + \ldots + n^3$.  To prove them requires skills that you probably won't learn until high school -- proof by induction is the simplest way.  The formulas for the integers and the squares are as follows:

[hide]$1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$
(so, for example, $1 + 2 + 3 + \ldots + 15 = \frac{15(15+1)}{2} = 15\cdot8 = 120$)
$1^2 + 2^2 + 3^3 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$
(so, for example, $1^2 + 2^2 + 3^3 + \ldots + 8^2 = \frac{8(8+1)(2\cdot8+1)}{6} = \frac{8(9)(17)}{6} = 4 \cdot 3 \cdot 17 = 204$
[/hide]

For the cubes, see if you can come up with the formula yourself by looking for a relationship between the pairs of numbers $1 + 2$ and $1^3 + 2^3$, $1 + 2 + 3$ and $1^3 + 2^3 + 3^3$, $1 + 2 + 3 + 4$ and $1^3 + 2^3 + 3^3 + 4^3$ and so on.

For this problem, it's worth noting that the first 15 squares are all pretty small numbers -- $15^2$ is only 225.  So it's not too difficult just to list out the numbers and add them up.  You might be less likely to make a mistake adding then you would putting so many things into a calculator.

*Edited to make Latex legible.

[quote="JBL"][quote="LuCy4EvA"]can you tell me what it is and try to explain it?[/quote]

There are nice formulas for the sum of the positive integers, $1 + 2 + 3 + \ldots + n$, for the sum of squares, $1^2 + 2^2 + 3^3 + \ldots + n^2$, and for the cubes, $1^3 + 2^3 + 3^3 + \ldots + n^3$.  To prove them requires skills that you probably won't learn until high school -- proof by induction is the simplest way.  The formulas for the integers and the squares are as follows:

[hide]$1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$
(so, for example, $1 + 2 + 3 + \ldots + 15 = \frac{15(15+1)}{2} = 15\cdot8 = 120)$
$1^2 + 2^2 + 3^3 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$
(so, for example, $1^2 + 2^2 + 3^3 + \ldots + 8^2 = \frac{8(8+1)(2\cdot8+1)}{6} = \frac{8(9)(17)}{6} = 4 \cdot 3 \cdot 17 = 204
[/hide]
For the cubes, see if you can come up with the formula yourself by looking for a relationship between the pairs of numbers $1 + 2$ and $1^3 + 2^3$, $1 + 2 + 3$ and $1^3 + 2^3 + 3^3$, $1 + 2 + 3 + 4$ and $1^3 + 2^3 + 3^3 + 4^3$ and so on.
For this problem, it's worth noting that the first 15 squares are all pretty small numbers -- $15^2$ is only 225.  So it's not too difficult just to list out the numbers and add them up.  You might be less likely to make a mistake adding then you would putting so many things into a calculator.[/quote]

sorry, but I can't see your formula for the sum of the perfect squares,  your latex is not showin...

[quote="JBL"]Fixed in original post.  Sorry about that.  (By the way, it usually isn't necessary to quote an entire long post like that when you are the first person responding to it -- it end up filling up a lot more space than is necessary.)[/quote]
how do u quote only part of the message?

[quote="JBL"]Fixed in original post.  Sorry about that.  (By the way, it usually isn't necessary to quote an entire long post like that when you are the first person responding to it -- it end up filling up a lot more space than is necessary.)[/quote]

don't know if it's the computer's fault or what, i'm just not seeing the formula for perfect squares.... I can see everything else tho. i'm sooo sorry, maybe I'll look it up my self on something else.

Delete the part you don't want.  When you hit "quote," it copies the entire text of the message you are quoting into a new message, surrounded by the tags [code][quote="username"] TEXT [/quote][/code].  So, for example,[quote="JBL"]This says that I am quoting myself, even though I've never written it before.  That's because I put in the quote tags and then typed these two sentences in between them.[/quote]
You can then go back and delete the parts of things you quoted that aren't relevant.  Also, it means you can break the text you are quoting into more than one part by writing your own tags.

Also, of course, you can always respond without quoting anyone.

I don't know why it isn't appearing for you -- I can see it just fine.  I'll put here again, just in case:
[hide]$1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$[/hide]

thank you, now it's appearing  :D  :)  :D  :)  :D  :lol:  :lol:  :D  :lol:  :)  :blush:

[quote="LuCy4EvA"][quote="RC-7th"][hide="1"]Let the base of the solid be the trapezoid formed by the shallow and deep end, the surface of the water, and the bottom of the pool. The height is then 25, and the average of the bases is 3. 25*3=75. Then 75*15=1125[/hide][/quote]

that's what I would put, but the correct answer was 1125000 for some reason, how many liters are in a meter?[/quote]

[hide="Hint"]A meter is 10 decimeters. One cubic decimeter is equal to one liter. 1125 cubic meters=1125000 cubic decimeters.[/hide]

For a non-induction proof of the formulas JBL mentioned, go here: [url]http://www.usamts.org/Gallery/G_Gallery.php[/url]

[quote="lucy"]3. Billy Bob is reading a volume of the Britannica Encyclopedia. He notices that he is on page 238. How many digits are in the book up to and including the page he is on if the book starts on page 5? [/quote]

[hide="this is my short answer to number 3"]there are 5 one digit numbers
there are 90 2 digit numbers
there are 138 3 digit numbers
5*1+90*2+138*3=599[/hide]

[quote="lucy"]6. It takes Adam five hours to solve a certain math problem alone. It takes Vicky four hours to work the same problem alone. Vicky works on the problem alone for one hour before Adam joins her. How many hours does it take them to finish the problem together? [/quote]

[hide="this is my answer"]1/5+(1/4+1/5)x=1
9x/20=4/5
45x=80
x=16/9 hours[/hide]

[quote="lucy"]7. Coach Pratt lined his students up in a row. He said to Sally, "Pick the 12th and 13th kids in the line for they will be today's team captains". sally replied, "Coach, from which end do I start?" "Sally," replied Coach, 'It does not matter." How many kids were in line? [/quote]

[hide="my answer"]they must be in the middle
there is 11 before 12
there is 11 after 13
13+11=24[/hide]

[quote="lucy"]8. If 387,a42 is divisible by 3, then how many possible values of a are there? [/quote]

[hide="answer"]26+a/3= a whole number
a=1,4,7
there are....3 values!![/hide]

[quote="lucy"]10. If w is increased by 400% of w and the result is decreased by 80% (of the result) then 54 is the final value. What is 80% of w?[/quote]

[hide="ans"]w=54
[b]43.2[/b][/hide]

[quote="jli"][quote="lucy"]2. what is the number of square units in the area of the region encosed by the graphs of the lines represented by teh equations y = x, y = -x, and y = 3? [/quote]

[hide="this is my short answer to problem number 2"]the height is 3
the base is 6
3*6/2=9[/hide][/quote]

how did u get that the base is 6?

let me say this to all of you

this problem can be tricky

watch out for liter part

1000 liters is 1 cubic meter

kinda unbelievealbe, this is for those of you who aren't familiar with liters and meters

[there are 5 one digit numbers
there are 90 2 digit numbers
there are 138 3 digit numbers
5*1+90*2+138*3=599 ]

shouldn't there be 139 3-digit numbers => 238-100+1 = 139


Diophantine equation
number theory unsolved
number theory

hi,
find all squares in the sequence :
a_m = (m+1)*m /2

It's equivalent to $ (2m\plus{}1)^2 \minus{} 8x^2\equal{}1$. Now see Pell-equations.

[quote="ZetaX"]It's equivalent to $ (2m \plus{} 1)^2 \minus{} 8x^2 \equal{} 1$. Now see Pell-equations.[/quote]
i dont get it.
pell-equations doesnt sound nice anyway...

[quote="ZetaX"]As it is equivalent to a Pell-equation, you can't avoid it  :wink:
The minimal solution of $ a^2 \minus{} 8b^2 \equal{} 1$ is $ a \equal{} 3,b \equal{} 1$. This gives that all solutions are of type $ a \plus{} b \sqrt 8 \equal{} \pm (3 \plus{} \sqrt 8 )^k$. Now independent from $ k$, $ a$ is always odd, so we can find $ m$ such that $ a \equal{} 2m \plus{} 1$.[/quote]
so your solution is .. ?
i would like to have a explicit formulae for all solutions(squares).

\[ a\plus{}b\sqrt 8 \equal{} \\
\equal{}\pm (3\plus{} \sqrt 8)^k \equal{} \\
\equal{} \pm \sum_{0 \leq i \leq k} \binom{k}{i} \sqrt 8^i 3^{k\minus{}i} \equal{} \\
\equal{} \pm\left( \sum_{0 \leq 2i \leq k} \binom{k}{2i} 8^i 3^{k\minus{}2i} \plus{} \sqrt 8 \cdot \sum_{0 \leq 2i\plus{}1 \leq k} \binom{k}{2i\plus{}1} 8^i 3^{k\minus{}2i\minus{}1} \right).\]

Thus \[ a\equal{} \pm \sum_{0 \leq 2i \leq k} \binom{k}{2i} 8^i 3^{k\minus{}2i} \equal{} \pm 3^k  \sum_{0 \leq 2i \leq k} \binom{k}{2i} \left(\frac{8}{9}\right)^i.\]

[quote="ZetaX"]
\[ a \plus{} b\sqrt 8 \equal{} \\
\equal{} \pm (3 \plus{} \sqrt 8)^k \equal{} \\
\equal{} \pm \sum_{0 \leq i \leq k} \binom{k}{i} \sqrt 8^i 3^{k \minus{} i} \equal{} \\
\equal{} \pm\left( \sum_{0 \leq 2i \leq k} \binom{k}{2i} 8^i 3^{k \minus{} 2i} \plus{} \sqrt 8 \cdot \sum_{0 \leq 2i \plus{} 1 \leq k} \binom{k}{2i \plus{} 1} 8^i 3^{k \minus{} 2i \minus{} 1} \right).
\]
Thus
\[ a \equal{} \pm \sum_{0 \leq 2i \leq k} \binom{k}{2i} 8^i 3^{k \minus{} 2i} \equal{} \pm 3^k \sum_{0 \leq 2i \leq k} \binom{k}{2i} \left(\frac {8}{9}\right)^i.
\]
[/quote]
its ok , i was confused because i actually searched b^2 not m.
a recursive is given by (a_n = n. square in the sequence)
a_0 = 0 a_1=1
a_(n+2) = 34 a_(n+1) - a_n +2


algebra
polynomial
linear algebra
matrix

Let $ A\in M_{n}(\mathbb{R})$ such that $ A^{2006}=A$ and $ \text{rank}(A)=1$. Prove that $ |I-A|=0$.

Since $ A^{2006}=A$, we know the minimal polynomial of $ A$ divides $ \lambda^{2006}-\lambda$.  This means $ A$ can only have eigenvalues of $ 0$ and/or $ 1$.  Since the $ \text{rank}\hspace{1 mm}A=1$, we must have that $ A$ has one eigenvalue of $ 1$ with the rest being $ 0$ and no Jordan blocks since $ \text{rank}\hspace{1 mm}A=1$.  (If they were all zero with one $ 2\times 2$ Jordan block, the matrix would be nilpotent).  Therefore there exists a matrix $ C$ such that $ C^{-1}AC=D$ where $ D$ is diagonal with all zeros and a one.

Now $ \det(I-A)=\det(C^{-1}) \det(I-A) \det(C)=\det(I-C^{-1}AC)=\det(I-D)=0$.

I think the condition must be $ A \in M_{n}(\mathbb R)$

If not,consider $ a_{11}=t;a_{ij}=0$,otherwise.
where $ t$ is any root of $ x^{2005}=1$,but $ t \neq 1$

Zhaobin, you are correct.  In my proof above, I was obviously assuming the real field.


[quote="ajai"]For how many integer values of x does the expression $x^{2}+2x-19$ have a negative value?[/quote]

[hide]$x^{2}+2x-19=(x+1)^{2}-20$

Therefore, there are $\boxed{9}$ values.[/hide]

[hide]Completing the square, we get $x^{2}+2x-19=(x+1)^{2}-20$, and $9$ values of x make $(x+1)^{2}-20$ negative.
[/hide]

[color=blue][size=75]kyyuanmathcount: Please remember the hide tags.[/size][/color]

[hide]This can be written as $(x-1)^{2}-20$.  There are nine values that make this negative, particularly, -3,-2,-1,0,1,2,3,4,5.

$9$[/hide]

[quote="usaha"]dont u mean -5,-4.., 2,3 and u can just see that x^2 + 2x has to be less than 19 and u cna tell thats 9 terms[/quote]
Yes, mathgenius completed the square wrong. :D

This one is actually pretty easy to do just by using guess and check...  :maybe:


Find all nonnegative pairs (x,y) such that
$x^{3}+8x^{2}-6x+8 = y^{3}$

It's easy to check that \[x^{3}< x^{3}+8x^{2}-6x+8 < (x+3)^{3}= x^{3}+9x^{2}+27x+27\] holds for all non-negative integers. Therefore, $x < y < x+3$ and we just have to solve two cases: $y=x+1$ and $y=x+2$. Plug that in in $y^{3}= x^{3}+8x^{2}-6x+8$, solve, and find $(x,y) = (1,3),(9,11)$, if I didn't make any mistake.

$(1,3)$ is not a solution. It should be instead $(0,2)$. You can create stricter bounds where it only gives you $y=x+2$.

Masoud Zargar


What do you predict the cutoff for nationals in FL will be?
Remember to consider the difficulty of chapters.
Last year's cutoff was a 35, for reference.
NOTE: If you have been to states mathcounts already, don't vote.

then again, many 8th graders who were good last year have left

[quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]i say 34.  test will be harder i predict, but competition is fiercer.  so only at little less.[/quote]
Why do you say competition is fiercer?
Last year Sam Lim, Philip Bauer, and Brynmor Chapman were there.[/quote]

yeah, good point.  but i think people have gotten alot smarter in general. idk.  im still saying 34.[/quote]
I also forgot Alex Anderson, Kushal, and Gene.
I personally predict a 33, since I think this year's will be harder.

[quote="bpms"][/quote][quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]i say 34.  test will be harder i predict, but competition is fiercer.  so only at little less.[/quote]
Why do you say competition is fiercer?
Last year Sam Lim, Philip Bauer, and Brynmor Chapman were there.[/quote]

yeah, good point.  but i think people have gotten alot smarter in general. idk.  im still saying 34.[/quote][quote="bpms"]
I also forgot Alex Anderson, Kushal, and Gene.
I personally predict a 33, since I think this year's will be harder.[/quote]

that'd be cool and all...but i am a junior in high school from illinois...i was not interested in math when i was eligable for mathcounts...

[quote="Altheman"][/quote][quote="bpms"][/quote][quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]i say 34.  test will be harder i predict, but competition is fiercer.  so only at little less.[/quote]
Why do you say competition is fiercer?
Last year Sam Lim, Philip Bauer, and Brynmor Chapman were there.[/quote]

yeah, good point.  but i think people have gotten alot smarter in general. idk.  im still saying 34.[/quote][quote="bpms"]
I also forgot Alex Anderson, Kushal, and Gene.
I personally predict a 33, since I think this year's will be harder.[/quote][quote="Altheman"]

that'd be cool and all...but i am a junior in high school from illinois...i was not interested in math when i was eligable for mathcounts...[/quote]
Believe it or not, there was a different Alex Anderson on the FL nats team last year. When I saw the FL teams I was like wtf, how is he on the B team I was better than him last year (barely), then I found out that your name was Alex Anderson too.

ok haha

The message is too small. Please make the message longer before submitting.

[quote="bpms"][/quote][quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]i say 34.  test will be harder i predict, but competition is fiercer.  so only at little less.[/quote]
Why do you say competition is fiercer?
Last year Sam Lim, Philip Bauer, and Brynmor Chapman were there.[/quote]

yeah, good point.  but i think people have gotten alot smarter in general. idk.  im still saying 34.[/quote][quote="bpms"]
I also forgot Alex Anderson, Kushal, and Gene.
I personally predict a 33, since I think this year's will be harder.[/quote]

kushal got 26th last year.  :|

i beat him.  :rotfl:

[quote="Walk Around The River"][/quote][quote="bpms"][/quote][quote="Walk Around The River"][quote="bpms"][quote="Walk Around The River"]i say 34.  test will be harder i predict, but competition is fiercer.  so only at little less.[/quote]
Why do you say competition is fiercer?
Last year Sam Lim, Philip Bauer, and Brynmor Chapman were there.[/quote]

yeah, good point.  but i think people have gotten alot smarter in general. idk.  im still saying 34.[/quote][quote="bpms"]
I also forgot Alex Anderson, Kushal, and Gene.
I personally predict a 33, since I think this year's will be harder.[/quote][quote="Walk Around The River"]

kushal got 26th last year.  :|

i beat him.  :rotfl:[/quote]
He got 9th the year before :|
He had a bad day last year.

[quote="mathgeniuse^ln(x)"]I don't see how that is funny.

The cutoff in any state is going to be 35 in my opinion.[/quote]

its funny for two reasons...

1. he was in 8th, i was in 7th
2. i was EXTREMELY stupid in 7th grade, and im not that smart now, if that says anything

[quote="diophantient"][/quote][quote="ragnarok23"][/quote][quote="diophantient"][/quote][quote="ragnarok23"][quote="Walk Around The River"][quote="Klebian"]Hey, TX 5th place got a 41 last year.[/quote]

yeah, well, ckck is stupid, so....

 :rotfl:

jkjk chris.

but yeah were no geniuses, so we have the right to be nervous. except pratik [ragnarok] he doesnt. cause hes smarter than us.

 :ninja:[/quote]

???

i am thoroughly and utterly confused by your statement sir[/quote][quote="diophantient"]
We all bow down to your beastliness.[/quote][quote="ragnarok23"]
you sir, are quite strange[/quote][quote="diophantient"]
Dude, if you don't get 1st in FL I'll be utterly shocked.[/quote]

I have the strange suspicion that you are under the effects of crack

[quote="LawOfSigns"]Hmm...just did last practice test.  Was the 2002 state easier/harder than the 2006 state (should I lower/raise my 2002 score to get my expected placing this year, and by how much), or does anyone know the cutoffs in FL in 2002?

Are Preet and Pratik their real first names? I'llbewatchingattheCDroundmycomputerisweird

ong pwnzorz, i don't have school tomorrow and neither does our team![/quote]

i dont remember 2002, but id say this yr ull get lower just cause...

2002 wasnt terribly hard, i would say 06 is harder

After some in depth research, as in typing his name into google and clicking on the first link, i found out liam got 45th at the national spelling bee last year and made it again this year.

math and spelling.  smart dude.

somehow i found out that 45th place got a 25.  :|

I was really nervous and bombed it (according to my expectations) with a 26...got 39th.  :o  I looked over the solutions and it seems that I easily could have had an 18-20 sprint and 12-16 target (expecting more specifically a 19/14, I know 4th got 33 and I would have been horrible at tiebrakers so it doesn't seem as if I ruined a chance for the nationals) I got a 37 on the 2002, btw...at least no one from my school beat me at states. The 6th place individual in my chapter got 5th in state!...wow.  (she got 1st in chapter last year- would have been really bad to have not made it to state as an 8th grader, but as a 7th grader...nice to have a strong team to help you, but you all didn't need that ;) :o)  She also said the state test was easier than chapters, and the other people I beat at chapters were ranking  in the teens.  Very surprising year...at least I improved from 7th to 8th. (had I not ranked 1 below the state cutoff at chapters last year, I wonder if I would have had 38th or better as a 7th grader)

I had top 10 in the city as a 6th grader in the spelling bee, but due to extreme lack of talent and interest among most all students, there is no spelling bee at my middle school (darn)

[quote="mathgeniuse^ln(x)"]I don't see how that is funny.

The cutoff in any state is going to be 35 in my opinion.[/quote]

south dakota's was 22 i believe, for last year.

I heard it was 24 (!) in north dakota last year.

wow...22...you don't even have to get better than 50%.

[quote="bpms"][quote="indianamath"]guess what? i know who bpms is!!! :D[/quote]
guess what, I do too[/quote]

DO you have two accounts?

What were the top 4 scores?  I know 4th got 33.

[quote="pab92"][quote="jhredsox"]you didn't get at least a forty bpms?

i have states tomorrow

based on chapter, CT will probably be a 32ish (we had 35 last year)

you guys have eli though

and preet

etc.

wait, didn't the highest scorer (i thought it was bauer) come from FL last year (for states)

-jorian[/quote]

im Philip Bauer from FL and somehow i had no idea...and im too stupid to find old standings on my own...can someone give me a link to FL 06 standings or does anyone at least know what my score was?

btw...yeah im new. i just stumbled across this googling myself[/quote]
I was on your team that year and IIRC you got a 40.

Venkatramaanvaas Govindaswamisherma?  :maybe:


geometry
geometric transformation
reflection
analytic geometry

Find the length of the shortest path from the point (3,5) to the point (8,2) that touches the $x$-axis and also touches the $y$-axis.

[hide="hint"]what happens if you reflect the points over the axis[/hide]

[hide="Answer"]It must travel $|-3|+8=11$ units to the sides so that it can touch the $y$-axis then reach the $x$-coordinate of the second point, and $|-5|+2=7$ units to touch the $x$-axis and then the $y$-coordinate of the second point, so we have $\sqrt{7^2+11^2}=\boxed{\sqrt{170}}$.[/hide]


ratio
real analysis
limit
calculus
calculus computations

I was reading through Rudin and saw that basically the root test was a more "powerful" test for convergence than the ratio test. However, in the examples he showed a series where $ \lim\sup\left(\frac{a_{n\plus{}1}}{a_n}\right)\equal{}2$ but $ \lim\sup \sqrt[n]{a_n}\equal{}\frac{1}{2}$.

the series is $ \frac{1}{2}\plus{}1\plus{}\frac{1}{8}\plus{}\frac{1}{4}\plus{}\frac{1}{32}\plus{}\frac{1}{16}\plus{}\frac{1}{128}\plus{}\frac{1}{64}\plus{}\dots$

I don't see how the ratio test doesn't apply here could someone please explain it to me.   :oops:

Stolz-Cesaro.

What  is the general term of that though?

[quote="jmerry"]The ratio test says "inconclusive" in this case, because some of the ratios are less than 1 and others are greater than 1 no matter how far out you go. It does apply, but you get no information. The root test effectively filters out the multiplication by $ 1,4,1,4,\cdots$ and sees the underlying exponential decay.[/quote]

Ahh I see now. Thanks. I guess I had misread the assumption for divergence with the ratio test.


A, B, C paint a long line of fence-posts. A paints the first, then every ath, B paints the second then every bth, C paints the third, then every cth. Every post gets painted just once. Find all possible triples (a, b, c).

I got [hide]3,3,3[/hide] but I don't know about the rest...

hmm, thats a good start

but theres more...

there's also (4,2,4). I can't prove the nonexistence of any others though....

Between them, they have to paint all the fences. What does this say about a, b, and c? Can you find any relation at all among them and then get a few cases that you can check?

Just think about who [hide] paints the 4th fence[/hide].

That works too. I was trying to hint that 1/a+1/b+1/c=1, since each fence must be painted.

Take any fence. The probability that the first guy painted it will be 1/a, the probability that the second guy painted it will be 1/b, and the probability that the third guy painted it will be 1/c. Since those are the only guys painting the fence, and the probabilities need to add up to 1, there you go.

Ok, with your hint, MMM, i think (3,3,3) and (4,2,4) are the only solutions. Let me think a bit longer though...

yup...those are the only solutions...none of a, b, or c can be 1, since that would force the other two to be 0, and I think they all have to be positive integers. Okay, now, c cannot paint #4 since that would make c=1. So now, either a or b paints #4. If a does, that makes a=3. now, that also means a paints #7. Consider who paints #5. If c does, that makes c=2, and that means c will also paint #7, which is BAAAAAD. So, we have b painting #5, which makes b=3. We easily find c=3. This gives the triple (3,3,3).

If b paints #4, that makes b=2. Now, if c paints #5, a will have to be 0, which isn't allowed. So a paints #5 which makes a=4, and we easily find c=4.

Therefore, the only possible two triples, corresponding to whether a or b paints #4, are (3,3,3) and (4,2,4).


I'm unable understand the equation of the work doe by a gas in irreversible process...
books say that W=P(external) * dV.....
and the explanation that they give is as follows...
workdone =force*displacement
              =Pressure*(change in volume)
however the pressure is the pressure of the gas...and not the external pressure...

then how is the equation
W =P9external) *dV valid?....

In general, the work done on an irreversible process cannot be computed with thermodynamics. However, this can be done if we suppose that the irreversible process is made in the following way: suppose we want to find the work when the volume of a gas inside a cylinder is suddendly (and so irreversibly) increased by pushing out the piston until its external pressure is $ P_{ext}$. Now, the process is irreversible because the pressure outside the piston was instantaneously dropped, and the gas inside the cylinder will not be in equilibrium - turbulence, temperature gradients, and pressure gradients will develop - and so $ P_{ext} \neq P_{gas}$. However, in this process the pressure ($ P_{ext}$) is constant, and so what was the energy necessary to push the piston? That was $ P_{ext} \Delta V$. And what was the work done by the gas? Well, that was $ \minus{} P_{ext} \Delta V$. For an infinitesimal process, $ \delta w \equal{} \minus{}P_{ext} dV$.

[quote="Carcul"]In general, the work done on an irreversible process cannot be computed with thermodynamics. However, this can be done if we suppose that the irreversible process is made in the following way: suppose we want to find the work when the volume of a gas inside a cylinder is suddendly (and so irreversibly) increased by pushing out the piston until its external pressure is $ P_{ext}$. Now, the process is irreversible because the pressure outside the piston was instantaneously dropped, and the gas inside the cylinder will not be in equilibrium - turbulence, temperature gradients, and pressure gradients will develop - and so $ P_{ext} \neq P_{gas}$. However, in this process the pressure ($ P_{ext}$) is constant, and so what was the energy necessary to push the piston? That was $ P_{ext} \Delta V$. And what was the work done by the gas? Well, that was $ \minus{} P_{ext} \Delta V$. For an infinitesimal process, $ \delta w \equal{} \minus{} P_{ext} dV$.[/quote]

Thanks a lot for your reply!!
P(external)*(volume change) is the work done by the atmosphere on the piston.So that much energy is required to push the piston isn't it?
so how can we conclude that the work done by the gas would be exactly the same as the energy required?...it could also be more?

is the equation an approximate one?


How many different triangles are there if all the sides are integer lengths and the perimeter is 48.

Can you please tell me how you got that many solutions?
If you need the answer here it is, I need an explanation on
how you got all the solutions.

[hide]The answer is 48.[/hide]

[quote="bubble"]using casework, 65[/quote]You probably forgot the triangle inequality.

We need $a+b+c = 48$ and $a+b>c$ and $1\le a\le b\le c$. This is equivalent to $a+b+c=48$ and $1\le a\le b\le c<24$.
[code] c   the number of (a,b)
23   11
22   10
21    8
20    7
19    5
18    4
17    2
16    1[/code]
Total = 48


Let $ m\geq 1$ and $ n\geq 1$ two integers such as $ \sqrt{7}\minus{}\frac{m}{n} > 0$. Prove that $ \sqrt{7}\minus{}\frac{m}{n} >\frac{1}{mn}$

Let $ m\equal{}kx, n\equal{}ky$ for coprime $ x,y$, then $ \frac{1}{xy} \ge \frac{1}{mn}$, so we only need to prove the case that $ m,n$ coprime.

Assume that there exist $ m,n$ satisfy the condition and $ \frac{m}{n}\plus{}\frac{1}{mn}>\sqrt{7}$ (equality can't hold because $ \sqrt{7}$ is irrational),
then $ \frac{(m^2\plus{}1)^2}{(mn)^2}>7>\frac{m^2}{n^2}$

The inequality on the right gives $ 7n^2>m^2$, as $ m,n$ coprime, we have $ 7n^2 \ge m^2\plus{}3$ by mod 8 .
The inequality on the left gives $ (m^2\plus{}1)^2>7(mn)^2 \ge (m^2)(m^2\plus{}3) \ge (m^2\plus{}1)^2$, impossible, done.

Suppose there are such integers, then
$ \frac{m}{n} \plus{} \frac{1}{mn} > \sqrt{7} > \frac{m}{n}$
$ m^2\plus{}2\plus{}\frac{1}{m^2} > 7n^2 > m^2$.
So we either have $ 7n^2\equal{}m^2\plus{}1$ or $ 7n^2\equal{}m^2\plus{}2$
The first is false by mod 4, the second by mod 16.


I need to rearrange the following:

h = 0.003LV2 / 2dg

so that V is the subject of the formula.
then find its value when:
h = 0.614,
 L = 168,
 d = 0.25
and g = 9.81

Any help would be much appreciated.

Why can't you just substitute all the variables first? If you do want to make V the subject of the equation, multiply or divide both sides of the equation by all the other variables so that only V is on one side.


The path is the unit circle, repeated 49 times. The integrand (which can also be written as $\coth(z)$) has poles at the points $k\pi i,\ k\in\mathbb{Z}.$ Of those, only the origin lies inside the curve. The singularity at the origin is a simple pole with residue $1.$ The winding number of the curve about that pole is $49.$ Hence by the residue theorem, the answer is

$2\pi i\cdot49\cdot1=98\pi i.$


I have no idea why you think such a thing would have a closed-form expression.

http://www.research.att.com/~njas/sequences/A003101 is the sequence at OEIS

[quote="miyomiyo"]$\sum_{n=1}^{k}n^{k-n}$[/quote]
The expression does not work for the $a_{k}$ term.
It it did then $a_{3}= 1^{3-1}+2^{3-2}+3^{3-3}= 1+2+1 = 4 \not= 8$
It should be $\sum_{n=1}^{k}n^{k+1-n}$
EDIT Change to $k+1-n$

Try again: you "fixed" the wrong thing.  (Note that the sum of the base and exponent in each summand for $a_{k}$ should be $k+1$.)


Denote S_j the average of all possible products of a_1,a_2,..,a_n taken j at time. Prove the three inequalities

1) (S_1) 2  \geq S_2

2) (S_k) 2  \geq (S_k-1)(S_k+1)

3) (S_k)^(k+1) \geq (S_k+1)^k

Oh yeah... symmetric means inequalities, this is really well-known, and in Shklarsky / Chenzov / Yaglom there is a proof without analysis... but I've never had the patience to read this proof.

  Darij

another one in the solved problems section and the problem has no solution....  :?  :maybe:


trigonometry
geometry
trig identities
Law of Sines

Let ABC be a triangle satisfying that 3A+2B=180
Prove that $\ AB^2-BC^2=AB.AC$ :lol:

$AB=c, BC=a, AC=b$
prove: $(c+a)(c-a)=bc$
given: $3a+2b=180$
so, $b=90-\frac32 a$
$(c+a)(c-a)=90c-\frac32 ac$
It doesn't seem like there is enough information...

EDIT: oh wait, is that angles?
then we have
$3\angle A+2\angle B=180$
we have by law of sines $\frac{a}{sinA}=\frac{b}{sinB}=\frac{b}{90-\frac32 A} \implies b=\frac{a\cdot cos(\frac32 A)}{sinA}$
etc. etc. I'm tired maybe tomorrow...


quadratics
Diophantine equation
number theory proposed
number theory

Finally I solve a problem proposed by Lhvietbao!!
Suppose x<y.
x(y+1)=(u+v)^2/4
(x+1)y=(u-v)^2/4
Then y-x=uv so
x(x+uv+1)=(u-v)^2/4.
Regardimg as a quadratic in x discriminant is perfect square so (u^2+1)(v^2+1) is PS
u^2=ds^2-1 v^2=dt^2-1.
So (u,s),(v,t) are solutions to a generalised Pell Equation and so
u=((m_1+n_1e)(m_0+n_0e)^k-(m_1-n_1e)(m_0-n_0e)^k)/2 and other similar
3 equatios for s,v,t
Note: e^2=d, m_0,n_0 are primitive solutions for equation x^2-dy^2=1
m_1,n_1 are primitive solutions for equation x^2-dy^2=-1
Then a routine computation shows
x=((m_0-n_0e)^(k-l)+(m_0-n_0e)^(k-l))/4-1/2.
Since u,v have same parity, so do k,l
Then routine computation solve our problem.

NOTE: Since x(y+1) y(x+1) are perfect squares x=a^2b y+1=c^2b x+1=u^2v
y=w^2v(b,v-square-free)
Replacing u,v by m,n we get
mx^2-ny^2=1,-1 has solutions in integers.
Since the problems are equivalent we solved the following theorem:
If mx^2-ny^2=1,-1 both have integer solution, then m or n is 1

The problem proposed by Lhvietbao : Suppose x,y are positive integers such that both x(y+1) and y(x+1) are squares. Prove that exactly one of the x,y is a perfect square.

After I read the solution of [size=100][size=150]iura[/size][/size], i have a question  : find all pairs (x, y) st that hypothesis ? What do you think ?[/quote]

[quote="nntrkien"]i have a question  : find all pairs (x, y) st that hypothesis ? What do you think ?[/quote]

Yes, it might be interesting. :)

Well, at least, we have : $(x, y) = (1, 8)$.  :P

EDIT: After all, it's not that interesting. Indeed, there are an infinity of solutions...


The square root of $2x+2\sqrt{x^2-1}$ is.....(in terms of x...to be more specific like $\sqrt{a}+ or - \sqrt{b}$

[hide="my method 1"]The equation can be writtten as
$x+1+2\sqrt{(x+1)(x-1)}+x-1$
So ,sqroot is :-

$\sqrt{x+1}+\sqrt{x-1}$

Mthod 2:-
Setting the equation as follows:-

$\sqrt{2x+2\sqrt{x^2-1}}=\sqrt{a}+\sqrt{b}$
squaring(boy too much latex for me... :D )
$L.H.S=a+2\sqrt{ab}+b$
now after this i'm not too sure to do.....
 $x^2-1=ef$....(?)
$(x+1)(x-1)=ef$
now what,directly equate them and get the answer?
or else again write he equation as:-
$x+1+2\sqrt{(x+1)(x-1)}+x-1$ and then directly equate to get a and b?[/hide]

Please correct me on my second method.

P.S-after 600+posts,this has been the most latex I've ever written in any post :D
 :D  :D  :D  :P

[hide]The square root is defined for $x^2\geq 1 \land x+\sqrt{x^2-1}\geq 0$, which sums up to $x\geq 1.$ In that case
$2x+2\sqrt{x^2-1}=(x-1)+(x+1)+2\sqrt{x-1}\sqrt{x+1}=(\sqrt{x-1}+\sqrt{x+1})^2\implies \sqrt{2x+2\sqrt{x^2-1}}=\sqrt{x-1}+\sqrt{x+1}$[/hide]

[b]Prove that[/b] $\boxed {\ c\ge 0,\ a^2-b^2c=d^2\Longrightarrow \sqrt{a+b\sqrt c}=\sqrt{\frac{a+|d|}{2}}+(sgn\ b)\cdot \sqrt{\frac{a-|d|}{2}}\ }\ .$
[b]Examples.[/b]
$\blacktriangleright\ 1^{\circ}.\ x\ge \frac 12\Longrightarrow \sqrt{x-\sqrt{2x-1}}=\sqrt{\frac{x+|x-1|}{2}}-\sqrt{\frac{x-|x-1|}{2}}\ .$
$\blacktriangleright\ 2^{\circ}.\ x\ge 1\Longrightarrow \sqrt{x+2\sqrt{x-1}}=\sqrt{\frac{x+|x-2|}{2}}+\sqrt{\frac{x-|x-2|}{2}}\ .$


Does anyone know a forum where I can ask/discuss about English grammar, vocabulary, etc.?

there are lots of english forums.
But what is your native language ?

[quote="roh3x2n"]there are lots of english forums.
[/quote]

Show me some good ones. I can't find

Slightly off-topic, this is an excellent site if you want to learn Dutch.  Not only is there a forum and chatbox where people can write all sorts of things (serious or just for fun) and other people correct it....there are also very decent basic explanations about the spelling and grammar :
[url]http://dutchgrammar.com[/url]

here is one, but it is not an english one.
http://www.uni-protokolle.de/foren/viewf/23,0.html

http://www.babelboard.de/index.php?s=82a5e26018acd04cf83ce20c6f1fb0c2

http://forum.wordreference.com/

Try that, although there are probably better options.

Also, feel free to PM me with any English language questions or if you want to discuss something--I'd be delighted. I doubt anyone would mind if you posted specific queries on this thread, either.

I was just discussing this with my sister : English is pretty easy if you just wanna read and write.  It has no genders like French, no cases like German, no complicated system of conjugations like French,....
But if you wanna get all the pronunciations right.... :o  you'll discover that the spelling makes no sense at all!

I've been learning English for 9 years or something and I still make mistakes.

You can try to watch English shows and movies subtitled in your native language, but the problem is that that still won't point out subtle errors in your pronunciations.  I think only a native speaker, who tells you whenever you do something wrong, can solve this.

[quote="akkaren"]Living in the Southern US... people tend to pronounce things however they want. I have a slight speech impediment and have read/written more than I've talked/listened, so I still have difficulty with my native language when I'm supposed to speak proper English--although that means different things to different people; there are so many dialects around here, not to mention people with very low literacy levels (so, for example, a native speaker mispronounces the word "sure" when reading aloud). People speaking strongly accented English as a second language are also common enough that most intelligent people are used to it and don't have a problem.

(Do you think there really IS a right way to speak English?)[/quote]
Yeah I know, there are a lot of dialects as well (I can hear the difference between Southern American, American, British and Australian).
There might not be a "correct English" but some things are still plain wrong.

If I would have learned English without ever hearing it, I would probably pronounce "to wear" like the "ea" in weak.  And I think all native speakers would consider that very very weird. :ninja: And if I would ask you if it is correct you'd probably just say "no" instead of telling me about dozens of dialects existing somewhere. :(

Honestly? Pronouncing wear like that (and in general using the "ee" sound too often) is the stereotypical mistake of foreigners. People would know English wasn't your first language, but they wouldn't blink an eye at it.

Also, that's one of the words I had problems with when I was younger. I couldn't distinguish between many of the vowel sounds, and I remember my mom lecturing me about that word. I still can't get my tongue to form the right vowels all the time.

And I'm not talking about dozens of dialects existing somewhere... I'm talking about them existing right here where I live (which happens to be metropolitan Atlanta).

Yeah, some things are wrong... but you'd be hard put to come up with a mispronunciation I haven't heard before.

A bit off topic, but that brings up a funny story...

When I was very young, I saw the word "garage" while reading something.  I had no idea what it meant.  So I asked my parents "What does gah-rah-gee mean?"  They had no idea what I was talking about until I wrote it down and they started laughing.

Even though it's my first language, I have to say the English language makes no sense...and yet it's the most widespread language in the world?

Having roots in every other language in the world is part of the problem...


geometry
perimeter
incenter
geometry theorems

I'm looking for a proof that the perimeter of the orthic triangle of an acute triangle is no greater than twice the shortest height.

Let $ABC$ be the triangle, with orthic triangle $DEF$, and $AD$ the shortest height.  Reflecting $D$ in $AB$ to $D'$, and in $AC$ to $D''$, we find that the perimeter of the orthic triangle is equal to the length of the broken line $D'EFD''$.  In fact, because the orthocentre is the incentre of the orthic triangle, that broken line is a straight line, so the perimeter of the orthic triangle is the length of the segment $D'D''$.  This is no more than the length of the broken line $D'AD''$, and as $AD'=AD''=AD$ (the shortest height) we are done.

(Proof stolen from discussion of "Fagnano's problem" in Coxeter's [i]Introduction to Geometry[/i].)


If $\sum^{n}_{i}a_{i}=1$, prove or disprove:

$\sum^{n}_{i}a_{i}^{2}\geq \frac{1}{n}$

 :wink:

Sorry, now i can prove with cauchy!
Please moderator, delete this problems! :(

Use  Cauchy's ineq  we have :
$a_{i}^{2}+\frac{1}{n^{2}}\geq \frac{2a_{i}}{n}$$( i=1\to n)$
  Add $n$ ineq this : we done completed


AMC
USA(J)MO
USAMO
MATHCOUNTS
AIME
AMC 12
AMC 10

Try asking on the olympiad forums too, if you haven't already.

...somehow I could barely get an a in algebra last grading period and yet make it to nats so......I would guess no, except in rare cases.

[quote="karatemagic7"]HAS ANYONE COMPILED ANY STATS ON HOW MANY NAT. MATHCOUNTS QUALIFIERS THIS YR ALSO MADE USAMO? Anyone know or think there is any relationship between how well someone does at Nats and if they qualified to take USAMO? I wonder if past Nat winners fell into this category. They are very different tests afterall.... any past Nat winners care to weigh in?[/quote]

A correlation: ALmost all of the top 12 at Nats have made USAMO at some time or another. And I actually do have stats:

22 People qualified for USAMO that are going to Nationals this year.
Out of those 22, 5 made the top 57 in Nationals last year.

[quote="shentang"]USAMO only because getting to nationals in Massachusetts is hard.[/quote]

Um.... What does that mean? Do you mean USAMO is harder or easier then MC?

[quote="l Bob l"]There's some above-MC level stuff in aime and amc12, I think that's the only major difference. Of course, people who make nats generally already know above-MC level topics. but I only know a little :) That's why I didn't make USMAO (I was 3.5 points off).

I wonder though: Which is harder to make in 6-8th grade: MC or USAMO? Please don't say somthing like "it depends"; If you [b][i][u]HAD[/u][/i][/b] to choose between one, which one would you pick?[/quote]

It seriously depends. How good am I? What, specifically, grade (6th grade = two more chances)? How have I done in previous years? How do starfish reproduce?
Actually, come to think of it, that last one doesn't really matter as much. But the previous questions are pretty important if I were deciding.

[quote="Yongyi781"]shentang, you seem to be contradicting yourself...

If you said getting to nationals in Massachusetts is hard, and you live in Massachusetts, then Mathcounts should be harder.[/quote]

I think he means getting to Nationals(not in Mathcounts) (AKA USAMO)

in Massachusettes is hard.

thanks uldivad - I was up till 2 am trying to figure out what you told me! (I should have gone to sleep ... :oops:

Sorry I thought he said which one is easier.  :blush:
Mathcounts in Mass.-hard
USAMO-easier.

Erm...
Seeing as there was only one 8th grader and one 7th grader who made USAMO in Mass, while there were 4 people making Nats, I don't see how you came up with that...

And also, USAMO is unquestionably a lot harder than Nats.
It's making USAMO that might almost be comparable to making Nats.

Yes, and you were the only 8th grader who made USAMO :o

The only reason I didn't make USAMO was because I didn't qualify for AIME! :(

[quote="alkjash"]It's making USAMO that might almost be comparable to making Nats.[/quote]
Hmm, in Texas there were 4 8th grader USAMO qualifiers, two of whom made USAMO last year and three of whom made nats this year.

I made USAMO but not nats :( .

Eh, well Texas is different ok...
I mean, it's a big state...
With a high azn population...

2/4 hoosier natsers made USAMO. i think.

my state had 0 ppl on nats team qualify for USAMO.

I was the only one who qualified for AIME, and my AMC-10 index wasn't high enough 120+60  :mad:

Our MI team had one qualifier. Actually, MI in general only had one middle school USAMO qualifier. Coincidentally, that person was me. (Or maybe not as coincidentally)

we had none. Only one of them did amcs at all. He did get a perfect on AMC 8 though.

[quote="PI-Dimension"]3/4 of the NC nats team made USAMO, if you want to know. so did #5... lol

1,2,3,5 from NC made it XD.[/quote]

w00t  :D

someone on the NC team got a perfect on the AMC 10.

The one who didn't qualify got a [hide="shameful score"] 100.5 [/hide]

Someone on the Indiana team got a perfect on the AMC 10 also.

Someone on the GA team did fairly well on the AMC 10 and failed on the AIME but still managed to be the only one to qualify for USAMO in middle school.

Guess who it was :wink:  :wink: ?  (wink, wink, nudge, nudge)

[color=white][hide][hide][hide]Wow, I can't believe you're still clicking to find out, cuz it was me...[/hide][/hide][/hide][/color]

[quote="dnkywin"][quote="PI-Dimension"]I must note that: please do not use almost all caps[/quote]
Meaning...[/quote]

Go to the first post  :censored: .

[quote="alkjash"][quote="uldivad9"]Someone on the Michigan team failed the AMC10.
Note: that was me.[/quote]

What was your score?
Can't possibly be worse than my 138... :blush:[/quote]

Nah, it can't possibly be worse than my 112.5... :blush:

[quote="alkjash"][quote="uldivad9"]Someone on the Michigan team failed the AMC10.
Note: that was me.[/quote]

What was your score?
Can't possibly be worse than my 138... :blush:[/quote]

Thanks for making me feel bad, I got 133.5 on the 10B. I'm going to go to the corner and cry now.
Note to gullible people: I'm not.

[quote="mathcrazed"]Thanks for making me feel bad. I'm going to go to the corner and cry now.
Note to cynical people: I am.

Ok so anyways.............................. david lu pwns.[/quote]

Right, because my 133.5 definitely beat your 150.

Your 10 beat my 9. I hate my idiotic mistakes of idiocy.

But your index is higher, so you did better in total.

are the aimes from this year available?

if so, where?

Try the AoPS wiki.  It's funny how David Lu and I are trying to make each other feel smarter than the other. xD

Or go to this site: http://www.mathlinks.ro/resources.php?c=182&cid=45
Yeah, that is pretty funny, but the USAMO Index historically has correlated to higher USAMO scores than the AIME scores. Plus, getting questions on the AIME without really knowing exactly how they work does not mean skill.

eh please don't tell me that you "intuitively" got them, because then I would be pretty mad. xD

[quote="mathcrazed"]AIME shows true skill (kinda...).[/quote]

Nothing shows 'true skill.'


Let $ p$ be a prime, $ p \ge 5$, and write

$ 1 \plus{} \frac{1}{2} \plus{} \frac{1}{2} \plus{} \dots \plus{} \frac{1}{p} \equal{} \frac{r}{ps}$

Prove that $ p^3 \mid (r\minus{}s)$

I think you mistyped the third term, $ \frac12$ should be $ \frac13$. [url=http://www.mathlinks.ro/viewtopic.php?p=1416506#1416506]Solution by lajanugen[/url]


calculus geometry trigonometry AMC AIME USA(J)MO USAMO

Not in any particular order...

1. Do problems.
2. Have a good background and understanding in what's known. Know the theorems in every subject, and more importantly understand the ways in which they can be used.
3. Do problems.
4. Learn silly tricks. Sorry, I don't like this part either.
5. Do problems.

AoPS books help for beginning. Then I guess ACoPS and do problems

What if I happen to complete every book listed here http://www.artofproblemsolving.com/Wiki/index.php/Math_books
and I thoroughly understand everything?

You would probably make it to IMO if you had the determination (which you hypothetically proved by achieving that task...) and didnt let nerves get to you.

I can get all of those books done by the AMC 12 if I just go all out.  Need to get my mom to hide the computer and tv for that to happen though.

Don't bother with the calculus ones, as they don't help.

I guarantee you won't get those 100ish books done in 6 months. Also, I doubt that any amount of work could get you to Alex Zhai's level unless you have a great deal of talent to go along with it. His performance is the culmination of many years' training.

So let's point out the books on that list that are worth getting through. Here's a list of my favorites, with the number indicating approximate difficulty (for comparison only), and in the order I liked them:

[list](3) [i]Problem Solving Strategies[/i] by Arthur Engel - 1300 problems with solutions to ~90%; work on all of them and you'll be set. It takes about half a summer with nothing else to finish, though. And it can get kind of boring; 170 number theory problems and you're sick of number theory (I was).

(5) [i]Problems in Plane and Solid Geometry[/i] by Viktor Prasolov - also over 1000 problems with solutions to ~90%, this is a free PDF that's been floating around. You might be able to get a copy from someone.

(3) [i]Mathematical Olympiad Challenges[/i] by Titu and Razvan - 3 sections, 10 strategies/topics per section, and 9-12 problems per section. It's nice to be able to practice certain strategies but also has some very hard problems.

(5) The four "10_ Problems in..." books by Titu and Zuming. I just used the trig book, and it made me stronger at trig than about half the black group.

(4) [i]Geometry Revisited[/i] by Coxeter and Greitzer. This book, while it's a bit sparse on problems, really helps with depth of knowledge in geometry.

(6) IMO shortlists. As Zuming put it, they usually have 26-27 very good problems, so you know you'll get something worth studying. Kalva has a lot of these; they also give them out at MOP.

(6) [i]360 Problems for Mathematical Contests[/i] is pretty good, but there isn't much more than problems and solutions, and it's kind of boring. I was a little intimidated by it; don't take it lightly. They're hard problems.[/list]

No formula can produce talent, though, but lots of preparation can get you close to the top. Go ahead, try it out. It's hard, and it takes a lot of motivation to get through any one of those books. I started all of those (except the 101, 102 and 104), got through around 2/3, and then started on another book. Geometry Revisited was the only one I finished completely, but that's because it's a content book, not a problem book. So if you're willing and have the motivation, I'm not going to try to discourage you.

Also, as a general rule of thumb, when you can't solve more than half the problems, it's probably a bit too hard, and you won't get as much out of it as you would an easier book.

One thing that hasn't been mentioned that I find important is that there's no point in working so hard if you don't really enjoy what you're doing. For one, if you don't enjoy it, you'll end up halfassing everything (unless you are an uncommonly diligent person) and not getting where you would be if you did enjoy it. You shouldn't be working on whatever for the sake of getting better or going to the IMO, you should be doing whatever for the sake of having fun. Getting better will fall out from it. It shouldn't feel like homework.

[quote="kysuke"]I'm talking about the kids who make it to the imo.  They can solve just about any problem.  How can I get to the level people like Alex Zhai are on?[/quote]

The real answer, of course, is you can't.  "People like Alex Zhai," is an absurd turn-of-phrase: three people scored a 42 at this year's IMO, and he was one of them.  That's three out of like... more than a billion people of school age in the world.  A better question would, how can I get to the level people like MOPpers are on?

And Palmer's answer is pretty darn good.

which books do you guys use for number theory, and trig?

Does anyone have 101 problems in algebra?

[quote="not_trig"]There are a combination of things: talent (which can only get you so far), determination, and training. Talent is probably the least important (unless you want to be like Alex Zhai... very few people can.) But determination and training can take you at least to the IMO given that you have at least a little talent - but you're on AoPS because you have at least a little, right? :wink:

The difference between someone who is in honors math and someone who makes AIME is very interesting. Those who make AIME already must have some talent to make it that far. They have trained to some extent, and have the problem-solving mentality needed. But to make USAMO and beyond, you have to have a stronger will, a stronger drive. You have to be able to put in 1,2 hours a day (AT LEAST). And you have to love what you're doing... a lot. Otherwise you can't force yourself to do that.[/quote]

I'll work for 4 hours a day then.  I've finished AoPS volume 1 and my friend has 2 so I'm going to finish that before summer ends.

[quote="kysuke"]I'll work for 4 hours a day then.  I've finished AoPS volume 1 and my friend has 2 so I'm going to finish that before summer ends.[/quote]
dude trust me...it's easier said than done. You may be able to do that for a few weeks or so, but it gets hard afterwards, especially when you get to more difficult stuff.

[quote="SigmaTheta"]There is an online site with a ton of past AIME problems: http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html

The old kalva site is sort of dead but with the help of the wayback machine everything is fine! Also AoPS has a ton of problems already with solutions.[/quote]

Rumours say it will be up later this year or early next year at the domain kalva.org. Math olympiad books can be seen here for example:

- [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]AoPS Bookstore[/url]
- [url=http://olympiads.win.tue.nl/imo/books.html]Tom Verhoeff's IMO Site[/url]

[quote="mathemonster"]thank you everyone your comments have been very helpful. However, I have observed that almost everyone good at math has done aops v1 and v2. After that stage there seem to be a few different approaches: acops,pss,working through old problems, and just learning from the site. How do you guys tell what your weak in? I can solve a 1-5 aime geo problem but not a #13...does that mean im weak in Geo...because im only about a 7 problems on the aime level anyways.....
@kysuke: what books have you done?[/quote]

It's not like any particular books are necessary to get good.  There's no formula to get better at math.

ACoPS is always a good book.  For your current level, if you're going to get an AoPS book (I can't say much about them, but they're supposed to be quite good with material), then you should probably get AoPS 2.  However one benefit which I know ACoPS has is that it tells you a good way to approach these math books, and what kinds of things to do to get better at math.  I first tried to systematically go through every chapter in a math book.  That doesn't work so well.  You more or less want to more spontaneously look through a chapter or two, and re-read chapters and retry problems.  It helps sometimes to come back to a problem you've already solved.

Also, at this point, even if you find you have a weakness, you want to keep improving in other areas as well.  So don't think "I'm only going to do geometry".  Right now, I think the mains kinds of problems you want to be focusing on are AIME10s-15s.  You probably can't do them systematically, but by working on some and seeing the solutions to them, you can pick up techniques and get better at them.  Also, you could look at some easier Olympiad problems.  Do what problems you find interesting.  :D

[quote="K81o7"]

Also, at this point, even if you find you have a weakness, you want to keep improving in other areas as well.  So don't think "I'm only going to do geometry".  Right now, I think the mains kinds of problems you want to be focusing on are AIME10s-15s.  You probably can't do them systematically, but by working on some and seeing the solutions to them, you can pick up techniques and get better at them.  Also, you could look at some easier Olympiad problems.  Do what problems you find interesting.  :D[/quote]

hmm, do you think for Olympiad level, systematic working is good? (It seemed to be for some people... and I don't really do it, hence I kind of suck)

[quote="jmerry"]ACoPS is the Art and Craft of Problem Solving, by Paul Zeitz. It includes some college material, and is generally targeted at a fairly advanced audience.[/quote]

How do get better at writing proofs?

[quote="jb05"]That's all you really can do at olympiad level, and it's certainly what I've been doing for a couple years now.[/quote]

Really?  I almost never work systematically, and by that I mean I almost never work on olympiad math solely for the sake of improving.  I just try to solve whatever problems I find interesting, which is why I'm terrible at geometry - doing geometry just feels like too much of a chore.

@Damien: I think the answer to your question is that if you want to get better in as little time as possible, obviously working "systematically" would be better, but you can still improve without doing so.  I don't, and I get the sense from Krishanu's post that he doesn't either.

[quote="pythag011"]How do get better at writing proofs?[/quote]

USAMTS is really good for that, especially if you compare your proofs with the "official" proofs - you will be able to see what parts of your proof were insignificant enough to leave out and what parts you should have explained more.  Other than that, just solve problems that involve proving something, and write up your solutions.  If you can, give them to someone to read, but it can be hard to find someone who can evaluate your proof - that's why posting solutions on these forums is good.

[quote="matt276eagles"][quote="jb05"]That's all you really can do at olympiad level, and it's certainly what I've been doing for a couple years now.[/quote]

Really?  I almost never work systematically, and by that I mean I almost never work on olympiad math solely for the sake of improving.  I just try to solve whatever problems I find interesting, which is why I'm terrible at geometry - doing geometry just feels like too much of a chore.[/quote]

Oh, I guess we just interpreted the question differently, because I agree completely with you. I am intending to *try* to work on geometry systematically this year though. I suspect it will just turn into even more of a chore and I will stop :maybe: .

Okay working systematically is helpful but difficult.

When you're good at a subject and enjoy it, there is no reason to work systematically; you can just do random problems and have fun.

Sometimes, some sort of system or set of goals in necessary to improve if you really feel you are not good at a subject. Right now, for me, this is combinatorics, and whenever I look at an olympiad, my mind jumps straight to the number theory and geometry problems immediately. So I need to say "woah okay I need to do a bunch of combinatorics problems from here." Then, I was able to solve some combo problems successfully. And when you're successful, its fun :D .

But meanwhile I'm still being "spontaneous" looking at other problems while I'm not doing that.

[quote="matt276eagles"][quote="jb05"]Oh, I guess we just interpreted the question differently, because I agree completely with you. I am intending to *try* to work on geometry systematically this year though. I suspect it will just turn into even more of a chore and I will stop :maybe: .[/quote]

Ah, OK, sorry for misinterpreting what you were saying.  I feel exactly the same way as you about geometry, but I've heard it gets fun when you get good at it, although I don't believe it :lol:[/quote]

Yes, it becomes MUCH more fun and MUCH more rewarding when you get better.  I have to say this seems particularly true with geometry, I don't know.

Well okay, perhaps I was a little systematic.  When I found out at MOP last year that I was terrible at geometry, I decided I wanted to get better at it.  Thankfully, at that time, I found geometry problems somewhat interesting, so I actually did them without it being a chore.  Sure it was annoying and maddening at first, but it's really the case that the more you do, the better you get, and the more fun it gets, so it's a self-reinforcing loop.

It would be less maddening and easier to start on geometry if you used Sketchpad.  That way, whatever geometry problem you see, or happens to pop into you mind, you can easily construct and modify on your computer

[quote="yisun88"]I think it's important to work systematically in a weak subject so that you see the main ideas and techniques.  It can be a bit demoralizing to try to improve an area but to be unable to solve almost any problems you try because you don't have a grasp of the basic building blocks.  For a long time I was blocked on number theory for this reason (didn't know orders somehow), and grinding out some basic but focused problems really helped.[/quote]

See, I had an easy way out of this.  When I felt completely stuck and didn't know what to do when working on geometry, I asked Alex Zhai for hints.  :)

And man, was he helpful and patient with me...


Choose one or more of the properties listed below to account for the experimentally determined behavior of gases mentioned in each question.

Properties:
I. A particle is made up of a single atom or a combination of the same or different atoms so that particles of different gases may have different masses.
II. The particles must be perfectly elastic and must be in constant random motion.
III. Equal volumes of different gases compared at the same temperature and pressure contain equal numbers of particles.

Questions:

1. Some gases have color while others do not.
a) I     b) II      c) III     d) I and III     e) II and III

2. For a given gas, the pressure volume product is a constant when the amount of gas and temperature are constant.
a) I     b) II      c) III     d) I and III     e) II and III

3) Equal volumes of dihydrogen gas and dichlorine gas, compared at the same temperature and pressure, have different masses, but react with each other in a volume ratio of 1:1 to form HCl
a) I     b) II      c) III     d) I and III     e) II and III

With these questions:

For the first one, I have worked out the answer to be statement I or choice A as it seems to be the only choice that accounts for gas color (I think mass is important when considering gas color...)

I think that the answer to the last question is statement III or choice c as it explains why there are equal numbers of chlorine and hydrogen particles.

I have not worked out an answer to the second question as of yet.


linear algebra
matrix
vector
linear algebra unsolved

In each case, construct a matrix with the requisite properties or explain why no such matrix exists:

a) The column space contains $ \left(
\begin{array}{ccc}
1  \\
1  \\
1  \end{array}
\right)$ and $ \left(
\begin{array}{ccc}
0  \\
1  \\
1  \end{array}
\right)$ and the nullspace contains $ \left(
\begin{array}{ccc}
1  \\
0  \\
1  \\
0  \end{array}
\right)$ and $ \left(
\begin{array}{ccc}
1  \\
0  \\
0  \\
1  \end{array}
\right)$

b) The column space has basis $ \left(
\begin{array}{ccc}
1  \\
0  \\
1  \end{array}
\right)$, $ \left(
\begin{array}{ccc}
0  \\
1  \\
1  \end{array}
\right)$, and the row space has basis $ \left(
\begin{array}{ccc}
1  \\
1  \\
1  \end{array}
\right)$, $ \left(
\begin{array}{ccc}
2  \\
0  \\
1  \end{array}
\right)$

For part b, it would have to be a $ 3 \times 3$ matrix, but then the rank plus the nullity would have to be $ 3$, not $ 4$.

For part a, perhaps you could start with the two vectors which must be in the column space as columns, and then pick the remaining two columns somehow so that the null space comes out right.

a) $ \left(\begin{matrix}1&0&\minus{}1&\minus{}1\\1&1&\minus{}1&\minus{}1\\1&1&\minus{}1&\minus{}1\end{matrix}\right)$
b) $ \left(\begin{matrix}1&1&1\\2&0&1\\3&1&2\end{matrix}\right)$

What the rank-nullity theorem says about (a): The matrix in question must be $ 3\times 4.$ Since its column space contains the two independent columns shown, we must have that its rank is $ \ge 2.$ But the null space also contains two independent vectors, hence the nullity is $ \ge 2,$ hence the rank is $ \le 2.$ So the rank must be exactly $ 2.$

olorin's solution is the simplest, but it is by no means unique. For instance, we could have had
\[ \begin{pmatrix}0&3&0&0\\\minus{}2&0&2&2\\\minus{}2&0&2&2\end{pmatrix}\]

Of course it's not unique.

The general solution for a) is $ (\begin{matrix}v&w&\minus{}v&\minus{}v\end{matrix})$
with the columns $ v,w$ forming any basis of the span of $ \left(
\begin{matrix}1 \\1 \\1 \end{matrix}\right),\left(
\begin{matrix}0 \\1 \\1 \end{matrix}\right)$.

The general solution for b) is $ (\begin{matrix}v&w&{v\plus{}w\over 2}\end{matrix})$
with the columns $ v,w$ forming any basis of the span of $ \left(
\begin{matrix}1 \\0 \\1 \end{matrix}\right),\left(
\begin{matrix}0 \\1 \\1 \end{matrix}\right)$.


Consider all triangles $ ABC$ satisfying the following conditions: $ AB \equal{} AC$, $ D$ is a point on $ \overline{AC}$ for which $ \overline{BD} \perp \overline{AC}$, $ AD$ and $ CD$ are integers, and $ BD^2 \equal{} 57$. Among all such triangles, the smallest possible value of $ AC$ is

$ \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 13$
[asy]defaultpen(linewidth(.8pt));
dotfactor=4;

pair B = (0,0);
pair C = (5,0);
pair A = (2.5,7.5);
pair D = foot(B,A,C);
dot(A);dot(B);dot(C);dot(D);
label("$A$", A, N);label("$B$", B, SW);label("$C$", C, SE);label("$D$", D, NE);
draw(A--B--C--cycle);draw(B--D);[/asy]

[hide]
Since $ AD,CD \in \mathbb{Z}^\plus{}$, we have that $ AB\equal{}AC$ are also integers.

By the Pythagorean Theorem, \[ AB^2\minus{}AD^2\equal{}57 \implies (AB\minus{}AD)(AB\plus{}AD)\equal{}57\]

Since $ AB,AD$ are integers, there are two cases.
Case 1: $ AB\minus{}AD\equal{}1, AB\plus{}AD\equal{}57$
This means that $ AB\equal{}29, AD\equal{}28$.

Case 2: $ AB\minus{}AD\equal{}3, AB\plus{}AD\equal{}19$
This means that $ AB\equal{}11,AD\equal{}8$.

So, the smallest possible value of $ AC$ is $ 11$. $ \boxed{C}$
[/hide]

[hide="Hint (another method)"]Set $ AB\equal{}a$.  We can quickly solve for $ BC$ in terms of $ a$ using right triangles and the fact that $ AB\equal{}BC$.  Now, we have all the side lengths in terms of $ a$.  Time for the triangle Inequality.[/hide]


we have the current circuit
we have to find $E_1$ know that $I_1=I_2$

[hide="hint"]use kirchoff's rules. (you know that the current in the bottom part is $2I$) go from there. [/hide]


geometry calculus function videos rectangle

trivialness011: (2:48:05 PM) don't go to
trivialness011: (2:48:08 PM) University of Phailures
trivialness011: (2:48:10 PM) math classes:
tehbeastlyone: (2:48:11 PM) i wont
trivialness011: (2:48:14 PM) Undergraduate:
trivialness011: (2:48:18 PM) 1. Review of arithmetic
trivialness011: (2:48:21 PM) 2. Review of algebra
trivialness011: (2:48:25 PM) 3. Geometry
trivialness011: (2:48:29 PM) 4. Calculus
trivialness011: (2:48:36 PM) 5. Cool Calculator tricks
trivialness011: (2:48:40 PM) 6. Magic tricks
tehbeastlyone: (2:48:43 PM) lol
trivialness011: (2:48:44 PM) 7. Card tricks
trivialness011: (2:49:01 PM) 8. How to guess on the SAT (which you won't take since your in coolllge)
trivialness011: (2:49:08 PM) 9. How to guess on PSAT
tehbeastlyone: (2:49:33 PM) lol
trivialness011: (2:49:40 PM) 10. Get hypnotized to love math
trivialness011: (2:49:51 PM) 11. Why nerds suck (only good class :P)
trivialness011: (2:50:00 PM) 12. Why nerds are cool
trivialness011: (2:50:13 PM) 13. Memorize the list of best careers. (OMG, 1 is mathematicisn)
trivialness011: (2:50:22 PM) 14. Are mathematicians cool?
trivialness011: (2:50:28 PM) 15. Biography of Erdos
trivialness011: (2:50:54 PM) 16. Erdos number. (People who take corrobate with a professor with erdos number 7, getting an erdos number 8. Cool!)
tehbeastlyone: (2:51:12 PM) lol
trivialness011: (2:51:17 PM) 17. How to multiply 100 digits numbers!
trivialness011: (2:51:47 PM) 18. What is a continous function? (in this calss, we will learn what a continuous function is)
trivialness011: (2:51:59 PM) 19. Logic is worse than emotion.
trivialness011: (2:52:10 PM) 20. Why proof sucks (end with proof that proof sucks(
trivialness011: (2:52:19 PM) 21. Why tehbeastlyone sucks.
trivialness011: (2:52:28 PM) 22. Why olympiads suck
trivialness011: (2:52:33 PM) 23. Why math sucks
trivialness011: (2:52:42 PM) 24. Why math is useful, but still sucks
trivialness011: (2:52:48 PM) 25. A long list of math topics
trivialness011: (2:52:53 PM) 26. A long list of math jobs
tehbeastlyone: (2:53:27 PM) ...
trivialness011: (2:53:27 PM) 27. Math in games. (Every day, you play a video game, followed by a 5-minute session about math in it. (See, math is in video games, there are rectangles in video games!))
tehbeastlyone: (2:53:32 PM) lol
trivialness011: (2:53:34 PM) 28. Why math is better than physics
trivialness011: (2:53:42 PM) 29. Why physics is better than chess
trivialness011: (2:53:50 PM) 30. Why you should go to this university
trivialness011: (2:53:57 PM) 31. Why Zeb pwns everybody at go
trivialness011: (2:54:06 PM) 32. Why computers can do math for you!
trivialness011: (2:54:11 PM) 33. How to use a calculator
trivialness011: (2:54:22 PM) 34. How to use mathematica
trivialness011: (2:55:06 PM) 35. Cool math pictures (you look at pictures of the mandelbrot set in each class. However, each class has a different size! for example, on elcas is x1, one class is x2, etc...)
trivialness011: (2:55:40 PM) 36. Are you ready for graduate school? (You answer a lot of arithmetic quesitons. If you get all of them correct, you go to graduate school
trivialness011: (2:55:44 PM) Graduate school:
trivialness011: (2:56:01 PM) 1. Review. We review every thing you have learned
trivialness011: (2:56:10 PM) 2. We learn about various math websites
trivialness011: (2:56:19 PM) 3. We discuss which ones ib est.
trivialness011: (2:56:43 PM) 4. We write a thesis. Possible thesis topics:
trivialness011: (2:56:47 PM) 1. Arithmetic
trivialness011: (2:56:50 PM) 2. Magic tricks
trivialness011: (2:56:59 PM) 3. Philosophy
trivialness011: (2:57:03 PM) 4. A biography of a mathematician
trivialness011: (2:57:21 PM) requirements to get master degree in each category:
trivialness011: (2:57:32 PM) Arithemetic: add two 10 digit numbers
trivialness011: (2:57:52 PM) Magic trick: It must actually work at least 1 time in 1000.
trivialness011: (2:58:00 PM) Philsophy: It must be in English
trivialness011: (2:58:06 PM) Biography: Mathematiicna must exist
trivialness011: (2:58:10 PM) PhD requirements:
trivialness011: (2:58:17 PM) Arithemetic: muliply two 5 digits numbers
trivialness011: (2:58:29 PM) Magic trick: It must actually work at least 1 time in 10.
trivialness011: (2:58:38 PM) Philsophy: It must be at least a page llong
trivialness011: (2:58:51 PM) Biography: Mathematician must have a theorem named after him
trivialness011: (2:59:00 PM) don't you want to go to the University of Phailures?
trivialness011: (2:59:13 PM) DO YOU WANT TO GO TO THE UNIVERSITY OF PHAILURES?
tehbeastlyone: (2:59:47 PM) no
tehbeastlyone: (2:59:50 PM) not really
trivialness011: (2:59:55 PM) whynot?
trivialness011: (2:59:56 PM) isn't like
trivialness011: (3:00:00 PM) the math department
trivialness011: (3:00:03 PM) so cool and fun?
trivialness011: (3:00:43 PM) don't ou agree?
tehbeastlyone: (3:00:49 PM) not really
tehbeastlyone: (3:00:49 PM)

Umm... seriously, stop posting my aim conversations with youl...


in aops v2, they mention the following two combinatorial identities separately:
$ \dbinom{n}{k}\equal{}\dbinom{n\minus{}1}{k}\plus{}\dbinom{n\minus{}2}{k\minus{}1}\plus{}\dbinom{n\minus{}3}{k\minus{}2}\plus{}...\plus{}\dbinom{n\minus{}k\minus{}1}{0}\\
\dbinom{n}{k}\equal{}\dbinom{k\minus{}1}{k\minus{}1}\plus{}\dbinom{k}{k\minus{}1}\plus{}...\plus{}\dbinom{n\minus{}1}{k\minus{}1}$
my question is, arent these identities actually pretty much the same, because you can apply pascal's identity on the first one, and ull get an identity of the same form as the second one?

A lot of identities are direct consequences of other identities; it doesn't make them the same identity.  In any case, the two have a very different flavor; note that the bottom of the combination is the same in the second identity throughout but decreases in the first identity.


When will there be more AMC-10 resources? It would be great if there were. :)

Go to the USA competitions forum and look at my thread. THere are a few links there.

Also, I hope you signed up for the AMC 12 or 10 classes if you want more practice.

I couldn't find it. Could you link to it?  :maybe: