Can someone tell me how to make a lewis structure for NO[size=84]3[/size]?  (Not the polyatomic ion: just NO3 w/ no charge).  I got it to be kind of a linear structure w/ O - N - O - O as the skeleton structure and a free radical on the N.  Please tell me if this is correct! Thanks.

yes.  or at least my teacher thinks so...b/c it was part of our recent test and i want to check if i got it right...so anyone w/ the answer???

I thought it was a structure with three resonance structures with a double bond one O and singles on the rest?

On second thought, that'd be NO3-. I guess it'd be a localized model with a missing electron...very unstable.

yes, NO[size=84]3[/size] will be a very unstable polyatomic ion... but has anyone come up with the lewis structure for it yet???

$\text{NO}_3$ isn't a polyatomic ion (although $\text{NO}_3^-$ is), it's a molecule.  While it would be very unstable, it would most likely set up with the nitrogen atom in the middle and three oxygen atoms around it.  The odd electron will probably be involved in antibonding, although I was never very good with molecular orbitals.  I don't think a simple Lewis structure exists.

i am pretty sure that the lewis structure diagram for NO3 would be a double and two single $N-O$ bonds. But it is just that we refer to each bond as $1 \frac 1{3}$

ohh yea my teacher said the structure for it would be a N in the middle w/ three O's and the free radical would be attached to one of the O's....we're just getting into more molecular orbital theory (antibonding, hybrids). thanks!


number theory
prime numbers
relatively prime
number theory unsolved

Hello everyone

Determine how many lists of prime numbers $ (p,q,r)$ is such that: $ p \plus{} q \plus{} r \equal{} 2005$ and $ pqr \plus{} 1$ is a perfect square.

Greetings  :D

assume $ (p,q,r)$ is good.
$ pqr \plus{} 1 \equal{} k^2\iff pqr \equal{} (k \plus{} 1)(k \minus{} 1)$ - the last product is divisible by 4 or odd.
first case gives $ \{p,q,r\} \equal{} \{2,2,2001\}$ - not good.
so $ 2|k$, then $ k \minus{} 1$, $ k \plus{} 1$ relatively prime, so one of them is a prime, the second one is a product of two primes


LaTeX
real analysis
real analysis unsolved

Hey all,

My friend keeps bugging me about this eigenvalue problem:
x^2*u''+3x*u'+λ*u=0 for 1<x<e
and with u(1)=u(e)=0 and for lambda>1

I understand that I should be looking for a solution in the form of
u=x^m.

Thanks for any help,

Jay

Use LaTeX, and all will be prettier:

[quote="jpar"]Hey all,

My friend keeps bugging me about this eigenvalue problem:
$x^{2}\cdot u''+3x\cdot u'+\lambda \cdot u=0$ for $1<x<e$
and with $u(1)=u(e)=0$ and for $\lambda>1$.

I understand that I should be looking for a solution in the form of
$u=x^{m}$.

Thanks for any help,

Jay[/quote]


calculus
integration
calculus computations

Hey guys, I'd like to say hi!

But on a not so good note, lol, I would also like to say that as of recent, I have been struggling immensely with these questions our prof gave us as homework. I've been trying them for a few days now with no luck.

Anyways, my mind is tired of this. I've made the questions worse as a result of my overthinking and was hoping for some light lol. Here are some of what I've been having difficulties with:

1. f(x) = -x-1     if -3<= x <=0     (<=, greater/less than or equal to)
f(x) = -(1-x^2)^1/2   if o<=x<=1

evauluate intl-3 to 1 f(x)dx by interpreting the integral as a difference of areas...I dunno

2. If f is continuous and int 0 to 9 f(x)dx = 4, find int 0 to 3 xf(x^2)dx

3.If f(0) = g(0) = 0 and f'' and g'' are continuous. Show that int 0 to a f(x)g''(x)dx = f(a)g'(a)-f'(a)g(a) + int 0 to a f''(x)g(x)dx

And the the last one...

4. Evaluate the integral.

int cot^5#sin^4# (# = thetha).

Thanks in advance for any and all your help. [/code]

For 1, you can separate the interval:

$\int_{-3}^1f(x) dx= \int_{-3}^0 f(x) dx + \int_0^1 f(x) dx$

For 3:

$\int_0^a f(x) g''(x) dx= f(a) g'(a) - f'(a)g(a) + \int_0^a f''(x) g(x)dx$

By derivating left part:

$\frac{d}{da} \int_0^a f(x) g''(x) dx = \boxed{f(a)g''(a)}$

By derivating right part:

$\frac{d}{da} \left [f(a) g'(a) - f'(a)g(a) + \int_0^a f''(x) g(x)dx \right]= f'(a)g'(a) + f(a)g''(a) - f''(a)g(a) - f'(a)g'(a) + f''(a)g(a) = \boxed{f(a)g''(a)}$
 (which is the same as right part)

Thanks so much, Nazaf, especially for number 3.

4. $\int cot^5(\theta) \cdot sin^4(\theta) \ d\theta = \int \frac{cos^5 (\theta)}{sin^5(\theta)} \cdot sin^4(\theta) d\theta = \int \frac{cos^5(\theta)}{sin(\theta)} d\theta = \int \frac{(1-sin^2(\theta))^2 \cdot cos(\theta)}{sin(\theta)} d\theta$

let $u = sin(\theta) \implies du = cos(\theta) d\theta$

$= \int \frac{(1-u^2)^2}{u} du = \int \frac{1+u^4-2u^2}{u} du$

$\boxed{= ln|sin(\theta)| + \frac{1}{4}sin^4(\theta) - sin^2(\theta) + C}$

and finally

2. $\int_0^3 x \cdot f(x^2) dx$

 let $u = x^2 \implies du = 2x dx$

 $= \int_0^9 \frac{f(u) \cdot x}{2x} du = \frac{1}{2}\int_0^9f(u)du = \frac{1}{2}\int_0^9 f(x)dx = \frac{1}{2} \cdot 4 = 2.$


$ \sum \frac{\sqrt{n\plus{}1}\minus{}\sqrt{n}}{n}$ converges or diverges? I know that $ \sum \sqrt{n\plus{}1}\minus{}\sqrt{n}$ diverges because it's a telescoping sum with $ n$-therm $ \sqrt{n\plus{}1}\minus{}\sqrt{1}$. Hints?

Try to find $ A$ and $ \alpha$, such that $ \frac{\sqrt{n\plus{}1}\minus{}\sqrt{n}}{n}<\frac{1}{An^{\alpha}}$ and use the comparison test.

[hide]$ A\equal{}2$ and $ \alpha\equal{}\frac{3}{2}$[/hide]

Thank you guys! i was led to think this diverges...


How many natural-number factors does $ \textit{N}$ have if $ \textit{N} \equal{} 2^3 \cdot 3^2 \cdot 5^1$?

Using the common technique for finding the amount of factors: $ (3\plus{}1)(2\plus{}1)(1\plus{}1) \equal{} 4 \cdot 3 \cdot 2 \equal{} 24$

An easy way to find the number of factors a number has using prime factorization is by adding 1 to each of the exponents in the prime factorization and multiplying the new exponents. Therefore, N has $ 4\cdot 3\cdot 2\equal{}24$ factors.

Why do you add one to each of the exponents and multiply?

Suppose the number was $ 2^5\cdot 3^4\cdot 7^2$.

The factorscan have 2^0, 2^1, 2^2, 2^3, 2^4, or 2^5 as a factor (there are 6 of them). Then we move onto the 3s. The factors 3^0, 3^1, ..., 3^4 in the factor (5 of them). We do the same thing with the sevens, so the number of factors in $ 2^5\cdot 3^4\cdot 7^2$ is the product, or $ 6\cdot 5\cdot 3\equal{}90$.


calculus
integration
quadratics
limit
function
complex numbers
real analysis

t a real number consider x_1(t), x_2(t) the complex roots of x^2+x+t=0
f:R-->R given by f(t) = |x_1(t)| + |x_2(t)|
1) Prove f admits primitive on R
2) Compute  \int _[0,1]f(t)dt
-------------------------------------------------------------------
1)
The discriminant of the quadratic is D=1-4t

If t>1/4
x^2+x+t=0 has two complex conjugate roots
f(t)=|-1/2 + i/2 \sqrt (4t-1)|+|-1/2 + i/2 \sqrt (4t-1)|
=2 \sqrt t

If t=1/4 only one root 1/2 with multiplicity two
f(t) = 1

If t<1/4  the quadratic has two reals roots
f(t)= |-1/2 + 1/2 \sqrt (1-4t)| +  |-1/2 -1/2 \sqrt (1-4t)|=
1/2|-1+ \sqrt (1-4t)| + 1/2 + 1/2\sqrt (1-4t)

 \lim_(t->1/4) f(t) = 1

f is continous function, it has primitive on R.

2) For the integral [0,1]=[0;1/4] \cup [1/4;1]
\int _[0,1]f(t)dt =  \int _[0;1/4]f(t)dt +  \int _[1/4;1]f(t)dt
and consider the  expression of f on [0;1/4] and [1/4;1]

moubi, be more carefull. your answer is not correct.
there are more cases to consider.

nope. :)

the function looks like this  f(t) =

\sqrt { 1-4t} , t \leq 0
1, t \in (0, 1/4]
2\sqrt t , t > 1/4.

Obivously it is continuos and the integral on [0,1] gives 17/12.


If $a,b,c,d>0$ then the following inequality holds
$\frac{a^3+b^3+c^3}{4d^3-abc}+\frac{b^3+c^3+d^3}{4a^3-bcd}+\frac{c^3+d^3+a^3}{4b^3-cda}+\frac{d^3+a^3+b^3}{4c^3-dab}\ge4$
(the denominators are strictly positive !)


function
real analysis
real analysis solved

Let $f,g:\mathbb R\to\mathbb R$ be two real functions s.t. $f(x)<g(x),\ \forall x\in\mathbb R$. Prove that there is an infinite subset $M\subset\mathbb R$ s.t. $f(x)<g(y),\ \forall x,y\in M$.

For all $x$ we can squeeze some rational number $q(x)$ in between $f(x)$ and $g(x)$. So we can get some function $q: \mathbb{R} \rightarrow \mathbb{Q}$ such that $f(x) < q(x) < g(x)$, $\forall x \in \mathbb{R}$. Define $S_x = \{a \in \mathbb{R}:\ q(a) = x\},\ \forall x \in \mathbb{Q}$. Obviously $\cup_{x \in \mathbb{Q}}{S_x} = \mathbb{R}$. If $S_x$ is finite, $\forall x \in \mathbb{Q}$, then $\mathbb{R}$, which is an uncountable infinite set, could be represented as a countable set (i.e. the union of countably many countable sets), contradiction. So one $S_x$ must be uncountable, and $M = S_x$ works. Correct?


function
algebra
binomial theorem
number theory unsolved
number theory

In (1 + x + x^2 + 2x^3 + x^4 + x^5 + x^6)^n

the coefficient of x^4 is?

Plz explain the entire method ?

If its easier for you to handle you can consider that to just be $ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)^n$ for your question.
What about the binomial theorem don't you get?
Any chance you know generating functions?

Plz post a solution to the above qs

I din't get u're point abt BINOMIAL Theorem
I just named the topic as that

hello, see also here
http://en.wikipedia.org/wiki/Multinomial_theorem
Sonnhard.


it suffices to show \[ 2S\minus{}2T\equal{}\sum\frac{a_{1}\minus{}a_{3}}{a_{2}\plus{}a_{3}}\ge0\Rightarrow\sum\frac{a_{1}\plus{}a_{2}}{a_{2}\plus{}a_{3}}\ge5,\]
which follows after AM-GM. :lol:


geometry
integration
algebra solved
algebra

calculate \[\sum_{k=0}^{2n}(-1)^k a_k^2\]  where the $a_k$ are given by :
\[(1-\sqrt2 x+x^2)^n=\sum_{k=0}^{2n}a_k x^k\]

I believe this is not from IMO 1987, and it's not a Geometry problem, either.

yes l'm sorry l don't know why l put this problem here but this is really a problem from imo 1987. if l said that, why do you don't believe me ?

Maybe you mean IMO Short List or Long List?
In my opinion, in any case moderators must move this problem to Advanced Section.

so everyone is satisfied.
does someone knows how to put this problem on the right section? mister Vornicu?

but the problem is in 'algebra' section. it is one day in 'geometry' section and the other one it is in 'algebra' ?
that's very weird....

I realize that following solution is not applicable for IMO. It is pure university solution.

Let $f(x)=x^2-\sqrt{2}x+1$.

It is easy to show that $f^n(x)+f^n(-x)=2\sum\limits_{k\ is\ even}a_kx^k$ and $-f^n(x)+f^n(-x)=2\sum\limits_{k\ is\ odd}a_kx^k$.

So if $a(t)=e^{2ti}-\sqrt{2}e^{ti}+1$ and $b(t)=e^{2ti}+\sqrt{2}e^{ti}+1$ then
1) $\frac{1}{2\pi}\int_0^{2\pi}|a^n(t)+b^n(t)|^2dt = 4\sum\limits_{k\ is\ even}a_k^2$
2) $\frac{1}{2\pi}\int_0^{2\pi}|a^n(t)-b^n(t)|^2dt = 4\sum\limits_{k\ is\ odd}a_k^2$
Since $|z|^2=z\cdot\overline{z}$ we can easy show that $|a^n(t)+b^n(t)|^2-|a^n(t)-b^n(t)|^2=4(e^{2ti}+e^{-2ti})^n$.
It follows that
\[\sum\limits_{k=0}^{2n}(-1)^ka_k^2=\frac{1}{2\pi}\int_0^{2\pi}(e^{2ti}+e^{-2ti})^ndt=\frac{1}{2\pi}\int_0^{2\pi}\sum_{k=0}^n C_n^ke^{2(n-2k)ti}dt=A_n\].
It is obvious $A_n=C_n^{n/2}$ if $n$ is even and $A_n=0$ if $n$ is odd.

$\sum (-1)^k a_k^2
= [x^0] (\sum a_k x^k) (\sum a_k (-\frac1x)^k)
= [x^0] (1-2\sqrt2 x +x^2)^n (\frac{1}{x^2} +\sqrt2 \frac 1x + 1)^n
= [x^0] \frac{(1+x^4)^n}{x^{2n}}
= [x^{2n}] (1+x^4)^n
= C_{\frac n2}^n \text{if n is even or 0 if n is odd}$.

I think this approach is more appropriate for the Olympiad level.

congratulations !
michael's solution is very similar to mine

That solution may be good, but I think we can have another simply solution, I will try it. By the way, do you know how I can get the IMO long list ? I've found them for long time and get nothing. Thanks.

[quote="nntrkien"] By the way, do you know how I can get the IMO long list ? I've found them for long time and get nothing. Thanks.[/quote] I have some IMO shortlists and IMO longlists as .pdf-files. I generated these files from scanned images. Mostly these files represents the problem materials which were distributed to the jury annually.  :)


It doesn't write it to the c746.css file when I save the custom CSS.

XP, ML, FF2, but I doubt that matters.

You're right. It doesn't. Just a permission issue that I'll take care of right now.


How can we define a mathematics research to be original? And, what does it mean for a math research paper to be not original?

Come up with something that hasn't been done.  A new theorem or conjecture, an alternative or simpler proof, a new idea.


ARML
analytic geometry
graphing lines
slope
quadratics

[Note: A tangent line to the curve $ y\equal{}x^2$ at the point $ (a,a^2)$ will have slope $ 2a$.]

The angle formed by the tangents to $ y\equal{}x^2$ from the point $ (r,s)$ in Quadrant II is bisected by the line through $ (r,s)$ with slope $ 1$.  Compute $ s$.

Isn't there only one tangent to $ y \equal{} x^2$ from the point $ (r,s)$? Is the question wrong, or am I missing something obvious?

Ah, I was assuming $ (r,s)$ was on $ y \equal{} x^2$.

[hide]Let the two points of tangency be $ (a,a^2)$ and $ (b,b^2)$.  Plug the slopes in to whatever $ tan$ slope formula you choose to get $ (2a)(2b) \equal{} 1$ or $ ab \equal{} 1/4$.
Now express $ (r,s)$ as the intersection of the tangent lines $ y\minus{}a^2 \equal{} 2a(x\minus{}a)$ and $ y\minus{}b^2 \equal{} 2b(x\minus{}b)$.  Isolating $ y$ and equating the RHS's, we have
$ y \equal{} 2a(x\minus{}a) \plus{} a^2 \equal{} 2b(x\minus{}b) \plus{} b^2$, so
$ 2ax \minus{} a^2 \equal{} 2bx \minus{} b^2$, whic in turn gives
$ 2x(a\minus{}b) \equal{} (a\minus{}b)(a\plus{}b)$, and because we have two distinct points of tangency, we may divide through by $ a\minus{}b$ to obtain
$ x \equal{} \frac{a\plus{}b}{2}$.
Plug this in $ y\minus{}a^2 \equal{} 2a(x\minus{}a)$ or the other equation to get $ y \equal{} ab\equal{} 1/4$.[/hide]

[hide]Let the points at which the tangents touch $ y \equal{} x^2$ be $ (a,a^2)$ and $ b,(b^2)$. The tangents must have reciprocal slopes in order for the angle they form to be bisected by a line with slope $ 1$. Thus, $ 2a \equal{} \frac{1}{2b}$, so $ b \equal{} \frac{1}{4a}$.

Consider the slopes of the tangent lines:
$ \frac{a^2\minus{}s}{a\minus{}r} \equal{} 2a$
$ a^2\minus{}s \equal{} 2a^2\minus{}2ar$
$ a^2\minus{}2ra\plus{}s\equal{}0$
$ a\equal{}\frac{2r\pm\sqrt{4r^2 \minus{} 4s}}{2}\equal{}r\pm\sqrt{r^2 \minus{} s}$

$ \frac{\frac{1}{16a^2}\minus{}s}{\frac{1}{4a}\minus{}r} \equal{} \frac{1}{2a}$
$ \frac{\frac{1 \minus{} 16sa^2}{16a^2}}{\frac{1\minus{}4ra}{4a}}\equal{}\frac{1}{2a}$
$ \frac{1\minus{}16sa^2}{4a(1\minus{}4ra)}\equal{}\frac{1}{2a}$
$ 1\minus{}16sa^2\equal{}2\minus{}8ra$
$ 16sa^2\minus{}8ra\plus{}1\equal{}0$
$ a\equal{}\frac{8r\pm\sqrt{64r^2\minus{}64s}}{32s}\equal{}\frac{r\pm\sqrt{r^2\minus{}s}}{4s}$

These two quadratics must have the same roots, $ a$ and$ \frac{1}{4a}$. Thus, either $ r\plus{}\sqrt{r^2 \minus{} s}\equal{}\frac{r\plus{}\sqrt{r^2\minus{}s}}{4s}$, in which case $ 4s\equal{}1$ and $ s\equal{}\frac{1}{4}$, or $ r\plus{}\sqrt{r^2 \minus{} s}\equal{}\frac{r\minus{}\sqrt{r^2\minus{}s}}{4s}$ and $ r\minus{}\sqrt{r^2 \minus{} s}\equal{}\frac{r\plus{}\sqrt{r^2\minus{}s}}{4s}$. In this case,  $ r\plus{}\sqrt{r^2 \minus{} s}\equal{}\frac{r\plus{}\sqrt{r^2\minus{}s}}{16s^2}$, so $ s\equal{}\pm\frac{1}{4}$. Since $ (r,s)$ is in quadrant II, $ s$ is postive, so $ s\equal{}\frac{1}{4}$ in both cases.

The second case occurs only if $ (r,s)$ is $ \left(\frac{1}{2},\frac{1}{4}\right)$, which is on $ y \equal{} x^2$. This caused me to notice that, had I decided to try to solve the problem anyway thinking that $ (r,s)$ was on $ y\equal{}x^2$, I would have gotten the correct answer much more easily:

Since the two tangent lines are the same, the angle is a 180-degree angle. Thus, the tangent line must be perpendicular to a line with slope $ 1$, i.e., have slope $ \minus{}1$. Thus $ 2r\equal{}\minus{}1$, so $ s\equal{}r^2\equal{}\frac{1}{4}$.[/hide]


I was doing more problems from my book, and I stumbled upon this one:

SHow that if A and B are real constants, and X a real variable, the fraction:

(Ax^2+BX-1)/ (X^2 +X)

can take all real values if and only if a-b is greater than or equal to 1.

Sorry that I didn't use latex, I'm still trying to learn how to do that, I tried to solve this question, setting the fraction equal to Y, and then getting rid of the denominator of the fraction, and forming a quadratic equation using Y, from which you then take the discriminant and set it equal to 0 (or greater). However, I dont know why a-b has to be greater than 1, But i would appreciate if I could listen to any suggestions or hints you may have.

Thank you

sincerely

JHM

i think that a-b has to be more than 1 for the quadratic to be rational based on the quadratic formula, but im probably wrong

[hide="Hint 1"] Rewrite the function as

$f(x) = A+\frac{ (B-A)x-1}{x(x+1)}$ [/hide]
[hide="Hint 2"] Then consider the behavior of $f(x)$ around $x = 0, x =-1$. [/hide]

should that second hint be $x =-1$ as it seems that is where the function approaches infinity?


linear algebra
matrix
algebra
polynomial
inequalities
linear algebra unsolved

[b]
Suppose that $A$ is  a real  matrix $n\times n$ , $(n\ge 3)\; ,\;$ having the characteristic equation
\[ x^n-Bx^{n-2}+c_3x^{n-3}+\ldots+ c_{n-1}x +c_n = 0 \; \; \; ,\; \; B>0\; .  \]
[color=blue] If  $A$ has an eigenvalue in the interval  $\left( \sqrt{\frac{2(n-1)}{n}B\; } \; ,\: +\infty \right)\; ,\;$ prove that  $A$ is not symmetric.
[/color]
[/b]

If A was symetric then the characteristic polynomial would have n real roots :${a_1,...,a_n}$ let's consider $a_n$ the greatest root ,then $a_n>\sqrt{2(n-1)B/n}$.
we have $\sum_{i=1}^{n-1}a_i=-a_n$ and $\sum_{i<j\neq n}a_ia_j=a_n^2-B$
we have $(\sum_{i=1}^{n-1}a_i)^2-2(\sum_{i<j\neq n}a_ia_j)=\sum_{i=1}^{n-1}a_i^2$
and by using Cauchy Schwarz inequality $(n-1)(\sum_{i=1}^{n-1}a_i^2)\geq (\sum_{i=1}^{n-1}a_i)^2$
so $(n-1)[a_n^2-2(a_n^2-B)]\geq a_n^2$ Thus $a_n\leq \sqrt{2(n-1)B/n}$ :)

[quote="MoHaMeD"]If A was symetric... then the characteristic polynomial would have$n$ [b]real roots[/b] ${a_1,...,a_n}$  .......let's consider $a_n$ the greatest root ,then $a_n>\sqrt{2(n-1)B/n}$. ......Thus $a_n\leq \sqrt{2(n-1)B/n}$ ......[/quote] Nice, thanks for interest/proposer


integration
trigonometry
logarithms
calculus
inequalities unsolved
inequalities

Show that $ \sqrt{3} \le \text{exp}\left(\displaystyle\int_{\pi/6}^{\pi/2}\frac{\sin{x}}{x}dx\right) \le 3$.

[hide="solution"]$ \frac {1}{2}\leq \sin x\leq 1$ on $ \left [\frac {\pi}{6},\frac {\pi}{2}\right ]$ so $ \exp \left( \frac {1}{2}\log 3 \right) \le \text{exp}\left(\displaystyle\int_{\pi/6}^{\pi/2}\frac {\sin{x}}{x}dx\right) \le \exp \left( \log 3\right)$ as desired.

This seems like it should be in the calculus section.[/hide]


$ \varpi$ $ \varpi$
$ \bigcup$ $ \bigcup$
is that cool or what?!?!?!?
 :rotfl:  :ninja:  :rotfl:

Now that I'm back, it's time to start cleaning up this trash.


function
algebra
polynomial
binomial theorem

Find the coefficient of the $x^3$ term in the expansion of $(x^2+x-4)^7$

[quote="coffeym"][hide]$\mbox{-34048}$[/hide][/quote]

Whoa.  How did you get that?

Use binomial theorem to expand the function, keeping in mind that only some pieces are relevant.  Find the parts of the expansion which contains $\mbox{x^3}$, evaluate them, and collect like terms  :) For using the binomial theorem of $(\mbox{a+x})^n$, simply let $\mbox{a}$ be $\mbox{x^2}$ and replace $\mbox{x}$ with $\mbox{x-4}$.

[hide="Easier explanation"]Put all the parentheses next to one another: $\underbrace{(x^2+x-4)(x^2+x-4)\dots (x^2+x-4)}_{7}$. Since multiplying polynomials comes to "multiplying each with each of the others", the problem is how to count the cases where the product is $x^3$, and those are:

1. Choose $x^2$ from one of those, $x$ from another and $-4$ from the rest, where the coefficient would be $(-4)^5$;
2. Choose $x$ from one of those, $x^2$ from another and $-4$ from the rest, where the coefficient would be $(-4)^5$;
3. Choose $x$ from three of those and $-4$ from the rest, where the coefficient would be $(-4)^4$.

Counting the cases and taking the coefficients into account, you get ${7\choose 5}(-4)^5+{7\choose 5}(-4)^5+{7\choose 4}(-4)^4=-34048$[/hide]


I've decided to make our own Impossible Quiz, with questions written by members of this forum.

I am open to suggestions for questions, minigames (like "Don't touch blue/green/pink/red"), music, etc.

The questions can be as random as you want.  I may not accept all questions, but I'll probably accept most of them.  Also, I don't mind making references to the original quizzes.

I'm also considering adding a section of the quiz dedicated to "classics," or questions from the original Impossible Quizzes that I liked.  You may nominate questions from either the first or second ones.

EDIT: These questions can, but don't have to be, math-related.  Also, if you make up a question, please give the answer.

can it / is it supposed to be maths orientated?

I've made some changes to the requirements.  I no longer require a logical explanation to questions.  So it's pretty much anything goes.

You're going to create an SWF...?

How about

"Simplify $ \sqrt[4]{(\minus{}4)^2}$.
a) 2
b) i
c) 2i (CORRECT)
d) Arfies."

Because $ \sqrt[4]{(\minus{}4)^2}\equal{}(\minus{}4)^{2/4}\equal{}(\minus{}4)^{1/2}\equal{}\sqrt{\minus{}4}\equal{}2i$

[quote="Temperal"]You're going to create an SWF...?

How about

"Simplify $ \sqrt [4]{( \minus{} 4)^2}$.
a) 2
b) i
c) 2i (CORRECT)
d) Arfies."

Because $ \sqrt [4]{( \minus{} 4)^2} \equal{} ( \minus{} 4)^{2/4} \equal{} ( \minus{} 4)^{1/2} \equal{} \sqrt { \minus{} 4} \equal{} 2i$[/quote]

Except that that is simply incorrect by parenthetization. Make it $ \sqrt[4]{\minus{}4}^2$.

Well, if I put that in, I'll just throw in some ambiguity, so you can't tell whether or not the exponent is in the radical or not.

Just a reminder: although this is a math forum, [b]the questions do NOT have to be math-related.[/b]  They can be, but they don't have to.

I've got one:

Is the answer to this question no?

A: Yes it is.     B: Of course it isn't.

C: Maybe.(correct)       D: No.

Keep 'em coming!  I will also take mouse minigames, so if you have any ideas for those, tell me.

Hmmm...on my computer right clicking freezes the flash thing, so it didn't detect me mousing over the bad part, even if the right click menu wasn't there.

Darn

oh well its more practical I guess then...

[quote="tjhance"]Maybe there could be a mouse minigame where the spot  you're trying to reach
[hide="spoiler about the original quiz here"]
is too far away for the right-click menu to reach...
[/hide]
then the only way to get there would be to write a computer program to automatically move your mouse  :D

This is actually what I did for the problem on the original impossible quiz.

This may not be practical though, because, depending on who would end up playing this, may not know how to make such a program...

Another question:

What is 3+3
1) -9
2) -6
3) +6
4) +9

The answer will be +9 because if + is treated as a variable then we have 3 times + times 3 which is + times 9.[/quote]

If the thing was too long for the right click menu, this is what you could do. Alt+tab, and then move your mouse to the place on the screen where the thing you have to reach is, and then alt+tab back.

[quote="lingomaniac88"][quote="tjhance"]This is actually what I did for the problem on the original impossible quiz.[/quote]
Wow.  Here's what I did for that one:
[hide]Alt-Tab.[/hide]
I plan on using that concept for a mouse minigame.[/quote]

oh yeah? i have a good cheat for those mouse games.

[hide="this is a crucial part"]Find the shortcut for minimization of a window using the keyboard. Then  get another window, and then call up the impossible quiz. Then click the other window, and move your mouse over the goal on the IQ. Click twice, then presto![/hide]

Hm... I'll try another one.

Are you nowhere?

(a)Maybe
(b)Cow
(c)No I am nowhere
(d)No

[hint]
sdrow no nup.

[hide="Answer"]
C.
[/hide]

Find x.
1) 1
2) 2
3) 3
4) 4
After you click on something, the number has a slash through it.
e.g
O) 0
after you click on O it becomes the null symbol.
[hide="solution"]
click on any of the choices.
Now click on it again.
Those two slashes form an X. Click on the X.
[/hide]

69.  Can you pass the henway?

(A) What's a henway?
(B) Do you mean salt?
(C) About 5 pounds.
(D) *I can't think of a 4th choice*

You don't know the joke?  You say, "Can you pass the henway" at the dinner table.  Then somebody says, "What's a henway?" and you say, "About 5 pounds."  So (C).

I know that joke.  I just wasn't sure what the answer was.

#whatever

The smiley  :starwars:  pops up. A text message pops up:  "Who will win?"

A: Darth Vader
B: Obi-Wan Kenobi
C: Luke Skywalker
D: Darth Maul

[hide="answer:"]If you know Star Wars well, you may notice that the good guy has a green lightsaber, and continuously fights the other guy. Therefore, Obi-Wan Kenobi and Darth Maul are voted out.

When Luke had a green lightsaber, he defeated Darth Vader. So C.[/hide]

!!!!

A: !
B: !
C: !
D: !

and in the middle of the screen put !!!! and if you hover over it, you can see it, and click the !!!!

yes, this is stupid, but it IS the ImAoPSible Quiz.

[quote="1=2"]After question 339, go to

1337.

Did you even THINK that we wouldn't put this level on?

(a) No.
(b) Yes.
(c) I am ecstatic.
(d) LEET YEAH!

[hide="answer"](b) or (d) are the right answers.[/hide][/quote]

I could actually put that as #337, and have a 1 next to it, making it question #1337.

I know, I haven't had time to work on the quiz in a long time.  But I managed to get a 69th question in.

As of now, no.  This is mainly because there are still some things I need to work out.

And Where is all of this going to be going on because I want to join :)

Here's one: Type the luckiest number.
The answer is:
[hide]you have to type the phrase "the luckiest number."  if they type a number, make it game over!!!!!!!![/hide]


probability
combinatorics unsolved
combinatorics

Disclaimer: I'm not sure if this should be posted here, but i don't know where else to post probability questions that require Poisson modelling, etc. I'll let a mod move it if necessary.

In order to get money from a cash dispenser I have to punch in an identification number. I have forgotten my identification number, but I do know that it is equally likely to be any one of the integers $ 1, 2, \cdots , n$. I plan to punch in integers in order until I get the right one. I can do this at the rate of $ r$ integers per minute. As soon as I punch in the first wrong number, the police will be alerted. The probability that they will arrive within a time t minutes is $ 1\minus{} e^{\minus{}\lambda t}$ where $ \lambda$ is a positive constant. If I follow my plan, show that the probability of the police arriving before I get my money is \[ \sum^n_{k\equal{}1} \frac {1 \minus{} e^{\minus{}\lambda (k\minus{}1)/r}}{n}\]

Simplify the sum.

On past experience, I know that I will be so flustered that I will just punch in possible integers at random, without noticing which I have already tried. Show that the probability of the police arriving before I get my money is \[ 1 \minus{} \frac{1}{n \minus{} (n \minus{} 1)e^{\minus{}\lambda r}}\]

[quote="aidan"]Disclaimer: I'm not sure if this should be posted here, but i don't know where else to post probability questions that require Poisson modelling, etc. I'll let a mod move it if necessary.[/quote]

The [url=http://www.artofproblemsolving.com/Forum/index.php?f=498]Probability and Statistics Forum[/url] would be a good place in the future -- definitely better than either intermediate topics or the combinatorics forum.


function
ratio
algebra
polynomial
geometric sequence

The following two questions completely stumped me, so I was wondering if anyone of you could possibly set me on the right track:

$1. \text{ Prove that } 2x^3 -3x^2 +x +31 \geq 0.$

$2. \text{ Find the result of dividing } x+x^2+...+x^{100} \text { by } x^{-1} + x^{-2} + ...x^{-100}.$

Number 1 is certainly false without more conditions.  Consider, say, x = -100000000000 as a counter-example.  For the second one, [hide="Hint"]try to express those things as rational functions (ratios of polynomials) so that taking a quotient is easier.[/hide]

Yes, 1 is definetly false.

[hide="hint on 2nd"]
They are both geometric sequences.  Use the formula for sum of geometric sequences.
[/hide]

[hide="ans for 2nd"]
Formula for sum of geometric sequence:
$\frac{a(1-a^n)}{1-a}$ where a is initial value and n is number of terms we're adding.
So the top of the fraction will be:
$\frac{x(1-x^{100})}{1-x}$
And the bottom will be:
$\frac{x^{-1}(1-x^{-100})}{1-x^{-1}}$
Simplifying was not fun, but I simplified to
$\frac{x^{100}-1}{x^{100}(x-1)}$
Then dividing we get:
$\frac{x^{101}(x-1)(x^{100}-1)}{(x-1)(x^{100}-1)}=x^{101}$
[/hide]

hey, that was a pretty slick trick on the second one

sorry for the mess up on the first one, how about this one:

$x^{12} - x^9 + x^4 - x + 1 > 0$ for all real $x$.

(edit)

I got it down to $(x)(x^8+1)(x^3-1) + 1 > 0$

My reasoning was that if $x \geq 1$, then all three terms in the polynomial would multiply to something either greater than one or equal to 0, but then the +1 at the end would make it greater than 0.  If $0<x<1$, then the product of the three terms would also be >-1 and <0, but the +1 would make it positive. If x was negative, then the $x^3 - 1$ and the x would make it positive.

Is this sufficient enough?

whoa really? is my answer close to the real one?

Im sure you can make your method work, but in the solution to this, first its obvious for negative x, so assume x is greater than 0. Then divide by x^3 and i forget what happens after this....

How do you prove the bit 'if x is less than 1?

what do you mean by greater less than 1?


Prove that:
$1)rank(A+B)\leq rank(A)+rank(B)$
$2)rank(A-I_n)+rank(A+I_n)=n$
$3)rank(A)+rank(B)-n \leq rank(AB) \leq \min \{rank(A),rank(B)\}$

(take $A=0$)

[quote="spix"]$2)rank(A-I_n)+rank(A+I_n)=n$[/quote]

What was actually meant here? Was it $\geq n$?

For the other two, see here: http://www.mathlinks.ro/Forum/viewtopic.php?t=60203 .

\begin{eqnarray*} \text{Im}(A+B)&\subseteq&\text{Im}(A)+\text{Im}(B) \\ \text{dim}(\text{Im}(A+B))&\leq&\text{dim}(\text{Im}(A)+\text{Im}(B)) \\ &\leq&\text{dim}(\text{Im}(A))+\text{dim}(\text{Im}(B)) \end{eqnarray*}
By definition, this means that
\[ \text{rank}(A+B)\leq\text{rank}(A)+\text{rank}(B)  \]

blahblahblah, what do you think about the second one?

I think it makes no sense as stated; it should probably be $\geq n$.


let ABC be an equilateral triangle.p is any point inside the triangle.from which PF,PEand PD are perpendiculars from the point p to AB,AC, and BC.prove that the ratio (PD+PE+PF)/(BD+CE+AE) does not depend on the position of P and remains unchanged for all positions of P.
please help !

i think it will be AE?why?nothing seems wrong???

No, if AE is what you mean, then AE+CE=AC, if by feet of perpendiculars, you mean the intersection of the altitudes  from P with the sides of the triangle.

It should be AF, by symmetry
[hide="Solution"]
Note that $ [PF \cdot AB \plus{} PE \cdot AC \plus{} PD \cdot BC]/2$ is the area of the triangle. Since $ AB \equal{} BC \equal{} CA \equal{} s$, and the area is $ \frac {s^2 \sqrt {3}}{4}$, the numerator of your fraction is $ \frac {s \sqrt {3}}{2}$

For the denominator, draw an altitude AY of $ \triangle ABC$, extend $ EP$ to intersect it at K. Let X be the foot of the altitude from K to AB. Since $ PF \parallel KX$ and $ PD \parallel KY$, we have $ \angle XKA \equal{} \angle AKE \equal{} 60$, so $ \angle XKP \equal{} 120$ and $ \angle FPK \equal{} 60$. Similarly $ \angle KPD \equal{} 60$. Thus, $ FX \equal{} YD$. So we have $ AF \plus{} BD \plus{} CE \equal{} (AX \minus{} FX) \plus{} (BY \plus{} YD) \plus{} CE \equal{} AX \plus{} BY \plus{} CE \equal{} \frac{3s}{2}$.

Thus, the fraction you gave equals $ \frac {\sqrt {3}}{3}$[/hide]

[quote="earth"][b]An easy extension.[/b] Let $ ABC$ be an acute triangle and let $ P$ be an interior of $ \triangle ABC$ . Denote the projections $ D$ , $ E$ , $ F$ of point $ P$

on sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively. Prove that $ \frac {a\cdot BD \plus{} b\cdot CE \plus{} c\cdot AF}{a\cdot PD \plus{} b\cdot PE \plus{} c\cdot PF} \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{4S} \equal{} \sum\cot A$ (constant).[/quote]
[hide="Indication."]
Denote $ \left\|\begin{array}{c} BD \equal{} x \\
\ CE \equal{} y \\
\ AF \equal{} z\end{array}\right\|$ and apply the [u][b]Carnot[/b]'s lemma[/u] : $ \sum x^2 \equal{} \sum (a \minus{} x)^2\implies 2(ax \plus{} by \plus{} cz) \equal{} \sum a^2$ .

Observe easily that $ \sum a\cdot PD\equal{}2S$ and $ 4S\equal{} \left(b^2 \plus{} c^2 \minus{} a^2\right)\cdot\tan A\implies 4S\cdot\sum\cot A \equal{} a^2 \plus{} b^2 \plus{} c^2$ a.s.o.[/hide]

Wait, I thought there was something about intersection of altitudes with the sides... sorry if this is spam.


search
pigeonhole principle
combinatorics unsolved
combinatorics

Given $ 10^{6}$ points in the space ,show that the set of pairwise distances of given points has at least $ 79$ elements

this is IRAN 2005 try to search it,it has been posted before..
the main idea is pigeonhole,you should draw some sphares...  :wink:
try to search for a complete solution

for example take a look at [b]myth[/b]`s post here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=34948[/url]


geometry
rectangle
parallelogram
LaTeX
pigeonhole principle
counting
distinguishability

Hello,

This is my first thread here, sorry if you don't think this is the right section.
I want to post some problems looking for help here, but I'm having trouble finding the right place.

Please let me know how difficult you think these are, for me to post in the right sections in the future.
If you haven't seen these anywhere else in the forums and want to go ahead and try to figure them out, please do so.

1. Find all primes p, q so that
p 2 + q = 37 q 2 + p .
Note: p > 1; q > 1

2. A game's pieces are squares of side length 1, and their sides are painted in 4 different colors: blue, red, yellow and green, in such a way that every piece has a different color for each side. There're pieces with every possible combination of colors, and the game has one million pieces of each class. With them, rectangular puzzles are formed, without any holes or juxtapositions, so that two pieces that share a side have the same color in that side.
Determine if with this procedure it's possible to come up with a rectangle of size 99x2007 with each side of the same color. What about 100x2008? And 99x2008 ?

3. ABCD is a parallelogram of sides AB, BC, CD, AD, so that AB > AD and AC/BD = 3. Let r be the line symmetric to AD in respect to AC, and s the symmetric line of BC in respect to BD. If r and s intersect in P, calculate the value of PA/PB

4. We'll say that a positive integer is [i]lucky[/i] if 31 divides the sum of its digits. What's the maximum possible difference between two consecutive [i]lucky[/i] numbers ?

5. For every positive integer [b]n[/b] we consider the progression of 2004 integers:

[img]http://mt15.quickshareit.com/share/8fc1b3e51c0d7f0348.png[/img]

Determine the smallest integer [b]n[/b] so that the 2004 integers of the progression are 2004 consecutive integers.
Note: the brackets indicate the integer part

6. Find all positive integers a, b, c, d so that
a>b, a^2c=b^2d and ab+cd=2^99+2^101

7. The pentagon ABCDE has AB=BC, CD=DE, angles ABC = 120, CDE = 60 and side BD = 2. Find the area of the pentagon.

Sorry for not using LATEX, I'll look into it right now.
Thanks in advance

[hide="1"]
$ p^2 \plus{} q \equal{} 37q^2 \plus{} p$ $ \implies$ p(p-1)=q(37q-1) $ \implies$ p-1=nq and 37q-1=np $ \implies$ p=nq+1 $ \implies$ 37q-1=n(nq+1) $ \implies$ $ q \equal{} (n \plus{} 1)/(37 \minus{} n^2)$. We see n=6 is the only possibility, hence q=7 and p=43.[/hide]
Not too hard in my opinion, but a nice problem. This is the right section.
For the last problem, BD is not a side, do you mean diagonal BD, side CD, or side BC?

[hide="5"]
[url=http://www.research.att.com/~njas/sequences/A000037]This[/url] might help

[/hide]

[hide="4"]
If take the last two digits, say a and b, and turn them into a+1 and b-1, the sum will be the same.
This will always be an addition of 9 unless b is 0.

49990 --> 50899 is a 909 difference, which is the maximum if I'm correct

[/hide]

Is there really enough information for #7?

Even if it was side BC, CD, or diagonal BD, there should be more than one possible area of the pentagon. And since you only gave a length to one side, it can really be as long or short as you want in comparison to the pentagon. Like, I've drawn many pentagons that satisfy the condition, but obviously have different areas.

[hide=4]

let a be a digit smaller than 9.

_ ... _ a 9 next lucky is _ ... _ (a+1) 8 with a difference 9

_ ... _ a 9 9 next lucky is _ ... _ (a+1) 9 8 with a difference 99

_ ... _ a 9 9 9 next lucky is _ ... _ (a+1) 9 9 8 with a difference 999

_ ... _ a 9 9 9 9 next lucky is _ ... _ (a+1) 0 0 0 4 with difference 5.

Max diff is 999 (Unless we are not in decimal representation of numbers)

[/hide]

thoughts on #2

pigeonhole for the 1 million pieces in however many different types of pieces there are... prehaps $ 4!/4 \equal{} 6$, or can you flip pieces over in which case there are only $ 3$ distinguishable?

Anyway, if there are $ n$ types of pieces we have $ 1,000,000/n$ rounded up for the least possible number of the most common piece that we have.  Since the colors have no significance in the problem, we can just assume, without loss of generality, the arrangement of colors on the most common piece to be something specific.

After that, it is just figuring out if we have enough tiles of each type to guarantee that a tiling is possible.

But I've had enough of tilings for about a month.  ARML.  Anyway, perhaps I'll come back after I get some sleep, but if anyone wants to work off here, feel free.

Thoughts for 5

only possible if $ \sqrt {n \plus{} 2003} \minus{} \sqrt {n} < 1$ because otherwise the square root must cause the integer part to change.  Therefore $ 2n \plus{} 2003 \minus{} 2\sqrt {n(n \plus{} 2003)} < 1$ and thus $ (n \plus{} 1001)^2 < n(n \plus{} 2003) \iff 2002n \minus{} 2003n \plus{} 1002001 < 0$ which has the solution set $ n > 1002001$, so our desired answer is at the least $ n \equal{} 1002002$, but this fails.

Can anyone prove whether the number has to be square or not?  because $ n \equal{} 1004004 \equal{} 1002^2$ is the first successful case I can find.  Also, $ 1002^2 \minus{} 1001^2 \equal{} 2003$ so adding $ 2003$ gives an exactly perfect square, but $ 1003^2 \minus{} 1002^2 \equal{} 2005 > 2003$ so adding $ 2003$ makes no difference with $ 1004004$... Thoughts?

I misread on #2.  there are 1 million pieces of each type.  This makes the problem much easier.

The picture attached uses 2 of the different possible pieces.  it can be repeated as much as needed just by placing such blocks of pieces side by side, without rotation.  If an odd width or height is needed for the rectangular puzzle, only a portion of one of these squares may be used to fill in the edges.  Without loss of generality, we can always start by positioning one of these 2 by 2's in the bottom left corner of the puzzle.

Let at least one of $ m$ and $ n$ be even integers.  Then, since one of the dimensions is even, to tile an $ m$ (rows) by $ n$ (columns) rectangle, we must have $ mn/2 \le 1000000 \implies mn \le 2000000$ because each piece is used once in every other time in the puzzle.

Let both $ m$ and $ n$ be odd.  Then we have that the bottom left tile is used a total of $ \lceil\frac{mn}{2}\rceil \equal{} \frac{mn\plus{}1}{2} \le 1000000$ times, so $ mn \le 1999999$

All cases that were specified come nowhere close to violating either inequality, so all have a possible tiling with the rules described.

6) Given: $ a>b$, $ a^{2c}\equal{}b^{2d}$, $ ab\plus{}cd\equal{}2^{99}\plus{}2^{101}$, and all numbers are integers.

thoughts:

$ a^c\equal{}b^d$ so we know that $ a \equal{} b^{\frac{d}{c}}$,  which tells us that $ c|d$.  Let $ d\equal{}cn$, for some natural number $ n$.  Then we have $ a\equal{}b^n$.

$ ab\plus{}cd \equal{} b^{n\plus{}1} \plus{} nc^2\equal{}2^{99}\plus{}2^{101}$, so perhaps we have lots of numbers which are powers of 2?

[quote="Guiller"]
3. ABCD is a parallelogram of sides AB, BC, CD, AD, so that AB > AD and AC/BD = 3. Let r be the line symmetric to AD in respect to AC, and s the symmetric line of BC in respect to BD. If r and s intersect in P, calculate the value of PA/PB[/quote]

My approach for this very hard and nice geometry problem:
W.L.O.G, we can "give" some additional data as: BD=10,AC=30,AB=19,AD=12 with consideration that it satisfies every triangle inequalities to make sure that this parallelogram does exist. Now, applying Laws of cosines we will get:
FC^2=FA^2+AC^2-2FA.AC.cosFAC=FA^2+900-60FA.cosFAC=FB^2+BC^2-2FB.BC.cosFBC=FB^2+144-24FB.cosFBC
(note that cos(FBC)=cos(2DBC)=1-2cos^2(DBC),which is easily calculated by using additional data above. Similar to cos(FAC). This is the first equation)
FD^2=FA^2+144-24.AF.cos(FAC)=FB^2+100-20.FB.cos(FBD)(again, we can easily calculated those values of cos(FAC) and cos(FBD) by using the same method as we did for cos(FBC) and cos(FAC).This is the second equation)
From two equations above, we see that it can be solved easily by adding 1st and 2nd equation to use the substitution. Or doing by other ways as long as we get the quadratic equation for one variable(FA or FB).Calculate it and easy to find the other(Done!)
PS: This is just theoretical solution! I'm too lazy to do the rest :D . Anybody interests in solving this problem can try doing it in this way, or find a better solution for this, which I can't think of.

[quote="CircleSquared"][

If take the last two digits, say a and b, and turn them into a+1 and b-1, the sum will be the same.
This will always be an addition of 9 unless b is 0.

49990 --> 50899 is a 909 difference, which is the maximum if I'm correct

[/quote]
The counter-example to your result is: 5000899 and 4999000, which I'm sure that is two consecutive lucky numbers and its difference is 1899, which is greater than 909. I'm almost sure that this problem needs to have some restrictions on the number of digits of the lucky numbers.If not, it won't be possibly solved! :wink:

[quote="facis"]$ a^c \equal{} b^d$ so we know that $ a \equal{} b^{\frac {d}{c}}$,  which tells us that $ c|d$.[/quote]

$ 10 \equal{} 100^{\dfrac{1}{2}}$, yes?

If $ c$ and $ d$ are relatively prime, $ b \equal{} k^c$ for some $ k \in \mathbb{Z}$. But we don't know that they are.

You also have the problem wrong. It's not $ a^{2c} \equal{} b^{2d}$, it's $ a^2c \equal{} b^2d$.


Two circles and three straight lines lie in the same plane.  If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection.

[hide]17[/hide]

*Could someone please post a solution to this problem, a diagram if necessary?

**Also, could someone teach me a shortcut for problems like these in general?  Is there a better way than to just keep guessing and checking by drawing a bunch of  diagrams?

Thank You!

[quote="abacadaea"][hide]
2 circles can intersect at a maximum 2 points. 2
the first line can intersect each circle at most 2 points each. 2+(2*2)=6
the second line can intersect each cricle and the other line. 6+(2*2+1)=11
the third line can intersect the circles and the other 2 lines 11+(2*2+2)=17
[/hide][/quote]

Does this approach work for every problem of this kind?  Thanks.

[hide="solution"]there is an equation, so

$ \binom{2}{2}\times{2}$ + $ \binom{3}{2}$ + $ 2\times{3} \times{2}$ = $ \fbox{17}$

the equation is $ \binom{c}{2} \times{2}$ + $ \binom{l}{2}$ + $ c\times{l}\times{2}$, where c represents number of circles and l represents number of lines.   [/hide]


group theory
abstract algebra
pigeonhole principle
superior algebra
superior algebra unsolved

1. Let G be a group and H and K are two subgroups of G. Let a,b Є G. Show that either Ha ∩ Kb = Φ or Ha ∩ Kb = (H ∩ K)c for some c Є G.

2. Let G be a group and H and K are two subgroups of G of finite indices. Show that (H ∩ K) is of finite index.

thank you.... :)

I see that way:

2) We prove by taking the contrapositive. Assume $ [G: H\cap K]$ in infinite,
then there exist infinitely many $ g_1, g_2, \cdots \in G$ such that, for every
$ 1 \le i \neq j$, $ (H \cap K) g_i \neq (H \cap K) g_j \iff g_ig_j^{ \minus{} 1} \not\in H\cap K$.
By the infinite pigeonhole at least one of $ H,K$, wlog let it be $ H$, must be s.t.
there exist infinitely many indices $ i_1, i_2, \cdots$ such that for every $ 1 \le k \neq j$
$ g_{i_k}g_{i_j}^{ \minus{} 1} \not\in H$, so $ Hg_{i_k} \neq Hg_{i_j}$. This implies the thesis.

So here is another question related:
If $ H,K \le G$ are of finite index, find an upper bound on the index of $ H\cap K$.

thank you for the solution sir....
can this problem be done in any different way?

and please help me out in the first problem anybody

thank you.... :)

(1) Decompose $ H,K\leq G$ into a disjoint union of right cosets with regard to $ H\cap K$, i.e $ H\equal{}\bigcup_{i\in I}(H\cap K)h_i,K\equal{}\bigcup_{j\in J}(H\cap K)k_j$, then we calculate for arbitrary $ a,b$:

$ Ha\cap Kb\equal{}\bigcup_{i\in I,j\in J}[(H\cap K)h_ia\cap (H\cap K)k_jb]\equal{}\bigcup_{v\in V}(H\cap K)g_v$ where the last disjoint union is defined for some appropriate index set $ V$ and representatives $ g_v\in G$. Now we have to verify $ |V|\in\{0,1\}$. If we observe $ (H\cap K)h_{i_s}a\equal{}(H\cap K)k_{j_s}b$ for $ s\equal{}1,2$, then there exist $ u_1,u_2$ in $ H\cap K$ with $ u_sh_{i_s}\equal{}k_{j_s}ba^{\minus{}1}$ and this will imply

$ h_{i_2}^{\minus{}1}u_2^{\minus{}1}u_1h_{i_1}\equal{}k_{i_2}^{\minus{}1}k_{i_1}\Longrightarrow k_{i_2}^{\minus{}1}k_{i_1}\in H\cap K\Longrightarrow k_{i_1}\equal{}k_{i_2}$
$ \Longrightarrow h_{i_2}h_{i_1}^{\minus{}1}\in H\cap K\Longrightarrow h_{i_1}\equal{}h_{i_2}$

2. consider $ G/(H \cap K) \to G/H \times G/K$.

would you explain little elaborately sir?
thank you....


AMC
USA(J)MO
USAMO
AIME
trigonometry
AMC 12
ARML

well, obviously Arnav/Jeremy made it

and Berman

Vitek?
Misha?
Vivek?

I did, I think

Bryce?

and who else

Well, you should know if you definitely made it (AIME I index above 220 is definitely safe, dunno about AIME II index, AIME I floor 7 is definitely safe, AIME 2 floor 9 is definitely safe). I think it's pretty certain that 6 is the floor (with tiebreakers) for AIME I.

From what I've heard from Yasha it sounds like he's definitely in... but I guess that's not really a surprise. What about the three Enloe people?

Did you purposefully exclude yourself, not_trig, or was that an oversight?

[quote="not_trig"]

I did I think

[/quote]

um no berman, i did not exclude myself... :wink:

thanks for the imposed humility though i need more of that

oops. that is a very awkward and easily skipped over sentence, in defense of my foible.

EDIT: ah. much better.

misha - yeah, most likely - 132 AMC12 + at least 110 AIME = 242, maybe 252 which puts him in.

vivek - probably not - 146.5 AMC12 + 70 aime = 216.5 index, maybe not especially for AIME II

me (vitek) - probably - 123 AMC12 (bleah) + 110 AIME II = 233 index, but floor score should put me in (hopefully).

2/3 enloe is not bad.  maybe it will inspire vivek to work for usamo next year as he will almost certainly make it.

Yeah I made it... unless major misbubbling  :maybe:

Welcome to NC, Zhai. Glad to have you on our ARML team.

233 is definitely a safe index. Even without, they'd have to give 15 #1's for the floor to be as high as 11 (with 500 qualifiers especially). So... you're fine.

When the results come out, AIME I scores are going to look severely artificially deflated. Oh well.

[quote="jb05"]Welcome to NC, Zhai. Glad to have you on our ARML team.
[/quote]

did we abduct him or did he come by himself? :D

btw good job vitek

Also, to add to the list, I'm pretty sure Jenny Chen got 9 on the AIME I, so she qualified.

lol the 2007 state stuff is "up" on Mr. Reiter's site but not really

 :roll:

all give 404 messages... hopefully will be up soon?

Hmm... he's a state coordinator, right? So does that mean he gets a state report? Otherwise, I don't see how he could get any information up there. If that's true, then stats will probably be up next week (or sooner).

Last year it took like half a week between the time he updated the site and the time the scores came up.

But really... I don't care this year, because AIME I scores are going to look dreadful and the AMC12 read my erasures wrong...

? read your erasures wrong?

i'm just interested in this kind of stuff... i guess i'm kinda vain or something

I have to explain myself? Well, everybody's interested in this stuff. I meant that my recorded AMC score was a problem less than my actual AMC score, due to the fact that I got a space off on my bubbling and had to erase everything and fill it back in in the last five minutes. So... I'll be pretty low on the lists, which is why I'm not too interested.

[quote="b-flat"]You mean that the cutoff has been posted officially?

Yes, I'm pretty sure that I made it, unless the cutoff is 10 or I miss bubbled a lot (I checked over to make sure I bubbled correctly).[/quote]

I really hope that the cutoff is 7 for floor?

Man, I can't believe you guys are already making some bigtime predictions.

Ouch... beaten by a 7th grader  :P

Overall, I'm guessing there will be fewer NC qualifiers this year than there were last year  :(

On the other hand, I wouldn't be surprised if the average NC score on the USAMO goes up

[quote="Teki-Teki"]jenny chen got in?!?  impressive...

what is she, 7th grade? 8th grade (I think the latter)

maybe I'm putting a little too much into this, but that's very impressive.  maybe in 3-5 years she will be our 'Arnav'.  hahaha.

seriously.  maybe even the next melanie wood.  she'll probably go to IMO.

 - daniel[/quote]

You are surprised that Jenny Chen did well?  She's probably the top girl in the state now that Amy Wen (can't remember last name; she was on the A team last year from the school of Science & math) is in college.

She's in 8th grade with me (I've been competing with her & Brendan since 6th grade).

@n^4+4: she's in 8th!

[quote="N^4+4"]
Overall, I'm guessing there will be fewer NC qualifiers this year than there were last year Sad
[/quote]

You might be surprised.  Expect a lot of new people on the USAMO qualifier lists and a few previous qualifiers who are on the bubble this year.

An average state with roughly 3% of the US population, ought to produce 15 USAMO qualifiers with expansion this year.  NC should produce at least 20.  Many will be 8th and 9th graders.

Well, I guess I was wrong about Jenny Chen, because she was not on the list.

[hide="14 people!!"]$\begin{tabular}{|c|c|c|c|}\hline Tian-yi Jiang &9 & Cary Academy & Cary \\\hline Andrew Hertzberg &10 & Chapel Hill HS &Chapel Hill\\\hline John Pardon &12 &Durham Academy Upper School &Durham\\\hline Jeremy Hahn &11 &East Chapel Hill HS &Chapel Hill\\\hline Arnav Tripathy &12 &East Chapel Hill HS &Chapel Hill\\\hline Bryce Taylor & 8 &Hanes MS &Winston-Salem\\\hline John Berman &10 &John T Hoggard HS &Wilmington\\\hline Yakov Berchenko-Kogan &12 &Needham B Broughton HS& Raleigh\\\hline Joseph Lozier &12 &North Carolina School Sci/Math &Durham\\\hline Ray Wang &11 &North Carolina School Sci/Math &Durham\\\hline Seoungjun Lee &10 &Ravenscroft School &Raleigh\\\hline Vivek Bhattacharya &11& William G Enloe HS &Raleigh\\\hline Mikhail Lavrov &12 &William G Enloe HS &Raleigh\\\hline Daniel Vitek &10 &William G Enloe HS &Raleigh\\\hline \end{tabular}$[/hide]

looks like I was the only 8th grader to qualify for USAMO but not mathcounts Nationals...

Yay! Once again, we can have an ARML team's worth of present and past USAMO qualifiers.

[quote="b-flat"]looks like I was the only 8th grader to qualify for USAMO but not mathcounts Nationals...[/quote]

Well, that puts you above the 150+ who qualified for mathcounts nationals but not USAMO. :wink:

not sure if I understand completely...there are only 14 people from NC who made USAMO.  Then again if Hyunji Bae from east chapel hill comes (he made USAMO last year as a sophomore but not this year as a junior) we will have 15 people who are USAMO qualifiers.  yay!!!

edit: the only trouble is that two of them (hertzberg and seongjun) probably don't know what ARML is.

Great job Bryce!  USAMO in the 8th grade is pretty cool  :wink:

Hyunji Bae is not a "he"

Rather, she is Jane, and was on the ARML B Team last year.

Unless she does excellently on the state math contest I'm guessing she won't be on the A team  :(

Enloe has a well-deserved claim to more USAMO qualifiers than East.  You folks seem to consistently produce very good people, but East's recent success has been mostly a matter of luck.  Arnav, Noah, and I all wound up at the same school semi-randomly.  Our math team's coach has given us lots of wonderful opportunities to take interesting tests, but doesn't really provide much training for either us or the younger students.  Now Noah is gone, and Arnav soon will be too.

Does anyone know Hertzberg or Lee?  Have they done well on any other competitions?

Now the question is: How many will make MOP?  :)

They updated it with 17 new people and Kevin Lang was one of them! We now have an ARML team's worth of USAMOers!

okay - sorry to 'jane'.  i am trying to get in contact with lee and hertzberg, maybe i can get them to come to the may 19/20 arml practice.

just figured out who probability1.01 is...yay, that makes 2 imoers and if yasha/jeremy makes it, then that makes 3.  wow.  that would be beast.

umm...enloe has had 7 USAMO qualifiers since like 2002 and none before 2004 that I know of...so I dunno.  misha and vivek and I all happened to go to the same school sort of randomly too...

but dude arnav - noah should have stayed.  then next year you guys could have had a fighting chance against beating chapel hill's total.

hmm...actually, now that zhai is here, that makes 16 USAMOers and 17 if we include jane.  let's see...Jeremy, Arnav, Yasha, and Alex should make MOP, and Berman and us 3 Enloe kids have a fighting chance.  That makes 4, probably 5 as Misha is pretty good and looking to avenge not making USAMO last year, maybe 6 or 7.  7 NCers (even if 1 is an 'immigrant', per se) would be amazing.  dude.  we could like REPEAT!!! and easily, too.

Hahahaha :rotfl:.

Since when did Zhai move?

[quote]Welcome to NC, Zhai. Glad to have you on our ARML team. [/quote]

Umm...am I missing something here?

oh i see. :blush: w/e.  i definitely did NOT pick up on that.

the only trouble is that two of them (hertzberg and seongjun) probably don't know what ARML is.[/quote]

hey, I am hertzberg, I was just showed this forum by a friend. I just wanted to say that you are right and i dont know what ARML is. So could someone please tell me?

@whatthedeuce:  If David Mermin still hanging around Chapel Hill HS (perhaps teaching and/or coaching math teams), ask him about ARML.  He's one of the ARML coaches.

If that fails, try the [url=http://www.arml.com/]ARML website[/url] and the [url=http://courses.ncssm.edu/goebel/statecon/Inform/ARML2006/ARML2006.htm]NC State Contest ARML page[/url].

yeah - or there's probably someone who remembers arnav coming over from middle school - they might have a more personal insight.

anyway, welcome to the NC forum!  good job!

[quote="Teki-Teki"]okay - sorry to 'jane'.[/quote]

Apology accepted  :)


Good luck to everyone on this year's ARML team. Let's have fun and do well.

If you have any more comments like this regarding on everyone, post. But if you want to just cheer for one team, go somewhere else.  :rotfl:  :rotfl:

In that case...I'll cheer on 3 teams  :P

GO TEXAS GOLD!!!!!

GO TEXAS SILVER!!!!!!!

GO TEXAS BRONZE!!!!!!!!!!

Good luck and sorry to all the teams I sabotaged!  (I didn't say that...)

Well, SOMEBODY will win  :P

Go both teams from NC!

[quote="Sly Si"][quote="b-flat"]Well, SOMEBODY will win  :P [/quote]

I don't know about that... What if no one gets any points?  Does everyone win, or does no one win?[/quote]

lol!

we're all winners no matter if we "win" or not.

I've been at the western site the last three years as a grader, but I can't make it this year- I have to grade calculus finals instead.

Go Southern California!
Go Washington!

yay for arml, good luck to everybody...don't make careless errors

Good luck and have fun!

Go Chicago A!
Go Chicago B!
Go Chicago C!
Go Chicago D!
 :lol:

Go Bay Area!  Go Minnesota!

And don't forget to have fun!

[quote="MathChauffeur"][quote="MysticTerminator"]GO BLACK MOP TEAM!!![/quote]The only thing worse than having a BLACK MOP ARML TEAM would be  the BLACK MOP ARML TEAM losing at ARML  :D[/quote]

yeah there was a pretty big debate on one point about whether USACO team would beat BLACK MOP team or not (I would tend to think . . . NOT)

Wait... er... Arnav you're not serious about the Black MOP Team are you??? Because that's really ...  :wacko:  :?

No offense but this is something I see a lot from many of you.. I said:

[quote="Silverfalcon"]Good luck to [b]everyone[/b] on this year's ARML team. Let's have fun and do well.

If you have any more comments like this regarding on everyone, post. [b]But if you want to just cheer for one team, go somewhere else.[/b][/quote]

I hope you guys realize that I didn't mean to make this thread to be funny. I wished everyone good luck at ARML for real but it turned out that some people do not know when to be serious. It would've been at least OK if you say "good luck to everyone" and then cheer for your own teams. But some people just cheered for their own team and great, that's just great.  :|

[quote="jmerry"]

Go Washington![/quote]

especially team A :)

Yes, good luck to everyone at ARML!

I'm on a team that has a $\frac{1}{10^{10^{10^{10}}}}$ probability of winning, so I have no qualms about cheering on other teams.

Also good luck to the ARML scorers on getting all the tests graded in time! :rotfl:


[quote="VANCHANH123"][color=orange]16-X^4=(2^X-2^{-x})^4[/color][/quote]

I think you wanted to write solve:

$16-x^4=(2^x-2^{-x})^4$

am I not right?  :)

I think the equation can be solved by estimate two hand

If equation is
$16=x^4+(2^x+2^{-x})^4$
it is solved obviosly x=0, because $2^x+2^{-x}\ge 2\Longrightarrow x^4+(2^x+2^{-x})^4\ge 16$.


calculus
derivative
function
limit
calculus computations

If $ g(x)\equal{}0$ for $ x<0$ and $ g(x)\equal{}1$ for $ x\ge 0$, prove that there does not exist $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that $ f'(x)\equal{}g(x)$ for all $ x\in\mathbb{R}$.

Is it possible to show this by applying the definition of the derivative for $ g(x)$ as $ x\rightarrow0^{\minus{}}$ and $ x\rightarrow0^{\plus{}}$ and then equating that with the definition of the derivative for $ f(x)$ to obtain a contradiction?

What's a less confusing way to approach this problem?

I don't know how to say but existence of antyderivetive  needs that function must be "Parts" continuity.in your case $ g(x)$ is not "Parts" continuity.. yes you can solve by left and right limit

does anyone have an alternate way of showing that there doesn't exist an $ f(x)$?

[hide="By definition..."] We say that $ f'(a) \equal{} L$ when it is true that

$ \lim_{h \to 0^{\plus{}}} \frac{f(a\plus{}h) \minus{} f(a)}{h} \equal{} \lim_{h \to 0^{\plus{}}} \frac{f(a) \minus{} f(a\minus{}h)}{h} \equal{} L$

In other words, the left and right derivatives must be equal.  This is clearly not the case. [/hide]

Let's say I m not explicitly given the definition of the left and right hand limit and my definition of the derivative is simply for $ c$ a cluster point in $ D$ then

$ \forall \epsilon >0 \exists\delta>0$ such that $ |x\minus{}c|<\delta$, then

$ |\frac{f(x)\minus{}f(c)}{x\minus{}c}\minus{}L|<\epsilon$ (1)

Could I expand (1) for $ g(x)$ as $ x\rightarrow 0$

$ 0\minus{}\epsilon < \frac{g(x)\minus{}g(c)}{x\minus{}c}<1 \plus{} \epsilon$

and then somehow....say that since $ g(x)$ is not continuous then if there were such a $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that $ f'(x) \equal{} g(x)$ then $ f(x)$ couldn't be continuous contradicting the premise?

please suggest a better way i could approach this, thanks!

[quote="t0rajir0u"]In other words, the left and right derivatives must be equal.  This is clearly not the case.[/quote]Hmm, I don't find the "clearly" so obvious.  Could you elaborate?

[quote="mstrip"]Let's say I m not explicitly given the definition of the left and right hand limit and my definition of the derivative is simply for $ c$ a cluster point in $ D$ then

$ \forall \epsilon > 0 \exists\delta > 0$ such that $ |x \minus{} c| < \delta$, then

$ |\frac {f(x) \minus{} f(c)}{x \minus{} c} \minus{} L| < \epsilon$ (1)[/quote]

Pick $ c \equal{} x \plus{} \delta$ then $ c \equal{} x \minus{} \delta$.  This definition is equivalent.

how about the way I m choosing to show it. Is it correct? or should i use some other way of explaining it?

The standard argument here is the mean value theorem: if such an $ f$ exists, we must have $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}0$ for any unequal $ x,y\le 0$ and $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}1$ for any $ x,y\ge 0$. This makes $ f(x)\equal{}\begin{cases}a& x\le 0\\ x\plus{}b& x\ge 0\end{cases}$; by existence of $ f$, $ a\equal{}b$.
Once we have this, we can easily show that $ f$ is not differentiable at zero, contradicting our assumptions.

A stronger theorem: If $ f$ is differentiable on an interval $ I$, $ f'$ has the intermediate value property ($ f(J)$ is an interval for any subinterval $ J$ of $ I$).

Proof: Consider the difference quotient function $ g(x,y)\equal{}\frac{f(x)\minus{}f(y)}{x\minus{}y}$ for $ x>y$. Since $ f$ is continuous, $ g$ is as well. By the generalized intermediate value theorem, $ g(T)$ is connected and must be an interval, where $ T$ is the triangle $ x<y,x\in J,y\in J$.
In addition, $ f'(J)\supseteq g(T)$ by the mean value theorem; if $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}c$, there is some $ z\in (x,y)\subseteq J$ with $ f'(z)\equal{}c$. Conversely, if $ w\in J$ and $ f'(w)\equal{}c$, $ \frac{f(w_n)\minus{}f(w)}{w_n\minus{}w}$ converges to $ c$ whenever $ w_n\in J$ converges to $ w$ (without ever being equal); this means that $ c\in \overline{g(T)}$, and $ f'(J)\subseteq \overline{g(T)}$.

For any interval $ g(T)$, there are at most four sets between $ g(T)$ and $ \overline{g(T)}$. All four are intervals, and thus $ f'(J)$ is an interval.

The mean value theorem is [b]the[/b] way to go from information about a derivative on an interval to information about the original function.

[quote="mstrip"][color=darkred] If $ g(x) \equal{} 0$ for $ x < 0$ and $ g(x) \equal{} 1$ for $ x\ge 0$, prove that there does not exist $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that $ f'(x) \equal{} g(x)$ for all $ x\in\mathbb{R}\ .$[/color]  [/quote]
[color=darkred][b][u]Theorem,[/u][/b] If the function $ g: I\rightarrow\mathcal R$ admits a primitive, where $ I$ is an interval, then it has the [b]Darboux's property [/b](with other words, if the function $ g$ hasn't the [b]Darboux's property[/b], then it hasn't primitives).[/color]

[color=darkblue]In the our case the functin $ g$ hasn't the [b]Darboux's property[/b]
because $ Im\ g\equiv g(\mathcal R) \equal{} \{0,1\}$ which isn't interval.[/color]


Bob and Laquisha have volunteered to serve on the Junior Prom
Committee. The names of twenty volunteers, including Bob and
Laquisha, are put into a bowl. If two names are randomly drawn from
the bowl without replacement, what is the probability that Bob’s name
will be drawn first and Laquisha’s name will be drawn second?

[hide="hint"]
find the probability that Bob;s name will be drawn first. Then, based on the names left, find the probability that Laquisha's name will be drawn next. Then multiply the probabilities together.
[/hide]

This is the best hint given for one of posted questions thus far.


induction
calculus
inequalities
algebra solved
algebra

If a,b,c three positive numbers satisfy the ralation:

\[ a^2+b^2+c^2> \sqrt {2(a^4+b^4+c^4)} \] then a,b,c are the sides of a triangle!!

thank you!!

Solution:

We have that: (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)>0 which gives us:
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)>0

Now, because a,b,c>0 then at least three of the four factors of the ineq are positive and because the product of four terms is positive then every term is positive. From this reason we have that: a+b>c,a+c>b,b+c>a.

cheers!

Okay I also have the generalization of this thing (I think it works :D ) :

Generalization: If a_1,a_2,..,a_n satisfy the realation:

(a_1+a_2+..+a_n)^2>(n-1)(a_1^4+a_2^4+...+a_n^4) with n natural >=3 then any triplets of these n numbers can be the sides of a triangle!

Hint: it's a simple induction. n=3 true so you only need to show that it is true for n-->n+1

cheers!!

this is problem 22, from the RoMOP homework I gave this year - but I seem no to remember where I took it from - any sugestions are welcomed :)

Induction works just fine! :D

n=3 it's done! :D

we consider the result true for n=k. Let k+1 be positive numbers so that they satisfy the ineq:

(a_1^2+a_2^2+..+a_(k+1)^2)^2>k(a_1^4+..+a_(k+1)^4).
For an easier calculus we denote:

X=a_1^2+a_2^2+..+a_k^2 and Y=a_1^4+..+a_k^2 now we have:

(X+a_(k+1)^2)^2>k(Y+a_(k+1)^4) which gives: X^2+2Xa_(k+1)^2+a_(k+1)^4>k(Y+a_(k+1)^4) so:
X^2-(k-1)Y+2Xa_(k+1)^2>Y+(k-1)a_(k+1)^4 from which we get:
X^2-(k-1)Y+2Xa_(k+1)^2>2sqrt(Y)*sqrt(k-1)*a_(k+1)^2 which is equivalent to:

(X-sqrt(k-1)*sqrt(Y))(X+sqrt(k-1)*sqrt(Y)+2a_(k+1)^2)>0. From this we get that:

X>sqrt(k-1)sqrt(Y) which gives:

(a_1^2+a_2^2+..+a_k^2)^2>(k-1)(a_1^4+..+a_k^4) and we get what we need. We show the same that: (a_1^2+a_3^2+..+a_(k+1)^2)^2>(k-1)(a_1^4+a_3^4..+a_(k+1)^4) so any triplet from
{a_1, a_3,.., a_(k+1)} can form a triangle. Now , following this result we can extend it to any triplets from a_1, a_2 .. a_{k+1}.

kinda ugly but it works! :D
cheers!!

hey
actually this problem was proposed by murray klamkin,the famous problemist.and it also featured in 1988 china(the proof is very well known)
best regards

galois i haven't heard of this guy up untill now, could you tell us more?

hey
i guess the americans must be knowing him.he has been a prominent contributor to "mathematics magazine" (published by MAA ,i guess) problems section.and also there used to be a section of "klamkin's quickies" (problems with very short solution but requiring plenty of insight)".and since i've said sumthing abt him ,i must leave with another one of his problems :D
find the maxima and minima of
 \sqrt (a^2x^2 +b^2y^2 +c^2z^2) +  \sqrt (a^2y^2 +b^2z^2 +c^2x^2)
 +  \sqrt (a^2z^2 +b^2x^2 +c^2y^2) where a,b and c are non-negative constants and x^2 + y^2 + z^2 =1
best regards

This problem is easy and perhaps is not the best representative of Klamkin's problems.  It was proposed recently in the Mathematics Magazine and solutions are due Sept. 1, 2003. So maybe it is better for obvious reasons (copyright e.t.c) to submitt a solution here after the deadline. (I have already solved the problem and submitted a solution).

Murray Klamkin is a well-known, respectable problem solver and problem poser. He is in his eighties now, an emeritus math professor at the University of Alberta, and he has done A LOT! He has solved and proposed numerous problems; some of the problems of past IMO's are actually his.
He has been the leader of the USA team in the 80's, he has written two books on Mathematical Olympiads, one on the USA Math Olympiads and one on the IMO from 1978 to 1985 published by the MAA. Also, he has been problems editor of many journals in the US and Canada, e.g. he was the editor of the Olympiad Corner of Crux Mathematicorum before R.Woodrow, problems aeditor of Mathematics Magazine, American Mathgematical Monthly, SIAM review e.t.c. The USA has named an award after him and Greitzer for the best students at the USA math olympiad. He himself has received many awards e.t.c. e.t.c. e.t.c.

By the way, he was the person who awarded the special prize to Nikolay Nikolov of Bulgaria in the 1995 IMO in Canada.

you are right achilleas - didn't knew that. topic is locked until 1st of september.

for the first problem - the opener of this topic moubi tells us that a reference is

[quote]M.S.Klamkin, "Simulatneous triangle inequalities",Math.Magazine (1987),
p.236-237 [/quote]

Now I guess that it is okay to post solutions!!

it's the 14 of September! :D :D

cheers!

yeah right :D ups didn't noticed that ... boy how time flyies :D

The minimum value is $a+b+c$ and the maximum value is $\sqrt{3(a^{2}+b^{2}+c^{2})}$. Indeed, we first put $A=\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}$, $B=\sqrt{b^{2}x^{2}+c^{2}y^{2}+a^{2}z^{2}}$ and $C=\sqrt{c^{2}x^{2}+a^{2}y^{2}+b^{2}z^{2}}$, and note that since
$x^{2}+y^{2}+z^{2}=1$, it is $A^{2}+B^{2}+C^{2}=a^{2}+b^{2}+c^{2}$.
Then our claim about the maximum follows immediately from Cauchy's inequality,
\[
    (A+B+C)^2 \leq 3(A^{2}+B^{2}+C^{2})=3(a^{2}+b^{2}+c^{2}).
\]
(Note that it is achieved when $x=y=z$.) Now again from Cauchy's inequality we get
\[
    AB=\sqrt{a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}} \sqrt{b^{2}x^{2}+c^{2}y^{2}+a^{2}z^{2}}
    \geq ab x^{2}+bc y^{2}+ca z^{2}
\]
and similarly $BC \geq bc x^{2}+ca y^{2}+ab z^{2}$ and $CA \geq ca
x^{2}+ab y^{2}+bc z^{2}$. Hence
\[
    AB+BC+CA \geq (ab+bc+ca)(x^2+y^2+z^2)=ab+bc+ca,
\]
and so
\[
    (A+B+C)^2 = a^2+b^{2}+c^{2}+2(AB+BC+CA)\geq a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2,
 \]
from which our claim about the minimum follows immediately. (Note that it is achieved when one of the $x,y,z$ is 1 and the other two are 0.)


geometry
geometric transformation
rotation
symmetry
AMC
AIME

Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c.$

No problem! All comments have been removed.  :)

There seems to be some discrepancy of the answer.  I put 227, and I know a lot of people have different answers, so does anyone want to back me up (or tell me it's wrong)?

[quote="mysmartmouth"]Thanks for posting all the problems!

I'm adding them into the Resources section.  It will be automatically compiled into a PDF, so can you (or a mod) please remove your comment at the beginning of the problem, so it will look nice all compiled together?

Thanks![/quote]

where is the answer anyways?

Me and one of the other people from my school got 227, I'm pretty sure that's right.

My answer was 179. How did you guys get 227?

The final answer for me was  $80\pi+96\sqrt{3}$ .

It's kind of hard to figure out who's right when you both don't post solutions.  :roll:

We don't need solutions.  We just need to vote. :D jk

I got 179 as well, same solution as the others.

[quote="SamE"][hide="I got 179"]So normally the tube would just go its circumference in distance: $144\pi$. But the smaller tube gets in the way a bit. Basically, you have to know the fact that whenever a cylinder rolls without slipping, the distance moved on the circumference is equal to the distance moved by the center of the tube. This path deviates from the straight line for an arc length, replacing a distance, so we subtract the arc length and add the distance to our calculation. The radius of the arc is the sum of the radii of the tubes, or 96. Since this arc starts at a point 48 above the center of the arc (which is the center of the smaller tube) and ends at the other point at the same height, it goes through $120^\circ$ of rotation, or 1/3 of a circle. The distance is therefore $96\sqrt3$ and the arc length $64\pi$, so we get $144\pi-64\pi+96\sqrt3=80\pi+96\sqrt3$, for an answer of $80+96+3=\fbox{179}$.[/hide][/quote]

Hmm I think people got 227 because they subtracted 1/3 of the smaller circle's circumference from the larger circle to get 144-16=128. Then there is the 96 and the 3 so 128+96+3=227. Thats what I did when I saw the problem and I do believe that is correct. But dont take my advice i am a novice at AIMEs.

[quote="SamE"][hide="I got 179"]So normally the tube would just go its circumference in distance: $144\pi$. But the smaller tube gets in the way a bit. Basically, you have to know the fact that whenever a cylinder rolls without slipping, the distance moved on the circumference is equal to the distance moved by the center of the tube. This path deviates from the straight line for an arc length, replacing a distance, so we subtract the arc length and add the distance to our calculation. The radius of the arc is the sum of the radii of the tubes, or 96. Since this arc starts at a point 48 above the center of the arc (which is the center of the smaller tube) and ends at the other point at the same height, it goes through $120^\circ$ of rotation, or 1/3 of a circle. The distance is therefore $96\sqrt3$ and the arc length $64\pi$, so we get $144\pi-64\pi+96\sqrt3=80\pi+96\sqrt3$, for an answer of $80+96+3=\fbox{179}$.[/hide][/quote]

That "fact" is, in fact, false.
[hide]
instead of subtracting $64\pi$, which is the distance the center moved, subtract $(2\pi*24)/3=16\pi$, which is the distance moved on the small cylinder. It should be $128\pi+96\sqrt3$
$128+96+3=\fbox{227}$
[/hide][/hide]

[quote="SamE"][hide="I got 179"]it goes through $120^\circ$ of rotation, or 1/3 of a circle. The distance is therefore $96\sqrt3$ and the arc length $64\pi$[/hide][/quote]

Since this is the step in which people differ, you would do well to explain how you got $64\pi$.  Based on what you say here, I see no reason why the arc length should be $64\pi$.  If anything, your previous sentence suggests $\frac{144}{3}\pi=48\pi$

The circumference of a circle with radius 96 is $192\pi$, and 1/3 of that is $64\pi$. As for my fact, it's something I've learned from physics; that's why when you rotate one penny about another fixed penny, it goes around two times per revolution rather than one (try it!).

[hide="a third one that I missed"]
take a cross section of the figure and we have two circles and a line.

Let's start with the large circle tangent to the line and the small circle. The sum of the radii of the circles is 96, and the difference is 48. Dropping perpendiculars from the centers of both circles to the line and connecting their centers gives us a right trapezoid. Translate the line up 24 units to get a right triangle with leg length 48 and hypotenuse 96. This is recognized as a 30-60-90 triangle, so the lateral distance between the centers now is $48\sqrt3$. If the large circle has center O and is tangent to the line and circle at A and B, the measure of angle AOB=60.
Thus, arc AB=24pi.
Now, consider O on the other side of the smaller circle but tangent to the line at C and to the smaller circle at D. By symmetry, arc CD=24pi. The angle formed By B, the center of the smaller circle, and D is 120, so O revolves 16pi over the smaller circle to get to D, so arc BD=16pi. This means major arc CA=80pi, and that's how much it has left to roll. So with symmetry, the answer is 2(48rt3)+80pi=80pi+96rt3 giving us 197. The key is that 48pi pf the circle never touched the line.

[/hide]

I'm mad because several people got the wrong answer (128pi+48rt3) but the right answer...

Here's the diagram, with everything scaled down by 8. It indicates that the answer is 179. If anyone disputes my lengths or anything, please tell me. The bold sections are where the large cylinder actually touches a surface.

ok, based on the image five posts above, the second circle from the right starts to roll over the smaller circle with a 60 deg. arc never touching the ground. Then it rolls over a 120 deg. arc of the smaller circle, and touches the ground with another 60 deg. arc never touching the ground.

This is a total of $144\pi * 120/360+48\pi * 120/360 = 48\pi+16\pi = 64\pi$, so the distance the larger circle traveled while on the ground is $144\pi-64\pi = 80\pi$.

The blue triangle is a 30-60-90 triangle, so the horizontal distance the larger circle traveled while rolling over the smaller circle is $96\sqrt{3}$, as indicated in the image.

$80\pi+96\sqrt{3}$ is the distance the larger circle traveled, so $80+96+3 = \fbox{179}$.

By looking at the decimal approximations of both answers, it is easy to see which is correct and which is wrong. The answer must be less than that of a simple horizontal path, so the answer must be less than $144\pi$ .

Yeah I realize my mistake the answer seems to be 179. To make this problem seem easier just start with the two cyclinders touching. Lablel their point of intersection, and draw the large cyclinder again after the roll over the other cyclinder, notice that part of the circumference never actually touches the surface.

EDIT: I just realized that if you had 2 pencils you could kinda acutally experiment with this. But it may not work because the pencils will be the same size.

omg, i think i'm gonna cry... i did all that stuff, but made a careless mistake!! i forgot to simplify something(my guess)!! i am a RETARD!

yeah i got 179 why are there so many 227? 0.o

I got a different answer than both of the two currently being discussed, and though I'm sure its wrong, could someone please explain why?

[img]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=8122[/img]

Using this picture, I said the large circle starts 48 root 3 on the right of the little circle it will stop rolling over it when its 48 root 3 on the other side, meaning its been displaced by 96 root 3.  To make this roll it had to do 120 degrees, so two thirds of its circumference is left.  So (2/3)(2*pi*72)=96pi.  96 + 96 +3=195

[quote="SamE"]... that's why when you rotate one penny about another fixed penny, it goes around two times per revolution rather than one (try it!).[/quote]

I remember seeing this in the AoPS Volume 2, but could never find it since.  Could someone tell me what section it is in? (not page number, as there are many different editions)

Well, the official answer was 179.
But I got 227. =(

[quote="abcdefghijklmnopq"]To make this roll it had to do 120 degrees, so two thirds of its circumference is left. So (2/3)(2*pi*72)=96pi.  96 + 96 +3=195[/quote] Your problem is that the radius of the path of the center of the large cylinder is 96, not 72. So you can't say that it rolls through 1/3 of its circumference, since we only really care about the distance moved by the center of the cylinder. You have to do $2\pi\cdot72-(1/3)(2\pi\cdot96)=80\pi$, and $80+96+3=179$.

This was a very confusing problem and unfortunately I know nothing about physics. I got 227 like many others. The reason is because we're looking at how much the circumference of the larger circle is traveling, not the center (I still don't get why it's the center). So when the larger tube rolls from one side to the other of the smaller tube, it touches 1/3 of the smaller tube's circumference, which is 16pi, so a point on the larger tube's circumference would travel 16pi.

The thing is, though, that when you have a rolling object, not all points on the circumference move at the same rate. The point(s) of contact don't move at all (that's what is meant by without slipping), while the points opposite those move the fastest. You simply have to take the average of all of those velocities, which happens to be the center of mass, in this case the center of the tube.

Just because there's contact doesn't mean there's movement. Without slipping, contact means a lack of movement, so you can't simply measure the distance on the circumference.

Here's another way to think about it, or get the right answer: The distance on the smaller tube is the distance the outside rolls. But suppose the same point was in contact with the small tube the entire time, negating this distance. The whole tube still rotates since the point of contact is in a different position on the tube. It used to be on the leading edge, and now it's on the back. Refer to worthawholebean's picture above to visualize this. If you add up those two effects, you get the right answer.

After reading through this topic
I've realized
that I could've just straightened out the circumference in order to solve how much it rolled over the small cylinder

Those diagrams helped alot too :)
Thanks undefined117 for posting up the very helpful picture

Yeah I get it now, thanks you guys.

I have half a mind to complain about this question but I won't, I think it's just my own stupidity.

The people who got 227 (me being one of them) just forgot to subtract the section of the circumference that was "skipped over" and will never touch the table. It's a quick fix because if you got 227, you have already found the 30-60-90 triangle, so using the same triangle you can see the the two sections of the circumference that were skipped over are 60/360 of the circumference. There are two of those, so just subtract 144pi/3 = 48pi from your answer and you get the correct answer.

Well because it was bumped, is this a relatively easy question for the AIME? It didn't seem hard to me, yet it's #11.

I'd say it's pretty easy for an AIME #11 especially since it's geometry. I usually don't even read the geometry problems past 10. Maybe they just expected people would make the same dumb mistake (which a lot of people like me did) and forgot about the circumference that gets skipped.