limit
inequalities
LaTeX
real analysis
function
real analysis unsolved

[img]http://sites.google.com/site/asdfasdf23135/ra8.JPG[/img]

By definition,
a_n->a iff
for all ε>0, there exists an integer N such that n≥N => |a_n - a|< ε

But other than stating and looking at the definitions, I really have no clue how to proceed on this problem.

Any help is very much appreciated!

Since $ a_n$ and $ b_n$ are nonnegative, you can use $ \sup\{a_k b_k : k\ge n\}\equal{}\sup\{a_k: k\ge n\}. \sup\{b_k : k\ge n\}$ which, I think, is true.

Why?
How can we prove that? And how will this help us?

thanks.

[quote="JBL"]I don't think you want to go about using false statements as key steps in your proof ;)  (Think about the situation $ a_1 \equal{} 2, a_2 \equal{} 1$ and $ b_1 \equal{} 1, b_2 \equal{} 2$.)  There is a simple orthographical change that can be made to limsup's statement to make it true, though, and perhaps even useful.

Probably it would be helpful for you to think about what the question is telling you, and choose how to proceed based on that.  In fact, though, there isn't a lot of freedom here.  Let $ a \equal{} \lim_{n \to \infty} a_n$ for convenience.  We're going to have to show on one hand that the left side is no smaller than the right side, and on the other hand that the left side is no larger than the right side.  At some point, you're probably going to have to estimate $ a_n$ for large $ n$ in terms of $ a$ and some preselected $ \varepsilon$.  So, start putting these pieces together![/quote]
[img]http://sites.google.com/site/asdfasdf23135/ra14.JPG[/img]

Yes, that statement is true -- you should consult the definition of supremum to see why.  This gives you (with a very small argument) that for all $ \varepsilon > 0$, we have $ \limsup a_n b_n \leq (a \plus{} \varepsilon)(b \plus{} \varepsilon)$.  What can you conclude from this?  Now how about the other inequality?

It would be really nice if you would use text and LaTeX rather than images (which frequently are hosted on other sites and die after some amount of time, leaving the thread unintelligible).  To learn about [[LaTeX]], see that link and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=327395]this one[/url].  Also, for posterity (in case the image in the first post does disappear), I note that the question is to show that if $ (a_n)_{n \equal{} 1}^\infty$ and $ (b_n)_{n \equal{} 1}^\infty$ are nonnegative real sequences then $ \limsup a_n b_n \equal{} \lim a_n \cdot \limsup b_n$, provided the former is a real number.

[quote="JBL"]Yes, that statement is true -- you should consult the definition of supremum to see why.  This gives you (with a very small argument) that for all $ \varepsilon > 0$, we have $ \limsup a_n b_n \leq (a \plus{} \varepsilon)(b \plus{} \varepsilon)$.  What can you conclude from this?  Now how about the other inequality?
[/quote]
OK, since it's for ALL epsilon>0, we can put epsilon=0 and the inequality still holds, so we proved one direction.
1) But WHY do we need the hypothesis that a(n) and b(n) are [b]nonnegative[/b]? I don't see where in my proof is it necessary to assume it.

2) Also, how can I prove the other direction (limsup a(n)b(n) ≥ ab) ??

Thanks for helping!

[Sorry, I don't know how to use latex, and I put it as an image because of all those subscripts. It would be much nicer than typing a_n,b_n, etc. Maybe I'll use function notation a(n) instead of a_n]

[quote="kingwinner"]OK, since it's for ALL epsilon>0, we can put epsilon=0 and the inequality still holds, so we proved one direction.[/quote]  No, actually, you can't put $ \varepsilon \equal{} 0$ because you can only choose $ \varepsilon > 0$.  The right way to approach this is with a small proof by contradiction.  (A professional analyst would see that inequality and say something like, "We can make the right side arbitrarily close to $ ab$, so the result holds."  However, I think it's good practice for students of analysis to avoid this sort of handwaving -- it's much more important to understand [i]why[/i] you can do this handwaving than to be able to emulate the handwaving.)

Things need to be nonnegative because inequalities behave badly when you multiply things if you don't know that everything is positive -- in particular, you used the fact that $ w \leq x$ and $ y \leq z$ implies $ wy \leq xz$, but this is not true for all of $ \mathbb{R}$.  For a counterexample to the original statement, consider the sequences defined by $ a_n \equal{} \minus{}1$ for all $ n$ and $ b_{2n} \equal{} 0$, $ b_{2n \plus{} 1} \equal{} \minus{}1$ for all $ n$.

For the other direction, you should be able to proceed in a similar way -- now you need some lower bounds instead of upper bounds.

[quote="JBL"][quote="kingwinner"]OK, since it's for ALL epsilon>0, we can put epsilon=0 and the inequality still holds, so we proved one direction.[/quote]  No, actually, you can't put $ \varepsilon \equal{} 0$ because you can only choose $ \varepsilon > 0$.  The right way to approach this is with a small proof by contradiction.  (A professional analyst would see that inequality and say something like, "We can make the right side arbitrarily close to $ ab$, so the result holds."  However, I think it's good practice for students of analysis to avoid this sort of handwaving -- it's much more important to understand [i]why[/i] you can do this handwaving than to be able to emulate the handwaving.)[/quote]
OK, and yes to prove it I realized that it's a proof by contradiction. Suppose limsup a(n)b(n) > ab, then there exists some epsilon>0 s.t. $ \limsup a_{n}b_{n}\leq (a \plus{} \varepsilon)(b \plus{} \varepsilon)$ would not hold, hence contradiction.

[quote]Things need to be nonnegative because inequalities behave badly when you multiply things if you don't know that everything is positive -- in particular, you used the fact that $ w \leq x$ and $ y \leq z$ implies $ wy \leq xz$, but this is not true for all of $ \mathbb{R}$.  For a counterexample to the original statement, consider the sequences defined by $ a_n \equal{} \minus{} 1$ for all $ n$ and $ b_{2n} \equal{} 0$, $ b_{2n \plus{} 1} \equal{} \minus{} 1$ for all $ n$.

For the other direction, you should be able to proceed in a similar way -- now you need some lower bounds instead of upper bounds.[/quote]
For the other direction:
Given ε>0.
a(n)->a => there exists N1 s.t. if n≥N1, then a(n)>a-ε. <-----this is true for sure
b=limsup b(n) => there exists N2 s.t. if n≥N2, then [color=red]b(n)>b-ε[/color] <------but is this true? Why or why not? If not, what else should I do?

Thanks!

a(n)->a
b=limsup b(n)

Right now, I'm trying to prove that [b]limsup a(n)b(n) ≥ ab[/b], but I'm still stuck. In particular, I am not sure what lower bound to use for b(n).

Can someone help me, please?

Here is my try...

Given ε>0.
a(n)->a => there exists N1 s.t. if n≥N1, then a(n)>a-e.
b=limsup b(n) => there are infinitely many n's s.t.then b(n)>b-e

Case 1: (a,b>0)
So for infinitely many n's, we have a(n)b(n)>(a-e)(b-e) for ALL e>0.
=> for infinitely many n's (not necessaily of the form n≥N), we have a(n)b(n)≥ ab.

But now I DON'T see how we can use this to prove that sup{a(n)b(n): n≥N} ≥ab for some N (and hence our result limsup a(n)b(n) ≥ ab).

Can somebody help, please?


linear algebra
matrix
linear algebra unsolved

Let $ M\equal{}P\plus{}iQ \in \mathrm{GL}_n(\mathbb{C})$, with $ P,Q \in \mathcal{M}_n(\mathbb{R})$.
Prove that there exists $ R,S \in \mathrm{GL}_n(\mathbb{R})$, s.t. $ RMS \equal{} \begin{pmatrix} A & 0 \\
0 &I_{n\minus{}r}\end{pmatrix} \plus{} \begin{pmatrix} I_{r} & 0 \\
0 & B\end{pmatrix}$, with $ A \in \mathcal{M}_r(\mathbb{R})$, and $ B \in \mathcal{M}_{n\minus{}r}(\mathbb{R})$ is nilpotent.

... there might be an $ i$ missing in the statement. ;) Or am I missing something here?

You are right Third Edition, it should be:

Prove that there exists $ R,S \in \mathrm{GL}_n(\mathbb{R})$, s.t. $ RMS \equal{} \begin{pmatrix} A & 0 \\
0 & I_{n \minus{} r}\end{pmatrix} \plus{} i \begin{pmatrix} I_{r} & 0 \\
0 & B\end{pmatrix}$, with $ A \in \mathcal{M}_r(\mathbb{R})$, and $ B \in \mathcal{M}_{n \minus{} r}(\mathbb{R})$ is nilpotent.


calculus
integration
analytic geometry
geometry unsolved
geometry

let L denote the set of all lattice points of the plane (points with integral coordinates)
show that for any three points A,B,C of l there is a fourth point d,different fromA,B,C,
such that the interiors of the segments AD,BD,CD contain no points of L.is the statmant true if one consider four points of L instead of three?

Since there are no lattice points between $ (x,y)$ and $ (x+a',y-b')$ on the line $ ax+by+c=0$, where $ a'=a/(a,b),b'=b/(a,b)$, we are looking for integers satisfying the following relations
\[ x_D=x_A\pm u_1=x_B\pm v_1=x_C\pm w_1\\ \\
y_D=y_A\mp u_2=y_B\mp v_2=y_C\mp w_2\]
with $ (u_1,u_2)=(v_1,v_2)=(w_1,w_2)=1$; here $ (x_K,y_K)$ denotes the point $ K$. But the existence of such integers seems kinda obvious... :maybe:


Find the sum of the smallest and largest prime factors of 10,101.

3,7,13,37 so 40. rofl i think there is a leet influence here?

In case MewMew wasn't entirely clear, the first few numbers he gave were the [b]prime factorization[/b] of the given number. You can see the smallest and largest prime's sum is $ 3 \plus{} 37 \equal{} \boxed{40}$.


Let f : R-->R satisfying :
 $ f(x+y) \leq f(x) + f(y) $ for all x,y
 Prove that :
  For all $ x\in R $ , $ n\in N $ :
    $ f(nx) \leq \sum_{i=1}^{n}\frac{f(ix)}{i} $

Oh no,It 's a problem in APMO and can solve by induction.
[Moderator edit: See http://www.mathlinks.ro/Forum/viewtopic.php?t=79787 ]

In fact, I claim there exists a Pigeon Hole solution.  :)


limit
logarithms
integration
calculus
calculus computations

Find $\lim_{n \rightarrow \infty}\left( \prod_{k=1}^{n}\frac{k}{k+n}\right)^{e^{\frac{1999}{n}}-1}$

[quote="N.T.TUAN"]${\lim_{n \rightarrow \infty}\ln{(\left( \prod_{k=1}^{n}\frac{k}{k+n}\right)^{e^{\frac{1999}{n}}-1})}=\lim_{n\to\infty}[(e^{\frac{1999}{n}}-1})\sum_{k=1}^{n}\ln{\frac{\frac{k}{n}}{\frac{k}{n}+1}}]=\lim_{n\to\infty}\frac{e^{\frac{1999}{n}}-1}{\frac{1}{n}}\int_{0}^{1}\ln{\frac{x}{x+1}}dx$. Now $\lim_{x\to 0}\frac{e^{x}-1}{x}=1$, we have done. :)[/quote]

lol, i don´t believe what i did!!.I did every think like you, but the easiest party i missed!!!!!I simple forgot $\lim_{x\to 0}\frac{e^{x}-1}{x}=1$!!!!!!!!!!!!!! :blush:  :rotfl:  :mad:

[quote="gabriel ponce"]

lol, i don´t believe what i did!!.I did every think like you, but the easiest party i missed!!!!!I simple forgot $\lim_{x\to 0}\frac{e^{x}-1}{x}=1$[/quote]
$\lim_{n\to\infty}\frac{e^{\frac{1999}{n}}-1}{\frac{1}{n}}=\lim_{n\to\infty}1999.[\frac{e^{\frac{1999}{n}}-1}{\frac{1999}{n}}]=1999$ :D

yes, i did it hehehe.My answer is $e^{1999(1-ln4)}$.It si right??

[quote="N.T.TUAN"][quote="gabriel ponce"]But i don´t know what i´m doing wrong now, because my answer is less than 0! :huh:  :huh:[/quote]
Post your solution, please! I will check it !  :wink:[/quote]

Sorry, i edited my last post, because it was not clean.I was saying that $\int_{0}^{1}ln\left(\frac{x}{x+1}\right)=1-ln4 <0$, and i was confused if its right.
Your answer is $e^{1999(1-ln4)}$???

[quote="gabriel ponce"]
Your answer is $e^{1999(1-ln4)}$???[/quote]
Yes, that's right :D

Thanks for your patient!! :D  :lol:
Just an ask.Do you have some preparation material to university olimpyads level???


modular arithmetic
group theory
abstract algebra
superior algebra
superior algebra unsolved

I found this problem in Lang, Chapter II Problem 8. I was reviewing for the midterm and just saw this problem... However, I have no idea how I should proceed in the problem. I will appreciate any inputs from you guys! Thanks.

Let $ p$ be a prime number, and let $ A$ be the ring $ \mathbb{Z}/p^r \mathbb{Z}$, where $ r \in \mathbb{N}$. Let $ G$ be the group of units in $ A$, i.e., the group of integers prime to $ p$, modulo $ p^r$. Show that $ G$ is cyclic, except in the case when $ p \equal{} 2, r \geq 3$, in which case it is of type $ (2, 2^{r\minus{}2})$ (Hint: in the general case, show that $ G$ is the prouct of a cyclic group generated by $ 1\plus{}p$ and a cyclic group of order $ p\minus{}1$. In the exceptional case, show that $ G$ is the product of the group $ \{ \pm1 \}$ with the cyclic group generated by the residue class of $ 5$ mod $ 2^r$.)

The hint is a good one; what issues are you having following it?  (The cyclic group mentioned in the first hint is a copy of the group of units $ \bmod p$.)

Show $ x\in (1 \plus{} p^k\mathbb{Z})\setminus (1 \plus{} p^{k \plus{} 1}\mathbb{Z})\Rightarrow x^p \in (1 \plus{} p^{k \plus{} 1}\mathbb{Z})\setminus (1 \plus{} p^{k \plus{} 2}\mathbb{Z})$
for all primes $ p$ and $ k\in\mathbb{N}$ except for $ (p,k) \equal{} (2,1)$.
For that calculate $ (1 \plus{} tp^k)^p\pmod{p^{k \plus{} 2}}$ for any $ t\in\mathbb{Z}\setminus p\mathbb{Z}$ and use $ p|\binom{p}{2}$ for $ p > 2$.

Using this we get $ 5 \plus{} 2^r\mathbb{Z} \equal{} 1 \plus{} 2^2 \plus{} 2^r\mathbb{Z}$ of order $ 2^{r \minus{} 2}$ in $ (\mathbb{Z}/2^r\mathbb{Z})^\times$ for $ r\ge 2$.
Then $ \langle \minus{} 1, 5\rangle \equal{} \langle \minus{} 1\rangle \times\langle 5\rangle$ has order $ 2^{r \minus{} 1}$ in $ (\mathbb{Z}/2^r\mathbb{Z})^\times$ with $ |(\mathbb{Z}/2^r\mathbb{Z})^\times| \equal{} \varphi(2^r) \equal{} 2^{r \minus{} 1}$,
so $ (\mathbb{Z}/2^r\mathbb{Z})^\times \equal{} \langle \minus{} 1\rangle \times\langle 5\rangle$ of type $ (2,2^{r \minus{} 2})$.

Also for $ p > 2$ we get $ 1 \plus{} p \plus{} p^{r}\mathbb{Z}$ of order $ p^{r \minus{} 1}$ with $ |(\mathbb{Z}/p^r\mathbb{Z})^\times| \equal{} \varphi(p^r) \equal{} (p \minus{} 1)p^{r \minus{} 1}$,
so $ \langle 1 \plus{} p\rangle$ is the unique Sylow $ p$-subgroup of the abelian group $ G \equal{} (\mathbb{Z}/p^r\mathbb{Z})^\times$,
and $ G$ has a normal Hall $ p'$-subgroup $ H$ with $ G \equal{} \langle 1 \plus{} p\rangle\times H$.
But the canonical epi $ G\to (\mathbb{Z}/p\mathbb{Z})^\times,x \plus{} p^r\mathbb{Z}\mapsto x \plus{} p\mathbb{Z}$ has kernel $ N \equal{} \{1 \plus{} pt \plus{} p^r\mathbb{Z}: t\in\mathbb{Z}\}$ of order $ p^{r \minus{} 1}$,
so we must have $ N \equal{} \langle 1 \plus{} p\rangle$ and $ H\cong G/N\cong (\mathbb{Z}/p\mathbb{Z})^\times \equal{} \mathbb{F}_p^\times$ cyclic of order $ p \minus{} 1$.
So $ G \equal{} \langle 1 \plus{} p\rangle\times H$ is cyclic of order $ (p \minus{} 1)p^{r \minus{} 1}$.


Find all primes p, q, r such that $ pqr|2^{p}\plus{}2^{q}\plus{}2^{r}$. :)

i think u mean $ pqr|2^p\plus{}2^q\plus{}2^r$ otherwise the problem would be obvious...


Columbia calculus integration rotation function trigonometry vector

There are given n segments in the plane whose sum of lenghts is at most 1. show that there exist a line such that the sum of the projections of the segments in that line is smaller than 2/pi.

When $\pi$ comes along in problems such as this one it's almost certain that we must use some integral :). We rotate a line around the segments and integrate the function of the angle made by the line with another fixed line that gives the sum of the lengths of the projections. We have to find $\displaystyle \int^{2\pi}_0 (\sum \ell_i\cdot |\cos (x+t_i)|)dx$, where $t_i$ are real numbers which appear due to the different inclinations of the segments wrt the line and $\ell_i$ is the length of the segment $i$. That integral is equal to $\displaystyle 4\sum \ell_i$, so there is a line s.t. the sum of the projections of the segments on it is less than $\frac {\displaystyle 4\sum \ell_i}{2\pi}\le \frac 2{\pi}$

Another approach :
If there is 1 segment, it is trivial, so let's assume that $n \geq 2$.
Now translate each of the segments to give them the same midpoint, say $O$.
Let $A_1,...,A_n,A'_1,...,A'_n$ be the endpoints of the segments in the clockwise sense.

Let $P_0 = O$, and for $i \leq n$ let (as vectors) $P_{i-1}P_i = OA_i$ and $P'_{i-1}P'_i = OA'_i$ (where $P'_0 = P_n$).
Thus the points $P_1,...P_nP'_1...P'_n$ are the vertices of a convex $2n$-gon whiwh has a center of symmetry (the opposite sides are parallel and with the same length), say $ \Omega$.

Let's consider a pair of opposite sides whose mutual distance is minimal, say $d$. Then, the circle with center $ \Omega$ and diameter $d$ is tangent to these sides at $T$ and $T'$, and is contained into the polygon.
Let $l_1,...,l_n$ be the lengths of the initial segments. Thus :
$ \pi d \leq $ perimeter od the polygon $ = \sum_{i=1}^n l_i \leq 1$
which leads to $d \leq \frac {1} {\pi }$.
It follows that the projections of the polygon on the line $(TT')$ has a total length a most $2d \leq \frac {2} { \pi }$, and we are done.

Pierre.


Find all positive integers $ n$ such that $ n^6\plus{}206$ is divisible by $ n^2\plus{}2$.

[hide="Solution"]
If $ \dfrac{n^6\plus{}206}{n^2\plus{}2}\in \mathbb{Z}$ then so is $ \dfrac{n^6\plus{}206}{n^2\plus{}2}\minus{}n^4\equal{}\dfrac{206\minus{}2n^4}{n^2\plus{}2}$, and from here we can see that $ \dfrac{206\minus{}2n^4}{n^2\plus{}2}\plus{}2n^2\equal{}\dfrac{4n^2\plus{}206}{n^2\plus{}2}$ is also an integer. Furthermore, we can see that $ \dfrac{4n^2\plus{}206}{n^2\plus{}2}\minus{}4\equal{}\dfrac{198}{n^2\plus{}2}$ is also an integer.

In other words $ \dfrac{n^6\plus{}206}{n^2\plus{}2}\in \mathbb{Z}\implies \dfrac{198}{n^2\plus{}2}\in \mathbb{Z}$. Thus, we need to find the prime factorization  of $ 198$ and check which factors are such that when subtracted by $ 2$ and then taken the square root of remain positive integers, and those are our answers. After some computations we see that the only positive $ n$'s that work are $ 1,2,3,4,8,14$.
[/hide]

[hide="Alternate Solution"]
Just use long division: $ \frac{n^{6}\plus{}206}{n^{2}\plus{}2}\equal{}n^{4}\minus{}2n^{2}\plus{}4\plus{}\frac{198}{n^{2}\plus{}2}$. Thus, for $ n^{2}\plus{}2$ to divide $ n^{6}\plus{}206$, it is both necessary and sufficient for $ n^{2}\plus{}2$ to divide $ 198$.  From here, we can obtain the same solutions as tonypr...[/hide]


On Tuesday, October 30, 2007, Matt McGann, Associate Director of Admissions at the Massachusetts Institute of Technology (MIT), and Kiran Kedlaya, Assistant Professor of Mathematics at MIT and former US IMO team member, will discuss math at MIT and the admissions process.  This Math Jam will begin at 7:30 PM ET (4:30 PM PT).

[quote="emjay285"]For those of us who didn't attend, Can anyone say something about what was discussed?[/quote]

There's a transcript up: Up top click "Math Jams", then "Transcripts" and find the MIT Math Jam transcript.


modular arithmetic
number theory proposed
number theory

Determine all positive integers $n$ such that $3^{n}+1$ is divisible by $n^{2}$.

Let $p$ the smallest prime divisor of $n$. (for n>1)
we have that $p|3^{2n}-1$ and $p|3^{p-1}-1$.
so, we will have that $p|3^{gcd(p-1,2n)}-1 = 3^{2}-1 \Longrightarrow p = 2$.

$n^{2}|3^{n}+1 \Longrightarrow p^{2}|3^{n}+1 \Longrightarrow 4|3^{2k}+1$, impossible

[quote="e.lopes"]
so, we will have that $p|3^{gcd(p-1,2n)}-1$ [/quote]

why?   :oops:

He used the following lemma :
$(a^{p}-1, a^{q}-1) = a^{(p,q)}-1$

However, $n = 1$ is a solution.

Remarks : There's an easier problem in 104 NT Problems to prove that $n \not| 3^{n}+1$ for odd $n > 1$..  :wink:

[quote="e.lopes"]Let $p$ the smallest prime divisor of $n$. (for n>1)
we have that $p|3^{2n}-1$ and $p|3^{p-1}-1$.
[/quote]
 Can you explain more clearly?

$3^{n}\equiv-1\pmod{p}$ therefore $3^{2n}\equiv 1\pmod{p}$.

$p|3^{p-1}-1$ by Fermat's theorem .

[quote="RDeepMath91"]He used the following lemma :
$ (a^{p} \minus{} 1, a^{q} \minus{} 1) \equal{} a^{(p,q)} \minus{} 1$

However, $ n \equal{} 1$ is a solution.

Remarks : There's an easier problem in 104 NT Problems to prove that $ n \not| 3^{n} \plus{} 1$ for odd $ n > 1$..  :wink:[/quote]
Yes it is easy .
If $ n|3^n\plus{}1$ then $ n$ must be even.


1/640 = x/10^y

for this problem i need to find x y intergers. Also, what is the lease values of X

so far i have 10^y = 640x
                     2^y * 5^y

so how do i find the lease vaules of x????

Thanks

You already posted this in the GS forum and I already posted a solution. ;)


how can I curtly prove that every closed subset of Vitali set has (outer) measure 0 ?

By saying that the Vitali set has [i]inner[/i] measure $ 0$ (or, if you want to go into details of the construction, that $ F\plus{}q$ , $ q\in \mathbb Q\cap[0,1]$, are disjoint, contained in $ [0,2]$, and have the same measure, where $ F$ is your closed set).


And all of the titles have at least one O!  :o  :o  :o

wow....i never realized that before. great observation!!

[quote="math92"][quote="keta"]Has anyone ever realized that the titles of book 1-3 parallel those of 4-6?

1. Sorcerer's Stone (an object)-> 4. Goblet of Fire (an object)

2. Chamber of Secrets (a location/organization) -> 5. Order of the Phoenix (organization)

3. Prisoner of Azkaban (person) -> 6. Half-Blood Prince (person).

Just thought this was interesting. Who knows what book 7 will be called? (probably back to an object), if the pattern persists.

Thoughts?[/quote]
that's a very interesting observation! :D[/quote]

[quote="math92"]that is a good observation too shobber!!![/quote]

 :spam:  :spam:  :spam:  :spam:  :spam:  :spam:

he was talking about two different things..........

I think it's spam to just say "That's neat", even though math92 did use complete sentences to say that. These comments do not contribute to the conversation.
[url=http://www.amazon.com/exec/obidos/tg/detail/-/0972322108/qid=1122058659/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-1357601-9764044?v=glance&s=books&n=507846]Did you read this?[/url]
I came across this book during my summer at HCSSiM, a few months after [u]Harry Potter and the Order of the Pheonix[/u] came out. I skimmed through it in the bookstore. I think the author mentioned these relationships and also talked about hidden meanings and symbols and so on. If Harry Potter was widely used as a book to be read in class, this book would be an equivalent to SparkNotes.

I meant that they were both spam. Not that one was spam because of the other.

[quote="shobber"]There is another interesting thing:
Try to find the complete names of the founders of the four houses. You shall see their names like that:
H....Hufflepuff
R....Ravenclaw
G....Gryffindor
S....Slytherin

And there is also a lot of guys with this fact, like
Minerva McGongagall
Severus Snape
Peter Pettigrew

And even:
Duddly Dursly
Cho Chang
......[/quote]
This form of alliteration in names is very commonly found in literature. Though, I don't see any specific purpose behind using them in Harry Potter.

Anyone notice how all the books start with

"Harry Potter and..."

Conspiracy?...maybe

Spam?....definitely

Here's the mathnerd314 Conjecture:

The first, third, and fifth books begin with info about what happened to Harry's parents, etc. etc. etc. Background info. The second, fourth, and sixth books don't. I bet the seventh will.

But correct me if I'm wrong.

[quote="mathclass"][quote="byrdgyrl"]Also, if you line the books up in order from first to last so you can see their covers you will notice that the covers become much darker as the series progresses.  I am betting on a black cover for the 7th.[/quote]

that's the first reasonable guess I've seen here :P It makes sense... harry's slowly going towards the dark side. just like luke skywalker, until he met mara! [or something like that...][/quote]

Well, not REALLY... he was just overusing the Force...  *grumble* :lol: mathclass, you've read Specter of the Past and Vision of the Future, right?  There's a reference of Luke going to the Emperor on Byss... Did you read the comic where that happened?

PS: Sorry, I know this is really off-topic.

Maybe all of those are just coincidence.... Or maybe another Riddle --> Voldemort game in Book 2.

P.S I [b]don't[/b] think that math92 is spamming here.


function
floor function
modular arithmetic
inequalities
number theory
Diophantine equation
algebra

Suppose $r$ is a real number for which \[ \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.  \] Find $\lfloor 100r \rfloor$. (For real $x$, $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.)

[hide="Solution"]
Let $s = \lfloor r \rfloor$. Let $\{ r \} = r - s$ be the fractional part of $r$. Notice that we must have \[ \left\lfloor r + \frac{91}{100} \right\rfloor - \left\lfloor r + \frac{19}{100} \right\rfloor \le 1.  \] Hence, we can write \[ k\cdot s + (73-k)(s+1) = 546  \] for some $k$. ($k$ is the number of terms in the sum that round down to $s$ and $73-k$ is the number of terms that round down to $s+1$.) Expanding, we get

\begin{eqnarray*} ks + 73s - ks + 73 - k &=& 546\\ 73s - k &=& 473\\ \end{eqnarray*}

We now have a diophantine equation in $s$ and $k$. Modulo 73, we get

\begin{eqnarray*} -k &=& 473 - 73s\\ -k &\equiv& 35 \pmod {73}\\ k &\equiv& 38 \pmod {73}. \end{eqnarray*}

Since $1 \le k \le 73$, the only possible $k$ is $38$, which gives $s = 7$. If 38 of the terms round down to 7, then $\left\lfloor r + \frac{19+38}{100} \right\rfloor$ must round up to 8. Hence, $r = 8-.57 = 7.43$, and $\lfloor 100r \rfloor = \boxed {743}$.
[/hide]

[hide]

Well I put $[]$ as floor function here.

Suppose $[r+0.19]=k$ , then assume there are $73-i$ such $k$ where $[r+(91-i)/100]=k$ . And the remaining terms would be all equal to $k+1$ .

Hence it is equivalent to

$k(73-i)+(k+1)i=546$ $\implies 73k+i=546$

Since $i\leq 73$ so $k=7,i=35$

So 38 of them are 7 , while 35 of them are 8 .

Since the 38th term wuold satisfy

$7\leq r+0.56<8$ or $6.44\leq r<7.44$...(i)

And the 39th term satisfy

$8\leq r+0.57<9$ or $7.43\leq r<8.43$...(ii)

(i),(ii) gives us $7.43\leq r<7.44$ so $[100r]=743$
[/hide]

[hide]Denote $r=R+\alpha$ where $R$ is integer and $0\leq\alpha<1$. The given sum has $73$ terms: for some of those the integer part will be $R$ and for the others it will be $R+1$ (note that $r+{91\over 100}=R+\alpha+{91\over 100}<R+2$, hence no term can have an integer part $R+2$). Let there be $n$ of those whose integer part is $R$ and $(73-n)$ of those whose integer part is $R+1$. Now

$nR+(73-n)(R+1)=546\implies 73R-n=473\implies R={{n+473}\over 73}$

Now $n\leq 73\implies R\leq{{473+73}\over 73}\implies R\leq 7$ and $n\geq 1\implies R\geq{{473+1}\over 73}\implies R\geq 7$. From these results it follows $R=7$, hence $n=38$.

Since the first $38$ terms have the integer part $R$, and the $39th$ has the integer part $R+1$, we have

$r+{56\over 100}<R+1\land r+{57\over 100}\geq R+1$

Solving for $r$ gives \begin{eqnarray*}R+{43\over 100}\leq r<R+{44\over 100}\implies 100R+43\leq 100r<100R+44\implies \lfloor 100r\rfloor=100R+43=\boxed{743}\end{eqnarray*}[/hide]

Farenhajt, your solution is almost exactly the same as mine. :)

EDIT: Nevermind, everyone's solution is exactly the same. :D

[quote="jimhu"]My Solution is way different but answer still same:
[hide]
We denote of of the $73$ terms that add to $546$ as $x_n$. $x_n$ can only have $2$ different values because the increments only range from $\frac{19/100}$ to $\frac{91/100}$. Let $x$ and $x+1$ denote that 2 values respectively. Let $n$ and $73-n$ denote the number of those values, respectively.
We have: $xn+(73-n)(x+1)\Rightarrow 73(x+1)-n=546$ The only value of x in which allows n to be in the range $0-73$ is $x=7$. So $n=38$.
Knowing its easily seen that $r=743/100$.
Therefore $100\left\lfloor r \right\rfloor=\boxed{743}$.
[/hide][/quote]

actually $r$ is not $7.43$ only but is in the interval $[7.43,7.44)$ :)

[quote="chess64"][quote="jimhu"]My Solution is way different but answer still same:
[hide]
We denote of of the $73$ terms that add to $546$ as $x_n$. $x_n$ can only have $2$ different values because the increments only range from $\frac{19/100}$ to $\frac{91/100}$. Let $x$ and $x+1$ denote that 2 values respectively. Let $n$ and $73-n$ denote the number of those values, respectively.
We have: $xn+(73-n)(x+1)\Rightarrow 73(x+1)-n=546$ The only value of x in which allows n to be in the range $0-73$ is $x=7$. So $n=38$.
Knowing its easily seen that $r=743/100$.
Therefore $100\left\lfloor r \right\rfloor=\boxed{743}$.
[/hide][/quote]

Actually that's the same solution as mine...[/quote]]

O wait.. yea   :rotfl:  stupid me..

Let $ r \equal{} \frac {k}{100}$, we can easily see that the first 38 terms are 7 and the last 35 terms are 8. Thus, we have $ \lfloor \frac {k \plus{} 56}{100}\rfloor \equal{} 7$ and $ \lfloor \frac {k \plus{} 57}{100}\rfloor \equal{} 8$. We can easily see $ r \equal{} \frac {743}{100}$ is a solution.

[quote="Chronos"]What's wrong with:

$7\le r+\dfrac{40}{100}< 8 \le r+\dfrac{41}{100}<9$

so we have:

$660\le 100r < 760$ and $759 \le 100r < 900$ so $100r=759$???[/quote]
How did you get these inequalities?

[hide=Solution]There are $73$ terms. Note that $\frac{546}{73}$ rounded down is $7$, so some terms are $7$ and the others are $8$. Let there be $y$ terms of $7$ and $z$ terms of $8$. Then $7y+8z=546$ and $y+z=73$, so $y=38, z=35$. This means the $7$ terms end at $r+\frac{56}{100}$ and the $8$ terms begin at $r+\frac{57}{100}$, so $r+\frac{56}{100} < 8 \leq r+\frac{57}{100}$, or $\frac{743}{100} \leq r < \frac{744}{100} \implies \boxed{743}$.[/hide]

Nice problem, indeed. There are $73$, terms, the floor of $\frac{546}{73}$, is $7.$ So the terms are either $7$ or $8$, let there be $x$ terms of $7$ and $y$ for $8$'s. Now  $7y+8z=546$ and $y+z=73$, so $y=38, z=35.$ So $r=\boxed{743}.$


inequalities
vector
trigonometry
trig identities
Law of Cosines
geometry proposed
geometry

Prove that if the hexagon $A_1A_2A_3A_4A_5A_6$ has all sides of length $\le 1$, then at least one of the diagonals $A_1A_4,A_2A_5,A_3A_6$ has length $\le 2$.

WLOG, we may assume that both $ A_1A_3$ and $ A_1A_5$ are less than or equal to $ A_3A_5.$
Now, applying [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109578]Ptolemy's inequality[/url] on $ A_1A_3A_4A_5$ we get :
$ A_1A_4 \cdot A_3A_5 \leq A_1A_3 \cdot A_4A_5 \plus{} A_1A_5 \cdot A_3A_4 \leq A_1A_3 \plus{} A_1A_5 \leq 2A_3A_5,$
we're done. :)

Among the vectors $ \overrightarrow{A_{1}A_{2}}$, $ \overrightarrow{A_{3}A_{4}}$, $ \overrightarrow{A_{5}A_{6}}$ there are two for which the angle between them is less than $ 60^{\circ}$. WLOG, assume that these two are $ \overrightarrow{A_{1}A_{2}}$ and $ \overrightarrow{A_{3}A_{4}}$ and let $ X$ be a point such that $ \overrightarrow{A_{2}X} \equal{} \overrightarrow{A_{3}A_{4}}$. In this case, $ XA_{4} \equal{} A_{2}A_{3} \leq 1$ and since $ \angle{A_{1}A_{2}X} \leq 60^{\circ}$, $ A_{1}A_{2} \leq 1$, $ A_{2}X \leq 1$, by the law of cosines it follows that
\[ A_{1}X^{2} \equal{} A_{1}A_{2}^{2} \plus{} A_{2}X^{2} \minus{} 2A_{1}A_{2} \cdot A_{2}X\cos{A_{1}A_{2}X} \leq 1.
\]
Hence, $ A_{1}A_{4} \leq A_{1}X \plus{} XA_{4} \leq 2$.


Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


topology
search
advanced fields
advanced fields unsolved

Consider the connected sum $\mathbb{C}P^{n}$#$\mathbb{C}P^{n}$

1. Find the cohomology ring.
2. What about the Chern classes?

The first one I can be somehow convinced by my answer, but I really need some hints for the second one.  Thanks

for the first question..
let me call $L=K=\mathbb{P}^{n}(\mathbb{C})$, and $B$ the ball you use for the connected sum $M = L \# K$.
just use mayer-vietoris sequence for the triple $(M;L,K)$. you know what are the homologies for $L,K,B$, so you can construct the homologies of $M$, and from this (using the decomposition of torsion and free components of $H^{*}$ in terms of the ones of $H_{*}$) you can get what you search for.
at least, i hope so :D

as regarding the second question.. sorry, i don't know what chern classes are :p not yet, at least :)

Yeah, you can get the cohomology groups immediately this way, but the key part is the ring structure, that is, how does the elements multiply each other.

[quote="ma_go"]for the first question..
let me call $L=K=\mathbb{P}^{n}(\mathbb{C})$, and $B$ the ball you use for the connected sum $M = L \# K$.
just use mayer-vietoris sequence for the triple $(M;L,K)$[/quote]

Isn't the ball $B$ supposed to be missing from the connected sum?
Then how do you apply Mayer-Vietoris to that triple?

since you can make the ball small enough, so the missing ball has a boundary $S^{n-1}$, the neighbourhood of the boundary is homeomorphic to $S^{n-1}\times I$ which does not affect the homology after attaching to each component after cutting off.


The length of one leg of the largest right triangle is 12cm, and the length of its hypotenuse is 20cm. What is the number of centimeters in the length of the altitude $ h$? Express your answer as a decimal to the nearest tenth.

[asy]import olympiad;
pair A,B,C,D;
A = dir(0); B = dir(74); C = dir(180); D = foot(B,A,C);
draw(D--B--A--C--B); rightanglemark(A,B,C, 1/2);
label("h",B--D,W); label("12",A--B,NE); label("20",A--C,S);[/asy]

By the Pythagorean Theorem the other leg of the triangle is $ \sqrt{20^2\minus{}12^2}\equal{}16$.

The area of the triangle is $ \frac{12\cdot 16}{2}\equal{}96$ which is equal to $ \frac{20h}{2}\Rightarrow h\equal{}\boxed{9.6}$


Since this is a general question, assume the answer will be the same no matter where you start. Take the easiest $ 3x3$ array, the top left one.
The product of the top right and lower left numbers here is 45, and the top left and lower right is 17.
$ 45\minus{}17 \equal{} \boxed{28}$


Just curious, how many of you have AoPS shirts?

(And should I add an "I don't where shirts" choice)

uh....no.  and i don't really want to know where AoPS shirts are available....

well, for those of you that want to know
[hide]its on the website in the bookstore[/hide]

EDIT: Whoops, it's actually in [hide]under the "community" heading, there's a T - shirt section[/hide]

[quote="Prime factorization"]you'll sure land lots of girls (or even friends for that matter) by advertising your adulation for mathematics.[/quote]

Hehe what if your friends love math?  :D

I wanna get an aops shirt soon, partly because i already filled number 4 on the list, partly because i like navy shirts?

[quote="Prime factorization"]you'll sure land lots of girls (or even friends for that matter) by advertising your adulation for mathematics.[/quote]

yo the girls love my linear algebra textbook

sadly i think the textbook gets more love than me

[quote="Lazarus"]sadly i think the textbook gets more love than me[/quote]

 :rotfl:

Yay, im gonna get one! I don't get the listing of the numbers though, cuz if it were in base 12 then it would translate to 10 12 11 9 8 7.....

smack

thats my one of my stupidest mistakes. I accidently forgot 10_12 = 12 + 0


geometry
circumcircle
angle bisector
geometry unsolved

In triangle $ ABC$ ($ \angle B>\angle C$), let the tangent to the circumcircle at $ A$ meet $ BC$ at $ K$. Let the bisector of $ \angle AKC$ meet $ AB,AC$ at $ D,E$ respectively. Let $ L\equal{}BE\cap CD$. Suppose $ K,L,O$ are collinear.

Find $ \angle A$.

[quote="outback"]In triangle $ ABC$ ($ \angle B > \angle C$), let the tangent to the circumcircle at $ A$ meet $ BC$ at $ K$. Let the bisector of $ \angle AKC$ meet $ AB,AC$ at $ D,E$ respectively. Let $ L \equal{} BE\cap CD$. Suppose $ K,L,O$ are collinear.

Find $ \angle A$.[/quote]
Just curious, where is this problem from?  I have seen this before, but I never knew where this came from.
[hide="Solution"]
Rename $ L$ to $ Q$ and $ K$ to $ P$ (this solution is copied and pasted from one I had written before.)  Extend $ AQ$ to meet $ BC$ at $ R$ and let the midpoint of $ BC$ be $ M$.  Notice that $ \triangle APC\sim \triangle BPA$, so $ \frac {AB}{AC} \equal{} \frac {PA}{PC} \equal{} \frac {PB}{PA}$, from which we see that, by the Angle Bisector Theorem, $ \frac {AD}{BD} \equal{} \frac {PA}{PB} \equal{} \frac {AC}{AB} \equal{} \frac {PC}{AP} \equal{} \frac {EC}{AE}$.  Then, Ceva's Theorem on cevians $ CD$, $ BE$, and $ AR$ gives that $ \frac {BR}{RC}\cdot \frac {EC}{AE}\cdot \frac {AD}{DB} \equal{} 1$, so $ \frac {BR}{RC} \equal{} \frac {AE}{EC}\cdot \frac {DB}{AD} \equal{} \frac {AB^2}{AC^2}$, so $ AR$ is the $ A$-symmedian.  This means that\[ \angle QAP \equal{} \angle BAP \plus{} \angle QAB \equal{} \angle MAC \plus{} \angle ACB \equal{} \angle AMB\]This implies that $ \angle QAO \equal{} 90 \minus{} \angle AMB$.  Yet, $ \angle OAP \plus{} \angle OMP \equal{} 90 \plus{} 90 \equal{} 180$, so $ OAPM$ is cyclic, so $ \angle AMB \equal{} \angle AOQ$.  This implies that $ \angle AQO \equal{} 90$ from $ \triangle AQO$.  Yet, $ \angle OAP \equal{} 90$, so $ \triangle OAQ\sim \triangle OPA$, so $ PB\cdot PC \equal{} PA^2 \equal{} PQ\cdot PO$, so $ QOCB$ is cyclic.  This means that $ Q$ is the intersection between the $ A$-symmedian and the circumcircle of $ \triangle OBC$.  This implies that $ Q$ is actually $ F$ from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1116181#p1116181.  Thus, from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247506#p1247506, we have that $ Q$ lies on the circumcircle of $ \triangle ADE$ since $ \frac {DB}{AD} \equal{} \frac {AE}{EC}$.  Then,\[ 180 \equal{} \angle BAC \plus{} \angle DQE \equal{} \angle BAC \plus{} \angle BQC \equal{} \angle BAC \plus{} \angle BOC \equal{} 3\angle BAC\]so $ \angle BAC \equal{} 60$
[/hide]

Thanks for the solution!

It's from an AMY sheet...


modular arithmetic
Euler
function
number theory
totient function

Find the last three digits of $ 3^{2004}$

If it asked for just the last digit, I would figure out a bunch of powers of three and look for a pattern. Is there a better way to do this problem? If we look at $ 3^{2004}$ in mod 1000, is it illegal to replace 2004 with numbers congruent to it in mod 1000?

[quote="zolojetto"]i dont understand the first part of the solution like the part that says o(1000) = 100 or something like that.  What does that mean???[/quote]
That refers to the Euler Phi Function:
[url]http://en.wikipedia.org/wiki/Euler%27s_totient_function[/url]
Euler's Theorem states that $ a^{phi{\left(n\right)}}\equiv {1} \pmod{n}$ if $ a$ and $ n$ are relatively prime.
[url]http://en.wikipedia.org/wiki/Euler%27s_theorem[/url]

I'm guessing they used Euler's Theorem: http://en.wikipedia.org/wiki/Euler%27s_theorem

My mistake. $ \phi{\left(1000\right)}\equal{}400$. But anyway, the answer still holds true. I have edited my solution post to change my errors.


symmetry
geometry
3D geometry
sphere
function
geometric transformation
rotation

A spherical, 3-dimentional planet has center at $(0,0,0)$ and a radius of $20$. at any point $(x,y,z)$ on the surface of this planet, the temperature is $T(x,y,z)=(x+y)^2+(y-z)^2$ degrees. What is the average temperature of the sureface of this planet.

This appears in the section on symmetry but i don't see how it relates.

Because it's a sphere, we know $x^2 + y^2 + z^2 = 400$.  Thus, our goal is to kill off the $xy$ and $yz$ terms in the temperature function without changing the average temperature.

Because of the symmetry of a sphere, a sphere will temperature function $T(x, y, z)$ has the same average temperature as a sphere with temperature function $T(y, z, x)$ or $T(z, x, y)$, and as a result the same average temperature as a sphere with temperature function $\frac13 (T(x, y, z) + T(y, z, x) + T(z, x, y))$.  In our case, this is beautiful: $T(x, y, z) + T(y, z, x) + T(z, x, y) = (x + y)^2 + (y - z)^2 + (y + z)^2 + (z - x)^2 + (z + x)^2 + (x - y)^2 = 3(x^2 + y^2 + z^2)$, so our sphere has the same average temperature as a sphere with temperature function $x^2 + y^2 + z^2$, which is just an average temperature of 400.

Another thing you might try is to compare our sphere's temperature to that of a sphere of constant temperature $400 = x^2 + y^2 + z^2$.  The difference in their temperatures is $y^2 - 2yz + 2yx = y(2x + y - 2z)$.  If you can find a simple symmetry (rotation or whatever) of the sphere which reverses the sign of that expression, then you're done.  I don't know if this works -- I actually started down this path first, before knowing the answer, and couldn't find a symmetry that worked and so gave up on it and came up with the method above.

Other, probably simpler versions of the same question, with some different tricks: give the sphere temperature function $T(x, y, z) = x^2$ or $T(x, y, z) = xyz$ or $T(x, y, z) = xy$ or $T(x, y, z) = x^2 + (x + y)^2 + (y - z)^2 + z^2$.


modular arithmetic
IMO Shortlist
number theory unsolved
number theory

determine the minimum value of $ n$ be positive integer . Such that : there exist $ a_1,a_2,...,a_n \in N^*$ and:

$ a_1^4\plus{}a_2^4\plus{}...\plus{}a_n^4\equal{}2009$

The answer is 10.
First look mod 16. Since fourth powers are either 1 or 0 and since $ 2009\equiv9\pmod {16}$, the answer must be at least 9.
Suppose it is 9. Then 2009 would be the sum of 9 odd fourth powers, which you can check is not possible (first note that the number of 625s must be 3, then that no matter how you split the 81s and 1s, you can't get 2009).

But, $ 2009 \equal{} 6^4 \plus{} 5^4 \plus{} 3^4 \plus{} 7\cdot1^4$

[quote="v235711"]there is a extrmely similar problem at imo shortlist 2002.
(well,if it´s not 2002 ,it´s at another year).
the idea is exactly as caffeineboy posted.
What test is this from?Just a school test ?
best regards
V.[/quote]

it just my school test. my solution is $ 10$ but in bad and long way


geometry
circumcircle
parallelogram
trapezoid
geometry proposed

[quote="mr.danh"]$ ABCD$ is a cyclic quadrilateral. $ O$ is the circumcenter of $ (ABCD)$. $ E,F$ are two points satisfying $ \widehat{EAD} \equal{} \widehat{EBC} \equal{} \widehat{FDA} \equal{} \widehat{DCB} \equal{} 90^o$. Prove that $ E,O,F$ are collinear.[/quote]
Is it $ \widehat{EAD} \equal{} \widehat{EBC} \equal{} \widehat{FDA} \equal{} \widehat{FCB} \equal{} 90^o$  small type mistake  :P

Proof:
Denote G intersection of AD and BC, it is easy to find F,D,C,G lie on one circle, E,B,A,G lie on one circle.
DF ∥AE ,BE∥CF, triangle EBA is similar to triangle FDC,
Further, triangle AGH is similar to triangle EGB, triangle FDG is similar to triangle CXG.
So OO2GO1 is a parallelogram and OO2=1/2EG.
Thus we have EOF is collinear

[quote="mr.danh"]$ ABCD$ is a cyclic quadrilateral. $ O$ is the circumcenter of $ (ABCD)$. $ E,F$ are two points satisfying $ \widehat{EAD} \equal{} \widehat{EBC} \equal{} \widehat{FDA} \equal{} \widehat{DCB} \equal{} 90^o$. Prove that $ E,O,F$ are collinear.[/quote]
[hide="Solution"]
Denote $ M$ as the midpoint of $ AD$, $ O'$ that of $ EF$, and $ N$ that of $ BC$.  Since $ EA\perp AD$ and $ FD\perp AD$, we can see that $ EADF$ is a trapezoid with median $ MO'$.  Similarly, $ EFCB$ is a trapezoid with median $ O'N$.  It follows that $ O'M\perp AD$, so $ O'$ lies on the perpindicular bisector of $ AD$.  Similarly, $ O'$ lies on the perpindicular bisector of $ BC$.  Yet, this means that $ O'\equal{}O$, so $ O$ lies on $ EF$, and more specifically, $ O$ is the midpoint of $ EF$. [/hide]


geometry
circumcircle
incenter
projective geometry
geometry unsolved

Show that the axis of perspectivity between triangle ABC and it's
    Pelletier triangle is tangent to incircle of ABC.

    Maj.   Pestich

What is the Pelletier triangle of a given triangle? I am not unexperienced with triangle geometry but I never heard about this triangle... :s

Here's another question: what is Google and how do you use it? :) I had no idea what that was either, but now I do :):

[url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Pelletier_p.html[/url]. It defines the Pelletier points and the Pelletier triangle.

Dear Mathlinkers,
can this nice problem can be solved synthetically...
Sincerely
Jean-Louis

I could only prove the Pelletier triangle and the main triangle are perspective.But this problem can be rephrased in following way:-

$ABC$ is a triangle.$DEF$ is its intouch triangle.$\triangle P_aP_bP_c$ be the Pelletier triangle of $\triangle ABC$.Let $P_a'$ be the harmonic conjugate of $P_a$ wrt $EF$.Similarly define $P_b', P_c'$.Let $A_1B_1C_1$ be tht triangle formed by the rays $AP_a', BP_b', CP_c'$.Then $A_1D, B_1E, C_1F$ concur on the incircle of $\triangle ABC$.

[b]Lemma 1:-[/b]

Given a triangle $ ABC $ and two point $ P,Q $, suppose, $ P_aP_bP_c $ and $ Q_aQ_bQ_c $ are their cevian triangles. $ R_a=P_bP_c\cap Q_bQ_c $, similarly define $ R_b,R_c $. Then $ ABC $ and $ R_aR_bR_c $ are perspective and the center of perspectivity is the intersection point of the tri-linear polars of $ P,Q $ wrt $ ABC $.

[b]Proof:-[/b]

By a projective transformation if we take $ P $ to the centroid of $ ABC $, then the problem directly follows from Property 2 [url=http://www.artofproblemsolving.com/blog/66415]here[/url].

[b]Lemma 2:-[/b]

In a triangle $ ABC $ if $ S $ is the symmedian point, then the tri-linear polar of any point $ P $ wrt $ ABC $ passes through $ S $ iff $ P $ lies on the circumcircle of $ ABC $.

[b]Proof:-[/b]

Suppose, $ P $ lies on the circumcircle and $ P_aP_bP_c $ be its cevian triangle. $ A'B'C' $ be the tangential triangle of $ ABC $. By Pascal's theorem we can see that $ P_aP_b $ passes through $ C' $ and similar for the others. If $ P_a'P_b'P_c' $ be the tri-linear polar of $ P $ and $ S_aS_bS_c $ be the cevian triangle of $ S $, then note that, $ (AB,P_cP_c')=-1,(AS_a;A'S)=-1,(AC,P_bP_b')=-1 $. So $ P_c'P_b' $ passes throug $ S $ and similar for the others. So done.

[b]Back to main proof:-[/b]

Applying lemma 1, we can see that the Pelletier triangle of $ ABC $ is perspective to $ ABC $ wrt the intersection point of the tri-linear polar of its incenter and Georgenne point. Suppose, $ A_1B_1C_1 $ is the intouch triangle of $ ABC $, easy to see that the polar(wrt $ \odot A_1B_1C_1 $) of tri-linear polar(wrt $ \Delta ABC $) of Georgenne point is the symmedian point of $ ABC $ and polar of the tri-linear polar of incenter is the centrois of $ A_1B_1C_1 $, so if $ G,S $ are its centroid and symmedian point, then note that its enough to prove that the point whose tri-linear polar wrt $ A_1B_1C_1 $ is $ GS $ lies on $ \odot A_1B_1C_1 $ and clearly this follows from Lemma 2. So done.

Dear Mathlinkers,
this nice problem has now a synthetic proof
see for example
http://perso.orange.fr/jl.ayme vol. 9 Le triangle réfléchi p. 16
Sincerely
Jean-Louis


Consider a circle $ (O,R)$ and it's diameter $ AB$ and $ C$, $ D$ are points on the tangent at $ B$ on opposite sides  of $ B$.If $ AC$ and $ AD$ meet the circle again at $ E$ and $ F$ respectively and also $ CF$, $ DE$ meet the circle again at $ G$ and $ H$ respectively.Show that $ AG\equal{}AH$

$ \triangle ACF$ and $ \triangle AED$ are similar cause $ \frac{\overline{AE}}{\overline{AF}}\equal{}\frac{\overline{AC}\sin^2\angle ACB}{\overline{AD} \sin^2\angle ADB}\equal{}\frac{\overline{AD}}{\overline{AC}}$. Then $ \angle AFG \equal{} \angle AEH \ \Longleftrightarrow \ AG \equal{} AH$

[quote="galois01"]Consider a circle $ (O,R)$ and it's diameter $ AB$ and $ C$, $ D$ are points on the tangent at $ B$ on opposite sides  of $ B$.If $ AC$ and $ AD$ meet the circle again at $ E$ and $ F$ respectively and also $ CF$, $ DE$ meet the circle again at $ G$ and $ H$ respectively.Show that $ AG \equal{} AH$[/quote]
[hide="Solution"]
Note that since $ AB$ is a diameter, $ \angle BEA\equal{}90$, so $ EB\perp AC$.  Also, $ FB\perp AD$.  Since $ AB$ is a diameter, we notice that $ AB\perp CD$ and it follows that $ AE\cdot AC\equal{}AB^2\equal{}AF\cdot AD$, so $ FDCE$ is cyclic, which means that \[ \angle AGH\equal{}\angle AEF\equal{}180\minus{}\angle CED\equal{}180\minus{}\angle CFD\equal{}\angle AFG\equal{}\angle AHG\] from which we conclude that $ AG\equal{}AH$. [/hide]


Hello. I'm kind of stuck on these problems. When solving could you show me step by step on how to arrive at the answer. thank you :lol:

A projectile is fired at an upward angle of 45.0 from the top of a 265 m cliff with a speed of 185 m/s. What will be its speed when it strikes the ground below? (Use conservation of energy).

A driver notices that her 1000 kg car slows down from 90 km/h to 70 km/h in about 6.0s on the level when its neutral. Approximately what power is needed to keep the car traveling at a constant 80 km/h?

A shot-putter accelerates a 7.3 kg shot from rest to 44 m/s. If this motion takes 2.0, what was the average power that was developed?

A pump is to lift 8.00 kg of water per minute through a height of 3.50 m. What output rating (watts) should the pump motor have?

I'd say it is correct. The rest of the problems can be solved using the same principle.


Problem.Give $ u = (a_1;a_2);v = (b_1;b_2)\in R^2$ such as $ a_1b_1 + a_2b_2 = 1$ and $ a_1^2+a_2^2=b_1^2+b_2^2=1$. Prove that
$ (u,v)$ is establishment of $ R^2$. Find ordinate of $ w = (a,b)\in R^2$ in establishment $ (u,v)$

By "$ (u,v)$ is establishment of $ R^2$" do you mean that vectors $ u,v$ span $ R^2$? If yes, then that's false, since those vectors are linearly dependent, as its Gram determinant is zero. And what exactly is the meaning of $ w \equal{} (a,b)^2$?


function
limit
calculus
calculus computations

Let $ f(x)$ be a differentiable function with respect to $ x$, find the following limit.

$ \lim_{n\to\infty}\left(\frac{n}{k}\right)\left\{f\left(\frac{x}{n}\right)-f(0)\right\}^{k}$,

where $ \left(\frac{n}{k}\right)$ denotes the number of combinations of $ r$ objects chosen from set of $ n$ objects.

Especially find the limit for $ f(x)=(x-\alpha )(x-\beta )$ ?

[quote="kunny"]Let $ f(x)$ be a differentiable function with respect to $ x$, find the following limit.

$ \lim_{n\to\infty}\sum_{k=0}^{n}\left\{\dbinom{n}{k}\left[f\left(\frac{x}{n}\right)-f(0)\right]^{k}\right\}$,

where $ \dbinom{n}{k}$ denotes the number of combinations of $ k$ objects chosen from set of $ n$ objects.

Especially find the limit for $ f(x)=(x-\alpha )(x-\beta )$ ?[/quote]

is there a [i]summation [/i]somewhere ?  :|

It's just $ \frac{(xf'(0))^{k}}{k!}$. $ \binom{n}{k}$ is approximately $ \frac{n^{k}}{k!}$ for large $ n$, and the difference is approximately $ \frac{xf'(0)}{n}$.


Let's take $\ x,y,n$ positive real numbers so that $\sqrt x+\sqrt y=2$.Find $\ all$ $\ n$ such that the number
$\sqrt{x+n}+\sqrt{y+n}$ is natural, $\ n>2.$

If $n$ is a positive [b]real[/b] then I think the expression can take on a lot of natural number values!   :P

:huh: What do you want to say?$\ x$ is not equal with $\ y$.

It's very likely that he wants to ask whether it isn't better to have $n$ an integer ;)


There used to be an excellent set of solutions written up for the 2005 AIME here: http://students.imsa.edu/~albertni/AIME%202005/AIME1%202005%20Solutions.pdf  .  However, the link seems to have since been disabled, and Google doesn't seem to have archived it at any point.  Does anybody have a copy of this document, or a link to a similar set of solutions?

You can go to the Math Jam transcript for the 2005 AIME.  The solutions aren't exactly written up, but you can piece together the parts of it.

[quote="chess64"]Those are actually darktreb's solutions :P I have them...

http://www.artofproblemsolving.com/Forum/files/aime1_202005_20solutions_172.pdf[/quote]
Yay, thank you!  I'm going to save it on my computer now so I won't lose it again.

Haha thanks I was actually looking for a copy of my old solutions after my high school wiped all the student servers after graduation :).


Couldn't figure this out:
If log A(base B)= log B(base A), and A isn't equal to B, and A or B aren't equal to 0, what is AB?

If $\log_B{A} = \log_A{B}$, then by the change of base forumla we have $\frac{\log_A{A}}{\log_A{B}} = \log_A{B}$, thus $(\log_A{B})^2 = 1$. So $\log_A{B} = \pm 1$. If $\log_A{B} = 1$, then $A=B$, which you stated was not allowed. If $\log_A{B} = -1$, then $\frac{1}{A} = B$. Thus $AB = A\left(\frac{1}{A}\right) = 1$.

You could also use the "switcheroo" rule: $\log_x{y} = \frac{1}{\log_y{x}}$.

Could you explain how you transitioned through the steps in more detail?

Explaining in further detail: $\log_AB = \log_BA$, therefore, $B^{\log_BA} = A$ and $A^{\log_AB} = B$, substituting the first into the second we obtain $B^{({\log_BA})*({\log_AB}) }= B$, or, because $\log_AB = \log_BA$, $B^{({\log_BA})*({\log_BA}) }= B$, which simplifies to $B^{{(\log_BA)^2}}= B$.  Therefore, $(\log_AB)^2 = 1$, which yields $\log_AB=\pm 1$. Follow JS from here.

if log_A B = log_B A then log B / log A = log A / log B so log A = - log B so AB = 1


Frobenius
modular arithmetic
number theory
Additive Number Theory
Additive combinatorics
IMO
IMO 1983

Let $a,b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc-ab-bc-ca$ is the largest integer which cannot be expressed in the form $xbc+yca+zab$, where $x,y,z$ are non-negative integers.

This is a three variable expansion of the chicken mcnugget theorem.
we can show easily that given positive integral relatively prime $a,b$ the highest number which can't be constructed as $ma + nb$ is equal to: (hope I remember this correctly) $ab-a-b$. The proof is relatively simple (if you think about it long enough, the problem doesn't need a proof). To show something like this, just use similar reasoning (but since its an IMO problem I'm guessing there will be some tricks).

See : http://www.kalva.demon.co.uk/imo/isoln/isoln833.html .
If someone is interested in another solution, just ask.

Also, one can try to generalize this problem with $n$ integers $a_1, a_2, \ldots, a_n$ pairwise relatively prime and show that  :
\[ a_1a_2 \ldots a_n \left(n-1-\sum_{k=1}^n \frac 1{a_k} \right) \] is the greatest integer that can't be written as $\sum_{k=1}^n x_k \prod_{l \ne k} a_l$.

 :)

[hide="Yay"]

First off, I prove $2abc-ab-bc-ca$ is un-achievable.  Also, assume WLOG $a\ge b\ge c$.

Assume $2abc-bc-ca=xbc+yca+zab$, then take the equation $\pmod{ab}$ to give us $(x+1)bc+(y+1)ac\equiv 0\pmod{ab}$.  By CRT and $\gcd(a,b)=1$, we take this equation mod a and mod b to give us $y+1\equiv 0\pmod{b}$, and $x+1\equiv 0\pmod{a}$ (and using all numbers are relatively prime pairwise).  Substituting this back into the original equation gives us $z\le -1$, contradiction (this is where the $2abc$ part comes in).

Now, let $2abc-bc-ca-ab+k=xbc+yca+zab$, and if a solution exists where $bc-1\ge k\ge 1$ then we are done because we can add $1$ to $x$ always.

Consider the set $x=a-m$, where $m\in \{1,2,\cdots, a-1, a\}$ and $y=b-n$ where $n\in \{1,2,\cdots, b-1, b\}$.

Therefore, we get $2abc-bc-ca-ab+k=(a-m)bc+(b-n)ca+zab$.  This is the same as $bc(m-1)+ca(n-1)+k=(z+1)ab$ after doing some simplifying.  By CRT, this must hold mod a and mod b and because $\gcd(a,b)=1$.  Mod $a$ gives us $bc(m-1)+k\equiv 0\pmod{a}$, for which a value of $m-1$ obviously exists mod $a$, which can be chosen from the set of values we have assigned for $m$.  Similar method shows a value of $n-1$ exists mod $b$ from the set we have given to $n$.

Now, we already know that $x,y\ge 0$.  Also, the LHS of the equation $bc(m-1)+ca(n-1)+k$ is at least $1$, therefore $z\ge 0$ and we are done.  [/hide]

[hide=Yay]
First we will prove $2abc-ab-bc-ca$ is unattainable, as such:
Suppose $xbc+yca+zab=2abc-ab-bc-ca$. Then, taking this mod $a$, we have that $xbc\equiv 0\pmod{a}$, so $x\equiv 0\pmod{a}$, and $a|x$. Similarly, $b|y$, and $c|z$, so $x\ge a$, $y\ge b$, and $z\ge c$. Thus, $xbc+yca+zac\ge 3abc$, so $2abc\ge 3abc$, which is a contradiction.

Now we will prove all $n>2abc-ab-bc-ca$ is attainable, as such: consider the integer $r$ such that $r\equiv \frac{n}{bc} \pmod{a}$ and $0\le r\le a-1$. Rearranging the equation $xbc+yca+zab=n$ yields $\frac{n-xbc}{a}=yc+zb$, so set $x=r$. We see that $xbc-n=rbc-n\equiv 0 \pmod{a}$, so $\frac{n-xbc}{a}$ is a positive integer (obviously $n-xbc>0)$. Now, note that since $x\le a-1$, we have that $2abc-ab-ac-bc=(a-1)bc+abc-ab-ac\ge xbc+abc-ab-ac$, so $n>xbc+abc-ab-ac$. Thus, $n-xbc>abc-ab-ac$ so $\frac{n-xbc}{a}>bc-b-c$, so by Chicken Mc-Nugget theorem, there exist $b, c$ that satisfy the equation and are now done. $\blacksquare$

[hide=Alternative Solution]
lemma - $bc - b - c$ is the largest number which cannot be written as $mb + nc$ with m and n non-negative
[hide=Proof to lemma]
$0, c, 2c, ... , (b-1)c$ is a complete set of residues $mod b$
If $r > (b-1)c - b$, then $r = nc (mod b)$ for some $0 <= n <= b-1$
But $r > nc - b$, so $r = nc + mb$ for some $m => 0$
That proves that every number larger than $bc - b - c$ can be written as $mb + nc$ with $m$ and $n$ non-negative
Now consider $bc - b - c$ It is $(b-1)c (mod b)$, and not congruent to any $nc$
where $0 <= n < b-1$
So if $bc - b - c = mb + nc$
then $n => b-1$
Hence $mb + nc => nc => (b-1)c > bc - b - c$
Contradiction.
[/hide]
$0, bc, 2bc, ... , (a-1)bc$ is a complete set of residues $mod a$
So given $N > 2abc - ab - bc - ca$ we may take $xbc = N (mod a)$ with $0 <= x < a$
But $N - xbc > 2abc - ab - bc - ca - (a-1)bc = abc - ab - ca = a(bc - b - c)$
So $N - xbc = ka$, with $k > bc - b - c$
Hence we can find non-negative $y, z$ so that $k = zb + yc$
So, $N = xbc + yca + zab$
Finally, we show that for $N = 2abc - ab - bc - ca$ we cannot find non-negative $x, y, z$ so that $N = xbc + yca + zab$
$N = -bc (mod a)$
so we must have $x = -1 (mod a)$ and hence $x => a-1$
Similarly, $y => b-1, and z => c-1$
Hence $xbc + yca + zab => 3abc - ab - bc - ca > N$
Contradiction
[/hide]


Is it a page file or a virus?

My hard drive space is originally 6.5 gb (because I use a lot of space) but every time I boot Windows it dramatically decreases to 3.5 as time goes by and then (since I have a dual boot) in the other partition the hard drive says it has 6.5 gb again.
Any help?
I'll try system restore but tell me if you know anything...

I couldn't really understand what you were saying. Could you rephrase it?

However, I doubt you have a 3GB virus  :roll:

[quote="mgao"]now the "virus (?)" is causing the computer to malfunction, the system restore is not working and I can't see the restore point I created earlier, remember doing it clearly :Mad:[/quote]

[quote="7h3.D3m0n.117"][quote="Treething"]Clearly you have been browsing far too much porn.[/quote]

 :rotfl:

lol have you tried scanning your computer with Norton Anti Virus, AVG, Spybot, etc.?[/quote]

Ad-Aware's great too.


Show that $ gcd(2^m \minus{} 1, 2^n \minus{} 1) \equal{} 2^d \minus{} 1$, where $ d \equal{} gcd(m,n)$

Thanks.

It's old problem and very well-known

$ (n^a \minus{} 1,n^b \minus{} 1) \equal{} n^{gcd(a,b)}$ ,where $ n,a,b \in \mathbb{N}$

[u]Proof[/u]

Let $ d \equal{} gcd (n^a \minus{} 1 , n^b \minus{} 1)$ and $ k \equal{} ord_{d}n$

It's easy to see that $ n^{gcd(a,b)} \minus{} 1 | d$

since $ n^a \equiv n^b \equiv 1 (mod d)$ ,Hence $ k|a , k|b$

so $ k | gcd (a,b)$

thus $ n^{gcd (a,b)} \equiv 1 (mod d) \rightarrow d | n^{gcd(a,b)} \minus{} 1$

so $ d \equal{} n^{gcd(a,b)} \minus{} 1$

Sorry... but what does $ ord_d n$ means???

It's the minimum non-zero integer "k" such that n^k - 1 is a multiple of d.


If the mole fraction of a solute is $ X$, what is the molarity of the solution (given density $ d$).

[hide="Answer"]If there is more than one solute, then we need more data to solve the problem (the mole fraction of the solvent, at least).
If there is only one solute with mole fraction X then the mole fraction of solvent will be 1 - X. Let's consider a sample of our solution with a total number of moles of 1: then the number of moles of solute will be X and the number of moles of solvent will be 1 - X. If $ M_s$ and $ M$ are the molar masses of solute and solvent, respectively, then the masses of solute and solvent present in the sample will be $ m_s \equal{} XM_s$ and $ m \equal{} (1\minus{}X)M$, and the total mass of the sample will be $ m_t \equal{} XM_s \plus{} (1\minus{}X)M$. Using the density, then this mass will correspond to a volume of solution of $ V \equal{} \frac{m_t}{d} \equal{}> V \equal{} \frac{XM_s \plus{} (1\minus{}X)M}{d}$. Assuming the density is given in units of g/mL, then V is in mL, and we have to convert it to L through division by 1000:

$ V(L) \equal{} \frac{XM_s \plus{} (1\minus{}X)M}{1000d}$.

Now, by definition, the molarity of a solute in a solution is the number of moles of the solute that exist in every 1L of solution. Hence, the molarity of X, [X], in our solution will be

$ [X] \equal{} \frac{1000dX}{XM_s \plus{} (1\minus{}X)M}$.[/hide]


A bag contains 6 red marbles and 4 green marbles. Joe reaches into the bag and draws out 2 marbles at the same time. What is the probability that they are both red? Express your answer as a common fraction.

This is equivalent to drawing 2 marbles at different times. Therefore the answer is (6/10)(5/9)=[b]1/3[/b].


geometry
circumcircle
inequalities
inequalities proposed

Given are $\triangle ABC$ with area $S$, three median $m_{a},m_{b},m_{c}$ and circumradius $R$ prove that
$\frac{m_{a}}{a}m_{b}+\frac{m_{b}}{b}m_{c}+\frac{m_{c}}{c}m_{a}\ge \frac{3S}{R}$  :)

because
$m_{b}m_{c}\geq p(p-a)$,etc,we needs to prove
$\frac{p(p-a)bc+p(p-b)ca+p(p-c)ab}{abc}\geq \frac{3rp}{R}$
<=>$p^{2}\geq 8Rr+11r^{2}$
since $p^{2}\geq 16Rr-5r^{2}$,then above inequality is holds.

oh if you use $m_{b}m_{c}\ge p(p-a)$ then in my inequality you need to prove
$\frac{p(p-a)}{b}+\frac{p(p-b)}{c}+\frac{p(p-c)}{a}\ge \frac{3pr}{R}$ which don't like yours! :)

[color=purple]It's easy to check that $m_{a}\geq\sqrt{ p(p-a)}$ so to prove your inequality we must prove
$\frac{ \sqrt{(p-a)(p-b)}}{a}+\frac{ \sqrt{(p-b)(p-c)}}{b}+\frac{ \sqrt{(p-c)(p-a)}}{c}\geq \frac{3r}{R}$
  By AM-GM inequality we have
$\frac{\sqrt{(p-a)(p-b)}}{a}+\frac{\sqrt{(p-b)(p-c)}}{b}+\frac{\sqrt{(p-c)(p-a)}}{c}\geq 3\sqrt[3]{\frac{(p-a)(p-b)(p-c)}{abc}}=3\sqrt[3]{\frac{r}{4R}}\geq \frac{3r}{R}$(because $\frac{r}{R}\leq \frac{1}{2}$)
  Done :D[/color]

ok a simple proof!  :D
Actually I make it weak for the beauty , the stronger one
$\frac{m_{a}}{a}m_{b}+\frac{m_{b}}{b}m_{c}+\frac{m_{c}}{c}m_{a}\ge \frac{\sqrt{\sqrt{27}S}}{2}$
Try it if you want (and if you can)!  :D


Simplify

$ \frac{\sqrt{3\plus{}\sqrt{5}} \plus{} \sqrt{4\minus{}\sqrt{15}}} {\sqrt{3\plus{}\sqrt{5}} \minus{} \sqrt{4\minus{}\sqrt{15}}}$

Have you tried multiplying by:

$ \frac{\sqrt{3\plus{}\sqrt{5}} \plus{} \sqrt{4\minus{}\sqrt{15}}} {\sqrt{3\plus{}\sqrt{5}} \plus{} \sqrt{4\minus{}\sqrt{15}}}$?

$ \frac {\sqrt {3 \plus{} \sqrt {5}} \plus{} \sqrt {4 \minus{} \sqrt {15}}} {\sqrt {3 \plus{} \sqrt {5}} \minus{} \sqrt {4 \minus{} \sqrt {15}}}$ $ \cdot \frac{\sqrt{3\plus{}\sqrt{5}} \plus{} \sqrt{4\minus{}\sqrt{15}}} {\sqrt{3\plus{}\sqrt{5}} \plus{} \sqrt{4\minus{}\sqrt{15}}}$ $ \equal{} \frac{\left (\sqrt {3 \plus{} \sqrt {5}} \plus{} \sqrt {4 \minus{} \sqrt {15}}\right )^2}{3 \plus{} \sqrt{5}\minus{}4\plus{}\sqrt{15}}$ $ \equal{} \frac{3 \plus{} \sqrt {5} \plus{}2\sqrt{(3\plus{}\sqrt5)(4\minus{}\sqrt{15})} \plus{} 4 \minus{} \sqrt {15}}{3 \plus{} \sqrt{5}\minus{}4\plus{}\sqrt{15}}$

$ \frac{7\plus{}\sqrt{5}(1\minus{}\sqrt3) \plus{} 2\sqrt{12\minus{}3\sqrt{15}\plus{}4\sqrt5 \minus{}5\sqrt3}}{\sqrt5(1\plus{}\sqrt3) \minus{}1}$

But then?

It's probably easier to start by [hide]multiplying by $ \frac{\sqrt{2}}{\sqrt{2}}$.[/hide]

let x = sqrt(3 + sqrt5) and y = sqrt(4 - sqrt15)
you can also proceed as x/x-y + y/x-y
then add and subtract y/x-y
which gives you x-y/x-y + 2y/x-y
= 1 + 2y/x-y
now you can multiply and divide y
= 1 + 2(y^2)/(xy - y^2)
now your equation will become something like this
 1 + 2(4 - sqrt15)/(8 + 4 sqrt5 - 5 sqrt3 - 2 sqrt15)
now factorize the denominator so that numerator gets canceled out
4(2 + sqrt5) - sqrt15(sqrt5 + 2) = (4 - sqrt15)(2 + sqrt5)
after doing this you will be left with
1+ 2/(2+sqrt5)
which can be easily solved

Or you could note that $ \sqrt{3\plus{}\sqrt{5}} \equal{} \frac{1\plus{}\sqrt{5}}{\sqrt{2}}$, $ \sqrt{4\minus{}\sqrt{15}} \equal{} \frac{\sqrt{5}\minus{}\sqrt{3}}{\sqrt{2}}$, as JBL pointed out, and then the expression immediately reduces to $ \frac{1\minus{}\sqrt{3}\plus{}2\sqrt{5}}{1\plus{}\sqrt{3}}$.


floor function
number theory solved
number theory

Find all primes $p \geq 3$ with the following property: for any prime $q<p$, the number
\[
p - \Big\lfloor \frac{p}{q} \Big\rfloor q
\]
is squarefree (i.e. is not divisible by the square of a prime).

Thus, p mod q is square-free for any prime q<p. Let us suppose p>15. Take any prime factor q of p-4. If q>3, then p mod q is 4, false. Thus, q=3. This means that $ p=3^k+4$ for k>2. In the same way we find that any prime factor of $ p-8=3^k-4 $ is 5 or 7 and any prime factor of $ p-9=3^k-5 $ is 2 or 7. Since 7 cannot divide both p-8 and p-9, the problem reduces to solving in naturals the equations
 $ 3^k-5=2^r $ and $ 3^k-4=5^r $. For the first one, reduce modulo 4 and find that k is even. Then reduce modulo 3 and find that r is even. Thus, we have the equation $ (3^a-2^b)(3^a+2^b)=5 $ and it's trivial. For the second one reduce modulo 4 and find that k is even. That is k=2a and so $ 3^a-2, 3^a+2$ are both powers of 5, not possible for a>1. Thus these equations have the trivial solutions and this implies p<15. And from here hand is an useful instrument that I won't use. Anyway, I'm sure my solution is incorrect, since it makes the problem too simple and I'm sure it's not the case.

Actually, your solution IS correct! This, surprisingly perhaps, is an absolutely *trivial* problem which was solved by all contestants. So, all solutions are:
\[
p=3,5,7,13.
\]

[quote="harazi"]It makes me happy to hear that in a bulgarian TST there was a problem even easier than that proposed by Marian Andronache and me in our TST. I really couldn't believe that![/quote] that problem wasn't that easy. Not all of our contestants solved it. Some even will go to IMO, and they didn't solve it.  ;)

Not all of them solved it because they had some other problems to solve, too. This is the only reason for which they didn't solve it.

The same with China Team Selection Test 1999, Day 1, Problem 2
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=239205#p239205[/url]

[quote=themathematicalmantra]p = 43, q =41 actually all twin primes satisfy the requirement[/quote]

Read the question carefully. You do not have any control of $q$ except for the fact that $q<p$. It has to work for all $q<p$.


calculus
derivative
function
calculus computations

let $ f: \mathbb{R}\longrightarrow [0,\plus{}\infty)$ be a derivative function and $ f(0) \equal{} 0$ and $ k > 0$ that for every $ x\in \mathbb{R}$ $ f'(x)\le kf(x)$ prove that $ f(x)\equiv 0$

[quote="Amir.S"]let $ f: \mathbb{R}\longrightarrow \mathbb{R}$ be a derivative function and $ f(0) \equal{} 0$ and $ k > 0$ that for every $ x\in \mathbb{R}$ $ f'(x)\le kf(x)$ prove that $ f(x)\equiv 0$[/quote]
You are wrong. Let $ f(x)\equal{}\minus{}xe^{kx}$, then $ f'(x)\equal{}\minus{}(kx\plus{}1)e^{kx}<\minus{}kxe^{kx}\equal{}kf(x)$ and $ f(0)\equal{}0$.

Suppose that $ f(x) \not=0$ then $ f(x)>0$ and $ g(x)=f(x)e^{-kx} \Longrightarrow g^'(x)=(f^'(x)-kf(x))e^{-kx}\leq 0 \Longrightarrow g(x)$ is decreasing function ,then for $ x>0$ ,$ g(x)\leq g(0)=0$ and it's contradicton,because $ g(x)>0$ when $ x>0$.

[quote="ehsan2004"]Suppose that $ f(x) \not = 0$ then $ f(x) > 0$ and $ g(x) = f(x)e^{ - kx} \Longrightarrow g^'(x) = (f^'(x) - kf(x))e^{ - kx}\leq 0 \Longrightarrow g(x)$ is decreasing function ,then for $ x > 0$ ,$ g(x)\leq g(0) = 0$ and it's contradicton,because $ g(x) > 0$ when $ x > 0$.[/quote]
You are wrong because $ f(x)$ doesn't great than 0 when $ x>0$

[quote="TTsphn"][quote="ehsan2004"]Suppose that $ f(x) \not = 0$ then $ f(x) > 0$ and $ g(x) = f(x)e^{ - kx} \Longrightarrow g^'(x) = (f^'(x) - kf(x))e^{ - kx}\leq 0 \Longrightarrow g(x)$ is decreasing function ,then for $ x > 0$ ,$ g(x)\leq g(0) = 0$ and it's contradicton,because $ g(x) > 0$ when $ x > 0$.[/quote]
You are wrong because $ f(x)$ doesn't great than 0 when $ x > 0$[/quote]

Yes,you're right, the problem must be edited and $ f$ must be defined in $ [0,+\infty) \rightarrow [0,+\infty)$and it's wrong in $ \mathbb{R} \rightarrow \mathbb{R}$.

sorry.It should be $ f: \mathbb{R}\longrightarrow [0,\plus{}\infty)$.


function
inequalities
calculus
derivative
topology
inequalities theorems

I'm not sure if this is the proper place to post this, but it's related to an inequality theorem (Jensen's Inequality, specifically) so hopefully it's relevant.  Can somebody explain to me how to determine convexity for a multivariable function?  I'm aware of the rule for a univariate function $f(x)$ (i.e. that $f(x)$ is convex on [a,b] iff $f''(x) \geq 0$ on [a,b]), but I'm not even sure what it means to be convex for functions in more than one variable, and Mathworld isn't being helpful at all.  Thanks to anybody who responds for helping me out - I'm rather new to advanced inequalities, and have been trying to bone up on some basic theorems when I ran across this difficulty.

[quote="blahblahblah"][quote="zanttrang"]I'm aware of the rule for a univariate function $f(x)$ (i.e. that $f(x)$ is convex on [a,b] iff $f''(x) \geq 0$ on [a,b]), [/quote]

In a more general sense, this is not true.  A convex function is not necessarily differentiable even once (even though all convex functions are continuous)

It can be proven that if $f$ is twice differentiable, then your statement is true.  However, the only real definition of continuity is that for every $a+b=1$, $a,b\in (0,1)$, then $f(ax+by)\leq af(x)+bf(y)$[/quote]
Ah, thanks for the clarification.  How would you apply this definition in a specific case to prove convexity, though?  Aren't you just creating another inequality that must be proven true (or false)?  Also, how can this definition be extended to multivariable functions?

Well, the classical definition is awkward to apply.  But you can show that, for example, if $f$ is twice differentiable, then it's good enough to show that $f''\geq 0$.  Or if $f$ is just continuous, then its good enough to show that $f([a+b]/2)\leq [f(a)+f(b)]/2$

[quote="blahblahblah"]
It can be proven that if $f$ is twice differentiable, then your statement is true.[/quote]

Can you show the proof ? Or at least give link/reference to some paper about it?

For clarification: so if multivariable function is twice differentable and every second partial derivative of it is non-negative then the function is convex, right ?

Define Hessian of a twice differentiable $f$:
\[H(X)=
\left[\begin{array}{cccc}
\frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1\partial dx_2} & \ldots & \frac{\partial^2 f}{\partial x_1\partial x_n}
\\ \frac{\partial^2 f}{\partial x_2\partial x_1} & \frac{\partial^2 f}{\partial x_2^2} & \ldots & \frac{\partial^2 f}{\partial x_2\partial x_n}
\\ \vdots & \vdots & \ddots & \vdots
\\ \frac{\partial^2 f}{\partial x_n\partial dx_1} & \frac{\partial^2 f}{\partial x_n\partial x_2} & \ldots & \frac{\partial^2 f}{\partial x_n^2}
\end{array}\right]\]

Then the function is convex/concave on the open set $S$ iff $H(X)$ is positive/negative semidefinite for all $X\in S$. If $H(X)$ is positive/negative definite for all $X\in S$ then the function is strictly convex/concave.

[quote="Megus"][b]Bojan[/b]: can you give any links or reference about it ? I'd like to read something more about it.[/quote]
Try this: http://www.chass.utoronto.ca/~osborne/MathTutorial/CVNF.HTM

[quote="Megus"][quote="blahblahblah"]
It can be proven that if $f$ is twice differentiable, then your statement is true.[/quote]

Can you show the proof ?[/quote]

Sure.  I'd write it out, but I happen to know of a website which has the solution (it's an exercise from Rudin.  See the solution to Ex.14 here:

http://www-math.mit.edu/~rbm/18.100-F02.HMW/HMW7s.pdf

OK thanks, but I thought you were talking about multivariable functions :D - I knew this exercise before as I have Rudin's book


function quadratics calculus integration ratio algebra

1. The volume of a gas varies inversely with the amount of pressure.  If the volume of a gas is $70in^{3}$ when the pressure is $30 lb/in^{2}$, what is the volume of the gas when the pressure is $15lb/in^{2}$?

2. Write the function of least degree that has x-intercepts at $-7$ and $3$.

3.  What is the solution set for $(x+5)^{2}-13=0$?

4.  Write the quadratic equation, with integral coefficients, that has solutions $x = 4/5$ and $x = 1/4$.

answer to #1[hide]35, because you would set it up in a ratio form:
70/30= x/15   x=35, as you'ed divide both sides by 2
[/hide][/hide]

[quote="GlenthemanN"][hide="1"]
$140in^{3}$
Since the relationship between volume and pressure is inversely proportionate, when the pressure halves, the volume doubles. This is the case.
[/hide]

[hide="2"]
since a line doesnt have more than one x intercept we are dealing with a quadratic.
we know the roots so developing a function is easy
$(x+7)(x-3) = 0$
$x^{2}+4x-21$
[/hide]

[hide="3"]
$(x+5)^{2}-13=0$
$(x+5)^{2}= 13$
$x+5 = \pm\sqrt{13}$
$x_{1,2}=-5\pm\sqrt{13}$

Simply put 13 over to the right side and find the square root of both sides, isolate x and theres your solution set.
[/hide]

[hide="4"]
we are given both roots, so it is a matter of simply making one side of the equations given equal to zero in order to create our quadratic. (this is similar to #2)
$x = \frac{4}{5}\Rightarrow 5x-4 = 0$
$x = \frac{1}{4}\Rightarrow 4x-1 = 0$
Thus our equation is:
$(5x-4)(4x-1) = 20x^{2}-21x+4$
[/hide][/quote]

whoops.

[quote="GlenthemanN"][hide="1"]
$140in^{3}$
Since the relationship between volume and pressure is inversely proportionate, when the pressure halves, the volume doubles. This is the case.
[/hide]

[hide="2"]
since a line doesnt have more than one x intercept we are dealing with a quadratic.
we know the roots so developing a function is easy
$(x+7)(x-3) = 0$
$x^{2}+4x-21$
[/hide]

[hide="3"]
$(x+5)^{2}-13=0$
$(x+5)^{2}= 13$
$x+5 = \pm\sqrt{13}$
$x_{1,2}=-5\pm\sqrt{13}$

Simply put 13 over to the right side and find the square root of both sides, isolate x and theres your solution set.
[/hide]

[hide="4"]
we are given both roots, so it is a matter of simply making one side of the equations given equal to zero in order to create our quadratic. (this is similar to #2)
$x = \frac{4}{5}\Rightarrow 5x-4 = 0$
$x = \frac{1}{4}\Rightarrow 4x-1 = 0$
Thus our equation is:
$(5x-4)(4x-1) = 20x^{2}-21x+4$
[/hide][/quote]

i got the same  for 3.

[quote="Algebra 1 ©"]1. The volume of a gas varies inversely with the amount of pressure.  If the volume of a gas is $70in^{3}$ when the pressure is $30 lb/in^{2}$, what is the volume of the gas when the pressure is $15lb/in^{2}$?

2. Write the function of least degree that has x-intercepts at -7 and 3.

3.  What is the solution set for $(X+5)^{2}-13=0$?

4.  Write the quadratic equation, with integral coefficients, that has solutions $x=4/5 and x=1/4$.[/quote]

1.
[hide]$V \cdot P = k$
$70 \cdot 30 = k$
$k = 2100$

$V \cdot P = 2100$
$V \cdot 15 = 2100$
$V=\boxed{140 in^{3}}$[/hide]

2.
[hide]$x=-7 \text{ and }x=3$
$x+7=0 \text{ and }x-3=0$

$\Rightarrow f(x)=(x+7)(x-3)$
$\Rightarrow f(x)=\boxed{x^{2}+4x-21}$[/hide]

3.
[hide]$(x+5)^{2}-13 = 0$
$x^{2}+10x+25-13 =0$
$x^{2}+10x+12 =0$
$x= \frac{-10 \pm \sqrt{100-4(1)(12)}}{2 \cdot 1}$
$x= \frac{-10 \pm \sqrt{52}}{2}$
$x= \frac{-10 \pm 2 \sqrt{13}}{2}$
$x= \boxed{-5 \pm \sqrt{13}}$[/hide]

4.
[hide]By Viete's formulas:
$r_{1}+r_{2}=-\frac{b}{a}$
$-\frac{b}{a}= \frac{4}{5}+\frac{1}{4}$
$\frac{b}{a}=-\frac{21}{20}$

$r_{1}\cdot r_{2}= \frac{c}{a}$
$\frac{c}{a}= \frac{4}{5}\cdot \frac{1}{4}$
$\frac{c}{a}= \frac{1}{5}= \frac{4}{20}$

$\Rightarrow a=20, b=-21, c=4$
$\Rightarrow \boxed{20x^{2}-21x+4}$[/hide]

Sorry if this is redundant, I started before any replies were posted but I did part 4 much differently.

[quote="Algebra 1 ©"]3.  What is the solution set for $(x+5)^{2}-13=0$?

4.  Write the quadratic equation, with integral coefficients, that has solutions $x = 4/5$ and $x = 1/4$.[/quote]
[hide="3"]$(x+5)^{2}-13=0$

$x^{2}+10x+25-13=0$

$x^{2}+10x+12=0$

$x=\frac{-10\pm\sqrt{100-48}}2=\frac{-10\pm2\sqrt{13}}2=-5\pm\sqrt{13}$[/hide]
[hide="4"]The sum of the solutions is equal to $-\frac{b}{a}$, so $-\frac{b}{a}=\frac{21}{20}$.

Let $a=20$ and $b=-21$. Plugging $a$ and $b$ into the quadratic equation gives $c=4$.

The equation is $20x^{2}-21x+4=0$.

If $c$ turned out to be non-integral, the entire equation would have to be multiplied by some number to make it an integer.[/hide]

That is impossible it is a law from chemistry.
Pressure, $P_{n}$, and Volume, $V_{n}$,
$P_{1}V_{1}=P_{2}V_{2}$
In this problem
$P_{1}= 30$
$V_{1}=70$
$P_{2}=15$
Solve for $V_{2}$ after substituting and the answer is $140$.

[quote="Algebra 1 ©"]Correct!  #1 varies inversely though.  tridecalogism_lover's answer of 35 is right.  The answer isn't $140 in^{3}$.[/quote]
yes it varies inversely, that is exactly why it is 140 and not 35.

Elaborating from GlenthemanN , the formula $P_{1}V_{1}= P_{2}V_{2}$ is an inverse relationship.

Oops, my mistake. The answer is $140 in^{3}$!  I was using y=k/x and forgot about the P1 x V1 = P2 x V2 formula.  Thanks for posting the right answer and explanation!


vector
algebra
polynomial
linear algebra
matrix
symmetry

A,B,C are $ n \times n$ matrices, and $ CA\equal{}AC, CB\equal{}BC, C^2\equal{}I, AB\equal{}2(A\plus{}B)C$.
a) prove that :$ AB\equal{}BA$.
b) if $ A\plus{}B\plus{}C\equal{}0$, prove that: $ rank(A\minus{}C)\plus{}rank(B\minus{}C)\equal{}n$

We may assume $ C\equal{}diag(I_p,\minus{}I_q)$ , $ A\equal{}diag(U_p,V_q),B\equal{}diag(W_p,X_q)$ where $ p\plus{}q\equal{}n$. $ AB\equal{}2(A\plus{}B)C$ gives $ UW\equal{}2(U\plus{}W),VX\equal{}\minus{}2(V\plus{}X)$.
a) We must show $ UW\equal{}WU,VX\equal{}XV$.
   Assume $ x$ is a non zero vector s.t. $ Wx\equal{}2{x}$. Then $ UWx\equal{}2{U}x\equal{}2{U}x\plus{}2{W}x$ and $ Wx\equal{}0$, a contradiction. Thus $ W\minus{}2I_p$ is an invertible matrix. $ U(W\minus{}2I_p)\equal{}2W$ and $ U\equal{}2W(W\minus{}2I_p)^{\minus{}1}$. Finally $ U$ is a polynomial in $ W$. By the same way, $ V$ is a polynomial in $ X$.
b) $ A\plus{}B\plus{}C\equal{}0$ gives $ (U\minus{}I_p)(U\plus{}2I_p)\equal{}0, (V\plus{}I_q)(V\minus{}2I_q)\equal{}0$. Thus $ dim(ker(U\minus{}I))\plus{}dim(ker(U\plus{}2I))\equal{}p$ and $ dim(ker(V\plus{}I))\plus{}dim(ker(V\minus{}2I))\equal{}q$. Moreover $ A\minus{}C\equal{}diag(\minus{}I\plus{}U,V\plus{}I),B\minus{}C\equal{}diag(\minus{}U\minus{}2I,\minus{}V\plus{}2I)$. Then $ dim(ker(A\minus{}C))\equal{}dim(ker(\minus{}I\plus{}U))\plus{}dim(ker(V\plus{}I)),dim(ker(B\minus{}C))\equal{}dim(ker(\minus{}U\minus{}2I))\plus{}dim(ker(\minus{}V\plus{}2I))$.
Conclusion: $ dim(ker(A\minus{}C))\plus{}dim(ker(B\minus{}C))\equal{}p\plus{}q\equal{}n$.   QED

[quote="loup blanc"]We may assume $ C \equal{} diag(I_p, \minus{} I_q)$ , $ A \equal{} diag(U_p,V_q),B \equal{} diag(W_p,X_q)$ where $ p \plus{} q \equal{} n$. [/quote]
but why  ?

Hi Extremal,
$ C^2\equal{}C$ then $ C$ is a projection and there exists an invertible matrix $ P$ s.t. $ P^{\minus{}1}CP\equal{}diag(I_p,I_q)$ with $ p\plus{}q\equal{}n$. (of course we can have $ p\equal{}n$ or $ q\equal{}n$). Now let $ U\equal{}diag(U_1,\cdots,U_k)$ be a diagonal block matrix be s.t. for all $ i\not\equal{}j$ $ spectrum(U_i)\bigcap{s}pectrum(U_j)$ is void. Then $ UV\equal{}VU$ iff $ V\equal{}diag(V_1,\cdots,V_k)$ with for all $ i$, $ U_iV_i\equal{}V_iU_i$.

Obvious corrections:
$ C^2\equal{}I_n$, $ C$ is a symmetry and $ C$ is similar to $ diag(I_p,\minus{}I_q)$.

[quote="replay"]A,B,C are $ n \times n$ matrices, and $ CA = AC, CB = BC, C^2 = I, AB = 2(A + B)C$.
a) prove that :$ AB = BA$.
b) if $ A + B + C = 0$, prove that: $ rank(A - C) + rank(B - C) = n$[/quote]
  Simple Solution
a)   We have $ CA=AC,CB=BC,C^2=I_n$ then $ (A-2C)(B-2C)=AB-2AC-2CB+4C^2=AB-2AC-2BC+4I_n=AB-2(A+B)C+4I_n=4I_n$
 Hence $ (A-2C)(B-2C)=(B-2C)(A-2C)$ easy see that $ AB=BA$
b)   If $ A+B+C=0$ theorefore $ CA=AC,CB=BC,C^2=I_n$
we have $ (A-C)(B-C)=AB-(A+B)C+C^2=(A+B)C+C^2=-C^2+C^2=0$
 So use $ \forall X,Y \in \M_n{\mathbb{(C)}}$ we have $ rank(X+Y) \leq rank(X)+rank(Y) \leq rank(XY)+n$
we have $ rank(A+B-2C) \leq rank(A-C)+rank(B-C) \leq rank[(A-C)(B-C)]+n$
 So $ rank(A+B-2C)=rank(C)=n=rank[(A-C)(B-C)]+n$ easy have $ rank(A - C) + rank(B - C) = n$

Some observations:

Suppose A,B,C are in M(n,R) or M(n,C)
With the statement 2/
$ AB\equal{}2(A\plus{}B)C\equal{}\minus{}2C^2\equal{}\minus{}2I$

A,B are invertibles

If it is in M(n,C) since A,B,C commute two by two they are simultaneously trigonalisable.


When scholarships require you to be pursuing a major in some subject, do they actually confirm if you do major in that subject. Let's say that a requirement for some scholarship is that you major in, say, Environmental Science. Do they track you down over the next four years to see if you are really majoring in Environmental science? What if you decide that you want to do, say, Engineering instead. Do you then lose the scholarship money?

[quote="mathnovice"]When scholarships require you to be pursuing a major in some subject, do they actually confirm if you do major in that subject. . . . Do they track you down over the next four years to see if you are really majoring in [specified subject]. Do you then lose the scholarship money?[/quote]

I think that varies among scholarship programs. Some would probably have you fill out a renewal form year by year, on which you would affirm what you are doing, and to which you might have to attach a current college transcript. Other programs might not bother with any of this.

An important gotcha with many third-party scholarship programs is that sometimes they simply reduce your financial aid award. This varies from college to college depending on college policies. For some colleges, any third-party scholarship money you bring to the table is counted as part of your family's resources, and then the college offers less "financial aid" (that is, discount from list price) than it offers to someone else in otherwise identical circumstances. Some colleges, on the other hand, award MORE of their discretionary aid money, including their own in-house academic scholarships (again, that is discounts from list price) to applicants who show that they have received third-party scholarships.

[quote="tokenadult"]An important gotcha with many third-party scholarship programs is that sometimes they simply reduce your financial aid award.[/quote]

Right.  This is one of the most perverse aspects of financial aid.  A friend and I were both Intel finalists -- she finished in the top 10, earning a $20,000 scholarship, while I did not and won a $5,000 scholarship.  However, she recieves a lot of need-based financial aid, with the result that her $20,000 are worth about $5,000 to her.  I don't get financial aid (yet), so my $5,000 is worth the full $5,000 to me.

On the other hand, winning scholarship money can't [i]increase[/i] the amount you have to pay -- it might just not be worth very much to you.


Hi everyone  :)

There is a single path up a mountain. A climber starts at 6 A.M. and arrives at the top at 6 P.M. He spends the night on top of the mountain, and starts down the mountain the next day at 6 A.M. and arrives at the bottom of the mountain at 6 P.M. On each day, he travels at varying speeds - enjoying the scenery, stopping for lunch, etc. What are the chances that there was a spot on the mountain path that he passed at exactly the same time both days?

There are three common answers: very very slim, decently likely, and 100% likely. Please decide which of these three answers you agree to, and  [i]give sound reasoning[/i] for your answer.

I will be posting the correct answer and reasoning when there have been at least 5 posts from different people.

 :lol:

That's quite a classic problem, I think.

[hide]The probability is 100%. The situation is comparable to two climbers walking respectively towards the top and the bottom at the same time: of course they will meet each other in some point of the path.[/hide]
I must say this is not my own solution, I had already seen it somewhere, but no doubt it's nice.

[hide]I think the answer would be 100%; this is because if one were to make a graph of distance traveled over a given amount of time. The two plots of information would always intersect.[/hide]

bad explanation:[hide]100 p[ercent, for the same reason you cant go from 5 foot 2 to five foot 4 without being somewhere in between[/hide]

Pertaining to the comment about this being a classic problem:  I know that just about everytime I heard a promotion for Count Down this problem was used as an example.

You can also solve it by using the intermediate value theorem from Calculus:

Let $f(t)$ be the difference in the heights of the climber at time $t$, and $h$ be the height of the mountain.  Therefore, $f(6 A.M.)=h$, and $f(6 P.M.)=-h$.  Since $f$ is a continuous function, and goes from positive to negative, by the intermediate value theorem, there will be at least one $x$ such that $f(x)=0$, indicating that the climber is at the same height at the same time at the day before.


function
calculus
derivative
real analysis
real analysis solved

Prove that there is no function from [b]R[/b]x[b]R[/b] to [b]R[/b] which is injective of class [b]C[/b]1 in all the domain. (the first derivative is continuous).

Let $g$ denote the function $g :[0,2\pi] \to  R$  such that $g(t) = f(cos(t),sin(t)) - f(-cos(t),-sin(t))$
Manifestly $g$ is continuous  and $g(t+pi) = -g(t)$  for all $0 <t < \pi$, so there is $u(t)\in[t,t+\pi]$ such that
 $g(u)=0  \Leftrightarrow f(cos(u),sin(u)) = f(-cos(u),-sin(u))$ , so $f$ cant be injectve
More general, if n > m, , there is no continuous & injective  function  $R^n \to R^m$ , but the proof is more dificult .. :D
 :cool:  Do you know Brouwer  :cool:


modular arithmetic
function
number theory
totient function

Find this without the calculator
$ (97^{98} \plus{} 98^{97}) : 5$
can someone help me? :blush:

What are you supposed to do in this problem?

i think we have to use modular arithmetic
sorry i am bad in english but i will try to exlpain :
case 1)
x:y=z
[u]x-zy=0 [/u]

case 2)
x:y=z
x-zy=m$ \neq0$ and $ m$ is natural number

so we have to find $ m$ or $ z$ and i suppose that for this we need  modular arithmetic, but don't know about modular arithmetic so i am asking for you're help  :)

Now this isn't too hard.

[hide="solution"]
$ 97^{98} \equiv 97^{\;98 \pmod {\varphi (5)}} \equiv 97^2 \equiv 2^2 \equiv 4 \pmod 5$

$ 98^{97} \equiv 98^{\;97 \pmod {\varphi (5)}} \equiv 98 \equiv 3 \pmod 5$

$ 97^{98}\plus{}98^{97} \equiv 2 \pmod 5$[/hide]

You should check these links:

[url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url]
[url=http://en.wikipedia.org/wiki/Euler_theorem]Euler's theorem[/url]; [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Euler's totient function[/url]
[url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael's theorem[/url]
[url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url]


geometry
trigonometry
Pythagorean Theorem
AMC 10
trig identities
Law of Sines

A circle passes through the three vertices of an isosceles triangle that has sides of length 3 and a base of length 2. What is the area of this circle?

(A) 2π (B) 5π/2 (C) 81π/32 (D) 3π (E) 7π/2

[quote="patelparth"]A circle passes through the three vertices of an isosceles triangle that has sides of length $ 3$ and a base of length $ 2$. What is the area of this circle?

$ \mathrm{(A)}\ 2\pi \qquad\mathrm{(B)}\ 5\pi/2 \qquad\mathrm{(C)}\ 81\pi/32 \qquad\mathrm{(D)}\ 3\pi \qquad\mathrm{(E)}\ 7\pi/2$
[/quote]

[hide="Solution 1"]
Let $ \triangle ABC$ have vertex $ A$ and center $ O$, with foot of altitude from $ A$ at $ D$.
[asy]import olympiad;
pair B=(0,0), C=(2,0), A=(1,3), D=(1,0);
pair O=circumcenter(A,B,C);
draw(A--B--C--A--D);
draw(B--O--C);
draw(circumcircle(A,B,C));
dot(O);
label("\(A\)",A,N);
label("\(B\)",B,S);
label("\(C\)",C,S);
label("\(D\)",D,S);
label("\(O\)",O,W);
label("\(r\)",(O+A)/2,SE);
label("\(r\)",(O+B)/2,N);
label("\(h\)",(O+D)/2,SE);
label("\(3\)",(A+B)/2,NW);
label("\(1\)",(B+D)/2,N);[/asy]
Then by Pythagorean Theorem (with radius $ r$, height $ OD = h$) on $ \triangle OBD, ABD$
\begin{align*} h^2 + 1 & = r^2 \\
(h + r)^2 + 1 & = 9 \end{align*}
Substituting and solving gives $ r = \frac {9}{4\sqrt {2}}$. Then the area of the circle is $ r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \frac {81}{32} \pi \Rightarrow \mathrm{(C)}$.
[/hide]

[hide="Solution 2"]
By $ A = \frac {1}{2}Bh = \frac {abc}{4R}$ (or we could use $ s = 4$ and Heron's),
\[ R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}
\]
and the answer is $ R^2 \pi = \mathrm{(C)}$
[/hide]

[hide="Solution 3"]
By the Extended Law of Sines,
\[ 2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}
\]
Answer follows.
[/hide]

Edit: :wink: I just completely overkilled an AMC 10 question

can you show me the exact steps for substitution. Thanks.

Sure thing.
\begin{align*} h^2 + 1 & = r^2 \Longrightarrow h = \sqrt{r^2 - 1}\\
(h + r)^2 + 1 & = 9 \Longrightarrow (\sqrt{r^2-1}+ r)^2 = 8\end{align*}
Take square root of both sides:
\begin{align*}
\sqrt{r^2 - 1} + r &= 2\sqrt{2}\\
\end{align*}

Now move $ r$ to the other side and square:
\begin{align*}
r^2 - 1 &= (2\sqrt{2} - r)^2 = 8 - 4\sqrt{2}r + r^2 \\
4\sqrt{2} r &= 9
\end{align*}

And then you can easily solve for the circle's area.


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

Let $A,B\in M_n\mathbb({C})$ such that $rank(AB-BA)=1$.Prove that $(AB-BA)^2=0$ ;)

Yes, it's easy :).

Since $AB-BA$ has rank $1$, it's the product of an $(n,1)$ matrix and a $(1,n)$ matrix, or, in other words, it's the product $xy^t$, where $x,y$ are column vectors of size $n$. $\mbox{Tr}(AB-BA)=0$ simply means $y^tx=0$, so $(AB-BA)^2=(xy^t)(xy^t)=x(y^tx)y^t=0$.

easier: JordanForm of AB-BA has rank 1, and obviously all eigenvalues 0 (trace=0), therefor it has minimal polynomial x^2, which means (AB-BA)^2=0. :)


In the United States, coins have the following thicknesses: penny, $ 1.55$ mm; nickel, $ 1.95$ mm; dime, $ 1.35$ mm; quarter, $ 1.75$ mm. If a stack of these coins is exactly $ 14$ mm high, how many coins are in the stack?

$ \textbf{(A)}\ 7\qquad
\textbf{(B)}\ 8\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 11$

[hide="Click to reveal hidden answer"]$ \boxed{8}$ quarters.[/hide]

All you have to do is to add different numbers of quarters, pennies, dimes, and nickels until you get $ 14$.  Your answer is then obvious, $ 8$.


[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=638050#638050]Problem 1[/url]
Let f: \mathbb{R}^{+}\rightarrow \mathbb{R}+ be an increasing function. For each u\in\mathbb{R}^{+}, we denote g(u)=\inf\{ f(t)+u/t \mid t>0\}. Prove that:
(a) If x\leq g(xy), then x\leq 2f(2y);
(b) If x\leq f(y), then x\leq 2g(xy).

mods please delete this duplicated topics also how i can move these topics, turkish mathematical olympiad series, to [url=http://www.mathlinks.ro/Forum/resources.php?c=174]olympiad resources[/url]. or can u do it?


show (without a calculator) that 1/sin 45<cbrt 3
-edit- yes mathfanatic you are right

[hide]Uh, assuming that the 45 is supposed to be degrees, I don't think that's right.

sin 45 degrees = 1/:rt2:, so 1/sin 45 = :rt2:.

 ?

 ?

2^3 ? 3^2

8 < 9[/hide]

Here's my answer:

[hide]1/sin45=1/(1/:rt2:)=:rt2:

:rt2:_cbrt3  cube both sides:

rt8_3   then square:

8<9 so 1/sin45<cbrt3[/hide]

[hide]

So we can assume that

So we can tell that since  that [/hide]

Btw, how do you do cube roots in LaTeX?


Standing in a circle are 99 girls, each with a candy. In each move, each girl gives her candy to
either neighbour. If a girl receives two candies in the same move, she eats one of them. What
is the minimum number of moves after which only one candy remains?

May be you missed something   :huh:  :?:
 :wink: I doubt that it will ever achieve, if every girl in every move would give her candy to her right neighbor. So we can't certainly set minimum number of moves after which only one candy remains.

You can arrange the moves as you like, so arrange to get the minimum :wink:

Let the girls be labelled 1 to 99 in clockwise order. We firrst show that the task can be
accomplished in 98 moves. In each of the first 49 moves, if she still has a candy, the k-th girl
gives hers to the (k -1)-st girl for \[ 2 \leq k \leq 50\] and to the (k + 1)-st girl for \[ 51 \leq k \leq 99\]. The
1-st girl, who will always have a candy, gives hers to the 99-th girl. After 49 moves, only the
1-st and the 99-th girl have candies. These two candies can be passed, in opposite directions,
to the 50-th girl in another 49 moves. We now show that the task cannot be accomplished in
less than 98 moves. We will not allow the girls to eat the candies, but each must pass all she
has to the same neighbour. Our target is to have all candies in the hands of one girl. Consider
what happens to a candy in two consecutive moves. It either returns to the girl who has it
initially, or is passed to a girl two places away. Suppose the candies all end up with the 50-th
girl in at most 98 moves. The number of moves is not enough for the candy initially with the
50-th girl to go once around the circle before returning to her. It follows that the number of
moves must be even. However, in order for the candy initially with the 49-th girl to end up
in the hands of the 50-th girl in an even number of moves, it must go once around the circle,
and that takes 98 moves.


Determine all positive integers $n$ with the property that  $n = (d(n))^2$. Here $d(n)$ denotes the number of positive divisors of $n$.

Because n is square d(n) is odd.
1. One solution is n=1.
2. Let n>1 $n=p_1^{2k_1}\dots p_m^{2k_m}.$ Our equation equavalent to
$(2k_1+1)\dots (2k_m+1)=p_1^{k_1}\dots p_m^{k_m}$, were $p_i$ are odd primes.
Because $(2x+1)<3^x$ for x>1, we have only one solution (n>1):
$n=9=3^2$.

Now we know that $ n \equal{} 1$ is ok.

Suppose $ n\equal{}3,4,5,...$,
Let $ n \equal{} (2k \plus{} 1)^2(k \ge 1)$,
$ d(n) \equal{} 2k \plus{} 1$

All of n's fractions are:
$ {m,n/m},1 \le m \le 2k$,
and $ 2k \plus{} 1$;

So the number of '$ m$' is $ k$.
$ m \equal{} 1,3,5,7,9,...,2k \minus{} 1$.

Now we know that $ (2k \minus{} 1)|(2k \plus{} 1)^2$,
$ (2k \minus{} 1)|2,k \equal{} 1,n \equal{} 9$

Thus $ n \equal{} 1,9$


geometry
rhombus
rectangle
Pythagorean Theorem

On side AB and DC of rectangle ABCD , points E and F are chosen so that
AFCE is a rhombus. If AB = 16 and BC = 12 , what is the length of AF ?

hello, let $ \overline{DE}\equal{}x$ and $ \overline{FB}$ also $ x$ and $ \overline{AF}\equal{}\overline{AE}\equal{}\overline{EC}\equal{}\overline{CF}\equal{}16\minus{}x$ because $ AECF$ is a rhombus. After Pythagorean theorem we have $ 12^2\plus{}x^2\equal{}(16\minus{}x)^2$ and from here we got $ x\equal{}3,5$ and $ \overline{AE}\equal{}16\minus{}3,5\equal{}12,5$
Sonnhard.


function geometry parallelogram MATHCOUNTS

Who took the MAML level 1 today? I felt like I did pretty well despite the fact that I only got 2 hours of sleep last night... but I also think that this year's test was a lot easier than previous years.

Anyways, here's my answers; Does anyone know the answer to #28? That's the only one that I was not able to do...

[hide]
1. B
2. D
3. D
4. D
5. A
6. D
7. C
8. C
9. A
10. B
11. B
12. D
13. D
14. B
15. E
16. C
17. B
18. E
19. C
20. C
21. D
22. E
23. C
24. C
25. B
26. A
27. B
28. ???
29. C
30. B
[/hide]

I also took it today. I answered 26 of them, all of which were right (and probably should have gotten more, solved the other 4 during my next class).

The answer to #28 was D.

Here are my answers, all of which were checked:
[hide]
6.D
7.C
8.C
9.A
10.B
11.B
12.D
13.D
14.B
15.E
16.C
17.B
18.E
19.C
20.C
21.D
22.E
23.C
24.C
25.B
26.C*
27.B
28.D
29.C
30.B
[/hide]

(note that I solved 26, 28, 29, and 30 after the contest)[/hide]

hmm... i think i actually jus solved 28, i get functions:

f(x) = x^2-4 and g(x) = x^2 + 2x + 5

if you combine that and find the min value, you get 2/3... which is C

ahhh crap... i just realized that for #26, i forgot to count negative values of b...

wait... if i count negative answers, i still only get 40 for #26... where do the extra nine values come from? (i basically factored, then plugged in numbers, noting that at every integer, the expression jumped to another value, then counting the number of integers in each interval)

Does anyone recall the actual, numerical answers - not the letter choices - to #14, #28, and...whichever problem had the parallelogram of area K? I don't remember which letters correspond to which answers (or maybe you all could keep your test booklets, which wasn't done at my school).

Thanks.

[quote="sirorange"]Meh. . .I didn't do so well. 22 answered, all correct I'm pretty sure. And people on our team got for than 20 for number 26, someone was pretty sure that the answer was at least 48.[/quote]

Yeah, I also found that the answer was at least 48!

I just guessed it might be 49 (and remember, I figured this out after the test) so I put that in my answers above and put a * next to it. Otherwise, my answers should be completely correct (checked them all with Feiqi).

Concyclicity, #28 had an answer of $\frac{14}{3}$, and the K problem answer was $\frac{K}{6}$. Hehe, for that problem I said WLOG let it be a square and then it was much easier.

I made a large diagram and measured.

...well, it worked. And it was quick.

#19...eh I screwed that up, spent 25 minutes on it, ended up not getting the right answer.

21 answered, 1 wrong. 109. Not too shabby for an 8th grader. My goals were to get top 50, highest 8th grade, 2nd highest 9th grade and below.

[quote="123s"]Do you have any idea when the qualifiers will be posted for the MAML level 2?[/quote]
Mid to late December.  The rankings will be available before then, but the MAML meeting where the cutoff values are determined (and voted on) is in December.

I also think #26 is 48, and my calculator agrees with me (graphing fPart(x)^2 - x^2 shows exactly two ways to achieve every integer between -1 and -24 inclusive within the relevant domain).

[quote="Phelpedo"]Eh, I kinda suck at speed contests (Mathcounts was so long ago...). Answered 24 I think, and made at least several careless errors (forgot to add 1 for #3, did circumference = $r\pi$ for #7, etc). I shall blame this on lack of sleep.[/quote]

people from ct can do the maml?

Yes they can, if they live in Massachusetts  :wink:

(If the answers in this thread are all correct) I got 26 right, 2 wrong for a total score of 132.

Does anyone else think this year's was easier than previous years', and that the cutoff for level 2 will be much higher?  Or is this forum just the very top few high scorers?

I got a 122. Last years test was definitally harder.

The first 20 questions last year were pretty easy...the rest were pretty beastly :wink:

Yup, 23 correct.

Yah, I finished all the problems I could, and then I just sat there and checked because I couldn't do anymore of the problems.