topology
real analysis
real analysis unsolved

Prove that every uncountable closed set (on the real line) contains a Cantor set.

Definition of a Cantor set : Start by deleting an open middle interval $ (c,d)$ from the initial interval $ [a,b]$, leaving two intervals. (The only requirement to be a middle interval of $ [a,b]$ is that $ a < c < d < b$) Next, delete open middle intervals from each of the remaining intervals, leaving four intervals. Continue this process to infinity. all points in $ [a,b]$ that are not deleted at any step form a set which we will call a Cantor set.

Let $ A\subset\mathbb{R}$ be an uncountable closed set.

Definition 1 : $ x\in\mathbb{R}$ is a right accumulation point (resp. left accumulation point) of $ A$ if $ [x,x\plus{}\epsilon]\cap A$ (resp. $ [x\minus{}\epsilon,x]\cap A$) is uncountable for all $ \epsilon > 0$.

Definition 2 : [a,b], $ a < b$, is a good interval for $ A$ if $ a$ is a right accumulation point and $ b$ is a left accumulation point.

Observe that if $ A$ contains a Cantor set $ C$ then the remaining intervals from each step of construction of the Cantor set are good intervals. That's the idea.

Lemma 1 : If $ [a,b]\cap A$ is uncountable, there is a left accumulation point in $ (a,b]$. (proof : $ \inf\{ x : [x,b]\cap A\text{ is countable}\}$ is a left accumulation point.)

It follows that there is a good interval for $ A$. (first there is a right accumulation point x. and then there is a left accumulation point y with $ x < y$. Then [x,y] is a good interval.)

Lemma 2 : If [a,d] is a good interval then [a,b] and [c,d] are good intervals for some b and c with $ b < c$. Moreover, b (resp. c) can be chosen arbitrarily close to a (resp. d) (proof : it follows from Lemma 1.)

Now let's construct $ C$. Inductively for each $ n\in\mathbb{N}$ we construct a family $ \mathcal{C}_{n}$ of $ 2^{n}$ disjoint good intervals of length $ <\frac{1}n$ :
Suppose we have $ \mathcal{C}_{n}$, then delete an open middle interval from each [a,d] in $ \mathcal{C}_{n}$, (Lemma 2) leaving two good intervals of length $ <\frac{1}{n\plus{}1}$. The remaining $ 2^{n\plus{}1}$ good intervals form $ \mathcal{C}_{n\plus{}1}$.

Let $ C$ be the Cantor set corresponding to $ \{\mathcal{C}_{n}\}_{0}^{\infty}$. We want to show that $ C\subset A$: If $ x\in C$, for each $ n\in\mathbb{N}$ choose $ x_{n}\in A$ in the interval of $ \mathcal{C}_{n}$ containing $ x$. Since $ \{x_{n}\}_{0}^{\infty}$ is a sequence in $ A$ converging to $ x$, we have $ x\in A$.

(btw is this a well known result?  i came up with this result while trying to solve [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=166281]another problem[/url].)

Well, this is a small improvement upon the classical lemma that states that every uncountable closed set contains a perfect set. As to the other problem, you seem to be very close to solving it now, so I won't spoil the fun ;)

yes I think redkimchi should make the meaning of Cantor set is more clearer. It's wellknow that every closed uncountable set is union of a perfect set and a countable, this proof seems just proved something like that.


geometry
parallelogram
vector
trigonometry
geometric transformation
reflection
cyclic quadrilateral

Problem from IMAC 2008. Proposed by Moldova...

You are right, ahiles.A teacher has shown me the subjects.
I think this was a copied problem and not a new one, but I can't solve it.

are only the people who it is entitled are allowed to answer (AM i allowed :) )

OK here goes
first we construct a parallelogram BCMD on the outer side of AD
now angle EBL = angle EAN
and angle ENA = 90 = angle ELA
so triangle ELB is similar to ENA
so we have EN/EL=EA/EB
but EA/EB=AD/BC (triangle EBC is similar to EAD)
but also BC = MD so we have EN/EL=AD/DM
also by simple angle chasing we get angle ADM = angle NEL
so triangle NEL is similar to ADM
but EL is perpendicular to BC implies EL is perpendicular to DM
also EN is perpendicular to AD
so NL is perpendicular to AD
but KL is parallel to AD (midpoint theorem and consider triangle KXL where X is the midpoint of BC )

Let be the point $ F\equiv BC\cap AD$ and the points $ K,\ M,\ P,$ where $ P$ is the midpoint of the segment $ EF,$ are collinear, as well $ ($ from the complete quadrilateral $ FCEDBA$ $ ).$

Because of $ \frac {LB}{LC} \equal{} \frac {NA}{ND}$ we have that the points $ K,\ Q,\ M,$ where $ Q$ is the midpoint of the segment $ LN,$ are collinear, as well.

That is the line segment $ PK,$ bisects the segment $ LN.$

So, in the cyclic quadrileteral $ ELFN,$ with diameter $ EF,$ we conclude that the line segment $ PQ,$ is the midperpendicular of the cord $ LN.$

Hence, we conclude that $ KM\perp LN$ and the proof is completed.

Kostas Vittas.

Let $ E$ be the origin, and the lowercase letters denote the position vectors of the respective points. Then we have to show that $ (l\minus{}n)\cdot\left(\frac{a\plus{}b}{2}\minus{}\frac{c\plus{}d}{2}\right)\equal{}0$. Using $ l\cdot (b\minus{}c)\equal{}0$ and $ n\cdot (d\minus{}a)\equal{}0$ it remains to show that $ l\cdot (a\minus{}d)\equal{}n\cdot (b\minus{}c)$ i.e. $ EL\cdot DA\cos\angle EYA\equal{}EN\cdot CB\cos\angle EXB$; here $ Y\equal{}EL\cap AD,X\equal{}EN\cap BC$. But we have $ \angle EXB\equal{}90^{\circ}\minus{}\angle XEL\equal{}90^{\circ}\minus{}\angle YEN\equal{}\angle EYA$ and $ \triangle AED\sim\triangle BEC\implies\frac{EL}{EN}\equal{}\frac{BC}{AD}$. Therefore we are done.

(USA TST 2004) Let $ ABCD$ be a cyclic quadrilateral whose diagonals intersect at $ E$. Let $ M$, and $ N$ be the midpoints of $ AB$ and $ CD$, respectively. Let $ Q$ and $ R$ be the projections of $ E$ onto $ BC$, and $ DA$, respecitvely. Prove that $ QR$ is the perpendicular bisector of $ MN$.

My solution: Let $ F$ be the reflection of $ E$ in $ BC$. Since $ \triangle AED$ is indirectly similar to $ \triangle BEC$, and $ \triangle BEC$ is indirectly similar to $ \triangle BFC$, so $ \triangle AED$ is directly similar to $ \triangle BFC$. Note that $ \frac {1}{2} \triangle AED \plus{} \frac {1}{2} \triangle BFC \equal{} \triangle MPN$. It follows that $ \triangle MPN\sim\triangle AED\sim\triangle BEC\sim\triangle MQN$. But $ \triangle MPN$ and $ \triangle MQN$ share corresponding side $ MN$ so they are congruent. The conclusion follows.

Dear Mathlinkers,
let A' be the symmetric of A wrt N, B' the symmetric of B wrt L.
By a simple angle chasing, it follows that AB'=BA' ; then we applied "the little Thalès" i.e. the line joining the midpoint of two sides of a triangle from which follow that KLN is K-isoceles.
Mutatis mutandis, MLN is M-iscoceles.
We are done.
This problem has been proposed in OM Slovenia 1996.
Sincerely
Jean-Louis

This problem was posted before in the High School section at
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55836[/url].

A more general version (with 2 inscribed kites) also appears at
[url]http://www.mathlinks.ro/viewtopic.php?t=63589[/url].

A diagram for my solution (there is a typo; it is P;Q are projections of E onto BC, AD)

J,K,L,M is the midpoint of AG,DG,CG,BG and it is easy to proof that
$ \triangle HME\ \sim \equal{} \triangle EJI\$$ so EH=EI
also FI=FH so it is easy to proof HI$ \perp$ EF[/quote]

[color=darkblue][b]See[/b][/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=210729][color=red][b]here[/b][/color][/url] [color=darkblue][b]and another proofs and[/b][/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216738][color=red][b]here[/b][/color][/url] [color=darkblue][b]an its extension.[/b] [/color]

[quote="Inequalities Master"]The diagonals of the cyclic quadrilateral $ ABCD$ are intersecting at the point $ E$.
$ K$ and $ M$ are the midpoints of $ AB$ and $ CD$, respectively. Let the points $ L$ on $ BC$ and $ N$ on $ AD$ s.t.
$ EL\perp BC$ and $ EN\perp AD$.Prove that $ KM\perp LN$.[/quote]
[hide="Solution"]
Let $ P$ be the intersection between $ AD$ and $ BC$, let $ Y$ and $ X$ be the respective orthocenters of $ \triangle PAC$ and $ \triangle PBD$, let $ YA$ and $ PC$ meet at $ Q$, let $ YC$ meet $ PD$ at $ R$, and let the intersections between $ PY$ and $ AC$ and $ PX$ and $ BD$ be $ T$ and $ U$ respectively.  Now, since $ ABCD$ is cyclic, we have that $ \triangle PBD\sim \triangle PAC$, so $ \frac {PT}{PY} \equal{} \frac {PU}{PX}$, so $ TU\parallel XY$.  Furthermore, $ (TU, PD) \equal{} \angle TPN \plus{} \angle PTU$.  Since $ \angle ENP \equal{} 90 \equal{} \angle PLE \equal{} \angle PUE \equal{} \angle ETP$ (notice that since $ Y$ is the orthocenter of $ \triangle PAC$, we have that $ CT\perp PY$, and similarly, $ DU\perp PX$), we have that $ ENTPUL$ is cyclic, so $ (TU, PD) \equal{} \angle AEN \plus{} \angle PEB$ from $ \angle AEN \equal{} \angle APT$ and $ \angle PTU \equal{} \angle PEB$.  Yet, $ \angle AEN \equal{} 90 \minus{} \angle DAE \equal{} 90 \minus{} \angle EBC \equal{} \angle BEL$, so $ (TU, PD) \equal{} \angle PEL \equal{} \angle PNL$ from which we see that $ NL\parallel TU\parallel XY$.

Now, from the fact that $ Y$ is the orthocenter of $ \triangle PAC$, we have that $ \angle CRD \equal{} 90$ and $ \angle AQB \equal{} 90$.  Then, $ Q$ lies on the circle with diameter $ AB$ and $ R$ lies on that with diameter $ CD$.  Since $ \angle ARC \plus{} \angle AQC \equal{} 90 \plus{} 90 \equal{} 180$, we have that $ AQCR$ is cyclic, so $ YA\cdot YQ \equal{} YR\cdot YC$.  This means that $ Y$ has equal power with respect to the circles with diameter $ AB$ and $ CD$, so $ Y$ lies on the radical axis of these two circles.  Similarly does $ X$.  This means that $ XY$ is perpendicular to the line connecting the centers of these circles, $ KM$.  It follows that $ XY\perp KM$, so $ NL\perp KM$. [/hide]

Let’s change your M to P, since M is supposed to be the middle of CD as well.
It has been nicely proved that NL and AP are perpendicular, but KM is midline of triangle ABP (M is middle of BP as BCPD is parallelogram).
Another approach was to see that KLMN is a kite: take X and Y the middles of BE, AE respectively, see that triangles KXL and NYK are congruent (LX and NY being medians in right-angled triangles), hence KL = KN; similarly for KM = MN.

Best regards,
sunken rock


function
real analysis
real analysis unsolved

Let $f: [0,\infty)\to\mathbb R$ be a function such that for any $x>0$ the sequence $\{f(nx)\}_{n\geq 0}$ is increasing.

a) If the function is also continuous on $[0,1]$ is it true that $f$ is increasing?

b) The same question if the function is continuous on $\mathbb Q \cap [0, \infty)$.

"abordabila" -> "abordable" (Just like in French BTW)

[quote="perfect_radio"]This seems to have been the most "abordabila" (how do you say that in English?) problem from the 11th grade. I think that:

(a) No.

(b) Yes.

Are my answers correct? If so, I'll present a sketch of my solution.[/quote]

You are wrong ! If you answer "(b)Yes" then you have to answer "(a)Yes"
The statement is still right if $f$ is continuous on $S$, and $S$ is dense in $[0,\infty[$..
Do you follow me ?
 :cool:

I don't get it. Who is $S$?

BTW, the solutions are online now, and I'm right.

(a) Sorry , I made a mistake . I did'nt see that the function is continuous only on $[0,1]$ , and not on $[0,\infty[$ :blush:
So the function can be non-increasing : $f(x)= x$ if $x\in[0,1]$, $f(x)=x$ if $x\in Q_+$, $f(x)=\frac{2x}{x+1}$ if $(x\in ]1,\infty[)\& (x\notin Q_+)$
(b) If $f$ is continuous on  a sub-set $s\subset[0,\infty[$, dense in $[0,\infty[$ , then $f$ is increasing on $\bar s = [0,\infty[$
:cool:

[quote="Diogene"](a) Sorry , I made a mistake . I did'nt see that the function is continuous only on $ [0,1]$ , and not on $ [0,\infty[$ :blush:
So the function can be non-increasing : $ f(x) \equal{} x$ if $ x\in[0,1]$, $ f(x) \equal{} x$ if $ x\in Q_ \plus{}$, $ f(x) \equal{} \frac {2x}{x \plus{} 1}$ if $ (x\in ]1,\infty[)\& (x\notin Q_ \plus{} )$
(b) If $ f$ is continuous on  a sub-set $ s\subset[0,\infty[$, dense in $ [0,\infty[$ , then $ f$ is increasing on $ \bar s \equal{} [0,\infty[$
:cool:[/quote]

I suppose you've meant

(b) If $ f$ is continuous on a sub-set $ s\subset[0,\infty[$, dense in $ [0,\infty[$ and increasing on this subset, then $ f$ is increasing on $ \bar s \equal{} [0,\infty[$

for otherwise it's clearly wrong. It's wrong in this case as well: take $ f(x) \equal{} x$ for $ x \in S : \equal{} \mathbb R^\plus{}_0\backslash \mathbb Q$ and $ f(\frac{p}{q}) \equal{} \frac{p \minus{} 1}{q}$ for $ \frac{p}{q}\in\mathbb Q$, $ p\in\mathbb Z$, $ q\in\mathbb N$, $ (p, q) \equal{} 1$. Then $ f$ is increasing and continuous on a dense subset $ S$ but not increasing on $ [0, \infty)$.

My solution for part b) goes as follows: note that if $ 0 \le x < y$ and $ \frac{x}{y} \in \mathbb Q$ then $ f(x) < f(y)$ (easy to obtain from the given condition). Hence $ f$ is monotonic on $ \mathbb Q$ and, moreover, if $ x \in \mathbb R\backslash\mathbb Q$ and $ a \in \mathbb Q$ and $ x < a$ then in any neighbourhood of $ a$ there exists infinitely many $ y$ as above and by continuity $ f(x) \le f(a)$. Similarly for $ x > a$. Note that this already implies $ f(x) \le f(y)$ for all $ x < y$. (If $ x$ and $ y$ are both irrational find a rational $ a$ between them to establish this inequality.) But $ f(x) \equal{} f(y)$ is impossible for $ x < y$ since there are infinitely many rationals in $ [x, y]$ and $ f$ is increasing on rationals.


Find all the values of M (M is positive intergers) satisfying the following conditions:
for any $n=\overline{a_{1}a_{2}...a_{k}}$ (k is some positive integer),
such that $M\ |\ n$, then $n\ |\ \overline{b_{1}b_{2}...b_{k}}$, where $b_{1}b_{2}...b_{k}$ is any permutation of $a_{1}a_{2}...a_{k}$.

hmm I think the problem should be as the following

Find all positive integer $M$ such that: for any $n=\overline{a_{k}a_{k-1}\ldots a_{1}}$ in $10$ base which $M\mid n$ then we should have $M\mid \overline{a_{1}\ldots a_{k}}$.

The last is a famost one appeared in a book of Yaglom, the answer is $M\mid 9$.

Last problem posted before. I give solution: $M=k\frac{10^{k}-1}{9},k|(10-1)$ (k=1 or k=3 or k=9).

Can you post a link?  I'd like to see the solution.

[quote="Kab^2"]How about x=142857. Then, x divides ax where 1<a<=7. Furthermore, ax is a permutation of x (digits).[/quote]
$x=142857\not |758241$.
I solve problem if $M|x=a_{1}a_{2}...a_{k}$ then $M|a_{k}a_{k-1}...a_{1}$. Solution is $M=k\frac{10^{m}-1}{9}, k=1 \ or \ 3 \ or \ 9$.


http://www.telegraph.co.uk/earth/main.jhtml?xml=/earth/2007/08/16/scispeed116.xml

.... Battlestar Galactica is coming...

Shouldn't this go to the Astronomy forum?

Well, the topic was about physics, but the physics forums were for physics problems. And no astronomy was involved (I think), so I stuck it here.

This is Astro Physics. The theory of relativity deals with massive objects like the sun or earth.

.http://www.newscientist.com/article.ns?id=dn2796
.http://physicsworld.com/cws/article/print/482
.Definitely not the first time.
.Laws intact.
.Enough said.

I'm sorry, but I am a rather strong adherent to relativity, and this is sure to be flawed. The same mistake occurred before, with muons I think, which apparently travelled at eight times the speed of light, but in their own timeframe, the speed was smaller than c. (I'm afraid that's rather non-technical.)


geometry
parallelogram
geometry proposed
U sviti matematyky

Let $ ABCD$ be a parallelogram, $ P$ the projection of $ A$ on $ BD$, $ Q$ the projection of $ B$ on $ AC$. Let $ M,N$ be the orthocenters of triangles $ PCD,QCD$ respectively. Prove that $ PQNM$ is a parallelogram.

Rather simple!
AC cuts BD at O. The quadri APQB is cyvlic, so OP*OB=OQ*OA, then OP*OD=OQ*OC, the quadri DPQC inscribed a circle (E).
F is the midpoint of CD. Then, \vec{PM}=2\vec{EF}=\vec{QN}, hence result.


[size=150]A reptile crawl on a rope (the original length is $1m$) at the speed of $1cm/s$ from the beginning to the end , when it get over $1cm$ , the rope stretch suddenly by $1m$ . If the rope can stretch to infinity and the reptile is immortal . So ask can it reach the end of the rope ? if it can't , say the reason ; if it can done , calculate how much time it will need .[/size]

:maybe:
[hide]
The reptile never reaches the end for the end is always 100 times the distance the reptile travelled.
[/hide]

[quote="lingomaniac88"][hide]After the first second, the reptile travels 1/100 of the way.  After the 2nd one, it travels 1/200.  And so on.  After the $n$th one, it travels $1/(100n)$ of the way.  So infinitely, it travels:
$\sum_{i=1}^\infty \frac{1}{100i}$
Or...
$\frac{\sum_{i=1}^\infty \frac{1}{i}}{100}$
The numerator converges to infinity, so it travels an infinite distance, so it will reach the end.[/hide][/quote]Well, sure, the reptile will go infinitely far, but he clearly cannot reach the end of the infinite rope which remains always $100$ times as long as him.

[size=150]Sorry , my faults , I didn't make my mean clear
I mean the part which the reptile have travelled also stretch along with the rope stretch [/size]

[hide="Parametrics"]
$l = lizard$
$r = rope$
$t = time$

$l = t+100t = 101t$
$r = 100t+100$
We want $l = r$
$101t = 100t+100$
$t = 100$
After 100 time steps, the lizard reaches the end of the rope.
[/hide]

[hide]
I could be getting this wrong.
Then this should be a relative motion problem.
$1 \frac{cm}{s}-100 \frac{cm}{s}=-99 \frac{cm}{s}$
So relatively, $\frac{-100cm}{99\frac{cm}{s}}= \frac{-100}{99}s$
Though that does not seem right, so it is impossible  :maybe: .
[/hide]

This has appeared as an entrance question for some of the summer math programs, and doubtless other places.  I'll re-write it in the following form:
A kangaroo is tied to an elastic rope of length 1m.  A flea is at the far end of the rope from the kangaroo.  The flea makes one jump, of length 1cm.  Then the kangaroo makes a jump, of length 1m.  When the kangaroo jumps, the rope stretches uniformly, i.e. the flea is still 1/100th of the way to the kangaroo (although now the distance left to travel is much greater).  These two processes (flea jumping 1cm, kangaroo jumping 1m and rope stretching) continue to alternate.  Will the flea ever catch the kangaroo?

miyomo and astrophys, you haven't taken the stretching of the rope into account.

Lingomaniac already gave the solution.  (I assume s/he has already seen the problem before, as I have.)  The "trick" is to not think about the distance between the flea and kangaroo and instead to think about the fraction of the distance that the flea has moved towards the kangaroo.

For the people who already know the answer to this question, here are some follow-ups to think about:

The reason that this is counterintuitive is that for a long, long time, the kangaroo is getting very far away from the flea.  What's the largest distance between them at any point in this process, and how long does it actually take the flea to get that close?

How long, approximately, does it actually take the flea to catch the kangaroo?

Suppose that, over time, the flea gets more and more tired, so that on its $n$th jump, it is only able to move a distance $\frac{1}{\sqrt[100]{n}}$ -- will the flea eventually catch the kangaroo?  What if the flea is only able to move a distance $\frac{1}{\log_{2}(n+1)}$ on its $n$th jump?

(These questions (esp. the last one) may require some knowledge which many people don't acquire until calculus, unfortunately.  Sorry about that.)

[hide]
hmm...what happens to the fly's distance from the start from the beginning of its $n$th jump to the kangaroo's stretching.  if the fly has already traveled a distance $d_{n-1}$, then his distance at the end is $d_{n}= \frac{n}{n-1}\left(d_{n-1}+0.01\right)$

if we let $r_{n}= \frac{d_{n}}{n}$, then $r_{n}= \frac{1}{n-1}\left(r_{n-1}\left(n-1\right)+0.01\right) = r_{n-1}+\frac{0.01}{n-1}$

setting $r_{1}= 0$, we find that ${r_{n+1}= \frac{H_{n}}}{100}$ where $H_{n}$ is the $n$th harmonic number, defined as $H_{n}= \sum_{i=1}^{n}\frac{1}{i}$.

We want $r_{n+1}\ge 1$ implying that we want to find $N$ with $H_{N}\ge 100$.

using the fact that $H_{N}\le 1+\frac{1}{2}+\int_{3}^{N}\frac{dx}{x}= \frac{3}{2}+\ln{N}-\ln{3}$ we get the bound $N \le e^{100}$.  this can be substantially improved, but not without Mathematica (or equivalent).

----------------------------------------------------------

now, if the flea gets tired in the first manner, then $r_{n+1}-r_{n}= \frac{1}{n^{1.01}}$  if you have taken calculus, then it should be obvious that $S_{n}= \sum_{i=1}^{n}r_{i}= \sum_{i=1}^{n}n^{-1.01}$ is a convergent $p$-series.  the convergence can also be shown using integral (over)estimation or knowledge of the Riemann zeta function.  however, using integral estimation we have $S_{n}= \sum_{i=1}^{n}n^{-1.01}\ge \int_{1}^{n}\frac{dx}{x^{1.01}}= frac{x^{-0.01}}{-0.01}_{1}^{n}= 100\left(1-\frac{1}{n^{0.01}}\right)$

this should be greater than 1 at some point, but i have no idea when.

--------------------------------------------------------

we modify the second one so that it becomes $\frac{1}{\ln{n+1}}$ - this only changes the time by dividing by $\ln{2}$ and makes it easier to integrate.

we have $S_{n}= 0+\sum_{i=2}^{n}\frac{1}{n \ln{n}}\le \int_{2}^{n}\frac{dx}{x\ln{x}}$ making the substitution $u = \ln{x}$, we find that ${\int \frac{dx}{x\ln{x}}= \int \frac{du}{u}= \ln{u}+C = \ln\ln{x}}+C$.  hence $S_{n}\le \ln{\ln{n}}-\ln\ln{2}$ implying that $N \le e{}^{e}{}^{100}$ (this is an approximate bound, taking out nasty constants and whatnot - really it should be $\ln{2}\cdot e{}^{e}{}^{100-\ln{\ln{2}}}$ or something like that.)

[/hide]

FWIW
Another way a physicst would look at this problem is to assume the rope's lenght  remains 1 m (100 cm) all the time, but when JBL's Kangaroo jumps it changes the measuring rod.. that is  the reptile flea (is flea a reptile? :) )  traves 1 cm the 1st second, but it traves (1/2) cm the second second ..  and   next it travels (1/3) cm and so on...
After $n$ seconds it will travel $(1+1/2+1/3+\dots 1/n) \ cms$  distance etc...

(Brings back the memory of a nighmare I once had as a child, soon after my brother (a physicist) explained to me about the red shift  and the  expanding universe ...:)

[quote="AstroPhys"]Gyan, how does that approach show that the reptile (flea) will reach the end of the rope?[/quote]
Just mentioned above, the series is divergent, as long as n is larger enough the sum can be more than  100 (or  any given number)

When $n$ is large, the sum  (which one wants to be 100) is about $(\ln n)$ ..
 or IOW the time , needed is roughly $e^{100}$ seconds.

$n$ here, of course has to be a  very large  number (of the order of $10^{43}$ .. compare that with the age of the universe which is only of the order of $10^{18}$ seconds..

BTW ... a similar  old classic problem is "Book stacking problem" (google search would probably give more details) which has  a similar solution.

Oh okay, I thought that sequence had a limit.  :oops:

( I like those astrophysical allusions in your posts.)

Also FWIW

The time required is about $10^{43}$ seconds , compare this with the age of (the known)  universe   about $10^{18}$ seconds...

And the rope would be about $10^{43}$ meters long  after all that stretching .... compare this with the diameter of the known universe: about $10^{26}$  meters :)

:lol:
Gyan, nice allusion for the problem.


We have $m\times n$ board , at first  we colored the $k$ unit squares with black and we follow this method s.t. every unit square  which has at least two black neighbours (neighbour means have the same line in share) we color with black. Find the minimum $k$ in term of $m,n$ s.t. when we follow this method, all of the unit square have the black color.

Hay guys, i think it's hard and beautiful problem !!!!! :maybe:

I think it's  $\frac{m+n}{2}$ if $m$ and $n$ have the same parity and $\frac{m+n+1}{2}$ otherwise.

I think so, but ,plz write the complete solution.


Why this site is not working anymore? Did it move to another adress?

Thanks.

Please use the "Search" button.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208682[/url]


calculus
integration
real analysis
real analysis unsolved

prove that $ kk'^{2}\frac{\mathrm{d} K}{\mathrm{d} k}\equal{}E\minus{}k'^{2}K$ ,where k'²=1-k²

What definition of K are you using?  How can it help in this proof?

http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html

see (17) , (18)


I need to find examples of power series that

a) converge conditionally at x = -1 and diverges when x = 1

b) converge conditionally at both x = -1 and x = 1

I have no idea how to go about trying to pull random examples that fit these criterion out of a hat. Can someone point me in the right direction?

Thanks!

The most likely examples you'll be using are far from "random."  Think about well-known series.  Think about alternating series.  [hide="Think about..."] The harmonic and alternating harmonic series.[/hide]

For b), to get the same behavior at both ends, try using either an even function (nothing but even powers) or an odd function (nothing but odd powers). I've even got the perfect example on my "Seven Power Series to Memorize" handout (you can probably find that by searching here).

For a), you'd like a series that alternates at one end but not the other, so a series with terms for all $ n$ would be appropriate.


How many three-digit positive integers are there?

[quote="hidude"]idea:900 because 999-100=899+1+900[/quote]

If someone really needs the answer, they're not going to understand this...

The 3-digit integers are $ 100,101,102,\cdots,999$ By the formula, there are $ 999\minus{}100\plus{}1\equal{}\boxed{900}$ numbers in the list.

Ok. So, you know that the positive three digit integers are 100-999. 999-100 equals 899. Add one because you must count the one integer you left out. This give [u][b]900[/b][/u]

[quote=venice.parnell][quote=hidude]idea:900 because 999-100=899+1+900[/quote]

just saying, 999-100= 899 and not 899+1+900...[/quote]

What they really meant was by the formula, you have to add 1 for the first number, 100. Also, next time I suggest you don’t pull out a topic that has been already answered. I think hidude knew that it was 899+1=900 not 899+1+900.

[hide = Solution]

There are 9 choices for the first digit(the first digit cannot be 0), 10 for the second and the third. Therefore there are $9\cdot10\cdot10 = \boxed{900}$ choices.

[/hide]

[quote=mrmath0720][quote=ilovepizza2020][quote=forester2015][hide=Solution]$100, 101, 102 \cdots 997, 998, 999$

subtract $99$ from each number.

$1, 2, 3, \cdots 898, 999, 900$

We see there are $\boxed{900}$ integers.[/hide][/quote]

Finally a non-postfarm post from forester! Congratulations![/quote]

why are you jealous he's catching up?[/quote]

Catching up what?

[quote=forester2015][quote name="mich7math" url="/community/p18713808"]
[quote=forester2015][hide=Solution]$100, 101, 102 \cdots 997, 998, 999$

subtract $99$ from each number.

$1, 2, 3, \cdots 898, 999, 900$

We see there are $\boxed{900}$ integers.[/hide][/quote]

Please don't bump threads from an year ago...
[/quote]

whats wrong i just posted a solution[/quote]

exactly. People stop saying forester2015 is postfarming he's just being him


Greetings.

Uglavnom, kao što neki od vas (npr. Vedran) znaju, u Zagrebu se prije par tjedana našlo nas 9 (taj broj se u međuvremenu povećao) studenata-mentora i osnovali smo udrugu Mladi nadareni matematičari "Marin Getaldić". Tako smo ujedinili matematičke grupe u V. gimnaziji i MIOC-u (XV. gimnaziji). Mislimo da će to pomoći učenicima u pripremi za natjecanje, olakšati im rad i učiniti ga zabavnijim. E, sad se vi pitate o čemu ja uopće pričam i o čemu se tu uopće radi i što znači Balkan Student Mathematical Competition i kakve veze to ima s vama.

Kao jednu od aktivnosti/projekata koju smo zamislili kako bismo bolje pripremili učenike za natjecanja (kako unutardržavna, tako i međunarodna) je napraviti jedno neslužbeno, pripremno natjecanje među školama regije. Kako sam ja glavni organizator, meni je pripala čast da nagovaram ljude da sudjeluju.

Natjecanje se zove (radni naziv i subject to change, naravno - uostalom, ujutro se natjecanje zvalo drugačije  :D ) Balkan Student Mathematical Competition - BSMC [size=75](Matematičko natjecanje učenika Balkana (MNUB))[/size]. Održalo bi se negdje u sredini/drugoj polovici 11. mjeseca... za sada razmišljamo o 15.11.  4 zadatka u trajanju od 3 sata i tri dobne skupine: 1. razred, 2. razred i 3./4. razred. Ne moraju svi održati natjecanje na isti dan - mogu i par dana prije/poslije, nema veze, ionako je sve to dosta neformalno.

Organizacija bi bila jednostavna, a troškovi za grupe sudionice nikakvi: Nekoliko dana prije natjecanja organizator (to bi bio ja ili netko drugi iz Udruge) pošalje lokalnom organizatoru (čovjek koji u svojoj školi/matematičkoj sekciji/gradu organizira natjecanja) zadatke s rješenjima. On održi natjecanje, na kraju ispravi (ili, ako mu/im se baš ne da, skenira pa pošalje mailom u Zagreb ili nešto) te testove i pošalje nazad rezultate. Stvori se i objavi zajednička lista. I to je to.

Uglavnom, tražim zainteresirane da tako nešto održe u svojoj školi/matematičkoj grupi. Zbilja ću se potruditi da sve bude što jednostavnije organizirano - na kraju krajeva, ovo je ionako natjecanje kojem su glavni ciljevi motivacija, priprema i upoznavanje, ništa ozbiljno. Mislim da su dosta jasni plusevi održavanja takvog natjecanja i ne vidim zapravo niti jednu lošu stranu - troškovi ne postoje, a organizacija će biti što je jednostavnija moguća. Ako ste zainteresirani ili ako znate nekoga tko bi mogao biti zainteresiran, a da ne čita ovo, javite mi. E-mail znate ([img]http://img229c.imageshack.us/img229/1889/stringimageln6.png[/img]... nadam se da sam ovime izbjegao jednu određenu količinu spama), postojim i na fejsbuku, a možete i tu napisati (ili PM). Također, ako netko ima kakvih pitanja ili, još bolje, prijedloga, stojim na usluzi.

Evo, da nastavim obavještavati ljude. Po broju odgovora, reklo bi se da nitko ovo ne čita, ali imam kontraprimjer. (Hvala Goblin!  :lol: )

Dakle, za sada će (vjerojatno) sudjelovati 7 škola iz 4 države. Mislim da je to dobar odaziv koji nam daje motivaciju da se posebno potrudimo u vezi organizacije, a pogotovo kvalitete zadataka. Stoga bih htio zahvaliti svima koji su do sada poslali zadatke i svima koji su obećali da će ih poslati u bližoj budućnosti. Which brings me to my point of this evening...

Vjerojatno su neki od vas koji čitate ovo svojedobno smišljali neke zadatke. Ti zadaci sada čame u kutu vašeg prašnjavog hard drivea i čekaju svoj kraj. Ne dajte da se to dogodi! Donirajte zadatke za BSMC! Ozbiljno, ako netko ima neke zadatke koje je smislio, a za koje vjeruje da bi mogli biti prihvatljivi za ovo natjecanje i ne čuva ih za nikakvo drugo, "važnije" natjecanje, javite mi se.

Ima samo jedan uvjet: ti zadaci moraju biti relativno tajni (u značenju - nitko osim vas i još par ljudi koji su ionako prestari da se natječu ne zna kako glase. Dakle, niste ih nikome izlagali na grupi i niste zvali ljude u 2:33 ponoći i pričali im kakav ste divan zadatak smislili.  :ninja: )

Evo, to bi bila moja večerašnja poruka narodu... uglavnom, donirajte zadatke (preferabilno s rješenjima, ali može i bez). You will give so much to so many. Jamčim vam da će, ako vaš zadatak bude odabran, kraj njega pisati vaše ime i prezime.

Moj mail je bio i ostao imeprezime@gmail.com (vjerojatno znate kako se zovem, ako ne, pogledajte ovogodišnji HRV5  :D ) i na njega mi možete poslati gorespomenute zadatke, spam ili jednostavno hate-mail. Ako ste slučajno u ovom trenutku shvatili da ste sposobni i održati natjecanje, ponovno, javite se. Nikad nije prekasno. :)

Ukupni rezultati će biti kad mi Salem dostavi potpune rezultate iz Sarajeva. Bilo je planirano da budu ovaj vikend, no kao što vidiš... ajde Saleme!

Evo, napokon je sve gotovo.

Rezultate sam organizatorima i sudionicima poslao danas ujutro - sve: i rezultate i zadatke i rješenja i statistike (koliko ja volim raditi statistike, to se ne da riječima izreći...  :D ) možete pronaći u zip fileu na http://www.2shared.com/file/4401926/280ec27f/BSMC.html

Sve linkove na zadatke i još neke moje mudre opservacije možete naći i tu: http://www.mathlinks.ro/viewtopic.php?t=243614 (pretpostavit ću da ipak čitatelji ovog podforuma znaju engleski dovoljno da im ne treba prevoditi - ako treba, nema problema, javite).

Na kraju još jednom bih htio zahvaliti svima koji su sudjelovali u organizaciji natjecanja (i učenicima i studentima i profesorima), a posebno Adrianu i Marku za njihove zadatke i prijedloge. Naravno, "moji" članovi organizacijskog tima se ne računaju - njima sam već obećao svoju zahvalnost i poštovanje do groba. :D

Čestitke idu prvacima. U prvim razredima to su Matija Bucić i Igor Spasojević, u drugima Matko Ljulj i Grgur Valentić, a u trećim i četvrtim Luka Milićević.  :first:

Mogao bih vas tu daviti godinama pričanjem o natjecanju, ali neću da ne izgubim barem ono malo poštovanja što neki ljudi ovdje imaju prema meni.  :D

Eto, za 2008. to je to, za 2009. ćemo vidjeti kako se stvari razvijaju. Ako nekog zanima više ili mi želi nešto reći o natjecanju ili želi iznijeti neke svoje dojmove, samo dajte.

Btw, obratite pozornost na 2. zadatak za 3./4. razrede - Krijan i ja se kunemo da je to najteže što smo ikad smislili. Barem za sebe mogu reći da je taj zadatak vjerojatno i vrhunac moje matematičke karijere.  :D

Da li zna neko kako se bira ekipa za ovo takmicenje?

Sorry Bugi, možda se iz mojih prethodnih nije dalo zaključiti - nema biranja ekipe ni ničega takvog.

Natjecanje smo zamislili da mu može pristupiti bilo tko - 9 škola (odnosno, 7 škola i Udruga) je organiziralo natjecanje kod sebe i u svakoj od njih došli su njihovi zainteresirani učenici i pisali test. 2 učenika iz još 2 škole su samoinicijativno došli pisati test s učenicima iz neke škole u blizini.

Tako je ukupno sudjelovalo 103 učenika iz 11 škola.

[quote="hsiljak"]Melkiore, sve čestitke tebi i Ivanu za NT-kombinatoriku. Zadatak je sjajan, mada pretežak  :lol: Po mom skromnom mišljenju, zadatak bi se mogao naći i u slijedećem izdanju PENa (u klasi O).

P.S. Pokušat ću naći neko jednostavnije rješenje... ako bude kakvih ideja, bićeš blagovremeno obaviješten, kao i ostali MathLinkeri.[/quote]

Evo da se i ja javim ovdje (Ivan  :)), hvala na pohvalama za zadatak  :).

Bilo bi super kada bi se našlo ljepše (jednostavnije) rješenje  :).

[quote="mornik"]
Na kraju još jednom bih htio zahvaliti svima koji su sudjelovali u organizaciji natjecanja (i učenicima i studentima i profesorima), a posebno Adrianu i Marku za njihove zadatke i prijedloge.[/quote]

Nema na čemu, Melkiore. I hvala vama što se izabrali čak 4 moja zadatka :)

[quote="hsiljak"]Melkiore, sve čestitke tebi i Ivanu za NT-kombinatoriku. Zadatak je sjajan, mada pretežak  :lol: Po mom skromnom mišljenju, zadatak bi se mogao naći i u slijedećem izdanju PENa (u klasi O).

P.S. Pokušat ću naći neko jednostavnije rješenje... ako bude kakvih ideja, bićeš blagovremeno obaviješten, kao i ostali MathLinkeri.[/quote]

Hvala na čestitkama. Da je težak, to je... što se i vidi po rezultatima  :D . Navodno postoji jednostavnije rješenje, samo što ga još nisam vidio. I'll believe it when I see it.  :D Stoga, keep working. :D

[quote="bugi"]Komentari za zadatke za I razred: geometrija i algebra su bile stvarno lake, teorija brojeva ok, kombinatorika teska[/quote]

Točno je tako i bilo zamišljeno. Zapravo, ako se dobro sjećam, bila je zamišljena algebra kao stvarno lagana, a geometrija kao lagana, ali ne toliko, a ispada po rezultatima da je bilo obrnuto.

[quote="djuro"]Nema na čemu, Melkiore. I hvala vama što se izabrali čak 4 moja zadatka[/quote]

Hej, da zadaci nisu bili dobri, ne bismo ih izabrali... stoga, nemoj se zahvaljivati.  :D U najmanju ruku, da ti nisi predložio zadatke, morali bismo ih tražiti među nepoznatim izvorima, a dvojim da bismo našli neke koji nam bolje odgovaraju.

Inače, samo da na kraju pohvalim kolegu JANKRIja i sebe: ukupno na 1. kolokviju imamo 418 od 420 mogućih bodova (Ja sam onaj s dva boda manje. Zapravo, budemo još to vidjeli nakon žalbi iz Elementarne - na jednom sam zadatku bio glup, ali oni će meni tvrditi da ja ne znam Euklidov algoritam...   :furious: Just kidding, i tu sam vjerojatno bio glup.)! Čestitke JANKRIju još jednom.

[quote="djuro"]I hvala vama što se izabrali čak 4 moja zadatka :)[/quote]

[quote="mornik"]Hej, da zadaci nisu bili dobri, ne bismo ih izabrali... stoga, nemoj se zahvaljivati. :D U najmanju ruku, da ti nisi predložio zadatke, morali bismo ih tražiti među nepoznatim izvorima, a dvojim da bismo našli neke koji nam bolje odgovaraju. [/quote]

Točno to, hvala tebi djuro na lijepim zadacima  :).

[quote]Inače, samo da na kraju pohvalim kolegu JANKRIja i sebe: ukupno na 1. kolokviju imamo 418 od 420 mogućih bodova (Ja sam onaj s dva boda manje. Zapravo, budemo još to vidjeli nakon žalbi iz Elementarne - na jednom sam zadatku bio glup, ali oni će meni tvrditi da ja ne znam Euklidov algoritam... :furious:  Just kidding, i tu sam vjerojatno bio glup.)! Čestitke JANKRIju još jednom.[/quote]

HVALA!  :)

Čestitke nam obojici i od mene! Melkiore, ta dva boda su ionako za moju dušu (interna fora  :D), ti si to namjerno, a i vjerojatno ćeš ih na žalbama dobit :).

Morniče, bravo za MNUB (POSEBNO, najviše si zaslužan za to što je sve odlično prošlo) i za ubijanje kolokvija :).

[quote="mornik"]Nakon današnjeg uvida u kolokvij iz Elementarne, zajedno imamo 419 od 420 bodova.  :lol: (Na ovaj drugi bod se nisam zapravo ni žalio - em znam da sam pogriješio i da mi ne bi dali bod, em zato da JANKRIjeva duša ostane zadovoljna  :D )[/quote]

Bravo, bravo! :)

Duša mi je hepi sad  :P   :rotfl:


quadratics
function
conics
parabola
analytic geometry
algebra
system of equations

Find a quadratic function that fits each set of data points exactly.

$(4,-4), (-2,5), (0,6)$

[quote="kunny"]You can use the formula of Lagrange's interpolation.[/quote]
thats not getting started, i don't even remember that crazy interpolation. what i do is assign variables to coefficients and then have a system of equations

Oh,sorry that I didn't confirm what forum this is. :blush:

To 4everwise:

You can set the equation of the parabola (quadratic equation) which pass through the given three points as $y=ax^2+bx+c\ (a\neq 0)$. :)

kunny

[hide]
from what Kunny told:

$\left\{\begin{array}{ccc}{16a+4b+c=-4}\\{4a-2b+c=5}\\{0\cdot a+0\cdot b+c=6}\end{array}\right.$

an easy system to solve, which gives us the function $f(x)=-\frac12x^2-\frac12x+6$[/hide]

[quote="xdl615x"]Oh the c=6 confused me

:blush:

and for anyone else who might be confused: He plugged in the coordinates (0,6) into the equation which gave him that answer, not the two equations above it.[/quote]yeah, sorry  :?
should've put a system sign in front  :blush:

edit-i added somethings :) now it should be clear for everyone ;)

I just quickly solved it with a 3x3 matrix.


[quote]Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4. If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?[/quote]

  [hide]  answer=  $\frac{9pi-4pi}{16pi}$=  $\frac{5pi}{16pi}$= $\frac{5}{16}$ [/hide]

[quote="G-UNIT"][hide] 5/16 [/hide][/quote]
Please show your work.

How he got the values is pretty obvious, but I guess he could just replace the 4s with 2 :^2: , the 9s with 3 :^2: , and the 16s with 4 :^2: ...

[hide]
We first want to find the "good" area.
That is given by $\pi(3^2-2^2)=5\pi$ .
The total area of the circle is $4^2\pi=16\pi$
$\frac{5\pi}{16\pi}=5/16$

[/hide]
I thought this was going to be about the dart shaped geometry figures.

[quote="IntrepidMath"][hide]
We first want to find the "good" area.
That is given by $3^2-2^2=5$ times pi. (I dont know how to do pi in $\LaTeX$)
The total area of the circle is $4^2pi=16pi$
$\frac{5pi}{16pi}=5/16$

[/hide]
I thought this was going to be about the dart shaped geometry figure.[/quote]

$\frac{5\pi}{16\pi}$

[hide="wat i think"]area of largest circle is 16pi (4^2 * pi) and the circle os radius 3 is 9pi (u know) and cirle with radius 2 is 4pi. so you subract 4pi from 9pi because u want it to be out of the raius 2 circle and in the radius 3 circle and so u get 5pi and you put that over 16pi. its pi over pi so you cross them out and since there is no more to simplify, you get 5/16

5/16 YAY!!!! :rotfl:
[/hide]


$ABC$ trebuie sa fie echilateral, nu oarecare :). Daca se da un triunghi $ABC$ care nu e echilateral, se poate gasi un punct $M$ pentru care nu e valabil.

Iar pentru ABC echilateral, iata o demonstratie "fara cuvinte"

Asa e , nu am observat acest amanunt important. Imi cer scuze :blush:

[quote="rockymarcianocosmin99"]Fie ABC un triunghi oarecare si un punct M in planul triunghiului diferit de cercul circumscris. Aratati ca MA , MB , MC pot fi laturi de triunghi .[/quote]

This is clear (for ABC equilateral, as is required) from the inequality form of Ptolemy's theorem; equality holds if and only if A,B,C,M are on a circle.


The radius is 2/3 of the median of the eq. triangle. The median is $ \frac{9}{2}\sqrt{3}$, so the radius is $ 3\sqrt{3}$. The area of the circle is $ (3\sqrt{3})^2\pi\equal{}27\pi$

[quote=Aequilipse]The radius is 1/3 of the median, not 2/3. So the radius is $\frac{3\sqrt{3}}{2}$, and the area is $\frac{27\pi}{4}$[/quote]

It's 2/3, not 1/3

The radius is $3 \sqrt{3}$, so the area is $(3 \sqrt{3})^2 \cdot \pi \rightarrow 27\pi$

[quote=GameBot]A circle is circumscribed about an equilateral triangle with side lengths of $ 9$ units each.  What is the area of the circle, in square units? Express your answer in terms of $ \pi$.[/quote]

[hide=The Hard Way]The radius of this circle is $\frac{\text{product of side lengths}}{4 \cdot \text{area}}$. The product of side lengths is $729$, and the area is $\frac{81\sqrt3}{4}$. Doing the computation and rationalization gives a radius of $3\sqrt{3} \implies \boxed{27\pi}$.[/hide]


I'm very interested in the solution of this problem, since the solution I gave to it is allmost impossible to find. I would like very much to see if there exists a simple solution.

This inequality is very nice and very hard to solve by conventional methods so I am very interested in your solution, Harazi, that will be certainly ingenious and genial.

Thank you very much.

Ok, but the following is very artificial (though nice), so I would be very happy is someone found a more natural solution:
   Suppose that (a_2-a_1)sqrt(2)+...+(a_n-a_n-1)sqrt(n)>=0. Since
 (a_1-a_2)sqrt(2)+(a_2-a_3)sqrt(3)+...+(a_n-1-a_n)sqrt(n)+a_n*sqrt(n+1)-a_1=a_1*(sqrt(2)-1)+a_2*(sqrt(3)-sqrt(2))+...+a_n*(sqrt(n+1)-sqrt(n)), it follows that sum a_i*(sqrt(i+1)-sqrt(i))<a_n*sqrt(n+1).
   But sum a_i*(sqrt(i+1)-sqrt(i))=sum a_i/(sqrt(i)+sqrt(i+1))>=(sum a_i)^2/( sum a_i*sqrt(i)+a_i*sqrt(i+1))=1/(sum a_i*sqrt(i)+sum a_i*sqrt(i+1)). Apply again Cauchy to find that (sum a_i*sqrt(i))^2<=(sum a_i)(sum i*a_i)=2 and (sum a_i*sqrt(i+1))^2<=(sum a_i)(sum a_i*(i+1))=3. So we must have sum a_i(sqrt(i+1)-sqrt(i))>=sqrt(3)-sqrt(2). Consequently, a_n>(sqrt(3)-sqrt(2))/sqrt(n+1). On the other hand, 2=1+a_2+2a_3+...+(n-1)a_n and so a_n<1/(n-1). Thus 1/(n-1)>(sqrt(3)-sqrt(2))/sqrt(n+1), which is false for n>12.
    A much more difficult problem is in my opinion to find all values of n for which the inequality is true.


A particle moves randomly along a number line in single unit jumps starting at 0, and is equally likely to move left or right for a move.  What is the probability that it is back at 0 after 4 moves?

[hide]There are 6 ways to go back to 0 after 4 moves. You can go right 2 moves and 2 moves left. You can go one move right, one move left, then either right or left then go back to zero. This totals 3 different ways from going right for the first move. Since you can go left in the first move too, you just multiply by 2 which 3x2=6. There is a total of 2^4 different possibilities since you have 2 choices for 4 moves, which equals 16. 6/16=3/8 possibility[/hide]


[url=http://www.usamts.org/Tests/USAMTSProblems_20_1.pdf]Something[/url] tells me this question should not be answered.

The amount under each radical is negative. But in general, there is a nice solution to these problems. I don't see this posted under the USAMTS link. Can I show a general solution?


How do you center equations in the forum?

Use \ [ ... \ ] (without the spaces) instead of dollar signs. For example, \[ 2\plus{}2\equal{}4\]

And the system of equations, where the { is on the left?  Thanks a lot, stevem, it's just that I haven't been using TeX in a long time.

Use the cases environment
\[ |x| \equal{}
\begin{cases}
x & \text{if } x \geq 0 \\
\minus{} x & \text{if } x < 0
\end{cases}\]
or possibly
\[ |x| \equal{}
\begin{cases}
\hphantom{\minus{}} x & \text{if } x \geq 0 \\
\minus{} x & \text{if } x < 0
\end{cases}\]


Last time i posted on this forum i was told my question was too easy for that particular part of the forum, but well, i'm a dane and don't really know where to put the posts since we have a different systems. Not even sure the university levels are the same. But well, plz advise if i'm on the totally wrong part of the forum.

Today's question is a smaller one.

x=3-3y ^ 6y=6-3x

When calculating on this i got to this:

5/6x=0 <=> x=0

Is this currect and what is the solution? If i put it into the starting equations i don't get the right results, which i believe is because i can't set x to be 0 or it would be the same as saying 5/6x=x, which of course isn't true. Is the solution then just to say 'Not possible'?

Hm?

Is the ^ meaning "and"?

The only solution that I got is x=0, y=1.

Yes, that's an 'And' symbol, at least in danish. But if you calculate it on the left side it is x=0 and y=1, but not on the right side. Did i do something wrong, or is it because you can't make such an equation where x=0?

Isn't the right side also that?
6(1)=6-3(0)

<off-topic>I think this should be in the 'getting started' forum :) </off-topic>

Well, it said 'Below High School' while i consider 'Getting Started' as the High School itself. But let me just finish it off and we will have no more to discuss anyway :)

You're right, can't find out why i didn't see that one earlier, but i for one reason or another considered 6-3y to be 3 (obviously couldn't do a 3x0) :D But well, then it is settled.

This is getting started level.  Someone should move it.

Well, doesn't matter now, i'm done with this post as i have the answer to the result already, so let's just wait a little and it will have plumeted down the list in this area of the forum.


calculus
integration
function
calculus computations

Calculate $ \int\limits_0^1 \sum\limits_{n\equal{}1}^\infty \frac{1}{(n\plus{}x)^2}\, dx$.

[quote="Tyl"]Calculate $ \int\limits_0^1 \sum\limits_{n \equal{} 1}^\infty \frac {1}{(n \plus{} x)^2}\, dx$.[/quote]

$ \int\limits_{0}^{1}\sum\limits_{n\equal{}1}^\infty\frac{1}{(n\plus{}x)^{2}}\, dx \equal{}\sum_{n\equal{}1}^{\infty}\int_{0}^{1}\frac{1}{(n\plus{}x)^2} d x\equal{}\equal{}\sum_{n\equal{}1}^{\infty}\frac{1}{n(n\plus{}1)}\equal{}1$

Let $ f_{n}\equal{}\frac{1}{(n\plus{}x)^{2}}$.  Since $ f_{n}$ is nonnegative and measurable, the Monotone Convergence Theorem lets us change the order, so we have $ \int_{0}^{1}\sum_{n\equal{}1}^\infty\frac{1}{(n\plus{}x)^{2}}\, dx\equal{}\sum_{n\equal{}1}^{\infty}\int_{0}^{1}\frac{dx}{(n\plus{}x)^{2}}$.


A triangle has side lengths 9,10, and 11. What is the length of the longest altitude of this triangle?


function
algebra
polynomial
algebra unsolved

Hello!

I can't solve the following equation:

Find all functions $ f: \mathbb{R}\to\mathbb{R}$ for which $ 3f\left(2x\plus{}1\right)\equal{}f\left(x\right)\plus{}5x$.

I found that if the function is a polynomial then $ f\left(x\right)\equal{}x\minus{}\frac{3}{2}$.

To find other functions, we set $ f(x)\equal{}x\minus{}\frac{3}{2}\plus{}g(x)$

$ 3(2x\plus{}1\minus{}\frac{3}{2}\plus{}g(2x\plus{}1))\equal{}x\minus{}\frac{3}{2}\plus{}g(x)\plus{}5x$

$ 3g(2x\plus{}1)\equal{}g(x)$

From which we can define g arbitrarily on $ [1,3)$ and all other points will follow by the recursion.

Let $ g(x)\equal{}f(x\minus{}1)\minus{}x\plus{}\frac{5}{2}\to f(x)\equal{}g(x\plus{}1)\plus{}x\minus{}\frac 32$.
Then $ 3g(2x)\equal{}g(x)$ or $ g(0)\equal{}0,g(x)\equal{}|x|^{\minus{}log_2(3)}\phi(x), x\not \equal{}0$, were $ \phi(2x)\equal{}\phi(x)$ or $ \phi(x)\equal{}f_1(log_2|x|)$ and $ f_1$ is periodic $ f_1(y\plus{}1)\equal{}f_1(y)$.

[quote="Rust"]
Then $ 3g(2x) \equal{} g(x)$ or $ g(0) \equal{} 0,g(x) \equal{} |x|^{ \minus{} log_2(3)}\phi(x), x\not \equal{} 0$, were $ \phi(2x) \equal{} \phi(x)$ or $ \phi(x) \equal{} f_1(log_2|x|)$ and $ f_1$ is periodic $ f_1(y \plus{} 1) \equal{} f_1(y)$.[/quote]
Can you detail, Rust. I can't understand? :(


calculus
integration
logarithms
limit
derivative
function
calculus computations

Compute $ I \equal{} \int_{0}^{\infty}\frac{(\ln x)^2}{e^x\plus{}1}\,dx$.

Is there anyone who can evaluate $ c \equal{} \lim_{n\to\infty} \left( \frac {1}{2}\log^2 n \minus{} \sum_{k \equal{} 1}^{n} \frac {\log k}{k} \right)$? This clearly converges if we interpret this geometrically. Then $ I$ is given by

$ I \equal{} \zeta(2) \log 2 \minus{} \gamma^2 \log 2 \plus{} \frac {1}{3}\log^3 2 \plus{} 2c \log 2$.

If we put $ n \equal{} 5 \times 10^5$, we have $ c \approx 0.0728027$ (evaluated by Mathematica 6).

Using this approximation, we have $ I \approx 1.12117$, which is closer to the value $ 1.12119$ otained by direct numerical approximation of $ I \equal{} \int_{0}^{\infty}\frac {\log^2 x}{e^x \plus{} 1} \, dx$.

[hide="My Solution"]
[b]Fact (1)[/b] If we define $ \eta(s) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {( \minus{} 1)^{n \minus{} 1}}{n^s}$ for $ s > 0$, then $ \eta$ is infinitely differentiable and its derivatives are obtained by applying termwise differentiation.

[b]Fact (2)[/b] If $ f$ is a $ C^1$ function on $ [0, 1]$, then $ \sum_{k \equal{} 1}^{n} f \left( \frac {k}{n} \right) \frac {1}{n} \sim \int_{0}^{1} f(x) \, dx \plus{} \frac {f(1) \minus{} f(0)}{2n}$. This is an easy consequence of Mean Value Theorem.

[b]Fact (3)[/b] If we put $ I(s) \equal{} \int_{0}^{\infty} \frac {x^{s \minus{} 1}}{e^x \plus{} 1} \, dx$, then $ I(s) \equal{} \Gamma(s) \eta(s)$. To show this, use $ \frac {x^{s \minus{} 1}}{e^x \plus{} 1} \equal{} \frac {x^{s \minus{} 1}(e^{ \minus{} x} \minus{} e^{ \minus{} 2x})}{1 \minus{} e^{ \minus{} 2x}}$ and expand this as a geometric series. Apply Tonelli's theorem to exchange the order of integral and sum. Evaluating each term will give the desired result.

[b]Fact (4)[/b] Define $ \psi_{n}(s) \equal{} \tfrac{d^{n \plus{} 1}}{dx^{n \plus{} 1}}\ln \Gamma(s)$. Then $ \psi_{0}(1) \equal{} \minus{} \gamma$ and $ \psi_{1}(1) \equal{} \zeta(2)$, where $ \gamma \equal{} \lim_{n\to\infty} (H_n \minus{} \log n)$ and $ H_n \equal{} 1 \plus{} \cdots \plus{} \tfrac{1}{n}$ is the harmonic number.

Then the integral we want to know is equal to $ I''(1)$. Using (1), (3) and (4), we have $ I''(1) \equal{} (\psi_{0}(1)^2 \plus{} \psi_{1}(1))\eta(1) \plus{} 2\psi_{0}(1)\eta'(1) \plus{} \eta''(1)$.

So it remains to find the value of $ \eta(1), \eta'(1)$ and $ \eta''(1)$.

Put $ A_{r, n} \equal{} \sum_{k \equal{} 1}^{n} \frac {1}{n} \frac {\log^r (1 \plus{} \frac {k}{n})}{1 \plus{} \frac {k}{n}}$. Then one can confirm that

$ \sum_{k \equal{} 1}^{2n} \frac {( \minus{} 1)^k \log k}{k} \equal{} (H_n \minus{} \log n) \ln 2 \plus{} (\ln 2 \minus{} A_{0, n}) \ln n \minus{} A_{1, n}$.

$ \sum_{k \equal{} 1}^{2n} \frac {( \minus{} 1)^{k \minus{} 1} \log^2 k}{k} \equal{} \log^2 n (A_{0,n} \minus{} \log 2) \plus{} 2 \log n \left(A_{1,n} \minus{} \frac {1}{2} \log^2 2 \right) \plus{} A_{2,n}$ $ \minus{} \log^2 2 (H_n \minus{} \log n) \plus{} 2 \log 2 \left( \frac {1}{2} \log^2 n \minus{} \sum_{k \equal{} 1}^{n} \frac {\log k}{k} \right)$.

Applying (2) and taking $ n \to \infty$ gives

$ \eta'(1) \equal{} \gamma \log 2 \minus{} \frac {1}{2}\log^2 2$

$ \eta''(1) \equal{} \minus{} \gamma \log^2 2 \plus{} \frac {1}{3} \log^3 2 \plus{} 2c \log 2$.

This gives $ I''(1) \equal{} \zeta(2) \log 2 \minus{} \gamma^2 \log 2 \plus{} \frac {1}{3}\log^3 2 \plus{} 2c \log 2$.[/hide]
[hide="Edited: Comment on c"]
If we put $ \tilde{\zeta}(s) \equal{} \zeta(s) \minus{} \frac {1}{s \minus{} 1}$, then $ \tilde{\zeta}(1 \plus{} h) \equal{} \frac {k(1 \plus{} h)\eta(1 \plus{} h) \minus{} 1}{h}$, where $ k(s) \equal{} \frac {s \minus{} 1}{1 \minus{} 2^{1 \minus{} s}}$.

Then it is easy to see that $ \tilde{\zeta}(1 \plus{} h) \equal{} \gamma \plus{} c h \plus{} o(h)$. So $ c \equal{} \tilde{\zeta}'(1)$.

So, by the defintion of [url=http://en.wikipedia.org/wiki/Stieltjes_constants]Stieltjes constants[/url], we have $ c \equal{} \minus{} \gamma_1$.[/hide]

Yes, $ c \equal{} \minus{} \gamma_1$ and your answer is correct.


topology
advanced fields
advanced fields unsolved

$ M$ is a pure differential manifold of dimention $ n$. If there is a linear mapping from the space of

r-forms to the space of n-forms (r<n), i.e. $ f: \bigwedge^{r}T(M)^{*}\to \bigwedge^{n}T(M)^{*}$.

Is it true that there is a unique (n-r)-form $ \alpha$, such that for any r-form $ \beta$:

$ \beta \wedge \alpha=f(\beta)$?

[quote="mlok"]I don't know what a [b]pure[/b] differential manifold is... does it have to be orientable? On an orientable manifold the [url=http://en.wikipedia.org/wiki/Hodge_dual]Hodge star[/url] should do the trick.[/quote]
yes, it is orientatble. pure means dim of M at every point is same.
Hodge star is exactly what I need.

[quote="mlok"]The Wikipedia article also says
[quote]in the non-oriented case, one can define the hodge star of a $ k$-form as a $ n-k$ - pseudo differential form[/quote]
but "pseudo differential form" is a red link. Does anyone know what that is?[/quote]

I guess:
a pseudo-Riemmannian G-structure on $ M$ of signature (p,q). When signature is (n,0), it is Reimannian.

 "pseudo differential form" maybe vol form on an (oriented)  pseudo-Riemmannian G-structure on $ M$.


ratio
inequalities
SFFT
special factorizations

Here are two nice easy problems.

Problem 1 on Round 26
Assume that $x,y$ and $z$ are positive real numbers that satisfy the equations given on the right.

$x + y + xy = 8\\
y + z + yz = 15\\
z + x + zx = 35$

Determine the value of $x+y+z+xyz$.

[hide="Answer"]
281.75[/hide]

Problem 1 on Round 8
Prove that there is no triangle whose altitudes are of length 4,7,and 10 units.

[hide="Second One"]
Just use:

\[\frac{1}{h_a}+\frac{1}{h_b}>\frac{1}{h_c}.\]
[/hide]

1
[hide]Using Simon's Favorite Factoring trick

$(x + 1)(y + 1) = 9$

$(y+1)(z+1)=16$

$(z+1)(x+1)=36$

Dividing the third and last equations we see that

$\frac {(x+1)}{(y+1)}=\frac 94$

$y = \frac {4x - 5}{9}$

Substituting that into the first equation we have

$(x+1)(4x + 4) = 81$

$(x+1)^2 = \frac {81}{4}$

$x+1 = \frac 92$

$x = \frac 72$

Plugging the value of $x$ into $y = \frac {4x - 5}{9}$ we find that $y = 1$. Plugging that value of $y$ into the second equation we find that $z = 7$.

$x + y + z + xyz$ then equals $1 + \frac 72 + 7 + 7 \cdot \frac 72 = 36$

Testing the values of $x$, $y$, and $z$, into the first equations, we get the correct values. Did I do something wrong?[/hide]

Those values seem right.  It would have been easier to solve for x+1, y+1, and z+1, however.

First Problem:
[hide]We have:
$x+y+xy=8$

$y+z+yz=15$

$x+z+xy=35$

Adding $1$ to each equation and factoring gives us:

$(x+1)(y+1)=9$

$(y+1)(z+1)=16$

$(x+1)(z+1)=36$

Dividing the equations by each other gives us the ratios:
$x+1:y+1:z+1=9:4:16$.
However, these do not fit the original equation, so they must be multiplied by a scale factor, which we must find out.  Let the factor be $x$.

Since $(x+1)(y+1)=9$, we can write $(9\times x)(4\times x)=36x^{2}=9$, using the values from the ratio.  Solving for $x$ gives us $2$ or $-2$, but it must be $2$ because the values are positive.
So now we find the actual values of $x+1$, $y+1$, and $z+1$, are $4.5$, $2$, and $8$ by multiplying by the factor, subtracting $1$ to find $(x,y,z)$ gives us:

$(3.5, 1,7)$, which checks correctly.[/hide]

eryaman, it's really, [i]really[/i] bad form to let the same letter stand for two different things in your proof.  It would have been much nicer to use, say, r in place of the second x.  Otherwise, nice work, all.

Yeah, that's like using "it" all the time. Anyway...

[hide="2"]Then let the bases by a, b, and c. 4a=7b=10c, and c+b>a, so 4a/7+4a/10>a. Multiply by 70 to get 40a+28a>70a or 68a>70a. Since a is positive and 68>70 is false, this is impossible.[/hide]

For those of you who don't know the inequality joml88 posted, here's how you can also solve the second problem:

[hide]
Let the sides of the triangle be a, b, and c.  Then using the altitudes and the bases, we get 10a = 7b = 4c

So a = 2c/5 and b = 4c/7

So a = 14c/35 and b = 20c/35 and c = 35c/35

So a + b > c should be true, but 14/35 + 20/35 = 34/35 is smaller than 35/35 so there is no triangle with altitudes 4, 7 and 10.
[/hide]


[quote="Svejk"]If $ a,b,c>0$ and $ abc\geq 1$ prove that:
\[ \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq 1 \]
[/quote]
Let's assume $ a=ka',$ where $ a'bc=1.$ Hence, $ k\geq1$ and
$ \sum_{cyc}\frac{1}{1+a+b}=\sum_{cyc}\frac{1}{1+ka'+b}\leq\sum_{cyc}\frac{1}{1+a'+b}.$
It remains to prove that $ \sum_{cyc}\frac{1}{1+a+b}\leq1,$ where $ a,$ $ b$ and $ c$ are positive and $ abc=1,$
 which very known.

We only need to prove when $ a+b+c=3$.
But we have
$ \frac{1}{4-a}\le \frac{1}{3}+\frac{1}{9}(a-1)$
$ \frac{1}{4-b}\le \frac{1}{3}+\frac{1}{9}(b-1)$
$ \frac{1}{4-c}\le \frac{1}{3}+\frac{1}{9}(c-1)$
Then, summing up above three inequalities. QED

Oh, arqady, you so fast. :lol:  :blush:  Sorry because when I'm typing, arquady posted the solution!

[quote="arqady"]
$ \sum \frac{1}{1+a+b}\leq 1$

 which very known.[/quote]

I'm sorry but I haven't seen this before with this condition. :blush:

[quote="andyciup"]See here : http://www.mathlinks.ro/Forum/viewtopic.php?p=344767#344767[/quote]

Yes .it's exactly like the one posted by me.If only I had been born 2 years earlier  :(


I just checked my profile in USAMTS, and it says that I have not submitted my entry form. I submitted my entry form in with my Round 1 solutions. Should I resubmit my form?

Yeah, I had the same problem too.
I sent my entry form attached with my solutions via email, and now I wonder whether to resubmit it. Maybe there's something missing in the protocol they use to look at the emails. It seems like they didn't read the text I typed with my email.

Hmm... I would resend it, just to be sure.

And I have no idea if they read the text in the email  :maybe:

Re-sending the entry form is probably a good idea.

My mom re-faxed my entry form earlier this week. Twice. My Profile still says that I haven't submitted it. Is this a problem? Or is it just that the profiles haven't been updated recently?

EDIT: My USAMTS ID is 6120, in case that helps.

Neither is mine, and I sent it in the same envelope as the answers that were logged and graded.


When I click on the Suggested Video button, I get a message that says "Invalid Video" What does that mean?

Bug. I'll look into it Monday or Tuesday.

I don't even have a Suggested Video button. It disappeared last week.

You don't have any suggested videos available, meaning you have seen all the videos for which you have access to. The link has now been removed.

I haven't seen all the videos yet I can't access any at all. Also, my reports dont show problem types or badges. Why?

Is this still a problem? I can see your report OK.

I like videos......theyre really instructive!


Find $k$ min such that for all $a,b,c$ real positive numbers we have

$\displaystyle k\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)} \ge 1+k$

The minimal k such that the inequality

$k\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 1+k$

holds for any positive numbers a, b, c is k = 1. In fact, proving that every k smaller than 1 doesn't work is almost trivial (take a = 1, b = 1, c = 0), and the fact that every $k\geq 1$ will work if k = 1 works follows from $\frac{a^2+b^2+c^2}{bc+ca+ab}\geq 1$ (since $a^2+b^2+c^2\geq bc+ca+ab$). So it is enough to prove the inequality for k = 1. In other words, it is enough to prove the inequality

$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$

for any positive numbers a, b, c. Here is how I prove this inequality:

WLOG assume that $a\geq b\geq c$. Then, $a-b\geq 0$, and thus $a^2\left(b+c\right)-b^2\left(c+a\right)=\left(a-b\right)\left(bc+ca+ab\right)\geq 0$, so that $a^2\left(b+c\right)\geq b^2\left(c+a\right)$. Similarly, $b^2\left(c+a\right)\geq c^2\left(a+b\right)$. Hence, we have $a^2\left(b+c\right)\geq b^2\left(c+a\right)\geq c^2\left(a+b\right)$. Combining this with $a\geq b\geq c$, we see that we have the right conditions to apply the Vornicu-Schur inequality (Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9) with $x=a^2\left(b+c\right)$, $y=b^2\left(c+a\right)$, $z=c^2\left(a+b\right)$; thus, we get

$a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)\geq 0$.

Now, what does this have to do with the inequality in question? Well, some algebra shows

$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}-2$
$=\frac{a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)}{\left(b+c\right)\left(c+a\right)\left(a+b\right)\left(bc+ca+ab\right)}$;

thus,

$a^2\left(b+c\right)\left(a-b\right)\left(a-c\right)+b^2\left(c+a\right)\left(b-c\right)\left(b-a\right)+c^2\left(a+b\right)\left(c-a\right)\left(c-b\right)\geq 0$

implies

$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}-2\geq 0$,

and thus $\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$, and the proof is complete.

[[b]EDIT:[/b] Now I see that the inequality

$\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$

is known - look at http://www.mathlinks.ro/Forum/viewtopic.php?t=6403 !]

  Darij


If $ 1\geq a^{11} \plus{} b^{11}$ and $ 1\geq x^{11} \plus{} y^{11}$ prove that $ 1\geq a^{5}x^{6} \plus{} b^{5}y^{6}$

Can you help me on this?

[quote="Elio (n)"]Can you explain this please    :oops:[/quote]

Holder's inequality: Let $ a_i,b_i,p,q$ be positive real numbers such that $ \frac 1p\plus{}\frac 1q\equal{}1$. Then
\[ a_1b_1\plus{}a_2b_2\plus{}\dots\plus{}a_nb_n\le (a_1^p\plus{}a_2^p\plus{}\dots\plus{}a_n^p)^{\frac 1p}(b_1^q\plus{}b_2^q\plus{}\dots\plus{}b_n^q)^{\frac 1q}\]
Therefore since $ \frac{5}{11}\plus{}\frac{6}{11}\equal{}1$ we have
\[ a^5x^6\plus{}b^5y^6\le (a^{5\cdot\frac{11}{5}}\plus{}b^{5\cdot{\frac{11}{5}}})^{\frac{5}{11}} (x^{6\cdot\frac{11}{6}}\plus{}y^{6\cdot\frac{11}{6}})^{\frac{6}{11}}\equal{} (a^{11}\plus{}b^{11})^{\frac{5}{11}}(x^{11}\plus{}y^{11})^{\frac{6}{11}}\le 1\]

More on Holder's inequality [url=http://www.mathlinks.ro/viewtopic.php?t=167515&search_id=109109294]here[/url].


quadratics
algebra
number theory solved
number theory

Let $a$ and $b$ be two positive rational numbers such that $\sqrt[3] {a} + \sqrt[3]{b}$ is also a rational number. Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers.

If $a=b$ the statement is trivial. So suppose $a\ne b$.

Let $s=\sqrt[3]{a} + \sqrt[3]{b}$. It follows from the identity
$s^3 = a + b + 3 s \sqrt[3]{a} \sqrt[3]{b}$
that the product $p=\sqrt[3]{a} \sqrt[3]{b}$ is a rational number.
Therefore, $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are the roots of a quadratic equation with rational coefficients, namely, $x^2 - s x + p = 0$.
Then $\sqrt[3]{a}=u + \sqrt{v}$ and $\sqrt[3]{b}=u - \sqrt{v}$ (or vice versa), where $u=s/2$ and $v=(s/2)^2-p$ are rational numbers, and $v\geq 0$.

Since $a = (u+\sqrt{v})^3 = u^3 + 3uv + (3u^2 + v)\sqrt{v}$ is a rational number, $\sqrt{v}$ must be rational implying that $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational.

You can do it by considering two cases
(i) One of them is irrational while other is rational. Write the irrational number in terms of rationals and you will see the contradiction.
(ii) Both are irrational. Again write one of them in terms of other and then eliminate the cuberoot and you will then see the contradiction.


geometry
trapezoid
congruent triangles
geometry unsolved

Lemma 1:

Let the points opposite $ A$ on circles $ \omega_1$ and $ \omega_2$ be $ S_1$ and $ S_2$, respectively.  Then
$ S_1, S_2$ and $ B$ are collinear.

Proof:
$ m\angle{AS_1B}\equal{}\frac{1}{2}m\widehat{AB}_{\omega_1}$
$ m\angle{AS_2B}\equal{}\frac{1}{2}m\widehat{AB}_{\omega_2}$
$ m\angle{S_1AS_2}\equal{}m\angle{S_1AB}\plus{}m\angle{BAS_2}\equal{}\frac{1}{2}(m\widehat{S_1B}\plus{}m\widehat{BS_2})$
Then the sum of these three is $ \frac{1}{2}(m\widehat{S_1A}\plus{}m\widehat{AS_2})\equal{}180^{\circ}$ (where all arcs are taken going counterclockwise).  Therefore, $ B \in S_1S_2$, as desired.

Because $ m\angle{AN_1B}\equal{}m\angle{AS_1B}$ and $ m\angle{AN_2B}\equal{}m\angle{AS_2B}$, it suffices to prove that $ m\angle{N_1AN_2}\equal{}m\angle{S_1AS_2}$.  To show this, it suffices to show that $ m\angle{M_1AO_1}\equal{}m\angle{M_2AO_2}$.  We will prove that both of these angles are equal to $ m\angle{AOO_1}$, where $ O\equal{}P_1P_2 \cap O_1O_2$.

We have $ \triangle{O_1M_1P_1} \sim \triangle{O_1P_1O} \implies O_1M_1 \cdot O_1O\equal{}O_1P_1^2\equal{}O_1A$.  It then follows that $ \triangle{O_1M_1A} \sim \triangle{O_1AO}$.  Thus, $ m\angle{O_1AM_1}\equal{}m\angle{O_1OA}$.  We similarly have the same result on $ m\angle{O_2AM_2}$.  The result follows.

It has already been posted before. http://www.mathlinks.ro/Forum/viewtopic.php?t=139392

Kostas Vittas.

[quote="lasha"]$ w_1$ and $ w_2$ circles intersect each other in points $ A$ and $ B$. Their centers are $ O_1$ and $ O_2$ respectively. Line $ P_1P_2$ touches the circles at $ P_1$ and $ P_2$($ P_1$ is point on $ w_1$ circle, $ P_2$ is a point on $ w_2$ circle). $ M_1$ and$ M_2$ are two points on a line $ O_1O_2$ so that $ P_1M_1$ and $ P_2M_2$ are perpendicularly $ O_1O_2$. Line $ AM_1$ intersects $ w_1$ circle again at a point $ N_1$ and the line $ AM_2$ intersects $ w_2$ circle at a point $ N_2$. Prove that $ N_1$,$ B$ and $ N_2$ are on the same line.[/quote]
Very nice problem and nice solutions to everyone!

I have a different solution than anyone in [b]vittasko's [/b]link (I hope):
[hide] Say that $ AB$ and $ O_1O_2$ meet at $ M$.  We extend $ BA$ to reach $ P_1P_2$ at $ P$.  Notice that $ PP_1^2 \equal{} PA(PB) \equal{} PP_2^2$.  Thus, we obtain that $ P_1P \equal{} P_2P$.  Now, notice that $ P_1M_1$, $ PM$, and $ P_2M_2$ are all perpindiuclar to $ O_1O_2$.  So, we obtain that $ P_1P_2M_2M_1$ is a trapezoid with median $ PM$.  So, we have that $ M_1M \equal{} M_2M$.  Thus, $ AM$ is the perpindicular bisector of $ M_1M_2$.  So, we have that $ AM_1 \equal{} AM_2$.  Now, we call $ \angle AM_1M_2 \equal{} \theta \equal{} \angle AM_2M_1$.  We can thus say that $ \angle M_1AB \equal{} \angle M_2AB \equal{} 90 \minus{} \theta$.  So, we have that $ \angle N_1OB \equal{} 180 \minus{} 2\theta \equal{} \angle BO_2N_2$.  Thus, we can show that $ \angle O_1N_1B \equal{} \angle O_1BN_1 \equal{} \angle O_2N_2B \equal{} \angle O_2BN_2$.  From here, we can observe that $ \angle O_1M_1N_1 \equal{} \theta \equal{} \angle O_1MN_1$, giving us cyclic quaderilateral $ O_1M_1BN_1$.  We can also show that $ O_2M_2N_2B$ is cyclic.  So, we have that $ \angle N_1M_1B \equal{} \angle N_1O_1B \equal{} 180 \minus{} 2 \theta \equal{} \angle BO_2N_2 \equal{} \angle BM_2N_2$.  So, we can say that $ \angle BM_1M_2 \equal{} \theta \equal{} \angle BM_2M_1$.  We obtain congruent triangles $ AM_1M_2$ and $ BM_1M_2$, thus giving us that $ \angle M_1AM_2 \equal{} \angle M_1BM_2$.  So, we have that $ \angle M_1BN_1 \plus{} \angle M_1BM_2 \plus{} \angle M_2BN_2 \equal{} 180 \minus{} 2 \theta \plus{} \theta \plus{} \theta \equal{} 180$, giving us that $ N_1$, $ B$, and $ N_2$ are collinear.   [/hide]


Please, in future, post the problems (when possible using $\LaTeX$) as text into the forum; to my knowledge, it is not possible to see attachments when not logged in :(
Also attachments often get deleted after some time or otherwise get lost, the problem is lost then, too.

The problem is the Hadamard's Three circle Theorem and it's the question:
Suppose $f$ is holomorphic in an open annulus $\{z\in\mathbb C|r_1<|z|<r_2\}$ and continuous in the closed annulus. Let $M(r)$ denote $\sup \{|f(z)|: |z|=r\}$. Now prove that $M(r)\leq M(r_1)^\alpha M(r_2)^{1-\alpha}$ where $\alpha=\frac{\log(r_2)-\log(r)}{\log(r_2)-\log(r_1)}$.

Here is a brief solution. you could see that for each $\beta$ which $\beta\log|z|+\log|f(z)|$ is localy harmonic and take sits maximum at boundary. And there exist $\beta\log|r_2|+\log|M(r_2)|=\beta\log|r_1|+\log|M(r_1)|$. You could easily prove the inequality


There's absolutely no snow here at all-it really would snow, with the temperature at about 80 degrees outside. :(

the same in italy..it's all frozen :lol:

Here in Beijing, it is definitely under $0^o$, but with no snow...

Its been bouncing around 0 for a while...no snow!!!  *sigh*  What's Christmas without snow in Canada?!?!

yea its been about 80 degrees in texas it hasnt snowed since like last year 1 day and then like 4 years before...yea it like never snows

It does in the Sierra Nevada (normally), but this year, places like Truckee, CA (near Lake Tahoe, elev. 5600 feet) have been getting rain due to temperatures between 30 and 50 degrees Farenheit (spelling). Also, ski resorts have had to make their own snow due to lack of snow....T_T....No snow....

The weather here in BC purposely toys with us.  The beginning of December it snows in heapfuls - causing extensive traffic problems throughout the province.  The last few days leading up to Chrismas, it's dry enough to wipe a baby's bottom ( commend me on my superb choice of anologies ) .

hahaha. commended indeed.
Yeah, in TX, it's been nice and 70 degrees warm.

Here in Detroit, it's cold but not snowing. There was huge snow a week ago, and weather reports say it's expected to snow in 2 or 3 days.
Please please please snow! XD

I know that for $C$ and $F$ there is a linear formula: $F=kC+b$, but I don't know the exact numbers... Who can tell me what $k$ and $b$ really are?

[quote="easyas3.14159..."]i think k is 9/5 and b is 32
In southern california, people are usually grateful for it to get as low as 80.[/quote]
Thanks! So 80F is still a pretty high temp....

Well, up here in the Chicagoland area, it snows most days and the temperature is usually about 25-30 degrees Fahrenheit.

It has been warm lately, with temperatures in the 30's (compared to the lower single digits weeks earlier). About two inches of snow remains after a whole day of light rain.

[quote="Bictor717"]Lol, I seriously doubt you'll die.
[/quote]

I dunno...depends how long I stand out there.

[quote="yif man12"]26 inches, not bad for a day.[/quote] :o  :help:
That is the most Delaware has gotten in 10 years

OK, now we have snow.  Maybe 4 inches.  :P

I heard from the News that the north-east part America is snowing heavily, some of the areas even lose their electricity. Is that true?

Yeah, its pretty bad here.

In delaware we got about 1.5 feet but we have electricity :idea:

In North Carolina, we got the southern edge of all the snow so we got flurries!!!  Why in the world couldn't we have gotten the 26 inches other people got??

It has been mostly 50s and 60s, but several days where it was in the 70s.  Now, it is a little colder, in the 40s and it was in the 30s all day Saturday.

because the snow god hates north carolina

Its not fair!!! The bay area in california never gets snow. Our max was 1/4 an inch 7 years ago!!!

Oh be quiet. The Snow God refuses to acknowledge Texas.

...and getting lots of snow is good?  I suppose it makes things more interesting, but I personally miss southern CA now that i live in NY...

yes it is.
you can o skiing or snoboarding :nhl:

I live in California and it basically never snows here, except in Tahoe.
 :(  It was really hot a couple days ago but now its sort of cold

Yeah, the temperature dropped about 20 degrees in two days.


There are certain primes p for which if a five digit number ABCDE (each representing digits) is divisible by p, so are BCDEA, CDEAB, DEABC, and EABCD.  One example is 41, and another trivial case is 3.  Find the other prime p for which this "cyclic permutation" holds true.

Any idea how to start?

To motivate your thinking a little, try to think what 5-digit number ABCDE and its corresponding set of "cycles" {BCDEA,CDEAB,DEABC,EABCD} would be most easily analyzed. For example, what if all of the digits were the same?

[hide="Solution"]
Consider a five-digit number $ 10000a \plus{} 1000b \plus{} 100c \plus{} 10d \plus{} e$ and a prime number $ p$ such that $ p|10000a \plus{} 1000b \plus{} 100c \plus{} 10d \plus{} e$.

If $ p$ divides all of the "cycles", then we have:
$ p|10000a \plus{} 1000b \plus{} 100c \plus{} 10d \plus{} e$
$ p|10000b \plus{} 1000c \plus{} 100d \plus{} 10e \plus{} a$
$ p|10000c \plus{} 1000d \plus{} 100e \plus{} 10a \plus{} b$
$ p|10000d \plus{} 1000e \plus{} 100a \plus{} 10b \plus{} c$
$ p|10000e \plus{} 1000a \plus{} 100b \plus{} 10c \plus{} d$

Adding these must still yield a number divisible by $ p$, so we can write:

$ p|11111e \plus{} 11111a \plus{} 11111b \plus{} 11111c \plus{} 11111d$
$ \implies p|11111(a \plus{} b \plus{} c \plus{} d \plus{} e)$

If $ p\nmid a \plus{} b \plus{} c \plus{} d \plus{} e$, then clearly $ p$ must be a divisor of $ 11111$, and thus our solutions are $ 41$ and $ \boxed{271}$.
[/hide]


is there a function $f$ from the positive integers to the positive integers such that $g(n)=f(f(n))$
then $g(1)=5,g(2)=6,g(3)=4,g(4)=3,g(n)=n+2,n \geq 5$

Let $f(3) = k$ and $f(4) = j$.

$f(k)=4 \rightarrow ff(k) = j \rightarrow g(k) = j$
$f(j)=3 \rightarrow ff(j) = k \rightarrow g(j) = k$
$\rightarrow gg(k)=k$

But the only solution to $gg(x) = x$ is $3$ and $4$.

But if $f(3)=3$, then $g(3)=ff(3)=3$ Contradiction.
If $f(3)=4$, then as $g(3)=ff(3)=f(4)=4$, then $f(4)=4$ But that means $g(4)=ff(4)=4$ Contradiction.


[quote="williams"]a,b and c are two-digit nos.d unit digit of a is 7,d unit digit of b is 5 and d tens digit of c is 1. If ab+c equals 2006,fing d value of a+b+c.[/quote]

Set:
$ a \equal{} 10u \plus{} v$
$ b \equal{} 10w \plus{} x$
$ c \equal{} 10y \plus{} z$

where $ x, y, z$ are the unit digits, and $ u, v, w$ are the tens digits.

From the problem condition, we replace some of the digits with their actual values:

$ a \equal{} 10u \plus{} 7$
$ b \equal{} 10w \plus{} 5$
$ c \equal{} 10 \plus{} z$

The equation we are provided is:

$ (10u \plus{} 7)(10w \plus{} 5) \plus{} (z \plus{} 10) \equal{} 2006$
$ (10u \plus{} 7)(10w \plus{} 5) \plus{} z \equal{} 1996$

We conclude that $ z \equal{} 1$.

This simplifies to:
$ (10u \plus{} 7)(10w \plus{} 5) \equal{} 1995$

We know that 1995 factors to:
$ 1995 \equal{} 3 \cdot 5 \cdot 7 \cdot 19$

Which means:
$ 10u \plus{} 7 \equal{} 57 \equal{} a$
$ 10w \plus{} 5 \equal{} 35 \equal{} b$
$ z \plus{} 10 \equal{} 11 \equal{} c$

So the answer we want to find is

$ a \plus{} b \plus{} c \equal{} 57 \plus{} 35 \plus{} 11 \equal{} \boxed{103}$


superior algebra
superior algebra solved

Let $(A,+,\cdot)$ be a ring. If for every $x\in A$ there is a $n_x\in\mathbb{N}^*$ s.t. $x^{n_x}=x$ then $A$ is commutative. ($x^k=x\cdot x\cdot\ldots\cdot x$ )


calculus
integration
function
trigonometry
calculus computations

How do you integrate (2x^2+1)*e^(x^2) without just "seeing" the solution....like how can you make a substitution or use parts to do it?

You use a little bit of calculus lore :)

It's somewhat well known that $f(x)=e{}^x{}^2=\exp(x^2)$ is one of the simplest functions with no elementary antiderivative.  ($x^{-1}\sin x$ is another).  The form of the question suggests integration by parts, because we are multiplying together two 'different' functions.  So we know that we want to perform an integration by parts on

$\int 2x^2\exp(x^2)dx$

We are going to have to pick carefully - try using $u=x$, $dv=2x\exp(x^2)$ (one of the few times we can integrate something involving $\exp(x^2)$)

So

$\int 2x^2\exp(x^2)dx=x\exp(x^2)-\int \exp(x^2)dx$

and the result follows.

ook thanks for the reply; this is the same way i ended up solving it.


Inscribe a regular 15-gon in a circle using compass and straightedge.

Just do 360/15 = 24
Carve out arcs of 24* and then draw chords. right?

[quote="13375P34K43V312"]Explain how you cut out the $24^\circ$ arcs?[/quote][hide]Inscribe a pentagon in the circle, and then inscribe an equilateral triangle in the circle (both are well-known constructions) so that the triangle shares a vertex with the pentagon. Going clockwise from that vertex, the arcs cut off by the vertices of the pentagon and triangle will have measures $72^\circ, 48^\circ, 24^\circ, 72^\circ, 24^\circ, 48^\circ, 72^\circ$. Voila! $24^\circ$ degree arc! You are done.

Also, you could construct a pentagon and then construct an equilateral triangle from each of its vertices. Might be overkill though.[/hide]


What happens if I try to answer the questions for a topic but cant pass it before it is due? What happens if I get a past due topic for the alcumulus?

Don't worry about that too much for the Intro to Counting class.  Alcumus is currently in a test phase for integration with the Intro Counting class.  Just do you best; try to do at least a couple dozen problems in Alcumus each week and you'll be fine.

Thanks a lot! I was getting really worried there.


$ \text{ Find the possible solutions  a and b , where}\quad a, b \in \mathbb{Z}.$

$ \frac{a^2\minus{}ab\plus{}b^2}{a\plus{}b} \text{\equal{} }\frac{7}{3}$

[hide="solution"]
With $ a^2\minus{}ab\plus{}b^2\equal{}(a\minus{}\frac{1}{2}b)^2\plus{}\frac{3}{4}b^2\ge 0$,we see $ a\plus{}b>0$.

The eq is equivalent to $ 3\{(a\plus{}b)^2\minus{}3ab\}\equal{}7(a\plus{}b)$,

 by $ (3,7)\equal{}1$ then $ 3|(a\plus{}b)$,the LHS is dividible by $ 9$,so $ 9|(a\plus{}b)$ follows.Therefore we have $ a\plus{}b\ge 9$.

now for all $ a,b\in Z$,$ (a\plus{}b)^2\ge 4ab$ holds, so we have

$ \frac{7}{3}\equal{}\frac{(a\plus{}b)^2\minus{}3ab}{a\plus{}b}\ge \frac{(a\plus{}b)^2\minus{}\frac{3}{4}(a\plus{}b)^2}{a\plus{}b}\equal{}\frac{1}{4}(a\plus{}b)$

so we have $ a\plus{}b\le \frac{28}{3}$, so we see $ a\plus{}b\equal{}9$,and $ ab\equal{}20$,

thus we finally have the solutions $ \boxed{(a,b)\equal{}(4,5)(5,4)}$

[/hide]


A man has $5000$ coins in a safe, all of which are either dimes or nickels.  He repeatedly transfers 5 coins into his bank account.  After $n$ transfers, he has more money (in value) in his bank account than in his safe.  Compute the number of possible values of $n$.

[hide] The best he can do is to transfer $5$ dimes each transfer. That gives $50n > 5000 \implies n > 100 \implies n = 101$.  The worst he can do is tranfer $5$ nickels each time. Which gives $25n > 5000 \implies n > 200 \implies n = 201$.  So $n$ is any integer on the interval $[101,201]$ which includes $100$ possible values for $n$.[/hide]

No, because your inequalities assume that he has 5000 cents, not 5000 COINS as stated in the problem.

[hide="umm"]
so let's reduce by 5 so we get coins of 1 and 2 cents. let's see. I'm not sure, but if we take the max value of $n$ (1000, because we can always transfer all 500 coins), subtract the min value of $n$, and add 1, we should get the right answer. If the ratio of nickels to dimes is as close as possible to 2:1, we should have our optimal bank. because if we transfer all the dimes first, we get our min $n$. so we have 334*5 dimes and 666*5 nickels, meaning it takes 334 transfers minimum to get this accomplished, so our answer should be very close to $667$[/hide]


geometry
ratio
perimeter
geometry unsolved

Points  $ D$ and $ E$ are chsen on the sides $ AB$  and $ AC$ of triangle $ ABC$ such that $ DE$ is parallel to $ BC$ and $ DE$  is tangent to the incircle $ I$ of triangle $ ABC$
 [b]prove that[/b]    $ S_{ABC} \geq 8 S_{IDE}$

[img]http://i227.photobucket.com/albums/dd72/TPFC/hongquy-4.png[/img]
We have $ \frac{S_{ADE}}{S_{ABC}}\equal{}\frac{AK^2}{AH^2}\equal{}\Bigg(1\minus{}\frac{2r}{h_a}\Bigg)^2$
And $ \frac{S_{BDM}}{S_{ABC}}\equal{}\frac{BD.BM}{BA.BC}\equal{}\frac{HK.(p\minus{}b)}{HA.a}\equal{}\frac{2r.(p\minus{}b)}{h_aa}$
$ \frac{S_{CEM}}{S_{ABC}}\equal{}\frac{CE.CM}{CA.CB}\equal{}\frac{HK.(p\minus{}c)}{HA.a}\equal{}\frac{2r.(p\minus{}c)}{h_aa}$
Then
$ \frac{S_{ADE}\plus{}S_{BDM}\plus{}S_{CEM}}{S_{ABC}}\equal{}\Bigg(1\minus{}\frac{2r}{h_a}\Bigg)^2\plus{}\frac{2r.(p\minus{}b)}{h_aa}\plus{}\frac{2r.(p\minus{}c)}{h_aa}$
$ \equal{}\Bigg(\frac12\minus{}\frac{2r}{h_a}\Bigg)^2\plus{}\frac34\geq \frac34$
Then $ \frac{S_{MDE}}{S_{ABC}}\leq \frac14$
And $ S_{MDE}\equal{}2S_{IDE}$ then OK!
Equality when $ h_a\equal{}4r$

We know the ratio in question is the ratio between $ DE$ and the perimeter of the triangle.  Let $ AB$ be tangent to the incircle and A-excircle of the triangle at $ X$ and $ Y$, respectively.  Let $ p$ be the perimeter of the triangle.  Then

$ \frac{DE}{BC}\equal{}\frac{AX}{AY}\equal{}\frac{\frac{b\plus{}c\minus{}a}{2}}{\frac{p}{2}}\equal{}\frac{p\minus{}2a}{p}$
Thus,
$ \frac{DE}{p}\equal{}\frac{a(p\minus{}2a)}{p^2}$
Let $ a\equal{}kp$, where $ \frac{1}{2}>k>0$.  Then
$ \frac{DE}{p}\equal{}k(1\minus{}2k)\equal{}\frac{1}{8}\minus{}2\left(k\minus{}\frac{1}{4}\right)^2$.
which has a maximum value of $ \frac{1}{8}$, as desired.


For $ a,b,c$ positive reals, prove:
$ a^5\plus{}b^5\plus{}c^5 \plus{} abc(ab\plus{}bc\plus{}ca) \geq a^2b^2(a\plus{}b) \plus{} b^2c^2(b\plus{}c) \plus{} c^2a^2(c\plus{}a)$

[quote="hendrata01"]For $ a,b,c$ positive reals, prove:
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(ab \plus{} bc \plus{} ca) \geq a^2b^2(a \plus{} b) \plus{} b^2c^2(b \plus{} c) \plus{} c^2a^2(c \plus{} a)$[/quote]

It's easy
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2) \ge \sum ab(a^3 \plus{} b^3)$
So we need prove that:
$ \sum ab(a^3 \plus{} b^3) \plus{} abc(ab \plus{} bc \plus{} ca) \ge abc(a^2 \plus{} b^2 \plus{} c^2) \sum \ge a^2b^2(a \plus{} b)$
$ < \equal{} > \sum ab(a \plus{} b)(a \minus{} b)^2 \minus{} \frac {1}{2}abc[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \ge 0$
We have:
$ {S}_{c} \equal{} 2ab(a \plus{} b) \minus{} abc;{S}_{b} \equal{} 2ca(c \plus{} a) \minus{} abc;{S}_{a} \equal{} 2bc(b \plus{} c) \minus{} abc$
Assume:$ a \ge b \ge c$
$ {S}_{c}\ge 0;{S}_{b} \ge 0;{S}_{b} \plus{} {S}_{a} \ge 0$
We have done!  :P

[quote="hendrata01"][quote="caubetoanhoc94"]
$ {S}_{c}\ge 0;{S}_{b} \ge 0;{S}_{b} \plus{} {S}_{a} \ge 0$
We have done!  :P[/quote]

SOrry, is that legal? I thought you need that $ S_a \geq 0$ and not just $ S_a \plus{} S_b$?[/quote]
 :P .$ S_a \plus{} S_b \ge 0$
It's true :) by SOS

Isn't it $ \sum a(a^2\minus{}b^2)(a^2\minus{}c^2) \ge 0$ ?

Which holds clearly by Schur Inequality..


This problem was proposed by Laurentiu Panaitopol for a Romanian TST (forgot the year, sorry), and it is quite nice and not too hard:

Prove that, for any positive integer n,
\pi {k=1->n} k^(2k-n-1) is an integer.

It's quite easy. You look for every prime number p how often it comes up in the product. For each p you have some k in N with kp\leq n < (k+1)p. The prime p comes up -n-1+2p+(-n-1+4p)+...+(-n-1+p(k+1)) times. It is equal to k(-n-1+p(k+1)) which is positive or zero. Now it is done!

Discussion continues in http://www.mathlinks.ro/Forum/viewtopic.php?t=54194


function
inequalities
analytic geometry
number theory solved
number theory

heres a problem from iranian olympiad  1998 or 1999 (dont remember xactly) and i reall y need  help on this one
1)let P be the set of all points in R^n with rational coordinates.for A,B belonging to P ,one can move from A to B if the distance is 1.prove that every point in P can be reached from every other point in P by a finite sequence of moves iff n>=5
2)let f_1,f_2,f_3 be functions from the reals to the reals such that af_1+bf_2+cf_3
is monotonic for all a,b, c belonging to real numbers.prove that there exist real numberss x,y,z ,not all zero such that
xf_1(m) + yf_2(m) + zf_3(m) = 0 for all m belonging to the set of real numbers
thanx in advance

The 2nd problem goes something like this:

Let us consider all the functions f1(x)+af2(x), where a is a real number. The hypothesis tells us that such a function is always monotonous. Let A={k|  f1(x)+kf2(x) is decreasing} and B={k|  f1(x)+kf2(x) is increasing}. |A :cap B| is composed of all the constant fuctions. Let us show that |A :cap B| \leq 1. Indeed if we had f1(x)+af2(x)=k and f1(x)+bf2(x)=m, k and m constants, for a<>b, then f2 is x, a constant (by substracting), and so will be f1=y. Then we will have -xf1+yf2+0f3=0, and the problem would be over.

So A :cap B has at most one element
Let us observe that we cannot have an element of A between two elements of B (or the other way around). Suppose we have three numbers, a<b<c such that  f1(x)+af2(x) and f1(x)+cf2(x) are in A and f1(x)+af2(x) is in B, then for any two real x<y, we would have
 f1(x)+af2(x)<=f1(y)+af2(y)
f1(x)+bf2(x)>=f1(x)+bf2(x)
 f1(x)+cf2(x)<=f1(x)+cf2(x)

Because a<b<c, there exists a t with 0<t<1 such that b=at+c(1-t). So the first and the third inequalities above become:

 tf1(x)+taf2(x)<=tf1(y)+taf2(y)                         (1)
(1-t)f1(x)+(1-t)cf2(x)<=(1-t)f1(x)+c(1-t)f2(x)                      (2)

By adding we have exactly f1(x)+bf2(x)<=f1(x)+bf2(x), but we had exactly the opposite, that is f1(x)+bf2(x)>=f1(x)+bf2(x).

So, to have equality, (1) and (2) must be equalities for all x and y, so we would have 2 constant functions, but we had that |A :cap B|<=1.

Now, all of this was meant to show that no element of A could be between two elements of B (or the other way around). Since A \cup B=R, we must have A=(- \infty;a] and B=(a;+ \infty), for a certain a (of course we could have just as well that A=(- \infty;a) and B=[a;+ \infty), but that's just the same thing).

Now, what I want to do is to show that  the number a, which is in A and is the upper bound of A is also in B. Suppose a is not in B, so f1(x)+af2(x) is decreasing and non-constant, and f1(x)+(a+ \varepsilon )f2(x) is increasing for any positive  \varepsilon. We would have an x and a y with x>y such that

f1(x)+af2(x)<f1(y)+af2(y) (strict inequality), so we would have a positive  \alpha such that

f1(x)+af2(x)=f1(y)+af2(y)- \alpha
But for all  \varepsilon we have f1(x)+(a+ \varepsilon )f2(x)>=f1(y)+(a+ \varepsilon )f2(y). By substrabcting we obtain
 \varepsilon f2(x)>= \varepsilon f2(y)+ \alpha , for all positive  \varepsilon , which is impossible.

So we have f1(x)+af2(x) constant, and

f1(x)+af2(x)=k. In the same way we obtain
f2(x)+bf3(x)=L

So -Lf1(x)+(-aL+k)f2(x)+ kbf3(x)=0. QED

(oh, it's easy to see what happens if -L, -aL+k and kb were all 0, so this is not a problem)

About the first problem, I only managed to show that we can make the jumps if n>=5 (i didn't show that we couldn't for n=4).

So, instead of jumping from any A to any B, it suffices to show that we can jump from 0 to any X of rational coordinates. Let X=(x1,...,xn).  So we have to jump from (0,...,0) to (x1,...,xn).  Now bring all the coordinates of X to a common denominator, and make this denominator so large that all the numerators of the coordinates are even numbers. Let e be this denominator. So let X=(2a1/e,...,2an/e), where the ai's are integers.

A theorem of Fermat says that any positive integer can be written as a sum of four integer squares. So there exist integers a,b,c and d (some of them may be 0) such that e ^2 =1+a ^2 +b ^2 + c ^2+d ^2 (right here is the flaw for n=4. There exist integers that can't be written as the sum of three squares)

So from any point    (y1,...,yn) we can jump to  (y1 \pm 1/e, y2 \pm a/e, y3 \pm b/e, y4 \pm c/e, y5 \pm d/e, y6 ,...,yn), because the distance between the two points is 1.

So from all (y1,...,yn) we can jump to (y1 +1/e, y2 + a/e, y3 + b/e, y4 +c/e, y5 + d/e, y6,...,yn), and from this point on we jump to (y1 +1/e+1/e, y2 +a/e-a/e, y3 + b/e-b/e, y4 +c/e-c/e, y5 + d/e-d/e, y6 ,...,yn) which is the point (y1+2,...,yn) .

So we increased, after a valid sequence of moves ONLY one of the coordinates (it musn't necessarily be the first coordinate, it can be just as well any other) by 2/e (and we could decrease it by 2/e if we wanted). So it is obvious how we can jump from (0,...,0) to (2a1/e,...,2an/e).


logarithms
function
combinatorics open
combinatorics

Let $ E_n$ denote the number of permutations of $ n$ elements such that each cycle has even length, and similarly define $ O_n$.  Standard results imply that

$ \sum_{n \ge 0} \frac{E_n}{n!} x^n \equal{} \exp \left( \frac{x^2}{2} \plus{} \frac{x^4}{4} \plus{} ... \right) \equal{} \frac{1}{ \sqrt{1 \minus{} x^2} }$
$ \sum_{n \ge 0} \frac{O_n}{n!} x^n \equal{} \exp \left( x \plus{} \frac{x^3}{3} \plus{} \frac{x^5}{5} \plus{} ... \right) \equal{} \sqrt{ \frac{1\plus{}x}{1\minus{}x} }$.

It then follows from the identity

$ \ln 2 \equal{} 1 \minus{} \frac{1}{2} \plus{} \frac{1}{3} \minus{} \frac{1}{4} \pm ...$

that asymptotically $ O_n \sim 2 E_n$.  Can you give a heuristic argument as to why this is reasonable?  (I am not looking for an argument based on the generating functions.)

This is very, very rough. Taking $ n$ even since otherwise $ E_n \equal{} 0$ and things are blah.

Take an arbitrary element of $ E_n$. If there are an odd number of even cycles, pretend there is an extra $ 0$ cycle. Partition the cycles into pairs, and have the larger cycle "donate" an element to the smaller one. There's lots of ways to do these partitions or choose the element donated; just average over all possible choices and ideally the resulting overcount would balance things out. This gives us an element of $ O_n$.

The main issue is that we miss a lot of the elements of $ O_n$ with a lot of single-element cycles. For instance, we never count any where at least half of the elements are fixed points. So now we're going to do the same thing. But this time, we mark some element of the set we are permuting (average over all possible markings...). Now, once we've constructed our element of $ O_n$, we take the cycle containing the marked element and break it up into single-element cycles. This hopefully would balance out with what we obtained from above. So that's a 1-2 correspondence.

Actually, it seems that $ O_n \equal{} E_n$ when $ n$ is even.  For example, $ O_4 \equal{} E_4 \equal{} 9$ and $ O_6 \equal{} E_6 \equal{} 225$.

[quote="nsato"]Actually, it seems that $ O_n \equal{} E_n$ when $ n$ is even.[/quote]
You're right; in fact very plainly one generating function is $ 1 \plus{} x$ times the other, which implies two identities based on the parity of $ n$.  Now the question is to give bijective proofs of both.


A magic square is given below. The entries in it are the integers from 1 to 9 inclusive, and each row and each column sum to the same amount. (The diagonals do NOT sum to the same amount as the rows or columns.) With the 4 elements placed as below, what is the value of the element at the spot labeled A?

$\begin{tabular}{|c|c|c|}\hline 8&&6\\\hline &A&\\\hline 4&& 2\\\hline\end{tabular}$

(My 1500th post)

[hide="Not sure"]
I basically just wanted to maximize the value of the row of 4 _ 2 so I put a 9 in that spot. Then everything just fit!
8 1 6
3 5 7
4 9 2
$A=5$[/hide]
And why is the title "Bits"?

[hide="Quevvy is correct"]We still have to place the $1$, $3$, $5$, $7$, and $9$. From the middle row and the middle column, the sum of two of these odd numbers has to equal the sum of another two odd numbers. So trying it out,

$1+9=3+7$

The $1$ has to go between the largest two numbers, and the $9$ has to go between the smallest two.
$\begin{tabular}{|c|c|c|}\hline 8&1&6\\\hline &A&\\\hline 4&9& 2\\\hline\end{tabular}$

Trying out for the $3$ and $7$,
$\begin{tabular}{|c|c|c|}\hline 8&1&6\\\hline 3&A&7\\\hline 4&9& 2\\\hline\end{tabular}$

So $A=5$, with a square like this:
$\begin{tabular}{|c|c|c|}\hline 8&1&6\\\hline 3&5&7\\\hline 4&9& 2\\\hline\end{tabular}$

Nice problem  :) .[/hide]

[hide]the sum of all the numbers in the square is 45. so each row/column has a sum of 45/3=15. so the left middle is 15-8-4=3 and the right middle is 15-6-2=7. so A=15-7-3=5[/hide]

[quote="Walk Around The River"]yes, like pratik said, there is no need for trying different cases.

[hide="solution"]in any square like this, you just take the total of the numbers (45) and divide it by the number of rows (3), so 15 is the sum.  then subtract 15-8-6=1, and 15-4-2=9.  so 9+1+a=15, so a=5.
and for you [idiots] who insist on memorizing a formula for everything, the sum of each row/column in a $nXn$ square is $\frac{n^{3}+n}{2}$, and you can easily see how that is derived from the above.

[hide="for those people who cant figure this out on their own, i just added this since its the moems forum which is kinda basic"]the total amount of numbers in a $nXn$ square is $n^{2}$.  the sum 1+2+...$n^{2}$ is $\frac{(n^{2})(n^{2}+1)}{2}$ and then you divide that by the number of rows, $n$.  so we have $\frac{(n^{2})(n^{2}+1)}{2n}$ or $\frac{n^{3}+n}{2}$[/hide]
[/hide][/quote]
ahem I believe you are one of those idiots

[hide="solution"]

We have:

$\begin{tabular}{|c|c|c|}\hline 8&&6\\\hline &A&\\\hline 4&&2\\\hline\end{tabular}$

Since the sum of the first 9 positive integers is 45, and the rows have equal sums, each row has a sum of

$\frac{45}{3}=15$

So we now have

$\begin{tabular}{|c|c|c|}\hline 8&1&6\\\hline 3&A&7\\\hline 4&9&2\\\hline\end{tabular}$

The number missing is $\boxed{\underline{5}}=A$[/hide]

Didn't it say that the diagnols don't sum to the same as the rows and columns or am i missing something...

It was specifically stated in the problem that the diagonals don't have the same sum as the rows and columns, but it doesn't work any other way.

Isn't the middle # 5 in all magic squares?

Just to solve this problem, the numbers left are 1,3,5,7,9 and in order to make them fit, you have to take 4 numbers that the sum of two of each will give you the same sum.  (9+1; 7+3) Sorry if that last sentence was confusing.  Now, since you have all those numbers, take the number which is missing, 5, and that is your answer.


how may years have [size=150]you[/size] played an instrument?
post what you played below
i played the violin for 5 years  :D
cello for 1 year

I've played piano for 6 years, and violin for 4.

I've played trumpet and piano for about 2 years each.

i've played piano for about 3 years, but then i stopped.

[quote="aznmonkey1992"]alot of the chinese people i kno play piano, but i play violin! yay![/quote]

A lot of chinese people I know play violin too.  In fact, I only know about 3 who don't.  But then, I don't know a lot of Chinese people.

viola: 2
violin: 3
never had private lessons, though, so im not great at it.

It's well-known that you're not Chinese if you haven't played either piano or violin. For me, it was piano.

lol hey...i know...*thinks*...
fine nobody chinese hasn't played violin/piano lol

piano: 2 years
violin: 9 years
viola: 1 years

i have been playing the violin for 5 years

I played violin for 1 year-stopped cuz i switched schools and this one only has band not orchestra :(
alos played/play piano for around 3 years...

viola, 7 years

and um piano, 2ish years because my mom made me.  yes, i am asian  :P