algebra
polynomial
advanced fields
advanced fields unsolved

reducible (irreducible) means reducible (irre) on $ Z[x]$, polynomial will always refers to poly in $ Z[x]$.

1. there isn't such a polynomial $ h$, that for every $ a\in Z$, $ |a| > N$, $ N$ can large enough, that $ h(x) \plus{} a$ is reducible (irreducible).

2.for any $ h(x)$ and $ g(x)$, $ H_n(x) \equal{} H_{n \minus{} 1}\circ h(x)$, is there always a such integer $ m$, that $ H_m(x) \minus{} g(x)$ is reducible?

the 1st ques is trivil, but is there, for any h(x), an $ N$, if $ a\in Z \cdot |a|>N$ then $ h(x)\plus{}a$ is reducible $ \rightarrow$ $ h(x)\plus{}a\plus{}1$ is irreducible.


I made quite a few sigs and other stuff I would like to show off, so here we go:
#1[img]http://srishivananda.com/sumit/banner.png[/img]
#2[img]http://srishivananda.com/sumit/banner2.png[/img]
#3[img]http://srishivananda.com/sumit/zyron.png[/img]

Are they good? I had another one, but I'm only allowed to include 3 images.

The text on the flag in the first one doesn't really math the flag itself. You can tilt it, and put the first half of the text a bit higher and to the left, on the first half of the flag. Also, you could lift the entire flag up a bit to make it more centered, or even lift it over the center.

The second one is much better, and I don't think there are any faults in it. Not easily-seen ones, anyway.

The third one is also not so bad, but the text on the bottom is cut off and illegible. You can probably do something about the color also, but I don't remember how.

Were these images made in Photoshop and do they have layers? If so, it'll be a lot easier to edit them.

Uhh I just threw this together in around ten minutes

[img]http://www.lunchlady.org/lazarus/redivan.jpg[/img]

the first one has the obvious fault someone pointed out. between the other too, i like the bottom one.
good work, lazarus!

Haha I made one

[img]http://mathprob.com/_private/files/Anirudh/purple%20monster.png[/img]


[quote="kihe_freety5"]help me,please!!!
the inequality:
a,b,c>0
(cb+ac-ba)/(a²+b²)       +      (ab+ac-bc)/(b²+c²)     +   (ab+bc-ac)/(a²+c²)  [size=200]≥[/size]  a/(b+c)  +  b/(c+a)  +  c/(a+b)[/code][/quote]
Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}.$ :wink:


[quote="Pirkuliyev Rovsen"]Solve the equation : (x-2)/1998+√((x-3)/1997=(x-1998)/2+√((x-1997)/3
Kolmogorov tournament. :roll:[/quote]

It's impossible to understand your problem : $ 6$ left parenthesis and $ 4$ right parenthesis  :(

Click on the upper "Latex Help" button and please use Latex for your formulas.

I think the problem should be

\[ \frac{x\minus{}2}{1998}\plus{}\sqrt{\frac{x\minus{}3}{1997}}\equal{}\frac{x\minus{}1998}{2}\plus{}\sqrt{\frac{x\minus{}1997}{3}}.\]

However, it is very easy to see that the substitution $ t: \equal{}x\minus{}2000$ simplifies essentially the solution of the problem. Namely, you can write
\[ \frac{x\minus{}2}{1998}\equal{}\frac{t}{1998}\plus{}1,\quad\frac{x\minus{}3}{1997}\equal{}\frac{t}{1997}\plus{}1,\quad\dots .\]

After certain manipulations you get the answer that the only real solution is the value [hide]$ t\equal{}0$[/hide]. Now the resubstitution gives the final solution in the unkown $ x$.


help?

im feeling kind of depressed right now because i just got a horrible score on the 2001 states

could someone say how to do all of the targets except for the first one? (i got the first one)

-jorian

[hide]2. Just punch in the numbers in your calculator.$1+2*2+3*3+4*4+5*5+6*6+7*7+8*8+9*9+10*10=385$ $385/(1+2+3+...10=10(11)/2=5*11=55)=7$
Mean=7
Mode=10(obviously)
median=since there are 55 terms, the middle term would be the 27th term, which is 7.

Hope there arent any careless mistakes![/hide]

[quote="jhredsox"]ya, for #2 i messed up on the mean

...wow, im stupid!, i forgot to take the test WITH a calculator :P (...and got a 1!)
-jorian[/quote]

Something similar happened to me. For a practice state test, I used a calculator and got 30 on the sprint round. :rotfl:

[quote="jhredsox"]u do know i took it WITHOUT a calc...right?
-jorian[/quote]
Yeah.

it annoys me still because i wouldve needed an 8/8 on target to make my final score ok

(i got 14 on sprint.....12 out of first 15 and 2 out of last 15 :( )
-jorian

[hide]
3) Note that $x^{2}+y^{2}=4$ is the equation of a circle.

4) The second best player must be in the same bracket as the best.

5) Radius is perpendicular to the chord, so use the pythagorean theorem.

6) You kinda have to guess the optimal configuration...should end up being the 3 smaller cubes on top the big one.

7) $\binom{10}{3}$ ways to choose the boy and $\binom{8}{3}$ ways to choose the girl.

8) Substitute variables for digits. [/hide]

#3: what do u mean, "is the equation of a circle"?
#4: i get it, just i forgot that the 2nd best player took up one of the 16 spots :rotfl:
#5: i dont get chords AT ALL!
#6: i don't know where i messed up (my answer was wrong by 9)
#7: i was wondering what they meant, but didn't know whether to add $\binom{10}{3}$ and $\binom{8}{3}$ or wheter to multiply them

also, would u use the n! over p!(n-p)! formula for the separat boys and girls equations?
#8: where would u start?

-jorian


In a parlour game, the magician asks one of the participants to think of a three-digit number abc where a, b, c are digits. Then the magician asks the participant to add the five numbers acb, bac, bca, cab, and cba, and reveal their sum. Suppose the sum was 3194. What was abc originally?

[hide]
Denote the three digit number $ 100a\plus{}10b\plus{}c\equal{}<abc>$

We are given

$ 122a\plus{}212b\plus{}221c\equal{}3194$

Adding $ <abc>$ to both sides, we find

$ 222(a\plus{}b\plus{}c)\equal{}3194\plus{}<abc>$

$ 222 | 3194 \plus{} <abc>$

From here, we only have to test the four possible values of $ <abc>$, which are $ 136, 358, 580,$ and $ 802.$
$ \boxed{358}$
[/hide]


topology
advanced fields
advanced fields unsolved

Let $ X$ be a topology space,Prove that any derived set $ A'$ is closed if and only if $ \{x\}'$ is closed for any $ x \in X$

Ok,Nobody want to answer this problem,maybe you think it is too easy.
Here is my solution.

Proof,
For the necessary part is obvious.
For the sufficient part,
First for every $ x \in X$,$ \{x\}'$ is closed,thus $ (\{x\}')^c$ is open,
But obviously $ x \in (\{x\}')^c$.
Thus there exist an open neighborhood of $ x$ named $ U_1$,such that $ U_1 \subset (\{x\}')^c$.
Now we will prove that $ A'$ is closed,which means $ (A')^c$ is open.
If $ x \in (A')^c$,thus there exist an open neighborhood of $ x$ named $ V_1$,such that:
$ V_1 \cap (A \minus{} {x}) \equal{} \phi$
Let $ U \equal{} U_1 \cap V_1$ be an open neighborhood of $ x$,so for any $ y \ne x \in U\subset (\{x\}')^c$,
there exist a open neighborhood of $ y$ named $ V$ such that $ V \cap \{x\} \equal{} \phi$
Thus $ V \cap U \cap A \equal{} \phi$,but $ V \cap U$ is also an open neighborhood of $ y$,
Thus $ y \in (A')^c$,Thus  $ (A')^c$ is open which is equivalent to $ A'$ is closed.


Here go the tortoise (T and the hare (H) racing again, this time in opposite directions around a circular track 100 yards in diameter. They started at the same spot, but H did not move until T had already gone $1/8$ of the distance. H held such a poor opinion of T's racing ability that he sauntered along, nibbling the grass until he met T. At this point H had gone $1/6$ of the distance. How many times faster than he went once he started to move must H now go to die the race?

"die"???

or was that supposed to be "tie"???


When I type a bunch of text, and print it to a pdf, it keeps going past the end of the page, and the boxed answer doesn't show.  :?:

This makes little sense. You will have to give much more detail. Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please[/url].

I believe titan means that when he converts to pdf, instead of going onto a second page, it tries to cram everything into one page, and needless to say there is not enough room, so it doesn't get printed.

Are you sure you're not converting to .dvi instead of .pdf?

If that's the case we still need a minimal example to look at. Not sure why conversion to dvi would show such a problem. Maybe there's a page size problem or code errors on the second page or ... but there's little point guessing.


inequalities
quadratics
function
inequalities theorems

I've seen this technique referenced several times, but I haven't actually seen it in any texts on inequalities. As I understand it, the acronym stands for "sum of squares", and the idea is to reduce an inequality to $ \sum{a^2} \geq 0$... is this correct?

Can someone also please give an example of a problem and solution that implements this technique? Many thanks in advance.

For example try this simple sum:

Show that $ 2^x \plus{} 3^x \plus{} 6^x \minus{} 4^x \minus{} 9^x \leq 1$

 :)

Not the most elegant solution... but
[hide="I think it works"]
Letting $ a = 2^x$, $ b = 3^x$, and observing that a and b are positive, we have
\begin{eqnarray*}
a + b + ab - a^2 - b^2 &\leq& 1\\\Leftrightarrow
a^2 + b^2 - a - b - ab + 1 & \geq & 0 \\\Leftrightarrow
(a - b)^2 - a - b + ab + 1 & \geq & 0 \\\Leftrightarrow
(a - b)^2 + (a - b) + ab + a + 1 & \geq & 0 \\
\end{eqnarray*}

If we now let x = a-b, we have a quadratic in x:
$ x^2 + x + ab + a + 1 \geq 0$ with discriminant: $ 1 - 4(ab + a + 1) < 1 - 4(1) < 0$, so our quadratic is positive for all x and we're done.[/hide]
Can someone please confirm the correctness of that?

Nope, it doesn't work since what you have considered as a constant in your $ \Delta$ is actually a function of $ x$

kops723, i gave this as a question to your requirement :) try it by SOS method :)

Hmm.. I'm not seeing it >.>
Everything I've tried leaves a negative term.

well here's a soln :) Actually my friend did this.

multiply throughout by 2 and take everything to the RHS
We get:

$ ( 3^x \minus{} 2^x )^2 \plus{} ( 2^x \minus{} 1 )^2 \plus{} ( 3^x \minus{} 1 )^2 \geq 0$ which is true :)


How do I calculate the inverse of Euler's function?  What are all values of n such that $ \phi(n)\equal{}38$?  Thanks in advance.


algebra
polynomial
combinatorics unsolved
combinatorics

Let $ k$ and $ s$ be positive integers. For sets of real numbers  $ \{\alpha_1, \alpha_2, \ldots , \alpha_s\}$ and $ \{\beta_1, \beta_2, \ldots, \beta_s\}$ that satisfy

\[ \sum^s_{i\equal{}1} \alpha^j_i \equal{} \sum^s_{i\equal{}1} \beta^j_i \quad \forall j \equal{} \{1,2 \ldots, k\}\]

we write \[ \{\alpha_1, \alpha_2, \ldots , \alpha_s\} \overset{k}{\equal{}} \{\beta_1, \beta_2, \ldots , \beta_s\}.\]

Prove that if \[ \{\alpha_1, \alpha_2, \ldots , \alpha_s\} \overset{k}{\equal{}} \{\beta_1, \beta_2, \ldots , \beta_s\}\]  and $ s \leq k,$ then there exists a permutation $ \pi$ of $ \{1, 2, \ldots , s\}$ such that

\[ \beta_i \equal{} \alpha_{\pi(i)} \quad \forall i \equal{} 1,2, \ldots, s.\]

For $ s \le k$ the values of $ \sum_{i\equal{}1}^{s} \alpha_i^j, j \equal{} 1, 2, ... k$ uniquely determine the values of the entire sequence $ s_n \equal{} \sum_{i\equal{}1}^{s} \alpha_i^n$, which has a linear recurrence given by the coefficients of the polynomial $ P(x) \equal{} \prod_{i\equal{}1}^{s} (x \minus{} \alpha_i)$.  It follows that

$ \prod_{i\equal{}1}^{s} (x \minus{} \alpha_i) \equal{} \prod_{i\equal{}1}^{s} (x \minus{} \beta_i)$.


Prove the following inequality:
Given $ a,b,c > 0$ (Sums are cyclic)
$ \sum\frac {a^m}{b^n} \geq \sum\frac {a^p}{b^q}$
if
$ m \minus{} n \equal{} p \minus{} q\geq 0$
and $ m\geq p$
Generalize

[quote="Nemion"]Prove the following inequality:
Given $ a,b,c > 0$ (Sums are cyclic)
$ \sum\frac {a^m}{b^n} \geq \sum\frac {a^p}{b^q}$
if
$ m \minus{} n \equal{} p \minus{} q\geq 0$

Generalize[/quote]
Id est, $ \sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{a^2}{b^2}.$ Is it true? :wink:

[quote="arqady"]Id est, $ \sum_{cyc}\frac {a}{b}\geq\sum_{cyc}\frac {a^2}{b^2}.$ Is it true? :wink:[/quote]

Sorry argady, you are right, I omitted a condition. I've corrected the phrasing

By Chebyshev and AM-GM

\[ \sum\frac {a^m}{b^n}\ge \frac{1}{3} (\sum \frac{a^{m\minus{}p}}{b^{n\minus{}q}} )(\sum \frac{a^p}{b^q})\ge\sum \frac{a^p}{b^q}
\]

[quote="SplashD"]By Chebyshev and AM-GM
\[ \sum\frac {a^m}{b^n}\ge \frac {1}{3} (\sum \frac {a^{m \minus{} p}}{b^{n \minus{} q}} )(\sum \frac {a^p}{b^q})\ge\sum \frac {a^p}{b^q}
\]
[/quote]
Can you show me the way to rearange to use chebyshev?

[quote="manlio"]I don't think you can use Chebychev. :wink:[/quote]
Yes,I think like you too.


Determine the convergence or the divergence of the following series:

$\frac{n^{2}+1}{n^{3}+2n+4}$

where n=0 to infinity.

Diverges by comparison to $\frac{1}{n}$.


limit
trigonometry
real analysis
real analysis unsolved

Let $\{x_{n},\, n\ge1\}\subset\mathbb{R}$ be bounded sequence and $a$ be real number such that $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}x_{k}^{j}=a^{j},\ j=1,2.$ Prove that $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sin x_{k}=\sin a.$

consider $b_{k}=a_{k}-a$, that satisfies $\sum_{1}^{n}b_{k}=o(n)$, $\sum_{1}^{n}b_{k}^{2}=o(n)$, $|b_{k}|$ is bounded by a constant $M$. If one can shows that for every such sequence $b_{k}$ then $\lim_{n \to \infty}\frac{\sin(b_{k})}{n}=0$ and $\lim_{n \to \infty}\frac{\cos(b_{k})}{n}=1$, then the result will follows. But this is quite easy because one can find a constant $C$ such that $|x-\sin(x)|< C x^{2}$ and $|\cos(x)-(1-\frac{x^{2}}{2})|< C x^{2}$ for $|x|<M$ .


[quote="stephen38"]Prove that for all positive numbers $ a,b,c$,we have
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\geq\frac{3\left(a^{3}\plus{}b^{3}\plus{}c^{3}\right)}{2\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)}.$[/quote]
Use old inequality $ \sum\frac{a^{2}}{b\plus{}c}\geq\frac{a\plus{}b\plus{}c}{2}$
So we need prove that
$ \frac{a\plus{}b\plus{}c}{2}\geq\frac{3(a^{3}\plus{}b^{3}\plus{}c^{3})}{2(a^{2}\plus{}b^{2}\plus{}c^{2})}$
or $ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})\geq 3(a^{3}\plus{}b^{3}\plus{}c^{3})$
(Chebushev)

[quote="chien than"]
or $ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})\geq 3(a^{3}\plus{}b^{3}\plus{}c^{3})$
(Chebushev)[/quote]
This inequality is wrong. :rotfl:
Right version:
$ (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})\leq 3(a^{3}\plus{}b^{3}\plus{}c^{3})$
which is true by Muirherd.
To stephen38:
I think,that your inequality can be solved by S.O.S

$ LHS\minus{}RHS \equal{} 2(a^{2}\plus{}b^{2}\plus{}c^{2})\sum a^{2}(a\plus{}b)(a\plus{}c)\minus{}3(a^{3}\plus{}b^{3}\plus{}c^{3})(a\plus{}b)(b\plus{}c)(c\plus{}a) \equal{}$

$ \equal{} 2\left((a^{3}\plus{}b^{3}\plus{}c^{3})^{2}\minus{}(a^{4}\plus{}b^{4}\plus{}c^{4})(ab\plus{}bc\plus{}ca)\right)\plus{}\sum a^{3}(b\plus{}c)(a\minus{}b)(a\minus{}c)\geq 0$.

Hence, you only have to prove that

$ (a^{3}\plus{}b^{3}\plus{}c^{3})^{2}\geq (a^{4}\plus{}b^{4}\plus{}c^{4})(ab\plus{}bc\plus{}ca)$.

Does anybody have a nice prove of the last inequality?

[quote="xavi"]$ LHS\minus{}RHS \equal{} 2(a^{2}\plus{}b^{2}\plus{}c^{2})\sum a^{2}(a\plus{}b)(a\plus{}c)\minus{}3(a^{3}\plus{}b^{3}\plus{}c^{3})(a\plus{}b)(b\plus{}c)(c\plus{}a) \equal{}$

$ \equal{} 2\left((a^{3}\plus{}b^{3}\plus{}c^{3})^{2}\minus{}(a^{4}\plus{}b^{4}\plus{}c^{4})(ab\plus{}bc\plus{}ca)\right)\plus{}\sum a^{3}(b\plus{}c)(a\minus{}b)(a\minus{}c)\geq 0$.

You only have to prove that

$ a^{3}\plus{}b^{3}\plus{}c^{3}\geq (a^{4}\plus{}b^{4}\plus{}c^{4})(ab\plus{}bc\plus{}ca)$.[/quote]
If i right understand you said that:
$ \sum a^{3}(b\plus{}c)(a\minus{}b)(a\minus{}c)\geq 0$ for all $ a,b,c>0$.Why? :maybe:

It is Vornicu-Schur inequality, with
$ x\equal{}a^{3}(b\plus{}c)$,    $ y\equal{}b^{3}(c\plus{}a)$   and    $ z\equal{}c^{3}(a\plus{}b)$.

[quote="Erken"]
To stephen38:
I think,that your inequality can be solved by S.O.S[/quote]
I'll try it. :D

[quote="stephen38"]Prove that for all positive numbers $ a,b,c$,we have
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\geq\frac{3\left(a^{3}\plus{}b^{3}\plus{}c^{3}\right)}{2\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)}.$[/quote]
See also here ( page 2 ):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86201


function
real analysis
real analysis unsolved

Is there any general form of multiplicative periodic function $ f: \mathbb{N}\to\mathbb{R}$ ?

(I mean $ f: \mathbb{N}\to\mathbb{R}$ such that $ f(xy)\equal{}f(x)f(y)$ for all $ x,y\in\mathbb{N}$ and there exists $ T\in\mathbb{N}$ such that $ f(x\plus{}T)\equal{}f(x)$ for all $ x\in\mathbb{N}$)

Let's see ...

$ f(2T) \equal{} f(2)f(T)$ and $ f(2T) \equal{} f(T \plus{} T) \equal{} f(T)$, so either $ f(T) \equal{} 0$ or $ f(2)\equal{}1$.

Now consider whether $ T$ is prime or composite.

Just some thoughts.


AMC
USA(J)MO
USAMO
geometry
circumcircle
trigonometry
power of a point

[img]http://img367.imageshack.us/img367/4853/usamo6eb7.jpg[/img]

Took me like half an hour to draw that during the test lol.
So does anyone have a full solution?

are PA, QA, and the circumcenter collinear? I totally guessed that as a joke

It's colinear when ABC is equilateral. Thats how you prove equality.

umm don't they have to be colinear since both the circles are internally tangent to the circumcircle?

Another sketch, showing some colinear centers and points of tangency.  Unless I misread the problem, points $A,\thinspace P_{A},\thinspace Q_{A},\thinspace\text{and the circumcenter}$ are colinear, even if $\triangle ABC$ is not equilateral.

here is a quick proof that $A,P_{A},Q_{A},O$ are collinear.

Consider $\omega_{A}$ and $\Omega$.  It is obvious that they are tangent at $A$ (given).  Hence the radical axis of $\omega_{A}$ and $\Omega$ passes through $A$, is tangent to both circles, and is perpendicular to $P_{A}O$.  But a tangent to $\Omega$ at $A$ is perpendicular to $AO$, hence $P_{A}O$ and $AO$ are parallel and $P_{A}$ lies on $AO$.  an analogous proof using $\Omega_{A}$ and $\Omega$ shows that all four points are collinear.

i think this would suffice...

For an even quicker proof, they are all homothetic about $A$.

oh, It's too hard....
I spent almost 20 minute to draw a graph, then I found out it's too small that some segment intersect...
Another 30 minutes past.
It's too waste of time.
...
By the way, how many problems can you solve(round your answer to the nearest tenth...)

when drawing a diagram, it is not always best to draw everything since it is symmetrical it would suffice to draw one pair of the tangent circles, your diagram gets too confusing otherwise and you miss the important facts that can be gotten from a diagram, in particular, the collinearity of various circle centers

Can someone explain to me how the diagram can be constructed?

ok first, you can construct the incircle and the circumcircle

now draw AO

draw the line, L, perpendicular to AO and tangent to the incircle [the one farther away from A], tangent at T
[you should try and figure that out...it is relatively easy...]

consider an inversion centered at A passing through E,F [points of tangency of the incircle with AC,AB]

this maps w_A to L and the incircle to itself....let T' be the point that w_A touches the incircle, note that inversion preserves collinearity, so A, T, T' are collinear....and T' lies on the incircle, so intersect AT to intersect the incircle again to construct T'

the center of w_A is on AO and the circle passes through A, T', so take the perpendicular bisector of AT' intersect L, and that is your center

the other circle is constructed similarly; use the line perpendicular to AO on the other side of the incircle

Thanks, but how do you know that the line goes to the circle with the inversion? I'm not very good with geometry.

a circle passing through the center of inversion becomes a line under inversion

consider the circle in polar; $r=\cos\theta$

we have $OP*OP'=k^{2}$ where $k$ is the radius of the inversion circle

inversion in polar is relatively easy, $r'=\frac{k^{2}}{\cos\theta}\implies k^{2}=r\cos\theta$ which is a line [transform into cartesian]


AoPS would like to offer a few prizes to students who work on their schools' math team web pages.  However, we don't want to judge such a competition, so this is what we propose:

A student or a small panel of students judge a student contest for web page building.  AoPS will offer a $ UNIQ97841317b - QINU40 and two$\$25 gift certificates to the web design contest winners good for AoPS classes or books.  The goal should be relatively simply -- to build a single [b]practical[/b] math team page.

Any students who want to judge can create their own judging criteria, but we want the practicality aspect to drive the web page building.  Web pages should be linked from the school's website if at all possible and should contain information useful to all math team students.  They should be easy enough to read and navigate and not use backgrounds or graphics that make users avert their eyes in pain.  Popup ads should also be considered a major deduction because, well, they're irritating.

Note that AoPS will not host these pages, but schools might.

If students want to revamp old pages, it's probably a good idea to take before pictures of the site for comparison.

Are there any questions and is there a student willing to organize and judge this little community contest?

What was the motivation for this contest?

Edit--I meant for Mr. Crawford and Mr. Rusczyk to hold it in the first place.

[quote="MCrawford"]A student or a small panel of students judge a student contest for web page building.  AoPS will offer a $\$40 and two $\$25 gift certificates to the web design contest winners good for AoPS classes or books. [/quote]

That's the motivation I think, aside from bragging rights of course.

Anyways, I'd be willing to work on it after AP week is over. I'll have free time for the next several weeks and a bit in June.

Our mission is to build up the entire culture of problem solving, particularly through mathematics.  That means raising the profile of math students and math team is one of the most universal ways to do that.

Math team sites help parents and students know about advanced programs and that's a good thing in every community.  Most of these parents and students will not know what resources are available (I can't count the number of people I have helped find the AMC and MATHCOUNTS for instance).  Adding web connectivity between good students and good programs is essential to building a greater educational culture.

Also, building a webpage is also something virtually any computer savvy student should do at least once.  Many students here will be involved in computers in the future.  This is just an excuse for some people to learn or gain further experience.

When discussing this contest, we didn't want to make it too formal and don't expect a ton of submissions.  If a dozen students create math team pages, we'll consider that a whopping success and one that we're willing to incent with a few prizes.  On the other hand, if 50 students decided to build a page for their math team, we'd be surprised and overjoyed.

do we get a prize if our criteria gets chosen?

You just get the joy of lording over your peers as a judge.

Interested?

This is kind of intersting, I already discussed in making a math site in this forum. But it was mostly for the purpose of my own reasons, not really schools. Does it really have to be linked to our school website, I myself am sort of embaressed that I'm making a math website, not that I don't like math. It's just embaressing for me.

I actually sort of already have made my own math website:
https://207.234.184.170/lordsofperil.com/math/

My school doesn't really have a math club or team or anything that meets regularly so I don't think it would be very practical to create a website for them and I would be embarassed to do so like viscount.

Can it be a site devoted to mathematics, but maybe ++ points if it is for your school/club/team?

Edit: I will be unable to enter the competition...

Too bad people in our school aren't passionate about mathematics. Our principal canceled the Mathcounts here last year because our coach retired.

Richard can probably guess the school I'm in, since i'm in San Diego.

[hide]Marshall Middle School[/hide]

I've been working on a school Math [i]Club[/i] website for next year*, though it's probably quite cheesey and is by no means done.
*This year's stuff is on there, but I'm hoping it will allow for more natural use in the future (and will initially help).
[hide="If you really want to see it."][url=http://cc.usu.edu/~edrf]http://cc.usu.edu/~edrf[/url][/hide]
[Edit:} Sorry guys; this hasn't been up I think all summer.  I'll be looking at it so they can have things together for this year.  This month, the site should move as well. {]

I'd be glad to judge if you'd take more to do it.  What I'd be really interested in is seeing what other mathers do.  Even though our school is not overly math-savvy (mostly, I think, because of distance from anything but the Colorado State line) and its own website needs work, I'd love to leave a decent math page even if it did serve only links.  Maybe I could even use links to some of these other members' pages--which I'll take a look at straightaway.

Eddy

[quote="SnowStorm"]So does it have to be for a team or can it be about math in general?[/quote]

In this case it should be math team or math community based.  The idea behind the contest is to create more of a link between math teams/math programs/math students and any kinds of ideas/resources/math that might be helpful to them.  You can certainly include any kind of math you want, but making it functional for local students is a high priority.

[quote="MCrawford"]Rep123Max is going to be the judge.  You can talk to him about helping out if you like.[/quote]Cool; and thanks for that last explanation.  Good luck, all. [hide="""]:coolspeak: Let's just see if this works.  Those new smileys rock!  There's a hockey one even--!
 (But my linking a Hide to the first emoticon didn't work out.)[/hide]:o

[quote="SnowStorm"]Any idea what date this contest will end?[/quote]How's August?  I don't know if there's any official date set, but I'm thinking it should at least be ready (a site) by the time a person starts school--or maybe a few weeks later if membership, etc. updates will be available to give it a [i]feel[/i] of being put together for next year.  --That way, Mathcampers won't take any disadvantages for being busily away :P.

Yea, that sounds good.  The one I've been working on will probably take most of summer to complete.

Rep123Max has posted some information [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36399]here[/url].  He plans to accept submissions until August 19 and says he will provide more information after ARML.

Results of the contest can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=314548#p314548]here[/url].

Nooooo, this is over?  A friend and I are about to start a team in high school, and I rock at design.  Oh well.

Only about a year and a quarter over :P

Watch for post dates when reading recently unanswered posts, on the forum its realyl easy to bump such an old topic; I bumped a year old topic once by accident also.


geometry
geometric transformation
function
real analysis
real analysis unsolved

Suppose $ f\in L^p({{\mathbb R}})$ , and $ g: {{\mathbb R}}\to L^p$ be defined by
$ g(x) \equal{} f_x$ where $ f_x(y) \equal{} f(x \plus{} y)$ . show that $ g$ is continuous.

key phrase: translation is continuous in the $ L^p$ norm.  try the search.

The key idea in the proof: For $ 1\le p < \infty,$ $ L^p$ has "nice" dense subspaces. Pick such a nice dense subspace. Let's say, step functions (finite linear combinations of the characteristic functions of bounded intervals). Show by direct calculation that the result you want holds for such nice functions; then use the density to complete the proof.

Note: $ L^{\infty}$ does not have "nice" dense subspaces (this is essentially a manifestation of $ L^{\infty}$ not being separable) and this translation result is false for $ L^{\infty}.$


Who's received their AMY packet?
What subject is it?

I just got mine today, and I got the pidgeonhole packet. My friend got the pidgeonhole packet around a week ago.

I got mine about 2 weeks ago.  I haven't had a lot of time to do the problems, but they are very good ones.

I haven't gotten any packet at all :(
I got the tuition receipt confirmation.

I got mine, last week, on pigeonhole. I haven't had much time, only had a chance to do the "easy" problems that I could solve on sight.

I just got it, pigeonhole
These are from AMY

As bpms pointed out, the packet referred to here is for the AMY. AMSP packets will go out in May or June.

Well, yes, that's why the title is [quote]
So who got an AMY packet?[/quote]
And the first line of the post is
[quote]Who's received their AMY packet? [/quote]

I think there is something wrong with Problem 12 on Medium...

In response to number 12, I think that it may involve something like if you had 7 points, then the statement would be "at most sq(5)." Perhaps they mean that you can place them so that you are guaranteed to have all separated by a distance of at least sq(5).

That's what I was thinking. Like for that problem, you could just stick all the points near the corner so they're really close. And for #13, it doesn't hold for a regular pentagon.

I looked at it and I don't know what they mean.  Perhaps just exclude the pentagon, and try it that way.  If not, then I have no idea.


AMC
AIME
USA(J)MO
USAMO
FTW
THIS IS A RANDOM TAG

AIME and USAMO. You can add additions to this thread if you want.

I say A-I-M-E and U-S-A-M-O. My friends and my brother all pronounce the first one "Amy", though.

I pronounce it "Amy" and A-I-M-E but more often "Amy." Most people I know say A-I-M-E. All people I know (including myself) say U-S-A-M-O.

Addition: How to you say AoPS?

I say A-O-P-S, although I know people who say "Ay-ops."

I say A-I-M-E and Amy, for USAMO, I say U-S-A-M-O.

I usually say Ay-ops when talking to people. But if they then say "Huh?", I say A-O-P-S (thought they usually still don't know what I'm talking about)  :lol:

Amy, Usamo, Aops.

those are the pronunciations I use.

"Amy"

"usamo" as in the pronounciation of the nosotros form of "usar" in spanish (minus the 's' at the end)

"Ay-Ops"

[quote="7h3.D3m0n.117"]
"usamo" as in the pronounciation of the nosotros form of "usar" in spanish (minus the 's' at the end)
[/quote]

now you reminded me to do my Spanish homework..
but yeah.. Usar is a verb??

i used to say "kill-o-meter"
but now I say "kil-ah-me-tor"

[quote="kimmystar94"]
now you reminded me to do my Spanish homework..
but yeah.. Usar is a verb??[/quote]

usar - to use

Spanish III FTW  :P .

Aim. Us-ah-mo. Well, never even pronounced it though...Kih-lah-meh-ter.

How do you pronounce Tuesday? :huh: My teacher pronounces it as "Toos-dee"

[quote="pianoforte"]"Toos - dee"?  :rotfl:[/quote]

Well, he's also 51 years old so that could be his way of speaking or where he grew up and got that word pronunciation.

The Zuton Force, to me accenting the second syllable is far more natural. What puzzles me is the pronouciation of the "u" as "oo" and not "yoo". That sounds very un-English.

[quote="The Zuton Force"][quote="7h3.D3m0n.117"]"usamo" as in the pronounciation of the nosotros form of "usar" in spanish (minus the 's' at the end)[/quote]

Are you telling me you accent the second syllable?

Most people I know that pronounce it say YOU-suh-mow

Personally, I think that with a little practice, spelling it out can be done just as dynamically  :D[/quote]

My programming teacher says you-SAY-mo sometimes.

USAMO: Yoo-sah-mo
AIME: Aymee
Aops- ay-ops

[quote=reaganchoi][quote=reaganchoi]Please don't revive.[/quote][/quote]

Why not?

I say [b]A[/b]-O-[i]P[/i]-S, [b]A[/b]-I-[i]M[/i]-E  and [b]U[/b]-S-[i]A[/i]-M-[i]O[/i]
[i]italic[/i] is stressed
[b]bold[/b] more stressed than [i]italic[/i]
I pronounce the first syllable [i]neither[/i] like [i]knee[/i]
I also say [b]A[/b]-O-P-[i]S[/i]

[quote=andersonw]I say "aime", as in the french word for "like". Its pronounced like the letter m.

I spell out USAMO.

I just say "Art of Problem Solving".  :P

How do people here pronounce "kilometer"?[/quote]

kill-lawn-mehter


For a,b,c positive real number prove that

 (1/a + 1/b + 1/c)(1/(1+a) + 1/(1+b) + 1/(1+c)) \geq 9/(1+abc)

This is an interesting inequality of Walter Janous who gives a neat proof.

the solution for this one and a problem related to this one can be found here:
http://www.mathlinks.ro/phpBB/viewtopic.php?t=1339&highlight=

cheers! :D :D


Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ a\sqrt {b^2 \minus{} 3bc \plus{} 9c^2} \plus{} b\sqrt {c^2 \minus{} 3ca \plus{} 9a^2} \plus{} c\sqrt {a^2 \minus{} 3ab \plus{} 9b^2} \leq\ 9$
 :)
And the stronger seems trues but I haven't found the proof for this stronger:
$ a\sqrt {b^2 \minus{} bc \plus{} 9c^2} \plus{} b\sqrt {c^2 \minus{} ca \plus{} 9a^2} \plus{} c\sqrt {a^2 \minus{} ab \plus{} 9b^2} \leq\ 9$
(with the same  condition)  :maybe:
@can_hang: [hide]anh ko dc giai? 2 bai` nay` dau nhe', em sap' ra bai` manh. hon roi`  :oops: [/hide]

[quote="nguoivn"]And the stronger seems trues but I haven't found the proof for this stronger:
$ a\sqrt {b^2 \minus{} bc \plus{} 9c^2} \plus{} b\sqrt {c^2 \minus{} ca \plus{} 9a^2} \plus{} c\sqrt {a^2 \minus{} ab \plus{} 9b^2} \leq\ 9$
(with the same  condition)  :maybe:[/hide][/quote]
It is valid, nguoivn. My hint is to use Cauchy Schwarz and
$ \frac{1}{b\plus{}3c}\plus{}\frac{1}{c\plus{}3a}\plus{}\frac{1}{a\plus{}3b} \ge \frac{3(a\plus{}b\plus{}c)}{a^2\plus{}b^2\plus{}c^2\plus{}3ab\plus{}3bc\plus{}3ca}.$
;) :)

[quote="can_hang2007"]It is valid, nguoivn. My hint is to use Cauchy Schwarz and
$ \frac {1}{b \plus{} 3c} \plus{} \frac {1}{c \plus{} 3a} \plus{} \frac {1}{a \plus{} 3b} \ge \frac {3(a \plus{} b \plus{} c)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 3ab \plus{} 3bc \plus{} 3ca}.$
;) :)[/quote]
Very, very nice hint  :lol: .
@can_hang: [hide]Anh dung la` thien tai`, dang. bo? de` nay` cho ta rat nhieu` bai toan moi' :-j[/hide]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ a\sqrt {b^2 \minus{} 3bc \plus{} 9c^2} \plus{} b\sqrt {c^2 \minus{} 3ca \plus{} 9a^2} \plus{} c\sqrt {a^2 \minus{} 3ab \plus{} 9b^2} \leq\ 9$
 :)
And the stronger seems trues but I haven't found the proof for this stronger:
$ a\sqrt {b^2 \minus{} bc \plus{} 9c^2} \plus{} b\sqrt {c^2 \minus{} ca \plus{} 9a^2} \plus{} c\sqrt {a^2 \minus{} ab \plus{} 9b^2} \leq\ 9$
(with the same  condition)  :maybe:
[/quote]
Let $a,$ $b$ and $c$ are non-negatives such that $ab+ac+bc\neq0.$
By Cauchy-Schwartz we obtain:
$\left(\sum_{cyc}a\sqrt {b^2 \minus{} bc \plus{} 9c^2}\right)^2\leq\sum_{cyc}\frac{a(b^2-bc+9c^2)}{b+3c}\cdot\sum_{cyc}a(b+3c).$
Hence, it remains to prove that
$(a+b+c)^4\geq4(ab+ac+bc)\sum_{cyc}\frac{a(b^2-bc+9c^2)}{b+3c},$ which is equivalent to
$\sum_{cyc}(3a^6b+9a^6c+21a^5b^2+39a^5c^2+6a^4b^3-78a^4c^3+$
$+76a^5bc+85a^4b^2c+7a^4c^2b-108a^3b^3c-60a^3b^2c^2)\geq0,$
which is true by $AM-GM.$


geometry circumcircle trigonometry search AMC USA(J)MO USAMO

D lies inside the triangle ABC. ∠BAC = 50 degrees. ∠DAB = 10 degrees, ∠DCA = 30 degrees, ∠DBA = 20 degrees. Find the degree of angle DBC.

One of my friends did this problem like a while back using construction, but I cant seem to know what to construct/where.  When you get a problem like this, what is the first step you should generally take/what should you create/draw in in the picture to make the problem easier?

why don't you just go ask lyndon?  :)

i remember he made an equilateral or isoceles triangle.

Constructing auxiliary figures depend on what you need to prove or achieve something. You wouldn't want to draw unnecessary figures that would not be useful although it doesn't hurt to experiment by drawing a few experimental ones. When I get a problem like this, always map out what is given and what your goal is.

My way of solving this problem is the following:

Draw the triangle and the relevant segments from D.
Then, angle-chase to find $ \angle DAC\equal{}40^\circ$.
Use the theorem 'sum of all angles of a triangle is 180' to get $ \angle BDA\equal{}150^\circ$ and $ \angle CDA\equal{}110^\circ$ so that $ \angle BDC\equal{}100^\circ$.
At this point, it seems like a dead-end. Look closely what other information you can draw out of the figure. This is the hard part because the unexperienced may not be able to recognize geometrical facts that quickly. Below, I'm hiding the key step and the rest of the solution to solve this problem.
[hide="Hidden step"]Note that $ \angle BAC\equal{}50^\circ$ and $ \angle BDC\equal{}100^\circ$. What geometrical fact connects between these two angles, when the numbers are "right"? Notice that $ \angle BDC\equal{}2\angle BAC$? Does this look familiar? What does this suggest?
[hide="Rest of the solution"]D is the center of the circumcircle. Since $ DB\equal{}DC$, $ \angle DBC\equal{}40^\circ$.[/hide][/hide]

[img]http://myfiles.zapto.org/LotsofAngleChasing.gif[/img]

A bit of chasing evidently shows that $ \angle DAC \equal{} 40$. Therefore, $ \angle BDA \equal{} 150$, $ \angle CDA \equal{} 110\implies\angle BDC \equal{} 100$. Now, since $ D$ is the circumcenter, we can see that $ DB \equal{} DC$, and $ \measuredangle DBC \equal{} 40$.

EDIT: grr 10000th user beat me

wow. I wasn't even close  :rotfl: .

Would there be a way to do if the numbers were different?  In other words, if D was not the circumcenter?

Why is $ 2\angle BAC \equal{} \angle BDC$ sufficient to show that D is the circumcenter of the triangle?

uhh sorry to tell you guys but the answer is NOT $ 40$ degrees.

unimpossible: yea i could go ask lyndon (he would do it in 2 mins), but that would be in 2 weeks...unless I pm him here but this is a good technique that really helps for AIME/USAMO (so everyone should try it)...I tried equilateral triangles but it didnt work, so I was just wondering can anyone get me a start with the first construction you need to make (if it is indeed the equilateral triangle)
Also, i tried trig, and i spent 2 whole pages of paper for nothing...

[hide="Problem with the logic for D being circumcenter"]
1. If D were the circumcenter, all three of the triangles the three circumradiuses break ABC into must be isosceles.  Clearly, from the given angles, this is NOT the case.

2. Obviously, since DCA is 30 degrees, and DAC is 40 degrees, then CDA is 110 degrees.

If that were the case and D is the circumcenter, then CBA would be 1/2*110=55 degrees.  Meaning, DBC=55-20=35 degrees (DBC=CBA-DBA and DBA=20 degrees)

However, also if it were the circumcenter, then DB=DC, so from that, then DBC is 40 degrees.

Since 35 degrees isnt 40 degrees, D is NOT the circumcenter.
It is just an arbitary point.
Still saying all that, I think theres something special about D, but its not because its the circumcenter.[/hide]

Why is trig hard?

Use law of sines to find all sides in terms of one variable.

Then use law of cosines until you find something interesting about the triangle.

[quote="the future"]
yea i could go ask lyndon (he would do it in 2 mins), but that would be in 2 weeks...unless I pm him here but this is a good technique that really helps for AIME/USAMO (so everyone should try it)...[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=111601946&t=185447]Here[/url]  :)

whoa thanks
i didnt know he also posted the problem here
sorry for the double post.

Is this really USAMO like mathcrazed claims?

Also, has anyone found a different solution? Phelpedo said this problem is trivial using the trig form of Ceva's but I haven't found the answer yet...


calculus
derivative
function
real analysis
real analysis theorems

Here's a nice theorem:

If $f \colon \mathbb{R}^{n}\to \mathbb{R}$ has continuous second partial derivatives at any given point in $\mathbb{R}^{n}$, say, $(a_{1}, \dots, a_{n})$, then for $1 \leq i,j \leq n$, \[\frac{\partial^{2}f}{\partial x_{i}\, \partial x_{j}}(a_{1}, \dots, a_{n}) = \frac{\partial^{2}f}{\partial x_{j}\, \partial x_{i}}(a_{1}, \dots, a_{n}) .\] I'd like to see a proof of this  :D

Rather known as Schwarz theorem. You need only assume your function is two times differentiable.

Really, how do you prove such a nice result?

I was surprised by your question ? You can find a proof in any analysis book (this is rather tedious to type and uses the mean value theorem, but that's about it. You  need to approximate the second derivatives with the help of a symmetric expression. If you really cannot find a proof, I'll post one later (I'm in school now)).


Find a conformal maping from the region between the lines $ y\equal{}x$ and $ y\equal{}x\plus{}2$ to the upper half plane that also sends the 0 to 0.

rotate by $ \pi/4$ then scaly by factor of $ 2/\pi$ then apply $ e^z$.
I think you might have to subtract one to make the origin work.


geometry
trapezoid
perimeter
geometry solved

Consider Circle O.  AB is a diameter of circle O, and P is the midpoint of one of the arcs AB. A second circle is located so that its center is at P and the circumference of the circle contains the points A and B. If the radius of circle O is 20, then the area of the part of circle O that does not lie inside circle P is A. Find the greatest integer less than or equal to A.

Construct an isosceles trapezoid such that the lengths of all the sides of the trapezoid are integers, and the lengths of all the diagonals of the trapezoid are integers. If the perimeter of the trapezoid is 33 and the diagonals have length d, then what is the sum of all possible values of d?

[color=blue][b](1)[/b][/color] Let $r = OA = \frac{AB}{2} = 20$ be the radius of the smaller circle $(O)$, $R = PA = r \sqrt 2$ the radius of the larger circle $(P)$ and $a = AB = 2r$ the side of a square inscribed in the larger circle $(P)$. The area of $(O)$ not part of $(P)$ is

$A = \frac 1 2 r^2 - \frac 1 4 (\pi R^2 - a^2) =  \frac 1 2 r^2 - \frac 1 4 (2\pi r^2 - 4r^2) = r^2 = 20^2 = 400$

[color=blue][b](2)[/b][/color] Let $a \ge c$ be the parallel bases of the trapezoid, $b$ the non-parallel equal sides and $d$ its equal diagonals.  The perimeter of the trapezoid is $a + c + 2b = 33$. Since $\frac{a - c}{2} < b$, the maximum value of $a$  is $a < 2b + c = p - a = 33 - a$, $a < \left[\frac{33}{2}\right] = 16$. If $a$ is even, $c$ mus be odd and the other way around. Hence, the possibilities are:

[color=red][b](a)[/b][/color] $a = 2m$ is even, $c = 2n + 1$ is odd:

$2m + 2n + 1 + 2b = 33$
$m + n + b = 16$

Since $1 \le c < a \le 16$, the integers $m, n$ are bound by $0 \le n < m \le 8$.

[color=red][b](b)[/b][/color] $a = 2m + 1$ is odd, $c = 2n + 1$ is even:

$2m + 1 + 2n + 2b = 33$
$m + n + b = 16$

Since $1 < c < a < 16$, the integers $m, n$ are bound by $0 < n \le m < 8$

This gives a total of 64 possible combinations of $a, b, c$. By Ptolemy's theorem,

$ac + b^2 = d^2$

and since $a, b, c, d$ are all integers, the expression $ac + b^2$ must be a full square. The only combination that satisfies this condition is $a = 12, b = 6, c = 9$, which gives $d = 12$:

$a + c + 2b = 12 + 9 + 12 = 33$

$ac + b^2 = 108 + 36 = 144 = 12^2$

Two questions:

1.) So the the sum of all possible values of d is $144$?

2.) Say the perimeter was 66 instead of 33. Would the answer then be $288$ since $66 = 2 \cdot 33$, and $288= 144 \cdot 2$?

[quote="PeNtaGoNaL"]1.) So the the sum of all possible values of d is $144$??[/quote]
There is only one possible value $d = 12$. The sum of 12 is 12. The last 2 lines are just a check that the calculated integer values of $a, b, c, d$ satisfy the problem condition (the perimetr is 33) and Ptolemy's theorem (an isosceles trapezoid is always cyclic).

[quote="PeNtaGoNaL"]
2.) Say the perimeter was 66 instead of 33. Would the answer then be $288$ since $66 = 2 \cdot 33$, and $288= 144 \cdot 2$?[/quote]
If the perimeter was 66, then $\frac{a - c}{2} < 2b$ gives the upper limit $a < 2b + c < 66 - a$, $a < \frac{66}{2} = 33$. From the perimeter equation $a + c + 2b = 66$, it follows that $a, c$ have to be either both even or both odd.

[b][color=red](1)[/color][/b] If $a = 2m$, $c = 2n$ are both even:

$2m + 2n + 2b = 66$
$m + n + b = 33$

Since $1 < c \le a < 33$, the integers $m, n$ are bound by $0 < n \le m \le 16$.

[b][color=red](2)[/color][/b] If $a = 2m + 1$, $c = 2n + 1$ are both odd:

$2m + 1 + 2n + 1 + 2b = 66$
$m + n + b = 32$

Since $1 \le c \le a < 33$, the integers $m, n$ are bound by $0 \le n \le m < 16$.

This gives a total of 272 possible combinations of $a, b, c$. By Ptolemy theorem, $ac + b^2 = d^2$. Since $d$ is an integer, $ac + b^2$ must be a complete square. There are 6 possibilities out of 272, namely:

$a = 15,\ c =\ \ 9,\ b = 21:\ \ 15 \cdot\ 9 + 21^2 = 135 + 441 = 576 = 24^2\ \rightarrow\ d = 24$

$a = 21,\ c = 13,\ b = 16:\ \ 21 \cdot 13 + 16^2 = 273 + 256 = 529 = 23^2\ \rightarrow\ d = 23$

$a = 25,\ c = 21,\ b = 10:\ \ 25 \cdot 21 + 10^2 = 525 + 100 = 625 = 25^2\ \rightarrow\ d = 25$

$a = 12,\ c =\ \ 8,\ b = 23:\ \ 12 \cdot\ 8 + 23^2 =\ 96 + 529 = 625 = 25^2\ \rightarrow\ d = 25$

$a = 20,\ c = 12,\ b = 17:\ \ 20 \cdot 12 + 17^2 = 240 + 289 = 529 = 23^2\ \rightarrow\ d = 23$

$a = 24,\ c = 18,\ b = 12:\ \ 24 \cdot 18 + 12^2 = 432 + 144 = 576 = 24^2\ \rightarrow\ d = 24$

Hence, 3 different values $d = 23, 24, 25$ are possible. The answer is 72 and the Answer is [color=blue][size=200][b]42[/b][/size][/color].


geometry
circumcircle
geometric transformation
dilation

$ABC$ is a triangle.$Q$ is a point in the plane of it inwhich $\angle QAC=\angle ACB$.

from $Q$, draw a parallel line with $AB$.and from $B$, draw a paralell line with $AC$.and let them

intersect each other at $P$.

$R$ is a point on $BC$ is which: $\angle QRP=\angle ACB$.

prove that:

the circumcircles of triangles $PQR$ and $ABC$ are tangent to each other.

its a nice problem,enjoy it... ;)

[quote="Ashegh"][quote="JBL"]It is certainly not true as stated -- choose $Q$ outside the circumcircle of $\triangle ABC$, while $R$ is inside it, so the two circles cross.[/quote]

dear, i have been said that $Q$ is in the plane  of triangle $ABC$.

it seems that u didnt read it carefully... ;)[/quote]

JBL is saying that if $Q$ is outside the circumcircle it doesn't work; it's still in the plane of $ABC$. When I draw it on my geometry program, it doesn't seem to ever be tangent... unless I'm misunderstanding the problem.

[quote="Centy"]Maybe he means $Q$ is inside the triangle $ABC$.[/quote]

yes yes exactly what u said...

[quote="JBL"]Yes, Centy, that will do it.  In fact, here is a proof:

[hide]Note that when $Q = A$ we have $P = B$ and $R = C$.  Now, $Q$ is allowed to vary along a cevian of $\triangle ABC$.  Let this cevian intersect the circumcircle at $D$.  I propose that $D$ will always be the point of tangency of the two circles.  First, note that $\angle BDA \cong \angle ACB \cong \angle CAD$ where the first equality is because they share an arc and the second is by the defining property of $Q$.  But then we have that $BD \parallel AC$ (alternate interior angles) and so the point $P$ is in fact on $BD$.  Since $QP \parallel AB$, we have that $\triangle QPD$ is just a dilation of $\triangle ABD$ with center $D$, so the circumcircle of $\triangle QPD$ is a dilation of the circumcircle of $\triangle ABC$ and so is tangent to it at $D$.  Now, it remains to show that $R$ is also on the circumcircle of $\triangle QPD$.  However, this is obvious: the arc $QP$ has the same measure as the arc $AB$, so it is also the locus of points $X$ such that $m\angle PXQ = m\angle ACB$, and we're done.[/hide][/quote]

thanks dear JBL  ;)


how did they do this step??

$ x\equal{}$

$ \frac{C_1Ne^{Nat}}{1\plus{}C_1e^{Nat}}$=$ \frac{Ne^{Nat}}{C_2\plus{}e^{Nat}}$

HOW DID THEY CREATE A NEW CONSTANT C2 FROM THE PREVIOUS STEP?

$ C_2 \equal{} \frac {1}{C_1}$.  There is [i]never[/i] a good reason to use capslock.


geometry
3D geometry
analytic geometry
factorial
rate problems

This is the third contest. Please send your answers to either me or Rep123max. Thanks!

We have a new scoring system for Contest 3.

Right Answer: +5
Omit: 0
Wrong Answer: -2

Here are the problems:

1. If (-5/6)(a^m)(b^4) and (-32)(a)(b^(2n)) are like terms, find m^3 - n^4 + 8(m^2)(n^2).

2. If |8a-72| + |2b+44| + |3c-69| = 0, find a + b + c.

3. Find the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + ..... + 1/(15*16).

4. What is the value of  :sqrt: (13 -  4:sqrt: (10))? (Basically, simplify to where you do not have a square root under another square root sign)

5. Find the value of 151 :^2:  - 149 :^2:  + 201 :^2:  - 199 :^2: .

6. If you drive 300 miles at 30 miles per hour and 600 miles at 20 miles per hour, what is your average rate, in miles per hour, for the 900 miles driven?

7. The diagonal of a cube is 3x. What is the volume of the cube in terms of x?

8. A triangle has vertices with the coordinates A(0,15), B(0,0), C(10,0). Find the sum of x and y such that the point D(x,y) on AC makes the area of TriangleABD and TriangleBCD is equal.

9. How many digits are in the decimal representation of (8^23)(25^33)?

FACTORIAL CHALLENGES:
NOTE: All "!" in this section means factorial, not exlamation point, and only take positive factors into consideration.

10. How many 0's does the decimal representation of (4058!)/(950!*350!) end in?

11. How many odd factors does 13! have?

12. What is the number of perfect square and cube factors of 15!?

13. What is the sum of all perfect square and cube factors of 15!? Please give the answer in decimal representation.

CIRCLE CHALLENGES:
NOTE: If it says CircleM, that means its a circle with the center M.

14. CircleO meets CircleO' at A and B. AC is a diameter of CircleO. Extend CA to meet CircleO' at D. Extend CB to meet CircleO' at E. Connect DE. If CE = 15 and DE = AC = 9, find the length of the radius of CircleO'.

15. AB is a diameter of CircleO. C is a point on CircleO. From C, draw a line perpedicular to AB, meeting AB at D. Using CD as the radius, draw CircleC. CircleC intersects CircleO at p and Q. Connect PQ, which intersects CD at G. If GP = 12, and GQ = 36, find the length of CD.

Thanks, and Happy Holidays!
If you have any questions, please do post them on this topic!

HAPPY HOLIDAYS!!!

It works out nicely the way that it's stated.

MagnusRex wrote:Are you sure there is not a typo in number 9:
[hide]
Changing 8^23 to 8^22 makes the problem a lot lot nicer. Otherwise, it's quite nasty...
[/hide]

i'm tempted to say "so deal with it" but that would be too mean

ach, i am an idiot...i was trying to do it a harder way than i needed to

it's not a typo.

and on 2,

8a is not a two digit number

8a means 8 x a

so if a = 5, then 8a = 8(5) = 40.

These are good problems (even though I will probably do worse than usual....)

Thanks!

This contest will not be over until a week or so after break is over, since not many will like to do this over the break... so take your time.

ok, good, lots of time for me to check my work, lol

(Tare=last minute test taker :))
For 9 does the 0 count? (is it only the digits below the decimal point?)
also for 7 I assumed you were refering to the space diagonal, is that right?

decimal representation means base 10. in a cube when it says diagonal it assumes that you know that it means space diagonal unless it says surface diagonal

surface diagonal could also be called face diagonal.

Wait but when it goes like 0.02498... Do you include the 0 when you're counting the digits?

Tare, perhaps you're reading the problem wrong. It should not end in decimals, and if you're getting the idea that it is in decimals because i said decimal representation, ur wrong.

Neal's right (if you haven't read his post yet)

8 is the decimal representation of 2^3

50 is the decimal representation of 2*5^2

It does not have to have decimals

Oh stupid me I thought there was a slash between the two numbers...I got it now, thanks.


function calculus derivative trigonometry algebra polynomial logarithms

Can someone tell me what(which) family(ies) of functions satisfies the following properties:

It's derivitive has a value of 0 at x=0
It's derivitive is not of the same form as the original function (ie, lnx)
is not a trig function

I am trying to do some curve fitting and need something that satifies all these properties, but can't think of anything. Any help?

Polynomials satisfy the given conditions.

Not quite - the derivitive of a polynomial is still a polynomial. I need the derivitive to be of a different form altogether.

That's really a meaningless condition; we can always redefine the "form" to include the derivatives if we feel like it.

You know what I mean though, yes? Perhaps you could word it a little more clearly for me? :P

If I understood your conditions right then $ \ln{(x^2P(x) \plus{} c)}$ with $ P(x)$ a polynomial and $ c > 0$ would fit

So it does, many thanks :)

c can be < 0, just so long as c!=0, it works


group theory
abstract algebra
number theory
prime numbers
superior algebra
superior algebra unsolved

[quote="Hidden Scofield"]1) I need to find all maximal ideals of $ \mathbb Z_n$ ,I know I need to find all maximal subgroups, but I couldn't do it for $ \mathbb Z_n$. Could u give me any nice hint for it.

2) How to find all prime ideals of $ \mathbb Z_n$  :huh:

 :)[/quote]

Tell us first what you mean by $ \mathbb Z_n$ several algebra authors have several meaning for it. Do you mean $ \mathbb Z /n\mathbb Z$? or do you mean something else? If you mean the former. Then this easy.. as you wanted hint, I will just give a hint:

Know that:
1. Maximal ideals are prime ideals
2. What are the prime ideals of $ \mathbb Z$ .. do you see the relation between prime numbers and prime ideals here? ;) Do you see which of them are maximal?
3. Now the prime and maximal ideals of $ \mathbb Z/n\mathbb Z$ correspond in a natural way with those prime/maximal ideals of $ \mathbb Z$ that contain $ n\mathbb Z$

Sincerely,
Jose Capco


Four points have the coordinates $ A(3,3), B(5,7), C(8,7)$ and $ D(12,3)$. The points $ A'$, $ B', C'$ and $ D'$ are found by multiplying the abscissas of $ A,B,C,$ and $ D$ respectively by $ \minus{}1$. Find the difference in the numbers of square units in the areas of quadrilaterals $ ABCD$ and $ A'B'C'D'$.

abscissas=x coordinate :)
so this is just flipping all the points across the y acciss, so the area remains the same
so obviously, area difference is $ 0$


A two-digit prime number is randomly selected. What is the probability its digits sum to 9?

Any number whose digits add up to nine is divisible by 9, so the probability is $ \boxed{0}$.


algebra function domain AMC USA(J)MO USAMO

The computer security people at the University of Nebraska Lincoln, which hosts our web site, have contacted the AMC office about what appears to be hacking/cracking activity at our web site.  The activity is coming from the Art of Problem Solving domain and consists of repeated attempts to find "hidden" web pages by guessing URLs and addresses.   If this continues, it is possible that the computer security people would blacklist this site and prohibit access.

There are no hidden web sites.  The staff and I are working on the USAMO invitation list, but it requires time and care and cross-checking of several databases of info.  The list is not ready yet.  When it is ready, it will be promptly posted, and it will not exist as a hidden web site before then.

Activity such as this, and the prank web site counterfeited from last year's USAMO invitation list, slow down my ability to get the list done by requiring me to address this issue first.  If it continues, and I can trace it to an individual through weblogs,  I will disqualify that person from the USAMO.

Steve Dunbar
AMC Director

Sorry. [color=white]This is to make the message long enough[/color]

In connection with this, I have removed the offending thread.

Everybody please [b]calm down[/b].  Take a deep breath.  Posting "when will the list come out?" 50 times a day will not make it happen any faster.  Dr. Dunbar and his staff are doing an outstanding job in order to be certain that no mistakes are made, and they will release the list when, and only when, they are satisfied that it is accurate.

To repeat what Dr. Dunbar said, do not:
- post fake "the list is ready!" announcements or fake lists
- complain about how the list isn't ready yet
- look for "hidden" web pages on the AMC web site
- write scripts to check the AMC web site every n minutes for updates
- etc.

Finally, I'd like to apologize (on behalf of AoPS) to Dr. Dunbar and the AMC staff for not putting a stop to this behavior sooner, before things got this far.


geometry
circumcircle
combinatorics unsolved
combinatorics

Every point in the plane is colored in three colors.

(a) Must there be two points of the same color such that the distance between them is $ 1$ unit?

(b) Prove that there must be three points of the same color which form a triangle of circumradius $ 1$ unit and one angle is twice another angle or one angle is three times the other angle.

This is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200021]double post[/url].

I'm sorry. I now realize it's not that easy. Please delete the other post.

Can you solve this problem, JBL? (or any others)

Here is what I have.  It's kind of messy, so if somebody has a better way, I'd like to see it; especially Lemma 1.

Here is part (a).

Suppose the contrary.  Consider arbitrary points $ A$ and $ B$ of distance $ \sqrt{3}$ apart and points $ C$ and $ D$ such that $ ACD$ and $ BCD$ are equilateral triangles.  The $ C$ and $ D$ are both colored differently from $ A$ and differently from each other.  Thus $ B$ must be the same color as $ A$.  Therefore if any two points in the plane are of distance $ \sqrt{3}$ apart, they must be the same color.  It then follows that every point in the plane is the same color, a contradiction.

For part (b) I use a similar strategy.

Suppose the contrary.  We first have some intermediate results.  Let $ A_0, \dotsc, A_8$ be nine points in spaced evenly and in that order on a circle of radius 1.

[asy]defaultpen(.75);

pair A0=expi(0), A1=expi(2pi/9), A2=expi(4pi/9), A3=expi(6pi/9), A4=expi(8pi/9), A5=expi(10pi/9), A6=expi (12pi/9), A7=expi(14pi/9), A8=expi(16pi/9);

draw(circle((0,0),1));

dot(A0);
dot(A1);
dot(A2);
dot(A3);
dot(A4);
dot(A5);
dot(A6);
dot(A7);
dot(A8);

label("$A_0$",A0,E);
label("$A_1$",A1,NE);
label("$A_2$",A2,N);
label("$A_3$",A3,NW);
label("$A_4$",A4,WNW);
label("$A_5$",A5,WSW);
label("$A_6$",A6,SW);
label("$A_7$",A7,S);
label("$A_8$",A8,SE);[/asy]

[b]Lemma 1.[/b]  No four of the $ A_i$ are the same color.

[i]Proof.[/i] Suppose the contrary.  Let the four points of the same color be $ A_a, A_b, A_c, A_d$, with $ a<b<c<d$.

First, suppose no two points interecept an arc as small as $ 2\pi/9$.  Then the arcs that adjacent points intercept must be of lengths $ 4\pi/9, 4\pi/9, 4\pi/9, 6\pi/9$.  By excluding one of the points separating two consecutive arcs of length $ 4\pi/9$, we obtain a monochromatic triangle in which one angle is twice another, a contradiction.

Next, suppose that two, say $ A_a$ and $ A_b$, intercept an arc of length $ 2\pi/9$.  Then neither $ \widehat{A_aA_d}$ nor $ \widehat{A_bA_c}$ can have a measure of $ 4\pi/9$ or $ 6\pi/9$.  But one of these arcs, say $ \widehat{A_bA_c}$, must have measure less than $ 8 \pi/9$, so it must have measure $ 2\pi/9$.  Without loss of generality, let $ A_d$ be closer to $ A_c$ than to $ A_a$.  By considering triangle $ A_aA_bA_d$, $ \widehat{A_cA_d}$ cannot have length $ 2\pi/9$ or $ 4\pi9$; by considering triangle $ A_bA_cA_d$, it cannot have length $ 6\pi/9$.  Thus its length is at least $ 8\pi/9$, so $ A_d$ is in fact closer to $ A_a$ than to $ A_c$, a contradiction.  $ \blacksquare$

[b]Lemma 2.[/b]  If $ A_0$ and $ A_3$ are the same color, then $ A_6$ also shares this color.

[i]Proof.[/i] Let us say that the common color of $ A_0$ and $ A_3$ is blue.  Then no other point, except $ A_6$, may be colored blue, or we satisfy the problem's conditions, which is a contradiction.  If $ A_6$ is not colored blue, then among the 7 points other than $ A_0$ and $ A_3$, at least 4 must have the same color, which is a contradiction, by Lemma 1.  $ \blacksquare$

[b]Corollary.[/b]  In any equilateral triangle of side length $ |A_0A_3|$, either all vertices are the same color, or all vertices have different colors.

Now, let $ A$ and $ B$ be two arbitrary points such that $ |AB| \equal{} \sqrt{3} |A_0 A_3|$, and let $ C$ and $ D$ be points such that $ ACD$ and $ BCD$ are equilateral triangles.  By the corollary, either $ C$ and $ D$ are both the same color as $ A$, or they are both different---and distinct---colors, and either way $ B$ is the same color as $ A$.  It follows that any two points in the plane of distance $ \sqrt{3} |A_0 A_3|$ apart are the same color.  Hence every point in the plane is the same color.  Thus we obtain a contradiction, and we are done.  $ \blacksquare$

Where are these problems from?

It's from the Yugoslavia Olympiad

You can see some more (unsolved) problems from the competition here:

http://www.mathlinks.ro/viewtopic.php?t=200045
http://www.mathlinks.ro/viewtopic.php?t=199935
http://www.mathlinks.ro/viewtopic.php?t=199850

By the way, Boy Soprano II, your solution is very nice indeed  :)


$\{f_n(x)\}$ converges to $f(x)$ in measure on $[a,b]$.
$g(x)$ is countinuous on $\mathbb{R}$
Show that $\{g(f_n(x))\}$ converges to $g(f(x))$ in measure on $[a,b]$.

Doesn't this follow directly from the uniform continuity of $g$ on $[a,b]$? We need to show that for every $\varepsilon>0$, we have $m(\{x\in[a,b]\ |\ |g(f_n(x))-g(f(x))|>\varepsilon\})\to 0$ as $n\to\infty$, knowing that this holds for $(f_n)_n$ instead of $(g\circ f_n)_n$. This follows from the fact that for each $\varepsilon>0$, there is $\varepsilon'>0$ s.t. $|g(f_n(x))-g(f(x))|>\varepsilon\Rightarrow |f_n(x)-f(x)|>\varepsilon'$, so the sets of the type $\{x\in[a,b]\ |\ |g(f_n(x))-g(f(x))|>\varepsilon\}$ are included in sets of the type $\{x\in[a,b]\ |\ |f_n(x)-f(x)|>\varepsilon'\}$.

You can't use uniform continuity, grobber; the range of $f$ or of $f_n$ is not necessarily bounded.

Sorry... :blush: Forgot about that.

Could we do that (something similar to what I did above) on the sets $A_m=f^{-1}([m,m+1])$ for $m\in\mathbb Z$? I know we still know nothing about the range of the $f_n$'s, but on such a set $A_m$, the measure of the set where $f_n$ takes values outside $[m-1,m+2]$, for example goes to $0$ as $n\to\infty$, so it should work, separately for each $A_m$. We can then write $[a,b]=\cup_{m\in\mathbb Z}A_m$, and since each $m(\{x\in A_m\ |\ |g(f_n(x))-g(f(x))|>\varepsilon\})$ goes to $0$ as $n\to\infty$, we can conclude that the whole thing goes to $0$ from the fact that $[a,b]$ has finite measure, and we sort of "chop off" larger and larger parts of it.

Sorry for the lack of rigor. I think it's tedious to write rigorously, but I';m pretty sure something like that would work. Sorry if it turns out it doesn't :).


\[ \left\{ \begin{array}{l}
 \left( {3 \minus{} \frac{1}{{2x \plus{} 3y}}} \right)\sqrt {2x}  \equal{} \frac{{40}}{7} \\
 \left( {3 \plus{} \frac{1}{{2x \plus{} 3y}}} \right)\sqrt y  \equal{} \frac{{22}}{7} \\
 \end{array} \right.\]
 :P
[url]http://toancapba.com[/url]

This is very similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1247514#1247514]Vietnam NMO 1996/1[/url].


Let $ A$ be a 2 by 2 matrix.

(i) Express the definition of inverse matrix of $ A$.

(ii) According to the above definition, suppose the inverse of $ A\equal{} \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ exists, deduce the inverse matrix.

[hide](i) B is the inverse of A iff AB = BA = I, where I is the identity matrix of order 2.

(ii) $ B \equal{} \frac{1}{ad\minus{}bc}\left[\begin{array}{cc}d & \minus{}b\\ \minus{}c & a \end{array}\right]$.[/hide]


[quote="GameBot"]A Bronze Rectangle is one whose adjacent sides have ratio 1 to 1.6. Find the number of centimeters to the nearest tenth in the perimeter of the smallest Bronze Rectangle that can be constructed with 3.2 cm as the length of one of the sides. Express your answer as a decimal.[/quote]

The sides are $ 2,3.2$. Thus, $ 10.4$.


trigonometry
complex numbers
number theory unsolved
number theory

x + y + z = 6pi/7 (pi = 180 degree)) ve 2x = 3y = 6z

cosx . cosy . cosz = ?

Prepaired by İhsan YÜCEL

Yes, this problem has a lot to do with number theory...  :D

Actually, your equations $x+y+z=\frac{6\pi}{7}$ and 2x = 3y = 6z yield $x=\frac{3\pi}{7}$, $y=\frac{2\pi}{7}$, $z=\frac{\pi}{7}$. Thus, it remains to find $\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}$.

We will show that $\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}=\frac18$.

In fact, we will use complex numbers:

$8\cdot\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}$
$=8\cdot\frac{e^{i\frac{3\pi}{7}}+e^{-i\frac{3\pi}{7}}}{2}\cdot\frac{e^{i\frac{2\pi}{7}}+e^{-i\frac{2\pi}{7}}}{2}\cdot\frac{e^{i\frac{\pi}{7}}+e^{-i\frac{\pi}{7}}}{2}$
$=\left(e^{i\frac{3\pi}{7}}+e^{-i\frac{3\pi}{7}}\right)\cdot\left(e^{i\frac{2\pi}{7}}+e^{-i\frac{2\pi}{7}}\right)\cdot\left(e^{i\frac{\pi}{7}}+e^{-i\frac{\pi}{7}}\right)$
$=2+e^{i\frac{-6\pi}{7}}+e^{i\frac{2\pi}{7}}+e^{i\frac{-4\pi}{7}}+e^{i\frac{4\pi}{7}}+e^{i\frac{-2\pi}{7}}+e^{i\frac{6\pi}{7}}$
$=1+e^{i\frac{-6\pi}{7}}\cdot\left(1+e^{i\frac{\pi}{7}}+e^{i\frac{2\pi}{7}}+e^{i\frac{3\pi}{7}}+e^{i\frac{4\pi}{7}}+e^{i\frac{5\pi}{7}}+e^{i\frac{6\pi}{7}}\right)$
$\cdot\left(1-e^{i\frac{\pi}{7}}+e^{i\frac{2\pi}{7}}-e^{i\frac{3\pi}{7}}+e^{i\frac{4\pi}{7}}-e^{i\frac{5\pi}{7}}+e^{i\frac{6\pi}{7}}\right)$.

But since $e^{i\frac{\pi}{7}}$ is a primitive 7-th root of unity, we have $1+e^{i\frac{\pi}{7}}+e^{i\frac{2\pi}{7}}+e^{i\frac{3\pi}{7}}+e^{i\frac{4\pi}{7}}+e^{i\frac{5\pi}{7}}+e^{i\frac{6\pi}{7}}=0$, and thus

$8\cdot\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}=1+0=1$,

so that $\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}=\frac18$, and the proof is complete.

I have no time now to look for a better one...

[b]EDIT:[/b] I see the problem has almost been solved on ML: In http://www.mathlinks.ro/Forum/viewtopic.php?t=12775 post #4, Mildorf showed that $\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7}=-\frac18$; now, since $\cos\frac{4\pi}{7}=\cos\left(\pi-\frac{3\pi}{7}\right)=-\cos\frac{3\pi}{7}$, this becomes $\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\left(-\cos\frac{3\pi}{7}\right)=-\frac18$, so that $\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{3\pi}{7}=\frac18$. This proves our equality $\cos\frac{3\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{\pi}{7}=\frac18$ again.

  darij

hi,

it's a wonderful solutions...
İt's true.
Hope to me you!


number of ballers + number of soccerballers + number of mathers - number of ballers and soccerballers - number of soccerballers and mathers - number of ballers and mathers + number of ballers, soccerballers, and mathers= total

40+30+20-(two sporters)+8=60
98-twosporters=60
38 two sporters.

hope that helps.

Not quite. The number of times a person is counted is equivalent to the number of teams they are on. So, letting the number of people who play two sports be x and the number who play three be y, we have
60+x+2y=90. However, we already know that y=8, so our answer is 14.

I seem to remember this problem from Alcumus...is that where you found it?

[quote="nick42"]hey guys, this problem is confusing:

At a school, all 60 students play on at least one of three teams: Basketball, Soccer, and Mathletics. 8 students play all three sports, half the students play basketball, and the ratio of the size of the math team to the size of the basketball team to the size of the soccer team is $ 4: 3: 2$. How many students at the school play on exactly two teams?

i can deduce that there are 40 math guys, 30 basketball guys, and 20 soccer guys. but now what?[/quote]

Okay, there are 40 math guys, 30 basketball guys, and 20 soccer guys according to you.  Since there are 8 students that play all three sports, we now have:

Math: 32
Basketball: 22
Soccer: 12
All: 8

We subtracted 8 from each of the basketball, soccer, and math people.  Now if you see when we add them up: $ 32 \plus{} 22 \plus{} 12 \plus{} 8 \equal{} 74$, and our problem said there were only $ 60$ people.  Thus $ 74\minus{}60\equal{}\boxed{14}$ people do two activities.  To understand this more, go to the Venn Diagram Lessons in Alcumus.


summer program
PROMYS
floor function
function

Find an expression for the product of the divisors of a positive integer $n$ (including $1$ and $n$)

[hide="Solution"]
[b]Claim:[/b]  Let $d$ be the number of divisors of $n$.  Then the product of the divisors of $n$ is $n^{ \frac{d}{2}}$.

Proof (my PROMYS counselor would [b]kill me[/b] if he saw this proof :rotfl: ):  Suppose $n$ is not a square.  Then each divisor can be paired up with a unique, distinct other divisor such that their product is $n$.  Pairing up in this way, we have a number of $n$s equal to half the number of divisors, which will be even if $n$ is not square.

Suppose $n$ is a square.  Then it has an odd number of divisors, and the divisor $\sqrt{n}$ cannot be paired with another distinct divisor.  It alone is left out of the pairing, and we then have a product of $n^{ \lfloor \frac{d}{2}\rfloor}\cdot \sqrt{n}= n^{ \frac{d}{2}}$. [/hide]

[hide="seen it before"]
I think it's something like:

$(\sqrt{n})^{d(n)}$

where $d(n)$ is the function that gives the number of divisors of $n$.

Reason: we can pair up each divisor with it's complement that form the product $n$.  so we have $\frac{d(n)}{2}$ such pairs for general $n$.  If $n$ is a perfect square, then we cah pair up $\frac{d(n)-1}{2}$ such pairs, plus $n$'s square root, $\sqrt{n}$.  Multiply together we see that either case we arrive at the same formula.
[/hide]
Why would your PROMYS conselor kiil you??? It's all right... :maybe:

Because it's not complete at all.  I haven't proven that I can pair the divisors up the way I did, stuff like that.    :|

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394399#p394399

This problem uses a similar concept


algebra
polynomial
number theory unsolved
number theory

The positive integers $ \alpha, \beta, \gamma$ are the roots of a polynomial $ f(x)$ with degree $ 4$ and the coefficient of the first term is $ 1$. If there exists an integer such that $ f(\minus{}1)\equal{}f^2(s)$.

Prove that $ \alpha\beta$ is not a perfect square.

Is problem not true?
When f(-1)=0 then the problem is false.

the correct version of this problem is:

$ \alpha, \beta, \gamma$ are positive integers, 0, $ \alpha \beta, \beta\gamma, \gamma\alpha$ are the roots of a polynomial f(x) with degree 4 and  the coefficient of the first term is 1. If there exists an integer such that $ f(\minus{}1)\equal{}f^2(s)$, Prove that $ \alpha \beta$ is not a perfect square.

It follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150369]here[/url]


[quote="Treething"]One of Treething's songs on his iPod is a:bc long, where a, b, and c are consecutive integers. (a, b, and c represent digits.) He notices that the song is also exactly $x^{3}$ seconds long. Find x.[/quote]

I can't get x for some reason, I tried every possibility. Are you sure you didn't think 225 was a cube?

Actually, Splash D there is nothing wrong with the problem[hide]brute force leads to $x=7, \ 7^{3}=343=5: 43.$
Treething never said that they were consecutive as in a,b,c :wink: [/hide]

[size=75][color=darkred]Fixed the LaTeX
-nebula42[/color][/size]

[quote="bpms"][quote="SplashD"][quote="bpms"]Actually, Splash D there is nothing wrong with the problem[hide]brute force leads to $x=7 7^{3}=343=5: 43.$
Treething never said that they were consecutive as in a,b,c :wink: [/hide][/quote]

Oh, I thought $a=a$, $b=a+1$ and $c=a+2$, that's why I couldn't solve the problem.[/quote]
I realize that. Similar to a problem on the 2005 nats sprint, BUT mathcounts had made the error and was forced to accept three answers.[/quote]

Which question was that?

The product of three consecutive integers, a b c, divided by their sum is 341. Find a.

[quote="bpms"]Actually, Splash D there is nothing wrong with the problem[hide]brute force leads to $x=7 7^{3}=343=5: 43.$
Treething never said that they were consecutive as in a,b,c :wink: [/hide][/quote]

[hide]It is actually, $7^{3}$ isn't it?  But that is pretty clever.

I got $31$ for your other question.  2005 #21 I believe.[/hide]

[quote="mathgeniuse^ln(x)"][quote="bpms"]Actually, Splash D there is nothing wrong with the problem[hide]brute force leads to $x=7 7^{3}=343=5: 43.$
Treething never said that they were consecutive as in a,b,c :wink: [/hide][/quote]

[hide]It is actually, $7^{3}$ isn't it?  But that is pretty clever.

[/hide][/quote]

I believe he wrote $x=7$ and then separately $7^{3}=343$

[quote="mathgeniuse^ln(x)"]Okay, because it did look like 77.[/quote]

You should have realized that when you quoted him.

No it was still hard to realize, either way, I still believe he got it right and treething is a tricker.

I meant 7. Thanks to nebula42 for fixing the latex

Is that the only one I assume?  Is that the one Treething was looking for?

[hide="i love brute force"] i got 7 because it worked. the others didn't.[/hide]

Ignite, isn't it obviously easier to just cube numbers and see if they  work. :D.

Heh, Ignite stop being so literal, you know what the answer is.

[quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]There are lots of possibilities of answers for this problem, a couple include $\sqrt[3]{83}, \sqrt[3]{154}, \sqrt[3]{225}, 2\sqrt[3]{37}, \sqrt[3]{414}, 7, 2\sqrt[3]{34}, \sqrt[3]{201}, \sqrt[3]{130}$ etc.[/quote]

Why do you have to be so literal?  YOu know what I meant.[/quote]

Don't you mean Treething? He never specified x was an integer :P[/quote]  Mathcounts would have given you this wrong either way :D.


i took amc for the first time this year, so i don't understand the rules very well.
amc12a/108 (at our school)
amc12b/118.5 (at olympiad school)
amie 9 (at our school)
before march 18th, i asked a teacher which amie i should take, she told me there was not any different.
im in grade 10, and i would be in the first list if 118.5+9*10=208.5

Check the score report forms your teachers receive for the AMC and the AIME.  If your scores reported here are accurate, you should contact the AMC as described in the appropriate Sticky atop this forum.  It may be possible your AIME score got attached to the wrong AMC score, but you should consult the official score report forms your teachers receive to confirm your scores.

thanks, [b]rrusczyk[/b]
but can they add the scores from different schools() ?

i will try it anyway !

You should have qualified with a 208.5 index.


Not true...
Let $n = 2$ and let $x_1 = x_2 = \sqrt {5000}$


Hi, everybody. Well, I am thinking of studying in a college in the US. However, I don't know which universities I should apply to. I've heard from a professor that each university has its own - what word to use now? - style. I.e., each one likes students which characteristics A, B or C. But he also said the big/important universities have good departments in all areas, and maybe this style thing is not that important. Now I am kind of confused. I don't know how/where to find information about the general features of each university.

My "academic ideas": my main interest is pure mathematics. Also, I'd like to study some physics and some computer science in the next years. So, I would like a university in which I could be relatively free to study whatever I want (maybe all universities in the US have this property, I don't know). I also miss some motivation like "it is a good idea to study the theory x because the mathematician y and the physicist y' needed to develop it to solve problem z." For me, this looks like academic research (although I do not think it is that precise), which I would like to do. So, I am interested in how theories are built rather than in a beautiful, perfect, closed theory.

Also, I can't pay $ \$30,000$ a year. Therefore, I would need finantial help during the college (according to Google Translator, the word in English is "studentship").

That's it. Thank you for any help.

It is true that every university has its own distinct flavor.  For example, a research university is going to feel very different from a liberal arts school, and every research university has its own feel, as do liberal arts schools.  It depends what kind of environment you want.  We don't know what you like so we can't give examples of specific schools.

The term I think you want is "financial aid."  Different universities will give different financial aid amounts-you fill out the same information but each school evaluates it differently and they have different scholarship opportunities.  However, international students usually are less likely to receive financial aid, as the university usually prefers to, or is required to (for certain loans, for example) give it to US citizens or residents.  If you want low cost, public universities are the way to go.  They usually range from 10-25 K for out-of-state (which you would be), which is cheap when compared to 50-55 K for private schools.

One side issue regarding finances: I'm pretty sure that one of the requirements for obtaining a student visa in order to enter the U.S. to attend school is that you provide proof (to the U.S. government) that you are able to pay for the education in question. I've never seen or dealt with the paperwork that would be involved, and there has to be a way, since there are many people on student visas - but the requirement exists.

As for financial aid or scholarships (I think the word you were looking for was "scholarship" rather than "studentship"): many of the most common sources of aid comes with residency restrictions that make it unavailable to international students. Interestingly, some of the most famous and (at list price) most expensive institutions - and the hardest to get into - may actually have the most aid available for international students. Others who are more familiar than I am with these famous and hard to get into institutions might be able to supply more details.

As for my own state-supported institution (California State University, Long Beach), after the most massive one-year fee increase that we've ever had, in-state students taking a full-time load will pay \$4370 per year in fees, while out-of-state students (which includes international students) will pay \$15530 per year if they take 15 units per semester. To that you must add the cost of housing, meals, books, transportation, etc. - none of that is included. And of course, we're still cheap compared to many other places.

It is part of the general culture of U.S. universities to allow students quite a bit of choice in what they study. An undergraduate mathematics major will (in many cases) be a student in some very broad portion of the university - a College of Arts and Sciences, or some name like that - rather than part of some narrower grouping. There will be some requirements - breadth requirements, or general education requirements - for the university as a whole and there will be some specific requirements for the major, but there will often be a variety of ways to satisfy those.


inequalities
modular arithmetic
inequalities unsolved

For any positive integer $ n$ find the number of ordered n-tuples of integers $ (a_{1}, a_{2}, a_{3}, ..., a_{n})$ such that

$ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\geq n^{2}$
$ a_{1}^{2}\plus{}a_{2}^{2}\plus{}...\plus{}a_{n}^{2}\leq n^{3}\plus{}1$

EDIT: I think I got a solution, but I'm not sure of it.

[hide="solution"]
assume $ \sum_{i\equal{}1}^{n}a_{i}\geq n^{2}\plus{}1$ then
$ n^{3}\plus{}1\geq\sum_{i\equal{}1}^{n}a_{i}^{2}\geq\frac{1}{n}(\sum_{i\equal{}1}^{n}a_{i})^{2}\geq\frac{1}{n}(n^{2}\plus{}1)^{2}\equal{}\frac{1}{n}(n^{4}\plus{}2n^{2}\plus{}1) > n^{3}\plus{}2n > n^{3}\plus{}1$
which is a contradiction.
so $ \sum_{i\equal{}1}^{n}a_{i}\equal{}n^{2}$
assume there exists $ k$ such that $ a_{k}<n$.then $ a_{k}\equal{}n\minus{}l$ where $ l$ is an integer strictly greater than $ 0$.then
$ \sum_{1\leq i\leq n}a_{i}^{2}\equal{}a_{k}^{2}\plus{}\sum_{1\leq i\leq n,i\not\equal{}k}a_{i}^{2}\geq (n\minus{}l)^{2}\plus{}\frac{1}{n\minus{}1}(n(n\minus{}1)\plus{}l)^{2}\equal{}n^{2}\minus{}2nl\plus{}l^{2}\plus{}(n\minus{}1)n^{2}\plus{}2nl\plus{}\frac{l^{2}}{n\minus{}1}\equal{}$
$ n^{3}\plus{}l^{2}\plus{}\frac{l^{2}}{n\minus{}1}>n^{3}\plus{}1$
which is a contradiction. so all $ a_{i}\equal{}n$ and that is the only possibility and there is only one ordered n-tuple which is $ (n,n,...,n)$.
[/hide]
for B_Gasperov:napisi rjesenje pa cemo vidit...

Here's my solution.   :)

It's obvious that $ (n, n, n, ..., n)$ satisfy the condition. We have:

$ n^{2}\geq n^{2}$
$ n^{3}\leq n^{3}\plus{}1$

We can't reduce $ \sum_{i \equal{} 1}^{n}a_{i}$, let's keep it fixed. Then we have the following lemma which is easy to be proved. (We prove it by using deviations).

Lemma: If $ \sum_{i \equal{} 1}^{n}a_{i}$ is fixed then the minimum of $ \sum_{i \equal{} 1}^{n}a_{i}^{2}$ is attained for $ a_{i}\equal{} a_{j}$ for every $ i, j\leq n$. If there are different $ a_{i}, a_{j}$ then the value of $ \sum_{i \equal{} 1}^{n}a_{i}^{2}$ raises by at least $ 2$ and then the second condition is no longer satisfied.

By the analogy, the minimal value of $ \sum_{i \equal{} 1}^{n}a_{i}^{2}$ if $ \sum_{i \equal{} 1}^{n}a_{i}^{2}\geq n^{2}\plus{}1$ is always $ > n^{3}\plus{}1$.

If something seems to be obscure or not clear enough, I can provide further explaination.

@ pozdrav Goc i cestitke na 100.-tom postu

your idea of the solution is correct but in a competition i would rate this with a 1/7 beacuse there are no computational calculations and proofs. you didn't explain why the sum $ \sum_{i\equal{}1}^{n}a_{i}$ can't be increased(i know it's obvious but it should be written). the second thing is that you didn't prove the whole lemma(the second part). other than that your solution is clear. i know it's an easy problem but these are the kind of problems where you lose points in a competition for not being explicit enough.
there are nice ways to prove that $ \sum_{i\equal{}1}^{n}a_{i}^{2}$ raises by atleast $ 2$, i'd like to see yours.

[quote="goc"]there are nice ways to prove that $ \sum_{i \equal{} 1}^{n}a_{i}^{2}$ raises by atleast $ 2$, i'd like to see yours.[/quote]

Let $ \sum_{i \equal{} 1}^{n}a_{i}\equal{} n^{2}$ and let $ a_{i}\equal{} n\plus{}b_{i}$ where $ b_{i}$ stands for a deviation from $ n$ (either positive or negative). Then:

$ \sum_{i \equal{} 1}^{n}a_{i}^{2}\equal{}\sum_{i \equal{} 1}^{n}({n\plus{}b_{i}})^{2}\equal{} n^{3}\plus{}\sum_{i \equal{} 1}^{n}b_{i}^{2}\geq  n^{3}\plus{}2$

very nice since the sum of deviations is 0. also you can see that the parity of $ \sum_{i\equal{}1}^{n}a_{i}^{2}$ does not change if you change the $ a_{i}$ but keep the sum constant. and $ \sum_{i\equal{}1}^{n}a_{i}^{2}\equal{}n^{3}$ if and only if all of the $ a_{i}\equal{}n$ by the AM-GM inequality. so in any other case( since we are dealing with integers here) we have $ \sum_{i\equal{}1}^{n}a_{i}^{2}\geq n^{3}\plus{}2$ anyway, nice solution
hope to see you in the tournament of towns this fall...

[hide]
\[ 0\leq\sum_{k\equal{}1}^{n}(a_{n}\minus{}n)^{2}\equal{}n^{3}\plus{}\sum_{k\equal{}1}^{n}a_{k}^{2}\minus{}2na_{k}\leq n^{3}\plus{}(n^{3}\plus{}1)\minus{}2n^{3}\equal{}1\]
(since $ a_{k}^{2}$ is added and $ a_{k}$ is subtracted). Therefore, the integer
\[ \sum_{k\equal{}1}^{n}(a_{n}\minus{}n)^{2}\]
is equal to 0 or 1. In the first case, we require $ a_{k}\equal{}n$ for all $ k$, and this solution checks. In the second case, we need both of
\[ \sum_{k\equal{}1}^{n}a_{k}\equal{}n^{2},\ \sum_{k\equal{}1}^{n}a_{k}^{2}\equal{}n^{3}\plus{}1\]
for otherwise, there would be strict inequality in the first line. However, $ x^{2}\equiv x\pmod{2}$ for all integers $ x$ and $ n^{2}\not\equiv n^{3}\plus{}1\pmod{2}$, so this case never works.
[/hide]


geometry
3D geometry
complex numbers
combinatorics unsolved
combinatorics

Consider the string of length $ 6$ composed of three characters $ a$, $ b$, $ c$. For each string, if two $ a$s are next to each other, or two $ b$s are next to each other, then replace $ aa$ by $ b$, and replace $ bb$ by $ a$. Also, if $ a$ and $ b$ are next to each other, or two $ c$s are next to each other, remove all two of them (i.e. delete $ ab$, $ ba$, $ cc$). Determine the number of strings that can be reduced to $ c$, the string of length 1, by the reducing processes mentioned above.

There's some relationship here with complex numbers: those reduction rules are equivalent to the rules $ a \equal{}\omega$, $ b \equal{}\omega^{2}$ and $ c \equal{}\minus{}1$ with $ \omega$ a complex cube root of unity under the assumption that we don't have commutivity.  I don't know if this can be turned into an elegant solution, unfortunately.  Otherwise, the answer is small enough that this could succumb to a case-bash.  The nicest solution I've found involves building up from the bottom (there are no strings of length 2 that work, five strings of length 3, four strings of length 4, every string of length 5 arises from a string of length 3 or of length 4 and every string of length 6 arises from one of length 4 or length 5), but it's not exactly pleasant.

[hide="Ugly and shaky case bash"]First of all, realize the parity of the number of $ c$s in a string is invariant. So having an even number of $ c$s means we can't get down to $ c$.

Consider the case where there is only one $ c$ in the original 6-letter string.
[list][*]If $ c$ is the first letter or last letter, then we can use JBL's complex numbers idea to get that we must have four $ b$s and an $ a$ or the opposite (order does not matter, because $ a$ and $ b$ commute with each other), giving 10 possibilities. Double that since $ c$ can be on either side.
[*]If $ c$ is the second letter or fifth letter, nothing on either side of it can cancel. So 0.
[*]If $ c$ is the third or fourth letter, then the side with two letters must be $ ab$ or $ ba$. The side with three letters must be $ aaa$ or $ bbb$, also evident by JBL's idea. This gives 8 total ways: two positions for the $ c$ and two ways to do each of the two sides.[/list]
So $ 28$ total here.

Now if there are 3 $ c$s total, JBL's idea shows us that we must have the other three letters identical. Additionally, notice that the only thing that can be in between the span of these three letters is either nothing or two $ c$s in a row. So we can have: $ aaa$, $ accaa$, $ aacca$, and the same thing with $ b$ instead of $ a$. Then just throw $ c$s on either edge until the string is the right length. We get $ 3\cdot 2\cdot 2 \equal{} 12$ here.

If there are 5 $ c$s, the last $ a$ or $ b$ can't cancel. So nothing.

$ 40$ total.[/hide]

A possible idea...(for the general case of $ n$-length strings)

[hide="Lemma 1"]
Any string of $ a$'s and $ b$'s such that the product of the terms is $ 1$ (by JBL's system) can be reduced to the empty string via a sequence of moves. (easy to prove)[/hide]

[hide="an idea"]

If we consider all substrings of consecutive $ c$'s, notice that the parity of the number of $ c$'s in each one remains the same under any of these operations.  Then the only thing to worry about is that the region between two such substrings could be emptied.  Clearly, if the desired conditions are to hold, this must be possible for any two substrings of $ c$'s.  But then perhaps we can generally find the number of $ n$-length strings based on the number of substrings of $ c$'s...[/hide]

Woops. The four solutions that my case bash missed are $ caaacc$, $ ccaaac$, $ cbbbcc$, and $ ccbbbc$.

did anyone generalized this problem to string of length $ n$?


geometry
3D geometry
algebra
polynomial
algebra unsolved

Prove that 3x is not equal to the sum of cubes of polynomials with integer coefficients.

See this:

http://www.mathlinks.ro/Forum/viewtopic.php?t=183534

http://www.mathlinks.ro/viewtopic.php?t=184401

 :)


The arithmetic mean of four numbers is 15. Two of the numbers are 10 and 18 and the other two are equal. What is the product of the two equals numbers.


calculus
topology
trigonometry
complex analysis
advanced fields
advanced fields unsolved

I don't know if this question goes here or in calculus forum.

How can I show that any open subset of $R^{2}$ can be split into open, connected components? Is this splitting unique? If the set is unbounded, why must there be exactly one unbounded component?

These things are just stated in the text I am reading, so maybe they are trivial. I know very little topology, so I don't feel that I will be able to prove the statements. How is it done?

Define an equivalence relation on a set $U\subset\mathbb R^{2}$: $x\sim y$ if there is a connected subset of $U$ containing both $x$ and $y$. The definition of 'connected' must be used to verify that this is indeed an equivalence. By pure logic, the set $U$ splits into equivalence classes in a unique way. These classes are the connected components. It is not hard to prove that each connected component of $U$ is open if $U$ is open: just observe that an open disk (or ball) is connected.

There may be more than one unbounded connected component: consider $\mathbb R^{2}$ minus a line.

Thanks mlok. All your help has been valuable to me.

Are you saying that if you have any partition of $U$ into connected components, then this partition will always correspond to the given equivalence relation?

The correct assertion is that if the complement to $U$ is compact, then $U$ has exactly one unbounded component. I understand how this will be true under the partition of the equivalence relation, but not really why it is obviously true just given the statement that we have a partition of the set into open, connected subsets.

If "a connected component" means "a maximal connected subset" (as it does to me), then there is only [b]one[/b] partition of any set into connected components. And yes, it corresponds to the equivalence relation that I described (which is essentially a tautology).

Well, if we don't require the components to be open it's clear that the partition isn't unique. But if we want open components it seems quite clear to me (using the path-wise notion of connectedness) that there is only one partition (assuming now that $U$ is open for the existence of the partition into open components) . Am I wrong?

I'd like to emphasize the distinction between [b]connected components[/b] and [b]connected subsets[/b]. See, for example, [url=http://mathworld.wolfram.com/ConnectedComponent.html]here[/url]. Any topological space uniquely splits into connected components, without the assumption of openness. And in general connected components are not open. For example, in the set of rational numbers $\mathbb Q$ the connected components are one-point subsets.

Also, path-connectedness is not equivalent to connectedness in general. It's true for open subsets of $\mathbb{R}^{n}$, but there are plenty of counterexamples.
The "topologist's sine curve" is a standard counterexample: take the graph of $\sin\frac1x$ for $0<x\le 1$, together with the closed segment from $(0,-1)$ to $(0,1)$. This is closed in $\mathbb{R}^{2}$, and it's connected in the subspace topology. On the other hand, it's not path connected; the wave and the segment are separate path components.

Ok. By component I meant just a cell of a partition. I came up with the word component on my own; I guess it crashed with established notation.

I know that path-connectedness is stronger than connectedness, but in the course I need it for (complex analysis) we are mostly dealing with open sets. Although that's not really true when I think about it, in our definitions of simply and multiply connected sets we need connectedness of closed sets on the riemann sphere.

It's pretty frustrating to take the course with a poor understanding of topology, but I think my intuition is a little better now at least.


Prove or disprove that for $ x,y,z>0$ we have:
if $ x\plus{}y\plus{}z\equal{}1$ then $ x*ln(y) \plus{} y*ln(z) \plus{} z*ln(x)>\equal{} ln(x^{1/3})\plus{}ln(y^{1/3})\plus{}ln(z^{1/3})$


calculus
integration
function
logarithms
calculus computations

[hide=" 2"]

From the quotient rule, differentiating $ \frac {x}{ln x}$ gives the integrand and so we have $ \frac {x}{ln x} \plus{} c$.

[/hide]

[hide=" 1"]

Clearly we could expand and integrate term by term but there must be a neater way of doing it.

[/hide]

[quote="AndrewTom"][hide=" 2"]

From the quotient rule, differentiating $ \frac {x}{ln x}$ gives the integrand and so we have $ \frac {x}{ln x} \plus{} c$.

[/hide]

[/quote]

I never liked this kind of an answer. It's like you sad: "If we knew the solution, differentiating it we would see that it really is the solution!". How did you get to the point you chose to differentiate that exact function? By choice? In that case, it definitely isn't a solution. If not, show us your work..

[hide=" My solution"]

(1) There is no other way to do this but by expansion. We have:

$ \int x^8\left(x^5 \plus{} 1\right) ^{14} \, dx \equal{} \sum ^ {14} _ {n \equal{} 0} \binom{14}{n} \int x ^ {5 n \plus{} 8}  \, dx \equal{}\sum ^ {14} _ {n \equal{} 0} \binom{14}{n} \frac{x^{5 n \plus{} 9}}{5n \plus{} 9} \plus{} C$

(2)

$ \int \frac{1}{\ln x} \,dx \equal{} \frac{x}{\ln x} \plus{} \int x \frac{1}{x \ln ^2x} \,dx$ - partial integration where $ u \equal{} 1$ and $ v' \equal{} \frac{1}{\ln x}$.

Substituting thin in the original question we get:

$ \int \left( \frac{1}{\ln x} \minus{} \frac{1}{\ln^2x}\right)\,dx \equal{} \frac{x}{\ln x} \plus{} C$

[/hide]


algebra
polynomial
function
calculus
derivative
real analysis
real analysis unsolved

Prove that the equation $ 1\plus{}\frac{x}{1!}\plus{}\frac{x^2}{2!}\plus{}.......\plus{}\frac{x^n}{n!}\equal{}0$, $ n\in{N^{*}}$ has $ 0$ or $ 1$ real solutions(in function of $ n$)

it's in exacty Shturm theorem , and question is Algebraic than analysis. :blush:

Specifically, there are no real roots of $ P_n(x)\equal{}\sum_{k\equal{}0}^n \frac{x^k}{k!}$ if $ n$ is even, and exactly one real root if $ n$ is odd.
[hide]
If $ n\equal{}0$, it's obvious. For positive even $ n$, $ P_n$ is of even degree with positive leading coefficient, so it has an absolute minimum at $ a$, so that $ 0\equal{}P^\prime_n(a)\equal{}P_{n\minus{}1}(a)$. Then $ a\neq 0$ as $ P_{n\minus{}1}$ has constant term 1, so $ P_n(a)\equal{}\frac{a^n}{n!}>0$ as $ n$ even and $ a\neq 0$. As a result, $ P_n$ is everywhere positive for even $ n$.

For odd $ n$, $ P_n$ is guaranteed at least one real root. If it had at least two, Rolle's Theorem would imply a root of its derivative, $ P_{n\minus{}1}$. But $ n\minus{}1$ is even, so this is impossible as shown above. This means $ P_n$ has exactly one real root for odd $ n$.
[/hide]

Why $ P_n$ has an absolute minim at $ a$ ?

$ P_n$ has an absolute minimum somewhere. Then $ a$ is defined to be a point where an absolute minimum occurs.


[quote="maokid7"]how many factors of 729810 are not divisible by 6?

is there a way to generalized this process

like if $n=(2^a)(3^b)(p_3^c)(p_4^d)\dots$
could you find how many factors are not multiples of 6 or any combination of $p_np_m$[/quote]

[hide]Is it 48?
$729810=2\times 5\times 3^4\times 17 \times 53$. Take out a 2 or the threes and you have $5\times 3^4\times 17 \times 53$ and $2\times 5\times 17 \times 53$ which has lcm of $5\times 17\times 53$, which has 8 factors. So $(2)(5)(2)(2)+(2)(2)(2)(2)-8=\boxed{48}$?[/hide]

[hide]We have $729810=2\cdot 3^4\cdot 5\cdot 17\cdot 53=6(3^3\cdot 5\cdot 17\cdot 53)$. The factor inside paranthesis has $4\cdot 2\cdot 2\cdot 2=32$ factors, which is the number of factors of 729810 that are divisible by 6. Since there are $2\cdot5\cdot2\cdot2\cdot2=80$ total factors, our answer is $80-32=\boxed {48}$.[/hide]