If the average of 2 angles of a triangle is 75 degrees, what is the size of the remaining angle?

[hide]Call the angles x and y. If their average is 75, we know that $\frac{x+y}{2}=75$ so $x+y=150$. Thus, the other angle is 180-150=$\boxed{30\text{ degrees}}$
[/hide]

[quote="Teki-Teki"]not in hyperbolic (or spherical) geometry![/quote]

In that case, isn't this problem pointless?

yes - and your point is?

(not if you know some sides or the area)

Is it really important to talk about hyperbolic geometry in this problem?

[hide]So basically if the average of 2 of the angles is 75, then the sum of them is 2(75)=150.  The remaining angle is 180-150=30 degrees.[/hide]


M is a point in triangle ABC  as:
<ABC=50  & <MBC=10
<BCA=50 & <MCB=30

Find  <BMC=X=?

you wanna find $\angle{BMC}$ when $\angle{BCM}$ and $\angle{CBM}$ are given?

If you mean $\angle{BMA}$, then it is:
[hide]$70^{\circ}$[/hide]

[color=blue][b]Problem.[/b] Let ABC be a triangle such that < ABC = < BCA = 50°, and let M be a point inside this triangle such that < CBM = 10° and < BCM = 30°. Prove that < BMA = 70°.[/color]

[i]Solution.[/i] I didn't bother to solve this synthetically, since [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22551]the Ceva trick[/url] works really well:

We have < ABM = < ABC - < CBM = 50° - 10° = 40° and < ACM = < BCA - < BCM = 50° - 30° = 20°.

By the sum of angles in triangle ABC, we have < CAB = 180° - < ABC - < BCA = 180° - 50° - 50° = 80°. Let X be some point inside the triangle ABC such that < BAX = 70°. Then, < CAX = < CAB - < BAX = 80° - 70° = 10°.

Now, assume for a moment that we have shown that the lines AX, BM and CM concur. This means that the line AX passes through the point of intersection of the lines BM and CM, i. e. through the point M. Hence, < BAM = < BAX = 70°, and thus, by the sum of angles in triangle ABM, we have < BMA = 180° - < BAM - < ABM = 180° - 70° - 40° = 70°. Thus, the problem will be solved once we have shown that the lines AX, BM and CM concur.

By the trigonometric form of the Ceva theorem, in order to prove that the lines AX, BM and CM concur, it is enough to verify the equation

$\frac{\sin\measuredangle BAX}{\sin\measuredangle CAX}\cdot\frac{\sin\measuredangle ACM}{\sin\measuredangle BCM}\cdot\frac{\sin\measuredangle CBM}{\sin\measuredangle ABM}=1$.

But

$\frac{\sin\measuredangle BAX}{\sin\measuredangle CAX}\cdot\frac{\sin\measuredangle ACM}{\sin\measuredangle BCM}\cdot\frac{\sin\measuredangle CBM}{\sin\measuredangle ABM}=\frac{\sin 70^{\circ}}{\sin 10^{\circ}}\cdot\frac{\sin 20^{\circ}}{\sin 30^{\circ}}\cdot\frac{\sin 10^{\circ}}{\sin 40^{\circ}}$.

Since sin 70° = cos (90° - 70°) = cos 20°, $\sin 30^{\circ}=\frac12$ and $\sin 40^{\circ}=\sin\left(2\cdot 20^{\circ}\right)=2\sin 20^{\circ}\cos 20^{\circ}$, this becomes

$\frac{\sin\measuredangle BAX}{\sin\measuredangle CAX}\cdot\frac{\sin\measuredangle ACM}{\sin\measuredangle BCM}\cdot\frac{\sin\measuredangle CBM}{\sin\measuredangle ABM}=\frac{\cos 20^{\circ}}{\sin 10^{\circ}}\cdot\frac{\sin 20^{\circ}}{\left(\frac12\right)}\cdot\frac{\sin 10^{\circ}}{2\sin 20^{\circ}\cos 20^{\circ}}=1$,

and the problem is solved.

  Darij

Yes, my solution was the same as darij's. And after I got the answer, I found another way to solve the problem.

Construct equilateral triangle with base $BC$, and call the other vertex $X$, then it is easy to show that $\triangle{XAB} \cong \triangle{BMC}$. Hence $AB=BM$, which gives $\angle{BAM}=70^o$.

Shobber, your solution is nicely. Here are two solutions for this problem:

[u]The first method (trigonometrically).[/u]

$x=m(\widehat {AMB})\Longrightarrow m(\widehat {BAM})=140-x,\ m(\widehat {CAM})=x-60\Longrightarrow$
$\sin (140-x)\sin 10\sin 20=\sin (x-60)\sin 40\sin 30\Longleftrightarrow$ $2\sin (40+x)\sin 10=2\sin (x-60)\cos 20\Longleftrightarrow$
$\cos (30+x)-\cos (50+x)=\sin (x-40)+\sin (x-80)\Longleftrightarrow$
$\cos (30+x)=\cos (170-x)\Longleftrightarrow30+x=170-x\Longleftrightarrow x=70.$


1. BOGTRO
2. AIME15
3. jxl28
4. yaofan
5. isabella2296
6. cgyao15

$ \begin{tabular}{|c|c|c|c|c|c|c|} \hline X & BOGTRO & AIME15 & jxl28 & yaofan & isabella2296 & cgyao15 \\
\hline BOGTRO & X & & & & & \\
\hline AIME15 & & X & & & & \\
\hline jxl28 & & & X & & & \\
\hline yaofan & & & & X & & \\
\hline isabella2296 & & & & & X & \\
\hline cgyao15 & & & & & & X \\
\hline \end{tabular}$

[u][b]RULES:[/b][/u]
1. Good sportsmanship is expected.
2. One person post the results in this thread.
3. Best of 1, 3 or 5. The two people decide before they start playing.

[u][b]ROUND I[/b][/u]
1. BOGTRO v AIME15
2. jxl28 v yaofan
3. isabella2296 v cgyao15

[b]THIS ROUND IS DUE ON THE 24TH[/b]

Daaaaaaaaannnnnnnnnnnnnggggggggggg.
I get izzy again.
/me gesses that he will pm izzy so he can get pwned.

jxl28 defeats yaofan (best of 3) 2-1 (4-2, 1-4, 4-1)

Izzy pwns cgyao15 2-0(4-1,4-2) best of three.  :)

PM'ed BOGTRO.  He read it, but no response yet.

Which is because I don't know when I can play.

[quote="AIME15"]Are we allowed to play other matches if our opponent is online?[/quote]
Well I suppose.

$ \begin{tabular}{|c|c|c|c|c|c|c|}\hline X & BOGTRO & AIME15 & jxl28 & yaofan & isabella2296 & cgyao15\\
\hline BOGTRO & X & & & & &\\
\hline AIME15 & & X & & & &\\
\hline jxl28 & & & X & W &  &\\
\hline yaofan & & & L & X & &\\
\hline isabella2296 & & & & & X & W\\
\hline cgyao15 & & & & & L & X\\
\hline\end{tabular}$

jxl28 creams cgyao15 2-0 best of three (4-1,4-2)

jxl28 VS. BOGTRO

Best 2 out of three

jxl wins first game (4-1)
jxl wins second game (4-1)

jxl28 defeats BOGTRO (4-1, 4-1, best of 3)

Havn't played in a few months. Be quiet.

yaofan defeats cgyao15 (4-1,4-1,4-2) ithink


When will we be able to download the PDFs of contests instead of printing out the web page? It just gives me a file read error if i open the pdf.

It was a very convenient feature.

You should be able to do it now. If you don't then it's a bug :)

I've tried downloading them at work and at home. All I get is a blank page or a file read error.

It cannot read a file like this:

http://www.artofproblemsolving.com/Forum/resources/files/usa/45/USA-AIME-2004-45-182.pdf

Is this a problem on my side? I used to be able to view them.

Any idea when the .pdf glitch will be fixed?  :)

For now I am at the IMO, so it will be until at least Wednesday next week, sorry about that  :blush:

They're still not working if anybody cares.

Also, the PDF generator doesn't work either...

Some have a url in format of http://www.artofproblemsolving.com/Forum/download.php?id=xxxx

they don't work either... :maybe:

Well still doesn't work as of today :wink:

I downloaded the AMC10 PDFs at http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43 . The documents did not show the graphics but they displayed the laTex (?) code instead. How do I fix this problem. Thank you.


email
number theory
advanced fields
advanced fields unsolved

Recently I received an email (Spam?) from an unknown Korean guy who claims to have a simpler proof of FLT
(see attached file)

   Dear Sir.
   Please, see and make sure.
   Our proofs are perfect and plain. Andrew Wiles`s FLT proof is a guess.
   Thanks.
Sincerely yours. Jae Yul Lee and You Jin Lee.

I am not familiar with number theory, so I would like to hear some opinions from you guys, what do you think about his approach?
(despite his very incomprehensible English...)

i wouldn't ever believe a mathematician sending a .doc attachment about math :D

It's obviously garbage.  (I got the same spam, too -- I figured they got my e-mail by trolling through university math department contact pages.)  If FLT could be proved by wrong trivial algebraic manipulations, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205286]Don Blazys[/url] would have solved it already ;).

i got the same email, too. i actually just found this forum by googling the fucker's name. obvious nonsense.

first math spam i've ever gotten though. *chortle*

When they release me from the mental hospital I am going to become Don Blazys.

I got it too- it's a very widely distributed bit of crankery.


calculus
integration
trigonometry
limit
abstract algebra
function
algebra

Calculate $I=\int_0^{\pi} \frac{Im(B(e^{iz}))}{\sin{z}}dz$, where $B(z)$ is finite Blashke product $B(z)=\prod_{k=0}^n \frac{z-\alpha_k}{1-\bar \alpha_k z}$, $\alpha_k \in (-1,1)$.
By the word  "calculate" i meant that it would be great if someone could give some kind of the upper bound for this integral and to rpove that $\lim_{n \to \infty}I=\frac{\pi}{2}$

Are you sure you wrote the integral right? As it is, it diverges because $\frac 1z$ is not integrable near $0$ :?

I think i don't think something obvious, but,fedja, we have $B(e^{iz})$ in the numerator, why it diverges? For example if $\alpha_k=0$ then we have $I=\int_o^{\pi} z^ndz$ and it is ok.
Only what can be changed to become explicit formulation of the problem is the changment of the numerator $B(e^{iz})$ by $Im(B(e^{iz}))$
Or i don't see something?

[quote="eugene"] we have $B(e^{iz})$ in the numerator, why it diverges? For example if $\alpha_k=0$ then we have $I=\int_0^{\pi} z^ndz$ and it is ok.
[/quote]
$|B(e^{iz})|=1$ when $z$ is real for any Blaschke product. Or do you actually want $B(z)$ in the numerator as your example suggests? :?

Yes, fedja, thank you very much- now i've edited the numerator and the denominator(the last one i changed just to pose the initial condition).
If $\alpha_k=0$ and if we approximate $sinz$ by $z$ we have that $I=\int_0^{\pi}\frac{\sin{nz}}{z}dz \to \frac{\pi}{2}$ when $n \to \infty$.
Actually this integral came from the Dirichlet kernel of the rational analogue of Fourier series.

I'm afraid that something is still incorrect: just take one Blaschke factor with $\alpha\notin\mathbb R$. Then $Im\,B(1)\ne 0$ and the integral diverges at $0$. So, should we also assume that all $\alpha_k\in(-1,1)$?

Yes, fedja, let $\alpha_k \in (-1,1)$.
My ideas to solve this problem is using firstly this problem http://www.mathlinks.ro/Forum/viewtopic.php?t=78115 to calculate integral over a non-closed contour(if we change the variable $\omega=e^{iz}$ then the integral will be form $1$ to $-1$), the integrand will become rational function and the residues at finite poles of $B(z)$ and $0$ are easy to calculate.
I don't know how to work out the residue at infinity http://www.mathlinks.ro/Forum/viewtopic.php?p=456843#
Thank you again


Every time I want to write something like :

Problem 1. This is the problem you need to solve...

Solution:

This is how we solve this problem ...

, in LaTex, the text comes out,

Problem 1. This is the problem you need to solve...

Solution:

~~~This is how we solve the problem...

.  How to make it come out the way I want ?

Remark: ~~~ means empty space!

I assume you mean [i]This is how we solve the problem... [/i] is indented; this is the normal behaviour for the first line of a paragraph as happens in most books. If you don't want a particular line indented then put \noindent at the start of the line.

If you don't want [i]any[/i] indenting for the first line of pargraphs then put [code]\setlength{\parindent}{0pt}
\setlength{\parskip}{\baselineskip}[/code] at the start of the document.


Am si eu o intrebare. Care este numarul functiilor surjective f:A->B, card(A) = n, card(B) = k, si mai mult, cam cum s-ar face demonstratia?

At least we should have $n\geq k$. Then the answer is \[k^n-(k-1)^n{k\choose 1}+(k-2)^n{k\choose 2}-...+(-1)^{k+1}1^n{k\choose k-1}.\] (it comes from the inclusion-exclusion formulae).

I hope it is right.

Cred ca mere dak consideri pentru fiecare element $y_i$ din $B$ multimea $X_i=\{f:A\to B|y_i\not\in f(A)\}$.
$|X_i|=(k-1)^n$, $|X_i\cap X_j|=(k-2)^n$...
Acuma dak $X$ e setul tuturor functiilor de la A la B... atunci facand $X-\bigcup_{y_i\in B}X_i$ cu principiul includerii si excluderii ar trebui sa gasesti formula pe care a zis-o Myth.


real analysis
function
integration
limit
real analysis unsolved

Let $ \{ f_n \}$ be a sequence of Lebesgue-integrable functions on $ [0,1]$ such that for any Lebesgue-measurable subset $ E$ of $ [0,1]$ the sequence $ \int_E f_n$ is convergent. Assume also that $ \lim_n f_n\equal{}f$ exists almost everywhere. Prove that $ f$ is integrable and $ \int_E f\equal{}\lim_n \int_E f_n$. Is the assertion also true if $ E$ runs only over intervals but we also assume $ f_n \geq 0 ?$ What happens if $ [0,1]$ is replaced by $ [0,\plus{}\infty) ?$

[i]J. Szucs[/i]

First let's see $f$ is integrable. Suppose on the contrary $f\not\in L^1[0,1]$. We will construct $E$, and subsequence $n_1<n_2 <\ldots $ for which $|\int_E f_{n_k}| \ge k$.
Since $f\not\in L[0,1]$ and $f_n\to f$ a.e. , $\int_0^1 |f_n|$ diverges (by Fatou's lemma), thus there exists $n_1$ with $\int_0^1 |f_{n_1}| > 4$. Hence, $|\int_{E_1} f_{n_1}| > 2$ ,where $E_1$ is one of the sets where $f_{n_1}$ takes positive or negative values.  Suppose, we have already constructed $n_1<n_2,\ldots<n_{k-1}$ and $E_1,E_2,\ldots, E_{k-1}$. We choose $\delta\,,\, 0<\delta<2^{-k} $ s.t. $\int_{\Delta} |f_{n_i}| <2^{-k}, i=1,2,\ldots k-1$, whenever $m(\Delta) <\delta$.
One of the two possibilities holds for sure:

1) $\int_{[0,1]\setminus E_{k-1}} |f| =\infty$

2) $\int_{E_{k-1}} |f| =\infty$

If 1) holds, we choose a set $ \Delta_k $ with: $ \Delta_k  \subset [0,1]\setminus E_{k-1}\,,\,  m(\Delta_k)<\delta $ and so that $\int_{\Delta_k} |f|=\infty$. Then, with the same argument as above, there exists $n_k$ and $\Delta^*_k \subset \Delta_k$ such that $|\int_{E_{k-1}\cup \Delta^*_k} f_{n_k}| > k+1$. Now, we set $E_k=E_{k-1}+\Delta^*_k$.
If 2) holds, we choose $ \Delta_k$ with: $\Delta_k  \subset E_{k-1}\,,\,  m(\Delta_k)<\delta $ and so that $\int_{\Delta_k} |f|=\infty$. Using convergence of $\int_{E_{k-1}}f_n$, similarly, there exists $n_k$ and $\Delta^*_k \subset \Delta_k$ such that $|\int_{E_{k-1}\setminus \Delta^*_k} f_{n_k}| > k+1$. In this case, we set $E_k=E_{k-1}\setminus \Delta^*_k$.
Notice that correction of $E_{k-1}$, on each step, by some small set $\Delta^*_k$ doesn't affect much $\int_{E_{k-1}} f_j\,,\, j=1,2,\ldots,k-1$.
Let $E=\lim_{n\to\infty} E_n $. It can be checked that $|\int_E f_{n_k}| \ge k\,,\, k=1,2,\ldots $, a contradiction with the convergence of $\int_E f_n$. Therefore $ \int_0^1 |f| < \infty$.

It remains to see that $\int_E f_n \to \int_E f$. Suppose again it's not true. Set $g_n=f_n-f $. Hence, $g_n \to 0 $ a.e. and $\int_E |g_n|$ doesn't converge to $0$.
[b]Claim[/b]: There exists $\epsilon>0$, such that for any $\delta>0$ there exists $\Delta \subset E\,,\, m(\Delta)< \delta$ with $\int_{E\setminus \Delta} |g_n| \to 0$ but $\int_{\Delta} |g_n| > \epsilon$ for infinitely many $n$.
Proof: Using for example Egorov's theorem and the fact $\int_E |g_n|$ doesn't converge to $0$. $\square$

The idea is to construct, in a similar way as above, sets $E_k\subset E$ with $|\int_{E_k} g_{n_k}| >\epsilon/4$, when $k$ is even, and small enough, when $k$ is odd.
Lets choose $E_1$, $n_1$ with $\int_{E_1}|g_1|<\epsilon/8$. Suppose, $E_1,\ldots,E_{k-1}\,,\, n_1<n_2,\ldots <n_{k-1}$ are already chosen, where $k$ is even.
We choose $\delta\,,\, 0<\delta<2^{-k} $ s.t. $\int_{\Delta} |g_{n_i}| <2^{-k}, i=1,2,\ldots k-1$, whenever $m(\Delta) <\delta$. The above claim yeilds there exists $\Delta_k\,,\, m(\Delta_k) < \delta$, $\int_{E\setminus\Delta_k} |g_{n_k}|<\epsilon/8 \,,\,   \int_{\Delta_k} |g_{n_k}|>\epsilon$. Thus for some $\Delta^*_k\subset \Delta_k$, it holds $|\int_{E_{k-1}\cup\Delta^*_k} g_{n_k}|>\epsilon/4$. We set $ E_k=E_{k-1}\cup\Delta^*_k $.
The next step, for $ k+1 $, is similar: choose $ \delta\,,\, 0<\delta<2^{-k-1} $ s.t. $ \int_{\Delta} |g_{n_i}| <2^{-k-1}, i=1,2,\ldots k $, whenever $m(\Delta) <\delta$. For some $\Delta_{k+1}\subset E\,,\, m(E_{k+1})<\delta$, we choose $g_{n_{k+1}}$ with $\int_{E_k \setminus \Delta_{k+1}}|g_{n_{k+1}}|<\epsilon/8$ and set $E_{k+1}=E_k \setminus \Delta_{k+1}$.
Finally, we check out that for $E^*=\lim_{k\to\infty} E_k$ , the sequence $|\int_{E^*}g_{n_k}|$ alternatively is less than $\epsilon/8$ and greater than $\epsilon/4$, thus it cannot converge, a contradiction.

The assertion remains true if $[0,1] $ is replaced by $[0,\infty)$. We just apply the proven fact on $[0,N], N\to \infty$.
The assertion is false if $E$ runs over intervals and $f_n\ge 0$. As a counterexample one can take $f_n(x)=n\,,\, x\in [0,1/n]$ and $f_n(x)=0, x\in  (1/n,1]$.

[b]Comment[/b]. As a motivation to the above construction, one can consider a sequence of functions $f_n\,;\, f_n(x)=1/x, x\in [1/n,1]$ and $f_n$ being some negative constant on $[0,1/n]$ such that $\int_0^1 f_n=0$. Clearly $f_n\to 1/x \not \in L^1[0,1]$. Thus, if the problem claim is true, there must exist $E\subset [0,1]$ with $\int_E f_n$ not converging. Finding such $E$ gives us the basic idea of the above construction.


Hi,

1. A charged particle moves in a straight line through a particular region in space. Could ther be a nonzero magnetic field in this region?

2. If a moving charged particle is deflected sideways in some region of space, can we conclude for certain B is not=0 in that region?

the first one is possible if the magnetic field is present in the direction of the velocity of the charged particle hence the cross product $q{\vec{v}X{\vec{B}}}$
the second is possible to an electric field or a magnetic field or both , hence we cant come to any conclusion


calculus
integration
logarithms
trigonometry
function
limit
calculus computations

For the following integrals, state wether they are convergent or divergent, and give your reasons:

$1.\int^{1}_{0}\ln xdx$

$2.\int^{1}_{0}\sqrt{x}\ln xdx$

$3.\int^{\infty}_{1}\frac{\ln x.\cos x}{x^{2}+1}dx$

$4.\int^{\infty}_{1}\frac{\sin\sqrt{x}}{x+x^{4}}dx$

I have solved $1$,$3$ and $4$ but couldn't solve $2$, i tried many methods for $2$ but none of them worked, maybe there is an easy solution that i can't find...

#2 can be computed exactly - it's a straightforward integration by parts.

$\int_{0}^{1}x^{1/2}\ln x\,dx=\left.\frac23x^{3/2}\ln x\right|_{0^{+}}^{1}-\int_{0}^{1}\frac23x^{3/2}\frac1x\,dx$

$=0-0-\frac23\int_{0}^{1}x^{1/2}dx=-\frac49.$

It's not even an improper integral, as $\sqrt{x}\ln x$ can be extended to a continuous functions on $[0,1].$

For any $m>-1,\ \int_{0}^{1}x^{m}\ln x\,dx=-\frac1{(m+1)^{2}}.$

oh sorry :oops: thats right, i thought its gonna be a tough problem so i didn't even think about integration by parts.

but about number $3$, here's my solution:

[hide]let $\ln x=t$ so $\frac{dx}{x}=dt$ and:

$I=\int^{\infty}_{1}\frac{\ln x\cos x}{x\left(x+\frac{1}{x}\right)}dx=\int^{\infty}_{0}\frac{t\cos(e^{t})dt}{e^{t}+e^{-t}}=\int^{1}_{0}\frac{t\cos(e^{t})dt}{e^{t}+e^{-t}}+\int^{\infty}_{1}\frac{t\cos(e^{t})dt}{e^{t}+e^{-t}}=I_{1}+I_{2}$

$I_{1}$ is convergent obviously, but for $I_{2}$ by limit comparison test:

$\lim_{t\rightarrow\infty}\frac{\frac{t\cos(e^{t})}{e^{t}+e^{-t}}}{\frac{t\cos(e^{t})}{e^{t}}}=1$

so we have to discuss about $\int^{\infty}_{1}\frac{t\cos(e^{t})}{e^{t}}dt$ but again:

$\lim_{t\rightarrow\infty}\frac{\frac{t\cos(e^{t})}{e^{t}}}{\frac{1}{t^{2}}}=0$
and because $\int^{\infty}_{1}\frac{dt}{t^{2}}$ is convergent so $I_{2}$ is also convergent and at last $I$ is convergent.[/hide]

i hope it's correct :D

Since #3 does turn out to be absolutely convergent, it's a waste of time to hang onto $\cos x$ rather than just estimating it as $1.$

My "instant" evaluation of the integral as absolutely convergent goes something like this: "The cosine is just 1. The denominator looks like $x^{2}.$ The logarithm grows slower than any power of $x,$ so I can trade it off for a small fractional power of $x$ which still leaves me with 1 over a power of $x$ where the power is between 1 and 2; so it converges."

Here's another, more formal approach:

$\left|\int_{1}^{\infty}\frac{\ln x\cdot\cos x}{x^{2}+1}\,dx\right| \le\int_{1}^{\infty}\left|\frac{\ln x\cdot\cos x}{x^{2}+1}\right|\,dx \le\int_{1}^{\infty}\frac{\ln x}{x^{2}}\,dx$

Now integrate by parts.

$\int_{1}^{\infty}\frac{\ln x}{x^{2}}\,dx=\left.-\frac{\ln x}x\right|_{1}^{\infty}+\int_{1}^{\infty}\frac1x\cdot\frac1x\,dx$

$=0-0+\int_{1}^{\infty}\frac1{x^{2}}\,dx=1,$ converges, hence the original integral converges absolutely.

On #4 I hope you're not doing anything with the sine other than estimating it by 1. That integral converges absolutely by comparison to the integral of $\frac1{x^{4}}.$

that's right :oops: , wow nice solutions, thank you very much :)

is it always true that
$\left|\int_{a}^{b}f(x)dx\right| \leq \int_{a}^{b}\left|f(x)\right|dx?$

offcourse yes, do you have any counter example? :)

let
$f(x)_{(+)}=f(x)$ if $f(x)>0$ and $0$ if $f(x)\leq0$
$f(x)_{(-)}=-f(x)$ if $f(x)<0$ and $0$ if $f(x)\geq0$

so $f(x)=f(x)_{(+)}-f(x)_{(-)}$
    $\left|f(x)\right|=f(x)_{(+)}+f(x)_{(-)}$
and we have:

$\left|\int f(x)dx\right|=\left|\int f(x)_{(+)}dx-\int f(x)_{(-)}dx\right|$

and,

$\int\left|f(x)dx\right|=\int f(x)_{(+)}dx+\int f(x)_{(-)}dx$


trigonometry
inequalities
geometry
national olympiad

[size=150][b]International Mathematical Olympiad 2007
Hong Kong Team Selection Test 1[/b][/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107258]1[/url]. Let $p,q,r$ and $s$ be real numbers such that $p^{2}+q^{2}+r^{2}-s^{2}+4=0$. Find the maximum value of $3p+2q+r-4|s|$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107261]2[/url]. Let $A$, $B$ and $C$ be real numbers such that
(i) $\sin A \cos B+|\cos A \sin B|=\sin A |\cos A|+|\sin B|\cos B$,
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C-\sin A)^{2}+(\cot C-\cos B)^{2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107257]3[/url]. Let $ABC$ be a triangle with $AB=c$, $BC=a$ and $CA=b$. Show that
$\frac{\cos A}{a^{3}}+\frac{\cos B}{b^{3}}+\frac{\cos C}{c^{3}}\geq\frac{3}{2abc}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107259]4[/url]. (a) Given five points on a plane such that no three of the points are collinear. Show that among the triangles which are drwan using any three of these five points as vertices, at least three of the triangles formed are not acute-angled triangles. (An acute-angled triangle is one in which all the three interior angles are acute angles.)

(b) Given any 100 points on a plane such that no three of the points are collinear. SHow that among the triangles which are drawn using any three of these 100 points as vertices, at least 30% of the trinagles are not acute-angled triangles.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]5[/url]. The sequence $\{a_{n}\}$ is defined by $a_{1}=0$ and $(n+1)^{3}a_{n+1}=2n^{2}(2n+1)a_{n}+2(3n+1)$ for all integers $\geq 1$. Show that infintely many members of the sequence are positive integers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107260]6.[/url] Determine all pairs $(x,y)$ of positive integers such that $\frac{x^{2}y+x+y}{xy^{2}+y+11}$ is an integer.

A very weird test,right?Three inequality (If you count 4(b) as an inequality, then 3.5) with no pure Geometry problem.
I think I did $n$ questions in this test, which $2\leq n \leq3$. A poor result  :(


geometry
geometric transformation
reflection

Given a straight line $MN$ . Above $MN$ , there are two different points $A,B$ (assume A,M are at  left and B,N are at right) . Pick a point $P$ on $MN$ and connect $AP$ , $BP$ . Show that $AP+BP$ is minimised when $\angle APM=\angle BPN$

 :)

Reflect $A$ through $MN$, call the reflection $A'$. Then $A'B$ is the shortest distance. Because choose any $P$ on the segment, we always have $AP=A'P$, hence $A'B \leq A'P+PB$ with equality hold when $P = MN \cap A'B$.

The reflection shows that $\angle{APM}=\angle{BPN}$.


i lost the game http://www.losethegame.com

i lost two days back after a long time...

omg i havent lost for almost a month.
GAaa

I lost.

Maybe I should do lots and lots of math...


Σας παραθέτω το παρακάτω πρόβλημα το οποίο βρήκα με μια φανταστική λύση :)

Τοποθετούμε αστερίσκους σε κάποια κελιά (τετραγωνάκια) μιας $ m$ επί $ n$ σκακιέρας, όπου $ m<n$ (m είναι οι σειρές και n οι στήλες), έτσι ώτε να υπάρχει τουλάχιστον ένας αστερίσκος σε κάθε στήλη. Να αποδείξετε ότι υπάρχει τουλάχιστον ένας αστερίσκος τέτοιος ώστε να υπάρχουν περισσότεροι αστερίσκοι στην σειρά του απ' ότι στη στήλη του!

Δεν ξέρω αν αυτή είναι η λύση που έχει βρει ο Δημήτρης:

Χρησιμοποιώ επαγωγή στο n. Για n=2 ισχύει. Παίρνω τώρα οποιεσδήποτε m στήλες ας είναι $ c_1,\ldots,c_m$.

1) Αν μπορούμε να αντιστοιχίσουμε τις m σειρές στις m στήλες ώστε σε κάθε ζευγάρι να υπάρχει ένα αστεράκι: Μετρώ τα αστεράκια με δυο διαφορετικούς τρόπους. Από την μια, κοιτάζωντας την στήλες, έχουμε τουλάχιστον $ |c_1| \plus{} \ldots \plus{} |c_m| \plus{} (n\minus{}m)$ αστεράκια, όπου με $ |c_i|$ συμβολίζω τα αστεράκια της στήλης $ c_i$. Από την άλλα, αν η σειρά $ r$ έχει αστεράκι στην στήλη $ c_i$, τότε $ |r| \leq |c_i|$ αλλιώς τελειώσαμε. Άρα, λόγω της πιο πάνω αντιστοιχίας, κοιτάζωντας τις στήλες θα έχουμε το πολύ $ |c_1| \plus{} \ldots \plus{} |c_m|$ αστεράκια. Άτοπο

2) Δεν υπάρχει τέτοια αντιστοιχία. Από θεώρημα γάμου αυτό σημαίνει ότι υπάρχουν $ k \leq m < n$ στήλες οι οποίες έχουν αστεράκια μόνο σε  $ \ell < k$ από τις σειρές. Αλλά τότε τελειώσαμε από την επαγωγική υπόθεση.

Να βαλω και εγω μια λυση νομιζω to good to be true αλλα δεν βρισκω λαθος.

Βαζω συντεταγμενες στη σκακιερα(1,2,...,ν για στηλες και 1,2,...,μ για σειρες).

Τον αριθμο των αστερισκων που βρισκονται στη σειρα ρ και τη στηλη τ θα τον ονομαζω κ_ρ,τ υπο του περιορισμου οτι αυτος ο αριθμος των αστερισκων ειναι 0 ή 1. Συνολικα λοιπον θα υπαρχουν [size=150]Σ[/size]κ_ρ,τ=γ αστερακια

Εστω τωρα οτι η προταση προς αποδειξη δεν ειναι σωστη.

Τοτε θα προκυψουν καποιες ανισοτητες σαν και αυτη που θα ισχυε για το κελι στη στηλη 1 και στη σειρα 1. κ_1,1+κ_2,1+...κ_μ,1≥κ_1,1+κ_1,2+...+κ_1,ν(αν θελετε να τις γραψω explicitely πειτε μου)

Εεε, αν τις προσθεσουμε ολες κατα μελη θα παρουμε μγ≥νγ, αρα μ≥ν, ατοπο(απο υποθεση).

Αν βρειτε λαθος ενημερωστε με...

Υ.Γ. Αν η λυση ειναι σωστη τοτε το προβλημα γενικευται και για περισσοτερους αστερισκους... :wink:
 :wink:

Κατ' αρχάς να διευκρινίσω ότι το πρόβλημα το "βρήκα με μια φανταστική λύση"! Δεν βρήκα εγώ τη λύση (πολύ θα το 'θελα βέβαια :P )!

Αναφορικά με τη λύση σου Δημήτρη, δεν την καταλαβαίνω ομολογουμένως :blush:(αν και είμαι σίγουρος ότι η επαγωγή σου είναι σωστή :wink: ).  Πρώτον την αντιστοιχία που κάνεις στο 1 μπορείς να την εξηγήσεις λιγάκι περισσότερο? Σε ποια "ζευγάρια" αναφέρεσαι και πώς προκύπτει επίσης το $ |c_1| \plus{} \ldots \plus{} |c_m| \plus{} (n \minus{} m)$?
Παρεπιμπτόντως, μήπως θα μπορούσες επίσης να εξηγήσεις τι είναι το θεώρημα γάμου?

Filosofimene, η λύση μου φαίνεται πραγματικά πάρα πολύ ωραία και σωστή (εκτός κι αν δεν βλέπω τίποτα, στην οποία περίπτωση συγχωρέστε με...)! Η λογική της είναι κατά βάση ίδια με αυτή που έχω δει!

Ας αναφέρω τώρα και τη λύση αυτή που είδα και με ενθουσίασε!

Δημιουργούμε μια σκακιέρα Β αντικαθιστώντας κάθε αστερίσκο με τον αριθμό $ 1/k$ όπου $ k$ ο αριθμός των αστερίσκων στη σειρά του (στην αρχική σκακιέρα). Ετσι στη Β το άθροισμα των αριθμών σε κάθε σειρά, είναι $ 0$ ή $ 1$ ανάλογα με τον αν η αντίστοιχη σειρά στην αρχική σκακιέρα περιέχει τουλάχιστον έναν αστερίσκο ή οχι. Ετσι το άθροισμα όλων των αριθμών στη Β είναι ένας αριθμός μικρότερος ή ίσος του $ m$.
Φτιάχνουμε μετά μια αντίστοιχη σκακιέρα Γ αντικαθιστώντας κάθε αστερίσκο με τον αριθμό $ 1/\ell$ όπου $ \ell$ ο αριθμός των αστερίσκων στη στήλη του. Το άθροισμα σε κάθε στήλη πρέπει να είναι ακριβώς 1 αφού κάθε στήλη περιέχει τουλάχιστον έναν αστερίσκο. Ετσι το άθροισμα όλων των αριθμών στη Γ ισούται με $ n$.
Tώρα αφού $ m < n$ το άθροισμα όλων των αριθμών στο Β είναι μικροτερο αυτού στο Γ. Αυτό σημαίνει ότι θα υπάρχει τουλάχιστον ένα τετραγωνάκι στο Β που να έχει μικρότερο αριθμό από το αντίστοιχο τετραγωνάκι του στο Γ. Δηλαδή αν $ k, \ell$ είναι τα πλήθη των αστερίσκων στη σειρά του και στη στήλη του αντίστοιχα θα ισχύει:$ \frac {1}{k}<\frac {1}{\ell}\iff k > \ell$ που είναι αυτό που θέλαμε!

[quote="filosofimenos"]
Αν βρειτε λαθος ενημερωστε με...
[/quote]

Και εγώ στην αρχή κάτι παρόμοιο σκέφτηκα. Το πρόβλημα είναι ότι δεν έχεις όλες τις ανισότητες που έγραψες. Οι μόνες ανισότητες που μπορούμε να υποθέσουμε είναι αυτές για τις οποίες κ_ρ,τ =/= 0. (Θέλουμε στο τετραγωνάκι που θα βρούμε να υπάρχει ο αστερίσκος.)

Δημήτρη, η λύση που έχεις δει είναι πάρα πολύ όμορφη. Και εγώ θα ήθελα να την είχα βρει.

Για την δική μου λύση. Παίρνω τισ στήλες $ c_1,\ldots,c_m$. Yποθέτω πως για κάθε i, υπάρχει σειρά r_i (όλες διαφορετικές μεταξύ τους) ώστε να υπάρχει αστεράκι στο κοινό τετράγωνο των στηλών c_i κai r_i (για κάθε i). Μετράω τώρα αστεράκια. Η στήλη c_i έχει |c_i| αστεράκια (για $ 1 \leq i \leq m$.). Όλες οι υπόλοιπες στήλες έχουν από ένα τουλάχιστον αστεράκι. Σύνολο τουλάχιστον $ |c_1| \plus{} \ldots \plus{} |c_m| \plus{} (n\minus{}m)$ αστεράκια. Από την άλλη αφού η σειρά r_i έχει κοινό αστεράκι με την c_i, τότε πρέπει να έχει το πολύ |c_i| αστεράκια. Σύνολο, το πολύ $ |c_1| \plus{} \ldots \plus{} |c_m|$ αστεράκια, άτοπο.

Άρα δεν μπορώ να βρω σειρές r_i όπως πιο πάνω.

Θεώρημα Γάμου: Έχουμε μ άντρες και μ γυναίκες. Κάθε άντρας έχει μια λίστα με τις γυναίκες που δέχεται αν παντρευτεί. Μπορούμε να τους παντρέψουμε ώστε κάθε άνρας να παντρευτεί μια από τις γυναίκες που επιθυμεί; (Ελπίζω να μην κατηγορηθώ για ρατσισμό αλλά οι γυναίκες εδώ δεν έχουν λόγο.)

Αν κάποιος άντρας δεν του αρέσει καμία γυναίκα, τότε φυσικά δεν μπορούμε.
Αν υπάρχουν δύο άντρες που τους αρέσει η ίδια γυναίκα (και καμία άλλη) πάλι δεν μπορούμε.

Ομοίως αν υπάρχουν κ άντρες στους οποίους συνολικά αρέσουν το πολύ κ-1 γυναίκες δεν μπορούμε.

Το θεώρημα του γάμου λέει ότι αυτή η συνθήκη εκτός από αναγκαία είναι και ικανή.

Αφήνω ως άσκηση την χρήση του θεωρήματος στην περίπτωση 2 της λύσης μου.

[quote="Demetres"][quote="filosofimenos"]
Αν βρειτε λαθος ενημερωστε με...
[/quote]

Και εγώ στην αρχή κάτι παρόμοιο σκέφτηκα. Το πρόβλημα είναι ότι δεν έχεις όλες τις ανισότητες που έγραψες. Οι μόνες ανισότητες που μπορούμε να υποθέσουμε είναι αυτές για τις οποίες κ_ρ,τ =/= 0. (Θέλουμε στο τετραγωνάκι που θα βρούμε να υπάρχει ο αστερίσκος.)

[/quote]

Ναι, ρε Δημητρη δικιο εχεις, πως δεν το ειδα.?.. :wallbash: Τεσπα, απογοητευτηκα τωρα :(

Οταν προσπαθουσα την ασκηση ειχα κι 'αλλη μια προσεγγιση στο μυαλο μου η οποια ομως θελει αρκετο σπαζοκεφαλιασμα και ειναι και ασχημη. Θα το προσπαθησω και ετσι και αν βγαλω λυση θα την ποσταρω...


[quote="Pirkuliyev Rovsen"]Solve in the equation

$ \{x \} \equal{} \frac {x \plus{} [x]}{2009}$

_______________________________________
Azerbaijan Land of Fire  :roll:
 :ninja:[/quote]

Let $ [x]\equal{}n$ and $ {x}\equal{}a$.

Then the equation turn out to $ n\equal{}1004a$.

So, $ x\equal{}\frac{1005n}{1004}$, where $ n$ is a interger.

[quote="Stephen"]

$ x \equal{} \frac {1005n}{1004}$, where $ n$ is a interger.[/quote]

I don't understand your solution....

If $ n\equal{}1004 \Rightarrow x\equal{}1005$, but $ \{1005\} \ne \frac{1005 \plus{} \lfloor 1005 \rfloor}{2009}$
:o

In Stephen solution, $ x\equal{}n\plus{}a$, a$ \in[0,1)$ and $ 1004a\equal{}n$, so $ n\in[0,1003]$

Hence the result : $ x\equal{}\frac{1005n}{1004}$ for $ n\in[0,1003]$

[quote="pco"]In Stephen solution, $ x \equal{} n \plus{} a$, a$ \in[0,1)$ and $ 1004a \equal{} n$, so $ n\in[0,1003]$

Hence the result : $ x \equal{} \frac {1005n}{1004}$ for $ n\in[0,1003]$[/quote]

OK, I see it now. In Stephen solution is $ \{x\}\equal{}a$ and not $ x\equal{}a$. Thank you.


function
integration
calculus
calculus computations

Given a real differentiable function $ f(x)$, is it true in general that $ \int_0^1 f(2x) \equal{} \int_0^2 f(x)$?

There is some case, say, for $ f(x)\equal{}0$ or $ f(x)\equal{}\frac{1}{2}$.

That second case doesn't work.  The correct statement is $ \int_{0}^{1} f(2x) \equal{} \frac{1}{2} \int_{0}^{2} f(x)$, which is a matter of substitution.

if you dont give $ dx$ none of them works :rotfl:  :D  :P

Lol.. but essentially I forgot the factor of $ \frac{1}{2}$. Thanks!


geometry
Support
geometric transformation
search
group theory
abstract algebra

Given that we currently have 7813 problems from 125 competitions listed in the [url=http://www.mathlinks.ro/resources.php]Mathlinks resources section[/url] that cover regional, national and international school olympiads. But also competitions at the undergraduate level. Thus I thought some difficulty level categorization cannot hurt. There could be a sticky discussion thread and/or an AoPS wiki entry (because everybody can edit it) that discusses the difficulty level range, or just classifies it on a rough interval list going from 1 to 10 where 10 indicates the hardest level. I know there has been kind of related effort in the AoPS wiki that gives a small number of problems according topic area as algebra etc. and level of sophistication as intermediate. Maybe some extra comments would also be helpful that may discuss on other attributes as time/years, e.g. the International Mathematical Olympiad (IMO) got significantly harder over time. Discussion should be explicit to the actual competition, e.g. not [url=http://www.mathlinks.ro/resources.php?c=37]China[/url] in general but its olympiads it actually hosts:

[b]- China:[/b]

[i]-> Team Selection Test [/i]

score, comments etc.

[i]-> National Olympiad[/i]

score, comments etc.

[i]-> Western Mathematical Olympiad [/i]

score, comments etc.

[i]-> China Girls Math Olympiad [/i]

score, comments etc.

[i]-> South East Mathematical Olympiad [/i]

score, comments etc.

[i]-> Northern Mathematical Olympiad [/i]

I think this would be an excellent start for new members to explore the huge variety of posted problems and to contribute their ideas. This classification helps not to drown in the "impossibly" big numbers of problems but actually to get some decision support on which problems they may want to focus on and direct their energies on right now. It is kind of the same confusion that you sometimes feel in the supermarket deciding on products to choose, and recently nicely described in an [url=http://www.economist.com/science/displaystory.cfm?story_id=12792420]economist article.[/url] Members' help creating this classification system would be appreciated very much. They also can take the opportunity to discover exciting pearls whose discussion may yield some unexpected insight and/or treasure, and they would have otherwise missed.

[quote="greentreeroad"]So do you mean we rate the difficulty for each problem or just for each test?[/quote]

On a problem basis would be excellent as it is done by the Russian problem collection [url=http://problems.ru/]Problems.ru[/url], or look here for the [url=http://translate.google.com/translate?u=http%3A%2F%2Fproblems.ru%2F&sl=ru&tl=en&hl=en&ie=UTF-8]Google translation,[/url] and sorted by category. But I guess this effort would be infeasible.

Thus I thought about it on a competition basis, and because one is ranking competitions and not problems rather about a difficulty level interval. For example China TST could be ranked from 8-10.

But, each individual has his/her own perspective as to the rating... For example, a 9th grader from a less prominent country might rate a problem extremely difficulty, like a 10, and a 11th grader from say Romania would deem the problem as not-so-hard, like a 7. Since there are discrepancies between opinions, I see that you want to use the 'interval' method but it is also useful to add in other statistical measures like a simple mean on the ratings. After all, it is statistically sound that the mean 'averages' in differences. My opinion: the ratings that users say 'easy' or 'hard', please people, don't say them, they are USELESS. Thank you.

The categorization of problems is really a nice idea because there are a lot of times when some problems are just really hard to search that are related to certain topics of interest due to the large amount of problems posted (there are even some that are unsolved too, have a look!). Even the Search feature might not yield the desired results when the posts do not contain the keywords that are put on the search, mostly the problems are solved with equations or mathematical formulas.

To make Olympiad Section more useful, members should make more efforts. Some problems are not answered and I am sure this is not because no one is able to solve them. I don't like it.

It is easy to find examples:

http://www.mathlinks.ro/viewtopic.php?t=247057
http://www.mathlinks.ro/viewtopic.php?t=247109 (I would reply this, but for some reason I don't remember I lost my counts)
http://www.mathlinks.ro/viewtopic.php?t=247002
http://www.mathlinks.ro/viewtopic.php?t=247367
http://www.mathlinks.ro/viewtopic.php?t=247613
http://www.mathlinks.ro/viewtopic.php?t=245807
http://www.mathlinks.ro/viewtopic.php?t=247479

The worst is "posted before" without links.

@feliz: I don't quite understand your issue. I rather think that providing a difficulty classification of the competitions in the resources section makes it more likely that those problems are seen by more users and hence are more likely to receive a solution.

@10000th User: I know it is relative. But you may want to check and browse through the site problems.ru as mentioned above to get an idea. You probably realize that you also can search the site giving a difficulty range which is very helpful for students to get access to the problems at the appropriate difficulty level.

It says you are from New Jersey, so I guess you are familiar with the [url=http://www.mathlinks.ro/resources.php?c=182]USA System.[/url] Let me suggest to set AMC 8 at level 1-2 and USA TST at 8-10 for example. Then accordingly you should sample a few problems from certain competitions: look at them, think about them and solve of them to get the idea and classify them in correspondence with the US set difficulty intervals. Some time ago there was a somehow related thread in the [url=http://www.mathlinks.ro/viewtopic.php?t=241217]AMC forum.[/url]

I guess the first step now is to set up the AoPS wiki page. Let me know your ideas please.

[quote="orl"]
On a problem basis would be excellent as it is done by the Russian problem collection [url=http://problems.ru/]Problems.ru[/url], or look here for the [url=http://translate.google.com/translate?u=http%3A%2F%2Fproblems.ru%2F&sl=ru&tl=en&hl=en&ie=UTF-8]Google translation,[/url] and sorted by category. But I guess this effort would be infeasible.[/quote]

I think even so we should still value single problem instead. Because in a same competition and even in a same test, the difficulty may vary in a great scale. While problem 1 in IMO may be an excellent exercise for beginner, some problem 3 are way too difficult for an experienced solver. I don't think it really matters if we can't rate all the problems quickly, because there will certainly be enough problems rated at any time for people to try.

The biggest issue in practicing is that when I got stuck on a problem but don't want to look at the solution. Now, I don't know whether this problem is way too hard(so that I can never get it without clues) or just one step away. Merely rate the difficulty of an Olympiad can't distinguish between these two type of problems.(e.g. IMO 2005 number 6 is far more reachable than IMO 2007 number 6)

@greentreeroad

Clearly as indicated by the statistics I agree that problem 6 of [url=http://www.imo-official.org/year_statistics.aspx?year=2007]IMO 2007[/url] is more challenging than that of [url=http://www.imo-official.org/year_statistics.aspx?year=2005]IMO 2005.[/url]

But I want to emphasize that perceived difficulty does not only depend on the contestant's skill but also what is called luck. Part of luck is what you kind of problems and articles you have seen before. IMO 2008, Problems 6, gets a lot easier if you happen to know the [url=http://en.wikipedia.org/wiki/Restricted_sumset]Restricted sumset / Combinatorial Nullstellensatz.[/url] On mathlinks for example it was covered in a thread on [url=http://www.mathlinks.ro/viewtopic.php?t=19496]Combinatorial Nullstellensatz by sam-n.[/url]

Knowing that several IMO contestants already work on advanced math problems they may have come across the lecture notes for [url=http://www.math.cmu.edu/~pikhurko/AlgMet.ps]Algebraic Methods in Combinatorics, chapter 11 by Oleg Pikhurko.[/url] It was also described in "Задачи Санкт-Петербургской олимпиады школьников по математике 2005 г." on page 97 and the following, by Fedor Petrov and Vesselin Dimitrov. But apparently it did not help [url=http://www.imo-official.org/team_r.aspx?code=RUS&year=2007]Russia's 2007 IMO Performance[/url] much.

As I said above ranking on an individual problem's basis would be best. And Valentin may adopt the post quality ranking to a difficulty ranking for the problems posted to the resources section. But Val first needs to agree to do it. Even assuming Val's work it poses the question on who should be eligible to rank the problems? For less than advanced users let me say USA MO, Problem 1 or TST, Problem 6 does not make a difference they are all perceived as 10. Then as said before who is doing all the work of ranking the problems. Many times I have seen users with great ambitions to post Kvant problems, Tournament of Towns, you name it but this ambition mostly ended rather quickly. Same case here. Many users would benefit from it but few are capable of doing it reliably or are interested in it I guess.

I know a ranking on an individual's problem basis drives down the variance in the difficulty estimation process with the single problem in view but again is it feasible? But as discussed above the type of problems a rater has seen before and his level of problem-solving skills adds quite some variance to it, even with ranking each problem separately.

But the competition-wise ranking would be available for all to edit. For example for competitions that exist for a long time and are very popular you may subgroup in years and problems, e.g.

[i]- IMO 1959-1968

-- Problem 1/4:
--- difficulty level, comments etc.

-- Problem 2/5:
--- difficulty level, comments etc.

-- Problem 3/6:
--- difficulty level, comments etc.

IMO 1969-1968
...[/i]

For problems which are unusually easy or hard as [url=http://www.mathlinks.ro/viewtopic.php?p=365171#365171]IMO 1996, Problem 2[/url] one might put the idea in the comments.

Again I would like to have further input to my ideas.

How about suggest/force a screen for rating each problem every time you want to reply a topic? or even read it?

By the way, how can I do to send problems for that section?, for example: here are the missing ones for the iberoamerican competition:

[url]http://www.oei.es/oim/problemas.htm[/url], do you need a translation from Spanish to English?

or here the Argentinean competitions [url]http://www.oma.org.ar/enunciados/index.htm[/url].

[quote="mszew"]How about suggest/force a screen for rating each problem every time you want to reply a topic? or even read it?[/quote]

Again, do you really want that any user can rank? This would methodology would yield a ranking, alas a very noise of what you are hoping for. Then what is the point of the ranking result? To whom will it be useful? Will anybody care about it then? Ranking should not end up in the implementation in yet another useless feature.

IMO we should strive for a sufficiently reliable and consistent ranking which may require some longer term ML/AoPS members because they got more opportunity to see what they ranking should be like. And forcing anybody to rank problems may annoy certain users and seduce them to enter arbitrary choices to continue with their reply.

[quote="mszew"]
By the way, how can I do to send problems for that section?, for example: here are the missing ones for the iberoamerican competition:

[url]http://www.oei.es/oim/problemas.htm[/url], do you need a translation from Spanish to English?

or here the Argentinean competitions [url]http://www.oma.org.ar/enunciados/index.htm[/url].[/quote]

Well, you post the translated problems to the appropriate topics in the olympiad section. Collect the URLs and post an overview thread in the "National Olympiads" forum and/or pm a moderator who adds them to the resources section. We are always eager to add new problems to the resources section. All your suggestions are welcome.

But this thread is more about on how to make the resources section more useful, i.e. given a certain ML/AoPS user how to generate a subset of problems for him/her appropriate for him/her. A first step in this direction IMO is to describe the available competitions in the resources section in terms of difficulty within the AoPS wiki. Though surely the idea of tagging individual problems with properties as topic (rough topic classification in olympiad section is obvious) and difficulty level, as suggested by other users above, shall also be considered.

What about ratings for problems like post ratings, but available for only those users who have more than x posts, or are here longer than y days?

Resources section - rate the competitions with the easiest problem difficulty and with hardest problem difficulty. For example, if IMO 2008 Problem 1 or 4  is 6, and IMO 2008 Problem 3 or 6 is 8, then competition would be 6-8.

[quote="beyrem"]well I think orl suggestion is very interesting. I mean, look to the number of problems in this website !! we need some order here !  ( and not only a classification with countries ) ..
so please admins , think about it , some new thnigs for the new year ...  :roll:[/quote]

But to be honest I do not like your attitude. This should be a joint effort of the admins and mods. Your post could also be interpreted differently. Just coming to the site as a new member and you second my proposal. But you do not propose any efforts yourself. Guys, it is for mutual benefit. Somebody could have created the wiki page to get things going...

Hi all,

I think that user-based rating is too complicated to implement, because you need an automated tool to establish the skill level of a given user, so that you know how much his/her opinion contributes to the difficulty rating of the problem (a 10 rating on an IMO problem 1 from an inexperienced user should have a very small weight).

Personally, in terms of usability for this website, I have three suggestions:
[list]
1) Simplify post rating: 6 possible ratings is way too complicated. I usually can appreciate a post as "useful" or "not useful". 3 items should be enough (bad, average, good).

2) Since you are striving to identify each topic with a single problem discussed, implement tagging for threads. Associate with each discussion thread tags such as "IMO 2008", "Combinatorics" and "double counting". Combined with auto-complete for the tags when entering them(similar to blogger interface) you should be able to improve the search in the website.

3) Finally, if you are really bored and have a lot of spare time, you can implement a tool to automatically tag all problems so far (I think a keyword-based approach should give pretty good results). Also, you can make the tags editable so that they can be corrected on the fly.
[/list]
Tagging would improve usage of the website because after learning a new technique it's usually useful to solve a few similar problems in order to make sure that you understood it really well.

Also, I think that the "difficulty" of a problem is relative, so it's not useful for a user to know such a rating. Perhaps it will discourage many users to stop working on difficult problems, since many people only like to do things they are good at.

Happy new year everyone, and keep up the good work with this forum!

PS: I hope no-one feels offended by this post, it's only [b]my opinion[/b] :)

I want to declare my idea too. In the first glance, classification for difficulty seems like a good idea. Because, you can know a problem's difficulty -more or less- and decide that if it worths solving or not. In my opinion, except the difficulty of making this classification, it may have also some bad results. Starting a problem with the information that it considered as really hard, turns solving this problem to a psychological war inside. Evaluating your ideas about that if they are really hard, evaluating yourself about that if you are really good enough to solve this hard one... etc. These happen because of the so-called difficulty of the problem, not because of the problem. I think starting a problem without knowing anything about it is the best.
I think stating unexpectedly good problems from some not well-known olympiads can be a good idea. Because, the probability that the Russian Olympiad -or any in this level- includes really bad problems, is really less, so everyone should solve it without any warnings or classifications. But for the others some suggestions can be helpful. (Maybe with an asterisk or something)
For example; the 6th question of this year's Turkey NMO is really nice.

Thanks for your account on that. But I think you got the idea in a slightly different view. If you are doing it on a competition-basis it is really worthwhile, e.g. how does the Singapore TST compare to the first round of the German Bundeswettbewerb Mathematik, Round 1 for example? I am pretty sure most members on this site cannot immediately compare the problem-solving difficulty of the Centro American competition with the South Africa National Mathematical Olympiad, can you? I doubt that even IMO gold medal winners as Umut can do it. I hope I am wrong about this view. But the issue here is not to drown in an unstructured stream of thousands of problems but to actively filter down suitable problems to make the resources section a worthwile area to prepare for competitions.

Let me say you are solving problems from the USA-competition AIME. Then I would like to know which other competition problems may I tackle that have rougly the same problem-solving difficulty. And then I could browse the wiki for difficulty ratings and comments. Try to solve some of the suggested competition problems, adjust the wiki evaluations possibly etc. in a feedback-loop.

This competition-based rating could still make people avoiding to solve IMO problems in the first place but this is not our issue here.  :D

And mathematical difficulty can roughly be measured in two dimensions. The first component is the problem-solving component as suggested above. The second part are the mathematical tools that are required to solve the problems, e.g. once you know let me say calculus the problem gets straightforward. For example in Germany you have kind of two orthogonal competitions:

1: Deutsche Mathematik Olympiade (DEMO): Low-Medium problem-solving complexity, class/grade-specific requirement of mathematical tools, time pressure, categories for class/grade 3 to grade 12 (13), depending on grade you participate in up to four rounds, [url=http://www.mathematik-olympiaden.de/archiv.html]Previous Taks of the Deutsche Mathematik-Olympiade in German[/url]

2: Bundeswettbewerb Mathematik: Medium-High problem-solving complexity, mostly middle-school requirements for the mathematical tools requirement, lots of time for tinkering around with the problems because it is a homework competition that allows for several weeks working time, only one category for problems independent on class, mostly targeted at students grade 9 to 12 (13), two rounds and a third round that features an oral one hour discussion (Kolloquium) involving a mathematician from a school or university, [url=http://www.bundeswettbewerb-mathematik.de/aufgaben/aufgaben.htm]Previous Taks of the Bundeswettbewerb Mathematik in German[/url]

Sorry for missing your point. My words were only for question-based classification. Competition-based classication can be really helpful and I would be grateful to help. We can continue with discussing the details of this classification. But, I need to go to classroom now. High school s*cks. :mad:

perhaps this is not completely relevant to what is discussing here, but I have a reservation concerning the Resources section, namely, I believe, that if next to each problem it would be written the number of solutions proposed to this problem, and if there doesn't exist any, the number of responses made in the corresponding thread, it would make the Resource section more convenient and readable.

now regarding and supporting the effort of Umut, what if our dear moderators get an opportnuity to evaluate the difficulty of each problem and to rank each problem ? I suggest using a system that would be somehow similar to the ranking of posts. Furthermore I'm convinced that increment of moderators number would favourably affect the quality of the section and of the whole forum.

Well, Erken you are right about your insights about the resources section. But it poses the question whether you are biased towards problems yet unsolved on the forum.

As of now we can enter the post id of you of up to three solutions for each problem. Checking all the solutions of each thread and then adding the post ids to the resources section is a very time-consuming process. Of course you can just go right ahead and add the post ids of three solutions. Though not sure it is a good idea to enter un-checked solutions. Again there are two sides: what is nice to have and who helps to make it true.

Of course to provide the number of replies a problem has received in its specific thread should be programmatically easy. But the issue will this number be useful? First certain threads have many replies discussing typos, random blabla as "wait for me until next week" etc. Given a problem was posted several times few mods care about joining these threads and clearing of spam. But this actually may mean the problem was discussed several times and has a "beard" to quote Darij. But the thread whose problem in the first post was added to the resources section may only have one reply that has the link to another thread instance where the problem was discussed. And at times these others thread also do not contain any significant solution advances. And at other times again users do not look for the old threads but only claim it was posted before in the thread whose first post was added to the resources section. Anyway noting 1 reply for a problem in the resources section suggests little discussion though there may be little or indeed lots of previous discussion. Does this help you then whether you want to give a shot at this problem?

And if Val agrees to implement the problem-specific ranking we need to find reliable ranking curators. I am not convinced to select people by the time they have been on the forum or the number of posts they made. I rather would select people manually. People may also contact the admins to aks for permission to rate problems. A few members without aim of completeness are:

[b]Albanian Eagle, April, cosinator, dgreenb801, Erken, feliz, greentreeroad, Johan Gunardi, lambruscokid, Mathias_DK, MellowMelon, pohoatza, randomgraph, Rust, Umut, Vasc, Virgil Nicula etc.[/b]

This is a selection very biased towards the Olympiad section. This may lead to very few ratings for problems from competitions as AMCs or AIMEs, and/or ranking eligible members who are active in the pre-olympiad forums or below may rate their problems at olympiad level.

Currently I am about the only mod who is adding problems to the resources section at a high number. So we need suitable new mods who are willing to contribute to the society by rating problems, adding problems to the resources section, proof-reading solutions / specifying which solutions should be added to the resources section, merging multiple discussions of the same problem, helping and giving advice to problems posted by fellow members, posting problems from competitions they have access to etc. If many people are helping with a small bit there is not much work involved. If you are eligible to rate problems you may or may not be a moderator.

I cannot do the job alone. Only collaboration leads to big improvement of the community. Mutual efforts lead to mutual symbiotic benefit. We need you. Let me know your ideas.

How about this, let's first select some people to do some experiments by rating the 6 problems in IMO 2008 to see whether the rating can appropriate reflect the problem difficulty.

Another suggestion:considering the huge gap in difficulty between math competition problems(e.g. AMC and USAMO), how about rating them separately. On this forum, people have different interest:some people only want to focus on AIME, AMC level or below, but some others only try Olympiad problems. I think rating separately will help people find what they want faster.

I don't think divide all problems into too many categories is good. How about divide them into two groups.
A----represent problem on or below AIME level(including fill-in blanks, multiple choices...)
B----represent Olympiad level problem(including even early IMOs though they might be easy)

Now within each category, we divide it into whatever number of levels(say 10). So now, we have $ A_1$, $ A_2$, ..., $ A_{10}$, $ B_1$, $ B_2$, ..., $ B_{10}$. A hard AIME problem can be A9 and an easy IMO problem can be B2.

As I mentioned before, I propose the idea of separation because I observe that most users tend to either almost spend all their time dealing with A problems or almost spent time dealing with B problems. Indeed, the step between A type problems and B type problems can't be underestimated. In all, doing the separation might prevent the situation that a less experienced solver rate 10 on an easy IMO problem.

I reall don't know which option is better, im not fully convinced about this rankings, yet I think that something CAN be done and as Orl have said, it will require everybody's effort
I want to help to make this site more helpful to everyone, so if any work has to be done you can count on me

Daniel

PS: I think it would be nice if there were a section where "beautiful" problems can be identified. Obviously it is very a subjective matter but if we can choose the nicest problems and collect them we will benefit hugely  all the ML community

Let's first stick to the options which are easier and will probably be done - grading competitions

Also, should we grade by classes? For example, AMC 8 could be 3 for 7. and 8. grade, and AMC 12 could be 3 for 11. and 12. grade (bad example). I thought of that because when someone goes to Resources and sees such 2 competitions graded equally but one doesn't have a proof in itself, and other asks you to use olympiad theorems...

Beautiful problems: lambruscokid, nice idea but then most of the problems will be clasified as beautiful. Half of ML won't even know what beautiful is, so will put problems that they don't know

[quote="April"]I think we should choose reliable ranking curators for each type of problems, as we have Moderators for each sections (for example, one group will be in charge of ranking Algebra problems, another group will be responsible for ranking Geometry problems, etc). Because I think that it's very hard to find a person who is capable of ranking all problems from Algebra, to Combinatoric, Number Theory,...

BTW, I think that I can help you to adding problems to the Resources section and posting problems from some competitions. I will also try to ask other members for translating and posting their country competitions which are still missing on the Forum.[/quote]

Combine April's idea with my previous idea about separation, I think we can rate problems separately in each section(this will not cost more time): Algebra, NT, Geo and Comb(maybe inequality can go to algebra). Now, there will be more people who can get fully control of the section they often go to.(say I look at a lot of geometry so I can judge the difficulty of geometry problems more easily)

If I understand you correctly, you suggest segregating curators into several groups, and that is really worth doing it. And I think we should pay attention to the number of curators assigned to each section. This number should not be too tiny, otherwise, differently designated ranks may lead to the incorrect and false classification. On the other hand, the number of curators should not be too large, because in this case several incompetent estimators might get involved as well.
So we have almost studied out how to handle this problem with ranking system, and the only hurdle left is an intake of curators. Who is going to complete estimators ? And who can be an estimator?
P.S:
Just out of curiosity: Is Valentin Vornicu aware of the contrivance?

hiya.... i just want to make a suggestion that PEN (Problems in Elementary Number Theory) would be a really good place to start this difficulty rating program.... Its already a great collection of problems and divided into sub-categories and furthermore its in compiled file format which will make it easier for individuals to rate the problems (if that makes any sense).....

[quote="Mathias_DK"]This might be a stuped question, but I'm not familiar with the Wiki on this site. The wiki site orl talked about in the first post, where is it? (The on with the ratings of different Olympiads, where he used China as the example)[/quote]

Thanks for your question, Mathias. You can find the AoPS wiki [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]here.[/url] For questions about the wiki there is also a dedicated forum [url=http://www.mathlinks.ro/index.php?f=416]here.[/url]

Regarding the curators IMO it is a good idea to give any of them general ranking permissions and expect them to use these privileges responsibly. Focussing ranking permissions to narrowly might lead to some bias of underestimation the difficulty level I feel.

And there is this implicitely mentioned trade-off of having enough votes vs. having possibly more accurate ranking estimations.

[quote="ZetaX"]

And I don't think that rating is a good idea for two reasons:
- most people will not contribute; just take a look at how many contribute at the user ratings.
- an olympiad can at best be seen by it's average difficulty; an easy problem from China could be much easier than a difficult one from Germany, so ratings depending on olympiads are not really usefull.
And another point is, that it may even be good to have no idea that it's difficult. If the problem already says "I'm horribly difficult", it will push away lots of people; but in fact, it would be good for them to try (and mostly fail) on hard problems, you learn a lot from that or the given solutions.

[/quote]

 :wink:

Clear case of not having read the discussion thread!  :D

The last point sounds valid in that it cannot hurt to get screwed from time to time but in general the difficulty level gives you a good idea whether there is possibly something very tricky going on or you just missed some basic thing.

I mean that China NMOs/TSTs can be quite hard at times is well-known but how do you judge their regional competitions, and/or compare them with Croatia NMO? I am sure it also takes you quite some time to find out.

Difficulty ratings should be provided by curators. User rating is more of a popularity thing and to punish users for spamming or great contributions but average ratings are rarely entered I assume.

I think orl's idea is very good.

In Indonesia, the national olympiad in 2002 was easier AIME (seriously). But in 2008, I think Indonesia's national olympiad is already comparable to some USAMO problems. So I suggest that we put different rating for different years in an olympiad. For example, Indonesia 2002 is 3-4, and Indonesia 2008 is 6-7.

I can't wait for this system to be implemented. :)

[quote="iandrei2"] [list] 2) Since you are striving to identify each topic with a single problem discussed, implement tagging for threads. Associate with each discussion thread tags such as "IMO 2008", "Combinatorics" and "double counting". Combined with auto-complete for the tags when entering them(similar to blogger interface) you should be able to improve the search in the website.

3) Finally, if you are really bored and have a lot of spare time, you can implement a tool to automatically tag all problems so far (I think a keyword-based approach should give pretty good results). Also, you can make the tags editable so that they can be corrected on the fly.
[/list]
Tagging would improve usage of the website because after learning a new technique it's usually useful to solve a few similar problems in order to make sure that you understood it really well.

[/quote]

Actually as you can see here this suggestion of [url=http://www.mathlinks.ro/viewtopic.php?t=248978]problem description tagging[/url] is already partially in place.

[quote="Johan Gunardi"]any update? :roll:[/quote]

Yes, there is! Please check [url=http://www.mathlinks.ro/weblog_entry.php?t=269600]here[/url] and leave your comment! Thanks!  :)

I created an [url=http://www.artofproblemsolving.com/Wiki/index.php/Competition_Rating]AoPS/ML wiki entry template[/url] to enter problem ratings now. Check out this [url=http://www.mathlinks.ro/viewtopic.php?p=1561941#1561941]thread post[/url] and this [url=http://www.mathlinks.ro/viewtopic.php?p=1565063#1565063]thread post[/url] on what you should enter when you are editing the wiki.

I think it is good to have a look at a different competition every a couple few days to give you a chance to play around with those problems and evaluate their difficulty level. So far most US people could have guessed the AIME problems may be harder than AMC and USAMO ones are more challenging than those from AIME. But this wiki page only starts go get useful when it has entries that people do not routinely consider for olympiad preparation because they are not familiar with this respective contest and thus discard its problems apriori.

On Wednesday this week the IMO delegation members departed from Bremen in Germany. Thus I thought it is a good idea to start with a German contest: [url=http://www.mathlinks.ro/resources.php?c=64&cid=36&year=2001]Germany Bundeswettbewerb Mathematik (BWM), Resources Section.[/url] Please edit the difficulty ratings for BWM in the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings#Germany_Bundeswettbewerb_Mathematik]wiki here[/url] please!  :)

Today we should start rating [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings#Balkan_MO]Balkan MO[/url] whose problems can be found in the [url=http://www.mathlinks.ro/resources.php?c=1&cid=19]resources section.[/url] The German BWM has not been rated yet! What is going on guys?


I had someone register to the site about a week ago and he checked the under 13 year box.  He is unable to log in and didn't receive anything in his mailbox.  What is the procedure to activate his account?  thanks.

Sorry about that - we've had some trouble with our automated email server during the last week - our emails to Yahoo accounts are not getting through (and possibly others, at this point we're only aware of problems with Yahoo).  If you're on a Yahoo account, you'll get the email eventually, but a faster method would be for you to simply send me an email at rusczyk@artofproblemsolving.com with the following information:

1) Student login
2) registration email.

I will then email you the required form for students under 13.  (Non-automated emails are getting through to Yahoo).

Sorry for the inconvenience.

What is the difference in the form for a student under 13?

Pursuant to COPPA regulations (a federal law), we must have parental permission to allow them to register and use the boards.


Ring Theory
superior algebra
superior algebra solved

Find a) $Spec(\mathbb{Q}[X]/ (X^5))$ and $Max(\mathbb{Q}[X]/ (X^5))$.
         b) $Spec(\mathbb{Z}[i])$ and $Max(\mathbb{Z}[i])$.

Is there anyone intrrested in these problems??? I know they are easy, but.... :)

It is my experience that people on this forum are not generally interested in solved problems  ;)

Anyway: $Spec \mathbb Q[X]/(X^5)=(X)$. This is because the ring is principal and because in this ring $\hat X^5=0$, so any prime ideal contains $\hat X$. $Spec \mathbb Q[X]/(X^5)=(X)=Max \mathbb Q[X]/(X^5)=(X)$.

For the second, $Spec(\mathbb Z[\imath])=Max(\mathbb Z[\imath])\cup (0)$.
$\mathbb Z[\imath]$ is euclidian hence nonzero prime ideals correspond to prime elements in $\mathbb Z[\imath]$.
It is known that $p=a+\imath b$ is prime in $\mathbb Z[\imath]$ iff it is a prime integer $\equiv 3(mod\ 4)$ or $a^2+b^2$ is a prime integer $\equiv 1(mod\ 4)$.

sorry for the stupid question... but what is spec?

[quote="puuhikki"][quote="Soarer"]what is spec?[/quote]
Spectrum of a ring.[/quote]

Thank you.


function
integration
superior algebra
superior algebra open

It is stated there that it has already been solved (I even don't know if it's corect), but because of lack of 'easy' solution I put it here:

Let be $\lambda$ an algebraical (over $\mathbb{Q}$) of degree $n$ and let be $\sigma_1=id , \sigma_2 , ... , \sigma_n$ the different automorphisms of $\mathbb{K} := \mathbb{Q} [\lambda]$ to $\overline{\mathbb{Q}}$ (the mappings that map $\lambda$ to it's algebraic conjugates).

Show that for all $\omega \in \mathbb{K}^*$ there exists an $j$ such that $\phi_{\sigma_j (\lambda)} (\sigma (\omega ) ) \not\in \sigma(\mathbb{K})$ where $\phi_{\lambda '} (z) := \sum_{k=0}^{\infty} \frac{z^k}{(\lambda ' +1)(\lambda ' +2)...(\lambda ' +k)}$.

This function has a lot to do with the exponential, and I think the techniques used to prove $e$ is irrational/transcendental would be useful here.

Let $f_a(x)=x^a\phi_a(x)$. Then $f'(x)=f(x)+ax^{a-1}$. We can solve this to obtain $f_a(x)=e^x\int_0^xat^{a-1}e^{-t}\, dt$ (for $a>0$). For negative values, use the recursion (much like the Gamma function).

I would just like to point out that  $\sigma$ does not have a subscript in $\phi_{\sigma_j (\lambda)} (\sigma (\omega ) ) \not\in \sigma(\mathbb{K})$. So it seems to me like some clarification is required.

Regards.


inequalities
function
quadratics
algebra
polynomial
inequalities proposed

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that
$ 1)$\[ a \plus{} b \plus{} c \plus{} 2(a^2b \plus{} b^2c \plus{} c^2a)\geq9
\]
$ 2)$\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} 17(a^2b \plus{} b^2c \plus{} c^2a)\geq54
\]

I have an idea for $ 1)$ when $ a,b,c > 0$:

$ (a^2 b \plus{} b^2 c \plus{} c^2 a)(a \plus{} b \plus{} c) \ge (ab \plus{} bc \plus{} ca)^2 \ge 9$

$ \Rightarrow a^2 b \plus{} b^2 c \plus{} c^2 a \ge \frac {9} {a \plus{} b \plus{} c}$

$ \Rightarrow (a \plus{} b \plus{} c) \plus{} 2 \cdot (a^2 b \plus{} b^2 c \plus{} c^2 a) \ge (a \plus{} b \plus{} c) \plus{} 2 \cdot \frac {9} {a \plus{} b \plus{} c} \ge 2 \cdot \sqrt {2 \cdot 9} \equal{} 6\sqrt {2}$

The only problem is that $ 6 \sqrt {2} < 9$. I wait to see a complete solution.  :D

[quote="moldovan"]I have an idea for $ 1)$ when $ a,b,c > 0$:

$ \Rightarrow (a \plus{} b \plus{} c) \plus{} 2 \cdot (a^2 b \plus{} b^2 c \plus{} c^2 a) \ge (a \plus{} b \plus{} c) \plus{} 2 \cdot \frac {9} {a \plus{} b \plus{} c} \ge 2 \cdot \sqrt {2 \cdot 9} \equal{} 6\sqrt {2}$

The only problem is that $ 6 \sqrt {2} < 9$. I wait to see a complete solution.  :D[/quote]
But  with ur solution   the equality doesn't hold.
 i think,
 Assuming $ a$  the bigest $ (a,b,c)$ we have: $ a(b \minus{} a)(c \minus{} a) \geq 0$ or $ 2(a^2b \plus{} b^2c \plus{} c^2a) \geq 2a(b^2 \plus{} bc \plus{} c^2)$

Hence, $ A \geq a \plus{} b \plus{} c \plus{} 2a(b^2 \plus{} bc \plus{} c^2)$
Let be $ b \plus{} c \equal{} X$ then $ aX \plus{} bc \equal{} 3 \to bc \equal{} 3 \minus{} aX$
$ A \geq a \plus{} X \plus{} 2a(X^2 \minus{} 3 \plus{} aX) \geq 9$
Research function $ f(X) \equal{} 2aX^2 \plus{} (2a^2 \plus{} 1)X \minus{} 5a$  with $ X \geq \minus{} 2a \plus{} \sqrt {12 \plus{} 4a^2}$
$ f(x)$  increase then $ f(X) \geq f( \minus{} 2a \plus{} \sqrt {12 \plus{} 4a^2}) \geq 9$
 I'm right???? :blush:

I know that...  :lol: That's why I said that I was waiting a complete solution.

My extremely ugly solution of problem $ 1)$ (with the same method problem $ 2)$ can be solved, too):

I will assume that $ a\leq b\leq c$. I am able to do this since $ a^2b \plus{} b^2c \plus{} c^2a\leq ab^2 \plus{} bc^2 \plus{} ca^2\Leftrightarrow (a \minus{} b)(b \minus{} c)(c \minus{} a)\geq 0$. I will substitute $ ab \equal{} z$, $ ca \equal{} y$, and $ bc \equal{} x$ which is equivalent to $ a^2 \equal{} \frac {yz}{x}$, $ b^2 \equal{} \frac {zx}{y}$, and $ c^2 \equal{} \frac {xy}{z}$. Here, I have to notice that if the least variable $ a \equal{} 0$, then $ b \equal{} c \equal{} \sqrt {3}$ (in the worst case) and the inequlity is obvious. So, we can assume $ a,b,c > 0$ and make the substitution.

Well, we will have for the new variables that $ x\geq y\geq z$ and $ x \plus{} y \plus{} z \equal{} 3$. The inequality will turn into $ (xy \plus{} yz \plus{} zx) \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2)\geq 9\sqrt {xyz}$.
Now - the ugly part. I fix $ x$ ($ 3 > x\geq 1$). We have for the other variables $ y \plus{} z \equal{} 3 \minus{} x$. Let $ \lambda \geq 0$ is such a number that $ y \equal{} \frac {3 \minus{} x}{2} \plus{} \lambda$ and $ z \equal{} \frac {3 \minus{} x}{2} \minus{} \lambda$. Since $ x\geq y\geq z$, then $ \lambda$ must be less than $ \frac {3}{2}(x \minus{} 1)$, otherwise $ y\geq x$. Therefore, we can say at the end that $ 0\leq \lambda \leq \frac {3}{2}(x \minus{} 1)$.

$ f(x,y,z) \equal{} xy \plus{} yz \plus{} zx \equal{} x(3 \minus{} x) \plus{} \bigg(\frac {3 \minus{} x}{2}\bigg)^2 \minus{} \lambda^2 \equal{} f(\lambda)$.
$ f(\lambda) \minus{} f(0) \equal{} \minus{} \lambda^2$.

$ g(x,y,z) \equal{} xy^2 \plus{} yz^2 \plus{} zx^2 \equal{}$ $ x\bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)^2 \plus{} \bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)\bigg(\frac {3 \minus{} x}{2} \minus{} \lambda\bigg)^2 \plus{} \bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)x^2 \equal{} g(\lambda)$.
$ g(\lambda) \minus{} g(0) \equal{} \lambda^3 \plus{} \frac {3}{2}(x \minus{} 1)\lambda^2 \plus{} \bigg[\frac {3}{2}(x \minus{} 1)\bigg]^2\lambda$.

$ \Rightarrow f(\lambda) \minus{} f(0) \plus{} 2(g(\lambda) \minus{} g(0)) \equal{} \lambda\bigg[ 2\lambda^2 \plus{} (3x \minus{} 4)\lambda \plus{} \frac {9}{2}(x^2 \minus{} 2x \plus{} 1)\bigg] \equal{} \lambda h(\lambda)$.
$ h'(\lambda) \equal{} 4\lambda \plus{} (3x \minus{} 4)$, so $ h_{\min}(\lambda) \equal{} h\bigg( \frac {4 \minus{} 3x}{4} \bigg) \equal{} \frac {35x^2 \minus{} 48x \plus{} 20}{8} > 0$.

Also, $ \sqrt {x\bigg(\frac {3 \minus{} x}{2}^2\bigg)^2}\geq \sqrt {x\bigg[\bigg(\frac {3 \minus{} x}{2}^2\bigg)^2 \minus{} \lambda^2\bigg]}$.

Therefore, if $ F(x,y,z) \equal{} (xy \plus{} yz \plus{} zx) \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2) \minus{} 9\sqrt {xyz}$, it will be validly that $ F(x,y,z)\geq F(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2})$.

Consequently, it's enough to prove the inequality, when two of the variables are equal and moreover - the least two. Bla-bla... and the problem is solved.

[color=blue]PS: I'm very interested to see the original [b]arqady[/b]'s solution, though there are already two others.[/color]

[quote="BG Yoda"]
Well, we will have for the new variables that $ x\geq y\geq z$ and $ x \plus{} y \plus{} z \equal{} 3$. The inequality will turn into $ (xy \plus{} yz \plus{} zx) \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2)\geq 9\sqrt {xyz}$.
Now - the ugly part. I fix $ x$ ($ 3 > x\geq 1$). We have for the other variables $ y \plus{} z \equal{} 3 \minus{} x$. Let $ \lambda \geq 0$ is such a number that $ y \equal{} \frac {3 \minus{} x}{2} \plus{} \lambda$ and $ z \equal{} \frac {3 \minus{} x}{2} \minus{} \lambda$. Since $ x\geq y\geq z$, then $ \lambda$ must be less than $ \frac {3}{2}(x \minus{} 1)$, otherwise $ y\geq x$. Therefore, we can say at the end that $ 0\leq \lambda \leq \frac {3}{2}(x \minus{} 1)$.

$ f(x,y,z) \equal{} xy \plus{} yz \plus{} zx \equal{} x(3 \minus{} x) \plus{} \bigg(\frac {3 \minus{} x}{2}\bigg)^2 \minus{} \lambda^2 \equal{} f(\lambda)$.
$ f(\lambda) \minus{} f(0) \equal{} \minus{} \lambda^2$.

$ g(x,y,z) \equal{} xy^2 \plus{} yz^2 \plus{} zx^2 \equal{}$ $ x\bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)^2 \plus{} \bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)\bigg(\frac {3 \minus{} x}{2} \minus{} \lambda\bigg)^2 \plus{} \bigg(\frac {3 \minus{} x}{2} \plus{} \lambda\bigg)x^2 \equal{} g(\lambda)$.
$ g(\lambda) \minus{} g(0) \equal{} \lambda^3 \plus{} \frac {3}{2}(x \minus{} 1)\lambda^2 \plus{} \bigg[\frac {3}{2}(x \minus{} 1)\bigg]^2\lambda$.

$ \Rightarrow f(\lambda) \minus{} f(0) \plus{} 2(g(\lambda) \minus{} g(0)) \equal{} \lambda\bigg[ 2\lambda^2 \plus{} (3x \minus{} 4)\lambda \plus{} \frac {9}{2}(x^2 \minus{} 2x \plus{} 1)\bigg] \equal{} \lambda h(\lambda)$.
$ h'(\lambda) \equal{} 4\lambda \plus{} (3x \minus{} 4)$, so $ h_{\min}(\lambda) \equal{} h\bigg( \frac {4 \minus{} 3x}{4} \bigg) \equal{} \frac {35x^2 \minus{} 48x \plus{} 20}{8} > 0$.

Also, $ \sqrt {x\bigg(\frac {3 \minus{} x}{2}^2\bigg)^2}\geq \sqrt {x\bigg[\bigg(\frac {3 \minus{} x}{2}^2\bigg)^2 \minus{} \lambda^2\bigg]}$.

Therefore, if $ F(x,y,z) \equal{} (xy \plus{} yz \plus{} zx) \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2) \minus{} 9\sqrt {xyz}$, it will be validly that $ F(x,y,z)\geq F(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2})$.

Consequently, it's enough to prove the inequality, when two of the variables are equal and moreover - the least two. Bla-bla... and the problem is solved.

[color=blue]PS: I'm very interested to see the original [b]arqady[/b]'s solution, though there are already two others.[/color][/quote]
If $ x \geq\ y \geq\ z$, this ineq is obviously trues by Am-Gm.
The difficult case is $ x \leq\ y \leq\ z$, my friend.

No, it's the easier case -
when $ x\leq y\leq z$, then
$ xy^2 \plus{} yz^2 \plus{} zx^2\geq x^2y \plus{} y^2z \plus{} z^2x$
$ \Leftrightarrow xy(y \minus{} x) \minus{} z(y \minus{} x)(y \plus{} x) \plus{} z^2(y \minus{} x) \equal{}$ $ (y \minus{} x)(z^2 \minus{} yz \minus{} zx \plus{} xy) \equal{} (y \minus{} x)(z \minus{} x)(z \minus{} y)\geq 0$. :D
So, if we prove $ (xy \plus{} yz \plus{} zx) \plus{} 2(x^2y \plus{} y^2z \plus{} z^2x)\geq 9\sqrt {xyz}$ for $ x\leq y\leq z$ and $ x \plus{} y \plus{} z \equal{} 3$, which is the same with mine, we will have proved $ (xy \plus{} yz \plus{} zx) \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2)\geq 9\sqrt {xyz}$, too.

[b]nguoivn[/b], can you give the solution with Am-Gm, because I can't find it? :blush:
Indeed, it's obviously that $ xy \plus{} yz \plus{} zx \equal{} \sqrt {(xy)^2 \plus{} (yz)^2 \plus{} (zx)^2 \plus{} 2xyz(x \plus{} y \plus{} z)}\geq 3\sqrt {xyz}$, but what do we make with $ xy^2 \plus{} yx^2 \plus{} zx^2\geq 3\sqrt {xyz}$?

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ xy \plus{} yz \plus{} zx \equal{} 3.$ Prove that
$ 1)$
$ x \plus{} y \plus{} z \plus{} 2(x^2y \plus{} y^2z \plus{} z^2x)\geq9$

[/quote]
If $ xy^2 \plus{} yz^2 \plus{} zx^2\leq x^2y \plus{} y^2z \plus{} z^2x$, we only need to prove:
$ x^2y \plus{} y^2z \plus{} z^2x \geq\ 3$
But we have: LHS $ \geq\ \frac {xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x)}{2}$
Using Am-Gm: $ x^2y \plus{} xy^2 \plus{} 1 \geq\ 3xy$
=> $ \sum\ xy(x \plus{} y) \geq\ 3(xy \plus{} yz \plus{} zx) \minus{} 3 \equal{} 9$
 :wink:

:roll:
That's right but what to do when $ x\leq y\leq z\Leftrightarrow xy^2 \plus{} yz^2 \plus{} zx^2\geq x^2y \plus{} y^2z \plus{} z^2x$. Indeed, this is the hard part, and I've proved exactly this case
[quote="BG Yoda"]I will assume that $ a\leq b\leq c$. ... . I will substitute $ ab\equal{}z$, $ ca\equal{}y$, and $ bc\equal{}x$ . ... . Well, we will have for the new variables that $ x\geq y\geq z$ and $ x\plus{}y\plus{}z\equal{}3$. ...[/quote]
 :P

No, when you let $ yz \equal{} a..., x \geq\ y \geq\ z$ <=> $ a \leq\ b \leq\ c$

And we haven't found a nice proof for the hard case  :wink:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that
$ 1)$
\[ a \plus{} b \plus{} c \plus{} 2(a^2b \plus{} b^2c \plus{} c^2a)\geq9
\]
[/quote]

[quote="BG Yoda"]I will assume $ a\leq b\leq c$ [i](the hard case)[/i]. ... . I will substitute $ ab \equal{} z;ca \equal{} y;bc \equal{} x$. ... We will have for the new variables that $ x\geq y\geq z$ [i](it's still the hard case)[/i]. ...[/quote]

So, I've solved  the hard case, haven't I? :blush:
Notice that I've proved $ xy \plus{} yz \plus{} zx \plus{} 2(xy^2 \plus{} yz^2 \plus{} zx^2)\geq 9 \sqrt {xyz}$ for $ x\geq y\geq z$ and then $ xy^2 \plus{} yz^2 \plus{} zx^2\leq x^2y \plus{} y^2z \plus{} z^2x$, which is the hard case, isn't it ?! :wink:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that
$ 1)$
\[ a \plus{} b \plus{} c \plus{} 2(a^2b \plus{} b^2c \plus{} c^2a)\geq9
\]
$ 2)$
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} 17(a^2b \plus{} b^2c \plus{} c^2a)\geq54
\]
[/quote]
Using quadratic polynomial, we can prove that the best constant for
\[ a \plus{} b \plus{} c\plus{} k(a^2b \plus{} b^2c \plus{} c^2a) \ge 3k \plus{} 3
\]
is $ k \equal{} 1 \plus{} \frac {2}{\sqrt {3}}$, and the best constant for
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} k(a^2b \plus{} b^2c \plus{} c^2a) \ge 3k \plus{} 3
\]
is $ k \sim 17. 61$.
:)
This shows that the inequality of Arqady is nice and rather tight. Will your way work for the best constant inequalities, arqady? :)

[quote="can_hang2007"]Will your way work for the best constant inequalities, arqady? :)[/quote]
Yes, of cause!   :lol:

[quote="can_hang2007"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that
$ 1)$
\[ a \plus{} b \plus{} c \plus{} 2(a^2b \plus{} b^2c \plus{} c^2a)\geq9
\]
$ 2)$
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} 17(a^2b \plus{} b^2c \plus{} c^2a)\geq54
\]
[/quote]
Using quadratic polynomial, we can prove that the best constant for
\[ a \plus{} b \plus{} c \plus{} k(a^2b \plus{} b^2c \plus{} c^2a) \ge 3k \plus{} 3
\]
is $ k \equal{} 1 \plus{} \frac {2}{\sqrt {3}}$, and the best constant for
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} k(a^2b \plus{} b^2c \plus{} c^2a) \ge 3k \plus{} 3
\]
is $ k \sim 17. 61$.
:)
This shows that the inequality of Arqady is nice and rather tight. Will your way work for the best constant inequalities, arqady? :)[/quote]

I am very eager to see how these constants are found... Is it "pqr" method?
Because the number $ 1 \plus{} \frac {2}{\sqrt {3}}$ is itself interesting.
Please show us how did you obtain it. And also, can you find best constant for the following:
\[ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} k(a^2b \plus{} b^2c \plus{} c^2a) \ge 3k \plus{} 3
\]
I want to learn general way to solve this kind of inequalities. Thank you very much for reading my post.

Regards.


algebra
polynomial
number theory solved
number theory

Let P(x)=ax^2+bx+c .
If the values of P(x) at every n (n element of Z) is a square of another element of Z ,
show the existences of d,e (d,eEZ) such that
P(x)=(dx+e)^2 holds for all x

Of course, a>=0. The case a=0 is simple, Take (x_n)^2=an^2+bn+c for all n, where x_n is a natural number. Then x_(n+1)-x_nis convergent, so we can find k and m such that for all n>m we have x_(n+1)=x_n+k. From here it's simple.

It can be generalized for P a polynomial of degree n, and P(k) a n-th power for all k in N.
Then P is the n-th power of a polynomial with integer coefficients.


\[ \text{Let}\; x,y,z\;\text{be nonnegative real numbers such that}\; xy+yz+zx=1,\; \text{ prove that} \]
\[ \frac1{\sqrt{x+yz}}+\frac1{\sqrt{y+zx}}+\frac1{\sqrt{z+xy}}\geq3. \]

[quote="romano"][quote="pvthuan"]When we remove the square root, it's true \[\text{Let}\; x,y,z\;\text{be nonnegative real numbers such that}\; xy+yz+zx=1,\; \text{ prove that}\]  \[\frac1{{x+yz}}+\frac1{{y+zx}}+\frac1{{z+xy}}\geq3.\] [/quote]
  Maybe you mean :
 $\frac1{{x+yz}}+\frac1{{y+zx}}+\frac1{{z+xy}}> 3$
  I tried to reduce to the same denominator and homogenize it (Maple forever  :lol: ) . It's not nice but true .[/quote]
Not $>$, it's $\ge$. The equality is $(0,1,1)$.


geometry
circumcircle
incenter
geometric transformation
homothety
ratio
geometry unsolved

Let $ O$ be the circumcircle of a $ \Delta ABC$ and let $ I$ be its incenter, for a point $ P$ of the plane let $ f(P)$ be the point obtained by reflecting $ P'$ by the midpoint of $ OI$, with $ P'$ the homothety  of $ P$ with center $ O$ and ratio $ \frac{R}{r}$  with $ r$ the inradii and $ R$ the circumradii,(understand it by $ \frac{OP}{OP'}\equal{}\frac{R}{r}$). Let $ A_1$, $ B_1$ and $ C_1$ the midpoints of $ BC$, $ AC$ and $ AB$, respectively. Show that the rays $ A_1f(A)$, $ B_1f(B)$ and $ C_1f(C)$ concur on the incircle.

[quote="trejos"]Let $ O$ be the circumcircle of a $ \Delta ABC$ and let $ I$ be its incenter, for a point $ P$ of the plane let $ f(P)$ be the point obtained by reflecting $ P'$ by the midpoint of $ OI$, with $ P'$ the homothety  of $ P$ with center $ O$ and ratio $ \frac {R}{r}$  with $ r$ the inradii and $ R$ the circumradii,(understand it by $ \frac {OP}{OP'} \equal{} \frac {R}{r}$). Let $ A_1$, $ B_1$ and $ C_1$ the midpoints of $ BC$, $ AC$ and $ AB$, respectively. Show that the rays $ A_1f(A)$, $ B_1f(B)$ and $ C_1f(C)$ concur on the incircle.[/quote]
Here is my solution, I am proving a stronger result.
Let $ \lambda {(P)}$ be the homothety that maps $ P$ with center $ O$ and ratio $ \lambda_k \geq 1$.
So, $ \lambda$ maps $ ABC\to \lambda{(A)}\lambda{(B)}\lambda{(C)}$. And again a homothety with ratio $ \minus{}1$ maps $ \lambda{(A)}\lambda{(B)}\lambda{(C)} \to f(A)f(B)f(C)$.
Now we consider the medial triangle $ A_1B_1C_1$ which is the homothety of $ ABC$ with center $ G$ and ration $ \minus{}\frac 12$. Which implies that $ f(A)f(B)f(C)$ and $ A_1B_1C_1$ are homothetic with same orientation. So, the result is true by Desargue's Theorem or by the fact that there exists a homothety center or by some similar triangle like argument.
:)

[hide]Probably it is enough to say that those triangles are homothetic.[/hide]

We have $ IF(A)\equal{}A'O\equal{}r$ then $ f(A)$ lies on $ (I)$.
Let $ Ax$ be the tangency of $ (O)$.
On the other side, $ IF(a)//AO$ then $ IF(A)\perp Ax (1)$
Let A'' be the intersection of AI and BC. Construct a tangency $ A''y$ of $ (I)$.
$ \angle AA''y\equal{}\angle AA''B\equal{}\frac{\angle A}{2}\plus{}\angle C\equal{}\angle xAA''$
Then $ A''y//Ax (2)$
From $ (1)$ and $ (2)$ we get $ Ax$ passes through $ F(A)$
Using [url=http://www.mathlinks.ro/viewtopic.php?t=283779]this problem[/url] $ \rightarrow QED$

In MoonmathPi's solution, he has said $ \lambda{A}\lambda{B}\lambda{C}$ maps to $ f(A)f(B)f(C)$ by ratio $ -1$? Can someone explain that ? What is the centre of homothety in that case? I request for a complete explanation in this regard please.

it is just similar to the geometric problem in IMO23.


Find all positive real solutions to the equation:

$x+\left\lfloor\frac{x}{6}\right\rfloor = \left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{2x}{3}\right\rfloor$

where $\left\lfloor t \right\rfloor$ denotes the largest integer less than or equal to the real number $t$.

Note that the floor functions are all integers, so it follows that x is an integer.

We need only consider the 6 values of x in $\mathbb{Z}_{6}$, and we do so case by case.

x=6k, LHS=7k, RHS=7k
x=6k+1, LHS=7k+1, RHS=7k
x=6k+2, LHS=7k+2, RHS=7k+2
x=6k+3, LHS=7k+3, RHS=7k+3
x=6k+4, LHS=7k+4, RHS=7k+4
x=6k+5, LHS=7k+5, RHS=7k+5

We see that LHS=RHS in all cases except x=6k+1, so the set of positive reals satisfying this equation is the set of integers excluding those of the form 6k+1.

[hide="Solution"]Put $x=6n+\varrho$, where $n\in\mathbb{Z}$ and $\varrho$ is a [u]real[/u] number satisfying $0\leqslant\varrho<6$.

Then the equation becomes

$6n+\varrho+n=3n+\left[{\varrho\over 2}\right]+4n+\left[{2\varrho\over 3}\right]$

or

$\varrho=\left[{\varrho\over 2}\right]+\left[{2\varrho\over 3}\right]$

Since RHS is integer, LHS must be too, therefore $\varrho\in\{0,1,2,3,4,5\}$. By checking we see that all the values except $1$ satisfy the equation. Therefore the initial equation's solution set is

$\mathcal{R}=\mathbb{Z}\setminus(6\mathbb{Z}+1)$[/hide]


If $a\geq b\geq c \geq 0$ and $a^{3}+b^{3}+c^{3}\leq a+b+c$ prove that
\[a^{2}+b^{2}+c^{2}\leq 1+a(b+c)\]
hope it's not trivial :lol:


[u]Individuals:[/u]

1 Max Ying
2 Surath Fernando
3 scai1
4 Derek Arbige
5 me  :|
6 Mengting Zhang
7 Lulu Zhang
8 Michael Menz
9 Jessica Li
10 Peter Brody
11 Bradley King
12 Jane Yu
13 Micah Margolis
14 Chan-Soo Kim
15 Robert Cheng
16 Jonathan Joo
17 Noah Arthurs
18 spiderboy95
19 Jeremy Fredette
20 Leo Tinone

[u]Teams:[/u]

1 Hopkins School (of course)
2 East Lyme Middle School
3 Mansfield Middle School
4 Sage Park Middle School
5 Roger Ludlow Middle School

Sorry if I used any names, PM me and I'll try to edit.

Ludlowe got 5th Oo  :o

We went on the same bus as them for a lot of things last year, and they absolutely sucked.

[quote="iasoaita"]Do you by any chance have the names of the top 40 individuals? I placed 39th and really want to know who got ahead of me in my school.  :)[/quote]

I'll PM you the stats I have from your school.

We haven't been called "Hopkins Grammar School" since the '80s. Back when I was on the team, we thought States should be renamed the Hopkins Invitational. :D

SCS resents this name.... :mad:

Even though we failed at chapter and only I made it to states. poor SCS. :(

[quote="spiderboy95"]Even though we failed at chapter and only I made it to states.[/quote]

Same!  I did get a trophy for 4th team though. :P

4th Team? I thought ur team didn't make it to states?

ME CONFUSED.

[quote="spiderboy95"]4th Team? I thought ur team didn't make it to states?[/quote]

Nope, we didn't, Top 3 Teams go to State.  But we still got a trophy out of it. :)

Well good job getting a trophy.

(btw...about the trophies...are those just modified deadly weapons? They look like something you might stab someone with.)

I actually cut myself on one of those trophies once.

I was having a sword fight with those trophies with a teammate at chapters last year.  :)  I had two swords so I outnumbered his one.

Well, congrats on having two trophies.....

Those things should be called trophyswords. That is a cool word.

TROPHYSWORDS

Actually four and a plaque, but I second your motion. Maybe sworphys?

Yeah, then we can call anyone who is good at Mathcounts "swordphy-ish"

LOL SWORDFISH

/me has 7!!!!! :D  :D  :D

Haha that's my inventory: (Although I had no team trophies since my school is a n00b :( )

6th grade:

6th Place Chapter

7th grade:

4th Place Chapter
6th Place State

8th grade:

3rd Place Chapter
2nd Place Chapter Countdown
2nd Place State
3rd Place State Countdown

Yay...

gauss1181- A final warning.  Don't post in our forum or we'll get our wolf (worthawholebean) after you.

Oh yes, my inventory:

6th grade: NOTHING YAY.
7th grade: 4th cd chapter.  NOTHING ELSE YAY.
8th grade: 2nd indiv 2nd cd 4th team chapter.  1st cd state.  FAIL.

Dude what's wrong with posting here if you're not from CT? This is a [b]public[/b] forum...so anyone can look at it...

Only [i]productive[/i] posts are welcome.  And so far, none of yours are.  Why can't you just start something up in the IL forum?  Or is that too boring. :P

Ernie, I think his posts are fine. As long as he's not spamming or being an idiot.

Anyways, you guys gotta remember that the first time I heard of a math competition was October of my 8th grade year and I had only made like 40 posts on this forum before then.  :)

My sworphys were

8th grade:
1st written chapter
1st team chapter
(lost at countdown, was so nervous :( )
2nd written state
1st countdown
no trophy but 5th at state.(my team was basically not good at anything :( I got 6 of our 9 chapter team questions and 5 out of our 8 state team questions. Also everyone was against the captain on a state question, she said ok then wrote down her answer. We were right, she was wrong and if we had gotten that right we would've been 3rd.) My teammates were 11th (2nd cd vs me) 41st, and 66th.

Generally, people don't post outside their own communities, but there is no *rule* against it. At the same time, if you're not from CT, I don't see any reason why you would want to discuss CT results.

/me has 9 sworphys
Sorry. I'm bored. My state forum phails. /me posts here.
6th grade- didnt do MC  :(

7th grade- (not on official school team cuz 5th school rd so no team trophies)  :maybe:
3rd Chapter
1st Chapter CD
2nd State

8th grade- (2nd school rd) (all awesome until i got 100 below my nats goal) :)
2nd Chapter
2nd Chapter CD
1st Team
1st State
2nd State CD
1st Team

hmm. by brother will prob end up with 18.

This forum fails too :(. No one ever posts here.

hey i remember the ct team from nats... sorta.

well, all i remember is hearing the 4th place guy go like "hey, uh sorry, but i'm gonna have to beat all of you" or something along those lines right before the test, then me and my teammate laughing hysterically (i doubt the ct team even heard us.. if they did... shoot).

How did we do this year? I was on the nats team last year and we got 19th.

[quote="Ihatepie"]How did we do this year? I was on the nats team last year and we got 20th.[/quote]

Well, no one told me, so pretty bad. :)

Oops, I meant 19th.  :lol:

/me quickly edits. Do you want to start the marathon here again?

A one question [hide=" "]
Shhh... here is the order for the first round.  If you talk about this publicly or privately, you will be MODKILLED!
cgyao15
spaceguy524
starwar
Mewto55555
maybach
tennis123
Xaenir
game.guy
dragon96
james4l
KING1
cf249
Bugi
MCManiac[/hide] marathon isn't really effective. :P

Hey, at least you guys did way better than we did...


Prove that $(ab)^k+(bc)^k+(ca)^k \le max(3,(3/2)^{2k})$

If $a+b+c=3,a,b,c\ge0$

This is Harazi's ineq?

[quote="Lovasz"]Try this...
http://www.mathlinks.ro/Forum/viewtopic.php?t=64926[/quote]

This isn't it  :(

I have post solution there:
http://mathnfriend.net/index.php?showtopic=488
P.S:are you Quan Vu,N.T.Tuan

[quote="spdf"]
P.S:are you Quan Vu,N.T.Tuan[/quote]

Those aren't me .


USAMTS
geometry
probability
expected value
number theory

Are there going to be only expected value questions, or lots of types of questions but an expected value question every month, or something else?

Philip Leszczynski

[quote="Philip_Leszczynski"]Are there going to be only expected value questions, or lots of types of questions but an expected value question every month, or something else?
[/quote]The latter -- there will be one expected value question per round, but the other questions will be the usual mix of USAMTS problems (algebra, geometry, number theory, counting, etc.).

Good point.

I was about to ask same question on that matter. I wish it's not too hard though (expected value).


What is the remainder when 13^13 +5 is divided by 6.

Note that $ 13 \equiv 1 \pmod{6}$, so $ 13^{13} \equiv 1^{13} \equiv 1 \pmod{6}$, so the desired number is $ 1\plus{}5 \equiv \boxed{0} \pmod{6}$.

Meh.

$ 13 \equiv 1 \pmod{6}$ is read as "$ 13$ is congruent to $ 1$ mod(ulo) $ 6$" and means that "$ 13$ has a remainder of $ 1$ when divided by $ 6$".  A property of mods is that if $ x \equiv y \pmod{a}$, $ x^n \equiv y^n \pmod{a}$, which is basically all you need to know to solve this problem.

[quote="jjx1"]I have no clue what any of those signs mean. mod the three line things,[/quote]

The three lines symbol, $ \equiv$, means "congruent to," but not congruent in the same sense as congruent is used in geometry. Here, $ a$ is congruent to $ b$ means that there is some integer, call it $ n$, that divides the difference of $ a$ and $ b$ exactly. $ (a \minus{} b)\div n$ has a remainder of $ 0$. The integer $ n$ is called the [i]modulus[/i].

When it is true that $ n$ divides the difference of $ a$ and $ b$ exactly then we say, " $ a$ is congruent to $ b$ [i]modulo[/i] $ n$, " which is written in symbols as:
\[ a\equiv b\pmod{n}\]

$ a\equiv b\pmod{n}$ if and only if $ a$ and $ b$ have the same remainder when each is divided by $ n$ individually.

Any number, and by number I mean integer, is congruent to its own remainder. Therefore when you see a statement like $ x\equiv 0\pmod{5}$ that tells you that the remainder when $ x$ is divided by $ 5$ is $ 0$ meaning that $ x$ is divisible by 5 with no remainder.

Likewise in your problem, when someone says $ 13\equiv 1\pmod{6}$ they mean that $ 13$ leaves a remainder of $ 1$ when divided by $ 6$, and that's true because $ 13 \equal{} 2(6) \plus{} 1$.

Now, the nice thing about congruences, or as others have called them, "mods," is that you can perform arithmetic on them just like you can regular, everyday, run of the mill, integers. You can add them, subtract them, multiply and divide (well maybe not divide - cuz that gets a little strange) but you can do arithmetic on them.

For example, suppose I asked you to find the remainder of $ (3\times 47)\div 13$ ?

You could multiply $ 3\times 47$ to get $ 141$. Then divide that by $ 13$ to get a remainder of $ 11$, but sometimes it is easier to say to yourself, $ 47\equiv 8\pmod{13}$, so $ 3\cdot 47\equiv 3\cdot 8 \equiv 24 \equiv 11\pmod{13}$. Either way you want to do it $ 3 \times 47$ has remainder $ 11$ when you divide it by $ 13$.

Back to your problem, given that you can multiply congruences,

$ 13\equiv 1\pmod{6}$ means $ 13\cdot 13\equiv 1\cdot 1\pmod{6}$, which means that $ 13\cdot 13\cdot 13\equiv 1\cdot 1\cdot 1\pmod{6}\ldots 13\cdot 13^{12}\equiv 1\cdot 1^{12}\pmod{6}$.

$ 13^{13} \plus{} 5\equiv 1 \plus{} 5\equiv 6 \equiv 0\pmod{6}$

I hope that clears things up for you.

okay, jjx, this is a mod:

really simply, $ 7\equiv 1 (\text {mod} 3)$ because when 7 is divided by 3, it leaves a remainder of 1. then, we proceed to powers, as aime said. hope that all of this stuff clears it up. (or just makes you more confused...)

Lol the mod doesn't understand mods..

Anyway, if a is congruent to b mod c, it means that a/c and b/c leave the same remainder.