For any positive integer $n>1$, let $p(n)$ be the greatest prime divisor of $n$. Prove that there are infinitely many positive integers $n$ with \[p(n)<p(n+1)<p(n+2).\]

Both solutions there seem wrong to me. Does someone have a correct solution?

[quote="Rust (in that topic)"]1. $ P(x)$ is not decrease, therefore exist $ x$, suth that $ P(x - 1) < P(x)$ (for example x is any odd prime).
If $ P(x + 1) > P(x)$ it give solution for one. If $ P(x + 1) < P(x)$ ($ P(x + 1)\not = P(x)$), then $ P(x^2 - 1) < P(x^2)$. If $ P(x^2 + 1) > P(x)$ we have solutione for one $ n = x^2 - 1$. If $ P(x^{2^l} + 1) < P(x) \ \forall l < k$ and ${ P(x^{2^k} + 1} > P(x)$ work $ n = x^{2^k}$. Because odd prime divisor $ P(x^{2^k} + 1) > 2^{k} > x$ if $ k > log_2 x$, then we find suth k for all x.
2. If $ S_k = P(2^k - 1)$ is not monotonik, exist k, suth that $ P(2^{k - 1} - 1) = P(2^k - 2) > P(2^k - 1) > P(2^k) = 2$.[/quote]Hmm... let me try to understand that.

Since $ P(x)$ is not decreasing, there is are infinitely many $ x$ with $ P(x - 1) < P(x)$, obviously (all odd primes).
If $ P(x + 1) > P(x)$ then we're done, so assume by contraposition that $ P(x) > P(x + 1)$, so by assumption $ P(x^2 - 1) < P(x^2)$. Again if $ P(x^2 + 1) > P(x)$ we're done, otherwise we have by assumption $ P(x^4 + 1) < P(x) = P(x^4)$.
Continuing like that we get either a valid $ n = x^{2^k}$ or we get $ P(x^{2^k} + 1) < P(x)$ for all $ k$.

So far I get it.
But then you finish the contradiction by claiming $ P(x^{2^k} + 1) > 2^{k}$. Why is this?

If odd prime $ p|x^{2^k}\plus{}1$, then (x,p)=1 and minimal period x mod p is $ T_p\equal{}2^{k\plus{}1}$, therefore $ 2^{k\plus{}1}|p\minus{}1$ or $ p>2^{k\plus{}1}>2^k$.

This was posted in another topic, I thought it was relevant here.

[quote="mavropnevma"]This is grobber's proof from one of the posts in the topics under the link given above. Peter seems to think it's wrong, but with some clarifications I will show it's right.

[quote="grobber"]
Let $ p>2$ be a prime, and let $ n_k\equal{}p^{2^k}\minus{}1$. We have $ P\left(n_k\plus{}1\right)\equal{}p$. We look at the sequence $ \left(n_k\plus{}2\right)\ \left\{k\geq 1\right\}$. We can see that any 2 distinct terms have 2 as their greatest common divisor, because for $ i<j$ we have $ \left(p^{2^j}\plus{}1\right)\minus{}2 \equal{} \left(p\minus{}1\right)\left(p\plus{}1\right)\left(p^2\plus{}1\right) ...\left(p^{2^i}\plus{}1\right) ...\left(p^{2^{j\minus{}1}}\plus{}1\right)$.

Since none of the terms of the sequence considered is divisible by 4, it means that the sequence $ P\left(n_k\plus{}2\right)$ contains arbitrarily large primes. Let $ k$ be the smallest integer for which $ P\left(n_k\plus{}2\right) > p$. Since $ n_k\equal{}\left(p\minus{}1\right)\cdot\left(p\plus{}1\right)\cdot n_1\cdot n_2\cdot ...\cdot n_{k\minus{}1}$ and all prime divisors of the number on the right are $ <p$, it means that $ P\left(n_k\right)<p\equal{}P\left(n_k\plus{}1\right)<P\left(n_k\plus{}2\right)$, and such a number $ n_k$ can be found for any prime $ p$.[/quote]
Let me first state [color=#FF0000]Kobayashi's Theorem[/color]:[color=#0000FF] If the set of prime divisors of the terms of an unbound sequence $(a_n)_{n\geq 1}$ is finite, then the set of prime divisors of the terms of any translate $(a_n + t)_{n\geq 1}$, for $t \in \mathbb{Z}^*$, is infinite.[/color]

With us, since the terms of the sequence $(n_k + 1 = p^{2^k})_{k\geq 1}$ have only $p$ as a prime factor, it follows the set of prime divisors of the terms of the sequence $(n_k + 2)_{k\geq 1}$ is infinite. grobber is actually proving this using Pólya's trick in showing Fibonacci numbers are pairwise coprime.

Let $ k$ be the smallest integer for which $ P\left(n_k\plus{}2\right) > p$. Now we'll fix a typo in grobber's proof: in fact we have $ n_k\equal{}\left(p\minus{}1\right)\cdot\left(p\plus{}1\right)\cdot (n_1 +2)\cdot (n_2 + 2)\cdot ...\cdot (n_{k\minus{}1} + 2)$, so $P(n_k) < p = P(n_k + 1) < P(n_k + 2)$.[/quote]

I found basically the same proof; just consider the sequence $p^{2^k}$. But when we need to show that $P(p^{2^k} + 1)$ gets arbitrarily large, I (am slightly ashamed to admit that I) invoked Stormer's Theorem (http://en.wikipedia.org/wiki/St%C3%B8rmer's_theorem), where mavropnevma used Kobayashi's Theorem, and where grobber, I believe, did the right thing in finding an easy, elementary proof (Polya's trick). By the way, this proof method leads to the the interesting statement that for every odd prime $p$ there exists exactly one positive integer $k$ for which we have $P(p^{2^k} - 1) < P(p^{2^k}) < P(p^{2^k} + 1)$. The fact that so far this method has been the only method posted, makes me wonder if a different easy argument can be found for this problem.

[quote=daoxun]Take $n=2^k$,and notice that the sequence ${p(2^k+1)}$ is not monotone increasing,And the rest is simple. :D[/quote]

I found the"notice that" is not quite easy to prove...I'll think a solution for it afterwards...

@above

Let $n = 2^k$ for some integer $k$. In your solution it ultimately came down to showing that there are an infinite number of integers $k$ such that $p(2^k + 1) > p(2^{k - 1} + 1)$. Suppose that number is finite and let $k$ be the last of such numbers. So
$$p(2^k + 1) > p(2^{k + 1} + 1) > p(2^{k + 2} + 1) > ...$$
By Zsigmondy's theorem, there exists another prime factor of $2^k + 1$ but doesn't appear in the factorization of $2^i + 1$ for $i < k$. Hence the number of prime factors is infinite and unbounded in the sequence
$\{2^i + 1\}$ for $i \ge k$. But $p(2^k + 1)$ is clearly bounded, so there must exist an $i$ such that $p(2^i + 1) > p(2^k + 1)$ for $i > k$. Hence the fact must hold that there are an infinite number of integers $k$ such that $p(2^k + 1) > p(2^{k - 1} + 1)$. (Note that $\gcd(2^k + 1, 2^{k + 1} + 1) = 1$ since $2^{k + 1} + 1 = 2(2^k + 1) - 1$)


calculus
integration
trigonometry
Euler
calculus computations

Evaluate:

([b]i[/b]) $ \int_{0}^{\infty}\left(\cos x \minus{} \frac{\sin x}{x}\right)\frac{dx}{x}$

([b]ii[/b]) $ \int_{0}^{\infty}\frac{\cos x \minus{} e^{\minus{}x^2}}{x}dx$

For #1:

$ \int\frac{(x\cos x\minus{}\sin x)}{x^2}dx\equal{}\frac{\sin x}{x}\plus{}C$

So (I think) $ \int_0^{\infty}\frac{(x\cos x\minus{}\sin x)}{x^2}dx\equal{}0\minus{}1\equal{}\minus{}1$

Using Mathematica I found that
$ \int_0^{\infty } \left(\frac{\text{Cos}\left[x^n\right]\minus{}e^{\minus{}x^m}}{x}\right) \, dx\equal{}\frac{ \gamma(n\minus{}m)}{nm}$ for all $ n,m >0$

But how to prove it :?:

For the second one, we can do like this:

Suppose $ p, q > 0$ and $ 1-p < r < \min (p, q) + 1$. Then consider the integral

$ I(r) = \int_{0}^{\infty} \frac{\cos (x^p) - e^{-x^q}}{x^{1+r}} \, dx.$

We are now going to determine $ I(r)$ in closed form with Gamma function. First, assume $ r \neq 0$. Then

\begin{eqnarray*}{ I(r)
& = & [ -\frac{\cos (x^p) - e^{-x^q}}{rx^r}]_{0}^{\infty} + \frac{1}{r} \int_{0}^{\infty} \frac{qx^{q-1} e^{-x^q} - px^{p-1} \sin (x^p)}{x^r} \, dx \\
& = & \frac{1}{r} [ \int_{0}^{\infty} x^{-r} e^{-x^q} \, (qx^{q-1} \, dx) - \int_{0}^{\infty} x^{-r} \sin (x^p) \, (px^{p-1} \, dx) ] \\
& = & \frac{1}{r} [ \int_{0}^{\infty} t^{-\frac{r}{q}} e^{-t} \, dt - \int_{0}^{\infty} u^{-\frac{r}{p}} \sin u \, du ] \qquad (t = x^q , u = x^p)\\
& = & \frac{1}{r} [ \Gamma ( 1 - \frac{r}{q} )- \frac{1}{\Gamma (r/p)} \int_{0}^{\infty} \int_{0}^{\infty} v^{\frac{r}{p}-1} \sin u e^{-vu} \, du dv ] \\
& = & \frac{1}{r} [ \Gamma ( 1 - \frac{r}{q} )- \frac{1}{\Gamma (r/p)} \int_{0}^{\infty} \frac{v^{\frac{r}{p}-1}}{1+v^2} \, dv ] \\
& = & \frac{1}{r} [ \Gamma ( 1 - \frac{r}{q} )- \frac{1}{\Gamma (r/p)} \int_{0}^{\infty} \tan^{\frac{r}{p}-1}} \theta \, d\theta ] \qquad (v = \tan \theta)\\
& = & \frac{1}{r} [ \Gamma ( 1 - \frac{r}{q} )- \frac{1}{2 \Gamma (r/p)} \Gamma (\frac{r}{2p} ) \Gamma ( 1 - \frac{r}{2p} ) ]\\
& = & \frac{1}{r \Gamma (1 + \frac{r}{p} )} [ \Gamma ( 1 - \frac{r}{q} ) \Gamma (1 + \frac{r}{p} ) - \Gamma (1 + \frac{r}{2p} ) \Gamma ( 1 - \frac{r}{2p} ) ]\\
\end{eqnarray*}

Now by continuity of $ I(r)$, we can evaluate $ I(0)$ by taking $ r \to 0$. Then we have

$ I(0) = \gamma ( \frac{1}{q} - \frac{1}{p} )$

where $ \gamma$ is the Euler-Mascheroni constant.


Can someone please post links to FREE Geometry Software (like Geometer's Sketchpad, without the cost)? Thanks.

I know two freeware programs:

[b]1.[/b] "Zirkel und Lineal" aka "Compass and Ruler" by René Grothmann, an Open Source program; see

  http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/index.html

[b]2.[/b] "Wingeom" by Peanut Software; see

  http://math.exeter.edu/rparris/

(among some other interesting utilities).

But I admit I don't use any of these two programs; instead I use "Euklid DynaGeo", which is a shareware program; see

  http://dynageo.de/

  Darij

Why can't you just use Paint (windows) or AppleWorks Drawing (mac)?

I can also suggest KSEG, available at
   http://www.mit.edu/~ibaran/kseg.html
(Windows and Linux versions, with source, GPL)

Is there anything like Geometer's sketchpad, in terms of quality?


search
function
combinatorics unsolved
combinatorics

On an infinite table,a horizontal line is chosen.Checkers may be placed below the line,at most one per square.The following move may be performed:A cheker jumped over is the removed.Determine the maximum number of sqrares that one checker can go across the chosen line.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=604280182&t=169621

Use the search function!


probability
pigeonhole principle
percent
complementary counting

2. A university has an undergraduate enrollment of $ 22,000$. There are a total of $ 366$ possible birthdays for these students including February 29. Knowing nothing of their birthdays, what is the largest number of these undergraduates that MUST share a birthday?

My thoughts were that for every 367 days, there must be two people who share a birthday. Therefore, I divided 22000 by 367, and came out with 59 and remainder 347. I thought 59 was the answer. I know I did something wrong, because the answer key says differently.

$ 8)$ What is the probability that W.C. will be yelled at during the school day if there is a 20% chance that Mr. Sturgeon will yell at him in the morning, a 25% chance that Miss Lutenbacher will yell at him by lunch, and a 50% chance that Miss Baker will yell at him in the afternoon?

On this problem, I don't see why it is not 95%, although that would make the problem way too easy if it really was.

[hide="2"] The problem with your reasoning is as follows.  Allow us to take another example.  Say that we wanted to fit 5 pigeons in 31 cages, what is the minimum number of pigeons that the most populated cage can have?  Okay, so we fit one pigeon in each cage.  We have 31-5=26 pigeons left.  Again, we fit another pigeon into each cage.  There are now two pigeons in each cage and 26-5=21 pigeons left.  Another pigeon in each cage.  16 pigeons left, 3 in each cage.  Another in each gives 11 pigeons left and 4 in each cage.  Another in each cage gives 6 left and 5 in each cage.  Put another in every cage and we obtain that there is one left and 6 in each cage.  However, we must put that one pigeon in a cage.  So, one cage will have 6+1=7 pigeons.  By your logic, however, every 6 cages will give 2 pigeons and there will thus be 31/6=5 pigeons, when there are really 7 pigeons in a cage.  This principle is important and is called the Pigeonhole Principle (yeah, because pigeons in cages...).

Now, back to the problem.  There are 22000 people.  So, 366 people have different birthdays.  Each birthday is repeated once.  Take another 366 people.  Two people have the same birthday now.  Okay, so we keep on doing this until we get that there are 60 people with a birthday and 40 people left.  So, again, put one person with a different birthday.  Thus, 40 birthdays will have 61 people and 326 birthdays with 60 people.  So, the minimum number of people that must share a birthday is $ \boxed{61}$.  [/hide]

[hide="8"] Here is what is wrong with your logic.  Again, let us take a different problem.  A child writes down the first 100 numbers, some colored red and the others colored black.  One third of the first 30 numbers are colored red, one fourth of the next 40 numbers are colored red, and one half of the final 30 numbers are colored red.  He tells his friend to pick a random number (the friend has not seen the child's number line).  What is the probability that the friend chose a red number.

By your logic, we have that probability is $ \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{2} \equal{} \frac {3}{4} \plus{} \frac {1}{3} \equal{} \frac {13}{12}$ .  However, we have that one third of 30 numbers are red.  So, 10 of these are red.  One fourth of the next 40 numbers are red.  So, 10 of these are red.  Finally, half of the final 30 numbers are red.  So, 15 of these are red.  So, there are 10+10+15=35 of the 100 numbers colored red.  So, the probability is $ \frac {35}{100} \equal{} \frac {7}{20}$.

Now, the problem with this question is that it doesn't specify how long lunch, the morning, and afternoon are.  So, I think that not enough information is given to solve the problem. [/hide]

[hide="2 generalized"]
The Generalized Pigeon Hole Principle states that if there are n pigeons and k holes, there must be a hole with at least $ \frac{n}{k}$, rounded up if it is not an integer.
[/hide]

[hide="complementary counting"]
The chance that W.C. will [b]not[/b] be yelled at in the morning is $ \frac{4}{5}$
The chance that W.C. will not be yelled at during lunch is $ \frac{3}{4}$
The chance that W.C. will not be yelled at in the afternoon is $ \frac{1}{2}$
So the chance he will not be yelled at is $ \frac{3}{10}$, or $ 30%
$
So the chance he [b]will[/b] yelled at is $ 70%
$

[/hide]

You have to put a \ before the percent sign,I think.

Thanks, you're right.  It should say 30% for the first error part and 70% for the 2nd error part.


Good afternoon

I need to do a cnc program, but the language don't have any function to do the inverse operations of sin and cos.

So, I need urgently to know how is possible to do the following without calculator:

cos-1(0.999946606)=0.592086533º
sin-1(0.055910543)=3.205109489º

I will appreciate your prompt help
Best regards

Pedro Ferrer

To do that without any compuational help should need some time...
But at least these can be calculated ba taylor series etc. See http://mathworld.wolfram.com/InverseSine.html for some possibilties.


function
real analysis
real analysis unsolved

Hi all, having trouble with the following:

1) Prove there is no norm on C([0,1]) st. convergence in this norm is equivalent to pointwise convergence in C([0,1])

2) Prove there is no norm on L^infty([0,1]) such that convergence in this norm is equivalent to convergence in measure in L^infty([0,1])

I don't really know where to start so any input would be appreciated. Thanks

If $ \|\cdot \|$ is a norm, then $ \|(f\plus{}g)/2\|\le \max\{\|f\|,\|g\|\}$. (You can derive this from the axioms of a norm).

Let's do 2). Let $ c$ be the norm of the function $ \chi_{[0,1]}$ (i.e. the function that is identically equal to $ 1$). We have $ \chi_{[0,1]}\equal{}\frac12(2\chi_{[0,1/2]}\plus{}2\chi_{[1/2,1]})$. Therefore, at least one of the functions $ 2\chi_{[0,1/2]}$ and $ 2\chi_{[1/2,1]}$ has norm not smaller than $ c$. Write this function as $ \frac12(\dots)$, and continue this process. You will get a sequence of functions all of which have norm at least $ c$. Yet, this sequence converges to $ 0$ in measure.


is the answer $(42-6\sqrt(4p^{2}-719))/p$.  This probalby can be simplified more..but trying out a specific case, it works..

All I did was use the angle bisector formula and a couple of similar triangles.

[quote="kohjhsd"][quote="GlenthemanN"]can you point to a proof for the fact that $\triangle{DBE}\sim \triangle{DCA}$?[/quote]

 :arrow: inscribed angles.[/quote]
 :rotfl:

I had been thinking of another way of approach, but maybe I was thinking too complicated?(it was from the top of my head)  This is part of my working.  The down sde to this is I have to do more to find the value of $p$ and put it back into the answer.
[hide]$7^{2}=12^{2}+16^{2}-2*12*16*cosACB$
$ACB=23.927$ and $ACB/2=11.96353$
$EAB=ECB=11.96353$ angles on same arc are equal.
$AEB=180-ACB=156.0729385$
by using the sin rule
$EB=3.5777$[/hide]

[quote="kohjhsd"][quote="worthawholebean"]Triangle $ABC$ is inscribed in a circle. $AC=12$, $BC=16$, and $AB=7$. The bisector of $\angle ACB$ is drawn and meets the circle at $E$ and $AB$ at $D$. If $CD=p$, find the length of $EB$ in terms of $p$.[/quote]

[hide]Applying angle bisector theorem, $BD=\frac{4}{7}\cdot 7 = 4$, $AD=7-BD=3$.

Since $\triangle DBE \sim \triangle DCA$, $\frac{EB}{BD}=\frac{AC}{CD}$, or $\frac{EB}{4}=\frac{12}{p}$, which gives $EB=\frac{48}{p}$.

 :?: [/hide][/quote]

uh you can forget my answer..but the reason why I got all confused is because you can actually find the numericla value of p and the exact value of EB, which turns out to be $48/\sqrt180$..so this is a bad problem finding in terms of p...(In my problem I forgot about the fact that DB has to be 4 and DA has to be 3.)

This contest generally isn't at a very high difficulty level; it doesn't expect you to know how to find an angle bisector (and quickly) (12 minutes, 6 people, 6 problems).


inequalities
Pythagorean Theorem
geometry

Given a triangle $ ABC$ with $ \angle BAC \equal{} 90$ degrees and $ BC\equal{}5$ find the maximum value of $ 3AB \plus{} 4AC$

By Pythagorean theorem, $ AB^2 \plus{} AC^2 \equal{} 25$. Then by the Cauchy-Shwarz inequality, we have

$ (AB^2 \plus{} AC^2)(3^2 \plus{} 4^2) \geq (3AB \plus{} 4AC)^2$

$ (25)(25) \geq (3AB \plus{} 4AC)^2$

$ 25 \geq 3AB \plus{} 4AC$

$ \therefore \max(3AB \plus{} 4AC) \equal{} \boxed{25}$

It can be attained when $ AB \equal{} 3$ and $ AC \equal{} 4$.


I got this crossword from school but I just need few more of them. I finished about 115 of them and here are only 3 that I couldn't get (assuming rest are right).

1. gaseous form of water

I know this is 5 lettered-word that is _ _ _ a _. I can't find other four.

2. A trumpet is made of the alloy ____

All I know is again, 5 lettered-word that is _ r _ s s. I read the section in my book ten times but it seem to be missing :mad: . Hope someone can help me. I'll talk with my teacher tomorrow though.

3. Greek name for salt-former

I got halogen but one thing I'm not sure is, is word "halogen" Greek? It doesn't sound like it  :P .

Thanks for all your help! This science forum rocks! :)

all i can think of for the first one is Steam

i dont get the second one

and for the third one, yes Halogen is a greek word.


AMC AIME USA(J)MO USAMO ARML geometry trigonometry

OK, fine. I admit it. All along, I've just wanted to qualify for Mock MoP.

That's why I took the Mock AMC. But since there's no Mock AIME, it looks like my hopes are simply going to die.

In all seriousness, though, is anyone really interested in doing a Mock AIME (as a follow-up to the Mock AMC) at some future date? (Especially those who "qualified" on the Mock AMC, but I'd let anyone do it anyway.) And is anyone interested in writing the questions? (Iff no one else wants to write the questions and there's interest in doing a Mock AIME, I'll write it, but I'd rather actually take the test, if it's an option.)

I'd want to do it but I didn't qualify.  I missed words in 4 problems.  Thus I got an 80.5 :( :( :(

[quote="white_horse_king88"]Even MR. Crawford would write excellent questions - I suppose. But I also didn't qualify by 0.5 points, so it doesn't really matter.[/quote]

Dude, that doesn't mean you can't take it. Actually, I'm desperate enough to have a Mock USAMO that I'll let anyone take the Mock AIME (whether or not it'll be "for real" remains to be seen.)

I am sure that Mr. Crawford and Mr. Rusczyk could write excellent questions.  The problem is that writing questions takes time.  Even for people with the kind of experience in problem that they have it won't be easy.  Maybe if a bunch of Advanced level problem solvers on this site could team up and write problems.

I have an idea for Mr. Rusczyk and Mr. Crawford however.  You can make a class that takes mock AMC's like these each week or so.  Or you could implement these into the problem series classes.  There are many options but the basic idea is that I think you should try to implement something of this sort into your classes.  I really liked this and I am sure that this concept would attract even more people to your classes.  Finally, my last idea for you.  You could make a contest like this and whoever gets the highest score gets to take one of your classes for free!!!  I'm just throwing out ideas that I think are good.

In that case, I'm in. How are we going to formulate the test though?

[quote="joml88"]Maybe if a bunch of Advanced level problem solvers on this site could team up and write problems.
[/quote]

We actually did something like that for our ARML practice one time. Which makes me think of a possible person who could write at least some of the problems...

I guess I'd take it... though I think AIME might be a little out of my league (I haven't seen it though so I don't know)

I'd be glad to fail a mock AIME. Count me in.

i'd like to fail an aime tooooo.

i'm in.

Umm...I doubt I would get any of the questions, but count me in.

Excellent.

Now all we need is someone to write the questions. Any volunteers?

No problem. We'll just retake the Mock AMC.  :)

There will be 15 problems to be solved in 3 hours.  All answers are integers between 0 and 999 inclusive, but many of these integers are obtained from formatting various non-integral expressions.  NO CALCULATORS ARE ALLOWED (to include slide rules and any other non-electronic calculation devices such as abaci).  Scratch paper (including graph paper) is permitted without limit as are rulers, compasses, and protractors, though these drawing devices are not necessary to solve any problems.

Scoring:  There are 15 questions, all weighted equally. Answer is given solely on the basis of the correct answer.  A correct answer receives one point, an incorrect answer receives zero points.  There is no penalty for guessing.  Hence, although it is unlikely to change your score, you should give an answer for each problem.

The rules and scoring of the Mock AIME will be exactly the same, with the following technical exception: On the actual AIME, you need to fill in lead 0's to make a three digit expression - e.g.For answers of 1 and 25 you would write 001 and 025 respectively.  I will not require this as I am not a machine and do not need to be reminded of how to distinguish 1, 10, and 100. :)

After your contest you will PM me your answers.  I will not take answers after a certain point, so try to be quick.  If for some reason PMing fails, as a backup you could email your answers to me at tmildorf@cox.net

Three questions:

1) Will you post high scorers?

2) Will you determine a USAMO qualifying score? If so, how?

3) Will we get our scores back individually following the contest?

Thanks! This is going to be fun.

* I will post statistics including high scorers, average score, easiest / hardest problem, # of solvers on each problem, score distribution.

* Is there a reason to send individual scores to people? The answers and solutions will be posted online and it will be very easy to figure out your score. (If you think I have made an error in grading your answers, which is quite possible as I will be hand scoring these problems, please let me know by PM or email and I'll recheck my scoring.)

* I think it will be necessary to make Mock USAMO (if we have it) invite only since it takes a substantial amount of time to grade 6 full solutions, especially if they are highly clever and unexpected. I guess we will use the formula

[tex]\displaymode{
Mock \,USAMO \,Index = \max\left(AMC_{JSRosen3}, AMC_{Rep123max}\right) + 10 \cdot AIME_{Mildorf}
}[/tex]

for qualifying (Maybe an AIME floor for $\le 10$-th graders?). I don't know exactly how many people will be invited to take a prospective Mock USAMO - it depends on the number of graders available.

[quote="crayshot97"]Hey Mildorf. It's your former Team Epsilon teammate from MOP this year, Illan. You know, one of the people who you had to explain all the problems to. I'd be happy to check over your answer key and your problems for clarity. And if that job is already taken, I'd also be glad to take your Mock AIME. I may not be able to solve all your problems, but I'm pretty sure I can make a good attempt to understand your solutions. Hope we can win the team contest again in MOP 2k5.[/quote]

BAAAAHHH it was all b/c of Jae Bae. you do realize both the teams that beat us had him on them

[quote="MysticTerminator"]BAAAAHHH it was all b/c of Jae Bae. you do realize both the teams that beat us had him on them[/quote]

That's just not true.

Believe it or not, I actually did a lot more of the problems than Jae.  :)  Of course, Jae and Yi did the black group homework so he was a big help in the homework contest, but the hard contests - especially the Romanian contest - were mostly my work.

[quote="Mildorf"]* I think it will be necessary to make Mock USAMO (if we have it) invite only since it takes a substantial amount of time to grade 6 full solutions, especially if they are highly clever and unexpected. I guess we will use the formula

[tex]\displaymode{
Mock \,USAMO \,Index = \max\left(AMC_{JSRosen3}, AMC_{Rep123max}\right) + 10 \cdot AIME_{Mildorf}
}[/tex]

for qualifying (Maybe an AIME floor for $\le 10$-th graders?). I don't know exactly how many people will be invited to take a prospective Mock USAMO - it depends on the number of graders available.[/quote]

If we use that formula, then the Mock AIME will only "officially" be open to those who took a mock AMC contest. Would it be possible to have qualification for the Mock USAMO determined solely by AIME score and grade? For example, if the test were average difficulty, then the qualifying score could be 6 or 7 for 10th grade and below and 8 or 9 for 11th and 12th, or something to that effect.

[quote="beta"]I am aiming for a double-digit score :), but I am probably only going to get a 3 or 4 because I think I will make a lot of careless mistakes.[/quote]

I'm aiming to get a question right.

But that question Rep posted seemed fairly easy.  Would it be possible to substitute 1.9972590695091477475689841168789 for [tex]$2\cos3^{\circ}[/tex]?  I thought you could, making the problem seem easy for me, but I'm pretty sure you can't.

[quote="interesting_move"]Perhaps we need a committee to write the questions.  Of course, nobody on the committee will be allowed to take the test...

-interesting_move[/quote]

Kinda like the AMC... ;) but seriously that's a good idea.  It does take a while to come up with good problems.

[quote="interesting_move"]Perhaps we need a committee to write the questions.  Of course, nobody on the committee will be allowed to take the test...

-interesting_move[/quote]

But the comitee should be able to take the mock USAMO.  If they can come up with questions, then they should be able to get at least a 10 on a AIME.

OK, so July 31, 2004, 3:00 PM to 6:00 PM is the official time?

3 PM EST tomorrow... it says all of this on the Mock AIME thread


abstract algebra
set theory
superior algebra
superior algebra unsolved

[quote="HilbertThm90"]You know the Five Lemma?

This is just a special case of the Five Lemma. Draw arrows down from the zeros and they are trivially isomorphisms, and everything still commutes. Apply the Five Lemma.

Or are you supposed to do this without the use of the Five Lemma?[/quote]

Ah yes, I didn't see that.  Thanks.

It also immediately pops out of the snake lemma, too, ($ 0 \to A \to 0$ exact means that $ A \equal{} 0$) not that it really matters ;).

You only need pure set theory here..
clearly $ f$ is injective because
\[ A\rightarrow B \stackrel{g}{\rightarrow} B' \equal{} A\stackrel{f}{\rightarrow} A' \rightarrow B'
\]
and the left-handside of the above equation is a composition of injections (I used exactness of the top row/sequence of your diagram and that g is a bijection)

To prove that $ f$ is a surjection is also not that difficult.

For simplicity name the horizontal maps taking note of exactness:
$ \phi : A\rightarrowtail B$ (map symbol for injective), $ \psi : B\twoheadrightarrow C$ (symbol for surjective), $ \phi' : A'\rightarrowtail B'$ and $ \psi : B'\twoheadrightarrow C'$

If $ b'$ belongs to the kernel of $ \psi'$, then by the surjectiveness of $ g$ there is a $ b\in B$ with $ g(b) \equal{} b'$ and this $ \psi(b) \equal{} 0\in C$ because $ h$ is also an injection
(use the commutativitiy of your diagram.. i.e. $ \psi'g(b) \equal{} h\phi$)

Thus
\[ g^{ \minus{} 1}(\mathrm{ker}(\psi')) \subset \mathrm{ker}(\psi) \equal{} \phi(A)
\]
and I think you can do the rest ;) .. feel free to ask more if its still confusing.

[quote="xianghu21"]Suppose

[b]see attachment[/b]

is a commutative diagram of modules over some ring $ R$. Suppose the rows are exact and $ g$ and $ h$ are bijective. Prove $ f$ is bijective.

Attempt:
I know the snake lemma and five lemma. This looks more like a diagram chasing question. In the five lemma and snake lemma I can use the commutativity of the diagram, but here, $ f$ is on the left, so I don't know how to use the commutativity of the diagram. Any hints for showing this would be nice. Thanks.[/quote]


Mporeite na deite ta apotelesmata se ayto to site: [url]http://www.imo-official.org/team_r.aspx?code=HEL&year=2008[/url]

[b]Sygxaritiria se olous![/b][/u]

Συγχαρητηρια σ'ολα τα παιδια! Συγχαρητηρια στα δυο μας μεταλλια!! Αλλα φυσικα συγχαρητηρια και στα υπολοιπα παιδια που ηταν πολυ κοντα στο μεταλλιο! Ελπιζω του χρονου να εχουμε 6 χρυσα!! :D

Συγχαρητήρια στα παιδιά για την προσπάθειά τους!
Σας εύχομαι ότι καλύτερο και στη συνέχεια.  :)

Συγχαρυτηρια και απο εμενα!!

Συγχαρητήρια και από μένα σε όλα τα παιδιά. Εύχομαι καλή πρόοδο, καλές σπουδές και το πιο σημαντικό από όλα: [b]να βοηθήσουν την επιστήμη να κάνει ένα βήμα παραπέρα.  [/b] Αυτό θα μείνει κληρονομιά στις επόμενες γενεές.

Τα αποτελέσματα ανά χώρα είναι εδώ:

http://www.imo-official.org/year_country_r.aspx?year=2008

Ευχάριστες εκπλήξεις αποτελούν το Ιράν, η Βόρεια Κορέα και η Ταυλάνδη.
Έκπληξη αποτελεί το γεγονός ότι η Ρουμανία πάτωσε και την πέρασε το Περού!

sygxaritiria kai apo emena paides!
kritike ti ennoeis?

[quote="kritikos"]Παιδιά τα αποτελέσματα της Ελλάδας δεν είναι με τίποτα αυτά που γράφει το site τα έχουν μπερδέψει. ΒΑσικά αυτό ισχύει για αρκετές χώρες[/quote]

Οι βαθμοί 36, 14, 2, 32, 1, 0 της Ελλάδας, είναι συνολικά ανά πρόβλημα (P1, P2,...) και όχι ανά διαγωνιζόμενο. Μήπως γιαυτό μπερδεύτηκες;


Is there some easy way to do this sort of problem?
\[ (1/x)+(1/x)^2 +(1/x)^3 ... \]

I'm not quite sure what you mean. I've seen a lot of problems like this and I haven't known what to do. In all these problems, x was an integer.

This is an infinite geomtric series with $a_1=\frac{1}{x}$ and $r=\frac{1}{x}$

The formula for the geomtric series is $a_1+a_1r+a_1r^2+...=\frac{a_1}{1-r}$ (if $|r|<1$).

Applying this formula, we get $\frac{\frac{1}{x}}{1-\frac{1}{x}} =\frac{1}{x-1}$.

For example, if $x=2$, then $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$.

I think for it to converge it must satisfy the condition: $-1 \leq \frac{1}{x} \leq 1$.

Strictly speaking, it should be $-1 <\frac{1}{x}<1$ because the series diverges if the common ratio is $1$.

Hey that's pretty neat! I got an extra question right because of it at a competition today :D

And just for kicks, to approximate this kind of series to a certain amount, say to $\frac{1}{x^{27}}$ you can plug in $\frac{a(1-r)^{26}}{1-r}$ where, in this case, $a=r=\frac{1}{x}$.


Find the value of x.

R and S are parallel.

[img]http://img200.imageshack.us/img200/4170/jjjsw.jpg[/img]

[hide="Hint"]
I drew a bunch of parallel and perpendicular lines. See if you can use them to find $ x$. Note: $ 4x \not \equal{} 90^\circ$ in my diagram. It denotes the entire angle, not a part of it.

[geogebra]077f948a2e13e61b4e319c12a6357f6b2c3eed92[/geogebra]
[/hide]

@luiseduardo : Your transversal construction was right. Get all angles in the lower triangle and use angle sum property in it. You get x = 30.

Can you explain how you arrived at that equation?

[hide="My Solution"]
Extend the perpendicular segment to line $ r$ until it hits line $ s$. You now have a quadrilateral with the angles $ 180^\circ \minus{} 4x, 2x \plus{} 10^\circ, 5x \minus{} 10^\circ, 90^\circ$. All these angles add up to $ 360^\circ$. Therefore,

$ 180 \minus{} 4x \plus{} 2x \plus{} 10 \plus{} 5x \minus{} 10 \plus{} 90 \equal{} 360\\
3x \equal{} 90\\
x \equal{} \boxed{30^\circ}.$
[/hide]

$ (4x\minus{}90^\circ)\plus{}(180^\circ\minus{}(5x\minus{}10^\circ))\equal{}2x\plus{}10^\circ \iff 90^\circ\equal{}3x \iff x\equal{}30^\circ$


calculus
derivative
LaTeX
algebra
function
domain
analytic geometry

Ths is an assignment question, so no solutions please. I just need some help understanding what is going on here.

Find the max and min values of $ f(x,y) \equal{} e^{xy}$ subject to the constraint $ x^3 \plus{} y^3 \equal{} 16$

So, taking the gradients, we get:

$ \nabla f(x,y) \equal{} (ye^{xy},xe^{xy})$
$ \nabla g(x,y) \equal{} (3x^2,3y^2)$

Setting $ \nabla f \equal{} \lambda \nabla g$
and solving each for $ \lambda$ and setting the expressions equal yeilds the condition $ y \equal{} x$
Plugging that into the constraint gives the point (2,2)

Now, the second derivitive test here says that (2,2) is a saddle point. So does this result mean that there is no max and min within this constraint? Or did I miss something when I was solving the system of equations?

$ \text{\LaTeX}$ tip: Spelling out words like "grad" in math mode is never a good idea. You could use \text{grad} but in this case, how about trying \nabla instead? $ \nabla f=\lambda\nabla g.$

Note: a saddle point in the "free" domain (an appropriate open set in $ \mathbb{R}^2$) does [i]not[/i] have to be a saddle point in the restricted domain (along the curve in question). Take a mountain pass: considered in a two-dimensional set, that's a saddle point for elevation. But considered along the road snaking through it, that is a local maximum of elevation.

Actually, you have another problem: $ (2,2)$ is not even a critical point for $ f$ in the free domain. $ \nabla f\ne 0$ there. That is typical for a constrained max-min problem. We don't even have the road going through the mountain pass; we just have the road angling partway up a slope and then angling back down again.

While there is a way to modify second derivative tests to deal with this, it's often simpler to consider extreme cases. Here, this is an unbounded curve; it tends off into the distance in the upper left and lower right. How does the function $ e^{xy}$ behave off in the distance of that curve?

I edited the latex, thanks. Wasn't sure what the symbol was called in Latex.

Anyway:

in the limits as x or y approach infinity, this constraint is essentially $ y \equal{} \minus{} x$ so the function tends to 0, meaning that (2,2) is a local maximum on the constraint.

That is intuitive, could you suggest a way to express this a little better in the context of the question?


function
calculus
integration
limit
real analysis
real analysis solved

It is a very beautiful and known problem that given a function f defined on a compact interval , if f has finite limit in each point then f is integrable Riemann. Prove that it is enough to assume that f has a limit (not necessarily finite) in a dense subset of the interval. This is a very very beautiful and difficult problem.

You mean that it's Riemann integrable, but the integral can be $\infty$? This is the case with $f:[0,1]\to\mathbb R,\ f(x)=\frac 1x,\ \forall x>0,\ f(0)=0$.

Yes, there's no finite limit at $0$, but it satisfies the conditions of the problem posted by Harazi.

Upps, sorry, I was taken by the weave. I am idiot today, as always. The correct statement is: if a real function has limit (possible infinite) in a dense subset then it is continuous in a dense subset. It is so strange, I must have been drunk when writing this. Oh, boy, I must take a rest. Please, sorry. But who knows, maybe with the condition that f is bounded results that f is integrable. Tough I really doubt. You know whats even strange, that Marian Tetiva and me ( the only role I had was to show that a student of the 11-th form can solve it) submitted this problem for the national Olympiad and many people did not believe that it is true!

nice one.
first we show that the thing with the infinite limit has no sense at all :). so suppose in some interval $[a; b]$, the infinite limits are dense.
first, choose $x_0$ as one point where we got an infinite limit. then there exists $\epsilon>0$ such that for all $x\neq x_0$ with $|x-x_0|<\epsilon$ we have $f(x)>1$. choose somewhere there $x_1$ which is a point with infinite limit. continuing that way, we find a sequence of intervals, each interval containing the next one, such that in the n-th interval, all values of f are $\geq n$. this, of course, is a contradiction(if $f$ gets arbritary small, then do the same with $-f$).
thus, the points with finite limit lie dense. now, we are doing quite the same as before. assume in $[a; b]$ there is no continuos point. choose $x_0$ as one point in $[a; b]$ with finite limit. thus, we find $\epsilon>0$ such that for all $x\neq x_0$ and $|x-x_0|<\epsilon$, $|f(x)-\lim_{x\to x_0}f(x)|<1$. choose $x_1$ with finite limit somewhere there. in that way, we may construct again a sequence of intervals, each interval containing the next, such that in the n-th interval, the values of f differ by at most $\frac 1n$. this, of course, defines a point where $f$ is continuos.

Peter

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16602[/url] for the question about Riemann integrability.


1) The set of all real numbers $x$ for which

$\log_{2004}({\log_{2003}({\log_{2002}({\log_{2001}{x}}}})))$

is defined as $\{x | x > c\}$. What is the value of $c$?

a) $0$
b) $2001^{2002}$
c) $2002^{2003}$
d) $2003^{2004}$
e) $2001^{2002^{2003}}$

the other is:

Square $ABCD$ has side length 2. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ instersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

[img]http://img59.exs.cx/img59/2290/am59fc.jpg[/img]

a) $\frac{2+\sqrt{5}}{2}$
b) $\sqrt{5}$
c) $\sqrt{6}$
d) $\frac{5}{2}$
e) $5-\sqrt{5}$

I didn't really understand these two  :blush:

Since we are looking for $x>c$, $c$ must be a minimum. Hence the following

\[\log_{2003}(\log_{2002}(\log_{2001}x))>0\]

because you can not take the $\log$ of a negative number in $\log_{2004}$.

Continuing with our previous result

\[\log_{2002}(\log_{2001}x)>1\]
\[\log_{2001}x>2002\]
\[x>2001^{2002}\]

Hence, $c=2001^{2002}$.

[quote="requiem"]1) The set of all real numbers $x$ for which

$\log_{2004}({\log_{2003}({\log_{2002}({\log_{2001}{x}}}})))$

is defined as $\{x | x > c\}$. What is the value of $c$?

a) $0$
b) $2001^{2002}$
c) $2002^{2003}$
d) $2003^{2004}$
e) $2001^{2002^{2003}}$[/quote]

O.K.The solution is as follows.

By the condition for which the logarithm is defined is that [b]anti-logarithm is  positive[/b],we have ${{\log_{2003}(\log_{2002}(\log_{2001}{x})})}>0\Longleftrightarrow\log_{2002}(\log_{2001}{x})>1$$\Longleftrightarrow\log_{2001}{x}>2002\Longleftrightarrow x>2001^{2002}$.
Therefore the answer is $c=2001^{2002}$.

[quote="requiem"]
Square $ABCD$ has side length 2. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ instersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

[img]http://img59.exs.cx/img59/2290/am59fc.jpg[/img]

a) $\frac{2+\sqrt{5}}{2}$
b) $\sqrt{5}$
c) $\sqrt{6}$
d) $\frac{5}{2}$
e) $5-\sqrt{5}$

I will give you a hint,requiem.

Let $O$ be the center of the semicircle and $T$ be the point of tangency between the tangent line to the semi circle from $C$ and the semi circle.

Set $\overline{ET}=x$,since $\overline{CT}=\overline{CB}$ and $\overline{EA}=\overline{ET}=x$,we have $\overline{DE}=\cdots$.
So you can see the triangle.Let's try again! :)

kunny

Here's the solution to the second one:
[hide]Let the point of tangency be F. since the length of two tangents from a fixed point to a fixed circle are trhe same, AE=EF and CB=CF. So let AE=EF=x. We know that CF=BC=2, so in right triangle CDE, we have lengths DE=2-x, CD=2, and CE=2+x. So we set up the equation
(2-x)^2+2^2=(2+x)^2
so x=1/2.
CE=2+x=5/2 D.[/hide]

*edit
nvm I was too late, someone already answered...


Let $ \theta_{1},\theta_{2},\ldots,\theta_{n}$ be real numbers (in the interval $ [0, 2\pi]$ if you like). Show that

$ \sum_{i\equal{}1}^{n}\sum_{j\equal{}1}^{n}\cos{(\theta_{i}\minus{}\theta_{j})}\ge 0$.

[hide="hint"] There is a one-line proof (maybe a little longer if you define some stuff). [/hide]

Solution:

[hide]
Let the double summation resolve to the value $ S$. Consider all unique pairs $ \theta_{a}$ and $ \theta_{b}$ in the above defined set. Then $ S=\sum{\cos{(\theta_{a}-\theta_{b})}+\cos{(\theta_{b}-\theta_{a})}+\cos{(\theta_{a}-\theta_{a})}+\cos{(\theta_{b}-\theta_{b})}=\sum{[2\cos{(\theta_{a}-\theta_{b})}+2]}\ge{0}}$.[/hide]

Wait, I don't think that quite works because you are using each $ \cos{(\theta_{a}\minus{}\theta_{a})}$ multiple times ($ n\minus{}1$ times, in fact). The inequality is stronger than that, but there is still a very short solution.

Use dot product of some unit vectors. The expression is the square of the norm of the sum of the vectors.

err can u actually show it alex im a lil confused

Consider the vectors $ x_{i}\equal{} (\cos{\theta_{i}},\sin{\theta_{i}})$ and their pairwise dot products.

[hide="no vectors needed"]
$ \sum_{i = 1}^{n}\sum_{j = 1}^{n}\cos{(a_{i}-a_{j})}=\sum_{i = 1}^{n}\sum_{j = 1}^{n}\cos{(a_{i})\cos{(a_{j})}+\sin{(a_{i})\sin{(a_{j})}}}$

$ =\left(\sum_{i = 1}^{n}\cos(a_{i})\right)^{2}+\left(\sum_{i = 1}^{n}\sin(a_{i})\right)^{2}$
all of the terms on the right side must be greater than 0
[/hide]


How many 3-git natural numbers n are there such that the ones and tens digit of n^2 are the same as thos of n?

--How do i [b]prove[/b] that these are 00,01,25,and 76??

[hide="Solution"]
That is equivalent to $ x^2 \equiv x \pmod {100}$, or $ x(x \minus{} 1) \equiv 0 \pmod {100}$. 5 cannot divide both x and x-1, so 25 must divide one of them, it is easy to check all possibilities from here, just try to make either x or x-1 a multiple of 25.[/hide]

I still dont get it. Can u kindly expound further?

Your set is not all there are numbers ranging from $ {100 to 900}$.
From$ {101 to 901}$.

[hide="Solution"]You want all $ n$ s.t. $ 100|n^2\minus{}n.$ Hence $ 100|n(n\minus{}1)$.

We take cases.

If$ n$ is odd, we know that one of the following applies:

1) $ 25|n$ and $ 4|n\minus{}1 \Rightarrow n$ ends in $ 25$
2) $ 5|n$ and $ 20|n\minus{}1 \Rightarrow n$ has no solutions, because two multiples of $ 5$ cannot differ by $ 1$
3) $ 100|n\minus{}1 \Rightarrow n$ ends in $ 01$

On the other hand, if $ n$ is even:

1) $ 25|n\minus{}1$ and $ 4|n \Rightarrow n$ ends in $ 76$
2) $ 5|n\minus{}1$ and $ 20|n \Rightarrow n$ has no solutions
2) $ 100|n \Rightarrow n$ ends in $ 00$

Hence the possible endings are 00, 01, 25, 76.[/hide]


Is it possible at any point in a game of chess for white pawns to be at a2, a3, a4, a5, a6, a7, and a8 (right before white promotes)

[hide]No, that'd require way too many captures. 21, in fact, when there are only 15 black pieces to capture (you can't capture the king)

You could, however, construct a game where all the pawns are stacked on the e-file, though, which would only require 12 captures. You could also stack the pawns on the a-file without having the a8 pawn (since it's not going to be a pawn anymore anyway), which will take 15 captures and leave black with just a king left (though you'd had to sacrifice your own pieces to set up enemy pawns in the right places for capture).
[/hide]


inequalities
function
calculus
derivative
geometry
3D geometry
vector

Hmm, I'm thinking that you can view it as a function, and since the second derivative is 0, it's a convex, bounded function, and then it will achieve it's maximum at the endpoints of a five cube?

edit> Actually, just take partial derivatives for all five variables, and it has to be orthogonal to the polytope $x_1+x_2+x_3+x_4+x_5=1$, and then you just set the vectors equal to $(c,c,c,c,c)$ and you're gonna get all $x_i$ equal as one solution and some extraneous solutions that will be less than $\frac{1}{25}$.

what is Rearrangement Inequality?

and could u please explain how Chebyshev is used here? i know u can only use it when the terms increase in order. but thats about it.

- Rearrangement Inequality:  a < b < c < d then the order:

cd + ab >= than other pair of combinations.

[url=http://planetmath.org/encyclopedia/RearrangementInequality.html]http://planetmath.org/encyclopedia/RearrangementInequality.html[/url]

For Chebyshev, I'm not really sure but I see people use it once so I copy :roll: Just look at how I did it then you could get a sense :oops:  The problem with most inequality is that people understand it but forgot the name.  I think it's not really Chebyshev, but a combination of AM-GM.

It's actually this one, I found it: [url=http://mathworld.wolfram.com/ChebyshevSumInequality.html]http://mathworld.wolfram.com/ChebyshevSumInequality.html[/url]


integration
calculus
trigonometry
geometric series
calculus computations

Evaluate :
1) $ \int_{0}^{\plus{}\infty}{\frac{x\minus{}sinx}{x^3}dx}$
2)$ \int_{0}^{\plus{}\infty}{\frac{1\minus{}cos(ax)}{x^2}dx}$,$ a \in R$
3)$ \int_{0}^{\plus{}\infty}{\frac{x^{m\minus{}1}}{x^n\plus{}1}dx}$  where n and m are positive integers and $ 0<m<n$
4)$ \int_{0}^{\plus{}\infty}{\frac{x}{e^x\minus{}e^{\minus{}x}}dx}$

For 1) and 2), just apply integration by part to reduce to Dirichlet integral.

For 3), use contour integration, or apply substitution $ x \equal{} \tan^{2/n} \theta$ to reduce to beta function.

For 4) Divide both denominator and numerator by $ e^{\minus{}x}$. Use geometric series to expand the integrand, and switch the order of integration and summation. Then you'll have a familiar series.

2) $ I(a) = \int_{0}^{ + \infty}{\frac {1 - cos(ax)}{x^2}dx}$
$ I(a)' = a\int_{0}^{ + \infty}{\frac {\sin u}{u}du} = \frac {\pi a}{2}$.
So that we have $ I(a) = \frac {\pi a^2}{4}$

We get $ \pi = I(2) \implies \int_{0}^{ + \infty} \left({\frac {\sin u}{u} \right)^2 du} = \frac{\pi}{2}$

[b]4)[/b] Using outline by sos440:

$ I = \int\limits_0^\infty \frac{x}{e^x - e^{-x}} \, dx = \int\limits_0^\infty \frac{x\, e^{-x}}{1 - e^{-2x}} \, dx =  \int\limits_0^\infty x\, e^x \frac{p}{1 - p} \, dx$ with $ p = p(x) = e^{-2x}$.

Now note that $ \frac{p}{1 - p} = \sum\limits_{k=1}^\infty p^k$ (obviously we have $ |p| < 1$ here), giving:

$ I = \int\limits_0^\infty [ x \, e^x \sum\limits_{k=1}^\infty p^k  ] \,dx  = \sum\limits_{k = 1}^\infty [ \int\limits_0^\infty x \, e^x e^{-2kx} \, dx  ]$
$ = \sum\limits_{k = 1}^\infty [  \int\limits_0^\infty x \, e^{(1-2k) x}\, dx  ] = \sum\limits_{k = 1}^\infty  [ [  \frac{x}{1-2k} e^{(1-2k) x}|_0^\infty - \int\limits_0^\infty \frac{1}{1-2k} e^{(1-2k)x} \, dx  ]$
$ = \sum\limits_{k = 1}^\infty  [ -\frac{1}{(1-2k)^2} e^{(1-2k)x}|_0^\infty  ] = \sum\limits_{k=1}^\infty \frac{1}{(2k - 1)^2} = \boxed{\frac{\pi^2}{8}}$ (Euler's series).

*: Still have to justify switching summation and integral.

For 1). Pick: $ J(a) \equal{} \int_{0}^{\plus{}\infty}{\frac{ax\minus{}sinax}{x^3}dx}$.
Then $ J'(a) \equal{} I(a)$, where $ I(a) \equal{} \int_{0}^{\plus{}\infty}{\frac{1\minus{}cos(ax)}{x^2}dx} \equal{} \frac{\pi a^2}{4}$ see problem 2.
Then $ J(a) \equal{} \frac{\pi a^3}{12} \plus{} C$, but C is zero.
We have:
$ J(1) \equal{} \int_{0}^{\plus{}\infty}{\frac{x\minus{}sinx}{x^3}dx} \equal{} \frac{\pi}{12}$


Let $ x,y,z>0$ and $ xyz\equal{}1$ Prove that
$ \sum\frac{1}{x^2\plus{}2z^2\plus{}3}\leq\frac{1}{2}$

[quote="Algadin"]Let $ x,y,z > 0$ and $ xyz \equal{} 1$ Prove that
$ \sum\frac {1}{x^2 \plus{} 2z^2 \plus{} 3}\leq\frac {1}{2}$[/quote]
$ \sum_{cyc} \frac{1}{x^2\plus{}2z^2\plus{}3}\equal{}\sum_{cyc} \frac{1}{(x^2\plus{}1)\plus{}(2z^2\plus{}2)} \leq \frac{1}{2}.\sum_{cyc} \frac{1}{x\plus{}2z}$
We need prove that $ \sum_{cyc} \frac{1}{x\plus{}2z} \leq 1$ with $ xyz\equal{}1$
I think it is similar as http://www.mathlinks.ro/viewtopic.php?t=198930 :P

[quote="Algadin"]Let $ x,y,z > 0$ and $ xyz \equal{} 1$ Prove that
$ \sum\frac {1}{x^2 \plus{} 2z^2 \plus{} 3}\leq\frac {1}{2}$[/quote]
  other solution :
 $ \sum\frac {1}{x^2 \plus{} 2z^2 \plus{} 3}\leq \frac{1}{2}(  \frac{xy}{xy\plus{}1\plus{}x} \plus{}  \frac{1}{xy\plus{}1\plus{}x} \plus{}\frac{x}{xy\plus{}x\plus{}1})  \equal{}\frac{1}{2}$  seem  it's  old


For $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}3$ prove that :

$ 3\plus{}\frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{2} \ge \frac{a\plus{}b^2c^2}{b\plus{}c}\plus{}\frac{b\plus{}c^2a^2}{c\plus{}a}\plus{}\frac{c\plus{}a^2b^2}{a\plus{}b}\ge 3$

[quote="Ronald Widjojo"]For $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that :

$ 3 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \ge \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}\ge 3$[/quote]

By cauchy-swarchz:
$ \frac{a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \sum \frac {b^2c^2}{b \plus{} c} \ge \frac {a \plus{} b \plus{} c}{2} \plus{} \sum \frac {b^2c^2}{b \plus{} c}$
By Am-Gm We have:
$ \sum (\frac {a^2b^2}{a \plus{} b} \plus{} \frac {a \plus{} b}{4}) \ge \sum ab \equal{} 3$
Done!  :P

[quote="Ronald Widjojo"]For $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that :

$ 3 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \ge \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$ [/quote]
By Am-Gm and Cauchy Schwarts, we can prove easily the stronger:
$ \frac{5(a^2\plus{}b^2\plus{}c^2)}{6} \plus{} \frac{1}{2} \geq\  \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{}
\frac {c\plus{}a^2b^2}{a \plus{} b}$
 :)

[quote="nguoivn"][quote="Ronald Widjojo"]For $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that :

$ 3 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \ge \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$ [/quote]
By Am-Gm and Cauchy Schwarts, we can prove easily the stronger:
$ \frac {5(a^2 \plus{} b^2 \plus{} c^2)}{6} \plus{} \frac {1}{2} \geq\ \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$
 :)[/quote]
Can you prove it hear  :maybe:

[quote="nguoivn"][quote="Ronald Widjojo"]For $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that :

$ 3 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \ge \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$ [/quote]
By Am-Gm and Cauchy Schwarts, we can prove easily the stronger:
$ \frac {5(a^2 \plus{} b^2 \plus{} c^2)}{6} \plus{} \frac {1}{2} \geq\ \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$
 :)[/quote]
The first, using my old result: $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{} \frac{1}{2} \geq\ \sum\ \frac{a}{b\plus{}c}$
Besides, by CS and Am-Gm: $ \sum\ \frac{a^2b^2}{a\plus{}b} \leq\ \sqrt{\sum\ a^2b^2. \sum\ \frac{a^2b^2}{(a\plus{}b)^2}} \leq$ $ \sqrt{\frac{1}{3}.(a^2\plus{}b^2\plus{}c^2)^2.\sum\ \frac{ab}{4}} \equal{} \sum\ \frac{a^2\plus{}b^2\plus{}c^2}{2}$
Add 2 inequalities, we have my stronger  :)

[quote="nguoivn"][quote="nguoivn"][quote="Ronald Widjojo"]For $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that :

$ 3 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \ge \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$ [/quote]
By Am-Gm and Cauchy Schwarts, we can prove easily the stronger:
$ \frac {5(a^2 \plus{} b^2 \plus{} c^2)}{6} \plus{} \frac {1}{2} \geq\ \frac {a \plus{} b^2c^2}{b \plus{} c} \plus{} \frac {b \plus{} c^2a^2}{c \plus{} a} \plus{} \frac {c \plus{} a^2b^2}{a \plus{} b}$
 :)[/quote]
The first, using my old result: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {1}{2} \geq\ \sum\ \frac {a}{b \plus{} c}$
Besides, by CS and Am-Gm: $ \sum\ \frac {a^2b^2}{a \plus{} b} \leq\ \sqrt {\sum\ a^2b^2. \sum\ \frac {a^2b^2}{(a \plus{} b)^2}} \leq$ $ \sqrt {\frac {1}{3}.(a^2 \plus{} b^2 \plus{} c^2)^2.\sum\ \frac {ab}{4}} \equal{} \sum\ \frac {a^2 \plus{} b^2 \plus{} c^2}{2}$
Add 2 inequalities, we have my stronger  :)[/quote]
Very nice;nguoivn  :P


Suppose that

$ f$ is continuous on $ [a,b]$,

$ |f(x_n)\minus{}f(y_n)|\geq\epsilon$, where $ x_n, y_n\in[a,b]$, and

$ x_{n_k}, y_{n_k}$ are convergent subsequences of $ x_n, y_n$ with the same limit $ c\in[a,b]$, respectively.

How do we get $ \left|\lim_{k\to\infty}f(x_{n_k})\minus{}\lim_{k\to\infty} f(y_{n_k})\right|\geq\epsilon$?

(please see the attachment)

Isn't this difference just the same as $ \lim_{k\to\infty}|f(x_{n_k})\minus{}f(y_{n_k}|$ ? ;)

Oops! I see.  :blush:
$ u_n\geq v_n$ for big $ n$ implies $ \lim_{n\to\infty}u_n\geq\lim_{n\to\infty}v_n$.


conics
ellipse
rotation
projective geometry
geometry
circumcircle
angle bisector

Let $ ABCD$ be a square in which the two diagonals intersect at $ E$.
A line through $ A$ meets $ BC$ at $ M$ and intersects $ CD$ at $ N$.
Let $ K$ be the intersection point of $ EM$ and $ BN$.Prove that $ CK \perp BN$

Let $ DM$ cut the line $ AB$ at $ L$ and let $ \mathcal E$ be an ellipse tangent to $ AB \parallel DC$ at $ B, C$ and tangent to $ DA.$ The ellipse is centered at the midpoint $ O$ of $ BC.$ By Brianchon theorem for the hexagon $ ABLNCD$ with main diagonals $ AN, BC, LD$ concurrent at $ M$, the line $ LN$ is also a tangent of this ellipse. The ellipse can be projected to a circle $ (O)$ by a parallel projection, which keeps its major axis in place. Its parallel tangents $ ABL \parallel DCN$ go into parallel tangents $ A'B'L' \parallel D'C'N'$ of the circle $ (O)$ with radius $ r \equal{} AB \equal{} CD \equal{} A'B' \equal{} C'D'$ and $ N'C' \equal{} NC, L'B' \equal{} LB.$ Let $ T'$ be the tangency point of $ (O)$ with $ L'N'.$ $ OL', ON'$ bisect the angles $ \angle B'OT', \angle C'OT'$ $ \Longrightarrow$ $ \angle L'ON' \equal{} 90^\circ$ is right, $ r^2 \equal{} OT'^2 \equal{} L'T' \cdot N'T' \equal{} L'B' \cdot N'C' \equal{} LB \cdot NC,$ hence, $ BC^2 \equal{} r^2 \equal{} LB \cdot NC$ and the right $ \triangle LBC \sim \triangle BCN$ are similar. Since their corresponding legs are rotated by $ 90^\circ,$ so are their hypothenuses and $ LC \perp BN.$ Let $ K$ by their intersections. By Pappus theorem, the intersections $ E \equiv AC \cap BD, M \equiv AN \cap LD, K \equiv LC \cap BN$ are collinear.

[quote="yetti"]Let $ DM$ cut the line $ AB$ at $ L$ and let $ \mathcal E$ be an ellipse tangent to $ AB \parallel DC$ at $ B, C$ and tangent to $ DA.$ The ellipse is centered at the midpoint $ O$ of $ BC.$ By Brianchon theorem for the hexagon $ ABLNCD$ with main diagonals $ AN, BC, LD$ concurrent at $ M$, the line $ LN$ is also a tangent of this ellipse. The ellipse can be projected to a circle $ (O)$ by a parallel projection, which keeps its major axis in place. Its parallel tangents $ ABL \parallel DCN$ go into parallel tangents $ A'B'L' \parallel D'C'N'$ of the circle $ (O)$ with radius $ r \equal{} AB \equal{} CD \equal{} A'B' \equal{} C'D'$ and $ N'C' \equal{} NC, L'B' \equal{} LB.$ Let $ T'$ be the tangency point of $ (O)$ with $ L'N'.$ $ OL', ON'$ bisect the angles $ \angle B'OT', \angle C'OT'$ $ \Longrightarrow$ $ \angle L'ON' \equal{} 90^\circ$ is right, $ r^2 \equal{} OT'^2 \equal{} L'T' \cdot N'T' \equal{} L'B' \cdot N'C' \equal{} LB \cdot NC,$ hence, $ BC^2 \equal{} r^2 \equal{} LB \cdot NC$ and the right $ \triangle LBC \sim \triangle BCN$ are similar. Since their corresponding legs are rotated by $ 90^\circ,$ so are their hypothenuses and $ LC \perp BN.$ Let $ K$ be their intersection. By Pappus theorem, the intersections $ E \equiv AC \cap BD, M \equiv AN \cap LD, K \equiv LC \cap BN$ are collinear.[/quote]

:clap2: i have no doubt to u dear yetti

u are really the one :wink:

let K' be feet of C on BN we have:$ \frac{CK'}{K'B}\equal{}\frac{CN}{BC}\equal{}\frac{CN}{AB}\equal{}\frac{CM}{MB}$ hence MK' is angle bisector of $ \angle BK'C$ hence K',M,E collinear hence $ K'\equiv K$

[quote="Amir.S"]let K' be feet of C on BN we have:$ \frac {CK'}{K'B} \equal{} \frac {CN}{BC} \equal{} \frac {CN}{AB} \equal{} \frac {CM}{MB}$ hence MK' is angle bisector of $ \angle BK'C$ hence K',M,E collinear hence $ K'\equiv K$[/quote]

I don't understand why $ \angle MK'B\equal{}\angle CK'M$ implies that $ E$, $ M$, and $ K'$ are collinear, could you please explain?

I had a similar solution...

Let the circumcircle of $ \triangle EBC$ intersect $ BN$ again at $ K'$. Then $ \angle EBC\equal{}\angle EK'C\equal{}\angle ECB\equal{}\angle EK'B\equal{}45$ so then $ \angle CKB\equal{}90$ then $ \triangle BKC\sim\triangle BCN$. Then $ K'E$ bisects angle $ BK'C$. Let $ K'E$ intersect $ BC$ at $ M'$. Then we have \[ \frac{BM'}{M'C}\equal{}\frac{BK'}{K'C}\equal{}\frac{BC}{CN}\equal{}\frac{AB}{CN}\equal{}\frac{BM}{MC}\]

Hence $ M\equiv M'$. But then $ K'$ is a point that lies on $ EM$, and $ BN$ so $ K\equiv K'$. But $ K'$ had the desired property, so we are done.

Dear mathlinkers,
another way...
let 1 be the circle with BC for diameter (E is on 1, and AB and CD are tangents to 1),
K' be the second intersection of BN with 1 and M' the intersection of EK' and BC.
Applie the Pascal's theorem in a particular case with tangents to BK'ECCBB.
It follows that N, M' and A are colinear...
then, M'=M, then K=K'
and finally CK is perpendicular to BN.
Sincerely
Jean-Louis

It is enough to prove that the point $ K\equiv EM\cap BN,$ lies on the circumcircle $ (O),$ of the right triangle $ \bigtriangleup EBC,$ with diameter the side-segment $ BC,$ of the given square $ ABCD.$

We can also consider the point $ K,$ as the intersection point of $ (O),$ from the line segment $ BN$ and it is enough to prove that the points $ E,\ M,\ K,$ are collinear.

Through the points $ A,\ N,$ we draw the lines tangent to $ (O)$ at points $ F,\ H$ respectively, which intersect the sidelines $ CD,\ AB,$ at points $ P,\ Q,$ respectively.

So, we have the configuration of the circumscribed quadrilateral $ APNQ$ and then, based on the [b][size=100]Newton Theorem[/size][/b], we have that the line segment $ FH,$ passes through the point $ M\equiv AN\cap BC$ $ ($ the line segment $ PQ,$ also passes through the point $ M,$ as the diagonal of $ APNQ$ $ ).$

Based now, on the bellow [b][size=100]Lemma[/size][/b], we conclude that the points $ E,$ $ M,$ $ K$ are collinear and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. – A circumscribed quadrilateral $ ABCD$ is given and let $ K,\ L,\ M,\ N$ be, the tangent points of $ AB,\ BC,\ CD,\ DA$ respectively, to its incircle $ (O)$ and we denote the points $ E\equiv (O)\cap AM,\ F\equiv (O)\cap CK.$ Prove that the points $ E,\ P,\ F$ are collinear, where $ P\equiv AC\cap BD$[/color][/size][/b].

Kostas Vittas.

PS. I will post here later the proof of the above [b][size=100]Lemma[/size][/b], I have in mind.

[size=134][color=darkblue][b]Very nice and old problem ![/b][/color][/size]

[quote="apollo"] [color=darkred][b]Let $ ABCD$ be a square in which $ E\in AC\cap BD$ . For a line $ d$ , $ A\in d$ denote

$ M\in BC\cap d$ , $ N\in CD\cap d$ and $ K\in EM\cap BN$. Prove that $ CK \perp BN$ .[/b] [/color] [/quote]

[color=darkblue][u][b]Proof.[/u][/b] Prove easily that the lower lemma :[/color]

[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle for which $ AB\perp AC$ . Then for a point

$ K\in BC$  exists the equivalence $ \frac {KB}{KC} \equal{} \left(\frac {AB}{AC}\right)^2\ \Longleftrightarrow\ AK\perp BC$ .[/color][/quote]

[color=darkblue]Denote the intersection $ P\in EM\cap CD$ . Observe that $ \frac {MB}{MC} \equal{} \frac {MA}{MN} \equal{} \frac {CB}{CN}$ . Apply the [b]Menelaus[/b]' theorem

to the line $ \overline {MPE}$ and $ \triangle ACN\ \ : \ \ \frac {MN}{MA}\cdot\frac {EA}{EC}\cdot\frac {PC}{PN} \equal{} 1\implies\frac {PC}{PN} \equal{} \frac {CB}{CN}$ . Apply the [b]Menelaus[/b]' theorem

to the line $ \overline {MPK}$ and $ \triangle BCN\ \ : \ \ \frac {MC}{MB}\cdot\frac {KB}{KN}\cdot\frac {PN}{PC} \equal{} 1\ \implies$ $ \frac {KB}{KN} \equal{} \left(\frac {CB}{CN}\right)^2$ $ \stackrel {(lemma)}{\ \Longleftrightarrow\ }$ $ CK\perp BN$ .[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension.[/u] Let $ ABCD$ be a rhombus in which $ E\in AC\cap BD$ .

For a line $ d$ , $ A\in d$ denote $ M\in BC\cap d$ , $ N\in CD\cap d$ and $ K\in EM\cap BN$.

Prove that the ray $ [CK$ is the $ C$ - symmedian in $ \triangle BCN$ , i.e. $ \frac {KB}{KN}\equal{}\left(\frac {CB}{CN}\right)^2$ .[/b] [/color] [/quote]


AMC
AIME
geometry
trigonometry
geometric transformation
rotation
trig identities

1. In an isosceles triangle $ ABC$ with base $ BC$ angle at vertex $ A$ is equal to $ 80^\circ$. Inside triangle $ ABC$ point $ M$ is taken so that $ \angle MBC \equal{} 30^\circ$ and $ \angle MCB \equal{} 10^\circ$. Find the value of angle $ \angle AMC$.

2. In an isosceles triangle $ ABC$ with base $ AC$ the angle at vertex $ B$ is equal to $ 20^\circ$. On sides $ BC$ and $ AB$ points $ D$ and $ E$, respectively, are taken so that $ \angle DAC \equal{} 60^\circ$ and $ \angle ECA \equal{} 50^\circ$. Find angle $ \angle ADE$.

For #1

[hide="hint"]
What is special about $ \triangle AMC$?

[/hide]

[hide="hint2"]

Suppose you make an equilateral triangle $ XAC$ with side $ AC$ and $ X$ on the same side as $ M$.  what can you say about $ \triangle XBA$?
[/hide]

I hope my hints are correct...

hmm all of these triangles are about a construction of a 60 degree trinalge (there was an old AIME like this-like 2000 or 2001 number 10), but I think I will bring a protractor to just brute force the angle..

Well after the equilateral triangle, you can find similarities using like ASA or something, then you get (usually), one of the triangles that the point splits up is isoscles...and yea then its easy...

or you can always use trig.

A friend recently gave me a problem similar to number 2, except that $ \angle ECA \equal{} 70$ instead of $ 50$, and the angle in question is $ \angle CED$ instead of $ \angle ADE$. I have worked on this problem for quite a while, but I have not been able to make any solid progress other than simple angle-chasing (once you're done with that, the remaining angles are much harder to find).

I would really appreciate it if anybody came up with a solution to this problem; I'm sure it's nice because brute force doesn't seem to be working.

Trig dominates both of these, but there is probably a more elegant solution, especially since the triangles are isosceles (rotations?)

[hide="1"]
Let $ \theta \equal{}\angle MCA$, then trig ceva says
\[ \frac{\sin \theta}{\sin (80^\circ\minus{}\theta)}\cdot \frac{\sin 20^\circ}{\sin 30^\circ}\cdot \frac{\sin 10^\circ}{\sin 40^\circ}\equal{}1\]
Write $ \sin 40^\circ \equal{}2\sin 20^\circ\cos 20^\circ$. The $ \sin 20^\circ$'s cancel, the 2 cancels the $ \sin 30^\circ$, and we're left with $ 1\equal{}\frac{\sin \theta}{\sin (80^\circ \minus{}\theta)}\cdot \frac{\sin 10^\circ}{\cos 20^\circ}$. Recognizing $ \cos 20^\circ$ as $ \sin 70^\circ$, we get $ \theta \equal{}70^\circ$ from which we find $ \angle AMC\equal{}\boxed{70^\circ}$ as well.
[/hide]

[hide="2"]
Forget about $ B$ and just look at quadrilateral $ ACDE$. Use the Law of Sines to get
\[ 1\equal{}\frac{AE}{ED}\cdot \frac{ED}{EC}\cdot \frac{EC}{AE}\equal{}\frac{\sin \theta}{\sin 20^\circ}\cdot \frac{\sin 30^\circ}{\sin (40^\circ \plus{}\theta)}\cdot \frac{\sin 80^\circ}{\sin 50^\circ}\]
First, $ \sin 80^\circ\equal{}2\sin 40^\circ\cos 40^\circ$, which helps cancel the $ \cos 50^\circ$ term. Then $ \sin 40^\circ\equal{}2\sin 20^\circ \cos 20^\circ$ and the $ \sin 20^\circ$'s cancel. We're left with
\[ 1\equal{}\frac{\sin \theta}{\sin (40^\circ\plus{}\theta)}\cdot 4\sin 30^\circ \cos 20^\circ\]
To deal with this, write $ \cos 20^\circ$ as $ \sin 70^\circ$. It looks like $ \theta$ had better be $ 30^\circ$ to cancel the $ \sin 70^\circ$. Because $ 4\sin 30^\circ \equal{}\frac{1}{\sin 30^\circ}$, $ \theta \equal{}\boxed{30^\circ}$ is in fact the solution.
[/hide]

[hide="1"]
Construct equilateral triangle $ XAC$ such that $ X$ is on the same side as $ M$.

Now, we note that $ \angle XCA \equal{} 60$, so $ \angle XCB \equal{} 10$ because $ \angle BCA \equal{} 50$.

Now, since $ XAC$ is equilateral, this means that $ XA \equal{} BA$, so $ XBC \equal{} 30$.

Now, we see that by ASA, $ XBA \equal{} MBC$.  So, $ XC \equal{} MC$.  But, $ XC \equal{} AC$ because we constructed the equilateral triangle $ XAC$. This means that $ MC \equal{} AC$, and we have an isosceles triangle.  Thus, $ \angle AMC \equal{} 70$.[/hide]


trigonometry
geometry
perpendicular bisector
algebra
linear equation
real analysis
real analysis unsolved

\[ z^n  \equal{} \left( {1 \plus{} z} \right)^n
\]

solve
thx

$ |z|^n\equal{}|1\plus{}z|^n\Longleftrightarrow |z|\equal{}|z\plus{}1|$ gives perpendicular bisector.

[quote="kunny"]$ |z|^n \equal{} |1 \plus{} z|^n\Longleftrightarrow |z| \equal{} |z \plus{} 1|$ gives perpendicular bisector.[/quote]

but this is a solution of these equation?

its the perpendicular bisector of the line joining (-1,0) and (0,0) in the Argand plane.

So solution is $ z \equal{} \frac{1}{2}$

[b]Nora.91[/b] Are you sure  :whistling: ?
$ z^n\equal{}(1\plus{}z)^n \Longleftrightarrow |z|^n\equal{}|1\plus{}z|^n$ isn't correct.

The correct solution is the following: it's clear that $ z \neq \minus{}1$
$ (z/1\plus{}z)^n\equal{}1$, make a substitution $ z/(1\plus{}z)\equal{}\xi$ (1), and  $ \xi$ is one of root $ \xi^n\equal{}1$- we know that they can be represented as $ \cos(\frac{2\pi k}{n})\plus{}i\sin(\frac{2\pi k}{n}),k\equal{}0,..,n\minus{}1$.Than you ought to solve the linear equation and note that $ \xi \neq 1$ so you'll have $ n\minus{}1$ roots which correlates well with Fundamental Theorem of Algebra [url]http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html[/url]

[quote="Tarasenya"]The correct solution is the following: it's clear that $ z \neq \minus{} 1$
$ (z/1 \plus{} z)^n \equal{} 1$, make a substitution $ z/(1 \plus{} z) \equal{} \xi$ (1), and  $ \xi$ is one of root $ \xi^n \equal{} 1$- we know that they can be represented as $ \cos(\frac {2\pi k}{n}) \plus{} i\sin(\frac {2\pi k}{n}),k \equal{} 0,..,n \minus{} 1$.Than you ought to solve the linear equation and note that $ \xi \neq 1$ so you'll have $ n \minus{} 1$ roots which correlates well with Fundamental Theorem of Algebra [url]http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html[/url][/quote]

I made up a similar solution. Let see:

\[ \begin{gathered}
  \left( {\frac{z}
{{z \plus{} 1}}} \right)^n  \equal{} 1 \hfill \\
  \frac{z}
{{z \plus{} 1}} \equal{} e^{\frac{{2k\pi i}}
{n}}  \hfill \\
  z \equal{} ze^{\frac{{2k\pi i}}
{n}}  \plus{} e^{\frac{{2k\pi i}}
{n}}  \hfill \\
  z \equal{} \frac{{e^{\frac{{2k\pi i}}
{n}} }}
{{1 \minus{} e^{\frac{{2k\pi i}}
{n}} }} \hfill \\
\end{gathered}
\]

I think so... Thats right?

z=-1/2 only if n is even, right? I'm confusing myself.


floor function
combinatorics unsolved
combinatorics

Let G be a graph with n vertices and more than n^2/4 edges. Prove that G contains a 3-cycle i.e. trianlgle.

Bomb

[quote="Igor"]It's Turan's theorem ...[/quote]

Actually it is Mantel's Theorem - special case of Turan's theorem  :D

And here's the general version:

[url]http://planetmath.org/encyclopedia/TuransTheorem.html[/url]

Here's a proof for this particular case:

If there is a vertex with at most $\left\lfloor\frac n2\right\rfloor$ neighbours, then we can eliminate this vertex and the edges adjacent to it. We're left with $n-1$ vertices and $>\frac{n^2}4-\left\lfloor\frac n2\right\rfloor$ edges, and we can use induction. We can thus assume that all vertices have degree $>\left\lfloor\frac n2\right\rfloor$. Given two adjacent vertices $x,y$, there are at least $\left\lfloor\frac n2\right\rfloor$ neighbours of $x$ different from $y$, and at least $\left\lfloor\frac n2\right\rfloor$ neighbours of $y$ different from $x$. Totally, we have at least $n+1$ vertices, a contradiction.


Prove:
$ \GCD(ac,bd) | \GCD(a,b).\GCD(b,c).\GCD(c,d).\GCD(d,a)$
for all positive integers

The product of four gcd's can be expressed in the form $ (as\plus{}bt)(bu\plus{}cv)(cw\plus{}dx)(dy\plus{}az)$, which expands into
\[ ac(sv(cw\plus{}dx)(dy\plus{}az)\plus{}sbuw(dy\plus{}az)\plus{}bt(bu\plus{}cv)wz)\]
\[ \plus{}bd(asux(dy\plus{}az)\plus{}t(bu\plus{}cv)x(dy\plus{}az)\plus{}t(bu\plus{}cv)cwy)\]
which is divisible by $ (ac,bd)$.

[hide="Solution"]If we prove $ (x,yz)|(x,y)(x,z)$ then $ (ac,bd)$ divides $ (ac,b)(ac,d)$, which in turn divides $ (a,b)(c,b)(a,d)(c,d)$, which is what we want to prove.

Now let $ x\equal{}\prod_{i}p_{i}^{\alpha_{i}}$, $ y\equal{}\prod_{i}p_{i}^{\beta_{i}}$, and $ z\equal{}\prod_{i}p_{i}^{\gamma_{i}}$.  Then $ (x,yz)\equal{}\prod_{i}p_{i}^{\min (\alpha_{i},\beta_{i}\plus{}\gamma_{i})}$, and $ (x,y)(x,z)\equal{}\prod_{i}p_{i}^{\min (\alpha_{i},\beta_{i})\plus{}\min (\alpha_{i},\gamma_{i})}$.

We want to prove $ \min (\alpha_{i},\beta_{i}\plus{}\gamma_{i})\le \min (\alpha_{i},\beta_{i})\plus{}\min (\alpha_{i},\gamma_{i})$, which is easy (but tedious) to do by casework on the relative sizes of $ \alpha_{i},\beta_{i},\gamma_{i}$.[/hide]

See my [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]old blog post[/url] for the technique to prove identities of this sort that matt276eagles uses.  The key concept here is [url=http://en.wikipedia.org/wiki/Additive_p-adic_valuation]p-adic valuation[/url].


number theory
relatively prime
Other problems

Prove that for every $ n \in \mathbb{N}$ there exists a list of $ n$ consecutive natural numbers none of which is a [b]Carmichael[/b] number.

- Remember that a [b]Carmichael[/b] number is a composite positive integer $ n$ that satisfies the congruence $ a^{n \minus{} 1}\equiv 1\mod{n}$ for all integers $ a$ which are relatively prime to $ n$.

All Carmichael numbers are squarefree, and given primes $ p_1, ... p_n$ the system of linear Diophantines

$ x \plus{} 1 \equiv 0 \bmod p_1^2$
$ x \plus{} 2 \equiv 0 \bmod p_2^2$
...
$ x \plus{} n \equiv 0 \bmod p_n^2$

has a nontrivial solution by CRT.

Good solution, dear [b]Qiaochu[/b].  Keep up the good work, [i]man[/i]... :)

[quote="harazi"]Are there two consecutive Carmichael numbers?[/quote]

That was a cool question to ask, esteemed [b]Harazi[/b].

It follows from the proof given by [b]Qiaochu[/b] (which is similar to the one I had in mind when I first submitted this proposal) that you can always find infinitely many pairs of consecutive positive integers none of which is a Carmichael number; however, I don't have a concrete answer for the original inquiry of yours. If you find a proof of the veracity/falsity of that result, let me know, alright?

[quote="coquitao"] That was a cool question to ask, esteemed [b]Harazi[/b].[/quote]

[url=http://www.ac-grenoble.fr/ugine/joomla/documents/maths/DMspeKorselt.pdf]This[/url] criterion provides us with a quick proof of the nonexistence of even Carmichael numbers. Nice going, [b]Harazi[/b].

Anyway, my proposal shows that there arbitrarily long gaps in the sequence of Carmichael numbers... Nevertheless, it would be cool to know now if one can find two Carmichael numbers [b]a[/b] and [b]b[/b] such that [b]|a-b|=2[/b]. ;)


ARML
geometry
3D geometry
email
articles

The title should explain most of what I am asking.  So...  I'm going.  Anyone else?

(you can assume i'll be going, since i'm replying to this) i think we should make a list of all the important stuff to bring: i think pretty much the most important thing is that rubik's cubes outnumber people there.  i'll be bringing 5 (yes, you read that right, i got a new one since nysml! :lol: )

oh yeah, and don't we have to make up an arml song??  i am pretty much useless at making up songs, but i can sing (hahaha, that's funny!!!!! :rotfl:   okay so i can't sing)

oh yeah, and of course, SET decks.  if we have 3 we can have some monster SET game.  i'll bring 1.

and someone should bring hitchhiker's guide to the galaxy movie to watch on our 400-hour long bus trip (assume that was written in some base that makes it accurate)

I will be bringing 2 rubik's cubes, including the one that is entirely black (I peeled all of the stickers off.)

And, of course, my set deck.

Song: First, we need to decide what song ours will be based on.  I was thinking "Hotel California" or "Mad World."  Any other suggestions.

And last, tragically, I don't have HHGG.

and how about food?  i sort of have to bring animal crackers, since it's tradition (so now doing something twice constitutes tradition???)  in fact, we could do more experiments to see how eating them affects your individual score.

[quote="perfectnumber628"](so now doing something twice constitutes tradition???)[/quote]

well, if you can have a "first annual" event, I'm pretty sure the answer to your question is "yes."

[quote="perfectnumber628"]and will you be bringing the mood ring?  it amuses people.  :huh:[/quote]

If I can find it.

and you sort of have to bring aops vol 1 and 2.  really you should carry those with you at all times.

(i am amused :D  because i am "watching" this topic and the upstate ny forum, so i get 2 emails when there is a reply!)

we need 1 more person with a SET deck.  too bad no one else has found this topic.

I will be going on the Upstate New York Team, with 4 Rubik's cubes.

As for the song, we should consider "Numa Numa" or "Knights of the Round Table."

[EDIT]: Sorry, no set decks here.

as for the songs, if we can't come up with anything good, we'll just say "we have discovered a truly marvelous song, which the time alotted to us is too small to contain."  :huh:

I'll be there.  Unfortunately I only have two Rubik's Cubes.  Actually I could probably also bring the 4x4x4 one that I don't know how to solve.

[quote="frankthetank"]Actually I could probably also bring the 4x4x4 one that I don't know how to solve.[/quote]
I would advise that.

yeah, bring all cubes you can find.

so, i have this thing called square-1, which is the same idea as a rubik's cube, sort of, but it can be turned so it's not a cube (nooo, hard to explain- here's the wikipedia article [url]http://en.wikipedia.org/wiki/Square-1[/url])  anyway, it is currently not solved, for, uh, obvious reasons.  actually, sometime last year my sister found a website that said how, then she solved it, then she mixed it up again. nice going there...

i suppose i should bring that to ARML?

[quote="perfectnumber628"]
i suppose i should bring that to ARML?[/quote]
most definitely.

I'll try to get a couple cases of [url=http://en.wikipedia.org/wiki/Bawls]Bawls[/url].  It's crazy expensive, though.

OH MY GOSH DO NOT LET ME FORGET TO BRING MY NYSML PICTURES! (uh...that didn't really warrant a whole sentence of capital letters...)  who wants to be in charge of the "picture-bringing-reminding committee"?  :rotfl:

I will, but someone will have to help me remember.  Maybe a picture-bringing-reminding-reminding committee?

we could get a whole infinite series of reminding committees...unfortunately, we only have a finite number of people

or i could just write it on my hand.  that's effective.  :lol:


Hey, can anyone help me with this question...

given |a-3|+(3b-1)^2=0......find the value of a+ab+ab^2+ab^3+ab^4

Thanks

|a-3| = -(3b - 1)^2 note that |a-3| is always positive or zero. and -(3b-1)^2 is always negative or zero. Thus we have that -(3b-1)^2 must be zero and we get b = 1/3. Then |a-3| = 0 gives that a = 3.


geometry
analytic geometry
rectangle
Pythagorean Theorem

In the figure below, the shaded rectangular region is of dimensions $3$ and $6$.

Find the radius of circle.

Let the center of the circle have coordinates $(r,-r)$.

We know the circle passes through the point (6,-3), so solve for r.

[hide]Draw r from center of circle to the corner of the rectangle that touches circle. Draw r to side of the square. Extend one side of the rectangle until it reaches the second line we drew.

We have a right triangle with hypotenuse r, and legs (r-3), and (r-6). Use pythagorean theorem and solve.
$r^2=(r-3)^2+(r-6)^2$
$r^2=r^2-6r+9+r^2-12r+36$
$0=r^2-18r+45$ which has the roots 3 and 15. Since the radius is clearly bigger than 3, the answer is 15.[/hide]


geometry
geometric transformation
reflection
trigonometry
vector
rotation
geometry solved

Let $\Delta ABC$ be a right isosceles triangle $\angle ACB=90$. Points $M$ and $N$ are on $AB$ so that $AM^2+BN^2=MN^2$. Find $\angle MCN$.

The answer is $\angle MCN=45^\circ$. It suffices to prove the converse, namely that if $\angle MCN=45^\circ$, then $AM^2+BN^2=MN^2$, because then, given an arbitrary position of $M$ between $A$ and the midpoint of $AB$, if we move $N$ from the position corresponding to $\angle MCN=45^\circ$, then $MN$ increases and $BN$ decreases, or vice-versa, and the equality cannot hold anymore.

Since $\angle MCN=45^\circ$, the reflection of $A$ through $CM$ and the reflection of $B$ through $CN$ coincide (denote it by $T$). A quick angle chase shows that $\angle MTN=90^\circ$, and since $MT=AM,NT=BN,\ MNT$ is a right-angled triangle (with $\angle T=90^\circ$), with legs equal to $AM,BN$, and hypothenuse equal to $MN$, hence the result: $MN^2=AM^2+BN^2$.

[color=red][b]The first (analytical) method.[/b] I suppose $CA=CB=1\ ,$ $CA\perp CB$ and the origin $C\ .$ Thus, $A(1,0)\ ,\ B(0,1)\ ,\ M(m,1-m)\ ,\ N(n,1-n)\ .$
$m(\widehat {MCN})=x\Longrightarrow \tan x=\left|\frac{\frac{1-m}{m}-\frac{1-n}{n}}{1+\frac{1-m}{m}\cdot \frac{1-n}{n}}\right|=\left| \frac{m-n}{2mn+1-(m+n)}\right|\ .$
$AM^2+BN^2=MN^2\Longleftrightarrow 2mn+1=2m\Longleftrightarrow \tan x=1\Longleftrightarrow x=45^{\circ}\ .$[/color]

[color=blue][b]The second (metrical) method.[/b] I suppose $CA=CB=2$ and denote the middlepoint $D$ of the hypotenuse $[AB]\ ,$ i.e. $DA=DB=CD=\sqrt 2\ ,$  $DM=x\ ,$ $DN=y$ and $m(\widehat {MCD})=u\ ,$ $m(\widehat {NCD})=v\ .$ We obtain $\tan u=\frac{x}{\sqrt 2}\ ,$ $\tan v=\frac{y}{\sqrt 2}\ ,$ $\tan \widehat {MCN}=$ $\tan (u+v)=$ $\frac{\tan u+\tan v}{1-\tan u\tan v}=$ $\frac{\frac{x}{\sqrt 2}+\frac{y}{\sqrt 2}}{1-\frac{xy}{2}}=$ $\frac{\sqrt 2 (x+y)}{2-xy}\ ,$ i.e. $\boxed {\tan \widehat {MCN}=\frac{\sqrt 2(x+y)}{2-xy}}\ .$ Therefore,
$BM^2+CN^2=MN^2$ $\Longleftrightarrow$ $\left(\sqrt 2-x\right)^2+\left(\sqrt 2-y\right)^2=(x+y)^2$ $\Longleftrightarrow$ ${\sqrt 2(x+y)=2-xy}$ $\Longleftrightarrow$ $\tan \widehat {MCN}=1$ $\Longleftrightarrow$ $m(\widehat {MCN})=45^{\circ}\ .$[/color]

[color=darkred][b]The third (syntetical) method.[/b] See the nice previous Grobber's proof.[/color]

[hide="Some remarks."][color=blue]Denote : the circles $m=C(M,MA)\ ,$ $n=C(N,NB)\ ;$ $T\in m\cap n\ ,$ the line $AB$ doesn't separate the points $T$ and $C\ ;$ the intersections $M'\in (AC)\cap m\ ,$ $N'\in BC\cap n\ ,$ $A'\in (AB)\cap m\ ,$ $B'\in (AB)\cap n\ ,$ $S\in A'M'\cap B'N'\ .$ If $MA^2+NB^2=MN^2\ ,$ then :
$1.\blacktriangleright TM\perp TN\ ,$ i.e. the circles $m\ ,$ $n$ are orthogonal .
$2.\blacktriangleright$ The points $B\ ,$ $B'$ are conjugated w.r.t. the points $A\ ,$ $A'\ ,$ i.e. $\frac{B'A}{B'A'}=\frac{BA}{BA'}\ .$
$3.\blacktriangleright T\in AN'\cap BM'\ ;\ SA'=SB'=ST\ ,$ $SA'\perp SB'\ .$
$4.\blacktriangleright m(\widehat {ATB})=135^{\circ}\ ,$ $m(\widehat {A'TB'})=45^{\circ}\ .$[/color][/hide]

[color=green]The fourth (vector) method.[/color]

$AM^2+BN^2=MN^2 \Longleftrightarrow \left(\frac{AM}{MN}\right)^2+\left(\frac{BN}{MN}\right)^2=1,$ thus we can set  $\frac{AM}{MN}=\frac{1-t^2}{1+t^2},\ \frac{BN}{MN}=\frac{2t}{1+t^2}\ (0<t<1).$

From $AM: MN=1-t^2: (1+2t+t^2)=1-t: 1+t,$ we have $\overrightarrow {CM}=\frac{(1+t)\overrightarrow{CA}+(1-t)\overrightarrow{CB}}{(1-t)+(1+t)}=\frac{(1+t)\overrightarrow{CA}+(1-t)\overrightarrow{CB}}{2}.$

Then $AN: NB=(1-t^2)+(1+t^2): 2t=1: t,$ we have $\overrightarrow {CN}=\frac{t\overrightarrow{CA}+\overrightarrow{CB}}{1+t}.$
Now let $|\overrightarrow{CA}|=|\overrightarrow{CB}|=k,$ by $\angle{ACB}=90^\circ,$ we have $\overrightarrow{CA}\cdot \overrightarrow{CB}=0.$ Use these we get

$|\overrightarrow{CM}|^2=\left|\frac{(1+t)\overrightarrow{CA}+(1-t)\overrightarrow{CB}}{2}\right|^2$

$=\frac{1}{4}\{(1+t)^2|\overrightarrow{CA}|^2+(1-t)^2|\overrightarrow{CB}|^2\}=\frac{k^2}{2}(1+t^2).$

$\therefore |\overrightarrow{CM}|=\frac{k}{\sqrt{2}}\sqrt{1+t^2}.$ Similarly we can obtain $|\overrightarrow{CN}|=\frac{k}{1+t}\sqrt{1+t^2},\ \overrightarrow{CM}\cdot \overrightarrow{CN}=\frac{k^2}{2}\cdot \frac{1+t^2}{1+t}.$ Consequently consider the inner product of two vectors $\overrightarrow{CM}$ and $\overrightarrow{CN},$ we have $\cos \angle{MCN}=\frac{\overrightarrow{CM}\cdot \overrightarrow{CN}}{|\overrightarrow{CM}||\overrightarrow{CN}|}=\frac{1}{\sqrt{2}},$ yielding $\angle{MCN}=45^\circ.$

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=371 .

  darij from munich

Indeed it is interesting!  Geometry seems to be dilsliked me, I am apt to use vector!
 :lol:

[quote="Altheman"]i don't seem to see this solution here, but i may be wrong...

(all degrees) rotate $90$  w.r.t. $C$, such that $A$ maps to $B$, $B \rightarrow B'$, $M\rightarrow M'$, etc

$\angle NBM'=90$, and $BM'=AM$, so using the pythagorean, $MN=NM'$
$MC=M'C$, and $\angle MCM'=90$ by rotation, it follows that $\angle MCN=\angle NCM'=45$[/quote]

wow , thats a nice and easy solution using rotation .   :lol:


how would u do this problem?

what is the probability of getting two heads and three tails on five flips of a coin?

[hide]Well, obviously, if you have two heads, you'll have three tails (unless the coin lands on its edge or something :roll:). Thus, the problem reduces to choosing two flips to be heads, out of five flips.
5 "choose" 2 (I don't know the LaTeX) is 10.[/hide]

Edit: D'oh, I forgot to add this part:
[hide]Therefore the probability is 10/32.[/hide]

[quote="mathnerd314"][hide]Well, obviously, if you have two heads, you'll have three tails (unless the coin lands on its edge or something :roll:). Thus, the problem reduces to choosing two flips to be heads, out of five flips.
5 "choose" 2 (I don't know the LaTeX) is 10.[/hide][/quote]

[hide] There are a total of $2^{5}=32$ ways. So the probability is $\boxed{\frac{5}{16}}$. :wink:

Another way: $\frac{5!}{3!2!}(\frac{1}{2})^{5}=\frac{5}{16}$ [/hide]

five choose two over two to the fifth

make sense of that tongue twister (it's basically saying you have to choose two of the five flips to be heads out of two to the fifth possibilities)

[hide]The answer is the number of ways you can arrange HHTTT, which is $\binom{5}{2}=10$.

This goes over $2^{5}=32$

So the answer is $\frac{5}{16}$.[/hide]

[quote="DanK"][quote="mathnerd314"][hide]Well, obviously, if you have two heads, you'll have three tails (unless the coin lands on its edge or something :roll:). Thus, the problem reduces to choosing two flips to be heads, out of five flips.
5 "choose" 2 (I don't know the LaTeX) is 10.[/hide][/quote]

There are 2 ways to do the LaTeX for that:

{x\choose y} or \binom{x}{y} both give $\binom{x}{y}$

And the other people are right about the probability, it's [hide]$\boxed{\frac{5}{16}}$[/hide][/quote]

so like Pikachu, $\binom{i}{u}$?

[hide]Every possible outcome has prob. (1/2)^5=1/32, and there are 5C2=10 ways to have 2 heads. 5/16.[/hide]

when you reply, above the writing box there's a button HIDE, between LIST and IMG.

And if you want to make it say something special, you go like hide=text in the first brackets, and you get [hide="text"]See?[/hide]

An elementary way would be to use a tree diagram (my math teacher likes to do that)
For the general case, where there are $n$ heads and $m-n$ tails, the formula is:
$\frac{\binom{m}{n}}{2^{m}}= \frac{\binom{m}{m-n}}{2^{m}}$.

[quote="vishalarul"]An elementary way would be to use a tree diagram (my math teacher likes to do that)
For the general case, where there are $n$ heads and $m-n$ tails, the formula is:
$\frac{\binom{m}{n}}{2^{m}}= \frac{\binom{m}{5-n}}{2^{m}}$.[/quote]
$\frac{\binom{m}{n}}{2^{m}}= \frac{\binom{m}{m-n}}{2^{m}}$ :wink:


You are participating in an alpha (maybe even pre-alpha) test of a new online game from The Art of Problem Solving.  Since this is a very early test, you should expect to see problems with the game.  It will also be periodically down for maintenance.  Please let us know of any issues you find after checking the [b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177811]Open Issues[/url][/b] topic.

The game requires you to have Flash installed.  You can download Flash [b][url=http://www.adobe.com/shockwave/download/download.cgi?P1_Prod_Version=ShockwaveFlash]here[/url][/b].  You can verify that you have Flash installed and working [b][url=http://www.macromedia.com/software/flash/about/]here[/url][/b].

You are in a pre-selected group of testers.  Please do not let others know about the game.

You can access the game [b][url=http://www.artofproblemsolving.com/Edutainment/g1]here[/url][/b].

You can view some documentation [b][url=http://www.artofproblemsolving.com/Edutainment/g1/Pages/Documentation]here[/url][/b].

As alpha testers, you currently have access to the game as though you were paid subscribers.

We are in need a good name for the game.  If you think of something good, let us know!

Have fun!

how does this "game" work sorry i might have missed something

EDIT: it wont let me in again

You can read the documentation [url=http://www.artofproblemsolving.com/Edutainment/g1/Pages/Documentation]here[/url].

Basically, you compete against others to answer questions correctly, and quickly.

How do you face yourself anyway? Nice job on making this game, though. Great idea, this should increase my MC scores.

I think that the game would be a little better if you could also play a computer with pre-programmed times (or something along those lines), because other users aren't always there.

I think that would be boring, i dont know why....

No. Technically you are facing the computer with the correct answers.

[quote="mihail911"]i played one countdown by myself. i didn't press a key for like 5 min. now when i try to go back in it says "sorry, you have been logged out." i try to refresh my screen by pressing F5. it doesn't work. help?[/quote]

Please post issues with the game in the Open Issues topic.

Remember, that this is an Alpha Test and weird behavior is expected.

Can you login now?

it works at home for me but not at school. why?

Maybe they don't have (or uninstalled) their Flash Player.
That is necessary to play it.

i checked that. we do have flash player. it says "could not establish server connection"

If the internet is working fine, I don't know.
Is it possible they edited the browser to say "could not establish server connection" when you enter a page with Flash?

I'd go ahead and try again since the server has to go down periodically for updates.

However, many schools block most ports on the Internet, which is probably what you are experiencing.  This is largely to prevent intruders into their networks, prevent downloading of music and movies, etc., but it would also hit a game like this.

Yeah, I was trying to load the game once and for about 30 minutes it kept on saying "Connecting...".
Now it works.
 :huh:
Probably the server?

Woops, I accidentally just reported an error on a problem. what should i do?

Nothing; Just post that you accidentally did (That way they don't spend time on nothing.)
[img]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/img]
Are you developing this to make it a paid feature? (Must pay to have access)
Also, doesn't PHP let you make only certain people access a page?  That way guests don't play it.

[quote="minicon"]Are you developing this to make it a paid feature? (Must pay to have access)[/quote]
We are not sure at the moment.

[quote="minicon"]Also, doesn't PHP let you make only certain people access a page?  That way guests don't play it.[/quote]
Yup, if we wanted to set it up that way we could.

[quote]I think it should be free. more people would play, too, and I don't know how many people would actually go to the trouble of paying unless they at least got to try it.[/quote]
[quote]As alpha testers, you currently have access to the game as though you were paid subscribers. [/quote]
what do you mean by "AS THOUGH you were paid subscribers".

sailingdog-i agree with you except for the fact that the highschoolers might come and own us just for fun. =(. lol

[quote="levans"][quote="abacadaea"]
[quote]As alpha testers, you currently have access to the game as though you were paid subscribers. [/quote]
what do you mean by "AS THOUGH you were paid subscribers".[/quote]
You currently have the same access that AoPS is planning to provide to paid subscribers.  So, essentially, right now, you are a paid subscriber without having paid anything.[/quote]

So we will have to pay in the future?

I kind of suspected that, but will a lot of people play this game, then? I mean, it's not too hard to get some friends, some buzzers, a timer, and some old MC countdown problems.

[quote="abacadaea"]sailingdog-i agree with you except for the fact that the highschoolers might come and own us just for fun. =(. lol[/quote]
Some of you guys are faster than me and I'm in college! I never did MATHCOUNTS though. I didn't start doing math contests until much later.

Also, it may be that "paid subscribers" include anyone who's paying to take a class.  Then there would be lots of players.

If there were to be a list of things to add to the game eventually, I would add:
a feature that was timed or against a simulated user so you could practice and gain points without playing other users
a tournament feature
something to keep the really good people from joining games the less experienced people set up so they don't uncontrollably loose points, or just a rating range for people who join that you set when you set up the game.
...
This game is possibly the best idea since that of sliced bread, so kudos to whoever thought of making it and to who actually did [b]: [/b])

Any decent rating system will not give much weight to a game between people who are far apart in rating. So if someone who's an 1800 beats someone who's an 1100, neither of their ratings will change much at all (because it's no surprised that the person who's 1800 won). On the other hand, if the 1100 wins then there should be a big swing for both of them.

edit: on the other hand, that feature may still be good just to prevent people who are much better from playing younger kids just for the sake of winning. That would be mean.

[quote="abacadaea"]if it will indeed be "pay to play", how much would it be to subscribe?[/quote]
Sorry, no information about the subscription option is available at this time.

Uh-oh. That would be a pity! :( I don't mind AoPS making money (actually, it is, probably the last organization in the world I would mind making money :)) but I'm afraid that the pay-to-play mode will reduce the interest to the game dramatically and restrict the pool of possible players to just a few countries, so this way not as much money will be made as harm.

As to the game itself, it is a nice one. I tried it a couple of times and I should say that I won only because I set the scoring according to the number of correct responses (if I had set it according to speed, I would come last: it seemed like the other players posted their answers before I finished reading the problems). The problems are [b]very[/b] uneven: there are some trivial ones when you see the answer before you finish reading the conditions and there are ones where you have either to spend 15 minutes figuring out the solution rigorously or just make an "educated guess" at the answer. On the other hand, there is nothing fundamentally wrong with that, after all, it is a game. There were also some problems where what to do was completely clear, but you had to multiply something like $ 153\times 12$ and you could easily make a stupid mistake doing mental calculations under pressure. I liked them least because they are the easiest to cheat on: just have your calculator at hand and learn to punch the buttons quickly.

Anyway, this would make a nice addition to AoPS, if it were free. As to whether it'll still be popular with the paid option, I'm not that sure. What I'm sure about is that AoPS won't make as much money on that as gameboy manufacturers anyway and the higher the price, the more kids or even their parents will prefer gameboy...

I like the concept.  ICC meets countdown.  Maybe there could be a feature such that you can do scratchwork in a space provided, saving the time of using scratch paper.  And maybe if you type "psi\", once you type it it turns into $ \psi$, so you can make symbols like in $ \text{\LaTeX}$, except the \ is at the end.

Like you type "boxed[4]\ is theta\ cdot\ 64", it appears as "$ \boxed{4}\ \text{is}\ \theta\cdot 64$". Note that I used brackets, so the shift key isn't required.

Just a suggestion. :lol:

I can play FTW on my computer at home  , but my school has older computers and I can not play FTW in my free time. :(  What are the system requirements for FTW  :?:

You need to have Flash installed and connected to the Internet. Your school's routers must allow access to ports 9898 and 9875.

[quote="isabella2296"]Will we ever have to pay to play FTW? Is it undecided or certain?  :maybe:[/quote]
it's undecided

I don't think we should have to pay. What's the reasoning behind it? I love playing the FTW, but I might not play if I have to pay for it...it just doesn't feel right! Isn't the point of AoPS to learn?

That's very true: we shouldn't pay to go on FTW!


parameterization conics hyperbola trigonometry

Eliminate the parameter and obtain the standard form of the rectangular equation

$Hyperbola:$

$x = h+a$ $\sec$ $\theta$
$y = k+b$ $\tan$ $\theta$

Can someone explain how to solve this problem?

I expect that the answer whould be  $\frac{(x-h)^{2}}{a^{2}}$ $-$ $\frac{(y-k)^{2}}{b^{2}}= 1$

I would greatly appreciate all your help.

[quote="tjhance"]The parameter is $\theta$ which we need to get rid of. A reasonable first step is to isolate it on one side of the equation:

$\frac{x-h}{a}=\sec\theta$
$\frac{y-k}{b}=\tan\theta$

A notable connection about the secant and tangent is the identity:

$\tan^{2}\theta+1=\sec^{2}\theta$ for all $\theta$

You should see how to proceed from here.[/quote]

Thank you for your help.
I finally figure out I need to use the Pythagorean Identity.