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ARML

Or just know someone who has? If so, how hard did they prepare for both AMC and AIME?  Is it possible for a rising Senior (me) to qualify for USAMO next year with intense studying?  I love math so I dont consider three hours a day excessive by any means.  I have only had the chance to complete Pre-calculus.  I have purchased both art of problem solving books, and I plan on finishing both prior to fall.  I also will use the practice AMC tests and another carefully chosen book.  Thanks for any response and don't be scared to tell me the truth even if it does hurt.  I am just sad I was introduced to this world of math this late.

I haven't qualified for the USAMO myself, but I understand your pain of being introduced to this world too late. I am a rising junior, and teh next AMC/AIME in 2004 will be the first time that I take the tests with some actual preparation. However, since I am going to be a junior, I don't get the benefits of freshmen and sophomores, who only have to meet the lowest AIME scores of the USAMO qualifiers, and thus sort of get an automatic 150 on the AMC.

Fierytycoon

I qualified for the USAMO three times. I don't think there's any reason why you can't make it there next year if you work hard. In my freshman year, I took the AMC12 for some unknown reason and got 86 and thus didn't qualify for AIME. But I worked really hard, and when I was a sophomore, I qualified for the USAMO.

One thing that might really help you if you think you might be close is to go through the articles page on this site. It's in resources, then click on the articles tab. I read them all very carefully before ARML, which was my last contest, since I really wanted to do well, and I think it helped me. I told myself that I would be very disciplined and use exactly one piece of paper (front and back) for each problem so that I could easily check to see if I did anything wrong. I found that it helped a lot, since I made some silly errors that I might not have found otherwise. This is especially useful on the AIME, in which I, at least, make quite a few careless errors each year.

As a senior, you will need a very high AMC12 score if you plan to make USAMO. Without at least 130, you will have a lot of trouble. I guess it won't hurt to go through past AHSME contests, but the styles of questions are significantly different, and the modern AMC12s seem to me to be quite a bit easier in general. Keep in mind when you work on the problems that each one can be solved. That's always very useful. Probably don't spend more than 3 minutes on any problem, even if you have more time because some of the easy problems took you under three minutes. Just move on after three minutes unless the end is in clear sight. Then come back to it.

If you tend to get really nervous during competitions, you can try to eat or drink something to calm you down. For example, during the last year, I brought in mineral water during the competition, and so instead of just overheating under the pressure of competition, I would periodically take a break to sip some water. I found this extremely useful.

On AIME, if you find that the time is a problem, don't look at the last two questions. They're usually quite hard, and you don't need a 15 or even 14 to qualify for the USAMO. I wasted far too much time on the last two problems this year. I got nine problems in the first hour, all of which were correct, and then in the remaining two hours, I solved four of the remaining six, but only one was right because I tried too hard on problems that I didn't know how to do.

Just go through as many problems as you possibly can and watch your scores increase! Have fun.

ComplexZeta,

How did you prepare for the competitions? What texts did you read to prepare? Do you think Paul Zeitz's book "The Art and Craft of Porblemsolving" is useful for AMC/AIME-level problems?

Fierytycoon

I just solved all the problems I could find. I read [u]The Art and Craft of Problem Solving[/u], [u]The Art of Problem Solving Volume 2[/u], [u]Problem-Solving Through Problems[/u], and probably some others. The first two are useful for AMC/AIME problems. The third has more proof-type problems and may be useful in preparing for the USAMO, but other books, such as [u]Problem-Solving Strategies[/u], are probably more useful for USAMO problems.

I qualified last year for the 2002 USAMO. Luckily that got me into MOSP 2002, since they allowed anyone in that qualified for the USAMO..

I got to take the USAMO this year.  I did not study for the AMC - 10, except with what I was doing with Mathcounts at the time.  For the AIME and USAMO I ordered copies of old tests, and found old USAMO tests online.  With the help of my parents and math teacher, I would go over those tests, not writing out proofs, but working out exactly how I would go about proving the question if I needed to.

I qualified for USAMO as an 8th grader, last year.  I'm sure that if you put your mind to it, anything is possible, so my answer is yes, you can certainly make it if you study hard enough.

The question is how much "enough" is.  Well, the key to qualifying for USAMO is to do really well at AIME.  I would project the qualifying USAMO index* will be around 220 or so, so you should shoot for at least a 130 on AMC12 and then a 9 on AIME.  If you do a little better on one or the other, it leaves more room for mistakes on the other.

Of course, it would also depend on what state you are from; if you're from an easy state (like Nevada, where I'm from), you can also get in by having the top index in the state.

As for what to study... well, I would definitely recommend the Art of Problem Solving - Volume I (most AMC topics are covered in this one) for the AMC and mayyyybe Volume II for AIME, but I can't be sure since I am still working on them myself.  But also get a bunch of old AMC tests by ordering them (if you order the AHSME problem books, you get about 6 tests AND the solutions) or look around on the net (i believe J. Douglas Faires' website has some AMC tests posted...which seems odd because they're selling them...oh well).  This is what I did, aside from training for mathcounts nationals during my 7th and 8th grade years.

Good luck!

mathfanatic :shock:

[quote]I would project the qualifying USAMO index* will be around 220 or so[/quote]

Why do you say this? Since the cutoff score was about 210 last year, and 226 this year (an increase), I would think the cutoff score would be at least 226 next year, probably even higher.

I remember being introduced to the AHSME (that was the name before AMC-12) and the AIME years ago.  I most became proficient by working through as many of the old tests as I could get my hands on.  If I can across a new concept, I read about it.  I also read this old problem solving journal called the Arbelos.  It's not a "total math trainer" like the AoPS are closer to being, but there were a lot of interesting and useful articles.

I ordered one of the issues of the Arbelos before the USAMO in my sophomore year, and I just found it way too difficult to be of much use to me. I'd do a lot better with it now, but I don't get to take the USAMO again.

Some of it is difficult, but I think students who are at least borderline USAMO candidates would get something out of many of the articles.  But you're right, it's definitely on the advanced side of problem solving for high school students.

[quote="Fierytycoon"] Why do you say this? Since the cutoff score was about 210 last year, and 226 this year (an increase), I would think the cutoff score would be at least 226 next year, probably even higher.[/quote]

Many have observed the rising trend in USAMO indices.  However, if you just compare between two years, it is hard to accurately predict next year's qualifying index.

I was just giving a general, ballpark figure but I do believe that it will be around 220 this year, mainly because the 2003 AMCs were easier than the 2002 AMCs, so the AMC question-writing committee will probably crack down and make 2004's AMCs harder, bringing the scores (and qualifying scores) down.  This happens a lot; sometimes you'll get a hard year, and so the next year the questions will be easier.

Besides, someone who gets a 130 on the AMC and a 9 on the AIME should just barely squeak by on to USAMO (they should have done better on AMC), while a score of 220 ought to weed out people who got 100 on the AMC, because they aren't likely to get more than a 12 at all.  I know that the numbers are set based on statistics (it's 250, or 2.5% of all AIME participants) - I'm just saying that a score like that would make sense.

But that's just my guess.  If you think 230 will be the cutoff next year, then just say so instead of arguing about it. :evil:

Just kidding about the evil face.  Smile!  :)  And good luck to everyone on next year's AMC contests.

mathfanatic

Hmm, that argument sounds fair enough. However, I do believe that back many years ago, the USAMO indices had been quite low compared to those nowadays--around 180-190 or so. Does anyone know the cutoffs for 2000 and 2001?

By the way, I personally really think that the AMC tests (particularly the first round) should be a lot harder so that the range of scores of even the top students would be more spread out. Just a few years ago, there were only a couple perfect scorers on the AMC12, but look at the tens of students who got perfect scores this year, not to mention the number of people who received 144's and 146.5's. I mean, the top 150 scorers received at least a 144. The best scores (of which only a few should be perfect) should not determine the students who make the least careless errors, but the ones who really know their stuff.

Fierytycoon

Well, I looked on [url]http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.html[/url] and dug up the good stuff:

For 1999: 10th graders 185, 11th graders 193, 12th graders 201
For 2000: 9th 164 (w/ 7 AIME) 10th 174 (w/ 8 AIME), 11th 204, 12th 212

So, by going through your mistakes and errors, you'll probably learn how to approach the problems in a more skillful way, right?

Sounds right to me.  And I have a lot farther to go than you do.  I had a 115 and a 1 this year (although it should have been a 4...), and I'm a sophomore...

So I really need all the help you guys can give me.  What annoys me is that I saw this site once like a year ago, and didn't join and get into it until now.  Now I have to work like crazy...

Ya, I've worked through both Aops books, lots of Paul Zeitz the Art and Craft of Problem Solving, and alot of Engel's prep book, but I still think the best thing for the AMC & AIME is taking a bunch of them before hand. I have like 30 AMC's at my house (not including contest books, lol) and 10 AIME's, and whenever I take one, after I finish I take some notes on ones I got wrong or that were interesting, so that I can actually learn new strategies and use them later, not just say "o cool" and forget about it.

I think I may have made USAMO this year as an undeserving eighth grader. I did no preparation whatsoever, and got 138 on amc10a and 144 on amc10b. I hope to ride my luck on AIME, and i know i got at least 7 right. So, if all goes well i make USAMO.

I heard that freshmen only need a positive score to qualify for MOP. Is that true, and if it is does that extend to 8th graders?

For the most part, freshmen only need positive scores, but then again, last year the cutoff was 7, because the questions were just that easy. Number four was basically an AIME problem for a general case.

I'm not completely sure, but I don't think the Red MOP applies to 8th graders? Is this correct?  :huh:

Ask Krustyteklown.  ;)   ( I believe the answer is no, it doesn't apply to 8th graders... poor guy..  :( )

I can understand the argument of maturity though. At the level we're talking about, middle schoolers may have strong math abilities, but their social and emotional levels are still that of a middle schooler, so it may be harder for them to adapt to the rigorous style of the camp.

yeah, 8th graders who qualify and score well on USAMO definitely don't get invited unless a change has happened.....

in 2004, i was in 8th grade and got a 7 on the USAMO, but didn't hear from MOP

i know of a 8th grader last year who got a 19 (quite amazing), but didn't get to go to MOP either, cuz the normal cutoff was 21(?)

[quote="darktreb"]It doesn't extend to eighth graders and in my personal opinion that makes very little sense to me.

From what I've heard, they're worried about "maturity" but it really isn't the biggest difference... there are plenty of mature eighth graders and immature ninth, tenth, eleventh, twelveth graders, college students....[/quote]

You have to draw the line somewhere. I suppose the AMC could've used the grant to take the 30 highest scoring freshmen and under, but maturity is a realistic issue and I'm glad they did it this way. Besides, younger qualifiers will be able to take advantage of the grant when they're in ninth grade (assuming that it continues) so the selection isn't unfair. The problem exists that younger kids are more likely to cause headaches than older ones. (Although I do recall one particular freshman moving trash bins from the cheerleading camp dorms onto the elevators...*COUGH*)

Anyway, officially Red MOP is for "approximately 30" freshman. The "only positive scores" whitehorse is referring to was in 2004, when there were 25 freshman qualifiers and sophomores had to be selected for the remaining spots. Due to this year's expanded USAMO, I suspect that far more than 30 freshman will qualify and from there, it'll be whoever gets the highest USAMO scores.

[quote="MithsApprentice"]Anyway, officially Red MOP is for "approximately 30" freshman. The "only positive scores" whitehorse is referring to was in 2004, when there were 25 freshman qualifiers and sophomores had to be selected for the remaining spots. Due to this year's expanded USAMO, I suspect that far more than 30 freshman will qualify and from there, it'll be whoever gets the highest USAMO scores.[/quote]
Last year, there were 38 usamo qualifiers from the freshman year. Among them, 25 made to MOP, one blue, 24 red. AMC could potentially fill 6 more spots from the remaining 13 freshmen qualifiers. But when they made the decision of the cut off score (7), no additional freshmen were selected.

What are average cutoff scores to make MOP?

[quote="probability1.01"]For red, no one really knows, because it has only been two years. I spent some time looking at USAMO stats, and for blue the cutoff seems to be around the 85th percentile, which usually means 15-20 depending on the specific year's difficulty.[/quote]

Heh, that's what I did, too. I think 20 (or above) would have to be a pretty easy test, though (like last year).

[quote="qvc122"]Last year, there were 38 usamo qualifiers from the freshman year. Among them, 25 made to MOP, [b]one blue[/b], 24 red. AMC could potentially fill 6 more spots from the remaining 13 freshmen qualifiers. But when they made the decision of the cut off score (7), no additional freshmen were selected.[/quote] Yeah, that was me. :D There was one 8th grader, Eric Larson of Oregon, who also got a 21, but otherwise, no 8th graders were there, even if some of them did make it above 7. And believe me, one Eric was enough... ;)


Factor by grouping

2x+2y+ax+ay

This is what I did

2x+2y+ax+ay

2(x+y) a(x+y)

2a(x+y)

I know this is wrong, but I dont know the awnser...

What do I have to do?

2x+2y+ax+ay

[color=red]2(x+y) a(x+y) [/color]
2a(x+y)

that is what u did

i highlighted one part above
u should not have multiplied them
u should have done [hide]2(x+y) [color=red]+[/color]a(x+y)[/hide]
which turns out to be [hide](2+a)(x+y)[/hide]

Ah, I see what I did, thanks.  I hate making those kind of mistakes.


Joshua flips three coins. What is the probability that they are neither all heads nor all tails? Express your answer as a common fraction.

Since the probability of him flipping all heads or all tails is 2/3^3, or 2/27, isn't the final answer (27-2)/27, or 25/27? I don't get why it's 3/4. Can someone explain this?? :huh:

The probability that he flips all heads is $ \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\equal{}\frac{1}{8}$.  This is because each time he flips, he has a $ \frac{1}{2}$ chance of landing on heads.  We multiply them together because the events are mutually exclusive (independent).  Because this is the same for tails, we do $ \frac{1}{8}\plus{}\frac{1}{8}\equal{}\frac{1}{4}$ to get the probability he [b]does [/b]get all heads or all tails.  This means that the probability that he does not get all heads or all tails is $ 1\minus{}\frac{1}{4}\equal{}\boxed{\frac{3}{4}}$.  I don't know what you did, so I cannot explain what you did wrong, sorry.  I hope this explanation helps though!

The only ways for the condition to not apply is if they are all tails or all heads. The probabilities of these are both 1/8. 1-2(1/8)=3/4


Firstly, I want to send my apology to Mathlinkers, especially JBL for not being calm and being so rude.
Secondly,thank to jvenezuela716.   ( :blush: )
And now I have an interesting problem for you:
Let a,b,c,d be positive integers; a>b>c>d and ac+bd=(b+d+a-c)(b+d-a+c)
S=ab+cd
Prove that S is not a prime.
(I'm having a solution but it's too long  :D )

This was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119217#119217]IMO 2001 #6[/url], so probably isn't suitable to this forum.


I don't really understand. If the x and y coordinates of all the points is 1, then the points will be $ (1,1)$, $ (1,1)$, and $ (1,1)$. Then the area is 0.

Sorry about that. I rewrote what the original problem said because it specified certain units in the coordinate system (which isn't really necessary since we specify units for coordinate systems nowadays). The problem is, I rewrote it badly.

It should read:

"A triangle is formed by joining three points whose coordinates are integers. The area of the triangle: "

[hide="Quick Solution"]
Applying the [url=http://en.wikipedia.org/wiki/Shoelace_formula]Shoelace formula[/url], it is easy to see that the answer is $ \textbf{(D)}$. (You end up getting the sum of the products of integers divided by two which is rational; it is not guaranteed that it will evaluate to an integer.)
[/hide]

[quote="sunehra"]A triangle is formed by joining three points whose coordinates are integers. If the $ x$-coordinate and the $ y$-coordinate each have a value of $ 1$, then the area of the triangle, in square units:

$ \textbf{(A)}\ \text{must be an integer}\qquad \textbf{(B)}\ \text{may be irrational}\qquad \textbf{(C)}\ \text{must be irrational}\qquad \textbf{(D)}\ \text{must be rational}\qquad \\
\textbf{(E)}\ \text{will be an integer only if the triangle is equilateral.}$[/quote]

is  it  multi optional  or single answer?

This is AHSME so it is single answer. You can assume it is a single answer if it doesn't specify otherwise.


On a desert island, five men and a monkey gather cocnuts all day, then sleep. The first man awakens and decides to take his share. He divides the coconuts into five equal shares, with on coconut left over. He gives the extra one to the monkey, hides his share, and goes to sleep. Later, the second man awakens and takes his fifth from the remaining pile; he too, finds one extra and gives it to the monkey. Each of the remaining three men does likewise in turn. Find the minimum number of coconuts originally present.

I think this is it, unless I misunderstand the question.

Let $ c$ represent the number of cocunuts.

$ (c/5) \minus{} 5 \equal{} 0$
$ (c/5) \equal{} 5$
$ c \equal{} 25$

If the question implies that the pile of coconuts is divided each time, then it is more complicated.
If I am correct, the number of coconuts should be divisible by 5. If so, then it would not make sense for 1 to be left over. Im obviously missing something.

[quote="coquitao"]ANSWER: n, where n is an integer that satisfies all of the equalities above:

\begin{eqnarray*}n & = & 5k_{1} + 1 \\
4k_{1} & = & 5k_{2} + 1 \\
4k_{2} & = & 5k_{3} + 1 \\
4k_{3} & = & 5k_{4} + 1 \\
4k_{4} & = & 5k_{5} + 1\end{eqnarray*}

Let us sum a 4 to each side of those eqns below. We get something like

\begin{eqnarray*}n + 4 & = & 5(k_{1} + 1) \\
4(k_{1} + 1) & = & 5(k_{2} + 1) \\
4(k_{2} + 1) & = & 5(k_{3} + 1) \\
4(k_{3} + 1) & = & 5(k_{4} + 1) \\
4(k_{4} + 1) & = & 5(k_{5} + 1)\end{eqnarray*}[/quote]

After multiplying we obtain

\begin{eqnarray*} (n + 4)(4(k_{1} + 1) )(4(k_{2} + 1))(4(k_{3} + 1))(4(k_{4} + 1)) & = & (5(k_{1} + 1) )(5(k_{2} + 1))(5(k_{3} + 1) )(5(k_{4} + 1))(5(k_{5} + 1)).\end{eqnarray*}

Canceling gives us that n must satisfy

$ (n + 4)4^{4} = 5^{5}(k_{5} + 1)$

Since 5 and 4 are coprime numbers we conclude that $ 5^{5}\vert(n + 4)$. The minimality condition allows us to conclude that $ n = 3121$.

HOPE IT HELPS.

wow thnx. That helps alot.

btw awesome solution. Im still pretty amazed at the simplicity of it.

Nice solution. I see where your coming from now. Thanks.


I wonder if it has to do with some incompetent FTW mods.

Hmm, maybe start with this discussion off the beastly debate.

levans and jjx1 vs. math154 and justinahmann ( me ) :ninja:

The Crucible is allegory that likens the events the Salem Witchcraft Trials to the current situation on FTW where jjx1=Abigail and levans=Danforth.

TYTIA

You guys are learning about the crucible in Comm. Arts?  :P


Can every positive rational number be written in the form

(a2 + b3)/(c5 + d7)

for certain positive integers a, b, c and d ?

Yes, it is true.

a=p3q2
b=p5q2
c=pq
d=p2q

How did you come up with that ? Is there a trick to find this solution ?

i guess that this problem required no trick - just some experience, some computations and a little bit of luck :)

Yes, of cause I used a trick.
Main idea is to find a,b,c,d such that
a2/c5 = b3/d7 = p/q !

Of course ... why didn't I think of that ! Thanks, Myth.

Look like every positive rational number can be written as

(a^3 + b^3)/(c^3 + d^3), where a,b,c,d are positive integer

http://www.kalva.demon.co.uk/short/soln/sh99n2.html


Find all functions $ f: \mathbb{Z}\setminus\{0\}\to \mathbb{Q}$ such that for all $ x,y \in \mathbb{Z}\setminus\{0\}$:
\[ f \left( \frac{x+y}{3}\right) =\frac{f(x)+f(y)}{2}, \; \; x, y \in \mathbb{Z}\setminus\{0\}\]

We''ll show that $f(x)=c$ for $x \in N$.
Denote $f(1)$ by $c$.
First of all:

by taking $x=1,y=2$ we obtain $f(2)=f(1)=c$

Now let's plug in $x=y=3$,in this case $f(3)=f(2)=c$.

Let's prove by induction that for every $x>1$ $f(x)=c$

We have $f(1)=f(2)=f(3)=c$. Assume that for every $1 \leq x \leq n$ (where $n \geq 3$) we have $f(n)=c$. Let us consider $f(n+1)$. Let $t$ be such number in $<1;n>$ that $(n+1)+t$ is a multiple of $3$. Then
\[f(\frac{n+1+t}{3})=\frac{f(n+1)+f(t)}{2}\] but as $t \leq n$ and $\frac{n+1+t}{3}\leq n$ we get $f(n+1)=c$ which means that \[f(x)=c\] for every $x \geq 1$.

Now let $k$ be a negative number. There exists $l>0$ such that $l+k>0$
and $l+k$ is a multiple of $3$. Puting $x=l,y=k$ in our equation leads us to $f(k)=c$.

Finally we got $f(x)=c$ for a constant $c \in Q$.


Please answer the following poll, as it is extremely important to sample the math/nerd demographic.

Yellow. It made half of people that play Pokemon camoflauge.

Is there an option for "I don't know anything about Pokemon other than there's this Pikachu thing?"  :P

Yes, it's called "I agree with the author."

[quote="worthawholebean"]Red/Blue by far. The new ones are just new Pokémon/trainers with a couple useless features added.[/quote]

I disagree, and I have to go with Diamond/pearl. Here are my reasons:

1. The main thing wrong with red/blue/yellow (which I will henceforth call R/B/Y) was that psychic pokemon were simply way too strong. Psychic was weak against bug and ghost, but there were no bug or ghost moves with any respectable power at all. In addition, one SPECIAL stat was used for both special attack and special defense. Add in the fact that Alakazam was psychic type, had a high speed and a high special stat, and had basically no weaknesses due to bug/ghost absolutely sucking, meant that the metagame was pretty much warped on "You have to include alakazam on your team, or else you lose" (assuming Mewtwo is banned already, which it should)

The unbalance of psychic pokemon was fixed with the addition of dark and steel types, which are immune to and resist psychic attacks, respectively, and adding some strong moves that are super effective against psychic pokemon, such as Shadow Ball.

2. More imbalances: Namely grass moves. The strongest grass move before diamond/pearl was Leaf Blade, and only one pokemon learned it. Besides that like the strongest grass attack you had was a Hidden Power set to Grass, and that was pretty sad.

Now you have Wood Hammer and Grass Knot in Diamond/Pearl. Much better moves.

3. The separation from physical/separation from an intrinsic property of the type is also a very good thing. Many pokemon have only a high attack stat or sp. attack stat, not both, and basically having a wider access to move types increases balance and strategical aspects of the game.

4. Diamond/pearl also sports many moves that overall balances the game. For example, in ruby/sapphire/emerald (R/S/E) a popular strategy was to use the pokemon Skarmory and Blissey together in order to stall the other team. Skarmory could take almost any physical attacks, and Blissey could take almost any special attacks. Together they were nearly invincible except for certain strategies designed to counter them (they usually used both physical and special). In diamond/pearl this strategy is still good, but is much easier to defeat because: 1) There are now physical fire and electric moves, allowing even physical pokemon to defeat Skarmory, 2) There are now better special attack boosting moves such as Nasty Plot that allow special attacking pokemon to defeat Blissey (and besides there are also special fighting moves like Aura Sphere).

5. Many pokemon from pre-diamond pearl that desperately needed to get some new life did in fact get new life (in the form of a second or third evolution) and are commonly used, such as Gliscor (evolved from Gligar), Tangrowth (evolved from Tangela), Mamoswine (evolved from Piloswine), Electivire (evolved from Electabuzz), Togekiss (evolved from Togetic), Magnezone (from Magneton), and Dusknoir (evolved from Dusclops - though Dusclops was pretty good too)

In addition, many old pokemon got additional moves that make them better. Typhlosion now has Eruption, which is a very powerful fire move under certain conditions. Crobat (evolved from Golbat) has Hypnosis. Gengar's shadow ball now uses special attack instead of attack (and his special attack is higher).

So, in fact, when they make new pokemon these days, they do not forget about the old ones.

6. Some of the new pokemon are in fact not repeats of the same old stuff. Of course, all of the bugs look about the same but no one cares about the bugs anyways. I'm talking about pokemon like Heatran (with the very interesting Fire/Steel type combo), Empoleon (water/steel), Drapion (Poison/dark), Bronzong (steel/psychic with levitate = only one weakness wtf hax)

7. BIBAREL IS A BEAVER! YOU HAVE TO GO TO DIAMOND/PEARL TO FIND A BEAVER! HOW CAN YOU ARGUE WITH THAT? (note: I may be slightly biased)

8. Cooler legendary pokemon, in general. Moltres isn't even playable in competitive battling. By the time you get to diamond/pearl all of the [new] legendaries see some light in competitive battling. (assuming they don't get banned of course...)

9. Increased biodiversity.... Last time I heard, more biodiversity is a good thing. Even if it means they have to make up a half-baked reason as to why suddenly 100 more pokemon got discovered. But really, having only 150(1) species is a very sparse world.

10. Hold items increase strategical possibilities. These were around for a while but the really cool ones didn't come out until later. (R/S/E - D/P)

Also, abilities (this is more of an R/S/E thing, but whatever). Dugtrio is actually useful once it has Arena Trap.

Side note: Besides, R/B/Y had that MISSINGNO glitch. Who the hell keeps a glitch in their released game that can absolutely DESTROY your savefile?

I will admit though there are some problems. Like Ash. Shouldn't he be like 16 now? (or at least 14...) and yet he still picks up 10 year old girlfriends?

[quote]
So, in fact, when they make new pokemon these days, they do not forget about the old ones.
[/quote]

I'm going to call that false...the only two of the pokemon listed that I recognized were electabuzz and togetic, and the only one I KNOW has roots in the original is togetic...although electabuzz might.

And the missingno glitch was quite fun. I never had it destroy my savefile o_O.

[quote="randomdragoon"]I will admit though there are some problems. Like Ash. Shouldn't he be like 16 now? (or at least 14...) and yet he still picks up 10 year old girlfriends?[/quote]
When the show started, he said he was 10.  If you find the year that the show started (I don't know this), you can figure out his age.

[quote="randomdragoon"]I will admit though there are some problems. Like Ash. Shouldn't he be like 16 now? (or at least 14...) and yet he still picks up 10 year old girlfriends?[/quote]

Lol, how old was he when the series started? I mean, pokemon is over 10 years old now, right? So shouldn't he be about 22 years old now? I mean, even randomly leaving for an "adventure" is already pretty far-fetched at 12 years old.
EDIT: Lingo beat me to the post. Still, he's about 20 years old now, or at least he's supposed to be.

Also, randomdragoon wins this thread. Period.

My opinion is very biased, since I have never played D/P (only enviously watched), and have barely played gold, and have only seen red and yellow, not played them. So yes, I'm not a veteran pokemon player (I was what, 3 years old when the series first came out?). I immediately favor Ruby/Sapphire and Diamond/Pearl, simply because they are for the SP and DS, which are much more playable than the GB pocket and color (IMO). Better lighting, larger screen, better graphics and artwork; it's just overall more pleasant to the eye. I have only played ruby with much depth (as in, collected all pokemon besides ones only in sapphire, explored, etc.), so I'm probably too biased judge whether Ruby/Sapphire or Diamond/Pearl is better. Personally, I feel the starter pokemon are really unbalanced in R/S; Blazekin is overpowered, so I tried playing another file using swampert. By my nooby pokemon skills, i ended up with only water and normal attacks, and 2 HM moves. All of swamperts moves seem too low SP and accuracy. Bleh. I haven't tried Grovyle yet, though.

Well, you do have to take into account the "screenwriter's license" that cartoon writers use.  If cartoon characters aged as fast as we do in real life, Bart and Lisa Simpson would be in their twenties by now (The Simpsons first aired around Christmas 1989).  If Ash was in his early twenties, kids may not feel as closely connected to the show as they would if he was 14 or 16.

Sometimes aging characters works, though.  It can show the vast changes that occur in a family.  Consider Full House, in which the Tanner children got older as the seasons progressed, creating new situations for future seasons, such as going into junior high school, taking the SATs, or getting kissed for the first time.  But of course, the Olsen twins aren't going to stay toddlers forever, so the producers didn't really have much of a choice, unless they wanted to get new actors every season.

Hmm, good point. Well, I don't watch the cartoon show anymore, since Cartoon Network stole it and effed it up (they did the same thing to yugioh, except even worse).

Hmm, about one new episode airs every week, and each episode seems to cover a span of about 1 - 3 days, so actually I guess Ash's age hasn't changed too much. Okay, so should we get back on topic now?

After R/B/Y when I was a little kid it was really annoying to memorize a new list of pokemon EVERYTIME a new game came out. Also after those the gameplay and the storyline became way too complicated with its random breeding stuff and watnot and I prefer the simplicity of the classics: Easy storyline, Maps aren't overtly complicated, you didn't have to go through the "other random world" crap and the cities were easy too. You had some really weird things going on in the other ones too.

Maybe my review would be more balanced if I had retried gameplay of Diamond/Gold/Silver/Ruby/Emereld/Whateverthehellnewonescameout in an older age instead of at 6, but I think they've basically beaten the franchise to pulp.

Well, you can only do so much until something gets old.  It's inevitable.

Wow, I never thought there would be a discussion about Pokemon on AoPS.

My thoughts:
I think that Emerald is the best Pokemon version.  The thing that sets it apart from all other versions (and especially its companions Ruby and Sapphire) is the Battle Zone.  Other versions have a Battle Park, but Emerald's Battle Zone gives you several different battle experiences that allow the game to never really be over (which I think is one of the most desirable features of a video game.  I was really disappointed when Diamond/Pearl didn't include this expanded Battle Zone.

Other reasons why I like the R/S/E series in general:
1. Contest halls.  R/S/E were the first versions to offer Pokemon Contests, and I believe their format is better than D/P.  (I think the dancing portion of the D/P contests is just pointless).  Contests provided a new and innovative feature to Pokemon that was badly needed.

2. The Trainer's Eyes/Match Call feature.  Trainer's Eyes allowed you to battle 70 different trainers you had previously battled.  Emerald expanded on this with the Match Call, which allowed you to call other Trainers (and them to call you) for battles, like in G/S/C versions.  This made it much easier to further train your Pokemon in preparation for the Elite Four after you had already beaten all of the trainers once.

3. Like randomdragoon mentioned, Pokemon abilities first appeared (and were most important in) R/S/E.  Again, this provided a new wrinkle to the "same old, same old" style of Pokemon battling.  Instead of just considering an opposing Pokemon's strengths and weaknesses, you also had to consider that Pokemon's specific ability.

4. Double Battles.  R/S/E introduced the Double Battle, and you could now cover up one Pokemon's weakness with another's strengths.  More strategy was required, which is always a plus.

[quote="Nerd_of_the_Ages"]Lol, how old was he when the series started? I mean, pokemon is over 10 years old now, right?[/quote]

Pokemon was first introduced in 1995.  Ash was 10 years old when the series debuted, so he would be 23 now.

[quote="Hamster1800"]I'm going to call that false...the only two of the pokemon listed that I recognized were electabuzz and togetic, and the only one I KNOW has roots in the original is togetic...although electabuzz might.[/quote]

Electabuzz appeared in the original 150 Pokemon, but Togepi/Togetic, although appearing in the original TV series (belonging to Misty), did not appear in the Pokemon video game until G/S/C.

randomdragoon's post made me wanna get back to pokemon...ex: I loved Magneton in Gold/silver/crystal...evolution of them would be cool

Wow...........  I'm not even going to say anything concerning how nerdy this thread is...  Dood, everyone who still plays pokemon should seriously get another hobby...

I completely second everything randomdragoon said. I actually secretly love pokemon...

Here are the games I played: R/B/Y(R+Y), G/S(S), then I stopped. Then, last year, a bunch of my friends decided to get D/P. (I couldn't tell you why...) So I got it, and it was actually very good. Well, for about a month anyway...That's the problem with Pokemon in general--It gets boring quickly. Like, had another game been released after that month, I wouldn't have gotten it. Now, I likely would(No I'm not ashamed :P ) Anyway, I like D/P more than the originals, for many of the reasons dragoon mentioned.

[quote="erabel"]Um... First of all, I seem to notice that you have missed (not that I like these, I just noticed):

1. Pokemon Snap (N64)
2. Pokemon TCG for GameBoy
3. Pokemon Stadiums 1+2 (N64)
4. Pokemon Pinball (both of them) (I think there are 2, both for GBA)
5. Pokemon Colosseum (NGC)
6. Pokemon XD: Gale of Darkness (NGC)
7. Pokemon Ranger (DS)
8. Pokemon Trozei (DS)
9. Pokemon Battle Revolution (Wii)

Second of all, my favorite has to be Pearl, even though I haven't played it. I only have 1 unsaid reason for my choice of Pearl over Diamond:

1. I am the BIGGEST Slowpoke fan ever (not because they're powerful or anything, I just like them), and without trading, you can't get them on Diamond.

The reason I said "unsaid reason" is because I agree with most of that very, very ling post by randomdragoon.[/quote]

I think the original poster was referring to the traditional games. Although among the ones you mentioned, pokemon colloseum was highly mediocre; pokemon snap was REALLY fun, believe it or not. What's your highest score non-mew photo?

Hi, me again.

You should not deduce Ash's age simply from the first release date of the first episode -- 1 year in real time does not necessarily correspond to 1 year in episode time. However, since Pokemon has a continuous storyline (as opposed to the Simpsons, for example, which is more individual episodes), we can do some analysis like the passing of seasons, etc. to deduce how many years have passed.

Though I'm going to do a rough estimate and say large Pokemon tournaments should happen about once a year, and there has been one at the end of each "generation" (R/B/Y, G/S/C, R/S/E, [D/P]), so Ash should be about 14 years old at the current time.

Still way too old to have 10 year old girlfriends though.

Oh, and if anyone wants a hilarious double battle team, try 6 gengars with shadow ball, explosion, destiny bond, and some other filler move. (not a solid team but hilarious nonetheless)
(for those of you that are slow: basically explosion hits all pokemon on the field for 250 power but ghosts aren't affected by explosion so only your opponents die ... and shadow ball is to take care of opposing ghosts.)

Also, erabel forgot to mention Hey You, Pikachu!
That was pretty lame...

Ahh, the good memories...

Let's go fishing today!
Somethings on the line!
"Yo, Pikachu, you suck ballz"
Pikachu, inspired by your words reels up a magikarp
WOAH, a big one!
"Pikachu, constipate on the toilet, not on the beach!"
You got a goldeen!

At least Goldeen can learn Megahorn now, so it's actually far from useless, unlike magikarp / half the bugs out there

@Nerd_Of...: I haven't actually played Snap.... and would anyone like to post their current pokemon team on any/all of their games? Can't remember mine right now and too lazy to check. :|

Dude magikarp was the best pokemon ever. Seriously, don't diss magikarp.

Ever.

No one has mentioned $ \binom{i}{u}$, Pikachu in this thread yet?

No, that's not cool enough.  Instead, it should be $ (\binom {I} {U} PIKACHU)(\binom {I} {U} PIKACHU\minus{}1)\cdot\cdot\cdot(3)(2)(1)$.  Wink, wink, nudge, nudge, get it? :P

complete pwnage

[quote="Klebian"]So, I herd u liek mudkips.[/quote]

Back to 4chan for you  :P

[quote="randomdragoon"]I have a good image parody of that meme but it has profanity so I can't post it here.[/quote]

PM me ;) I can appreciate /b/ to some extent...


[quote="mestav"]Prove that for every continuous function can be written in the form of sum of more than one even or odd functions[/quote]
f(x)=(h(x)+g(x))/2
with
h(x)=f(x)+f(-x) which is even function
g(x)=f(x)-f(-x) which is odd function. :)


Let a,b,c,d be four positive real numbers satisfy: a+b+c+d=1.
Prove that:6(a^3+b^3+c^3+d^3)>=a^2+b^2+c^2+d^2+1/8
I am seeking some nice solutions :)


[quote="countdownmath"]Find a one-to-one mapping from the interval $ (0,1)$ onto $ \mathbb{R}$ that is a homeomorphism and is a rational function.[/quote]
$ f(x) \equal{} \frac{x}{1\minus{}x^2}$ is a one-to-one mapping from $ (\minus{}1;1)$ to $ \mathbb{R}$. $ f(x) \equal{} f(y) \iff x\minus{}xy^2 \equal{} y\minus{}x^2y$ $ \iff$ $ x\minus{}y \equal{} \minus{}xy(x\minus{}y) \iff (xy\plus{}1)(x\minus{}y) \equal{} 0$. Since $ |x|,|y| < 1$ then $ x \equal{} y$. Then let $ g(x) \equal{} f(2x\minus{}1)$, so $ g(x) \equal{} \frac{2x\minus{}1}{\minus{}4x^2\plus{}4x}$.

It seems to work :)

$ g(x) \equal{} g(y)$ $ \iff$ $ (2x\minus{}1)(\minus{}4y^2\plus{}4y) \equal{} (2y\minus{}1)(\minus{}4x^2 \plus{} 4x)$ $ \iff$ $ \minus{}8xy^2 \plus{} 8xy \plus{} 4y^2\minus{}4y \equal{} \minus{}8x^2y\plus{}8xy\plus{}4x^2\minus{}4x$ $ \iff$

$ (xy \plus{} (x\minus{}1)(y\minus{}1))(x\minus{}y) \equal{} 0$. Since $ xy > 0$ and $ (x\minus{}1)(y\minus{}1) >0$ we get $ x\minus{}y \equal{} 0$ $ \iff$ $ x\equal{}y$ so $ g(x)$ is injective on $ (0;1)$.

$ \minus{}4x^2\plus{}4x > 0$ so $ g(x)$ is defined for all $ x \in (0;1)$, and it follows that $ g(x)$ is continous since it is a rational function. (And further more it is easy to see that it is monotomic)

Further more $ g(x) \to \minus{}\infty$ for $ x \to 0$ and $ g(x) \to \infty$ for $ x \to 1$, so since it is continous we get that it is surjective on $ \mathbb{R}$, and thats about it isn't it? :)

So a answer $ \boxed{g(x) \equal{} \frac{2x\minus{}1}{\minus{}4x^2\plus{}4x}}$


I got 47 in base 16 after converting to base 10, but after that, I didn't see an answer choice without the letter F in it, except for a few. I haven't done bases much before, and I got $ 4(16^{1})\plus{}7(16^{0}) \equal{} 71$ $ base$ $ 16$ What is F? Is there something different about converting to bases above 10?  :o

For base 16, 10=A, 11=B, 12=C, 13=D, 14=E, and 15=F, since the "single" digit "numbers" are 1-15 in base 10.

[hide="One way"]For any problem like this, you can divide up the binary number into strings of four numbers (putting 0s on the beginning if necessary). So in this case, we look at it as $ (0010)(1111)_{2}$. Then each string of four digits corresponds to a digit in the base 16 representation. As $ (0010)_{2}\equiv 2_{16}$ and $ (1111)_{2}\equiv F_{16}$, the answer is $ 2F_{16}$.[/hide]

[response to previous post, which came just before mine]
It sounds like you got a lot of things jumbled up. :( 47 is the base [b]10[/b] equivalent of $ 101111$, not the base 16 equivalent. So you converted from base 2 to base 10, put that number into base 16, and converted to base 10 again.

As for what letters in bases mean:
$ A$ is the digit for 10, used only in bases greater than 10,
$ B$ is the digit for 11, used only in bases greater than 11,
$ C$ is the digit for 12, etc.

So the digits in hexadecimal from $ 0$ to $ 15$ are $ 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F$.

Here's how you might approach the problem using a convert to base 10 approach. This method's advantage over the first is that it works when converting between any two bases. (The one above takes advantage of the fact that $ 16$ is a power of $ 2$):

[hide="Second way"]$ (101111)_{2}\equiv (47)_{10}$, as you found. Now we want to find it in base 16. If we divide 47 by 16, we get a quotient of $ 2$ and a remainder of $ 15$. So $ 47 \equal{} 2\cdot 16\plus{}15$. Since $ F \equal{} 15$ in base 16, the desired is $ 2F$.

Note that if you divide and get a quotient at least the base (here, anything 16 or greater), you have to do it again.

[hide="Example of this"]
Converting $ 10001001011$ from base 2 to base 16. Arithmetic shows that it is equal to $ 1099$ in base 10.
Divide 16 into $ 1099$. We get a quotient of $ 68$ and a remainder of $ 11$. So the last digit will be $ B \equal{} 11$, but we need to divide 16 into $ 68$ now.
Divide 16 into $ 68$. We get a quotient of $ 4$ and a remainder of $ 4$. So the next two digits are both $ 4$.
This gives a final answer of $ (44B)_{16}$.
[/hide]
[/hide]

Wow, thanks everyone for pointing out the jumble up. Therefore, I guess we kind of accept F and E as digits, so like 20 could be E6, etc.

I won't attempt the 1st way for now, since the 2nd way kind of would be how I would do it, and it works for all bases, like said previously.


[quote="Yimin Ge"]For positive reals $a,b,c$ prove
\[ \frac1a + \frac1b + \frac1c \geq \frac{b+c}{a^2+bc} + \frac{c+a}{b^2+ca} + \frac{a+b}{c^2+ab}  \][/quote]
 Source: Crux, 2580
It is impossible, as it is too easy! :D
1.04.?  :lol:

It's indeed easy but I can assure you it is a Crux-Problem
Btw, what do you mean by 1.04.?

My Solution:
[hide]
Multiplying everything out yields to the equivalent inequality
\[ \displaystyle\sum_{cycl} a^4b^4 \geq \sum_{cycl} a^4b^2c^2 \]
[/hide]

[quote="Yimin Ge"]
Btw, what do you mean by 1.04.?
[/quote]
1st of april - april's fools. :)
I have seen the following:
$\frac1a + \frac1b + \frac1c \geq \frac{b+c}{a^2+bc} + \frac{c+a}{b^2+ca} + \frac{a+b}{c^2+ab}\Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{a^3+abc}\geq0.$
The last inequality is obviously true. ;)

[quote="arqady"]
The last inequality is obviously true. ;)[/quote]

due to vornicu schur ineq??

arqady please explain...... :)

[quote="arqady"][quote="socrates"]
arqady please explain...[/quote]
I do it so:
Let $a\geq b\geq c.$
Hence, $\sum_{cyc}\frac{(a-b)(a-c)}{a^3+abc}=\frac{(a-b)(a-c)}{a^3+abc}+(b-c)\left(\frac{a-c}{c^3+abc}-\frac{a-b}{b^3+abc}\right)\geq0.$
This reasoning is stronger than vornicu - schur. ;)[/quote]

thanks ;)

We had this at http://www.mathlinks.ro/Forum/viewtopic.php?t=57975 and http://www.mathlinks.ro/Forum/viewtopic.php?t=43872 .

  darij

Whlle trying to solve arqady's problem, I didn't do my research and ended up trying to prove what I didn't realize was a weaker inequality.
Then I scrolled down and noticed that the inequality I mistakenly proved was the original problem for this thread. [hide="So here it is"]
\[ \sum\frac {b \plus{} c}{a^2 \plus{} bc} \equal{} \sum \frac {a^2 \plus{} bc}{a^3 \plus{} abc} \minus{} \frac {(a \minus{} b)(a \minus{} c)}{a^3 \plus{} abc} \equal{} \sum \frac {1}{a} \minus{} \frac {(a \minus{} b)(a \minus{} c)}{a^3 \plus{} abc}\]
Prove that:
\[ \sum \frac {(a \minus{} b)(a \minus{} c)}{a^3 \plus{} abc}\geq 0\]
Solution:
Solve the problem by adding the constraint $ abc \equal{} k$, and proving for each $ k > 0$.
We have that
\[ \sum \frac {(a \minus{} b)(a \minus{} c)}{a^3 \plus{} abc} \equal{} \frac {1}{abc\prod( a^3 \plus{} abc)}\sum [(b^2 \plus{} ac)(c^2 \plus{} ba)/a] (a \minus{} b)(a \minus{} c)\]
Let $ g(a) \equal{} ((abc)^2/a^3) \plus{} b^3 \plus{} c^3 \plus{} abc \equal{} f(a) \plus{} b^3 \plus{} c^3$.
Then $ f(a) \equal{} k^2/a^3 \plus{} k$ satisfies conditions for Vornicu-Schur to hold (where $ k$ is arbitrary but fixed value of $ abc$)
Hence,
\[ \sum f(a)(a \minus{} b)(a \minus{} c)\geq 0\]
Also,
\[ \sum (b^3 \plus{} c^3)(a \minus{} b)(a \minus{} c)\geq 0\]
holds because it is equivalent to
\[ \sum_{cyc}a^2b^3 \plus{} c^2b^3\geq 2\sum_{cyc}acb^3\]
which is just three applications of am-gm.
Thus
\[ \sum g(a)(a \minus{} b)(a \minus{} c)\geq 0\]
[/hide]
Edit: Arqady's inequality falls to [hide="Cauchy + Bunching"]
By Cauchy, we have\[ \sum \frac{b\plus{}c}{a^2\plus{}bc}\geq \frac{2(\sum a)^2}{\sum_{sym}a^2b}\]
So, we need only prove\[ 2(\sum a)^2\geq 3(\sum_{sym}a^2b)\]
That is,\[ 4(\sum a^2)(\sum a)^4\geq 27(\sum_sa^2b)\]
Expanding, it is equivalent to
\[ 4\sum_ca^6\plus{}\sum_sa^2b^4\plus{}10\sum_sa^2b^3c\plus{}16\sum_sa^5b\geq 22\sum_ca^3b^3\plus{}90a^2b^2c^2\plus{}6\sum_ca^4bc\]
This is apparent as we have \[ 14\sum_sa^5b\geq6\sum_ca^4bc\plus{} 22\sum_ca^3b^3\] to take care of all the strong enough RHS terms and the rest can be done in one Am-Gm.

[/hide]

[quote="arqady"]The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Prove that:
\[ \frac {b \plus{} c}{a^2 \plus{} bc} \plus{} \frac {c \plus{} a}{b^2 \plus{} ca} \plus{} \frac {a \plus{} b}{c^2 \plus{} ab}\geq3\]
[/quote]
By the Cauchy-Schwarz Inequality, we have
$ \sum \frac {b \plus{} c}{a^2 \plus{} bc} \ge \frac {4(a \plus{} b \plus{} c)^2}{\sum (b \plus{} c)(a^2 \plus{} bc)} \equal{} \frac {2(a \plus{} b \plus{} c)^2}{\sum ab(a \plus{} b)}.$
It suffices to prove that
$ 2(a \plus{} b \plus{} c)^2\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \ge 9\sum ab(a \plus{} b),$
or equivalently,
$ 2(a \plus{} b \plus{} c)^2 \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} 27abc \ge 9(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca).$
By Schur's Inequality, we see that
$ 27abc \ge 3(a \plus{} b \plus{} c)[2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)],$
and thus, it is enough to check the following inequality
$ 2(a \plus{} b \plus{} c)\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} 3[2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)] \ge 9(ab \plus{} bc \plus{} ca),$
or
$ 2(a \plus{} b \plus{} c)\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \ge 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca).$
This is true, because
$ 4(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)(a^2 \plus{} b^2 \plus{} c^2 \plus{} 3ab \plus{} 3bc \plus{} 3ca) \ge 0.$
The proof is completed.

Yes, Can_hang . This is also my solution ! How about Schur ineq ?

[quote=Yimin Ge]For positive reals $a,b,c$ prove
\[ \frac1a + \frac1b + \frac1c \geq \frac{b+c}{a^2+bc} + \frac{c+a}{b^2+ca} + \frac{a+b}{c^2+ab} \][/quote]

We have
\[LHS-RHS=\frac{1}{2(a^2+bc)(b^2+ca)(c^2+ab)}\sum{\frac{c^3(a^2-b^2)^2}{ab}}\ge{0}\]

[quote=xzlbq]Let $a,b,c$ be a sides of trianle,

\[{a}^{-1}+{b}^{-1}+{c}^{-1}\geq 5/2\left( {\frac {b+c}{3\,{a}^{2}+2\,bc
}}+{\frac {c+a}{3\,{b}^{2}+2\,ac}}+{\frac {a+b}{3\,{c}^{2}+2\,ab}}
 \right)\][/quote]
It's obviously true after using Ravi's substitution and full expanding: Schur and $uvw$.

[quote=arqady][quote=xzlbq]$a,b,c,d>0$,

${a}^{-1}+{b}^{-1}+{c}^{-1}+{d}^{-1}\geq {\frac {a+b+c}{ab+bc+ac}}+{\frac {
b+c+d}{bc+cd+bd}}+{\frac {a+c+d}{cd+ad+ac}}+{\frac {d+a+b}{ad+ab+bd}}
$
[/quote]
Because $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{3(a+b+c)}{ab+ac+bc}$.[/quote]

Thank you!
BQ

[quote=arqady]The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Prove that:
\[ \frac{b\plus{}c}{a^2\plus{}bc} \plus{} \frac{c\plus{}a}{b^2\plus{}ca} \plus{} \frac{a\plus{}b}{c^2\plus{}ab}\geq3\][/quote]

By AM-GM,we need to prove that
\[\sum{\frac{b+c}{a^2+bc}}\ge{\frac{9(a^2+b^2+c^2)+3(a+b+c)^2}{2(a+b+c)(a^2+b^2+c^2)}}\]
\[LHS-RHS=\frac{\sum{(ab^5+b^5c+5a^3b^3+6a^2b^4+5a^3bc^2+6a^4b^2+5c^2ab^3+a^5b+ca^5+5a^2b^2c^2)(a-b)^2}}{2(a^2+bc)(b^2+ca)(c^2+ab)(a+b+c)(a^2+b^2+c^2)}\ge{0}\]

[quote=arqady]The following inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Prove that:
\[ \frac{b\plus{}c}{a^2\plus{}bc} \plus{} \frac{c\plus{}a}{b^2\plus{}ca} \plus{} \frac{a\plus{}b}{c^2\plus{}ab}\geq3\][/quote]

[img]http://s16.sinaimg.cn/middle/0018zOAxgy70owTudPh1f&690[/img]


Consider the sequence of sets $S_n$ definied inductively by
$S_n = \{ s_d \cdot s_{n-d}\}$, for all $d$ and for all $s_d \in S_d$. We define $S_1 = {}, S_2={2}, S_3={3}$
If we wanted, we could continue this sequence
$\{4\},\{6\},\{8,9\},\{12\},\{16,18\}$
Clearly, any element of $S_n$ must be of the form $2^{x}3^{y}$
Claim: if $a \in S_n$ and $a=2^{x}3^{y}$, then $n=2x + 3y$
Proof: strong induction. Suppose it holds for $k < n$. Suppose $a \in S_n$. Then $a = s_d \cdot s_{n-d}$ for some $d$ and $s_d,s_{n-d}$ in $S_d,S_{n-d}$, respectively. Suppose $s_d=x^ey^f$ and $s_{n-d}=x^gy^h$. Then $a=x^{e+g}y^{f+h}$. By the induction hypothesis, $2e+3f=d$ and $2g+3h=n-3$. Adding, we get $2(e+g)+3(f+h)=n$. $\square$
Claim: converse of previous claim: if $2x+3y=n$, then $2^x3^y \in S_n$.
Proof: Induction. The claim holds for $n-2$ and $n-3$, and the result follows. $\square$
From the last two claims, we see that $S_n$ contains all numbers $2^x3^y$ where $2x+3y=n$.
It is clear that due to the construction of the sequence $a_1, a_2, \cdots$ $a_n$ is the largest element of $S_n$. Since $2^3 < 3^2$, $2^x3^y < 2^{x-3}3^{y+2}$, and $2x+3y=2(x-3)+3(x-2)$. It follows that $a_n$ is the element of $S_n$ that isn't divisible by $2^3$.
$1998=3 \cdot 666$, so the answer is $3^{666}$ :diablo:

I realize this could be proved MUCH more quickly and thoroughly with by just doing induction using the formulae for the separate cases $n = 0,1,2 \mod 3$, but I think the above idea better illustrates the coolness of the sequence.  :)

a few typos in ur solution i think but other than that cool proof!  :lol:


Reduce this quadric equation:

$ 3x^2 \plus{}2y^2 \minus{}z^2 \plus{}4yz\minus{}1 \equal{} 0$

There seems to be something wrong with the image.

hello, i have got
$ 6y^2\minus{}1\plus{}3x^2\minus{}(z\minus{}2y)^2\equal{}0$.
Sonnhard.


I don't know how to make a poll :blush: , but maybe you guys can just post who you think is the most promising new user, maybe even candidate to the MODs? by new users, I mean within 0~3 months.

this is a delicate topic...for us newbies at least. :D
FYI I'm a new user. Keep that in mind!  :lol: lol

PS can you vote for yourself?

When you create a new topic, if you scroll down to the bottom of the screen you will see the control panels that allow you to create polls.  (If you're replying to a message, it's replaced by the window that shows previous replies in the same thread.)

I'm not sure whether this is also the way it looks in the mathlinks shell.

I am not familiar with the rights and duties of a moderator completely, but are moderators chosen in a democratic way?  Can they be relieved of duty?  And one might wonder whether democracy is the best approach as people with almost no posts could have multiple accounts to boost themselves up in polls

[quote="JBL"]
I'm not sure whether this is also the way it looks in the mathlinks shell.[/quote]
It is exactly the same.

I don't think we can say who is the most promising new user now. Only time will tell.

[quote="fredbel6"]are moderators chosen in a democratic way?[/quote]
No.
[quote="fredbel6"]Can they be relieved of duty?[/quote]
Yes.

The moderators are asked to be moderators by the owners of the site.

The admins, if they need a mod, choose someone dependable and ask if they want to moderate. Or if an existing moderator feels overwhelmed, the existing mod might recommend someone for the admins' approval, who may or may not be accepted.

[quote="fredbel6"]I am not familiar with the rights and duties of a moderator completely, but are moderators chosen in a democratic way?[/quote]No democracy here my friend :D

Well Valentin that is just fine by me.  Because democracy is as churchill apparently said the worst way to run things, except all other ways.  Secondly as I pointed out, not all voting members would relate to consequences to their vote, as some may walk away any day if the forum goes bad (which normally in a nation you don't)
So surprisingly, I'm proposing dictatorship,  :blush: , this type of dictatorial yet benevolent competent rule is just fine by me flatter flatter (I am not kissing up to become a mod, I really am not, I couldn't handle it)

[quote="fredbel6"]So surprisingly, I'm proposing dictatorship,  :blush: this type of dictatorial yet benevolent competent rule is just fine by me [/quote]You are not proposing anything new: it's how the system currently works :D

Haha yup nothing we can do to change how things are run here.

I'm not exactly a new member anymore, and i dont come here as often as i used to, so no, not me..

[quote="Valentin Vornicu"]
You are not proposing anything new: it's how the system currently works :Mr. Green:
[/quote]

Indeed. Observe the Yoda avatar. That look is one of pure, iron-fisted malice.

i'd say i was the most promising new user a year or so ago :P

that reminds me, i haven't gotten into any arguments with the administrators for a while now... i'm pretty sure that's a bad thing.

[quote="Treething"]i'm pretty sure that's a bad thing.[/quote]It's a good thing :) Don't go to the Dark side son! (although I know they have cookies :P :D).


integration
inequalities proposed
inequalities

Do we need use the formula
$B_n=\frac{n(n-1)}{4(2^n -1)}\int_{0}^{1}E_{n-2}(x)dx$


calculus
integration
inequalities
real analysis
real analysis solved

Prove the following inequality.

\[\frac{2}{3}<\int^1_0 e^{-x^2}dx<\frac{\pi}{4}\]

and from $e^{-x^2} < 1-x^2+\frac{x^4}{2}$ we have $I < \frac{23}{30}$ and it is not difficult to show that $\frac{23}{30} < \frac{\pi}{4}$.

You are right. You can use $e^{x^2}>1+x^2$ for the second inequality.


1.  What would you expect to be the effect of putting mitochondria in a solution of DNP, which destroys the proton gradient across the inner mitochondrial membrane?

A) Oxygen no longer reduced to water.
B) No ATP would be made during transport of electrons down the electron transport chain.
C) Glycolysis will stop.
D) Mitochondria will switch from glycolysis to fermentation.

[hide]I was thinking B.[/hide]

2.  In some mammals, such as new born humans and hibernating animals, body temperature is raised by means of

A) the uncoupling of respiration by the protein thermogenin
B) an increase in rate of glycolysis
C) shivering
D) leakage of hydrogen ions across cell's plasma membrane
E) cytochrone reductase

[hide]I think C[/hide]

3.  When the supply of acteyl CoA being produced exceeds the demands of the citric acid cycle, some of the acetyl CoA is diverted to the synthesis of

A) pyruvate
B) NAD
C) proteins
D) fatty acids
E) lactic acid

4.  The citric acid cycle begins with

A) pyruvate
B) acetyl CoA

[hide]I was thinking A[/hide]

5.  Isocitrate dehydrogenase is located in the

A) mitochondrial matrix
B) cytoplasm
C) plasma membrane

[hide]I think A[/hide]

6.  Isocitrate dehydrogenase is stimulated by

A) H+
B) heat
C) oxygen
D) ADP
E) none of the above

[hide]number 1 - I believe DNP is a drug that used to be used for weight loss. It destroys part of the mitochondria, causing a lack of a significant concentration of H+ ions on one side. Since there isn't much diffusion, the cell has to spend a ton of energy to make ATP from the small amount of concentration difference. The result is hyperthermia and dangerous weight loss. As for the answers, B sounds good?[/hide]


advanced fields
advanced fields unsolved

Prove that every countably compact space (every countable open cover has a finite subcover) is weakly countably compact (=limit point compact) and that the converse is true in $ \mathrm{T}_1$ spaces.

Countably compact => weakly countably compact:

It suffices to show that every countably infinite subset has a limitpoint.
Suppose there exists a countably infinite subset $ S$ that does not have a limitpoint.
Then for every point not in S, there exists a neighborhood of it that does not intersect S.
This implies that S is closed, hence countably compact as well.
S doesn't have a limitpoint also implies that each point in S has an open neighborhood that intersects S only at itself.
These open neighborhood forms a countable open cover of S that does not have a finite subcover.

weakly countably compact + T1 => countably compact:
Let us prove the countably compactness via the finite intersection property, i.e.
If $ A_1, A_2,...$ are closed subsets such that the intersection of finitely many of them is nonempty, then $ \cap A_i$ is nonempty.
Without loss of generality, assume that $ A_i$ contains $ A_{i\plus{}1}$ for $ i \equal{} 1,2,..$ (else we may replace $ A_i$ by $ A_1 \cap ... \cap A_i$.
Pick $ x_i \in A_i \minus{} A_{i\plus{}1}$. These $ x_i$ would form a set $ X$.
1. If $ X$ is finite, it means that there is an element that occurs in infinitely many of $ A_i$, i.e. all of $ A_i$, so $ \cap A_i$ is nonempty.
2. If $ X$ is infinite, by weakly countably compactness there is a limitpoint $ x$ of $ X$. We will show that $ x$ is not in $ X$, then as $ \{x_i,x_{i\plus{}1},...\} \subset A_i$, we see that $ x$ is a limitpoint of each $ A_i$, hence in $ A_i$ as it is closed. Thus $ x \in A_i$ for any $ i \equal{} 1, 2,...$ i.e. $ \cap A_i$ is nonempty.

($ x \in X$ cannot be true. Otherwise, let $ x \equal{} x_i \in A_i \minus{} A_{i\plus{}1}$, so $ x$ is not contained in $ A_k$ for any $ k > i$. Call the whole space Y. Now consider the neighborhood $ Y \minus{} A_{i\plus{}1} \minus{} \{x_1,...,x_{i\minus{}1}\}$. This is open because $ Y$ is $ T_1$ so that $ \{x_1,...,x_{i\minus{}1}\}$ is closed. Now this neighborhood of $ x_i$ intersects $ X$ only at $ x_i$, contradicting to the fact that $ x$ is a limitpoint of $ X$.)


algebra
polynomial
inequalities
calculus
integration
absolute value
algebra proposed

Let $f$ be a polynomial with integer coefficients such that $|f(x)|<1$ on an interval of length at least 4. Prove that  $f=0$.

I bet ( but I have unfortunately not much time to "expand" this post) that this problem is based on the Polya theorem, related with the polynomial Czebyszew's theorem.

I'm probably wrong, but I think the norm of a monic polynomial of degree $n$ on an interval $[a,b]$ is at least $2\left(\frac{b-a}4\right)^n$ (with equality when the polynomial equals some nasty expression in terms of the $n$'th Chebyshev polynomial and $a$ and $b$ which I do not have the patience to derive right now :)), so that $1$ is too much :?. It would have sufficed to have $<2$ instead of $<1$, and it looks like all we needed from the fact that the polynomial has integral coefficients was the fact that the leading coefficient has an absolute value $\ge 1$.

I'm sure I'm wrong though, so I apologize in advance :).

And then how do you explain that there are polynomials of norm strictly smaller than 1 defined on intervals of length no matter how close to 4? ;)

If $n$ is large enough, $2\left(\frac{b-a}4\right)^n$ can be as small as you want, no matter how close $b-a$ is to $4$, so that's not the problem :).

I think you are right, Grobber, with both assertions.


geometry
circumcircle
parallelogram
geometric transformation
reflection
Euler
symmetry

$ AA_{3}$ and $ BB_{3}$ are altitudes of acute-angled $ \triangle ABC$. Points $ A_{1}$ and $ B_{1}$ are second points of intersection lines $ AA_{3}$ and $ BB_{3}$ with circumcircle of $ \triangle ABC$ respectively. $ A_{2}$ and $ B_{2}$ are points on $ BC$ and $ AC$ respectively. $ A_{1}A_{2}\parallel AC$, $ B_{1}B_{2}\parallel BC$. Point $ M$ is midpoint of $ A_{2}B_{2}$. $ \angle BCA \equal{} x$. Find $ \angle A_{3}MB_{3}$.

Let $ H \equiv AA_3A_1 \cap BB_3B_1$ be the triangle orthocenter, let the $ A_1A_2 \parallel AC,\ B_1B_2 \parallel BC$ meet meet at $ D \equiv A_1A_2 \cap \B_1B_2,$, and let these parallels meet the opposite altitudes $ BB_3B_1,\ AA_3A_1$ at $ F \equiv A_1A_2 \cap BB_1,\ E \equiv B_1B_2 \cap AA_1.$ The quadrilateral $ CA_2DB_2$ is a parallelogram with diagonal intersection $ M \equiv CD \cap A_2B_2.$ The angles $ \angle A_1EB_1 = \angle A_1FB_1$ are right, the quadrilateral $ A_1B_1EF$ with the diagonal intersection $ H \equiv A_1E \cap B_1F$ is cyclic, and $ HA_1 \cdot HD = HB_1 \cdot HE.$ The altitude feet $ A_3,\ B_3$ are midpoints of $ HA_1,\ HB_1 \Longrightarrow$ $ HA_3 \cdot HD = HB_3 \cdot HE$ and the quadrilateral $ A_3B_3DE$ is also cyclic, let $ \mathcal M$ be its circumcircle. Let $ U \in \mathcal M \cap A_1A_2,\ V \in \mathcal M \cap B_1B_2$ be the other ends of diameters spanning the right angles $ \angle A_1EB_1 \equiv \angle A_3EU = 90^\circ,$ $ \angle A_1FB_1 \equiv \angle B_3FV = 90^\circ.$ The quadrilateral $ A_3B_3UV$ is a parallelogram, its diagonals cutting each other in half in the center of the circle $ \mathcal M.$ This parallelogram is inscribed in the parallelogram $ CA_2DB_2,$ hence they have the same diagonal intersection M, identical with center of the circle $ \mathcal M.$ The $ \triangle A_1FH,\ \triangle B_1EH$ are right-angled, the altitude feet $ A_3,\ B_3$ being midpoints of their hypotenuses $ HA_1,\ HB_1.$ As a result,

$ \frac {\angle A_3MB_3}{2} = \angle A_3EB_3 \equiv \angle HEB_3 = 90^\circ - \angle HB_1B_2 =$
$ = 90^\circ - \angle A_1AC = \angle BCA = x$

$ \angle A_3MB_3 = 2x$

Nice problem.
Let $ H$ be orthocenter of $ ABC$ and $ X$,$ Y$ are points on $ BC,CA$ such that $ HX\parallel CA, HY\parallel BC$. Then $ HX\parallel CA\parallel A_{1}A_{2}$ and since $ H$ is orthocenter of $ ABC$, $ A_{3}$ is midpoint of $ HA_{1}$. Then $ HA_{2}A_{1}X$ is parallelogram. So $ A_{3}$ is midpoint of $ A_{2}X$. Since $ M,X$ is midpoint of $ A_{2}B_{2},A_{2}X$,  $ MA_{3}\parallel B_{2}X$. Since  $ HA_{2}A_{1}X, HXCY$ is parallelogram, $ A_{1}A{2}YC$ is parallelogram and similarly $ B_{1}B_{2}XC$ is parallelogram. Hence $ B_{2}X\parallel B_{1}C$. So $ MA_{3}\parallel B_{1}C$ and similarly $ MB_{3}\parallel CA_{1}$. Therefore, $ \angle A_{3}MB_{3} \equal{} \angle A_{1}CB_{1} \equal{} 2\angle C \equal{} 2x$

[quote="armpist"]Just  take the circumcircle of A3B3C and reflect in over A3B3 and follow

     where it's center lands.

    Thank you.
    M.T.[/quote]

Not a proof.

[color=red]Edited by moderator.[/color]

I propose a conjugate solution by reflecting midpoint of the segment A2B2 in the side A3B3 of the

   orthic triangle. It will surely end up in one of the three Euler centers, specifically

   the midpoint Ec of segment CH, which is the center of circle A3B3C, and this is wellknown.

  And central angle as we all know is exactly two times larger then the inscribed angle

  in the same circle, or (as Darij likes to say) in other words  <Ec = 2<C

 (which looks like it does not  have any  words in it at all).

  I hope this is better, not as concise as the earlier one.

  Thank you.
  M.T.

Let  $A_2', B_2'$ lie on the sides $BC$, $AC$  and  $A_2'A_3=A_3A_2$,  $B_2'B_3=B_3B_2$. ($A_2'\not= A_2 $ and $B_2'\not= B_2$)
We have $HA_2'A_1A_2$ and  $HB_2'B_1B_2$ are rombus.
So, $MA_3\| B_2A_2'$  and $MB_3\| A_2B_2'$. Let  $A_2B_2'\cap B_2A_2' = M'$.
Hence, $ \angle A_3MB_3 = \angle A_2'M'B_2' $.
We have  $HB_2CA_2'$ and  $HA_2CB_2'$  are  cyclic quadriterals. Then,  point $H$ is  Mikel's point  of the lines
 $B_2C, B_2A_2', A_2B_2', A_2C$.
Hence, $ A_3MB_3=A_2'M'B_2'=180^\circ  -2\angle B_2B_1B_3=2x $.

we know that $H$(orthocenter of $ABC$) is symmetric to $A_{1},B_{1}$ wrt $BC,AC$ so we have $\angle HB_{2}B_{3}=\angle HA_{2}A_{3}=x$ giving $HA_{2}A_{3}\sim HB_{2}B_{3}$ now reflect $A_{3}$ over $M$ at $N$. since $NB_{2}/B_{2}B_{3}=HB_{3}/HA_{3}$ and $\angle NB_{2}B_{3}=180-x$ i.e. $B_{3}B_{2}N\sim B_{3}HA_{3}$ and by spiral symmetry $B_{3}NA_{3}\sim B_{3}B_{2}H$ so(since $M$ is the midpoint of $A_{3}N$) $\angle A_{3}MB_{3}=2*\angle A_{3}NB_{3}=2*x$

Fact: Let T be intersection of lines B1B2 and A1A2 then,Point M is NİNEpoint-center of triangle TA1B1


HMMT
function
algebra
polynomial
binomial coefficients
binomial theorem

How is an expression like $ \binom {\minus{}4}{19}$ evaluated? In HMMT 2008 Combinatorics, they gave a generating functions solution to problem 10 and it said that $ \binom {\minus{}4}{19} \equal{} (\minus{}1)^{19} \cdot \binom {22}{19}$. How did they get that and how are negative binomial coefficients evaluated in general?

For a fixed positive integer $ k$, $ f(n) \equal{} \binom{n}{k}$ is just a polynomial in $ n$ and so can be evaluated at any real number.  For this polynomial, it's a simple matter to verify that $ f(\minus{}n) \equal{} (\minus{}1)^k f(n \plus{} k \minus{} 1)$.  The deeper fact about these polynomials is that they make the Binomial Theorem true for non-positive integer exponents.


we got numbers $ n$ and $ k$...
we consider binary numbers $ a_1,a_2,...$ in increasing order, which has $ n$ digits of $ 1$ and $ n$ digits of $ 0$
$ 0101$ is also a number, I mean $ 0$ is meaningful...
we have to find $ a_k$....
$ n < 200$
$ k$ has less then 150 digits...

The solution involves dynamic programming.
The dp state space is dparray(LEN,NUM) = dparray(LEN-1,NUM-1)+dparray(LEN-1,NUM)
each state counts the number of possibilities for LEN and NUM
note that this is essentially pascal's triangle
LEN is the current length of the sequence, and NUM is the current number of 0s.
You add up the two previous cases, where the new digit is either a 0 or a 1.
Now to find the final answer, you take the input value K, and work backwards, determining whether the next digit should be a 1 or a 0, using a kind of "reverse counting" algo.

That is the simplified algo, but you also have to take in account the fact that K can go to 400 choose 200, which is about 120 digits, so you have to implement a bignum object that you can implement an add function for

EDIT: i think you can use a combinatoric choose function to avoid dp, thus speeding the program up, but that involves division, which is annoying to implement in bignum


I proposed this exercise in the pre-olympiad section two days ago... :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=46762[/url]
 :)


There's a nice little plugin that lets you type latex formula in AIM windows just like the virtual classroom!!!! :lol:

Cool!

Are you talking about the same thing as this: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49746 ?

Joml88 said that you knew how to get it to work. :) But I dont' know how, and I would really like to be able to do that.

[quote="DPopov"]Yes, I am talking about the same thing. :lol:[/quote]

So, could you maybe tell how to use it?
Maybe? Thanks. :P

Take a look at my blog, I have given a link to the instruction page there.

:(  It still won't work.

Am I supposed to get an error msg when I enable it?  It tells me that I have an error w/toolchain not installed.  And then I get a chance to specify paths, but the next time I start Gaim it will disappear, and I have to enable it again, which gives me the same error.

I specified these paths:
C:\texmf\miktex\bin
C:\Program Files\ImageMagick-6.2.4-Q16

Are they right?

Those paths look right to me. After you specify them click on the plugin checkbox again and you should get a little watch to appear and after a few seconds the plugin should be enabled.

Thanks.
It still doesn't work.  I guess it's something else then.  I don't know.
I emailed the author, don't know if he'll reply though. Ideas?

Hmm, it should work, well it works for me at least but I also had to email him about the paths. I'll see what I can come up with and get back to you.

Hey DPopov, I got it to work!  :)
I reinstalled the ghostscript stuff and it worked, not sure why.
Thanks for your help!

For anyone else who wants this, the paths i used are posted above.

Edit-I tried with someone while directly connected and it didn't seem to work.  Not sure.

They must have it installed as well for it to work. My GAIM/AIM is Mitko101 if you want to test it out.

Question, do other users have to have Latex for them to see the latex stuff?

Another question: Does it only work for GAIM and aMSN as it says on the site?


We are now accepting enrollments for the Special AIME Problem Seminar.  This 6-hour course is designed for students preparing for the AIME.  The course consists of 2 three hour classes in which we tackle AIME-style problems and discuss test taking strategies. There will be a 30 minute break during each class, so there is a total of 5 hours of instruction.  Transcripts for the classes will be available to enrolled students, but there will be no message board for this special seminar course.  Both instructors of the course achieved perfect AIME scores.

Can we enroll after we get our results back or do we just have to assume we do make it?

The enrollment deadline is March 19, which is 4 days before the AIME.  You should have your results by then.

Yes - we will be cutting off enrollment at some point.  I will post here if the class gets near the enrollment limit.

I just saw the times on the classes. WOuld it be possible to make them a little later? Most people (CST and PST) haven't gotten out of church/are eating lunch at that time on Sunday. Perhaps an evening class would be better. I'm sure the AMC/AIME Problem Series class would consider rescheduling... :-)

Unfortunately, we can't move the AMC/AIME Problem Series.


Find $ \sqrt{3}$ with 20 numerals behind comma.

This can be achieved through the use of finding a square root by hand. And you did mean decimal point instead of comma right?  :wink:

in some countries  they use comma instead of a decimal point...

This can be achieved by what is known as the Babylonian square root approximation. At least, that's what I think it's called. The concept is that you take an initial guess at the square root (I suggest two) and divide 3 by your guess.

If your guess [i]is[/i] the square root, then quotient will be equivalent to your guess.
To hone in on the value, take the average between your guess and the quotient, then try again, and again....Do it 10 times, then get your fraction back into decimal format. That will probably constitute 20 digits of accuracy.

 :wink:

[hide="ORrrrrr"]
[url]http://en.wikipedia.org/wiki/Square_root_of_3[/url]
[/hide]
 :ninja:

OR YOU CAN USE NINJA POWERS AND STUFF AND BE HAPPY AND USE PROTON CANNONS AND DO NINJA AND BE HAPPY AND GO WHEEEEEEEE AND THEN SOLVE IT AND THEN BE MORE HAPPY AND IM ON DRUGS HELP ME


I'm new at tis forum and I need your help...

Here is the task:

You have a chessboard (with dimensions n x n) that is colored the same as the chess board. You can change color of one field - from black to white or white to black - and in the same time you change the colors of the fields that surround it. Is it possible to change the initial state of the board to a state where all fields are blank?

any ideas? Is it clear to you what is the problem about?


function calculus integration algebra polynomial

A cubic function f(x) with integral coefficients has the following properties f(3/2) = 0, (x-2) is a factor of f(x), and f(4) = 50. Determine f(x).

[hide]
Since $f(\frac{3}{2}) = 0$, we know that $\frac{3}{2}$ is a root of the polynomial, as well as $2$.

So we have $(2x - 3)(x-2)$ so far.

$f(4)$ must equal $50$; when we put in $4$ for the value of $x$ into the expression as it is right now, it equals $10$ So the other factor must equal $5$ when $x$ equals $4$. But there are multiple ways to do this: $(9-x)$, $(x+1)$...

Each of the corresponding polynomials fit the description:

$2x^3 - 5x^2 - x +6$
$-2x^3 + 25x^2 - 69x + 54$
[/hide]


I just signed up for Olympiad Problem Solving today, and it seems that I have access to the existing Olympiad Problem Solving forum (the one used for the previous class).  I judge this from the fact that there are already about 130 existing topics in that forum.  Is this a bug?  I'm guessing I shouldn't have access to the forum from a previous class...

Before I forget, about how much longer will the board and files for that class stay up? ( :blush: I'm not quite done saving it all.)


beautiful problem. Note my proof is not complete. :blush:

It only makes sense if all $a_{i}\ge0$. Otherwise put $a_{i+2}: =a_{i}-a_{i+1}$ and start with $a_{1}=N$ for odd $n$, or $a_{1}=-N$ for even $n$, and there is no maximum as $\sum_{i=0}^{n}a_{i}$ exceeds $N$.

So let's assume $a_{i}\ge0$. From $a_{i+2}\le a_{i}-a_{i+1}$ we see that the $a_{i}$ must be decreasing for $i=0,...,n-1$ (!), and we get a telescoping sum $\sum_{i=0}^{n}a_{i}\le a_{0}+a_{1}+a_{0}-a_{n-1}=2+a_{1}-a_{n-1}$.

Let $f_{i}$ be the $i$th Fibonacci number ($f_{0}=f_{1}=1$).
Choose $a_{1}: =a: =\frac{f_{n-3}}{f_{n-2}}$ and put $a_{i+2}: =a_{i}-a_{i+1}$. Then it is easy to see by the Fibonacci recursion that $a_{i}=(-1)^{i}(f_{i-2}-f_{i-1}a)$ and thus $a_{n-1}=0$. And $\sum_{i=0}^{n}a_{i}=2+a=\frac{f_{n}}{f_{n-2}}$. I'm pretty sure this is the maximal possible sum. (Note this tends towards $\frac{3+\sqrt{5}}2$ as $n\to\infty$.)

Starting with a bigger $a_{1}$, some of the $a_{i}$ must be reduced to avoid negative numbers. It remains to show that their overall sum cannot exceed the one above. Maybe it's very simple but right now I just don't see how to show that. :roll:


a ,b and c  in $ R$   such that a>0

let  P(x)= ax^2+bx+c

we suppose  M=b^2-4ac  and  b>=  1/8a .
prove  P(M) >= 0.

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


Let $a,k \in \mathbb{N}\cup \{ 0 \}$
$k \leq \frac{a^{2}}{n}< k+1$
$\Leftrightarrow k \cdot n \leq a^{2}< (k+1) \cdot n$
$\Leftrightarrow \sqrt{k \cdot n}\leq a < \sqrt{(k+1) \cdot n}$
Hence $\sqrt{k \cdot n}\leq a < \sqrt{(k+1) \cdot n}\Leftrightarrow [ \frac{a^{2}}{n}] = k$. Therefore,
$S_{2}\geq \sum_{k=0}^{n-2}k \cdot ([ \sqrt{(k+1)n}]-[ \sqrt{kn}] )$
$= (n-2) \cdot [\sqrt{(n-1)n}]-\sum_{k=0}^{n-2}[ \sqrt{kn}]$
$= (n-1) \cdot [\sqrt{(n-1)n}]-S_{1}= (n-1)^{2}-S_{2}$
q.e.d.

Number N(a) of integers a, suth that $\sqrt{kn}\le a<\sqrt{(k+1)n}$ is not equal $[\sqrt{(k+1)n}]-[\sqrt{kn}]$. For example n=8,k=2 N(a)=1.
However, $N(a)\ge [\sqrt{(k+1)n}]-[\sqrt{kn}]$. It give this result.

[quote="Rust"]
However, $N(a)\ge [\sqrt{(k+1)n}]-[\sqrt{kn}]$. It give this result.[/quote]
That's wrong: For example: $n=8, k=1: N(a) = 1 < [\sqrt{16}]-[\sqrt{8}]$.
$[\sqrt{(k+1)n}]-[\sqrt{kn}]$ is equal to the number of intergers $a$ which fulfill $\sqrt{kn}< a \leq \sqrt{(k+1)n}$. So some  summands of the $LHS$ of $S_{2}\geq \sum_{k=0}^{n-2}k \cdot ([\sqrt{(k+1)n}]-[\sqrt{kn}])$ became reduced by $1$ at the $RHS$.


geometry
function
geometric transformation
geometry unsolved

The squares $ ABCD$ and $ A'B'C'D'$ represent maps of the same region, drawn to different scales and superimposed. Prove that there is only one point $ O$ on the small map which lies directly over point $ O'$ of the large map such that $ O$ and $ O'$ represent the same point of the country.

Note: I'd like to try to solve this problem myself so could anyone give a hint first?

Central similarity (if $ ABCD, A'B'C'D'$ have the corresponding sides parallel) or spiral similarity (if not) takes $ A'B'C'D'$ into $ ABCD.$

Place the maps in the complex plane and define a function that transforms one of the maps into the other map.  Clearly, you want to find a (in fact, [i]the[/i]) fixed point of the function.


can someone please solve 3^(x^2) = 6^(x+4) without a calculatopr

(sorry I don't know how to use LaTeX)


inequalities
trigonometry
inequalities proposed

[color=blue][b] Problem. [/b]
Prove that in any triangle $ABC$, the followng inequality holds:
\[ (1-\cos A)(1-\cos B)(1-\cos C)\geq \cos A\cos B\cos C. \]

[/color]

Sorry, I'll edit it. ;)  I was drunk when I was writing . :blush:


What is the remainder when $ 2000!$ is divided by 2003.

Applying Wilson's Theorem, $ (p\minus{}1)! \equiv \minus{}1 \pmod{p}$ for $ 2003\equal{}p$, we have
$ 2002! \equiv \minus{}1 \pmod{2003}$
$ 2000!\times2001\times2002 \equiv \minus{}1 \pmod{2003}$
$ 2000!\times\minus{}2\times\minus{}1 \equiv \minus{}1 \pmod{2003}$

Can I multiply both sides by a negative integer?

$ 2000!\times2\equiv \minus{}1 \pmod{2003}$
$ 2000!\times2\equiv 2002 \pmod{2003}$
$ 2000!\equiv 1001 \pmod{2003}$\

So... is that it?


USAMTS
geometry
email
summer program
Mathcamp
Science Olympiad

I took this today.  I would also like to discuss.

Edit:  I think it ends on the 24th or something like that.

Be sure before you start discussion national competitions (which I suppose USNCO = US National Chemistry Olympiad stands for) that all the students have already taken the competition. If not for moral reasons, just think that you will help others score more than you have, thus lowering your ranking :D :D

Can anyone tell me what USNCO (other than what it stands for) is?

I'll be taking Chemistry and I'm actually quite excited (science + math :-D) so if I can, I want to attend competition of it.

If that's possible. ;)

i need to take the USNCO next year...hopefully i can get my school involved.

USNCO = United States National Chemistry Olympiad.  It's a 4.75-hour chemistry test used to decide who gets to go to the Chemistry Olympiad "Study Camp," basically MOP for chemistry (ChOP?).  The material is nothing that you won't learn in AP Chem, but you'll definitely need AP Chem to do well.

There are 3 sections, each around 1.5 hours long.  The first section is 60 multiple choice and is somewhat comparable to the multiple choice on the AP test, probably a bit harder.  The second part is similar to free-response on the AP, only much harder and with more problems(8).  The third part is a fun lab practical with two experiments.  For more info, see here:

http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education%5Cstudent%5Colympiad.html

I think we want the National test.

[quote="xxreddevilzxx"]is it administered by school???[/quote]

Your local ACS section does it.

that was waay too hard! much more demanding than the website made it sound... i did soo bad :(

I think I did quite well- I was tutored by Scott Rabin, who happens to go to my school, so I think I might have a shot at camp.  Now, to get into MOP and have to choose...

We shouldn't talk about this because I think there are still people taking the test.

why can't discuss a contest while other people are still taking it??
Is that against the US laws?

No, I think it's legal.  But for tests like the AMC, we don't talk about it until the period for taking it is over, to avoid giving information to people who haven't taken the test yet.  Even general stuff, like the difficulty.

I don't think the above made sense.

Inteluser!!! Do you happen to go to school with Scott? I know of him through Ben Au who I met online, lol.

Yep, just took the USNCO yesterday. I didn't think it was too bad; but one of the parts was a little more mind-blowing than usual.

Shooting for High Honors. What about everyone else?

please don't discuss at all (even something as seemingly innocuous as partially mindblowing or something) until we can

who is scott? is he the reid barton of chemistry or something?

I don't know Reid Barton; USNCO's my thing, not the AMCs. I never got past the AIME. Scott's one of the guys who went to camp the past two years; he's probably going to go to IChO this year. There's also Caleb from Libertyville, IL and Allen from Arcadia, CA who are probably going to camp and IChO.

oops, i meant dragonslayer, not hodge conjecture

lol... multiple choice... probably/maybe... but when i always think that on the USNCO exam... i always a miss a few here or there for some reason or another.  Umm... I think I did ok on free response... not full score though definitely.  And the lab... omg.. don't even go there... lol.

well, you got in, so you must have done well. how did your friend scott do?

They should have contacted you, if you made it.  I got mine by email from my local section coordinator.  I don't think they have a central list posted.

[quote="inteluser"]Hmm.... they're out?  I guess I must not have made it, since I didn't hear about it.

Where are they?[/quote]
My section coordinator said that the general results should be available sometime in mid-May.

Scott made the international team.  Just to let you know.

Scott made the US team.  Just to let you know.

How on earth did I get famous?  It took me 3 years to make the team, I'm definitely not the Reid Barton of Chemistry.  But yes, I did make the team and will be going to Taiwan in about a month.

For everyone interested in how they grade the national exam, lets run through the basics.

1) If you're not making above a 55/60 on the Multiple Choice, you may have some problems qualifying
2) If you're answering none of the higher-level (d through g) questions in the Free Response, you may have some problems qualifying
3) If you're absolutely terrible at labs, you'll have NO PROBLEM getting in.

As inteluser mentioned, our CaCl2 in H2O evolved a gas, which was definitely not right.  For the quantity I added to the water, there wouldn't have been enough heat released to vaporize any water and generate bubbles that way, but whatever.  Also, there was supposedly gas evolution when the two reagents were mixed, but no one in my section observed anything.  The qual scheme lab (identification of liquids) should have been relatively easy based on general knowledge of organic nomenclature and polarity/miscibility.

For everyone interested in going to the camp next year, I STRONGLY recommend taking all previous tests that are posted online.  That's how I prepared my first year, and how I helped inteluser prepare (who qualified but chose MOP over USNCO :P).  Also, if you think you have an exceptionally good chance at making it in, I recommend taking an organic class and lab at a local college.  That will prepare you for the absolute MINDJOB that is the organic lectures at camp.

Anyway, just thought I should chime in, since my name seems to appear a LOT in this post.  The other international participants (also listed on the website) are myself, Nick Sofroniew, Jake Sanders, and Allen Cheng; 1st alternate Andrew Freddo, 2nd alternate Michael Blaisse.

-Scott R.

that's cool, good luck you guys in Taiwan!!!  :clap:

You've always been famous to me, scotty boy

You too, N.D.

I'm still having problems convincing some friends it exists.  They say you lied to me.  :D

-Scott R.

you remember isaac shomer from two years ago?  hes up here on an REU... cool guy, showed me around his lab.  my parents (who don't hate you after all, well too much anyway) suggested that you should get an REU up here too, to prove to yourself our existance :).

do your parents really not like me? it was meant to be a joke. anyway, whatever an REU is, I don't think it'll pay enough to warrant a summer.

How's Isaac doin' for ya though? He probably doesn't remember me, but tell him I said hi anyway.

yes, they actually do like you, although you seem to have trouble believing that for some reason.  isaac actually does remember you.  i only met with him one day and went around his lab, cuz now im at a mathcamp in oregon, otherwise i would tell him that


inequalities geometry rectangle perimeter LaTeX

Well I was playing around with inequalities and decided to make some inequality questions:

$ 1)$ Under what conditions is the area of a rectangle the greatest?

$ 2)$ Show that if $ x_1\cdot x_2\cdots x_n\equal{}1$, then $ x_1\left(1\minus{}\dfrac{1}{x_1}\right)\plus{}x_2\left(1\minus{}\dfrac{1}{x_2}\right)\plus{}\cdots\plus{}x_n\left(1\minus{}\dfrac{1}{x_n}\right)\ge 0$

When is there equality?

For the next problems, $ x_1,x_2,\cdots,x_n > 0$
$ 3)$ Show that $ \left(\dfrac{x_1^2\plus{}1}{x_1}\right)\cdot \left(\dfrac{x_2^2\plus{}1}{x_2}\right)\cdots \cdot \left(\dfrac{x_n^2\plus{}1}{x_n}\right) > 1\plus{}2\plus{}4\plus{}8\plus{}\cdots \plus{}2^{n\minus{}1}$

$ 4)$ Prove that $ \dfrac{(x_1^2\plus{}1)\cdot (x_2^2\plus{}4)\cdot (x_3^2\plus{}9)\cdots \cdot (x_n^2\plus{}n^2)}{x_1\cdot x_2\cdots x_n}\ge 2^n\cdot n!$

When is there equality?

[hide="hints"]
[hide="hint 1"]
Assign variables to the sides of a rectangle and the sides of a square with the same perimiter, also use the AM-GM inequality.
[/hide]
[hide="hint 2,3,4"]
AM-GM inequality really helps.
[/hide]

[hide="hint 3,4"]
Try to seperate the products into different products which contain only the same variable.
[/hide]
[/hide]

Notes:
If you believe that the problems might have a typo or don't make sense at all just write it here, or pm me, it's my first time making inequality problems so there might be errors.

Any solutions that you write "hide" them, so that others can try the questions out.

For number 1, the greatest possible area of a rectangle occurs when it is a square, since a square is a rectangle.

Good now all your missing is to prove it.

Let $ x$ be the side of the square since we know that this creates the greatest possible area.

The perimeter of this square is $ 4x$ and the area would then be equal to $ x^2$

If we would take a rectangle with a length of $ x$ and a width of $ x\minus{}1$, we would see that the perimeter would be $ 4x\minus{}2$ and the area would be $ x^2\minus{}x$, smaller than the maximum.

Also, if we would make the width of the rectangle closer to x, the area of the figure would come close to but never reach $ x^2$

I'm not very fluent in LaTeX, so I'll write it out:
 Let $ x$ be the length of a rectangle and $ n$ be the width

As $ n$ approaches $ x$, the area approaches $ nx\equal{}x^2$ thus making the maximum area possible only when the sides of the rectangle are equal, i.e. a square

Uh, is this enough of a proof, not very pretty...proofs aren't my strong thing

For number 2,

[hide]

By AM-GM, $ x_1\plus{}\cdots \plus{} x_n \ge n$, which is a direct consequence of expanding the equation. Equality occurs when $ x_1, x_2, \cdots , x_n \equal{}1$

[/hide]

For 3,

[hide]

We have $ (x_1 \plus{}\frac{1}{x_1})(x_2 \plus{}\frac{1}{x_2})\cdots (x_n \plus{}\frac{1}{x_n})> 2^n\minus{}1$. By AM-GM, $ x_1 \plus{} \frac{1}{x_1} \ge 2$, so substitution gives $ 2^n>2^n\minus{}1$, which is clearly true.

[/hide]

#1 [hide]You have two sides L and W for a rectangle and side X for a square. Both must have the same perimeter so:

$ 4X \equal{} 2(L\plus{}W)$
$ 2X \equal{} L\plus{}W$
$ X\equal{}\frac{L\plus{}W}{2}$
Now, stating the arithmetic and geometric mean inequality we get:

(i) $ \frac{L\plus{}W}{2} > \sqrt{LW}$

We want to look for a contradiction, so lets say that $ LW>X^2$, then plug in X:

$ LW>X^2$
$ LW>(\frac{L\plus{}W}{2})^2$
$ \sqrt{LW}>\sqrt{(\frac{L\plus{}W}{2})^2}$
$ \sqrt{LW}>\frac{L\plus{}W}{2}$

Contradiction, the geometric mean is NOT greater than the arithmetic mean.[/hide]

at: rd5493, here's a hint to make a better concrete proof, use the AM-GM inequality

Here's the solution I have to #1:
[hide="Solution"]
Let $ a$ be the side of the square, and $ b,c$ are the sides of the rectangle whose sides are distinct. Both the square and rectangle have the same perimeter.

We know that $ 4a \equal{} 2(b \plus{} c)$, which means that $ a \equal{} \dfrac{b \plus{} c}{2}$.

We also know by AM-GM that $ \dfrac{b \plus{} c}{2} > \sqrt {bc}$ (also $ b\not \equal{} c$), substituting $ a$ for $ \dfrac{b \plus{} c}{2}$ you get:

$ a > \sqrt {bc}\rightarrow a^2 > bc$

This concludes the Solution, since $ a^2$ is the area of the square and $ bc$ is the area of the rectangle.

Looking at modularmarc101's(Marcos) solution, my solution seems to be the same, except that he complicated his solution with no need to.
[/hide]

To Lifeisacircle: Those solutions are correct. If you want a hint to number 4 here it is: [hide="hint"]Not much of a hint, but number 4 is extreamly similar to number 3[/hide]

[quote="tonypr"]
$ 1)$ Under what conditions is the area of a rectangle the greatest?
.[/quote]

When the sides are as big as possible.

Perhaps being a bit clearer would be helpful. For every x by x square, there is a 2x by x rectangle that is bigger. Specify that there is a constant perimeter. When you make problems, amiguity like this is not good.

Well, for 4,

[hide]

The LHS is $ (x_1 \plus{} \frac {1}{x_1})(x_2 \plus{} \frac {4}{x_2})\cdots (x_n \plus{} \frac {n^2}{x_n})$. By AM-GM, we have $ x_n \plus{} \frac {n^2}{x_n}\ge 2n$, and since $ 2(1)\times 2(2)\times 2(3)\times \cdots \times 2(n) \equal{} 2^n \times n!$, we have

$ \frac {(x_{1}^{2} \plus{} 1)\cdot (x_{2}^{2} \plus{} 4)\cdot (x_{3}^{2} \plus{} 9)\cdots\cdot (x_{n}^{2} \plus{} n^{2})}{x_{1}\cdot x_{2}\cdots x_{n}}\ge 2^{n}\cdot n!$

Equality occurs when $ x_n\equal{}n$, because $ x_n$ would equal $ \frac{n^2}{x_n}$

[/hide]

Woops, edited in equality case.

But the sides can be as big as you want, the point is that the condition that the rectangle has to meet to have the greatest area is that all the sides are the same.

I probably could have worded like this: Under what conditions does a rectangle with sides $ b,c$ have the greatest area compared to other rectangles with the same perimeter.

And to lifeisacircle:

That is correct.

Now when is there equality?

Here is the nicest proof that a square has greatest area:
Suppose the perimeter is 4x. Then
$ (x\plus{}n)(x\minus{}n)\equal{}x^2\minus{}n^2 \le x^2$
Equality when n=0!


factorial
search
calculus
calculus computations

Today I took my Algebra II exam, and one of the problems asked for this:

$\sum^{\infty}_{n=0}\displaystyle\left[\frac{(-1)^n}{(2n+1)!}\cdot\displaystyle\left(\frac{\pi}{3}\right)^{2n+1}\right]$

I'm absolutely clueless as how to do this. The factorial is really what messes it up; nothing I do can simplify that part of the expression. Any help?

well, this is just the Taylor series for $sin(\frac{pi}3)$, which is just $\frac{sqrt(3)}2$

Wow, just did a search about Taylor series, and saw that you're right.

Now I want to know why our teacher is putting Taylor Series on our Algebra II finals. :(

lol my algebra II final consisted of problems like: factor $2x^2-6x+4$.

When I took Algebra II, we did a bit of Taylor series, but not into so much depth. It was the honors course, and I don't think it was covered in the regular classes.
Unless you were told to memorize it, there's no way to do it, without a decent understanding of calculus.

Yeah...what the heck was that doing on an Algebra II test?  I'll move this to one of the Calculus forums.

yeah this one would probably require me to do the series tests and then see that it went to $\frac{\sqrt{3}}{2}$. no way this is alg 2 stuff


function
logarithms
inequalities
calculus
calculus computations

Prove that e^pi > pi^e

Can i have some more examples of this type?

Consider the function $ f(x)\equal{}\frac{\log{x}}{x}$.  We have $ f'(x)\equal{}\minus{}\frac{\log{x}}{x^{2}}\plus{}\frac{1}{x^{2}}\equal{}0\Rightarrow\log{x}\equal{}1\Rightarrow x\equal{}e$.  This is a maximum so $ f$ has its maximum at $ x\equal{}e$.  Then $ \frac{\log{e}}{e}>\frac{\log{\pi}}{\pi}$, so $ \pi\log{e}>e\log{\pi}\Rightarrow e^{\pi}>\pi^{e}$.

Lets define
$ f\left( x \right) \equal{} \frac{{\ln x}}{x},x \in \left[ {e,\pi } \right]$
we have
$ f'\left( x \right) \equal{} \frac{{1 \minus{} \ln x}}{{{x^2}}} < 0$
so f is strictly decreasing
$ e < \pi  \Rightarrow f\left( e \right) > f\left( \pi  \right) \Rightarrow ...$

Another example, try to prove $ \pi {e^{e \minus{} 1}} < {e^\pi }$

You can use the known inequality $ e^x\geq 1\plus{}x\ (x\geq 0)$.

can anyone recommend me some books where i can find such type of problems?