AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
analytic geometry
graphing lines

How many of you guys take the AMC and AIME without preparation and can make the USAMO?
And for people who don't prepare for the USAMO, what do you guys get for a score?
Obviously this isn't a great place for this considering if you've heard of AoPS you've probably done some preparation.
But yeah there's nowhere else.

If they target mathematical interest, wouldn't the USAMO just scare most non-preppers away from math?

Also, is it too difficult to return your papers and give you like a sentence on where your proofs went wrong?

I want to know why my #4 was wrong (I DONT DOUBT IT WAS WRONG, AT ALL).   Also, is "pairwise disjoint" very common terminology?

well...obviously we cannot give any feedback without your proof

I prepared mostly for MathCounts and the AMC 12/AIME and did very little to none proof practice, and as a first timer I got a 10 on the USAMO.

[quote]How many of you guys take the AMC and AIME without preparation and can make the USAMO? [/quote]

I took the AMC and AIME without preparation and mad the USAMO. I did not prepare because I had never really pursued math outside of school. Also, I did not know that the USAMO even existed untill after I qualified for it. I didnt find out about this site untill after I qualified for the USAMO either.

[quote]If they target mathematical interest, wouldn't the USAMO just scare most non-preppers away from math? [/quote]

For me it did the exact opposite. I never really considered math to be anything more than just a subject in school that I was really good at. But after I made it to USAMO I started reading about math that is more difficult forms of math and became very interested. I started studying a week b4 the test and got a 9. Even though this isnt a terribly good score, it inspires me to work on my math skills over this year and hopefully come back next year and do much better.

This year I didn't prepare for anything and i got a 144 AMC 12, 6 AIMe, and 10 USAMO as a first timer.

But I'd like to know what you did on your proof so that we can help you on it.

Also, I wasn't sure what pairwise disjoint meant, so I didn't even try #3

pairwise disjoint is fairly common.  if you are talking about a bunch of sets, then saying that that bunch of sets is 'pairwise disjoint' means that if you take any two, they contain no members in common.  you see this crop up quite a bit...

i did not prepare for EDIT: AMC/AIME b/c i had made USAMO before - i was not too worried.  I DID PREPARE FOR USAMO (quite a lot, too)

@Whatthedeuce: come to the NC forum more.  we can give you tips on usamo preparation.  btw, gj.  9 for the first year is not bad at all!

I prepared like crazy for AMC and AIME - over Spring Break, the two weeks before the AIME, I was doing upwards of 4 hours a day. I didn't make it.

To answer your general question, I would say that the majority of people that make the USAMO have prepared in some way, and an even larger majority of the people who make (non-red) MOP have also prepared in some way.

so by the time we get to like MOP level, it's not really a math interest thing, but more of a who is smarter and prepared well thing right?

worthawhilebean, that sucks.  good luck next year, or in college, or whatever.

Thanks everyone for replying.

This is my #4, although it occurred to me that it might not have been flawed.  So here's my #1 in case it's wrong and #4 is right.  Or maybe I just got 3-4s on both... although apparently that's like impossible.

My #2 i proved that three 5 radius circles tangent to each other couldn't cover all points, but nothing else, so I'm guessing that was like a 1-2.  #5 i got to the nasty composite, so like 1-2.  And I had 13 total... help me figure out where i went wrong lol.

Anyways,

4.
   Note that all dinosaurs with less than 4014 cells are primitive.  Let's attempt to construct a primitive dinosaur with more than 4013 cells.  Let's define this dinosaur as one central cell with 4 animal "limbs" coming off each side of the cell.  These animals can have zero cells, and can be connected.
   If a number of limbs with more than 2006 total cells is connected, we can bridge 2007 cells (a dinosaur) together without the central cell, and ALL the other cells will connect through the central cell.  This forms a second animal, and if we have more than 4013 total cells, we have another dinosaur.
   Thus, connected limbs have at most 2006 total cells in order for a dinosaur with more than 4013 cells to be primitive.  The maximum limb length for primitive dinosaurs is then 2006.
   So, the largest primitive dinosaur has 4 unconnected limbs of 2006 cells, giving a total cell count of $2006*4+1=8025$

1.
   Let $k$ be the x-coordinate, and $\sum_{x=1}^{k}a_{x}$ be the y-coordinate in a plane.  $\sum_{x=1}^{k}a_{x}$ is always a multiple of $k$, and thus we are limited to traveling on an infinite number of rays $y=px$, where $p$ is a positive integer.
  The slope is $a_{k}$, and we wish to prove it eventually becomes constant.  Since $a_{k}$ is always an integer, we wish to prove that we eventually follow one of the rays.
  We start at $(1,n)$ and move accordingly:
1. Move 1 unit to the right.
2. Move up until we hit the first ray, unless we are already on a ray, in which case we stay.
3. Back to step 1.
   Since the rays have positive slopes, we can never move to a ray with a greater p value.  When we begin to move along the ray with the same p value, we are done, since $a_{k}$ will always get us to the same ray.  Thus, to not have constant slope, we must keep switching rays.  However, since $(1,n)$ is on ray $n=p$, it only has $n-1$ rays to its right.  Thus, $a_{k}$ must become constant.

Okay, now that I think of it, my #1 isn't as strong as my #4, since you could stay on a ray for an infinitely long time, but still switch later... Or maybe my #4 has tons of flaws I don't see.

Whatever, 13 is good enough for me lol.  I didn't think I'd be within 10 points of blue MOP.

If that was the extent of your rigor for #4, I would say that warrants 1 point (1 more if you showed the case with 8025 cells). Proofs similar to this one have been discussed all over the place if you want to read about holes or missed cases in a proof like that.

For #1, that's a workable argument as long as you somehow explained that a nonincreasing, bounded sequence of integers must become constant, which isn't particularly hard.

comments for your proof of #4
a) you have the central idea

b) your proof doesn't work because you cannot go backwards; not every array can be made with a center and 4 branches [possibly 0], consider a "loop," if it is a loop, then clearly you can partition it, but what if it is a loop and not a loop? there is plenty here that needs to be addressed

bb) of course, this is relatively easy to fix, however, it is a significant gap in logic

overall, the proof is fine, i had to do a bit of thinking for certain parts...perhaps you should be more explicit with each step instead of "note," i have no idea how grading works on the usamo [this is why i posted that other topic asking the amc for proof examples with scores] i don't quite know what type of deduction that mistake would recieve...but using my nonexpert opinion, i would say you recieved 3-4 [is 4 possible???]

comments for your proof of #1
a) your proof is vague..."slope" is ill defined... i think you mean like a scatterplot and the "slope" of each little segment in the scatter plot; be more explicit about this, i doubt you would loose points for this, but it is best to be clear

b) it would be nicer if you had a more rigirous statement of your particle's movement, this would solidify the connection with the actual problem

c) what do you mean only n-1 rays to the right? i don't see how this is true...let p' where 0<p'<p be a rational number...this obviously passes to the right of y=px
i am not sure why the line it passes must pass through a lattice point between (1,n) and (1,0)
how do you know that when you go to a new point, there are not other lines to the "right"? you do not define what the criteria for the slope...

overall, the whole point of the problem is the part that you addressed in the last paragraph of your proof...i cannot see how you proved anything at the end there...since most of the proof is the introduction of a special notation [that does not seem to add a new insight], you would probably recieve like a 1 or 2 [there are some ideas in here...perhaphs i am just not understanding your idea]

obviously these points don't add up...i guess i am wrong...but w/e those are my comments, take them for what they are worth...i would like to see your official score distribution

--Alex

[quote="bloodysloth"]so by the time we get to like MOP level, it's not really a math interest thing, but more of a who is smarter and prepared well thing right?[/quote]
I'm not as smart as virtually anyone here, and I often slacked off on USAMO preparation to do other random things like try to prove properties of Boolean algebra with a single axiom (I usually failed). Is that what you're referring to by math interest?

The biggest extent of my preparation was writing good, rigorous solutions to some problems from books and the Mathcamp quiz. That turned out to be amazingly useful this year.

bloodysloth, here's the problem with your #4:

What if one of the limbs had, say, 8000 cells, and the central cell + the remaining limbs altogether had 26 cells? Then when you divide the animal into the large limb and remaining part, the remaining part will have only 26 cells, so it's not a dinosaur.

The key is that you can't just choose any cell to be your central cell. You have to use an algorithm/the extremal principle to find one such that the central cell + any three limbs always has at least 2007 cells.

I think interest is far more important than natural talent. If you are smart but aren't genuinely interested in math, then no matter how much effort you decide to put into these competitions, it will probably be somewhat stale because you will fail to see the beauty in some of these solutions, and may be more tempted to go through the problems too quickly for the sake of "training" rather than sufficiently enjoying all of them and trying to solve difficult ones.

On the other hand, no matter how interested you are, if you never pursue that interest then not much can be said. I'll just say that I did basically no preparation for math competitions until my Sophomore year in high school, at which point I decided to prepare for the AIME, and spend a good deal of time on this endeavour. I would say that I probably spent more time "training" then than enjoying the math, primarily because I felt that I would be a somewhat borderline USAMO qualifier, if at all, and I wanted to make the USAMO.

Having made the USAMO, I started focusing on olympiad style problems, and though I still took the tests in a timed environment for the sake of preparation, I think that at this point I was doing problems much more for the sake of enjoyment, perhaps because I didn't think I really had a reasonable chance at MOP. In retrospect, this was probably the period of largest growth for me mathematically, precisely because I didn't look at the problems so much as training as challenges or interesting problems to solve.

On the other hand, once a certain level has been reached, some form of training can be useful, but probably not what you have in mind. It is important to understand why you make the steps that you do. While human intuition can accrue remarkable amounts of forsight, from what I have seen, it is far more powerful when the person with the intuition understands where it comes from, and when it can be trusted. That is probably a really bad explanation, but it basically comes down to analyzing [b]why[/b] you made the steps that you did in solving a particular problem, which incorrect ideas could have been avoided beforehand, and how they could have been avoided (not necessarily all incorrect ideas can be avoided, as if you always could come up with the correct idea immediately, then math wouldn't be very interesting), and finally, how you can generalize the insight you gained into your thought process to other problems.

Here are a couple more lessons to bare in mind:

I like to say that there are four levels of understanding a problem:
1) You understand what the problem is saying. For example, many people got stuck at this point on USAMO 2005/6.
2) You understand the solution. Basically, it makes sense.
3) You understand the motivation behind the solution. Why did the solution use that method? For example, on USAMO 2006/6, what hint leads one to use spiral similarity?
4) You solve the problem yourself. This level is your goal.
If you don't understand the solution, you're never going to solve it on your own. If you don't understand the problem, you can't get anywhere. If you don't get the problem on your own, understand the solution and then look for the motivation behind the solution. Often, that motivation won't appear in the solutions and you'll have to try to understand it.

For optimization problems (like this year's #4), remember that what you think is the best strategy will not always be the best strategy. Yeah, you know that having four arms emanating from one cell is optimal, but you have to consider all other cases. This is kind of a lesson I need to learn, too; I tried a little more advanced technique dealing with "legs" and "body cells" but I don't think it works anyway.

Here's another example. At MOP 2005, there was a practice test problem where Nemo was trapped in a circular tank with a shark swimming only on the perimeter. The shark can swim 4 times as fast as Nemo, but the question asks if Nemo has a strategy to escape.

I wrote that the shark's best strategy is to be the nearest to Nemo. Nemo then swims in a circle of appropriate radius (slightly less than a quarter of the tank's radius) until the shark, who cannot keep up with him, is the farthest from Nemo, at which point Nemo darts for the edge and freedom. I showed that he makes it. They gave me a 5, but my solution really wasn't complete, because I shouldn't have assumed the shark would always try to get as close to Nemo as possible. My point is that you should not waste time trying to come up with the best strategy for your opponent and try to counter it (that's more like how you play chess), but rather come up with your own strategy that works no matter what.

This idea really only works for games, but you can think of most optimization problems as games with two parts. First, you come up with the optimal case, and get an answer. Next, you ignore your setup and pretend an opponent is creating a setup with a better answer. You find a way to manipulate his setup to find a contradiction, without referring to your setup.

Okay, I've said enough. BTW, when I say "you" in most of the above statements, it's only because "one" can sound awkward if overused.

jeez, thanks everyone

paladin - most of the holes people discussed, i felt my proof covered.  and yea lol i didnt say anything about that on #1, although maybe they wouldn't have really needed me to state it.

altheman - what do you mean by "what if it is a loop and not a loop"?  i'm not sure what you're saying about it either.  thanks - yeah i noticed how broken my thoughts were.  i should write them out better, without skipping steps.  yeah i should define my new things better as well.  p' would have to be an integer, not just rational, though.  the special notation that i used probably didnt make the problem any easier, but i couldn't see it any other way, and i didn't feel like introducing new variables and confusing my stupid self.  and yes.  I would very much like to see my score distribution as well lol.  part of the reason i made this thread haha.

thank you gaku liu!  i see where that's trivial now.  although it would've been very easy to take care of if i had noticed that.

mellowmellon and steinhardt - interesting posts.  i think i'm interested in math but for some reason don't prepare nearly enough.  i went to san diego math circle with richard rucyzyk himself around 2 years ago, and read his first AoPS book (which wasnt very useful, apparently i should read #2), and did a couple practice problems 2 days before each set of tests.  Between the AIME and USAMO i started using this forum.  however, last year, i didn't prepare at all for any of the tests (baaarely squeezed into the usamo).  i don't really have motivation to study for next year, since I'll be a junior (edit: SENIOR lol sorry about that) ... and black MOP = not me

samE - haha i like the nemo problem.  yeah if i make USAMO next year, i will rethink the way i approach the solutions.  instead of spending an hour getting 1 point on a different problem i can't solve, i'll rigorize an easy one... i'm understanding more what the graders are about now anyways.

Of course, if you can prove that your opponent's strategy is optimal, like "suppose he did something else, then he could just do this instead and be better" then that is fine, as long as you rigorously prove this. Of course, only do this if you actually are sure what you are doing, because as noted many beginners seem to mess this part up.

As for Black MOP, Blue MOP is also open to Juniors, so you actually only need to make that, not sure if that helps at all. Also keep in mind that MOP is not the only camp out there, there are plenty of other camps if you don't want to undergo the level of preparation that many olympiad students do. For example, there is MathCamp, HCSSiM, PROMYS, Ross, and AwesomeMath. I am told that these are all wonderful programs, and most try to cater to all areas of the spectrum in terms of prior experience + rigor of program, so I would recommend looking into some of them (if you hurry, you can probably apply to a few this summer even).

Well it's not really completely free, you have to pay the registration fee for the AMC =P.

A much more pertinent thing to note is that not everyone can get into MOP. In fact, only 30 non-freshmen can get into MOP, so it is not necessarily the most feasible summer plan.


inequalities
function
integration
calculus
real analysis
real analysis unsolved

Let $ (X,\mu)$ be a measure space, $ (f_n)$ a sequence in $ L^p(\mu)$, $ p\in(1,\infty)$, $ f \in L^p(\mu)$, and a sequence of measurable functions $ (g_n)$, $ g_n \ge 0$, be such that $ f_n \rightharpoonup f$ weakly in $ L^p(\mu)$, and $ g_n \to g$ a.e. Prove that
	\[ \int g|f|^p d\mu \le \liminf_{n\to\infty} \int g_n|f_n|^p d\mu.\]

[quote="volodja"]Let $ (X,\mu)$ be a measure space, $ (f_n)$ a sequence in $ L^p(\mu)$, $ p\in(1,\infty)$, $ f \in L^p(\mu)$, and a sequence of measurable functions $ (g_n)$, $ g_n \ge 0$, be such that $ f_n \rightharpoonup f$ weakly in $ L^p(\mu)$, and $ g_n \to g$ a.e. Prove that
\[ \int g|f|^p d\mu \le \liminf_{n\to\infty} \int g_n|f_n|^p d\mu.
\]
[/quote]

$ g_n$ and $ g$ should be in some space because if not, intgerals above may be do not make sense.  :D

Two integrals in question have sense because the integrands are nonnegative measurable functions (integrals are allowed to be infinite).  :wink:

yes volodja, I 'm so hurry  and careless at my previous post.  :blush:

BTW, it seems like [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167865]this[/url]

Indeed, it could be follow from that problem, because $ g_n\geq 0$ so we can write $ g_n\equal{}\big(|g_n|^{1/p}\big)^p$

Your statement would be true if we had that $ g_n^{1/p}f_n \rightharpoonup g^{1/p}f$ weakly in $ L^p(\mu)$. Indeed, it would be the case if $ (g_n)$ were bounded in $ L^\infty(\mu)$, but we do not have this boundedness condition on $ (g_n)$.


[quote="cezar lupu"]Let $x, y, z>0$ such that $xyz=1.$ Prove that

                                  \[ \frac{xy+yz+zx}{x+y+z}\geq 1+\frac{x+y+z-3}{xy+yz+zx}.  \][/quote]

It is not correct.

Take $x=1$, $y=2$ and $z=\frac1{2}$

[quote="zhaobin"]sorry.I didn't think your inequality is right.
because even:$xy+yz+zx \ge x+y+z$ is wrong
which is $\iff (xy+yz+zx)^3 \ge xyz( x+y+z)^3$
obvious wrong :)
or you may mean:
$\frac{xy+yz+zx}{x+y+z} \ge 1+\frac{3-x-y-z}{xy+yz+zx}$????[/quote]
that's most probably what he wanted to ask
the last one is obviously true.


I know about all the scholarships and prize money and stuff for the top mathletes/teams at nationals, but do we get anything just for going? I mean, getting to the top round is a great feat (at least in most states!)! Are national competitors getting any rewards?

t-shirts, a nice calculator, and the experience

A calculator? :?

I don't want to sound ungrateful :) I'm so excited for the experience to go to Detroit this year.

Enjoy the experience. Soak it all in. You will be sorry when it is all over after the 8th grade.

solafidefarms is right.  Mathcounts is a lot of fun actually.  They take you to fun places (except last year we kinda just walked around a lot).  It's like a vacation.  So at the very least, you get an all expenses paid trip to Detroit, which is probably worth like $500, all things considered.  Also, you get a calculator.  It will probably be TI-89, or TI-83+ silver.  Texas Instruments might have something better now, I don't know.

[quote="elementsofmathkid"][quote="Treething"]TI-84 possibly. I'm just going to return it to Best Buy's an buy an iPod (or a DS, can't decide) with my savings and that calculator.[/quote]Don't the calculators have the MATHCOUNTS logo on them? If they do, I don't think Best Buy would want them.  :) Nor that it would be legal to return it....[/quote]

Return the current calculator you have or sell it, and then use the one MATHCOUNTS gives you... Anyways, I have heard that they are giving out TI-89 Titaniums this year (yay!!!, even though I already have one...), I'm not sure though.

Mathcounts is an excellent experience and will help you continue to have wonderful experiences. Chances are, a lot of the 228 who go to national mathcounts will be able to go to the math olympiad summer program, the ARML, or other fun competitions. If you don't win something now, Mathcounts could help you win something later.

[quote="09husbri"]TI-89 Titanium??? Are you sure MathCounts can afford to give out 228 of those???

I have been enjoying MathCounts to a great extent for the past three years and I haven't even been to Nationals yet - that will change in a couple of weeks. I am thrilled just to be going - I didn't want to sound ungrateful or expectant in my post, only curious. Thanks for the inspirational booster anyway :)[/quote]

Texas Instruments sponsors MATHCOUNTS and they give out the calculators...

[quote="Monkey"]All of this year's awards are posted on this site: http://www.mathcounts.org/webarticles/articlefiles/536-2005%20Awards.pdf

Btw, 09husbri, regarding the title of this thread, if you want a ten-dollar bill, you could probably find one if you see an innocent guy on the street fumbling through his wallet to buy an ice cream and he accidentally drops a ten-dollar bill.  :D[/quote]

DARN IT!!! I wanted TI-89 Titaniums... :(

Nationals is so awesome!  I went there last year.  It is so exciting and fun.  It's the best thing in the world!

:( :( :( :( :( :( :( :( :( :( :( :(

I wanted a penguin!
Or a trip to Antarctica, not space camp.
At least it is colder in Detroit than here.
TI-84+Silver
That is just like the TI-83+Silver but it has a few more Apps on it.

Yes the 18,000 awarded to National Champion is quite attractive, but most attractive (to me at least) is the honor of being the #1 middle-school mathematical mind in the entire country. If they gave a thousand dollars to 228th and nothing to the National Champion, I would still want to be first.

[quote="09husbri"][quote="Monkey"]All of this year's awards are posted on this site: http://www.mathcounts.org/webarticles/articlefiles/536-2005%20Awards.pdf

Btw, 09husbri, regarding the title of this thread, if you want a ten-dollar bill, you could probably find one if you see an innocent guy on the street fumbling through his wallet to buy an ice cream and he accidentally drops a ten-dollar bill.  :D[/quote]

Or, Monkey, what if they gave out [i]plastic rubber duckies??!![/i] AUGH!!![/quote]

I second that. If Mathcounts gave out rubber ducks, Henry might as well go over the top with his "Worship It!" craze for rubber duckies. Oh, man, that'd give me a headache all night on May 5, and then I'll completely die on test day like I did last year at nats... :(

I've heard the term "die on test day" a lot, but not really sure what you mean by it. Did you get over-nervous, or just have a bad problem set?

How about this: I got a bad sleep the previous night, my brain shut down, and I basically screwed up on a lot of problems I shouldn't have. Also: I was too worried about the time limit.

Hopefully this will teach everyone something about how to do your best at nats.

I have a story like that about states in sixth grade :)

Anyway, I forgot to bring my calculator, and I woke up at about 6:30 AM in my hotel room with my dad and realized it the second my head left the pillow. We spent about two hours roaming downtown Columbus (stupid one-way streets), we two nervous wrecks, looking for a decent calculator. Finally, I got one at the bookstore of the college the competition was actually at, but that put a dent in my focus for the day. :lol:

[quote="sheepwarrior"]whats master round? and so if sum1 gets like first individual first written and first team they win 18000 dollars??? wow!!! i wish i made nationals last year... just the calc and experience would be cool enough[/quote]

Actually, you can win 20,000, if you also win Master's Round... That's what Gregory Gauthier got last year... $20,000... That's a lot... I just hope I can get the $2,000$ from being on the #1 team... (Go CA!!)

[quote="NoSoupForYou"]I want $20000$ dollars, but I would prefer $2000$ to zero.[/quote]

So would 99.999% of the sane population with a positive IQ.

I would think it would be 100% if those criteria were met...

[quote="NoSoupForYou"]1) Certain strict religious orders (i.e. people living in caves) prohibit materalism.
2) Some people would not take the money because they would assume that there is catch, which there is in most cases.
3) Some people make several thousand dollars from interest in the amount of time it would take to reach out their hand and take the money.[/quote]

1 and 2) the word used was [b]prefer[/b], not take
3) they still get the interest, and there is the [b]prefer[/b] thing again...

1) and 2) To prefer 2000 to zero means that you would rather take 2000 than not take 2000. In both cases, these people would prefer to not take the 2000 dollars.

3) Taking money requires effort, and if I were that rich I wouldn't care.

This is supposed to be a joke. You shouldn't take these things literally.

About the catch, maybe the 2000 dollars is tax deductable but for the 20000 you have to pay the government 95% - that wouldn't be good.

But you only pay taxes on the amount over the max amount that you don't have to pay taxes for. . .

For example, say the max amount that one doesn't have to pay taxes on is 5000, so if you win 20000, you only have to pay taxes on 15000

[quote="09husbri"][quote="Monkey"]All of this year's awards are posted on this site: http://www.mathcounts.org/webarticles/articlefiles/536-2005%20Awards.pdf

Btw, 09husbri, regarding the title of this thread, if you want a ten-dollar bill, you could probably find one if you see an innocent guy on the street fumbling through his wallet to buy an ice cream and he accidentally drops a ten-dollar bill.  :D[/quote]

Or, Monkey, what if they gave out [i]plastic rubber duckies??!![/i] AUGH!!![/quote]

oh darn... i wish i were there at nats... they are giving out baseball caps... jk lol  :lol: i'd prefer the experience by far

I have my money placed on the 20,000 dollar prize

[quote="Phelpedo"][quote="09husbri"]To quote NoSoupForYou, "This is supposed to be a joke. You shouldn't take these things literally."[/quote]

Sirorange is just looking at it from a mathematical point of view, in case a question comes up at Nats like "which will yield more money, a $2000 prize or a $20,000 prize?"[/quote]

I'm aware of that. That's exactly why I said that it's only a joke. You found my very reason. :)


On a scale drawing of a new school playground, a triangular
area has sides with lengths of 8 centimeters, 15 centimeters,
and 17 centimeters. If the triangular area located on the playground has a perimeter of 120 meters, what is the length of its longest side?

The total perimeter of the triangle in the drawing is 40 cm.

The scale is therefore 40 cm : 120 m = 40 cm : 12000 cm = $ \boxed{1 : 300}$

The longest side in the drawing is 17 cm. Using our scale, we find that the actual size is:

17(300) cm = 5100 cm = 51 m

It's because if you have two similar triangles, the perimeters are also proportionate, with the same proportion. (Did that make sense?)

Because it is a scale drawing, the perimeters are proportionate as well as the sides.

Like if the sides have a proportion of $ 1 : 2$, then so will the perimeters.

FIRSTLY,let the scale used in drawing he triangle on paper be [d].
then it is easy to see that the lenght of the longest side is[17d].
also,[17d+15d+8d]=120;d=3.hence the lenght of the longest side is 17.3=51 :)

FIRSTLY,let the scale used in drawing he triangle on paper be [d].
then it is easy to see that the lenght of the longest side is[17d].
also,[17d+15d+8d]=1200;d=30.hence the lenght of the longest side is 17.30=510
 [quote]CURIOSITY IS THE CAUSE OF THE HUMAN BRAIN[i]OR MAYBE NOT[/i][/quote]


How many arrangements of five letters can be made of the letters of the word PALLMALL?

[quote="kunny"]$ \frac{8!}{2!4!}=\frac{8\cdot 7\cdot 6\cdot 5\cdot 4!}{2!4!}=840$[/quote]

I got something totally different, as in $ 265$..

[quote="lokesh"][hide="The Solution"]
Case 1: You have 4 L's
This gives you $ \frac{5!}{4!}\cdot3$ combinations or $ 15$ combinations.
Case 2: You have 3 L's and 2 A's
This gives you $ \frac{5!}{2!3!}$ combinations or $ 10$ combinations.
Case 3: You have 3 L's and 2 other letters that don't repeat
This gives you $ \frac{5!}{3!}\cdot3$ combinations or $ 60$ combinations.
Case 4: You have 2 L's, 2 A's and one other letter
This gives you $ \frac{5!}{2!2!}\cdot2$ combinations or $ 60$ combinations.
Case 5: You have 2 L's and no other letters that repeat
This gives you $ \frac{5!}{2!}$ or $ 60$ combinations.
Case 6: You have 1 L and 2 A's
This gives you $ \frac{5!}{2!}$ or $ 60$ combinations.
If you add all of the cases up, you get $ 265$ combinations.
[/hide][/quote]

correct, nice job


the expression is the right hand approximation of $\int_{0}^{1}\int_{0}^{1}5x^{4}-18x^{2}y^{2}+5y^{4}\, dx\, dy$, this is easily determined with basic integration rules...


calculus
integration
limit
algebra
polynomial
college contests

I've got some troubles with this:

\[ \lim_{n\to\infty} \left(\frac{(2n+1)!}{(n!)^2}\right)^2 \int^1_0 \int^1_0 (xy(1-x)(1-y))^n f(x,y) dx dy,  \]

where $f$ is continous on $[0,1] \times [0,1]$.

Would Lebesgue's dominated convergence theorem be useful here?

The limit is $f(1/2,1/2)$.

Let us denote $a_{n}=\left(\frac{(2n+1)!}{(n!)^{2}}\right)^{2}\int_{0}^{1}\int_{0}^{1}(x(1-x)y(1-y))^{n}f(x,y)dxdy$.

A calculation shows that  if $f(x,y)=x^{k}y^{l}$ then

$\lim a_{n}=\frac{1}{2^{k+l}}$.

f is continuous, it follows it can be approximated by polynomials, hence for $\epsilon >0$ there is a polynomial $P$ such that

$\max_{x,y\in [0,1]}|P(x,y)-f(x,y)|<\epsilon$.

Now let $g_{n}(x,y)=\left(\frac{(2n+1)!}{(n!)^{2}}\right)^{2}(x(1-x)y(1-y))^{n}$.
Now, observe that ${a_{n}=\int_{0}^{1}\int_{0}^{1}g_{n}(x,y}f(x,y)dxdy$.
Then we have that

$a_{n}=\int_{0}^{1}\int_{0}^{1}\left(f(x,y)-P(x,y)\right)g_{n}(x,y)dxdy+\int_{0}^{1}\int_{0}^{1}g_{n}(x,y)P(x,y)dxdy$.

On the other hand,

$|\int_{0}^{1}\int_{0}^{1}\left(f(x,y)-P(x,y)\right)g_{n}(x,y)dxdy|\leq \epsilon \int_{0}^{1}\int_{0}^{1}g_{n}(x,y)dxdy=\epsilon$.

It sufficient then to estimate the second integral.

Let $P(x,y)=\sum_{k,l}a_{k,l}x^{k}y^{l}$ be the polynomial in question, i.e., the approximation of $f$.

It follows that

$\lim \int_{0}^{1}\int_{0}^{1}P(x,y)g_{n}(x,y)dxdy=\lim \sum_{k,l}a_{k,l}\int_{0}^{1}\int_{0}^{1}x^{k}y^{l}g_{n}(x,y)dxdy=\sum_{k,l}a_{k,l}\frac{1}{2^{k+l}}$.
Now,

$|P(x,y)-f(x,y)|=|\sum_{k,l}a_{k,l}x^{k}y^{l}-f(x,y)|\leq sup_{x,y\in [0,1]}|P(x,y)-f(x,y)|<\epsilon$,
in particular for $x=y=\frac{1}{2}$ we get that
$|\sum_{k,l}a_{k,l}\frac{1}{2^{k+l}}-f(1/2,1/2)|<\epsilon$.
The desired result follows.

Probably I should have writen, for the last part of my explanation , that

$\int_{0}^{1}\int_{0}^{1}g_{n}Pdxdy=\int_{0}^{1}\int_{0}^{1}g_{n}\left(P(x,y)-f(1/2,1/2)\right)dxdy+\int_{0}^{1}\int_{0}^{1}g_{n}f(1/2,1/2)dxdy$

$=\int_{0}^{1}\int_{0}^{1}\left(P(x,y)-f(1/2,1/2)\right)g_{n}dxdy+f(1/2,1/2)$.


find all triples (a,b,c) a,b,c are all integers.
and for any positive integer n a*4^n+b*6^n+c*9^n is a perfect square number.


Given $ O(0,0,0), A(0,2,4), B(4,4,8)$ and a point $ P$ such that $ \vec{AP}\cdot\vec{BP}\equal{}0$, find the maximum length of $ \overline{OP}$.

Let $ P\equal{}(x,y,z)$. Then $ \vec{AP}\equal{}<x,y\minus{}2,z\minus{}4>$ and $ \vec{BP}\equal{}<x\minus{}4,y\minus{}4,z\minus{}8>$.
$ \vec{AP}\cdot\vec{BP}\equal{}(x)(x\minus{}4)\plus{}(y\minus{}2)(y\minus{}4)\plus{}(z\minus{}4)(z\minus{}8)$
$ \vec{AP}\cdot\vec{BP}\equal{}x^2\minus{}4x\plus{}y^2\minus{}6y\plus{}8\plus{}z^2\minus{}12z\plus{}32$
$ \vec{AP}\cdot\vec{BP}\equal{}(x\minus{}2)^2\plus{}(y\minus{}3)^2\plus{}(z\minus{}6)^2\minus{}9$
$ 0\equal{}(x\minus{}2)^2\plus{}(y\minus{}3)^2\plus{}(z\minus{}6)^2\minus{}9$
$ (x\minus{}2)^2\plus{}(y\minus{}3)^2\plus{}(z\minus{}6)^2\equal{}9$

We want to maximize $ \sqrt{x^2\plus{}y^2\plus{}z^2}$. I found that letting $ x\equal{}3$, $ y\equal{}5$, and $ z\equal{}8$ leads to a result of $ \sqrt(98)$, but perhaps a higher value can be achieved.

[b]darkmage2009[/b], you did the right thing by finding that equation.

Basically, you have a sphere of radius 3, and you want the distance of the point furthest from the origin. Can you see how a simple geometric argument solves the problem from here?

While you're thinking about that, here's an alternate approach:

[hide="Algebraic Solution"]We have $ x^2 \plus{} y^2 \plus{} z^2 \minus{} 4x \minus{} 6y \minus{} 12z \plus{} 40 \equal{} 0$
Setting $ x^2 \plus{} y^2 \plus{} z^2 \equal{} OP^2 \equal{} K$, we get $ K \plus{} 40 \equal{} 4x \plus{} 6y \plus{} 12z$.

Squaring and applying Cauchy-Schwarz:
$ (K \plus{} 40)^2 \equal{} (4x \plus{} 6y \plus{} 12z)^2 \leq (4^2 \plus{} 6^2 \plus{} 12^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 196K$
$ K^2 \minus{} 116K \plus{} 1600 \leq 0$
$ (K \minus{} 16)(K \minus{} 100) \leq 0$
$ 16 \leq K \leq 100$
$ 4 \leq OP \leq 10$

We can go back and verify that the $ x,y,z$ from the equality case of Cauchy-Schwarz indeed produce maxima and minima of $ 4$ and $ 10$.
[/hide]


graph theory
inequalities proposed
inequalities

Hi...

I need a help with this problem. Actually I have some results, but I also have a little problem with a part of proof.

Let's consider n straight lines in 3D space (R 3) such that any 3 of them don't lies in the same plan and  any 3 of them don't intersect in the same point. How many twosomes of these straight lines at most could intersect.

My result:
I've found out that no more than [n 2 /4] of straight lines can intersect (it's easy to see this from graph theory - complete bipartite graphs). But I have problems with showing that it works (it's possible to find demanded straight lines) for [n 2 /4] twosomes.

Thanks

The example is from the same graph theory problem too.
Consider two parallel planes and choose  |_ n/2 _| points forming a convex polygon in the first and
 _      _
|  n/2  | points forming a convex polygon in the second.
Connecting each pair of points from the different planes by the line
we obtain |_ n^2/4 _| lines we need.

Hi..

I don't know if I understand you (your solution) properly. Because the task was to find n lines in the 3D space such that.... (from requirements of the task), and [n 2/4] twosomes of them should intersect in the same point (as our observation says). But your answer says "we obtain [n 2/4] lines we need." So I don't understand if we should look for n lines.

(Thanks for your solution of the chessboard problem, it's really very elegant proof - it looks very easy, but I think this problem isn't so easy.)

I'm sorry, I didn't understand the problem correctly.


Let A, B, C, D, E be five set such that(denote Card A=|A|):
 \[|A\cap B|\geq \frac{3}{4}|B|,\]
\[ |B\cap C|\geq \frac{3}{4}|C|,\]
 \[|C\cap D|\geq \frac{3}{4}|D|,\]
 \[|D\cap E|\geq  \frac{3}{4}|E|,\]
 \[|E\cap A|\geq  \frac{3}{4}|A|.\]
Prove that $A\cap B\cap C\cap D\cap E\neq \emptyset$.

Generally:
 \[|A_i\cap A_{i+1}|\geq  \alpha |A_{i+1}|.\]
Find $\alpha$ such that $\cap_{i=1}^nA_i\neq\emptyset$

[quote="pigfly"]Let A, B, C, D, E be five set such that(denote Card A=|A|):
 \[|A\cap B|\geq \frac{3}{4}|B|,\]
\[ |B\cap C|\geq \frac{3}{4}|C|,\]
 \[|C\cap D|\geq \frac{3}{4}|D|,\]
 \[|D\cap E|\geq  \frac{3}{4}|E|,\]
 \[|E\cap A|\geq  \frac{3}{4}|A|.\]
Prove that $A\cap B\cap C\cap D\cap E\neq \emptyset$.[/quote]
Let A={1,2,3,4}, B={1,2,3,5}, C={1,2,4,5}, D = {1,3,4,5}, and E = {2,3,4,5}. Then
$|A\cap B| = 3 = (3/4)|B|,$
$|B\cap C| = 3 = (3/4)|C|,$
$|C\cap D| = 3 = (3/4)|D|,$
$|D\cap E| = 3 = (3/4)|E|,$
$|E\cap A| = 3 = (3/4)|A|,$
but
$A\cap B\cap C\cap D\cap E = \emptyset$.

[quote="pigfly"]Generally:
 \[|A_i\cap A_{i+1}|\geq  \alpha |A_{i+1}|.\]
Find $\alpha$ such that $\cap_{i=1}^nA_i\neq\emptyset$[/quote]
Obviously, $\alpha>3/4$.

Erin Schram


modular arithmetic
ratio
geometric series
complex numbers

[quote="willydavidjr"][b]Let $i(=\sqrt-1)$ be the imaginary unit. Then $1+i+i^2+i^3......+i^{2004}=?$ What is the answer?[/b][/quote]

[hide="hint"] $i^k$, where $k$ is even, is equal to $1$[/hide]

The answer is $1$, since the $i$s repeat in a pattern of every $4$ blocks, and cancel everything out, with the exception of $i^{2004}=1$

think of it like mods

If you're adding up progressive powers of imaginary units, all you have to do is look at the first set of four.

$i + i^2 + i ^3 + i ^4 =$

$i + -1 + -i + 1 = 0$

Note that $i = i^{4k+1} , i^2 = i^{4k+2}, i^3=i^{4k+3}, i^4 = i^{4k+4}$

And the expression simplifies to

$1 + 0 + 0 + ... + 0 = 1$

Summing up the geometric series, we have
$\frac{i^{2005}-1}{i-1}=\frac{i-1}{i-1}=1$

[quote="willydavidjr"][b]Let $i(=\sqrt-1)$ be the imaginary unit. Then $1+i+i^2+i^3......+i^{2004}=?$ What is the answer?[/b][/quote]
[hide]
Remember to use HIDES!
$i=i$
$i^2=-1$
$i^3=-i$
$i^4=1$
The sum of $i^1+i^2+i^3+i^4$ is 0.
$i^{4k+1}=i$
$i^{4k+2}=i^2$
$i^{4k+3}=i^3$
$i^{4k}=i^4$
$2004=0\pmod{4}$
Thus, the entire series collapses except for the 1 in front which is our answer.
[/hide]

[quote="mathmanman"][quote="José"]
$i^k$, where $k$ is even, is equal to $1$[/quote]
This is wrong. See my post above. ;)  (counterexample : $i^2 = -1$...)[/quote]

You're right, my bad.

[quote="mathmanman"]Hi! :)
See : http://www.mathlinks.ro/Forum/viewtopic.php?t=74632 and use the formula for the sum of a geometric series to solve your problem. :) (And you can even generalize it..)[/quote]

that actually brings up an interesting point! how do you deal with complex ratios? the $|r|<1$ generally is not proven for complex numbers, but it does end up being true!, looking at that, the series is not convergent since $|i|=1$

[quote]how do you deal with complex ratios? the |r|<1 generally is not proven for complex numbers, but it does end up being true!, looking at that, the series is not convergent since |i|=1[/quote]
You may have noticed,  Formula for the sum of finitie number of terms, one need not worry about |r| < 1, (its valid  for all values of r)

So what is the formula for a finite geometric series?

[quote="Adunakhor"]finite geometric series, inital term $a$, common ratio $r$:

$S=\displaystyle \frac{a-ar^{n-1}}{1-r}$[/quote]Hmm, make sure you are learning the correct expression:

Correct expression: $S=a\frac{1-r^{n+1}}{1-r}$

What's wrong? Here it is:

$S=\displaystyle \frac{a-ar^{n-1}}{1-r}=\sum_{j=0}^{n-2}ar^j$

$S=a\frac{1-r^{n+1}}{1-r}=\sum_{j=0}^nar^j$

Oh, my bad, lost track of the number of terms. But I think that your expression includes one extra term (where the last term is $ar^n$ instead of $ar^{n-1}$) if we are talking about the sum of the first $n$ terms of a geometric series.

In which case, $S=\displaystyle \frac{a-ar^n}{1-r}$


How do you find the inverse of $ f(x) \equal{} x^3 \plus{} x \plus{} 1$?
I'm pretty sure it has an inverse, but I keep getting stuck.
I switched the x's and y's to get $ x \equal{} y^3 \plus{} y \plus{} 1$, but then get $ y^3 \plus{} y \equal{} x \minus{} 1$, and I can't find a way to isolate y to find the inverse.

Thanks!

edit to fix typo

Do you mean $ y^{3}\plus{}y\equal{}x\minus{}1$?

No one's answered yet. :(
I'm still stuck with the $ y^3\plus{}y\equal{}x\minus{}1$.
I tried factoring the $ y^3\plus{}y$, but that doesn't help at all.

Here is a somewhat condensed understanding about a cubic polynomial function (condensed because I'm afraid of jumping into generalizations and drowning into a sea of confusion in regards to HSB-level):

Not every cubic has an inverse. But your cubic do have an inverse. Cubics that do not have an inverse have 3 real roots. Cubics that do have an inverse only have one real root.

Cubics with inverses may not have a "nice form". In other words, you may not be able to express your inverse in a "nice-looking" formula. What if you wanted to do that anyways? There is a cubic formula that you can use, if the need arises.

So, hopefully, I have answered your question, maybe indirectly, and perhaps you can stop trying fruitlessly :).

I'm guessing what he meant was to do something of this:

$ y^3 \plus{} y \minus{} (x \minus{} 1) \equal{} 0$

And use this formula for Cubic Equation.

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

It is indeed ugly... :huh:


convert
acetaldehyde to butane 1,3 -diol
bromobenzene to 1-phenyl ethanol(2 steps)
benzaldehyde to $ \alpha$-hydroxy phenyl acetic acid(2-steps)

[hide="Answer"]1. "acetaldehyde to butane 1,3 -diol": NaHO (aldol reaction), followed by reduction with, for example, NaBH4.

2. "bromobenzene to 1-phenyl ethanol(2 steps)": first the preparation of the Grignard reagent with Mg/THF, followed by reaction with acetaldehyde (and an aqueous workup).

3. "benzaldehyde to $ \alpha$-hydroxy phenyl acetic acid(2-steps)": first reaction with NaCN and acid (catalytic) to obtain the cyanohydrine, followed by acidic hydrolysis to convert the cyano group into a carboxyl group.[/hide]


geometry
trapezoid
rectangle
trigonometry
MATHCOUNTS
LaTeX
Pythagorean Theorem

I think that the creators of this question have the answer wrong...

I get 56 as the area of this trapezoid and put "e" but the answers claim the answer is "b" which is 96, can anyone help me???

thanks

aaaaaahhhhh ok so i can do this

[hide="answer"]ok so we know that the top base is 11 and the height is four and the sides are both 5, we can then find through the pythagorean theorum that the bottom base is just 11+3+3=17 so plugging this into the formula we have (11+17)2 which is 56 so i think that the answer is wrong uless im being really dumb!!![/hide]

just as an aside, -- where is this from!!!

[quote="uldivad9"][hide]Note that it is an icoseles trapezoid, therefore the other leg is 5 as well. Since the height is 4, we can separate both of the legs with cuts perpendicular to the base, and we get one large rectangle with dimensions 11x4, and two congruent right triangles. Upon closer inspections, the right triangles are 3-4-5 triangles. 44 + 12 = 56, which definitely should be the answer. Huh.[/hide]

I got 56 too. Where was the problem from?[/quote]

It's from a competition in Kansas KCATM (Kansas city assocation of teachers of mathematics)

[hide="LaTeX/Clarified solution"]
Draw altitudes $ \overline{HX}$ and $ \overline{TY}$ from $ H$ and $ T$ to $ \overline{MA}$. Since $ HX \equal{} TY \equal{} 4$, by the Pythagorean Theorem, $ MX \equal{} AY \equal{} 3$. Thus $ MA \equal{} 11\plus{}3\plus{}3 \equal{} 17$.

The area of a trapezoid is $ \dfrac{h(b_{1}\plus{}b_{2})}{2}$, so we have the area as $ \dfrac{4(11\plus{}17)}{2} \equal{} \fbox{56}$.[/hide]

I agree.

Eh..., guess the teachers at Kansas can't write tests.

[hide="I pwn kansas teachers"]
we have the height is 4, so we can draw a rectangle by drawing a line from the two top vertices of the trapezoid to the base, making it perpendicular. We have a rectangle, and two right triangles. using the pyth thereom, we get the bases of those two right triangles to be 3. Therefore, the longer base is 6 more than the top base, so the bases are 11 and 17. The average of these is 14, and 14* the height= 14*4 =56
[/hide]

umm...isnt there that super long and complex formula to find the area of a trapezoid with only the side lenghts!!!!!

What was it!

[quote="sbanerjee"]Well, if you assume it's a cyclic quadrilatteral, you could use the extended heron's formula:
$ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$

Where a, b, c, and d are the side lengths and s is the semiperimeter.

But you don't even need it since you're given it's an isocolese trapezoid...[/quote]

Actually, because it's an isosceles trapezoid, you [b]know[/b] it is cyclic. But you still have to find that other base, zolojetto!

[url]http://mathworld.wolfram.com/Quadrilateral.html[/url]
Note in a cyclic quadrilateral, the term with the cosine in it goes away.
You probably don't need to know this for Mathcounts, I doubt they would just ask you to find the area of a random quadrilateral.

Are all isocolese trapezoids cyclic? What about one that has bases of 25 and 27 with the other two sides equal to 2? That doesn't seem cyclic to me... maybe it is though.

Yeah, all isosceles trapezoids are cyclic, since opposite angle add to 180 degrees.
EDIT: Hehe...yeah.

[hide="Explanation"]
Notice how, if $ \overline{AB}$ and $ \overline{DC}$ are the parallel bases of isoscleles trapezoid $ ABCD$, $ \angle A \cong \angle B$, $ \angle D \cong \angle C$ due to the fact that it is isosceles.

Also, $ \angle A$ is supplementary to $ \angle D$ and $ \angle B$ is supplementary to $ \angle C$. Clearly, $ \angle A$ is supplementary to $ \angle C$, and we have a cyclic quadrilateral.[/hide]

But it really doesn't matter for this problem.

Darn, I got beat again.

xpmath... isosceles usually means isosceles =D. I think you meant cyclic.

here the formula to find the area of the trapezoid with the side lenghts only:

along with $ \sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)\minus{}abcd cos^2\theta}$, an extension to non-cyclic quadrilaterals.
I spent a while on the proof... (too much tedious algebra...)

[quote="sbanerjee"]Well, if you assume it's a cyclic quadrilatteral, you could use the extended heron's formula:
$ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$

Where a, b, c, and d are the side lengths and s is the semiperimeter.

But you don't even need it since you're given it's an isocolese trapezoid...[/quote]
Actually, its just $ \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$

guys, don't bother memorizing these formulas until you've seen enough problems that use it

[quote="ZhangPeijin"]Eh, how do you prove the formula for the area of a trapezoid anyways?[/quote]

exactly my point :wink:

you can assign a bunch of variable and apply the pythagorean theorem, then use $ A\equal{}\frac{(b_{1}\plus{}b_{2})h}{2}$

[quote="alanchou"][quote="ZhangPeijin"]Eh, how do you prove the formula for the area of a trapezoid anyways?[/quote]

exactly my point :wink:

you can assign a bunch of variable and apply the pythagorean theorem, then use $ A \equal{} \frac {(b_{1} \plus{} b_{2})h}{2}$[/quote]
Yeah but $ A \equal{} \frac {(b_{1} \plus{} b_{2})h}{2}$ Is the area for a trapezoid, which essentially does not prove the formula.

[quote="ZhangPeijin"]
Yeah but $ A \equal{} \frac {(b_{1} \plus{} b_{2})h}{2}$ Is the area for a trapezoid, which essentially does not prove the formula.[/quote]

oh, I thought you meant the other one

[hide="one I found myself (I almost never do)"]
To make it simple, suppose $ ABCD$ is a trapezoid with $ AB\parallel{}CD$.  Draw $ AC$ and it's obvious from there
[/hide]

so i found that the longer base was 17 from the pythagorean theorem. 11::17=14
So apparently these kansas test writers think 4*14=96.
texas test writers are better...

Its called Brahmagupta's formula.

Heron's formula is the area of a triangle the root of s(s-a)(s-b)(s-c).  [sry no Latex]

Yes that is what the general case is. They called the cyclic quad case [i]extended[/i] Heron's, as I've heard it before. Stop discussing it here, it's not MC level.

And to answer sbanerjee's question, all isosceles trapezoids [u]are[/u] cyclic because opposite angles add up to 180 degrees. There is a theorem you can use to prove this; suppose you have a chord in a circle. Make a triangle using the endpoints of the chord and the center. Form another triangle by finding any point on the circumference outside the arc of the chord; the angle formed in the center of the circle will be twice the angle formed on the circumference.

A circle, like any shape, has 360 degrees in the center. In order to find the sum of the opposite points of a quadrilateral, divide 360 by 2. That gives 180 degrees. An isosceles trapezoid satisfies the requirement that in order for a quadrilateral to be cyclic, The opposite angles(of the vertices) must add to 180 degrees. Therefore, all isosceles trapezoids are cyclic.

Courtesy of Aritro Biswas :D

[quote="ernie"]Please do not revive threads more than a month old.[/quote]

Ahem 853596...

Is this really supposed to be in the MATHCOUNTS forum?

Er..my basic question was...I guess...but then the peeps started talkin about advanced techniques of mathematics not applicable in mathcounts (well not in chapter)


Sound travels at the rate of 1130 feet per second. How far away is a lightning strike if the sound of thunder reaches you 7 seconds after you see the lightning flash? Assume the speed of light is infinite. Round your answer to the nearest tenth of a mile.

We can use labels to change feet into miles, etc. so we don't division or multiplication mixed up. \[ (7\text{ seconds})\left(\frac{1130\text{ feet}}{\text{second}}\right)\left(\frac{1\text{ mile}}{5280\text{ feet}}\right)\approx\boxed{1.5}\text{ miles}\]


AMC
AMC 8
MATHCOUNTS
function
AIME
college
geometry

Why is there such a strong culture of modesty here on AoPS?

um, cuz most of us are asian and asians are modest? idk

I see people saying "I suck at math." That isn't modesty; that's self-insult.

I think it's because that we feel we have no right to say we are good at math, because there are ALWAYS people better than you in math, especially on AoPS.

Yeah...that's probably true...It's kind of weird to say your really good at math or something when there's so many other really good people. It would be kind of weird, and people would probably think you have a really big ego or something.

I think that when people say: "I sux at math" they want you to pity them and tell them oh you aren't bad at math blah blah. This type of pity-baiting is pretty common among some of my friends...its annoying.

Well, there are always the people that will get a 14 on the AIME's and say "man, I suck" just to bother other people.

I like to give people the benefit of the doubt though; AoPS is probably one of the few places where people who are good at math can express frustration over their own mistakes in math and have someone understand them.

yeah dude I think rrusczyk made a topic on this a while ago

anyways modesty is totally overrated

i mean, when I got a 23 on AMC8, I was the highest score in my school, but here, i was probably really sucky.

so I said "I suck compared to people on AoPS. but people revere me at school."

and I have high standards for myself.

You should be careful about using modesty. For example, if [t0rajir0u] said he sucked at math, this could be an indirect insult to all of us. "If [t0rajir0u] says he sucks at math, then what about us?!!" So by "being modest" you aren't really being modest at all.

Math is something where it is heavily appreciated on AoPS. Not because of the competitions, MATHCOUNTS or AMC, but because of it's underlying beauty. Some people on AoPS are not here for the competitions. Some are here just to appreciate mathematics. This so called "modesty" isn't the best way of acommodating everyone.

If we compare our current knowledge and ‘skills’ to that of other people, it may be warranted to say “I am proficient in mathematics”; nevertheless, if we gauge our current knowledge relative to the knowledge we [i]could[/i] have, every one of us would be found wanting. In that sense, we [i]do[/i] really “suck at math.”

Furthermore, saying “there is always someone better than you” is inaccurate: there is no definite method for measuring someone's abilities. You can argue that [Altheman] was the best competitor at IMO, based on certain criteria, but you can't really say that [Altheman] is the best mathematician of our generation. Perhaps, if we tried to measure these abilities a bit less, and focused more on strengthening them, we wouldn't feel insulted if [t0rajir0u] (or anyone “smarter” than us) criticized their own performance.

It is healthy to know what you are good at and tell yourself that. It isn't healthy to inflate it. I know that I am very good at math. I am not in the top tier, though. There are lots of people better than I.

When people say, "I suck at math," I just yell at them until they stop, unless they are a good friend of mine.

A strange, somewhat related conversation I remember

[b]Teammate 1:[/b] Dude, I got a 7 [out of 18]. I can't believe it! I suck at math
[b]
Teammate 2:[/b] Just because you got the lowest score on the team, and made some of the stupidest mistakes I've ever seen, and put us behind Westborough in the last minute, doesn't mean that you suck at math.

It is very interesting that almost nowhere on AoPS (I could be wrong) has a person been accused of boasting about results.

[quote="worthawholebean"]
When people say, "I suck at math," I just yell at them until they stop, unless they are a good friend of mine.[/quote]
 :rotfl:
I can't wait to try this out...

i am pretty sure it is one of those habits encouraged by the real world

like you do not see many intact math-type high-schoolers who accurately tell people how good they are at math; rather, they like misrepresent themselves in order to not look too freakish

and these habits kinda stay, even though you're on AoPS, not in school

Well... I know a lot of non-modest AoPSers...

And I think a higher IQ means that you are a better liar, or more likely, have the capability to be a better liar. I, for one, am an awesome liar. I take a bus home from school every day ad I lie to everyone all day on it. Its quite fun actually. I love to act completely serious and most people at my school know me for my random facts... I always do let them know when I am lying. But convincing someone that the president of Germany is Arnold Scwarzenegger is pretty fun. ( :evilgrin: )

[quote="dynamo729"]
Well... I know a lot of non-modest AoPSers...

 ...I, for one, am an awesome liar.... [/quote]

i know a lot of non-modest aopsers too....

I put that in there so someone could catch it. Yay! Someone caught it! I wanted to check that people still read all through posts.  :D

Here's what I think...

When you are actually good at something, you feel good about your abilities in it.  When people praise you for it, you feel no need to justify yourself.

In other words, when you already know that you are good and know people know you are good, there's no need to boast.  Instead you might downplay yourself.  For example "Oh, I'm only good because of this" or "Oh but I'm really nto that good, this person is much better" or something like that.

you don't need to boast; downplaying yourself is also stupid though

i've never told anyone that i suck at math
even though that i m the weakest on my MC team (3rd in illinois) i m still the smartest in my grade
so even though i m stupid compared to some of my teammates i m still better than everyone in my grade

i think this applies to everyone: if u give the time/effort/money to be on AoPS then u r probably smarter then most of the people @ ur school

[quote="RussianRocket"]
i think this applies to everyone: if u give the time/effort/money to be on AoPS then u r probably smarter then most of the people @ ur school[/quote]
everyone says that im the smartest kid in my grade.  i just say "REALLY?!?" and walk away.
force of habit.

[quote="JBL"]Does everyone also say that you make an awful lot of posts in this forum that suffer from poor grammar, punctuation, capitalization, etc., and (worse!) that offer no conceivable interest to anyone other than yourself?[/quote]
Maybe he's trying to be modest with his English skillz... :P

[quote="JBL"]Does everyone also say that you make an awful lot of posts in this forum that suffer from poor grammar, punctuation, capitalization, etc., and (worse!) that offer no conceivable interest to anyone other than yourself?[/quote]

nah...probably just you

why is grammar such a big deal, its the internet, relax dude...

Because this is a forum for serious discussion, and it's hard to have a serious discussion with someone who can't be bothered to express themselves in a readable fashion.  (Or, worse, who posts meaningless personal anecdotes about themselves as if they should have interest to the rest of us.)  The only reason I've not deleted myyellowducky's post is so that it can stand as an example of the kind of post that should never have been made.

I'm also perplexed as to why you think the fact that this is on the internet is relevant to anything, but not perplexed enough to let discussion along these lines continue.  If no one has anything else to say on the subject of this thread, please let it die in peace.

to be honest, i'm fine with anything between the extremes of modesty and arrogance, and i am sure that most people think the same way. personally, i'd say i'm like a half std more cocky than the average, and i don't think it's a major character flaw, just the way i am.

[quote="JBL"]Does everyone also say that you make an awful lot of posts in this forum that suffer from [b]poor grammar[/b], punctuation, capitalization, etc., and ([b]worse[/b]!) that offer no conceivable interest to anyone other than yourself?[/quote]

hello irony!

um... sorry, but what's "irony"? i'm pretty sure i've heard it before but my vocab suks... and i'm NOT trying to be modest. seriously, tho, that cant be considered being modest, can it? 'cause i genuinely suk at language arts! (and i still wanna kno what irony means.)

[url]http://pbfcomics.com/archive_b/PBF085-The_Tree_of_Irony.gif[/url]

probably the most concise way of defining irony

That is perfectly acceptable informal English, and this thread is now locked.


MATHCOUNTS
geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

Is there a pattern as to x:^3:-y:^3: if x-y is 1? If so what is it?

Yes, that is true, but will it necessarily help us?

I apologize for the way I wrote this problem, it was supposed to be something like "Is there a pattern as to the difference of cubes?" but I thought it'd be easier to undersatnd this way. Basically this problem is just asking for an easy formula to figure out 3490:^3:-3489:^3:.

dang it, my bad. didn't see that the 2 was missing :oops:

Well there is an easier way to calculate x^3 - y^3 where x-y = 1 instead of using x^2 + xy + y^2.. ml2 being rather close to getting that, actually ;)

Yup, (x-y)((x+y):^2:-xy) seems the easiest to me though it may not be worth memorizing...it's your call.

uh, but with numbers like 3489, thats still massive amounts of calculation...

*looks at insanely huge number.... pulls out my calc*

I love calculators... makes life much easier.

*sigh* What's the point of this whole problem if you can use a calculator? :(

If the numbers are that big... then why not??? multiplying and squaring their sum would be a pain

Well yeah, but why even try to make it simpler if you're going to just punch it into the calculator? In fact it'd be easier for you to punch in theoriginal equation then the transformed equation.

*sigh*

Personally, I would go for 3xy + 1 as my preferred form, since it involves much less calculation.  The way we get this is that
x:^2: + xy + y:^2: = x:^2: - 2xy +y:^2: + 3xy = (x - y):^2: + 3xy = 3xy + 1

Oy that works too, much better too, too true :lol:


I need help solving this problem.
If $ x \plus{} \frac{1}{x}\equal{}1$ what does $ x^3 \plus{} \frac{1}{x^3}$
Source: The Art of Problem Solving Volume 1: The Basics
by Sandor Lehoczky and Richard Rusczyk

$ \left(x \plus{} \frac {1}{x}\right)^3 \equal{} x^3 \plus{} 3x \plus{} \frac {3}{x} \plus{} \frac {1}{x^3}$
Thus, $ x^3 \plus{} \frac {1}{x^3} \equal{} \left(x \plus{} \frac {1}{x}\right)^3 \minus{} 3\left(x \plus{} \frac {1}{x}\right) \equal{} 1^3 \minus{} 3 \times 1 \equal{} \boxed{ \minus{} 2}$.

[quote="monkeygirl13"]I need help solving this problem.
If $ x \plus{} \frac {1}{x} \equal{} 1$ what does $ x^3 \plus{} \frac {1}{x^3}$
Source: The Art of Problem Solving Volume 1: The Basics
by Sandor Lehoczky and Richard Rusczyk[/quote]

The important thing is realizing how $ (x\plus{}\frac{1}{x})^3$ cubed generates $ x^3\plus{}\frac{1}{x^3}$
Although it gets longer when you try larger values of $ (x\plus{}\frac{1}{x})^n$ for large odd values of $ n$


Let $ a,b,c,d$  be the roots of $ x^4\minus{}x^3\minus{}x^2\minus{}1\equal{}0$ . Find $ p(a)\plus{}p(b)\plus{}p(c)\plus{}p(d)$, where $ p(x)\equal{}x^6\minus{}x^5\minus{}x^3\minus{}x^2\minus{}x\minus{}1$ .

[quote="NapoleonXIV"]Lets say $ f(x) \equal{} x^4 \minus{} x^3 \minus{} x^2 \minus{} 1$
Then $ p(x) \equal{} (x^2 \plus{} 1)f(x) \plus{} (x^2 \minus{} x)$
Then $ p(a) \plus{} p(b) \plus{} p(c) \plus{} p(d) \equal{} a^2 \minus{} a \plus{} b^2 \minus{} b \plus{} c^2 \minus{} c \plus{} d^2 \minus{} d$

By Vieta's formulas we know $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ and $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} \minus{} 1$

And $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} (a \plus{} b \plus{} c \plus{} d)^2 \minus{} 2(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd) \equal{} ( 1)^2 \minus{} 2( \minus{} 1) \equal{} 3$

Then $ p(a) \plus{} p(b) \plus{} p(c) \plus{} p(d) \equal{} a^2 \minus{} a \plus{} b^2 \minus{} b \plus{} c^2 \minus{} c \plus{} d^2 \minus{} d \equal{} 3 \minus{} 1 \equal{} 2$[/quote]
I don't think this is the true answer,try again please :) .


modular arithmetic
algebra
system of equations

How many whole numbers $ n$ between $ 1$ and $ 855$ comply that $ n(n\minus{}1)$ is a multiple of $ 34$?

[hide="solution"]
We know that $ 2|n(n \minus{} 1)$ always, and that in order for $ 34|n(n \minus{} 1)$, $ n \equal{} 17k$ or $ n \equal{} 17k \plus{} 1$.

Now we need to count the amount of $ n$ that satisfy this:

The lowest value that satisfies $ 17k$ is $ 17(1)$ and the highest is $ 17(50) \equal{} 850$ making already $ 50$ values for $ n$.

The lowest value that satisfies $ 17k \plus{} 1$ is $ 17(0) \plus{} 1 \equal{} 1$ and the highest is $ 17(50) \plus{} 1 \equal{} 851$ Another $ 51$ values for $ n$.

Making a total of $ \boxed{101}$ values.
[/hide]

$ 101$ is not in the answer choices... the closest one to that is $ 100$:

a) $ 81$
b) $ 88$
c) $ 92$
d) $ 96$
e) $ 100$

This is equivalent to asking for solutions to $ n^{2}\minus{}n\equiv 0\pmod{34}$.  Write $ 34\equal{}2\cdot 17$ so we need to solve the system of equations $ n^{2}\minus{}n\equiv 0\pmod{2}$ and $ n^{2}\minus{}n\equiv 0\pmod{17}$ by the Chinese Remainder Theorem.  By inspection, $ n\equal{}1$ is a solution to both, so let $ c_{1}\equal{}c_{2}\equal{}1$.  Now let $ n_{1}\equal{}17$ and $ n_{2}\equal{}2$.  It follows that the inverses of $ n_{1}$ and $ n_{2}$ are $ \tilde{n}_{1}\equal{}1$ and $ \tilde{n}_{2}\equal{}9$.  Hence a solution is $ X\equal{}c_{1}n_{1}\tilde{n}_{1}\plus{}c_{2}n_{2}\tilde{n}_{2}\equal{}1\cdot 17\cdot 1\plus{}1\cdot 2\cdot 9\equal{}17\plus{}18\equal{}35\equiv 1\pmod{34}$.  Hence one set of solutions is $ \{1,35,69,103,137,171,205,239,273,307,341,375,409,443,477,511,545,579,613,647,681,715,749,783,817,851\}$, which has 26 elements.
Going back to the original problem, $ n\equal{}0$ is also a solution so any integer congruent to 0 mod 34 will work (i.e. 34, 68,...) and there are 25 of those.  We can also add in all the numbers that are half-way in between, so that's another 25 solutions in the first set and 24 in the second, for a total of 26+25+25+24=100 solutions.

tonypr, I think you counted wrong somewhere.

[quote="JRav"]This is equivalent to asking for solutions to $ n^{2} \minus{} n\equiv 0\pmod{34}$.  Write $ 34 \equal{} 2\cdot 17$ so we need to solve the system of equations $ n^{2} \minus{} n\equiv 0\pmod{2}$ and $ n^{2} \minus{} n\equiv 0\pmod{17}$ by the Chinese Remainder Theorem.  By inspection, $ n \equal{} 1$ is a solution to both, so let $ c_{1} \equal{} c_{2} \equal{} 1$.  Now let $ n_{1} \equal{} 17$ and $ n_{2} \equal{} 2$.  It follows that the inverses of $ n_{1}$ and $ n_{2}$ are $ \tilde{n}_{1} \equal{} 1$ and $ \tilde{n}_{2} \equal{} 9$.  Hence a solution is $ X \equal{} c_{1}n_{1}\tilde{n}_{1} \plus{} c_{2}n_{2}\tilde{n}_{2} \equal{} 1\cdot 17\cdot 1 \plus{} 1\cdot 2\cdot 9 \equal{} 17 \plus{} 18 \equal{} 35\equiv 1\pmod{34}$.  Hence one set of solutions is $ \{1,35,69,103,137,171,205,239,273,307,341,375,409,443,477,511,545,579,613,647,681,715,749,783,817,851\}$, which has 26 elements.
Going back to the original problem, $ n \equal{} 0$ is also a solution so any integer congruent to 0 mod 34 will work (i.e. 34, 68,...) and there are 25 of those.  We can also add in all the numbers that are half-way in between, so that's another 25 solutions in the first set and 24 in the second, for a total of 26+25+25+24=100 solutions.

tonypr, I think you counted wrong somewhere.[/quote]

When you count the amount in between in the second set $ \{34,51,68,\cdots,850\}$ there are $ 24 \plus{} 25$ elements but, what happened to $ \{17\}$?

Edit: I agree with joeislittle, so my new answer is $ 100$. Now I believe that JRav counted in $ 1$ but not $ 17$ and I counted them both. The correct answer is $ 100$.

[quote="xcenario"]How many whole numbers $ n$ between $ 1$ and $ 855$ comply that $ n(n \minus{} 1)$ is a multiple of $ 34$?[/quote]

I think they mean that the allowable values for n are 1<n<855.

[quote="JRav"]This is equivalent to asking for solutions to $ n^{2} \minus{} n\equiv 0\pmod{34}$.  Write $ 34 \equal{} 2\cdot 17$ so we need to solve the system of equations $ n^{2} \minus{} n\equiv 0\pmod{2}$ and $ n^{2} \minus{} n\equiv 0\pmod{17}$ by the Chinese Remainder Theorem.  By inspection, $ n \equal{} 1$ is a solution to both, so let $ c_{1} \equal{} c_{2} \equal{} 1$.  Now let $ n_{1} \equal{} 17$ and $ n_{2} \equal{} 2$.  It follows that the inverses of $ n_{1}$ and $ n_{2}$ are $ \tilde{n}_{1} \equal{} 1$ and $ \tilde{n}_{2} \equal{} 9$.  Hence a solution is $ X \equal{} c_{1}n_{1}\tilde{n}_{1} \plus{} c_{2}n_{2}\tilde{n}_{2} \equal{} 1\cdot 17\cdot 1 \plus{} 1\cdot 2\cdot 9 \equal{} 17 \plus{} 18 \equal{} 35\equiv 1\pmod{34}$.  Hence one set of solutions is $ \{1,35,69,103,137,171,205,239,273,307,341,375,409,443,477,511,545,579,613,647,681,715,749,783,817,851\}$, which has 26 elements.
Going back to the original problem, $ n \equal{} 0$ is also a solution so any integer congruent to 0 mod 34 will work (i.e. 34, 68,...) and there are 25 of those.  We can also add in all the numbers that are half-way in between, so that's another 25 solutions in the first set and 24 in the second, for a total of 26+25+25+24=100 solutions.

tonypr, I think you counted wrong somewhere.[/quote]
How does the Chinese Remainder Theorem work in this case (I thought it is used to solve a system of (linear) congruences)?
Also, how come it won't yield the general solution (since we still have to look for numbers in between that are not given by CRT)?

[quote="tonypr"]Now I believe that JRav counted in $ 1$ but not $ 17$ and I counted them both. The correct answer is $ 100$.[/quote]
See my last paragraph.  I said "We can also add in all the numbers that are half-way in between...," which was me saying to include the multiples of 17.

In response to mathwizarddude, I turned out initial problem into a system of problems, since 17 and 2 are easier to work with than 34.  If you take everything mod 17 (instead of mod 34 like I did), you'll get everything in one shot.

[quote="JRav"]See my last paragraph.  I said "We can also add in all the numbers that are half-way in between...," which was me saying to include the multiples of 17.[/quote]  And which two numbers was 17 half-way in between?  (tonypr is right about what you counted ;).)

It's between 0 and 34, although 0 doesn't count of course.  Perhaps I was not clear enough.

Then that would make $ 25\plus{}25\equal{}50$ elements in the second set, making your total the same as my original $ 101$


So I'm pretty confident of taking AIME next year, and I did some practice problems and failed miserably. Is there a guide of things to keep in mind (all the symbols, factoring skills, formulas, geometry...)? Can someone tell me the basics? I want to have a chance next year :P.

There's far too much info for there to be a simple cheat sheet. Remember that for most AIME problems you're going to have to reframe the question somehow. If there's stuff you haven't learned, I suggest buying the AoPS books (volume 2 especially).

There's a big data base of problems here on this web site and many people say AoPS vol.2 is also very good to prepare for.

p.s. this has been asked many , many, many, .... times before

To do well on the aime, you should be doing just like a whole lot of math between now and the test. The AoPS books are good (vol. 1 and 2). Then ACoPS. Also the the AoPS classes are probably good too.


[quote="vuthanhtu_hd"][quote="tranquoc"]Solve
$ 2^x \plus{} 3^y \equal{} 4$
$ 5^x \plus{} 6^y \equal{} 7$[/quote]
Dear tranquoc ,We have x=0 and y=1[/quote]
Why x=0;y=1?

hello, consider the function
$f(x)=5^{x}+6^{\frac{\ln(4-2^{x})}{\ln(3)}}-7$
Sonnhard.

[quote="tranquoc"]Solve
$ 2^x\plus{}3^y\equal{}4$
$ 5^x\plus{}6^y\equal{}7$[/quote]

If one of $x, y$ is negative and the other is positive, we have the LHS in both problems to not be integers.

If both of $x, y$ are negative, simple divisibility argument shows that the $LHS$ is still not an integer.

So, $x, y \ge 0$

$\Longrightarrow 2^x, 3^y, 5^x, 6^y \ge 1$

From here, it is trivial.

For $2^x+3^y=4,$ we have $(x, y)=(0, 1), (2, 0)$

For $5^x+6^y=7,$ we have $(x, y)=(0, 1)$
 :)  :)  :)  :)


limit
IMO
polynomial
calculus
Fixed point
Sequence
IMO 1985

For every real number $x_1$, construct the sequence $x_1,x_2,\ldots$ by setting: \[ x_{n+1}=x_n(x_n+{1\over n}). \] Prove that there exists exactly one value of $x_1$ which gives $0<x_n<x_{n+1}<1$ for all $n$.

$x_{n+1}=x_n(x_n+\frac{1}{n})$
$x_n<x_{n+1}$ for all $n$ so:
$x_n+\frac{1}{n}>1$
$x_n>\frac{n-1}{n}$
$x_n<1$ for all $n$ so:
$x_n(x_n+\frac{1}{n})<1$
$x_n>0$ so from here we get:
$\frac{n-1}{n}<x_n<\frac{\sqrt{4n^2+1}-1}{2n}$
Now we will check a limit when $n\to \infty$
$\lim\limits_{n\to \infty} \frac{\sqrt{4n^2+1}-1}{2n}-\frac{n-1}{n}$ $=\frac{\sqrt{4n^2+1}-2n+1}{2n}$
$=\frac{n(\sqrt{4+\frac{1}{n^2}}-2+\frac{1}{n})}{2n}=0$
So we get that the limit of the diffrence of the limits  of $x_n$ is $0$!
Hence, there is one possible value for $x_n$ when $n$ is big enough so
there is one possible value for $x_1$ when $0<x_n<x_{n+1}<1$ for all $n$
and we are done.

Sorry to bump this but I'm not sure if the previous solution was very rigorous. I mean isn't that the same as doing $ 1\minus{}\frac{n\minus{}1}{n} \equal{} 1/n$ which approaches $ 0$ as $ n$ approaches infinity?.

Indeed, the above presumed proof proves nothing. For some terse solution, see John Scholes' proof at the [b]kalva[/b] cache.

I notice that no complete solutions have been given in this topic, so I offer mine below.

By recursive substitution, one can write $x_n=P_n(x_1)$ , where $P_n$  is a polynomial with non-negative coefficients and zero constant term.  Thus, $P_n(0)=0$, $P_n$ is strictly increasing in $[0,+\infty)$ , and $\displaystyle \lim_{x_1 \rightarrow + \infty} P_n(x_1)=+\infty$. We can therefore define the inverse $P_n^{-1}$ of $P_n$ on $[0,+\infty)$.  It follows that $x_1=P_n^{-1}(x_n)$, $P_n^{-1}(0)=0$, $P_n^{-1}$ is strictly increasing in $[0,+\infty)$, and $\displaystyle \lim_{x_1 \rightarrow + \infty} P_n^{-1}(x_1) =+\infty$.

Denote by $\displaystyle a_n=P_n^{-1}(1-\frac{1}{n})$ and $b_n=P_n^{-1}(1)$.  By the monotonicity of $P_n^{-1}$  we have $a_n<b_n$ for each $n$.  Note that:

(a) $\displaystyle x_n<x_{n+1} \Leftrightarrow x_n>1-\frac{1}{n} \Leftrightarrow P_n^{-1}(x_n)>P_n^{-1}(1-\frac{1}{n}) \Leftrightarrow x_1>a_n$;
(b) $\displaystyle x_n<1 \Leftrightarrow P_n^{-1}(x_n)<P_n^{-1}(1) \Leftrightarrow x_1<b_n$.

Thus, $0<x_n<x_{n+1}<1,\forall n$ holds if and only if $a_n<x_1<b_n,\forall n$, or $\displaystyle x_1 \in \bigcap_{n=1}^{+\infty}(a_n,b_n)$.  We need to show that $\displaystyle \bigcap_{n=1}^{+\infty}(a_n,b_n)$ is a singleton.  We have:

(c) if $x_1=a_n$, then $x_n=1-\frac{1}{n}$, which implies that $x_{n+1}=1-\frac{1}{n}<1-\frac{1}{n+1}=P_{n+1}(a_{n+1})$, and $x_1<a_{n+1}$.  It follows that $a_n<a_{n+1},\forall n$; and
(d) if $x_1=b_n$, then $x_n=1$, which implies that $x_{n+1}=1+\frac{1}{n}>1=P_{n+1}(b_{n+1})$, and $x_1>b_{n+1}$.  It follows that $b_n>b_{n+1},\forall n$; and

Thus, $a_n<a_{n+1}<b_{n+1}<b_n, \forall n$.  Therefore, the two sequences $\{a_n\}_{n=1}^{+\infty}$ and $\{b_n\}_{n=1}^{+\infty}$ converge, and their limits $a$ and $b$  satisfy $a \leq b$.  Hence, $\displaystyle \bigcap_{n=1}^{+\infty}(a_n,b_n)=[a,b]$ is non-empty, which demonstrates the existence of $x_1$.

Now, suppose that $a \leq x_1 \leq x_1' \leq b$.  We have $x_{n+1}'-x_{n+1} = (x_n'-x_n)(x_n'+x_n+\frac{1}{n}) \geq (x_n'-x_n)(2-\frac{1}{n}) \geq (x_n'-x_n)$ for each $n$, so that $x_n'-x_n \geq x_1'-x_1$ for each $n$.  However, $1-\frac{1}{n}<x_n \leq x_n'<1$, so that $0 \leq x_n'-x_n<\frac{1}{n}$, which implies that $\displaystyle \lim_{n \rightarrow +\infty}(x_n'-x_n)=0$.  Therefore, $x_1' \leq x_1$, proving unicity.

For each $n \ge 1$ let $I_n$ be the interval of real numbers $0 \textless x \textless 1$ such that if $x=x_1$, we will have $x_n \textless x_{n+1} \textless 1$. (That is, the values of $x_1$ which make $x_{n+1}$ "work"). We must prove these intervals intersect at one point.

First, I prove they intersect. Notice that as $x_1$ increases, $x_n$ and $x_{n+1}$ do too. So if $I_n=(a_n,b_n)$ it is easy to see $a_n=x_1$ will give $x_n=x_{n+1}=1-1/n$ and $b_n=x_1$ will give $x_{n+1}=1$ (the "extremal" values $x_{n+1}$ can take are given by the extremal values $x_1$ can take).

But if $a_n=x_1$ we see that $x_{n+1} \textless 1-1/(n+1)$ and so $x_{n+1} \textgreater x_{n+2}$ and similarly $x_1=b_n$ gives $x_{n+1}=1$ which gives a $x_{n+2}$ too big. So basically when $x_1$ only barely works for $x_{n+1}$, it won't work for $x_{n+2}$. So $I_{n+1}$ is a subset of $I_n$.

So we have an infinite sequence of open intervals, each contained inside the previous one. Therefore their intersection must contain at least one number. This guarantees the existence of $x_1$.

But, we must prove that $x_1$ can't take two different values. Suppose it could. Suppose $x_1=a,b$ both work. Let $x_i$ be the sequence generated by $a$ and $x_i'$ the sequence generated by $b$, and let $l_i=x_i'-x_i$ (wlog $a \textless b$). Then we see $l_{n+1}=l_n(2x_n+l_n+1/n) \textgreater l_n$ for $n \ge 2$ since $2x_n \textgreater 1$ since $x_n \textgreater 1-1/n \ge 1/2$. So there exists a constant $C$ such that $x_k' \ge x_k+C$ for $k \ge 2$. But since $x_k, x_k' \in (1-1/k, 1)$, this is impossible. So we're done.

Interesting real analysis problem!

[hide=solution]Consider, now, the following observations:

(a) If, at any point, $x_{n+1}\le x_n$, then from here we know that
\[x_n\le 1-\frac{1}{n},\quad x_{n+1}\le 1-\frac{1}{n} < 1-\frac{1}{n+1}\to x_{n+2}<x_{n+1}, \cdots  x_m\le 1-\frac{1}{n} < 1-\frac{1}{m}\to x_{m+1}<x_m, \forall m>n\]
and therefore the sequence $\{x_n\}$ will be monotonically decreasing after one point.

(b) If some $x$ does satisfy $0<x_n<x_{n+1}<1$ for all $n$, then $1-\frac{1}{n}<x_n<1$ for all $n$, and therefore by Squeeze's theorem $\lim_{n\to\infty}x_n = 1$.

(c) If $x_n\ge 1$ for some $n$, then $x_{n+1}=x_n(x_n+\frac 1n)>x_n^2\ge 1$ and so $x_{n+m}>(x_{m+1})^{2^{m-1}}$ which then gives $\lim_{n\to\infty}x_n=+\infty$.

(d) If $x_1=0$, then for each $n$, $x_n=0$. If $x_1\to\infty$ then for each $n$ (fixed), $x_n\to\infty$. Thus, we can denote a mapping $f_n:\mathbb{R}^{\ge 0}\to \mathbb{R}^{\ge 0}$ that maps $x_1$ to $x_n$, which is continuous and monotonically increasing, with $\lim_{x\to +\infty} f_n(x)=+\infty$ so $f_n$ is bijective.

Let's first show uniqueness. Suppose that $\{x_n\}$ and $\{y_n\}$ are both such sequences. We have $\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=1$ and suppose that $y_1>x_1$. Then for all $n$,
\[
\frac{y_{n+1}}{x_{n+1}} = \frac{y_n(y_n+\frac 1n)}{x_n(x_n+\frac 1n)}>\frac{y_n}{x_n}
\]
so inductively, $\frac{y_{n+1}}{x_{n+1}}>(\frac{y_{1}}{x_{1}})^n$ with $\lim_{n\to\infty}\frac{y_{n+1}}{x_{n+1}}=+\infty$.
However, $\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=1$ gives $\lim_{n\to\infty}\frac{y_{n+1}}{x_{n+1}}=\frac{1}{1}=1$, contradiction.
Hence, the $x_1$ that satisfies this must be unique.

Next, let's show existence. We have seen from above that, $y_1>x_1$ implies $y_n>x_n$, and that if $x_n>1$ for some $n$ then $\{x_n\}$ is monotonically increasing after some point. Suppose no such $x_1$ exists. Let
\[
A=\{x_1: \exists n: x_{n+1}<x_n\}\qquad B=\{x_1: \exists n: x_n>1\}
\]
then if $x\in A$, $y\in A$ for all $y<x$ and similarly $x\in B$, $y\in B$ for all $y>B$. Notice also an wasy fact that $1\in B$, so $A$ is bounded. Define, now, $c=glb(A)$. As we assumed $A\cup B=\mathbb{R}^+$, this $c$ implies that $x_1<c\to x_1\in A$ and $x_1>c\to x_1\in B$. It remains to ask whether $c\in A$ or $c\in B$.

If $c\in A$, then for this $x_1:=c$, $x_n\le 1-\frac{1}{n}$ and so $x_{n+1}<1-\frac{1}{n+1}$. Let $y_{n+1}$ be such that $x_{n+1}<y_{n+1}<1$. By above, there's exactly one $y_1$ with $f_{n+1}(y_1)<y_{n+1}$, and notice that $y_1>x_1=c$ by the monotonicity of $f_{n+1}$. This means that there exists $y_1>c\in A$, contradicting the definition of glb. A similar contradiction (but opposite direction) can also be established for the case $c\in B$.

Hence $c$ is neither in $A$ or $B$, means that $x_1=c$ should satisfy the problem condition. Q.E.D.
[/hide]


induction
combinatorics unsolved
combinatorics

[quote="Nbach"]Does there exist a permutation of $ 1,2,...,1992$ such that arithmetic mean of arbitrary two of the numbers isn't equal to
any numbers which is placed between these two numbers in this permutation.[/quote]

First consider that if a sequence $ \{a_{i}\}$ has the required property ($ 2a_{j}\neq(a_{i}+a_{k})$ $ \forall$ $ i<j<k$), then the two sequences $ \{2a_{i}\}$ and $ \{2a_{i}-1\}$ has too this property.

Then use induction :
a). For $ n=1$, the sequence $ \{1\}$ has the property.
b). If the property is verified for any $ k\in[1,n]$, $ n\geq 1$, then split the set $ \{1,\ldots,n+1\}$ in the subsets $ E$ and $ O$ of even and odd numbers.
Then, if $ p=|E|$ and $ q=|O|$, we have $ p\in[1,n]$ and $ p\in[1,n]$ (since $ n+1\geq 2$) and so we can find a permutation $ \{a_{i}\}$ of $ \{1,\ldots,p\}$ and another permutation $ \{b_{i}\}$ of $ \{1,\ldots,q\}$ having the property.

So the sequence $ \{2a_{i}\}\cup\{2b_{i}-1\}$ is a permutation of $ \{1,\ldots,n+1\}$ and obviously has the required property.

And so such permutations exist for any integer $ n$, including $ n=1992$.


I hope this is the correct forum...

I haven't done any maths for about 6 years and realised I don't really understand simplifying rules well enough. We are doing sequences, and specifically I am trying to get a recursive formula for a sequence from a formula for the general term. I am following this part, except for when I need to simpify. Here's the example I am trying to do:

general term:
\[t_n = 3 \cdot 5^{n-1}\]

my working out:
\[t_n - t_{n-1} = 3 \cdot 5^{n-1} - (3 \cdot 5^{n-1-1}) \]

and it's here I am stuck. I really just dont know what to do with the exponents. Don't know how to move them around or simplify them. I know the answer I want to get to, it's:

\[t_1 = 3,\] \[t_n = 5t_{n-1}\]

so simple, yet I don't know how to get there. Can anyone point me in the direction of information that can show me how to do/understand this?

EDIT: This is what I would have done, but the answer is so wrong.
\[t_n - t_{n-1} = 3 \cdot 5^{n-1} - (3 \cdot 5^{n-1-1}) \\
= 3 \cdot 5^{n-1} -3 \cdot -5^{n-1-1} \\
= 5^{n-1} -5^{n-1-1} \\
= 5^{n-1} -5^{n-2} \\
= {2n-3}
\]
[b]Note, the last statement should be an exponent (of nothing!?)[/b], but even still, this just seems dumb. Stuck!?!?!? Is there a tutorial or information that can explain what's going on here specifically related to the example above?[/b]

First, how to do the example. The key is not to look at the [i]difference[/i] between successive terms, but rather their [i]quotient[/i]. In other words, look at $\displaystyle\frac{t_n}{t_{n-1}}$ rather than $\displaystyle t_n-t_{n-1}$.
You should get
$\displaystyle\frac{t_n}{t_{n-1}} = \frac{3.5^{n-1}}{3.5^{n-2}} = 5$
and therefore
$t_n = 5t_{n-1}$.
There are several flaws in the algebra at the bottom of your post - if you'd like me to point them out, post and I'll get back to you.

Just building on Saul's post above...

the reason you find the quotient and not the difference is because it's a [b]geometric[/b] sequence, and not an arithmetic one. A geometric sequence is one where there is a common [b]ratio[/b] between successive terms...which is why you should find the quotient between two consecutive terms to find the ratio.

regarding your simplifying...here are the correct steps:

1) factor out the common factor of 3*5^(n-2) which gives:

3*5^(n-2)*(5-1)
= 3*5^(n-2)*4
=12*5^(n-2)

If you don't see how this works, try expanding the above term, you should get the original expression.

Hope this clears a few things up...cheers!

[quote="Saul"]First, how to do the example.[/quote]
Thanks, that makes sense now.

[quote="Saul"]There are several flaws in the algebra at the bottom of your post - if you'd like me to point them out, post and I'll get back to you.[/quote]
Please do, I would like to understand this.

[quote="mandy_pal"]Just building on Saul's post above...[/quote]
We actually covered this, but when pointed out in this context actually makes sense to me. Thanks.

[quote="mandy_pal"]regarding your simplifying...here are the correct steps:

1) factor out the common factor of 3*5^(n-2) which gives:

3*5^(n-2)*(5-1)
= 3*5^(n-2)*4
=12*5^(n-2)

If you don't see how this works, try expanding the above term, you should get the original expression.

Hope this clears a few things up...cheers![/quote]

I am not sure how to read the first statement, should it be:

[tex] 3 \cdot 5^{n-2} \cdot (5-1) [/tex] [b]or[/b] [tex]3 \cdot 5^{(n-2) \cdot (5-1)} [/tex] ?

Where did the (5-1) come from?

[quote]$\displaystyle t_n - t_{n-1} = 3 \cdot 5^{n-1} - (3 \cdot 5^{n-1-1}) \\
= 3 \cdot 5^{n-1} -3 \cdot -5^{n-1-1} \\
= 5^{n-1} -5^{n-1-1} \\
= 5^{n-1} -5^{n-2} \\
= {2n-3}$ [/quote]
Here are the flawed steps in your reasoning:
Between the second and third steps, you divided the whole by 3. This means that you need to put 3 outside the expression in brackets. You've also introduced a minus sign for no reason at step 2 - and then removed it afterward. Maybe this was just a typo, but if not it could have had a serious effect on some other caluculations. $-(3\cdot5^{n-1-1}) = -3\cdot5^{n-1-1}$. So in fact,
$t_n-t_{n-1}=3(5^{n-1}-5^{n-2})$
As for your last step - could you please post your reasoning for that? It seems to me to be a total [i]non-sequitur[/i].

[quote="Saul"]As for your last step - could you please post your reasoning for that? It seems to me to be a total [i]non-sequitur[/i].[/quote]

I had no idea what to do at this point, so I took a guess. What I did was combine (n - 1) with (n - 2) which gave me 2n - 3 and combined +5 and -5. So I guess what I should have written was [tex]0^{2n - 3}[/tex] :? Hehe, I knew this was wrong, so you see I didn't know how to simplify/deal with this. This is what I was hoping I could get a reference to so I know how to deal with this.

[size=59]I feel about this big.[/size]

Heh, don't worry about it- it happens to all of us. Me very frequently, actually.
So when you say you should have written $0^{2n-3}$, are you implying that $a^b-c^d=(a-c)^{b+d}$? That's not true at all - when you get a difference like that, the best thing you can do is take out any common factors and hope things clear themselves up.
In fact, it seems you know all the standard techniques for simplifaction - it's a matter of applying them.