AMC AIME function USA(J)MO USAMO geometry trapezoid

aimes have consistantly gotten harder, the people are just getting smarter at a much faster pace

Hmm, but how can we explain the huge increase in average score all of a sudden?

I heard that the AIME committee is trying to prevent people from getting 0's, since getting a 0 is discouraging. They're doing this by making the first couple of questions easier so people will get those right, which explains the huge jump in average score.

Bokagadha is probably right. Discouragement is one reason why they made the USAJMO.

Discouragement is also one of the reason for the AMC 10 formation.

Also the AIME has gotten harder. I took the 1986 last year and got a 9 but took the 2008 AIME I and got a 5??

[quote="Arachnotron"]Has anyone noticed that the AIMEs in recent years (2008/2009) have been much easier than the AIMEs of years past? I mean the average scores have doubled, the first few problems wouldn't even be worthy of hard SAT math...what gives?[/quote]

They aren't. Simple as that.

I agree with that. Can you give an example Arachnotron?

I don't think I'm ready for AIME yet.  I do see though it looks like Questions 1-3 are A LOT easier.  2008 AIME A Problem 1 was actually pretty simple!

But I think the late problems keep getting harder

Generally the problems 5 or 6 start getting nontrivial. At least to me.

None of the problems on the AIME are trivial.

I'm pretty sure that most people who take the SAT don't know how to apply their knowledge of math in a way that they can solve #2, especially.

Remember, some people still do poorly on the SAT, and that's saying something.

[quote="Arachnotron"]the first few problems wouldn't even be worthy of hard SAT math[/quote]

That's a little extreme.  #1 on last year's AIME is, imo, more difficult than the hardest problem you'll find on the SAT.
1. It's not multiple-choice.  I know people who have definite math ability, can get 800 on the SAT math in their sleep, and they still missed #1.  Had they had choices, they probably would have gotten it right.

Call it opinion, but the aime has NOT stooped to the level of the SATs.

I don't get it. What kind of SAT math are we talking about? I took a practice one and gee, I scored a perfect in 5 min.

I felt most questions weren't harder than Alg1

This problem is certainly more difficult than the two AIME problems I showed:

[img]http://img366.imageshack.us/img366/466/harderamclx8.jpg[/img]

[/img]

I very much disagree. The IMO nowadays, for instance, is much harder than it used to be.

[quote="Arachnotron"]I guess here's what's happening to contests:

AMC 12 - easier

AIME - easier at "easy" end, same at mid/high end

USAMO - fair bit harder than years past

IMO - more "easy" problems in the beginning but totally made up for by a few killers (3/6).[/quote] I disagree.

AMC 12: Discounting the first year, which was only hard because it was experimental, I think it's gotten steadily harder.

AIME - maybe slightly easier on the "easy" end, but harder for the rest

USAMO- agreed

IMO- harder

[quote="Arachnotron"]I guess here's what's happening to contests:

AMC 12 - easier

AIME - easier at "easy" end, same at mid/high end

USAMO - fair bit harder than years past

IMO - more "easy" problems in the beginning but totally made up for by a few killers (3/6).[/quote]

In my opinion, its like this:

AMC 12 - harder

AIME - easier throughout but requires more training

USAMO - much much much harder

IMO - I think it went up in difficulty until maybe a few years ago... after that it started decreasing in difficulty.

What do you mean, "requires more training". Does this imply that it's less about creative thinking and more about knowing techniques?

Doing things on a one-year or even three-year basis is just ridiculous. The committee comes up with problems, and if they like it, they put it on. It's definitely increasing in difficulty in a ten-year increment, but it's useless to look at things with two years as points of data and call it a trend.

The two hardest USAMO problems are like 1999 3 and 2000 6. Does this mean anything at all about difficulty by year? Why are we using just one question as a "trend-setter" here...?

[quote="Arachnotron"]Take AIME 2009 form I, problems #1 and #2:

[i] Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. [/i]

[i] There is a complex number z with imaginary part 164 and a positive integer n such that z/(z+n) = 4i
Find n [/i]

The first is just knowing what a geometric series is, and the second is mostly an easy manipulation...[/quote]

These problems may be simple to some people, but to others they can be hard.  Take number two, for example, you have to know what an imaginary part actually is and how to manipulate the given equation.

[quote="Arachnotron"]Take AIME 2009 form I, problems #1 and #2:

[i] Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. [/i]

The first is just knowing what a geometric series is, and the second is mostly an easy manipulation...[/quote]

I agree this one is easy concept-wise, but I don't agree that its that easy to get the right answer.  I got it wrong even because I didn't even consider 9,6,4 as a geometric sequence! And I got 10 others right, so it's not because of any incompetence on my part (I mean sure that was retarded of me but still...).  Some problems are hard because it's easy to miss something stupid.  Again this is how AIME's differ from (going back to this) the SAT's : SAT problems aren't tricky, so that region of error is eliminated.

Also, some of these "easier problems" are error-prone.

One of my friends put $ \boxed{839}$ for #1 because he thought $ 9,6,3$ was a geometric sequence

This is what happens when people write perfect papers. Stupid smart people. :P

Actually if these get harder, the standardized tests might get harder... which is good. Right???

[quote="not_trig"]lol. I know someone who got #2-15 but not #1 last year.

The AMC12 has become very slightly harder. The "lack of calculator" thing made them give more legit problems.
AIME: about the same? like pythag said, gotta know more stuff though
USAMO: harder, definitely. Except like 2005.[/quote]

I still don't think AIME has gotten easier or stayed the same.

Compare an AIME from 2000-2009 and an AIME from 1983-88

[quote="gauss202"][quote="AwesomeToad"]
I still don't think AIME has gotten easier or stayed the same.

Compare an AIME from 2000-2009 and an AIME from 1983-88[/quote]

What about the years in between, 89-99?[/quote]
I think gauss202 meant comparing an old one to a new one, not that those years necessarily matter.

I haven't seen those, except for '98.

But I thought 98 was sort of hard. I was younger back then, don't take my word for it...


ARML
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email
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so.. it seems the indiana forum's quieted down a bit since the last few months.. haven't checked the forums in like forever.

anyway, who's going to the rose-hulman contest? i transferred to a basically intelligence-deprived school in lafayette, so idk what kind of "team" i'll be on..

what are the tests like? the other post regarding the contest said it was about AMC 10/12 level.. any practice tests available?

OMG. I JUST FREAKING TYPED UP 3 PARAGRAPHS, AND THE STUPID THING LOGGED ME OUT.
Ok sorry.
http://www.rose-hulman.edu/~rickert/NovContest/
There's your site. Unfortunately, no previous tests are available, but you can probably ask the ARML coaches (e.g. Dr. Chi, Dr. Ji, and Dr. Rickert) if you actually did ARML.

Yes, the tests are about high school level, and can range from AMC 10 - the Last Few questions of AMC 12's, maybe even AIME.

In 2006:
Results were: Nathan, John Ryan, Tyler, Sachin, Peter, some other person, Lauren, Shawn, etc.

Nathan missed one, Sachin missed 4 (!), Lauren and Shawn both missed about 5~6. Ok, that should give you a scope of how difficult the test really is. The last few problems are tricky and difficult, so don't get your hopes up.

If you are who I think you are, which would be that one kid from ARML who was friends with Mikha and Anirudh, then I think you're probably gonna get maybe -7? Maybe? Of course, I'm not sure if I remember the problems missed correctly, because some of the harder problems are weighted more. So in reality, Sachin could have merely missed 2 or 3 problems while Lauren and Shawn missed 4.

The contests have different scoring systems, and I definitely don't know what Rose-Hulman's is. However, as the 9th grade class in Indiana is not too strong, with the exception of Jeffrey and some others, you'll probably get a pretty high rank.

[hide=" Read only if you have nothing better to do, Seriously."]My year sucks. So many damned good people, not. Jk. We have youkow, Justin, Michael Mardini, Andrew Keiser (? Maybe ?), Yuqing left (YES! LESS COMPETITION), hmm... there must be some others I haven't mentioned... Oh wait, maybe Marissa?? Dunno, apparently she's learning with Benjamin's in pre-calc now. W/e. [/hide]

Good luck, have fun :).

thanks for the info, and yeah, i'm soohyun, the one kid from ARML who hung around with mihika/anirudh

i'm just going to estimate -8 because
1. i STILL haven't picked up any math books other than my school math book, and
2. so i don't disappoint myself.

so much for dedication.

"do your best!" <_<

If you look only scores of 75+ (you will see rank, name and then score, but some scores for the people are NOT posted-they did nbot get 75+) are posted, thats for people that got atleast 16/20 right.

If you miss 4 thats a 76, if you miss 3 thats a 82...

If you dont do 4 and get the rest right thats a 80.

Its 5x-w, where you get 5 points for every right, -1 point for every wrong, none for leave blank.

clearly last years freshman class (this years sophomores) is the best returning group, they had the most peopel with 75+ and the best averages, i think 12th place 9th grade (10th this year)last year got a 68.  12th place for 10th grade last year (11th this year) got a 52, same with 11th grade last year..and plus they had the best average and they were only in 9th grade..

i dont know about this years 9th grade...i know this guy whos really good 9th grade.

anyways the questions 1-10 are fundamental grade probelms..  (they are fundamental problems, like find sin75-sojust use (sin(a+b))...ikeb 10-15 are amc types (they are like 2 step probablilityes), 16-20 suddenly becomes AIME problems....theres like no middle ground.

what kind of scoring do they have (eg # problems, credit for not answering, etc.)?

could i try out for the jeff team soo-hyun? i haven't heard anything from my school.

yeah, i don't think west side participates in this contest.. at least it didn't last year.

you have to enroll in jeff to join the jeff team. not worth it anyway; i'm not expecting any success with the jeff freshman team.

apparently anyone's elligible for individual awards, so you could email Dr. Rickert and see if you can register yourself.

se-hyoon, are you leaving before AMC 10.

*soo-hyun* is not leaving before the AMCs, thank you.

Soo hyun, don't underestimate yourself.

First off, the first 12 are ridiculously easy. Ridiculously, as in I can get them in 8 minutes (I'm not bragging, I end up failing the rest in the remaining 52 minutes and getting maybe 1 or 2, and I'm not even joking).

Second off, you beat Lauren and Shawn on the AIME last year by a good amount (7 vs. 5 and 4) -- I know there's no directly correlation, but I wouldn't be surprised if you beat them again this. Nothing against Shawn and Lauren -- just dishing out scores. I feel like a fantasy-math league owner. Haha that'd be hilarious -- draft math students, get points based on how they do on contests...

Third off, to whoever mentioned weighting questions, I don't think they do that based on the instructions I read for the 2004 and 2003 tests... but I could be wrong

Fourth off, is the system -1 for every one you miss, +5 for every one you get right, and 0 for those left blank? If so, that encourages guessing... (if you guess on all twenty, you're expected to get 4 right and miss 16, for a net score of +4. If you leave all 20 blank, your net score is 0)

ahh but they took the AIME I, which was apparently easier than the AIME II? and on your blog it says you got a 9 on the AIME II.. soooo.

and shawn/lauren probably practiced. i picked up a math book yesterday. i'm very proud of myself.

oh but if it's twenty questions (i'd thought it was thirty? idk.) i guess i can guess a -6

wellll see you next saturday/maybe we can play kemps again (:

Yea it should be 5,0,-1.25, not -1 for missing...to even it out (i think it should be -2 just to not encourage random guessing), but atelast 1.25

but they have some other probelsm with rankings

say person A gets a 100, person B gets a 95, person C gets a 94, and person D gets a 90.

Then the rakings would be

1.person A-100
2.person B-95
3.person C-94
4.person D-90

but lets say person A,B,and C get 100, while person D gets a 90.

then it would be

1.A-100
1.B-100
1.C-100
[b]2[/b].D-90

NOT

1.A-100
1.B-100
1.C-100
[b]4[/b].D-90

anyother competiton, the person D would get 4th place, but in the Rose-Hulman competiotn, he/she would get 2nd place, so if you dont get a top place, you hope the people in front of you tie, then you would get a better posotion.

if you dont believe me, go to the Rose-Hulman contest website and you'll see what I mean..I hope they changed it this year

And no I didndt study, I have no time wasting my time during the week with math...but I have done some AIME probelms during Sundays

1.     David Benjamin 94    Harrison
        Phil Thomas 94  Harrison
2.     Pardha Ponugoti 90   Terre Haute South

last years 10th grade top finishers, pardha shouldve gotten 3rd, not 2nd if this was anyother competion.

would i still have time to register myself as an individual or something if i really wanted to?

or is it too late?

website says
Novermber 2, 2007 (friday) was the last day to register

but you know dr rickert and dr. chi so i dont know..

BTW I dont know if our team has enough Freshman, if we dont, we are disqualified for team round scores because we dont have a full team of 16 (4 fresh, 4 soph...), even if we only have 15 people and get a 0 score (no person competes) for the 16th person and that is good enough for top 3 team score, they still wont count our team score because we dont have a full team..which is another dumb rule they need to change...
(along with the tie rule i discussed above)

thats like playing football 11 on 10 and the team with 10 people wins and they dont count it beucase the other team only had 10 players (even though the team with 10 players already are handicapped)-its not like we are using extra people, we are using less than required...(thats why it didnt say we got top 3 last year when we wouldve gotten either 2nd/3rd)

im pretty sure at ARML if you had a team with 14 people and they ended up winning, not only would they count that as a win (and not disqualification because they didnt have maximum of 15 competitors), but they would be praised even more overcoming the fact they had 1 less person (which obviously hurts relays and individuals)

i just took the 2005 test.

17 x 5 - 3 = [b]82[/b]

i'm going to be disappointed if i get anything less than that on saturday.

I just took the 2002 test
0x5-20= [b]-20[/b]

ill set a record if i get anything less than that.

you actually have to try to get -20

does anyone have solutions for numbers 19 and 20?

where can you find thhese tests? there arent any on wiki. also, where can you find the mandelbrot tests?

EDIT: besides buying them..

For #19, once you turn the problem into actual math, it isn't that difficult -- but turning into math is really fricken hard. I would explain, but it's hard as heck and I know I'd end up confusing most of you anyways.

For #20, assume the point is at a vertex of the face. Then you'll have (0+0+something)/(0+0+something else) and "something" and "something else" shouldn't be that difficult to find... I'll work on it again later.

-------
This year's test was geometry-tastic. Two 3-D geometry problems =( (and I missed both)

Oh, and I was about to have a heart attack when they called Soo-Hyun 3rd, then called some other Iowa City dude 2nd before they called me as co-2nd... thought they didn't grade my test!

Good job to everyone who was there, especially mihika for winning the freshmen comp!

i thought "d***" when they called me third place, but was shocked when they called you 2nd and mihika 1st then got really depressed that i'd counted wrong.. i actually got 78 cause I'M NEVER FREAKIN PERFECT

it frustrates me.

that other 3-D problem with the sphere in a cube was almost like the one with the square/circle on the 05/06 test (i don't remember which)

just had to replace $ \sqrt {2}$ with $ \sqrt {3}$..

i hate making mistakes. none of which will exist on my AMC 10 this year.

se-hyoon, i noticed you changed your goals for 13+ AIME to 9+ AIME and 140+ AMC12 to 132+ AMC12, why

as i've said, my original plan was to study math like crazy for the remainder of the summer and the rest of 2007, but i didn't pick up my math books until the last few weeks. of course my goals have gone down for now, assuming i'll work at my planned pace.

right now that pace is one AMC 10/12 each day + owning most problems on each test by end of week

owning all problems by end of winter break.

then practice AIMEs sometime after.

i'm just hoping that it'll work. heh.

what about you, [i]rudy[/i]?

[quote="Fanatic"]
For #20, assume the point is at a vertex of the face. Then you'll have (0+0+something)/(0+0+something else) and "something" and "something else" shouldn't be that difficult to find... I'll work on it again later.
[/quote]

the answer is just $ \csc\theta$ where theta is the angle two faces make with each other

and #19 was so annoying... i somehow got 800 miles and put e

[quote="Fanatic"]
Oh, and I was about to have a heart attack when they called Soo-Hyun 3rd, then called some other Iowa City dude 2nd before they called me as co-2nd... thought they didn't grade my test!
[/quote]

that was [i]awful[/i]. one of the worst couple seconds of my life.

[quote="Fanatic"]
For #20, assume the point is at a vertex of the face. Then you'll have (0+0+something)/(0+0+something else) and "something" and "something else" shouldn't be that difficult to find.[/quote]
???
to find the distance to a face, you would find the perpendicular distance from the point to the plane containing the face, right (that is if the perpendicular doesn't pass through the face itself)? i vaguely remember a formula that might've been of use finding the distance from each of the edges?

hjere are the results with the people with 75+ scores listed with their names and scores and the top 12 listed by name for each grade

http://www.rose-hulman.edu/~rickert/NovContest/novcont2007.html


floor function
logarithms
Euler
factorial

How many digits does the number $ 2^{100}$ have? And $ 2^{1000}$?

Please show your method... :)

The number of digits in a positive integer $ x$ is $ 1 \plus{} \lfloor log_{10} {x} \rfloor$. (Try to figure out why).

First of all, thank you dgreenb801!

Well... that's what I thought too, but then I wasn't able to find the effective value...

$ [100\log_{10} 2] \plus{} 1 \equal{}$ ?

(I can't use the calculator  :lol: !)

I also thought about
$ [10 \log_{10}1024] \plus{} 1$, so the result is around 30, but the off-by-one danger is killing me now :D

Suggestions? Hints? :lol:

hmmm this looks very similar to a programming contest problem (I believe project euler)

@batteredbutnotdefeated: [url=http://projecteuler.net/index.php?section=problems&id=16]This problem[/url]? I recall a similar problem involving the number of digits in a factorial from one of the ICPC judges I practice at.

@Gabuntu94: I wouldn't be very concerned with off-by-one error. We know that $ \log_{10}{1024/1000} \approx 0$, so $ \log_{10}{1024} \approx 3$ with a reasonable precision (probably $ \pm 0.01$). I think you're safe in saying that $ 2^{100}$ has $ 31$ digits.

However, I would definitely be more cautious about $ 2^{1000}$.


algebra
polynomial
real analysis
real analysis solved

Find countable disjoint sets in $\mathbb{R}^1$ such that each set is countable and dense in $\mathbb{R}^1$.

I think $ A_k={ k+\sqrt{p_k}\cdot Q }$ with $ k>0 $ is an example.Here $ p_k$ is the k+1-th prime and Q is the rationals.

Is my example right? It goes here.

if $x - y \in \mathbb{Q}$, then we put $x, y$ in the same set.
We choose the sets which contain $\sqrt{p_k}$ respectively where ${p_k\}}$ are primes.

Couldn't we take $A_i=Z[e^i]\setminus Z=\{m+ne^i|m\in Z,n\in Z^*\}$? Each such set is countable, dense, and if two sets have an element in common then it must mean that $e$ satisfies a polynomial equation in integer coefficients, which is false because $e$ is transcendental. You can easily produce an argument to prove the density of the sets only from the irrationality of $e$.

Yes, it's good. You choose in fact the sets $ A_k=\sqrt{p_k}+Q $ and they are good.

[quote="grobber"]Couldn't we take $A_i=Z[e^i]\setminus Z=\{m+ne^i|m\in Z,n\in Z^*\}$?[/quote]
Is $e^i$ a real number?

$i$ is just the index: it runs through the integers.


imi poate spune cineva th lui kroniker sau un link in lb. romana...pls

Teorema lui Kronecker: Daca $\alpha \in \mathbb R - \mathbb Q$ atunci multimea $\{ m\alpha + n \mid m,n \in \mathbb Z \}$ este densa in $\mathbb R$. Asta inseamna concret ca pentru orice $x \in \mathbb R$ si orice $\varepsilon >0$ exista $m,n \in \mathbb Z$ astfel incat
\[ | x - (m\alpha + n) | < \varepsilon . \]


induction
combinatorics solved
combinatorics

A country has 2001 towns. Each town has a road to at least one other town. If a subset of the towns is such that any other town has a road to at least one member of the subset, then it has at least k > 1 towns. Show that the country may be partitioned into 2001 - k republics so that no two towns in the same republic are joined by a road.

Hi,
in terms of graphs the problem can be formulated:

Let G = (V,E) be a graph with n vertices (n=2001), such that d(v)  \geq 1 for every vertex v. Suppose that for every subset S of vertices, if w has a neighbour in S for each w in V\S, then |S|  \geq k > 1. Then the chromatic number of G is at most n - k (i.e., G can be colored with n - k colors in such a way that two adjacent vertices have different colors).

Let N(v) be the set of neighbours of a vertex v.
First note that d(v)  \leq n - k for every v, otherwise we can take the subset S= V \ N(v), |S| < k, and every w in V \ S = N(v) is adjacent to v (so it has a neighbour in S).
The proof follows by induction in n.
For small n, it can be proved by hand.
Suppose that it holds for graphs with less than n vertices, and let G be a graph with n vertices.

If G has a vertex v with d(v) < n - k, Let G' be the graph induced by G-v.
Note that G' verifies the same property than G but with k-1. So, by inductive hypothesis it can be colored with n-1 - (k-1) = n-k colors.
Since v has in G degree less than n - k, it follows that the whole graph G can be colored with n - v colors.

Otherwise, all the vertices of G have degree equal to n - k.
Let v be a vertex of G, and let's consider the set S = V \ N(v).
|S| = k > 1. Suppose that there are two adjacent vertices w, w' in S.
By definition of S, w and w' are different from v. So, taking the set S \ w,
it verifies that every vertex not in the set is adjacent to a vertex in the set.
But this is not possible because |S \ w| = k-1. So S is a stable set.
On the other hand, since d(w) > 1 for every w, every w in S has at least a neighbour in V \ S, and |S| = k > 1. So, by the upper bounds in the degrees, the n - k vertices in N(v) can't form a complete subgraph. Therefore, they can be colored using less than n - k colors. Adding a new color for the vertices in S (they are a stable set), we have a coloring of G with  \leq n - k colors.

That's all.

bye!
fla

I have just read it. And it seems to be correct.
Are there other opinions?  :?

I have a question..

[quote="fbonomo"]

If G has a vertex v with d(v) < n - k, Let G' be the graph induced by G-v.
Note that G' verifies the same property than G but with k-1.
[/quote]

Are you sure it has the same property? What if, after removing $ v$, a vertex in $ G\minus{}v$ has degree $ 0$?

[quote="Albanian Eagle"]cant you just use this proof for a connected graph and then the problem would follow by applying the result to each connected component?[/quote]

How? Can you show some more details?

Consider the maximum subset $ S$ of towns such that no two of them are connected by a road. Each of the other $ 2001-S$ towns has a road with at least one of $ S$. This means $ S\geq k$ if $ S>k$ then, consider the $ 2001-k$ republics as $ S$ and each of the other $ 2001-S$ towns (each of them alone) making $ 2002-S$ republics, less or equal than $ 2001-k$ (we can divide $ S$ into smaller parts if we ended with less than $ 2001-k$ republics) . Then $ S=k$. Outside $ S$ there are $ 2001-k$ towns, if there are two towns in this $ 2001-k$ set not connected by a road, then we can make a $ 2-$republic with both of them, $ S$ and other $ 1999-k$ $ 1-$republics. This means every two of the $ 2001-k$ (say $ X=$no $ S$) are connected by a road. Each of the $ X_s$ is connected with at least one in $ S$. If there exists a town $ T$ of $ X$ such that it is connected with at least two towns in $ S$ then the set $ T\cup{}$all elements in $ S$ not connected with ${ T}$ has less than  $ k$ elements and each of the remaining towns has a road to it (a road going to $ T$). Then, all elements in $ X$ are connected with exactly one element in $ S$ for each element in $ S$ there is another in $ X$ such that both towns are not connected, otherwise one town in $ S$ would be connected with every town in $ X$ and each of the other towns in $ S$ would have no roads. Now we can construct the required republics by separating all $ 2001-k$ elements of $ X$ in different sets and adding each town of $ S$ to any of the republics that began being constructed with an $ X-$town  that has no road to it.


[quote="marko avila"]let n such that n-1 and n+1 are twin primes prove that the numbers 1 through n can be arranged in a row such that the sum of any two consecutive numbers is prime[/quote]

This is not number theory, this is combinatorics, and a really trivial combinatorics problem, by the way:

Since the numbers n - 1 and n + 1 are both prime, they are odd, so the number n must be even; now, define the permutation $\left(a_1;\;a_2;\;...;\;a_n\right)$ of the numbers (1; 2; ...; n) as follows: $a_{2k+1}=n-2k$ for any $k\in\left\{0;\;1;\;...;\;\frac{n}{2}-1\right\}$ and $a_{2k}=2k-1$ for any $k\in\left\{1;\;2;\;...;\;\frac{n}{2}\right\}$. Of course, the numbers $a_{2k+1}$ are the even numbers from the set {1; 2; ...; n}, while the numbers $a_{2k}$ are the odd numbers from this set. Here is how the permutation $\left(a_1;\;a_2;\;...;\;a_n\right)$ looks like:

$\left(a_1;\;a_2;\;...;\;a_n\right)=\left(n;\;1;\;n-2;\;3;\;n-4;\;5;\;...;\;2;\;n-1\right)$.

Now, if we arrange the numbers 1, 2, ..., n according to this permutation, then the sum of any two consecutive numbers will be prime, since it will be either n - 1, or n + 1; in fact,

$a_{2k}+a_{2k+1}=\left(2k-1\right)+\left(n-2k\right)=n-1$

and

$a_{2k+1}+a_{2k+2}=a_{2k+1}+a_{2\left(k+1\right)}=\left(n-2k\right)+\left(2\left(k+1\right)-1\right)=n+1$.

This solves the problem.

  darij

ohhh sorry darij!! yes it is very trivial indeed!! thanks!! :)


Find the greatest common factor of 9,009 and 14,014.

Is there an easier way to figure out that the answer is 1001?

use euclids algorithm because you can easily see that $14014-9009=5005$, and $9009-5005=4004$, and $5005-4004=\boxed{1001}$


integration
calculus
calculus computations

Use a substitution to prove that

$\displaystyle\int^{ab}_a \frac{1}{t}\,dt = \displaystyle\int^b_1 \frac{1}{t}\,dt$

It is indeed easy. Change $t$ to $at$. :)


Source: AHSME 1973 #7

The sum of all integers between 50 and 350 which end in 1 is
$\text{(A)} \ 5880 \qquad \text{(B)} \ 5539 \qquad \text{(C)} \ 5208 \qquad \text{(D)} \ 4877 \qquad \text{(E)} \ 4566$

[hide]We want $51+61+71+...+341$

There are 30 integers in that sequence so the sum is $\frac{30(392)}{2}=5880$.[/hide]

[quote="Farenhajt"][hide]Such numbers can be represented in the form $10n+1$ where $n$ runs from $5$ to $34$. Hence

\begin{eqnarray*}S &=& \sum_{n=5}^{34}{(10n+1)}\\ &=& 10\left(\sum_{n=5}^{34}n\right)+30\\ &=& 10\cdot\left(\frac{5+34}{2}\cdot 30\right)+30\\ &=& 5880\end{eqnarray*}[/hide][/quote]

Ryan22 would you kindly not duplicate posts word for word unless it's to post a comment or question about the poster's solution?


function
integration
complex analysis
complex analysis unsolved

$f$ is $C^{\infty}$ function on $R$ with compact support.

$Y_{z}(f)=\frac{1}{\Gamma(z)}\int_{0}^{\infty}x^{z-1}f(x)dx$

1.Show that $Y_{z}(f)$ is an entire function of $z$.

2.Show that $Y_{-k}(f)=(-1)^{k}D^{k}f(0)$ ($k$ times differential).

Can someone help, thanks.

I got a hint: also works for complex-valued $f(x)$,

use Taylor expantion of $f(x)=\sum_{k=0}^{\infty}\frac{D^{k}f(0)}{k!}x^{k}$

we can see that

$T_{z}(f)=\int_{0}^{\infty}x^{z-1}f(x)dx=\int_{0}^{1}x^{z-1}f(x)dx+\int_{1}^{\infty}x^{z-1}f(x)dx$

RHS second term is entire function of $z$,

first term=$\int_{0}^{1}x^{z-1}f(x)dx=\int_{0}^{1}\sum_{k=0}^{\infty}\frac{D^{k}f(0)}{k!}x^{z+k-1}dx$

$=\sum_{k=0}^{\infty}(\int_{0}^{1}\frac{D^{k}f(0)}{k!}x^{z+k-1}dx)=\sum_{k=0}^{\infty}\frac{D^{k}f(0)}{k!}\frac{1}{z+k}$

So $T_{z}(f)$ as a function of $z$,is meromorphic
and has simple poles at $0,-1,-2...$.

Since $\Gamma(z)$ is also meromorphic and also has simple poles at $0,-1,-2...$.
Thus the conclusion.

*******************************************************************
*******************************************************************

I want to know is it valid to write $\int_{0}^{1}\sum_{k=0}^{\infty}\frac{D^{k}f(0)}{k!}x^{z+k-1}dx=\sum_{k=0}^{\infty}(\int_{0}^{1}\frac{D^{k}f(0)}{k!}x^{z+k-1}dx)$?
Is it because  $\sum_{k=0}^{n}\frac{D^{k}f(0)}{k!}x^{z+k-1}$ converges uniformly on [0,1]? But $x^{z+k-1}$ may not be bounded on [0,1].
Or is it Leb dominated conv thm? But again, $x^{z+k-1}$ may not be bounded on [0,1].


linear algebra
matrix
abstract algebra
modular arithmetic
number theory unsolved
number theory

Write the positive integers in a spiral as follows:
\[ \begin{matrix}
				7 & 8 & 9 & 10\\
				6 & 1 & 2 & \cdots\\
				5 & 4 & 3 & \cdots
				\end{matrix}\]
		Place a white pawn on every square whose number is divisible by 3 or 5. Can a white queen, starting at 1, make a single move of a million squares? (A queen may move along any row, column, or diagonal for any distance, but may not move through a square occupied by a piece of her own color.)

$ \begin{matrix}
21 & 22 & 23 & 24 & 25\\
20 & 7 & 8 & 9 & 10\\
19 & 6 & 1 & 2 & 11\\
18 & 5 & 4 & 3 & 12\\
17 & 16 & 15 & 14 & 13
\end{matrix}$

From this diagram we can see the queen can't go
down right ($ 3|3$)
down ($ 5|15$)
down left ($ 5|5$)
left ($ 3|6$)
up left ($ 3|21$)
up ($ 5|45$)
up right ($ 5|25$)
and right will have $ 51$, which is divisible by $ 3$.
So no the queen can't move a million squares if a pawn was placed on squares divisible by $ 3$ or $ 5$.

O Noes !! Bleah, sorry :(

Anyways, then yes the queen can go straight up.

Through finite differences, the pattern going up can be found to be $ 4n^2 \minus{} 5n \plus{} 2$.

Taking this congruent to module 3, we get $ n^2 \plus{} n \plus{} 2\equiv 0 \pmod 3$
Since every integer may be written as $ 3k, 3k \plus{} 1, 3k \plus{} 2$ for an integer $ k$, the congruence has no solution.
Taking it $ \pmod 5$, we get the relation $ n^2\equiv 2 \pmod 5$.
Again, using the idea that every integer is expressed as $ 5k,...,5k \plus{} 4$, we can see no solution exists.

So yea, the queen can go up. I miscalculated earlier :(
Also the pattern is after the first spiral each new direction is increased by the recursion $ a_n \equal{} a_{n \minus{} 1} \plus{} 8$...if I wrote it write.
My idea is the +8 comes from the extra 8 from the sides beside the corners. All together its (at the corners) $ 12\minus{}4$ ("original ones")$ \equal{} 8$


inequalities
vector
inequalities unsolved

Let $x,y,z \in R$ and $ h_{1},h_{2}, h_{3} \in (0, \frac{1}{2})$ so that $ \sum  h_{1}=1$.
Prove that we have: $ h_{1}h_{2}h_{3}(x+y+z)^2 \geq \sum xyh_{3} (1-2h_{3})$. When do we have equality?

made by Mereni I.

cheers! :D :D

Well,the inequality is equiv. to $x^2+y^2+z^2\ge 2cos\alpha xy+2cos\beta yz+2cos\theta zx$ where $cos\gamma=\displaystyle\frac{h_i^2+h_{i+1}^2-h_{i+2}^2}{2h_ih_{i+1}},\sum\alpha=\pi$($\gamma=\alpha,...,\theta$)
Suppose that $X,Y,Z$ be vectors with angle between them $\alpha,\beta,\alpha+\beta$ in some order,then the above inequality becomes $X^2+Y^2+Z^2\ge 2XY+2YZ-2ZX\Leftrightarrow (X+Z-Y)^2\ge 0.$


limit
real analysis
real analysis unsolved

If $ \lim_{n \rightarrow \infty}a_{n}\equal{}0$

and for $ \forall$ $ i < j$, $ a_{i}>a_{j}$

In other words, $ \{a_{n}\}$ monotony[color=red](strictly)[/color] tends to zero as n tends to infinity

show that:

$ \sum_{n\equal{}1}^{\infty}(\minus{}1)^n\frac{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}{n}$

converges

$ \sum_{n\equal{}1}^{\infty}(\minus{}1)^n\frac{1}{n}$ converges

but $ {a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}$ can not always bounded

for example : $ a_{n}\equal{}\frac{1}{n}$

so I think Abel's Test is invalid......

If we consider Dirichlet's Test:

$ \{\frac{1}{n}\}$ is monotony tends to zero

but $ \sum_{n\equal{}1}^{\infty}(\minus{}1)^n\left(a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\right)$

It's $ \sum_{k\equal{}1}^{n}(\minus{}1)^k\left(a_{1}\plus{}a_{2}\plus{}...\plus{}a_{k}\right)$ will not always bounded.....

for example: $ a_{n}\equal{}\frac{1}{n}$

so Dirichlet's Test is invalid too.......

How can we beat down this problem....... :maybe:

Define $ S_n \equal{} \frac {a_1 \plus{} \cdots \plus{} a_n}{n}$, it can be shown that,
due to the fact the $ \{a_n\}$ is decreasing, we have $ \{S_n\}$ is also decreasing.

Now since $ \lim_{n\to \infty} a_n \equal{} 0$, we also have $ \lim_{n\to\infty} S_n \equal{} 0$,
thus by the Leibniz test (the alternating series test) we have $ \sum_{k \equal{} 1}^n ( \minus{} 1)^k \frac {a_1 \plus{} \cdots \plus{} a_k}{k}$ also converges.

Similar argument can be made to the series $ \sum_{k \equal{} 1}^n ( \minus{} 1)^k\frac {1}{k}$.


Can someone help me to solve this problem.
Obtain an appromaximation of the following function.
(See attached file)

Can you give us some more information?
Is this from physics?
Are you interested in one variable $A$, $b$ or $\omega$ in particular?
Can we assume something like $A\gg \omega$ or $b\ll A,\omega$?
Maybe some Taylor series can help.

T = 6(ln(A/b))/(ln(w/b))
T= ln((A/b)^6)/ln((w/b)^6)

Now, what do you want more?

[quote="neelnal"]$T = 6\frac{\ln\left(\frac{A}{b}\right)}{\ln\left(\frac{w}{b}\right)}$
$T=\frac{\ln\left(\frac{A}{b}\right)^{6}}{\ln\left(\frac{w}{b}\right)^{6}}$

Now, what do you want more?[/quote]
Actually, $T=\frac{\ln\left(\frac{A}{b}\right)^{6}}{\ln\left(\frac{w}{b}\right)}$ :wink:

And... $\log_{\frac{w}{b}}\left(\frac{A}{b}\right)^{6}=T$
$\left(\frac{w}{b}\right)^{T}=\left(\frac{A}{b}\right)^{6}$

Cause it is too complex. Therefore, is there an appromaxite to this question.
w and b are constant


The measures of the three angles of a triangle form an arithmetic sequence. What must be the measure, in degrees, of at least one of its angles?

Let the first term in the sequence be $ a$, and the common difference between the terms be $ d$. Then the terms are $ a$, $ a\plus{}d$, $ a\plus{}2d$. Thus, we also have

$ a\plus{}a\plus{}d\plus{}a\plus{}2d\equal{}180$

$ \implies 3a\plus{}3d\equal{}180$

$ \implies 3(a\plus{}d)\equal{}180 \implies a\plus{}d\equal{}60$.

Since one of the terms in the sequence is a+d, one of the aangles must be $ \boxed{60^\circ}$.


function
algebra
polynomial
algebra unsolved

[quote="pluristiq"]by Newton's Identities, the elementary symmetric functions of the $ x_i$ are uniquely determined, and thus the $ x_i$ (as a set) are uniquely determined. since $ (0,0,...,0,9)$ and its permutations are solutions, these are all solutions, and we're done.[/quote]
I think we must prove $ x_i \geq 0$ for every i=1,2,...9
Can you tell me some about Newton's Identities,
thanhks :)

if we have variables $ x_1, x_2, ..., x_n$, the elementary symmetric polynomials in these variables are

$ e_1(x_1,...,x_n) \equal{} x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} \sum_{i}x_i$

$ e_2(x_1,...,x_n) \equal{} x_1x_2 \plus{} x_1x_3 \plus{} \cdots \plus{} x_{n \minus{} 1}x_n \equal{} \sum_{i < j} x_ix_j$

$ e_3(x_1,...,x_n) \equal{} \sum_{i < j < k} x_ix_jx_k$

$ \vdots$

$ e_n(x_1,...,x_n) \equal{} x_1x_2\cdots x_n$

for $ m > n$ define $ e_n(x_1, ..., x_n) \equal{} 0$.

also define $ p_k(x_1, ...,x_n) \equal{} x_1^k \plus{} x_2^k \plus{} \cdots \plus{} x_n^k$.

Then Newton's Identities are

$ e_1 \equal{} p_1$

$ 2e_2 \equal{} e_1p_1 \minus{} p_2$

$ 3e_3 \equal{} e_2p_1 \minus{} e_1p_2 \plus{} p_3$

$ \vdots$

$ ke_k \equal{} e_{k \minus{} 1}p_1 \minus{} e_{k \minus{} 2}p_2 \plus{} \cdots \plus{} ( \minus{} 1)^{k \minus{} 1}p_k$.

Now, to your problem. $ x_1, ..., x_9$ are the roots of the polynomial $ f(x) \equal{} x^9 \minus{} e_1x^8 \plus{} e_2x^7 \minus{} \cdots \plus{} e_8x \minus{} e_9$. since $ p_k \equal{} 9^k$, plugging into Newton's Identities in order we get $ e_1 \equal{} 9$, $ e_2 \equal{} e_3 \equal{} \cdots \equal{} e_9 \equal{} 0$. thus $ f(x) \equal{} x^9 \minus{} 9x^8 \equal{} x^8(x \minus{} 9)$, whose roots are $ 0, ..., 0, 9$.


Suppose that $ x\plus{}\frac1y\equal{}y\plus{}\frac1z\equal{}z\plus{}\frac1x$, where $ x,y,z$ are different to each other. Find $ x^2y^2z^2$.

Let $ x \plus{} \frac {1}{y} \equal{} y \plus{} \frac {1}{z} \equal{} z \plus{} \frac {1}{x} \equal{} k$,

$ x \plus{} \frac {1}{y} \equal{} k\Longleftrightarrow x \equal{} \frac {ky \minus{} 1}{y}\ \cdots [1]\ \therefore \frac {1}{x} \equal{} \frac {y}{ky \minus{} 1}$.

$ y \plus{} \frac {1}{z} \equal{} k\Longleftrightarrow z \equal{} \frac {1}{k \minus{} y}\cdots [2]$.

$ \therefore z \plus{} \frac {1}{x} \equal{} k\Longleftrightarrow \frac {1}{k \minus{} y} \plus{} \frac {y}{ky \minus{} 1} \equal{} y$

$ \Longleftrightarrow (k^2 \minus{} 1)(y^2 \minus{} ky \plus{} 1) \equal{} 0$

$ \Longleftrightarrow k \equal{} \pm 1$ or $ k \equal{} y \plus{} \frac {1}{y}$.

Since $ x,\ y,\ z$ are distinct non zero numbers, $ k \equal{} \pm 1$, then  by $ [1],\ [2]$ we have $ x \equal{} \frac {y \minus{} 1}{y},\ z \equal{} \frac {1}{1 \minus{} y}$, yielding $ xyz \equal{} \minus{} 1$ for $ k \equal{} 1$,

or $ x \equal{} \minus{} \frac {y \plus{} 1}{y},\ z \equal{} \minus{} \frac {1}{y \plus{} 1}$, yielding $ xyz \equal{} 1$ for $ k\equal{}\minus{}1$. Thus desired answer is $ x^2y^2z^2 \equal{} |xyz|^2 \equal{} 1$.

Is there another method that we don't have to "solve" the equation(s)/system?


search
geometry
rectangle
combinatorics unsolved
combinatorics

How many way to cover the wall rectangular $ 3$x$ 2n$ by bricks $ 1$x$ 2$.

Let $ a_n$ be the number of ways to cover a $ 3\times 2n$ board by $ 1\times 2$ dominoes. Then consider the leftmost $ 3\times 2$ board-section. It can be covered in either one of the following ways:

1. One horizontal and two vertical dominoes

This can be done in 2 ways. The first with the horizontal domino in the bottom row and the second way with the horizontal domino in the top row.

2. Three horizontal dominoes

This can be done in 1 way.

Therefore we get the recurrence $ a_n \equal{} a_{n\minus{}1}\plus{}a_{n\minus{}1}\plus{}a_{n\minus{}1}\equal{}3a_{n \minus{} 1}$. Then noting that $ a_1 \equal{} 3$ we immediately get $ a_n \equal{} 3^n$.

No, you don't.  http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1779296039&t=126738

Or you could go read the thread I posted a link to, where there is a correct solution.

OK then sorry, I didn't notice your post as I have been writing up my solution for a long time. But can you please specify where my mistake is? :maybe:

You realized your first mistake: not every tiling is a tiling of the first $ 3 \times 2$ rectangle glued on to the rest.  It's also true that not every tiling is a tiling of the first $ 3 \times 4$ rectangle glued on to the rest, and so on.  So, for example, you missed the tiling of a $ 3 \times 6$ rectangle in which the bottom edge is covered with horizontal tiles, the left and right edges have one vertical tile each, and all other tiles are horizontal.  Such tilings exist for every $ n$, so your (corrected) method leads us to a recursion in which each term is dependent on all previous terms.  (It's solvable by this method, but the linked solution avoids this complication.)


[quote="don_e_thomas"]For a circle inscribed in a triangle, the lines joining the vertices of the triangle to the point of contact of the inscribed circle on the opposite side are concurrent.

Need to show that the tangents to the circumcircle of the triangle at the vertices of the triangle meet the opposite side of the triangle in three collinear points.[/quote]

Consider the triangle formed by these tangents. The circumcircle of the original triangle (it's hard to talk about triangles if they don't even have names...) is inscribed into this new triangle (at least it touches its three sidelines - however, it needs not be its incircle; it also can be an excircle). Apply the first result (which you hopefully can prove not only for the incircle, but also for the excircles), do the polar transformation with respect to the circumcircle, profit.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=73595 .

  darij


if k is a positve integer such that k+1, 2k+1 and 8k+1 have divisors 3, 5, and 7 respectively, determine the smallest possible value for k?

its not the good place for as exercices .k=83 :wink:


floor function
graph theory
combinatorics proposed
combinatorics

$ n$ Aborigenes are on an island and are pairwise either friends or foes.
The leader tells the citizens to wear a pearl necklace so that any two people have one pearl of the same color if and only if they're friends.
What's the minimal number of colors needed?

[hide="What does Bollobas have to say about this problem? -- Warning : spoiler inside."]
On page 137 of Béla Bollobás' [url=http://books.google.com/books?id=SbZKSZ-1qrwC&printsec=frontcover&dq=modern+graph+theory&hl=fr&source=gbs_summary_s&cad=0]Modern Graph Theory[/url] one can read the following problem :
22. (i) Show that the edges of a graph of order $ n$ can be covered with not more than $ \left \lfloor \frac {n^2}4 \right \rfloor$ edges and triangles.
(ii) Let $ G$ be a graph with vertices $ x_1, \ldots, x_n, \ \ n \ge 4.$ Prove that there is a set $ S, \ \ |S| \leq \left \lfloor \frac {n^2}4 \right \rfloor,$ containing non-empty subsets $ X_1, \ldots, X_n$ such that $ x_ix_j$ is an edge of $ G$ if $ X_i \cap X_j \neq \emptyset.$

Thanks to Darij for letting me know about this reference.
[/hide]

Is it a so-called edge-clique cover ? Why restrain yourself to triangles ?


Let $ a,b,c\geq0$ s.t. $ a\plus{}b\plus{}c\plus{}2\leq abc$.Prove that:
$ ab\plus{}bc\plus{}ca\geq2(a\plus{}b\plus{}c)$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a \plus{} b \plus{} c \plus{} 2\leq abc$.Prove that:
$ ab \plus{} bc \plus{} ca\geq2(a \plus{} b \plus{} c)$[/quote]
Let $ a \equal{} 2xt,$ $ b \equal{} 2yt$ and $ c \equal{} 2zt,$ where $ t > 0$ and $ x \plus{} y \plus{} z \plus{} 1 \equal{} 4xyz.$
Then $ a \plus{} b \plus{} c \plus{} 2\leq abc\Leftrightarrow t(x \plus{} y \plus{} z) \plus{} 1\leq4t^3xyz\Leftrightarrow$
$ \Leftrightarrow t(4xyz \minus{} 1) \plus{} 1\leq4t^3xyz\Leftrightarrow(t \minus{} 1)(4xyz(t^2 \plus{} t) \plus{} 1)\geq0\Leftrightarrow t\geq1.$
With that $ ab \plus{} bc \plus{} ca\geq2(a \plus{} b \plus{} c)\Leftrightarrow t(xy \plus{} xz \plus{} yz)\geq x \plus{} y \plus{} z.$
Since, $ t\geq1$ it remains to prove that
$ xy \plus{} xz \plus{} yz\geq x \plus{} y \plus{} z$ for positive $ x,$ $ y$ and $ z$ such that $ x \plus{} y \plus{} z \plus{} 1 \equal{} 4xyz.$
If $ x < 1,$ $ y < 1$ and $ z < 1$ then $ xyz < 1.$ But $ 4xyz \equal{} x \plus{} y \plus{} z \plus{} 1\geq4\sqrt [4]{xyz}.$
It gives $ xyz\geq1,$ which is a contradiction.
If $ x > 1,$ $ y > 1$ and $ z > 1$ than $ x \equal{} 1 \plus{} u,$ $ y \equal{} 1 \plus{} v,$ $ z \equal{} 1 \plus{} w,$ for positive $ u,$ $ v$ and $ w.$
Hence, $ 4xyz \equal{} x \plus{} y \plus{} z \plus{} 1\Leftrightarrow\sum_{cyc}\left(\frac {4uvw}{3} \plus{} 4uv \plus{} 3u\right) \equal{} 0,$ which is a contradiction.
Id est, we can assume that $ (x \minus{} 1)(y \minus{} 1)\leq0.$
We obtain: $ 4xyz \equal{} x \plus{} y \plus{} z \plus{} 1\Leftrightarrow z \equal{} \frac {x \plus{} y \plus{} 1}{4xy \minus{} 1}.$
Thus, we need to prove that $ xy \plus{} \frac {(x \plus{} y \plus{} 1)(x \plus{} y)}{4xy \minus{} 1}\geq x \plus{} y \plus{} \frac {x \plus{} y \plus{} 1}{4xy \minus{} 1}.$
But $ xy \plus{} \frac {(x \plus{} y \plus{} 1)(x \plus{} y)}{4xy \minus{} 1}\geq x \plus{} y \plus{} \frac {x \plus{} y \plus{} 1}{4xy \minus{} 1}\Leftrightarrow$
$ \Leftrightarrow4x^2y^2\minus{}xy\plus{}(x\plus{}y)^2\plus{}x\plus{}y\geq4xy(x\plus{}y)\plus{}1\Leftrightarrow$
$ \Leftrightarrow4x^2y^2\minus{}4xy(x\plus{}y)\plus{}(x\plus{}y)^2\geq xy\minus{}x\minus{}y\plus{}1\Leftrightarrow$
$ \Leftrightarrow(2xy \minus{} x \minus{} y)^2\geq(x \minus{} 1)(y \minus{} 1),$ which is obvious.

I was hoping for a proof which implies proving that:
$ f(a,b,c)\equal{}ab\plus{}bc\plus{}ca\minus{}2(a\plus{}b\plus{}c)$ attains its minimum when
$ a\plus{}b\plus{}c\plus{}2\equal{}abc$ and in that case we could make the substitution $ a\equal{}\frac{x\plus{}y}{z}$,etc.

[b]Allnames wrote:[/b]
[quote]th problem has a nice proof >We can use pqr and schur ineq,[/quote]

How, details, please.. :?:


Prove that if $n$ is a whole number and $\theta$ is suck $sin\frac{\theta}{2}=\frac{1}{2n}$ then:
$cos\frac{\theta}{2}+cos\frac{3\theta}{2}+...+cos\frac{(2n-1)\theta}{2}=nsin(n\theta)$

It is obviosly, because $sin\frac{\theta }{2}cos\frac{(2k-1)\theta }{2}=sink\theta-sin(k-1)\theta$.

Is well known that
$cosx+cos3x+...+cos(2n-1)x=\frac{sin(2nx)}{2sinx}$
Puting $x=\frac{\theta}{2}$ and using that $sin\frac{\theta}{2}=\frac{1}{2n}$ we got the identity.


Is there a passing score to get to the Math Olympiad Summer Program, or how do they select 12 USAMO participants from the 500??

-rts2007

Black is the top 12 people by score, usually about four problems completely solved (28 last year).

Blue is the approx. the next 18 non-seniors, usually about three problems solved (20 last year).

http://en.wikipedia.org/wiki/MOSP


Sorry if I've posted this before but I really need the answer.

[b]Seven people are to be photographed in a row. If two specific persons insist on standing together and two other persons refuse to stand together, how many arrangements are possible?[/b]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165593 for a related problem.


After the  beat in the Mathematics camp the  children bloged "Crosses and Nils", alt was known that there were 5  games "girl   againts  girl"  and   12 games "boy against boy" ,  and every participant  played with  girls one game less than with boys. How many children played "crosses and Nils"

____________________________
Azerbaijan Land of the Fir :rotfl: e

Once was enough:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=309980

Also, please choose better titles for your posts.


geometry
rectangle
analytic geometry
symmetry
geometric transformation
reflection
graphing lines

Here's a cute problem:

Let M and N be the midpoints of sides AD and BC in rectangle ABCD.  Point P lies on the extension of DC beyond D; point Q is the intersection of PM and AC.  Prove that NM bisects angle QNP.

Put the coordinate origin at A and the positive x-, y-axes along the rays AB, AD. Denote $AB = DC = a, AD = BC = b$ and $PD = p$. Let the line PM meet the line AB (the x-axis) at a point R. The triangles $\triangle PDM \cong \triangle RAM$ are congruent by ASA, because AM = DM (M is the midpoint of AD), the angles $\angle DMP = \angle AMR$ are vertical and the angles $\angle MDP = \angle MAR = 90^\circ$ are right. Hence AR = DP = p. The rectangle diagonal AC passing through the coordinate origin $A = (0, 0)$ and through the point $C = (a, b)$ has an equation $y = \frac b a\ x$. The line $PM \equiv MR$ passing throught the points $R = (p, 0),\ M = \left(\frac b2, 0\right)$ on the x- and y-axes has an equation $\frac x p + \frac{2y}{b} = 1$. Solving the set of 2 equations yields the coordinates of the intersection Q of these 2 lines:

$\frac{x_Q}{p} + \frac{2x_Q}{a} = 1,\ \ x_Q = \frac{1}{\frac 1 p + \frac 2 a} = \frac{ap}{a + 2p}$

$y_Q = \frac b a\ x_Q = \frac b a \cdot \frac{ap}{a + 2p} = \frac{bp}{a + 2p}$

The line QN passing through the points $Q = (x_Q, y_Q) = \left( \frac{ap}{a + 2p},\ \frac{bp}{a + 2p}\right)$ and $N = (x_N, y_N) = \left(a, \frac b 2\right)$ has an equation

$y - y_N = \frac{y_Q - y_N}{x_Q - x_N}\ (x - x_N)$

$y - \frac b 2 = \frac{bp - \frac b 2 (a + 2p)}{ap - a (a + 2p)}\ (x - a) = \frac{\frac{ab}{2}}{a(a + p)}\ (x - a) = \frac{b}{2(a + p}\ (x - a)$

Let the line QN meet the line AB (the x-axis at a point S. The y-coordinate of the point S is 0. Substituting this to the equation of the line QN yields the x-coordinate of the point S:

$-\frac b 2 = \frac{b}{2(a + p)}\ (x_S - a)$

$x_S = a - \frac{2(a + p)}{b} \cdot \frac b 2 = a - (a + p) = -p$

Consequently, the points $P = (-p, b),\ S = (-p, 0)$ are symmetrical with respect to the line $MN$ with an equation $y = \frac b 2$. It follows that the lines $PN,\ SN \equiv QN$ are symmetrical with respect to the line MN and consequently, the symmetry axis MN bisects the angle $\angle QNP$ formed by these 2 lines.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=14246 problem 2.

  darij

Reflect P across BC to P'.  Let S be the point on AB such that PS is perpendicular to AB.  Let R be the intersection of PQ and AB.  Note that AS = AR and CP' = CP.  Also, CPQ ~ ARQ, so SRQ ~ P'PQ.  It follows that S, Q, and P' (and N) are collinear, and you're basically done.

We can make the job with coordinates a little easier.
WLOG, let M=(1,0), D=(2,0), N=(1,p) C=(2,p) P=(2,q), then Q=($ \frac {2q}{2q \minus{} p}$, $ \frac {pq}{2q \minus{} p}$)
Now, MN bisects $ \angle$ QNP if the slopes from Q to N and P to N are negatives of each other, which is easy to check (they are p-q and q-p after simplification). I am still looking for a nice geometric proof beside Alex's (probability1.01).


LaTeX
number theory proposed
number theory

Hi,

I have found something (here is the picture).
I do not have the tools to go far on my calculations.
The picture is easy to understand. So if you can complete the sequence and if my conjecture is right please let me know it.

Thank you all.

So your conjecture would be:

If $ 2^a$ is the greatest exponent of $ 2$ that divides one perfect number, then $ a\plus{}1$ is the greatest odd divisor of predecessor perfect number?

[quote="Bugi"]So your conjecture would be:

If $ 2^a$ is the greatest exponent of $ 2$ that divides one perfect number, then $ a \plus{} 1$ is the greatest odd divisor of predecessor perfect number?[/quote]

My conjecture is that the next perfect one will be

(([b]2^(2^127 - 1[/b]))-1)*(2^216)

I do not know how to put in in latex.
where 2^127 - 1 = 170141183460469231731687303715884105727
and (2^170141183460469231731687303715884105727) - 1 is probably prime

Put dollar signs before and after the equation to use $ \text{\LaTeX}$.

You mean the number:
$ (2^{2^{127}-1}-1)\cdot2^{216}$ is the next perfect one?

[quote="lambruscokid"]Didn't you miss the number $ 496$ which is perfect and lies between $ 28$ and $ 8128$?[/quote]

I did not miss it but 496 which is equal to 16*31 lead you to 2^31 - 1 prime and (2^30)*(2^31 - 1) ts a perfect number.

So you have for each sequence a new start.
Maybe it will not lead each time to a prime number. I do not have the tools to check my conjecture.
Anyway if someone can prove the primality of the number above we will have a new record and maybe the biggest known prime.

[quote="Bugi"]Put dollar signs before and after the equation to use $ \text{\LaTeX}$.

You mean the number:
$ (2^{2^{127} - 1} - 1)\cdot2^{216}$ is the next perfect one?[/quote]

Yes. Exactly.
I just saw it the second part is not correct : it is 2^(2^126)


calculus
integration
limit
function
real analysis
real analysis unsolved

Let $ f(x)\ge 0$ be continuous on the interval $ [0,\infty)$, and suppose that $ \int_{0}^{\infty}f(x)dx <\infty$.  Prove that

$ \lim_{n\rightarrow\infty}\frac{1}{n}\int_{0}^{n}xf(x)dx\equal{}0$

Let $ F(x) \equal{}\int_{0}^{x}f(t)\,dt\minus{}\int_{0}^{\infty}f(t)\,dt.$ We can write this as $ F(x) \equal{}\minus{}\int_{x}^{\infty}f(t)\,dt.$ Note that $ F$ is negative, bounded, increasing, and tends to zero as $ x\to\infty.$ Integrate by parts.

$ \int_{0}^{n}xf(x)\,dx \equal{} nF(n)\minus{}\int_{0}^{n}F(x)\,dx$ so

$ \frac{1}{n}\int_{0}^{n}xf(x)\,dx \equal{} F(n)\minus{}\frac{1}{n}\int_{0}^{n}F(x)\,dx$

Since we know we have that $ \lim_{n}F(n) \equal{} 0,$ we concentrate on the integral. This is related to Cesaro summability.

Given $ \epsilon > 0,$ we can find $ m$ such that for $ x > m,\ |F(x)| <\epsilon.$ Then for $ n > m,$

$ \left|\frac{1}{n}\int_{0}^{n}F(x)\,dx\right|\le\left|\frac{1}{n}\int_{0}^{m}F(x)\,dx\right|\plus{}\left|\frac{1}{n}\int_{m}^{n}F(x)\,dx\right|$

Noting that the maximum value of $ |F(x)|$ is $ |F(0)|,$ we have that

$ \left|\frac{1}{n}\int_{0}^{m}F(x)\,dx\right|\le\frac{m|F(0)|}{n}$ while

$ \left|\frac{1}{n}\int_{m}^{n}F(x)\,dx\right|\le\frac{(n\minus{}m)\epsilon}{n}<\epsilon$

Both terms can be made arbitrarily small by choosing $ n$ sufficiently large.

Define $ a_{k}\equal{}\int_{k\minus{}1}^{k}f(x) dx$. After condition $ \sum_{k \equal{} 1}^{\infty}a_{k}$ converges absolutely.

Therefore $ k\cdot a_{k}\to 0$ and Cesaro mean $ \frac{1}{n}\sum_{k \equal{} 1}^{n}k a_{k}\to 0$ , too.

And $ \frac{1}{n}\int_{0}^{n}x f(x) dx \equal{}\frac{1}{n}\sum_{k \equal{} 1}^{n}\int_{k\minus{}1}^{k}x f(x) dx\le\frac{1}{n}\sum_{k \equal{} 1}^{n}k a_{k}$ .

[quote="Dr. Doe"]Therefore $ k\cdot a_{k}\to 0$[/quote]
This is not true. Consider a function $ f$ that is zero for long stretches with occasional positive bumps. We can arrange to have

$ a_{k}\equal{}\begin{cases}\frac{1}{j^{2}},&k\equal{}2^{j}\\0,&\text{otherwise}\end{cases}.$

Can you save your argument anyway? I'm not sure.

[hide="Yes, of course"]
Use summation by parts to get that $ \frac{1}{n}\cdot\sum_{k \equal{} 1}^{n}k a_{k}\equal{}\sum_{k \equal{} 1}^{n}a_{k}\minus{}\frac{1}{n}\sum_{k \equal{} 1}^{n\minus{}1}\sum_{j \equal{} 1}^{k}a_{j}$.
By Cesaro-Stolz the second part has the same limit as the first one.

I have the feeling that I've seen this argument before (posted by Kent Merryfield?).
[/hide]

... and if you think about it, that's an exact copy of the integration by parts argument, except that Kent included a proof of the relevant theorem rather than just quoting it.

Right, Kent. The conclusion $ \Rightarrow k\cdot a_{k}\to 0$ wasn't correct.

I committed a blunder again.  :rotfl:


Ok, this was inspired by a little looking around after reading mathnerds thread. The point of this game is to guess a number between 0-100 that will be [b]2/3 of the average[/b] of all the numbers that everyone else picked. Pm me with your choices, the results will be posted in a week, if enough people have participated. I will be participating, but I will chose my number first, to be fair. Feel free to talk about your choices before hand, there isn't any real need for secrecy in this, and the mind games you guys play should be amusing.

[quote="cf249"]Is this game scored the same as the previous 1/2 the average games?[/quote]

Could you link me to them?

Hmm, oh gee. Last time I played the 1/2 of average game, I completely forgot about it and lost really soon. So I'll have to think whether I'm joining or not.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173810]This game[/url] didn't finish but has links to all of the other games.

Okay, let's see:

Why would anyone ever pick a number greater than 66?

On that note, why would anyone ever pick a number greater than 44?

On that note, why would anyone ever pick a number greater than 14?

On that note, why would anyone pick a number greater than 4?

On that note, why would anyone pick a number greater than 1?

On that note, why would anyone pick a number greater than 0?

Guys, if we all choose zero we can all win! And even if a group tries to counter with a bunch of 100s, they will still fail!

Everyone could be sneakish and do .0000000001.  lol

This is almost exactly the same as my HTA games which I haven't had time to do recently.  If you want to use my rules, there are plenty people who have already played and I'm sure would be willing to play again.

By the way, did anyone catch my suddenly switching to the game "1/3 the average" halfway through my previous post? But the idea is still the same, it just requires more typing.

For what it's worth, we tried this game at RSI 2007 and multiple people picked $ 100$ (myself included).  Personally, I did it to demonstrate the futility of analyzing the game at all :P

[quote="Temperal"]Exactly 2/3? Are you serious?[/quote]

Whoever is closest wins


geometry
circumcircle
trapezoid
geometry proposed

Let two non-intersecting circles $ \omega, \phi$ be tangent internally to circle $ \Gamma$ at $ A,B$, respectively. Let $ X,Y$ be one of two intersections of internal common tangents of $ \omega$ and $ \phi$ with $ \Gamma$ respectively, $ X,Y$ lie on the same side as the center line of $ \omega$ and $ \phi$.  ( $ A,B,Y,X$ lie on $ \Gamma$ in clockwise order) Denote by $ P$ be the midpoint of arc $ \widehat{XY}$ (doesn't contain $ A$). Line $ BP$ cuts $ AX$ at $ C.$ Prove that if $ XC \equal{} AY,$ then $ AX\parallel BY.$

It is also necessary to assume that $ C$ is on the ray opposite to $ (XA,$ otherwise the proposition does not hold. See counter-example with $ C$ on the ray $ (XA.$

Let $ AP$ cut $ BY$ at $ D.$ $ \angle CAD \equal{} \angle XAP \equal{} \angle PBY \equal{} \angle CBD$ $ \Longrightarrow$ $ ABDC$ is cyclic with a circumcircle $ \Lambda.$ Parallel to $ AP$ through $ Y$ cuts $ \Gamma$ again at $ Q.$ $ \angle QAB \equal{} \angle QYB \equal{} \angle ADB$ $ \Longrightarrow$ $ AQ$ is tangent of $ \Lambda$ at $ A.$ Let parallel to $ AX$ through $ Y$ cut $ \Gamma$ again at $ B_0,$ let $ B_0P, AP$ cut $ AX, B_0Y$ at $ C_0, D_0.$ $ C_0$ is on the ray opposite to $ (XA.$ Let $ \Lambda_0$ be circumcircle of the isosceles trapezoid $ AB_0D_0C_0,$ which is tangent to both $ AQ, \Gamma$ at $ A.$ $ \angle B_0C_0X \equal{} \angle B_0C_0A \equal{} \angle C_0AD_0 \equal{} \angle XAP \equal{} \angle XB_0C_0$ $ \Longrightarrow$ triangle $ \triangle B_0XC_0$ is isosceles with $ XC_0 \equal{} XB_0.$ $ AXYB_0$ is also isosceles trapezoid $ \Longrightarrow$ $ AY \equal{} XB_0 \equal{} XC_0.$ As $ \Lambda$ is in pencil of tangent circles defined by $ A, \Lambda_0,$ $ AY \equal{} XC \Longleftrightarrow C \equiv C_0 \Longleftrightarrow \Lambda \equiv \Lambda_0 \Longleftrightarrow B \equiv B_0 \Longleftrightarrow BY \parallel AX.$


I'm trying to access the links posted in the 2003 IMO transcript, and I get the following error message:

[quote]
Forbidden

You don't have permission to access /Community/MJImages/MJ3/twodiags.gif on this server.
[/quote]

Now me sad. :(

Likely we did not port these over correctly when we moved to the new web server earlier this summer.

It's on the list of things to fix (but, honestly, fixing 3-year-old Math Jams is not going to be a high-priority item).

Well, math doesn't really change over 3 years, but I see what you're saying.  :D

Fixed, I think.  (Turned out to be a pretty easy fix.)  Let me know if you hit any more of these.


Let a,b,c be positive numbers. Prove that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{21(a^{2}+b^{2}+c^{2})-3(ab+bc+ca)}{2(a+b+c)^{2}}$

i guest we use:
 Assume :$c=min{a;b;c}$and let :$a+b+c=3.$
 and you influence ineq's S.O.S.  :blush:

I think we shouldn't use SOS method in here :P because this inequality is cyclic so we consider 2 case.  Maybe, we can solve it easily by the following technique:
Let $ a=min{a,b,c}$ and we put: $ b=a+q,c=a+p$. I will try ...


For a general reaction, $ HA <\equal{}> A^{\minus{}} \plus{} H^{\plus{}}$. If HA has a pka of 5.3 and the solution has a pH of 5.7, what percent of the molecule is in "acidic form" HA?

First, the reaction should be $ HA \plus{} H_2O \rightleftharpoons A^\minus{} \plus{} H_3O^\plus{}$.
Second, i believe you want this:

$ \alpha_0 \equal{} \frac{[HA]}{C} \equal{} \frac{[H^\plus{}]}{[H^\plus{}] \plus{} K_a}$,

where C is the "dissolved" concentration of the acid. By the way, you can also show that

$ \alpha_1 \equal{} \frac{[A^\minus{}]}{C} \equal{} \frac{K_a}{[H^\plus{}] \plus{} K_a}$.

For polyprotic acids we have similar expressions.


function
geometry
geometric transformation
reflection
3D geometry
sphere
complex numbers

Is anyone here interested in fractal art?  Post some of your discoveries!

http://carlosrd.home.comcast.net/fivepointedstar.png <-- found this morning

I made some fractals a while back, but I can't upload them, as they are in bitmap form.

No, it is Tierazon, a GREAT, legally downloadable, program!

Xaos is the most fun to use, but not too flexible. I was too lazy to figure out how to use Tierazon correctly, and it's kind of slow. I recommend http://www.chaospro.de/ or  http://www.eclectasy.com/Fractal-Explorer/

[quote="YSJeong"]what is fractal art, exactly??[/quote]

I am no expert, I am hardly even an amateur, but I would decribe it like this:

Fractals are mathematical patterns of nature expressed as art.

Find more information by clicking [url=http://math.rice.edu/~lanius/fractals/WHY/]here.[/url]

A fractal is simply a figure that is approximately self-symmetric. (short definition.  See mathworld.wolfram.com for a whole lot more...)

[quote="YSJeong"]what is fractal art, exactly??[/quote]

Most of the cool looking fractals are generated by plotting the results of a recursive function over complex numbers that exhibits chaotic behavir, (e.g. Z=Z^2+C, where C is the point being potted).

Not most, only a small part of them!

a lot are generated by some 'mappings' when iterated (e.g. moebius transformations) when looking at the image of some figure

then there are a lot (as my avatar) that are what you get when some constellation reflects itself on its surface over and over (e.g. some spheres)

my mistake then, it's just that those are the only fractals that I fully understand  :) .

And amid all this discussion, no one has posted any more examples!

http://carlosrd.home.comcast.net/3dish.pdf

I have a lot of fractals, all in jpeg, but not on this computer. I do know this website that has a lot of cool fractals though. [url=http://www.fractalartgallery.com]Ken Keller's fractals[/url]


inequalities
logarithms
inequalities proposed

Maybe I'm doing something wrong but it seems like inequality isnt true.

$(a^2+b^2+c^2)^3 \geq 9(a^3b^3+b^3c^3+c^3a^3)$
$a^6+b^6+c^6+3(a^4b^2+b^4c^2+c^4a^2)+6a^2b^2c^2 \geq 9(a^3b^3+b^3c^3+c^3a^3)$

Using Muirhead we get:
$6a^2b^2c^2 \geq 5(a^3b^3+b^3c^3+c^3a^3)$
which isnt true (from Muirhead again).

I've tried AM-GM but with the same result = finishing with some wrong inequality.
Show your solution please

[quote="nttu"]Prove that : with $ a, b , c \geq 0 $
     $ (\frac{a^2+b^2+c^2}{3})^3 \geq \frac{a^3b^3+b^3c^3+c^3a^3}{3} $[/quote]

Well, it's actually wrong for a = 1 and b = c = 3.

  Darij

Well , this ineq is wrong . Thank you Darij for showing mistakes .
   The original problem of this ineq :
   Prove that : with a, b , c >=0 :
   $ (a^2+2)(b^2+2)(c^2+2) \geq 9(ab+ac+bc) $
  Do you have any idea about this problem ?

[quote="nttu"]Prove that with $a, b , c  \geq 0$

\[ (a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca) \]
 [/quote]

It's from APMO 2004, probably posted on the forum somewhere, but here's a solution:

Expanding, we get

$a^2b^2c^2+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2+b^2+c^2)+8 \geq 9(ab+bc+ca)$,

This the sum of:

$a^2+b^2+c^2+a^2b^2c^2+1+1 \geq a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}$

$\geq \sum_{cyc} \sqrt[3]{a^4b^2}+\sqrt[3]{a^2b^4} \geq \sum_{cyc} 2ab = 2(ab+bc+ca)$,

from AM-GM + Schur + AM-GM,

$3(a^2+b^2+c^2) \geq 3(ab+bc+ca)$, and

$2(a^2b^2+1+b^2c^2+1+c^2a^2+1) \geq 4(ab+bc+ca)$,

where the former is well-known, and the latter is AM-GM.

[quote="zxingtan"]you are clever,THAzN1,
and now ,can the question be stronger
for example,is the following ineq right?
( :Sigma: a^2/3) :^3:  >= 9(ab+bc+ca)     for a,b,c>=0
and i think it's right[/quote]

Unless I'm mistaken, a=b=c=1 gives 1 >= 27, which is clearly false.

In this forum, Harazi proved that if $a^4b^4+b^4c^4+c^4a^4=3$, then $a^3+b^3+c^3\geq3$.
But the following more general result is valid:
If $ab+bc+ca=3$ and $m\geq2-\frac{\ln{4}}{\ln{3}}\approx 0.738$, then $a^m+b^m+c^m\geq3$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4830]This[/url] is harazi's proof of a stronger version of the problem.


Find the sum of all integers between 1 and 9999 that are divisible by 5 but not by 6 or 7.

This sounds very familiar... but I'll have a go :)
[hide]
The sum of all multiples of 5 from 1 to 9999, or 5 to 9995, or 5*(1 to 1999) is 9995000.
Then we leave out multiples of 5 and 6, or multiples of 30. Subtract 30 to 9990, or 30*(1 to 333), or 1668330.
We also leave out multiples of 5 and 7, or multiples of 35. Subtract 35 to 9975, or 35*(1 to 285), or 1426425.
But we left out multiples of 5, 6 and 7 (210) twice. So add 210 to 9870, or 210*(1 to 47), or 118440.

9995000 - 1668330 - 1426425 + 118440 = [b]7018685[/b]

[/hide]

Thank you for answering my question, but could you please give me a detailed explanation on how you got your answer? Much Appreciated-Infinity

I used the include-exclude principle for this kind of problem. If you count all the multiples of 5, those would include some multiples of 6 which we'd need to leave out. Numbers that are multiples of 5 and 6 are multiples of 30. We also need to leave out multiples of 5 and 7; multiples of 35. However there are some multiples of 30 that are also multiples of 35: multiples of 210.

So to find our total we calculate the sum of the multiples of 5, subtract the multiples of 30, subtract the multiples of 35, and add the multiples of 210.

What I did was find the highest multiple of each one less than 9999.
An important formula to remember is the sum of the first n integers: $ \frac{n(n+1)}{2} $
So first we want to find the sum of the multiples of 5. 5 + 10 + 15 + 20 + ... + 9990 + 9995. if we divide the whole sum by 5 we get 5*(1 + 2 + 3 + 4 + ... + 1998 + 1999), then $ \frac{5*1999*2000}{2} $ = 9995*1000 = 9995000.

Repeat the process for 30, 35, and 210, then crunch it to get [b]7018685[/b]. :)

Thanks for answering, but i dont understand properly. you said that the formula would work for 35, 30 and 210, but 35*1999*2000[u]2 doesnt equal the same answer as you said??!!![/u][/b][/code][/list]

Well look at it this way. If you wanted to find the sum of the first 70 multiples of 7:

The sum looks like 7 + 14 + 21 + 28 + 35 + ... 476 + 483 + 490.

But you can make this easier by doing the distributive property, reverse:

7 * the sum (1 + 2 + 3 + 4 + ... + 68 + 69 + 70.)

$ 1 + 2 + 3 + 4 + ... + n = \frac{n(n+1)}{2} $ is an important formula to remember! :)

7 * 70(71)/2

7 * 4970/2

7 * (5000-30)/2 (just so you can do it in your head)

7 * (2500-15)

17500 - 105

[b]17395[/b]

Oh -- Did you just mean I miscalculated a value?    To err is human.  :blush:
Figure it out yourself, check my work and see if we got the same thing.


function
inequalities proposed
inequalities

[quote="kunny"]Let $x,y,z$ be real numbers such that $0\leq x\leq 1,\ 0\leq y\leq 1,\ 0\leq z\leq 1$ and $x+y+z=1.$
Find the maximum value of $x^2y+xy^2+y^2z+yz^2+z^2x+zx^2.$[/quote]

  Yeah,i like this problem
See the parabolla: $\ f(x)= x^2(y+z)+x(y^2+z^2)+(y^2z+z^2y)$ treat y,z as parameters.
Because the coefficient of $\ x^2$ is positive due to the given condition,so the maximum value will be occured at boundary.
 Means f(x) max will be equal to f(0) or f(1).I think after that will be very easy repeat the above steps with y or z to reduce the given function to single variable only.Is that right Kunny? :D

denote by $f(x,y,z)=x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$

since $f(x,y,z)$ is symmetric on $x,y,z$ then we can assume wlog $x\geq y\geq z$

then, it's easy to verify that $f(x,y+z,0)\geq f(x,y,z) \Leftrightarrow yz(2x-y-z)\geq 0$, which is obviously true.

now, $f(x,y+z,0)=x^2(y+z)+x(y+z)^2=x(y+z)\leq \frac{(x+y+z)^2}{4}=\frac{1}{4}$

it can be verified that the equality case $x=\frac{1}{2}$ and one of $y,z$ equals $0$.


Sorry for not knowing lot of names in english, i'll try to fix it with names i do know.

Let p be a prime number, $ p>2$.
Let $ a_n\equal{}n$ for $ n \in N_0$, $ n<p$.
Let $ a_n\equal{}a_{n\minus{}1} \plus{} a_{n\minus{}p}$ for $ n \ge p$.

Tell the value of $ a_{p^3} (mod p)$.

anyone? that looks like a really hard problem


[quote="236factorial"]Conisder a semi-circle of radius 9 m. Let ABCD be inscribed in this semi-circle in such a way that D and C are on the diameter and A and B are on the semi-circle. Suppose in addition that AB= 2 BC . How many square meters, to the nearest whole number, are in the area inside the semi-circle but outside the rectangle? Use 3.14 as an approximation of $\pi$.[/quote]

[hide]
Suppose the center of segment DC is M and the midpoint of AB is N.  Then, the segment BM is also a radius of the semicircle and NBCM is a square.  Since $BC=MC$, $BC=MC=\dfrac{9}{\sqrt{2}}=\dfrac{9\sqrt{2}}{2}$.  AB would then be $9\sqrt{2}$.  Thus the area of the rectangle is $(9\sqrt{2})\left(\dfrac{9\sqrt{2}}{2}\right)=81$  The area of the semicircle is $\dfrac{81\pi}{2} \approx 127.17$

$127.17-81=\boxed{46.17}$
[/hide]

Too lazy to simplify

[hide="Answer"]
81pi/2 - 81
[/hide]


geometry
geometric transformation
reflection

"Something." compared by "Something". Why does one put the "." inside the quotation marks? In the same fashion for "hello," instead "hello", or like "are you there?!" instead "are you there"?!

Because if the phrase is terminating with "something,", then you stick the period inside.  As one of my instructors said, the period needs to "feel some love" from another letter or character so it tries to get in close at the end of the word.

For question marks and exclamation marks, stick the punctuation mark inside the quotation mark only if the quote itself is a question (or exclamation).  Otherwise, put it outside of the quotation mark.

Compare these two sentences:
Bobby said, "Are you going to the store?"
Did Bobby say, "I am going to the store"?

In the first sentence, the quote is a question, so the question mark goes inside the quotation mark.  In the second sentence, the sentence itself is a question (and the quote is a statement), so the question mark goes outside the quotation mark.

It is true that the convention for periods in quotation marks is unnatural and not nice and neat like our beloved parentheses. That's why we study math.  :)

Actually, the rules of grammar are pretty well-structured and rigid.  It's just that some of them don't seem natural so in order to use the rules correctly, you actually have to read a book on grammar to even find out about them.

There's an enormous difference between prescriptive, descriptive, and explanatory grammar.  "Prescriptive" grammar is a collection of rules governing socially acceptable and unacceptable language use.  One example would be this rule, which you've probably heard:

(1)  Don't end a sentence with a preposition.

"Descriptive" grammar generalizes what speakers can say in actual fact.  If a descriptive grammar were to come up with a rule like (1) governing when a preposition can or can't end an sentence, the grammar would take into account that a native English speaker could say (2a-c), but would find (2d) completely unacceptable (the * means "this sentence is unacceptable").  A descriptive grammar would have to explain what's difference between "play with Bob" and "cough up a hairball" so that moving the preposition to the front of the sentence is OK in the first but not in the second.

(2a)  Who were you playing with?  (answer:  Bob)
(2b)  With whom were you playing?
(2c)  What did the cat cough up?  (answer: a hairball)
(2d)  *Up what did the cat cough?

An explanatory grammar would take this explanation and expand it into a theory of how human language works in principle.

So, the rules of prescriptive grammar are well-structured and rigid in the same way that rules of etiquette are-- on some level it's completely arbitrary which fork you use to eat your radicchio at a swanky dinner, but only one fork is socially acceptable.  You wouldn't know that many rules of prescriptive grammar exist without looking them up in a book because they regulate the use of different forms which are in free variation in normal speech-- it's completely arbitrary which one should be "socially acceptable."

On the other hand, the rules of descriptive grammar should be the same rules by which you actually construct and process sentences in a language, and they should be well-structured and rigid because they are a direct reflection of your brain's natural language processing facility.


LaTeX
combinatorics unsolved
combinatorics

let $ A_1,A_2, \cdots ,A_n$ be $ n$ finite subsets of a set $ X$, $ n \geq 2$, such that

(i) $ |A_i| \geq 2$, for $ 1 \leq i \leq n$;
(ii) $ |A_i \cap A_j| \leq 1$ for $ 1 \leq i < j \leq n$.

Prove that the elements of $ A_1 \cap A_2 \cap \cdots \cap A_n$ may be colored with 2 colors so that no $ A_i$ is colored by the same color.

sorry for the bad latex.

Latex edit by pbornsztein.

\cap or \bigcap (the latter if you want to do something like $ \bigcap_{i \in {\bf N}} J_i$) gives the standard symbol for intersection.  \cup (and likewise \bigcup) gives the standard symbol for union.  [[LaTeX:Symbols]] has a dictionary of commonly-used symbols in Latex.

I don't understand the statement : do you color each element in $ X$?
Or are you asking for a bicolouring of the union of the $ A_i$'s? (you used 'intersection'...) with no monochromatic $ A_i$ ?

Pierre.


Solve this equation:

$ \sqrt{1\minus{}x^2}\equal{}(\frac{2}{3}\minus{}\sqrt{x})^2$
 :lol:

[quote]$ \sqrt{1\minus{}x^2}\equal{}\left(\frac{2}{3}\minus{}\sqrt{x}\right)^2$[/quote]

I'm not sure if there is an easy way. You could always multiply everything out and manipulate variables though.

try to use $ a^{2} \minus{}b^{2}\equal{}(a\minus{}b)(a\plus{}b)$

hello, after squaring and sorting we get
$ \frac{1}{81}(\minus{}65\minus{}96\sqrt{x}\plus{}216x\minus{}216x^{\frac{3}{2}}\plus{}162x^2)\equal{}0$
Making the substitution $ t\equal{}\sqrt{x}$ you have to solve
$ \minus{}65\minus{}96t\plus{}216t^2\minus{}216t^3\plus{}162t^4\equal{}0$.
Sonnhard.

[quote="enndb0x"]try to use $ a^{2} \minus{} b^{2} \equal{} (a \minus{} b)(a \plus{} b)$[/quote]

Transformation $ a^2\minus{}b^2\equal{}(a\minus{}b)(a\plus{}b)$ is often not useful under the radical.

[code]In[1]:= FindInstance[Sqrt[1 - x^2] == (2/3 - Sqrt[x])^2, x, Reals]
Out[1]= {{x -> Root[4225 - 37296 #1 - 15876 #1^2 + 23328 #1^3 + 26244 #1^4 &, 2]}}[/code]

Looks like there isn't a good answer... Definitely no rational answers.


Is it easy to see that $ x^5\plus{}2x^4\minus{}8x^3\plus{}2x^2\plus{}3x\plus{}7$ has no factor of deg. 1 $ \pmod{2}$?

Yes.
Think about what "having a factor of degree 1" means.

aha. $ f(x) \equal{} x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 3x \plus{} 7\equiv g(x)\pmod{2}$ is false for all $ g(x)\subset \mathbb{Z}$ because $ f(x)\subset \mathbb{Z}_{2^n}$ is false, even if you add terms of degree 0 and 1. Or more concisely said: $ 2|f$ is false, even if you add terms of deg. 0 or 1 to f. nuff said

I've no clue what you mean with $ \mathbb Z_2^n$ here, nor what you intended to show.
Further hint: A polynomial (of one variable) has a linear factor iff it has a root.

I mean, can't you see that $ f(x) \equal{} x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 3x \plus{} 7\equiv g(x)\pmod{2}$,where $ g$ is in $ \mathbb{Z}/2\mathbb{Z}$ is not true very apparent? For example, if $ g(x) \equal{} \minus{} x$, we still have $ x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 2x \plus{} 7\equiv 0\pmod{2}$. But $ 2|x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 2x \plus{} 7$ is obviously false here.

[quote="Los"]I mean, can't you see that $ f(x) \equal{} x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 3x \plus{} 7\equiv g(x)\pmod{2}$,where $ g$ is in $ \mathbb{Z}/2\mathbb{Z}$ is not true very apparent?[/quote]
I've already asked what you mean by "$ g$ is in $ \mathbb{Z}/2\mathbb{Z}$". And why not taking $ g\equal{}f \mod 2$...

[quote="Los"]For example, if $ g(x) \equal{} \minus{} x$[/quote]
Note that $ 1 \equiv \minus{}1 \mod 2$, so $ x \equal{}\minus{}x$ over $ \mathbb Z / 2 \mathbb Z$.

[quote="Los"]we still have $ x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 2x \plus{} 7\equiv 0\pmod{2}$[/quote]
As $ x$ is a variable, this is wrong. At least I don't get what you're doing.

[quote="Los"]But $ 2|x^5 \plus{} 2x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 2x \plus{} 7$ is obviously false here.[/quote]
Same as before...

Please start at the definitions.


As indicated by Lewis structures, which of the following would probably not exist as a stable molecule?

CH3Cl
CH3OH
C2H5OH
C2O3
CH2Cl2

I think  the answer is C2O3. Is it false?

Count the electrons Mr Math!
In order for the atom to be stable, all atoms must have a full octet and formal charges should be low. So C2O3 is clearly bogus as it would imply a structure that looks like:
O
1
C=$ C$-O
1
O
 or something along those lines, leaving the specified carbon without a full octet.

$ CH_3Cl$ : [url=http://en.wikipedia.org/wiki/Chloromethane]Chloromethane[/url]
$ CH_3OH$ : [url=http://en.wikipedia.org/wiki/Methanol]Methanol[/url]
$ C_2H_5OH$ : [url=http://en.wikipedia.org/wiki/Ethanol]Ethanol[/url]
$ CH_2Cl_2$ : [url=http://en.wikipedia.org/wiki/Dichloromethane]Dichloromethane[/url]

If $ C_2O_3$ mean $ C_2O_3 \cdot 3H_2O$, it is [url=http://en.wikipedia.org/wiki/Oxalic_acid]Oxalic acid[/url].

Well, not quite...

Oxalic acid is H2C2O4. An acid by bronsted definition is something that gives off a proton in solution.... This differs from a hydrated C2O3 molecule.

$ \ce{C2O3} \cdot 3\, \ce{H2O}$ doesn't look the same thing as oxalic acid. However, the correct answer is obviously that, as there is no such thing as $ \ce{C2O3}$. There is, on the other hand, such a thing as $ \ce{C3O2}$.

Oxalic acid was the carbon compound which anyone knew now, but it was written with $ C_2O_3$ in those days, and it was thought with an inorganic matter like carbonic acid.

C3O2 exists and is known as carbon suboxide.


How many intersections does $\frac{x \cdot |x|}{16}+\frac{y^2}{9}=1$ and $y=2x+3$?

Discussing the cases $x<0$ and $x\geqslant 0$, we find two intersection points.


The deck of a destroyer is 100 feet above sea level.  How far can a person see from the deck?  How far can a person see from the bridge, which is 150 feet above sea level?

HINT GIVEN:

Use the fact that the Earth's Radius = 3960 miles.

Well assuming that the sea is perfectly flat, and there's no weather hindering your vision, infinitely far, technically.

You have to give a certain angle that you're looking down at to be able to solve this problem.

umm...he said that the earth's radius was 3960 miles, so the sea is not flat, as was established like 3000 years ago.

Let $P$ be the point of the observer's location.  Let $O$ be the center of the earth, and let $\omega$ be a great circle on the earth's surface coplanar with $PO$.  Let $PO$ intersect $\omega$ at $S_{1},S_{2}$ with $PS_{1}< PS_{2}$.  Let the farthest point the observer can see on $\omega$ be $T$.  From the power of a point theorem we have $PT^{2}= PS_{1}\cdot PS_{2}$ so $PT = \sqrt{PS_{1}\cdot PS_{2}}$.  If $h$ is the height of the observer and $R$ is the radius of the planet, then $PT = \sqrt{h^{2}+2hR}$

you do the calculations.


Find the minimum of $ ab\plus{}ac\plus{}bc$
where $ a,b,c$ are nonegative real numbers satisfying
$ a^2\plus{}b^2\plus{}c^2\equal{}1$

The minimum value is $ 0$.Without using condition $ a^2\plus{}b^2\plus{}c^2\equal{}1$.,we can take $ b\equal{}c\equal{}0$.

[quote="Erken"]The minimum value is $ 0$.Without using condition $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.,we can take $ b \equal{} c \equal{} 0$.[/quote]

and $ a\equal{}1$  :wink:


limit
logarithms
function
calculus
calculus computations

let $ a>0$, then find the limit

$ \lim_{n\to\infty}{n^2(\sqrt[n]{a}-\sqrt[n+1]{a})}$

$ \sqrt[n]{a}\equal{}e^{\frac{\ln a}{n}}\equal{}1\plus{}\frac{\ln a}{n}\minus{}\frac{\ln^2a}{2n^2}\plus{}O(n^{\minus{}3})$

$ \sqrt[n\plus{}1]{a}\equal{}e^{\frac{\ln a}{n\plus{}1}}\equal{}1\plus{}\frac{\ln a}{n\plus{}1}\minus{}\frac{\ln^2a}{2(n\plus{}1)^2}\plus{}O(n^{\minus{}3})$

$ \sqrt[n]{a}\minus{}\sqrt[n\plus{}1]{a}\equal{}\ln a\left(\frac1n\minus{}\frac1{n\plus{}1}\right)\minus{}\frac{\ln^2a}{2}\left(\frac1{n^2}\minus{}\frac1{(n\plus{}1)^2}\right)\plus{}O(n^{\minus{}3})$

$ \sqrt[n]{a}\minus{}\sqrt[n\plus{}1]{a}\equal{}\frac{\ln a}{n(n\plus{}1)}\plus{}O(n^{\minus{}3})$

$ \lim_{n\to\infty}n^2\left(\sqrt[n]{a}\minus{}\sqrt[n\plus{}1]{a}\right)\equal{}\ln a.$

Hello

We can also use  " théorème des accroissements finis " for the function  $ f(x)\equal{}a^{\frac{1}{x}}$   in the intervalle   $ [n,n\plus{}1]$

We get the same result

In English, that would be the "mean value theorem."

If $ f(x)\equal{}a^{\frac1x}$ then $ f'(x)\equal{}\minus{}\frac{a^{\frac1x}\ln a}{x^2}$

So $ f(n)\minus{}f(n\plus{}1)\equal{}\minus{}1\cdot f'(\xi)$ for some $ n<\xi<n\plus{}1.$

$ n^2(\sqrt[n]{a}\minus{}\sqrt[n\plus{}1]{a})\equal{}n^2\cdot\frac{a^{\frac1{\xi}}\ln a}{\xi^2}$

as $ n\to\infty,$ $ a^{\frac1{\xi}}\to 1$ and $ \frac{n^2}{\xi^2}\to 1.$


You are given two segments of length $a$ and $b$. Construct using only a straightedge and compass a segment of lenght $\sqrt{ab}$

Place the two segments on a line (trivial).  Let the two segments be $AB$ and $BC$ with lengths $a,b$, respectively.  We are to construct the height of a right triangle with foot $AB\cap BC$ since the height of a right triangle dividing its hypotenuse into the line segments $a,b$ has height $\sqrt{ab}$ due to triangle similarity (why?).

First, we construct the center of $AC$: Draw a circle with center $A$ and radius $a+b$, and another with center $C$ and radius $a+b$.  The line connecting the two intersections of the circles bisects $AC$.  Draw the circle with radius $\frac{a+b}{2}$ and center $\frac{AB+BC}{2}$.

In order to construct a perpendicular line with foot $F=AB\cap BC$, draw a circle with with center $B$ and radius $b$ such that $\mathcal{C}(B,b)\cap AC=D$.  Then draw a circle with center $D$ and radius $2b$, and another circle with center $C$ and radius $2b$.  The line connecting the intersections is clearly the perpendicular bisection of $CD$ passing through $B$.  The intersection of this line with circle $\mathcal{C}\left(\frac{AB+BC}{2},\frac{a+b}{2}\right)$ gives a line segment with an endpoint $B$ such that the length is $\sqrt{ab}$.$\blacksquare$

Consider two segment line $AB=a, \ \ AC=b, \ \ BC=a+b$. Now, we construct the semicircle $(O)$ with diameter $AB$. From $A$ construct a line $d$ pependicular to $BC$, cut $(O)$ at $D$. Thus, $AD= \sqrt{ab}$.


Put $u={a\over x}, v={b\over y}$. Then the system becomes

$\begin{cases}au+bv=a+b\\{a\over u^{2}}+{b\over v^{2}}=a+b\end{cases}$

or

$\begin{cases}a(u-1)=b(1-v)\\ \frac{a(1-u^{2})}{u^{2}}=\frac{b(v^{2}-1)}{v^{2}}\end{cases}$

If $u=1$, then $v=1$ which is a solution, hence $(x,y)=(a,b)$. Therefore we'll assume $u,v\neq 1$. Dividing the last two equations we get

${u^{2}\over 1+u}={v^{2}\over 1+v}$, which yields

$u^{2}+u^{2}v=v^{2}+v^{2}u\iff u^{2}-v^{2}=uv(v-u)\iff u+v+uv=0\lor u=v$

The case $u=v$ yields $u=v=1$ which has already been discussed. Hence we assume $u+v+uv=0\iff v=-\frac{u}{u+1}$, which now can be plugged into $au+bv=a+b$ to get a simple quadratic in $u$. The rest of the proceedings can be left to the reader, IMHO.


Let $ a,b,c$≥$ 0$ satisfy $ a\plus{}b\plus{}c\equal{}1$
Prove that $ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)$≤$ \frac{1}{32}$

[quote="SUPERMAN2"]Let $ a,b,c$≥$ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 1$
Prove that $ (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$≤$ \frac {1}{32}$[/quote]
let $ f(a,b,c)\equal{}(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)$
let c=max(a,b,c);
we have
$ f(a,b,c)\le f(a\plus{}b,0,c)$(which is equivalent $ ab(\minus{}4abc^2\plus{}a^3b\plus{}ab^3\minus{}4a^2c^2\minus{}4b^2c^2\minus{}2c^4)\le 0$(true)
we will prove that $ f(a\plus{}b,0,c)\equal{}f(1\minus{}c,0,c)\le \frac{1}{2}$
which is equivalent to
$ \frac{1}{32}*(16c^4\minus{}32c^3\plus{}20c^2\minus{}4c\minus{}1))(\minus{}1\plus{}2c)^2\le 0$
remember that
$ 16c^4\minus{}32c^3\plus{}20c^2\minus{}4c\minus{}1\equal{}4(2c^2\minus{}2c\plus{}\frac{1\minus{}\sqrt{5}}{4})(2c^2\minus{}2c\plus{}\frac{1\plus{}\sqrt{5}}{4})\geq 0$ for every $ c\in[0,1]$


I am sure this is not the easiest way, but:
Moving $(x-1)^{2}$ to the LHS, squaring, and simplifying, we obtain:
$(2x^{4}-4x^{3}+4x^{2}-4x+2)\sqrt{x^{2}+1}=2x^{4}-4x^{3}+4x^{2}-4x+2$
The only way this can be true is if the polynomial on both sides are equal to zero, or if $\sqrt{x^{2}+1}$ is equal to $1$. The latter case yields $x=0$. We can factor the former into $(x-1)^{2}(x^{2}+1)$; we can immediately see from this that the roots to this expression are $1, i,-i$.
Hence the solutions are $\fbox{x=0, 1, i,-i}$.


trigonometry
inequalities
inequalities unsolved

Here is one interesting problem that appeared in crux and deadline for submitting solutions was 1.st October and since it finished I hope that problem can be discussed in this forum.

If $ A$, $ B$ and $ C$ are angles of triangle, and $ S\equal{}\cos\frac{A}{2}\plus{}\cos\frac{B}{2}\plus{}\cos\frac{C}{2}$ and $ P\equal{}\cos\frac{A}{2}\cos\frac{B}{2}cos\frac{C}{2}$
prove the following inequalities:

(i)  $ \frac{S}{P}\leq 2\sqrt{3}\max[\sec\frac{A}{2} , \sec\frac{B}{2} , \sec\frac{C}{2}]$

(ii)  $ \frac{S}{P}\geq 4\max[{\sec}^{2}\frac{B\minus{}C}{4} , {\sec}^{2}\frac{A\minus{}B}{4} , {\sec}^{2}\frac{C\minus{}A}{4}]$

Is there any solution or idea for this problem.

(Also, are there any similar inequalities?) :?:


Just to let everyone know, I will be sending out letters to those who made the SC team this weekend.  We will have our first practice on March 18th.
Also, I would like to start posting practice problems on this site.  Can someone tell me how to take a pdf file and place the text in a post?  I know how to add it as an attachment, but I want to know how to place it in the forum text.

i think you'll simply half to type out all the problems and in Latex, do you know how to post in latex mr. R? Just add dollar signs, and \frac{}, \sqrt{}, etc, can be learned from the latex forums. Other than that, you could just add the attachment itself.

Nice to see you on the forums Mr. R ;)

I think the easiest thing to do is this:

1.  Use the Snapshot Tool in Adobe Reader to select and copy all of the problems you want.
2.  Paste them in MS Paint (or a comparable program) and save the file as a .png or .jpeg or .gif (I prefer .png...but it doesn't matter).
3. Go to AoPS's [url=http://www.artofproblemsolving.com/Forum/album.php]Album[/url] which allows you to upload pics.
4. Upload your picture (I would suggest using "Your Personal Gallery").
5. Get the url for the picture.
6. Place the picture in a post by putting [ img ] and [/ img ] (without spaces) around the url.

For example:

[ img ] http://www.artofproblemsolving.com/Forum/album.php [/ img ]

produces

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=605[/img]

Oh, and btw, is there going to be a list of who all made the all state team posted online like in the past?

ok, never mind what I said, joe knows a lot more than i do about posting different things :blush:

Last year there was an option to arrive a day early to Penn State.  If that is still and option, I think we should try to do it.  That will give us all of Friday to practice (maybe play a little basketball/ ultimate frisbee)!  However, I have not seen an updated registration packet for this year, and I do not know if we will have the money.

[quote="prybarczyk"]Yes.  Figuring it out takes a little manipulation.[/quote]

 :rotfl:  :rotfl:


parameterization
geometry
advanced fields
advanced fields solved

Let $V=V(I)\subset\mathbb{A}_{k}^{3}$ be the algebraic set corresponding to the ideal $I=(x^{2}-yz,xz-x)$. Decompose $V$ into its irreducible components.

I tried decomposing by first letting $x$ vary over $k$ and then the same for $z$ (when you do $y$ you get the same as $x$). So $V=\{(t,t^{2},1): t\in k\}\bigcup\{(0,0,t): t\in k\}$. I'm not even sure if this is a correct decomposition (it seems to work), and if it is how I should show the components are irreducible. I thought maybe if I looked at the ideals from these components then showed they were prime I'd know that the components were irreducible. This seems like a round about way of doing it though.

$(0,t,0)$?

When you are a subvariety of $\mathbb A^{n}$ that is given by parameters, then it will generally be irreducible.
For example if you take $f: k\to k^{3}$ given by $f(t)=(t,t^{2},1)$, then $f$ is regular, $k=\mathbb A^{1}$ is irreducible. If its image were not irreducible, then you could pull the irreducible components back on $k$. But this is irreducible, so one of the pullbacks is $k$...

So in this case $V$ is decomposable into three irreducible sets each of which have a 1-1 and onto correspondence with $\mathbb{A}_{k}^{1}$. Is it true that $V=V(I)\subset\mathbb{A}_{k}^{n}$ is decomposable into sets $V_{i}$ where $1\leq i\leq n$ and each $V_{i}$ has a 1-1 and onto correspondence with $\mathbb{A}_{k}^{1}$? This is probably some sort of theorem later on or something, and I should just wait until we get to it...

[quote="HilbertThm90"]Is it true that
$V=V(I)\subset\mathbb{A}_{k}^{n}$ is decomposable into sets $V_{i}$ where $1\leq i\leq n$ and each $V_{i}$ has a 1-1 and onto correspondence with $\mathbb{A}_{k}^{1}$? This is probably some sort of theorem later on or something, and I should just wait until we get to it...[/quote]

Who is $I$ when you shift from $3$ to $n$?

$I$ is an ideal in $k[x_{1},x_{2},...,x_{n}]$. But I think I see what you mean: is it maximal, radical, prime, etc... Is there a case where a decomposition type theorem like that works?

It can happen that $X=V(I)$ is a surface, so there is no way you can decompose it into a finite union of curves isomorphic to $\mathbb A^{1}$.
Even if $X$ is 1-dimensional, it can be a union of how many irreducible components you may like, so you can rule out the condition that $V(I)$ has as many components as indeterminates.
Your question should be one of these:

1. Which (irreducible) subvarieties of $\mathbb A^{n}$ can pe parametrized by affine spaces? i.e. how do their ideals look like?

2. For which ideals of $k[X_{1},\ldots,X_{n}]$, $V(I)$ is a finite union of irreducible components isomorphic to $\mathbb A^{1}$.

3. For which ideals of $k[X_{1},\ldots,X_{n}]$ does $V(I)$ have $n$ irreducible components.

Question 3 is easy to handle. You look at a reduced primary decomposition of $I$ and you require that $n$ primary ideals appear.
If $I$ is a reduced ideal, you require that $n$ prime ideals appear in the reduced primary decomposition. This is equivalent to $k[X_{1},\ldots,X_{n}]/I$ having $n$ primary ideals.

Question 1 is not that hard to handle. To give a variety $X$ by parameters you have to give a morphism $f: k^{m}\to k^{n}$ which must have $X$ as its image. To give $f$ is the same as giving a rings homomorphism $F: k[X_{1},\ldots,X_{n}]\to k[t_{1},\ldots,t_{m}]$. It is not hard to see that $X=V(\ker F)$.

For question 2 it is enough to look for ideals of $k[X_{1},\ldots,X_{n}]$ that give varieties isomorphic to an affine line, or you can extend to an affine space, but I don't know if this problem is solved.
I know it is an open problem in algebraic geometry to decide whether a given variety is an affine space, but this concept of variety is wider and does not require embeddings in affine spaces.


ai  positive  real  numbers , i=1,...,5.
Prove that
Σ[ai/(a_i+1 + 2a_i+2 + 3a_i+3 + 4a_i+4)]  \geq 1/2
where  the  subscripts  ate  to  be  taken  mod5.

Strange that nobody has solved this simple problem since April 2004. Anyway, before I post my solution, I rewrite the problem in a form I would like it more: Prove that for any five positive numbers a, b, c, d, e, we have

[tex]\frac{a}{b+2c+3d+4e}+\frac{b}{c+2d+3e+4a}+\frac{c}{d+2e+3a+4b}+\frac{d}{e+2a+3b+4c}+\frac{e}{a+2b+3c+4d} \geq \frac12[/tex].

And here goes my solution:

At first,

[tex]\frac{a}{b+2c+3d+4e}+\frac{b}{c+2d+3e+4a}+\frac{c}{d+2e+3a+4b}+\frac{d}{e+2a+3b+4c}+\frac{e}{a+2b+3c+4d} = \frac{a^2}{ab+2ac+3ad+4ae}+\frac{b^2}{bc+2bd+3be+4ba}+\frac{c^2}{cd+2ce+3ca+4cb}+\frac{d^2}{de+2da+3db+4dc}+\frac{e^2}{ea+2eb+3ec+4ed}[/tex].

If we denote A = ab + 2ac + 3ad + 4ae, B = bc + 2bd + 3be + 4ba, C = cd + 2ce + 3ca + 4cb, D = de + 2da + 3db + 4dc and E = ea + 2eb + 3ec + 4ed, then we thus have

[tex]\frac{a}{b+2c+3d+4e}+\frac{b}{c+2d+3e+4a}+\frac{c}{d+2e+3a+4b}+\frac{d}{e+2a+3b+4c}+\frac{e}{a+2b+3c+4d} = \frac{a^2}{A}+\frac{b^2}{B}+\frac{c^2}{C}+\frac{d^2}{D}+\frac{e^2}{E}[/tex].

Now, after the [url=http://mathlinks.ro/Forum/viewtopic.php?t=15125]Cauchy-Schwarz inequality in the Engel form[/url], we have

[tex]\frac{a^2}{A}+\frac{b^2}{B}+\frac{c^2}{C}+\frac{d^2}{D}+\frac{e^2}{E} \geq \frac{\left( a+b+c+d+e\right)^2}{A+B+C+D+E}[/tex].

Thus,

[tex]\frac{a}{b+2c+3d+4e}+\frac{b}{c+2d+3e+4a}+\frac{c}{d+2e+3a+4b}+\frac{d}{e+2a+3b+4c}+\frac{e}{a+2b+3c+4d}[/tex]
[tex]\geq \frac{\left( a+b+c+d+e\right)^2}{A+B+C+D+E}[/tex].

And hence, in order to prove the inequality

[tex]\frac{a}{b+2c+3d+4e}+\frac{b}{c+2d+3e+4a}+\frac{c}{d+2e+3a+4b}+\frac{d}{e+2a+3b+4c}+\frac{e}{a+2b+3c+4d} \geq \frac12[/tex],

it will be enough to show that

[tex]\frac{\left( a+b+c+d+e\right)^2}{A+B+C+D+E} \geq \frac12[/tex].

But this is eqivalent to

$2 \left( a+b+c+d+e\right)^2 \geq A+B+C+D+E$.

An easy calculation shows that A + B + C + D + E = 5ab + 5ac + 5ad + 5ae + 5bc + 5bd + 5be + 5cd + 5ce + 5de, so that we can rewrite this as follows:

$2 \left( a+b+c+d+e\right)^2 \geq 5ab+5ac+5ad+5ae+5bc+5bd+5be+5cd+5ce+5de$.

Now,

$\left( a+b+c+d+e\right)^2 = a^2+b^2+c^2+d^2+e^2+2ab+2ac+2ad+2ae+2bc+2bd+2be+2cd+2ce+2de$,

and thus the inequality above rewrites as

$2 \left( a^2+b^2+c^2+d^2+e^2+2ab+2ac+2ad+2ae+2bc+2bd+2be+2cd+2ce+2de \right) \geq 5ab+5ac+5ad+5ae+5bc+5bd+5be+5cd+5ce+5de$,

or, equivalently,

$2 \left( a^2+b^2+c^2+d^2+e^2\right) \geq ab+ac+ad+ae+bc+bd+be+cd+ce+de$.

But this is almost trivial, since the inequality $u^2+v^2 \geq 2uv$ for any two positive numbers u and v implies

[tex]2 \left( a^2+b^2+c^2+d^2+e^2\right)[/tex]
[tex]= \frac12 \left( \left(a^2+b^2\right) + \left(a^2+c^2\right) + \left(a^2+d^2\right) + \left(a^2+e^2\right) + \left(b^2+c^2\right) + \left(b^2+d^2\right) + \left(b^2+e^2\right) + \left(c^2+d^2\right) + \left(c^2+e^2\right) + \left(d^2+e^2\right) \right)[/tex]
[tex]\geq \frac12 \left (2ab+2ac+2ad+2ae+2bc+2bd+2be+2cd+2ce+2de\right)[/tex]
[tex]= ab+ac+ad+ae+bc+bd+be+cd+ce+de[/tex].

And thus, everything is proven.

  Darij

Hmm... I've seen this before... but I remember a shorter sol. Will give it a try soon.

This inequality is weak.

Introduce variables $x_i, 1\leq i \leq 5$, with $x_i = a_{i+1} + 2a_{i+2} + 3a_{i+3} + 4a_{i+4}$, using cyclic subscripts.
Solving gives that $a_i = \frac{-9}{50}x_i + \frac{1}{50} (x_{i+1}+ x_{i+2}+ x_{i+3}) + \frac{11}{50} x_{i+4}$.  Therefore,

$\sum_{cyc} \frac{a_i}{x_i} = \sum_{cyc} (\frac{-9}{50} + \frac{1}{50} (\frac{x_2}{x_1} + \frac{x_3}{x_1} + \frac{x_4}{x_1}) + \frac{11}{50} \frac{x_5}{x_1} ) \geq \frac{-9}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{11}{10} = \frac{1}{2}$, by AM-GM (since the $x_i$ are positive).

I say the inequality is weak because the $x_i$ are much more than positive.  We've thrown away quite a bit.