ex.
A survey af 300 parks showed the fallowing:
15 had only camping
20 had only hiking trils
35 had only picknicking
185 had camping
140 had camping and hiking trails
125 had camping and picnicking.
210 had hiking trails
Find the number of parks that
a) had at least one of these features.
b) had all three features.
c)did not have any of these features.
d)had exactly two of these features.

I think this should perhaps be moved to the Beginner's section.

Try using PIE.

I'm not exactly sure on how to approach this problem, but I think that PIE might help.

[hide="hint"]Draw a venn diagram and fill it in[/hide]

With the venn diagram:  make sure you dont put things into the circles until you are sure they belong there (ie subtract correctly)  otherwise you will overcount.  write numbers (ie there are 25 in set A, this means at least in set A, could also be in other sets, so dont assume its just in A) outside the circles before you fill in.


given a set S={1,2,3...2001} . determine the minimum value of m such that . with arbitrry subset m elements of S has 2 elements u,v such that u + v = $ 2^n$ with n $ \ge$ 0

oh, yes when I thought this problem , I thought about it (your hint) ! but it was not  completed!
could you write your solution more clearly? :(


inequalities
trigonometry
inequalities unsolved

Let $ A$, $ B$, $ C$ be the three angles of the triangle $ ABC$. Show that for arbitrary positive reals $ x$, $ y$, $ z$, the following inequality holds:
$ x\sin{A}\plus{}y\sin{B}\plus{}z\sin{C}\le\frac{1}{2}\left(xy\plus{}yz\plus{}zx\right)\sqrt{\frac{x\plus{}y\plus{}z}{xyz}}$.

This is actually easier than it looks...

[color=blue][b]Problem.[/b] For any triangle ABC with angles A, B, C, and any three nonnegative reals x, y, z, prove the inequality

$ \sqrt {xyz}\cdot\left(x\sin A \plus{} y\sin B \plus{} z\sin C\right)\leq\frac12\left(yz \plus{} zx \plus{} xy\right)\sqrt {x \plus{} y \plus{} z}$.[/color]

[i]Solution.[/i] Upon squaring, this inequality becomes

[color=green][b](1)[/b][/color] $ xyz\cdot\left(x\sin A \plus{} y\sin B \plus{} z\sin C\right)^{2}\leq\frac14\left(yz \plus{} zx \plus{} xy\right)^{2}\left(x \plus{} y \plus{} z\right)$.

By the Cauchy-Schwarz inequality,

$ \left(x \plus{} y \plus{} z\right)\left(x\sin^{2}A \plus{} y\sin^{2}B \plus{} z\sin^{2}C\right)\geq\left(\sqrt {x\cdot x\sin^{2}A} \plus{} \sqrt {y\cdot y\sin^{2}B} \plus{} \sqrt {z\cdot z\sin^{2}C}\right)^{2}$
$ \equal{} \left(x\sin A \plus{} y\sin B \plus{} z\sin C\right)^{2}$,

so the inequality [color=green][b](1)[/b][/color] follows from

$ xyz\cdot\left(x \plus{} y \plus{} z\right)\left(x\sin^{2}A \plus{} y\sin^{2}B \plus{} z\sin^{2}C\right)\leq\frac14\left(yz \plus{} zx \plus{} xy\right)^{2}\left(x \plus{} y \plus{} z\right)$.

This inequality simplifies to

$ 4xyz\cdot\left(x\sin^{2}A \plus{} y\sin^{2}B \plus{} z\sin^{2}C\right)\leq\left(yz \plus{} zx \plus{} xy\right)^{2}$,

or, equivalently,

$ 4\left(zx\cdot xy\cdot\sin^{2}A \plus{} xy\cdot yz\cdot\sin^{2}B \plus{} yz\cdot zx\cdot\sin^{2}C\right)\leq\left(yz \plus{} zx \plus{} xy\right)^{2}$.

But this follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=590661#590661]http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 post #11[/url] Theorem 2 for the reals yz, zx, xy and the angles A, B, C such that A + B + C is a multiple of 180° (in fact, A + B + C = 180° since A, B, C are angles of a triangle). Thus, the problem is solved.

[b]PS.[/b]

[color=blue][b]Equivalent version of the Problem.[/b] For any triangle ABC with angles A, B, C, and for any acute-angled triangle XYZ with angles X, Y, Z, prove the inequality

$ \tan X\sin A \plus{} \tan Y\sin B \plus{} \tan Z\sin C\leq\frac12\left(\tan Y\tan Z \plus{} \tan Z\tan X \plus{} \tan X\tan Y\right)$.[/color]

  Darij


geometry
inequalities
trigonometry
number theory
complex numbers
\/closed

Which level of subject classes would you seriously consider taking during the Spring semester?

If you would like you can mention which classes you are most likely to take.  We are more likely to offer classes that generate a high level of response.

[quote="jeffrey"]By spring semester I would probably want to take an intermediate geometry, intermediate number theory (I don't know if you'll be able to offer these), or Olympiad courses in pretty much any other subject (algebra, inequalities, counting, etc.)[/quote]

We may have an Intermediate Number Theory course prepared by the Spring.  If not it will likely be prepared by the summer.

[quote="Rep123max"]I will want to be taking Olympiad courses most likely, except for an Intermediate Geometry and Number Theory. For Olympiad courses, it would be Olympiad Problem Solving and will there be an Olympiad Algebra?[/quote]

We will eventually have an Olympiad Algebra course, but it is several courses back on our current list of classes to write.

Are there any plans to have an Intermediate Geometry course in the spring (or anytime soon)?

We hope to have one by the end of 2005, but it won't be in the spring, unfortunately.

This semester we were seriously interested in the Complex Numbers and Trigonometry course, now in progress, but took another course on the same evening. Having that one again in the spring (on any night but Thursday) would be wonderful.

We have scheduled the Spring 2005 Complex Numbers/Trigonometry class for Tuesdays.  We will soon be making broader announcements about class listings.


A nursery employee wishes to plant 2 Golden Delicious apple trees and 5 Bartlett pear trees in one row. How many distinct arrangements are possible?

This question should have a clause stating that trees of the same type are indistinguishable.  Otherwise, the question is too confusing. :mad:

The way I see it is, if the trees were all distinguishable, then why would they take the time to specifically state that there are 2 Golden Delicious apple trees and 5 Bartlett pear trees?  This is just my opinion; if there is some other rule that applies to this or if I am wrong, someone please post/correct me.

There are 7 places for the trees, and 2 of these 7 must be Golden Delicious apple trees. The answer is 7 choose 2.

$ \binom72\equal{}\frac{7\times6}{2}\equal{}\boxed{21}$

the answer is 9 + 10 = 21 ( do u get it )

[hide = Solution]There are seven trees total, and we have to choose where the 2 Golden Delicious apple trees go. So, there are $\binom{7}{2} = \boxed{21}$ ways.[/hide]

[hide=Alternate Solution]
You can first solve for organizing 7 trees in 7 spaces, so you get $7!$. However, there are 2 repeats for the apple trees, and 5 for the pear trees. So, you get $\frac{7!}{5!2!}=\boxed{21}$.
[/hide]

In response to the posts about clarity, the Alcumus problem states the distinguishability of the trees:
[quote=Alcumus]A nursery employee wishes to plant 2 identical Golden Delicious apple trees and 5 identical Bartlett pear trees in one row. How many distinct arrangements are possible? [/quote]

There are $7!$ ways to arrange the plants if they are distinguishable. Since the are not, we must divide by $2! \cdot 5!$. Our answer is $\frac{7!}{2!5!}=21$


limit
trigonometry
calculus
derivative
calculus computations

This is a problem my calc teacher made up.

assume that $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$
Find:
a) $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x}$

a) $\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^3}$

Thanks in advance.

Perhaps there is a way to evaluate them using the given limit.  But using L'Hopital's Rule, we find the first limit is $0$, and using L'Hopital and Squeeze Theorem, we find the second limit is also $0$.  Although I dont' think you need the squeeze theorem...

when you look at the top of this board, it says to not put calculus problems here, there is a special board for calculus that you can post calculus problems in

I asked Mr. Crawford and he said it was okay to post a limits problem.
What is L'Hopital's Rule?

MODERATORS:FEEL FREE TO MOVE TOPIC.

This limit problem seems to be more advanced than the type I think Mr. Crawford would allow to be posted here.

But... L'Hopital's Rule is as follows:

If $\lim_{x \to a} \frac{f(x)}{g(x)}= \frac{0}{0}$ or $=\frac{\infty}{\infty}$, $\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}$

In other words, if a limit that you are evaluating is interdeterminate (check spelling) in both the numerator(f(x)) and the denominator(g(x)), you can evaluate the limit as the first derivative of the numerator divided by the first derivative of the denominator. This will give you the same answer.

 It's kinda hard to explain. If somebody else can do it in a more understandable way, please do so.

Multiply denominator and nominator with $1 + \cos x$ and you can just use that $\lim_{x\to 0}\frac{\sin x}{x} = 1$.

edit: Oh, you're supposed to use the thing you're supplied with. Well,
$\frac{1-\cos x}{x} = x\cdot\frac{1 - \cos x}{x^2}$ so it tends to zero.

$\frac{1-\cos x}{x^3} = \frac{1}{x}\cdot\frac{1 - \cos x}{x^2}$ so the limit doesn't exist.


No. My school doesn't even do Science Olympiad. :(

That sucks :(

We didn't get to go because we had a snow day Friday and our coaches said that if the roads were bad enough to close school it was too dangerous for us to drive to chicago!! We might go to another one though..


limit
integration
trigonometry
real analysis
real analysis unsolved

I was given such a problem and still thinking how to solve:

Find the limit  ${\lim_{n\to +\infty}({2\over {\ln(n+1)}} \int_{x=0}^{\pi\over 2}{{\sin^2 (n+1)x}\over {\sin x}}\,dx )}$ :ninja:

Let $I_n=\int_0^{\frac{\pi}{2}} \frac{\sin ^ 2 (n+1)x}{\sin x} dx,$ let's calculate $I_n-I_{n-1}.$

First step: $\frac{\sin (n+1)x}{\sin x}=e^{-inx}+e^{i(2-n)x}+\dots+e^{i(n-2)x}+e^{inx}$.

Now integrate each term (multiplying by the other $\sin ((n+1)x)$ and estimate.


The search button is your friend a quick search for image signature revealed this
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709190#709190[/url]


vector
real analysis
real analysis unsolved

proved that (u x v) x w = u x (v x w)  iff  (u x w) x v=0

using only a geometric argument...

u,v,w are all vectors.
u x v => the cross product of u and v

Assume $u,v,w$ are unit vectors from the origin satisfying $[[u , w] , v] = 0$. We will prove  \[[[u,v], w] = [u , [v , w]] \;\;\;\;\; \text(1) \]
 If  $[[u , w] , v] = 0$ , which is equivalent to $v \perp (u,w)$, then $[u , v] = u' \subset (u,w) $ ( because $u' \perp v$ ) and  $ \perp u$. Similarly we have $w'$. It is easy too see that the oriented angles $(w',w) =(u,u')$ , so the the resulting vectors in $(1)$ are the same .
 Now let \[[[u , v] , w] = [u , [v , w]]  = a \]  If $a = 0$ then $u,w,v$ are perpendicular to each other....
 It $a \not= 0$, it is easy to see that $a \perp (u,w).$  Then $[v , w] \perp a,v,w$ then $a,v,w$  lie on the same plane. Similarly $a,v,u$ lie on the same plane. Therefore if $a $ is not parallel to $v$ then $a,v,w,u$ lie on the same plane which is absurd because $a \perp (u,w)$. That implies $v \perp u,w$ hence $[[u , w] , v] = 0$

Use Jacobi Identity $U\times\left(V\times W\right)+V\times\left(W\times U\right)+W\times\left(U\times V\right)=0$


Two objects A and B of lengths 120 meters and 180 meters take 30 seconds to cross each when they travel in the same direction. However, if they were to travel in opposite directions, then they will cross each other in 7.5 seconds. If object A is faster than object B, then what is the speed of object A?

Kindly explain the solution. Thanks

I just set up a bunch of equations.

$30(a - b) = 120$

$7.5(a + b) = 300$

Solving for $a$ I get $a= 22$.

Hopefully that's correct  :|

Actually the answer is 90. I just dont know where did it come from.  :blush:

I got 22 also. Maybe we misarranged the first equation or the answer key is wrong.

The question is from this site:

http://free-quiz.4gmat.com/showquiz.php?quiz=22

What exactly does cross each other while they are travelling in the same direction mean ??

I think this is what it means "cross each other" means overtake since we are talking about the same direction( correct me if Im wrong).  For example objects A and B are trains. If two trains start at the same place, one will eventually overtake the other if they have different speeds. So if the other train is longer, it will take sometime for the shorter train to overtake it.

I think the problem itself is quite ambiguous.

Could you explain how you got  $30 (a - b) = 300$ ??

I can explain that. The two objects are going in the same direction. Therefore, the speed of the faster object decreased by the speed of the slower object gives you the rate at which the two objects are moving past each other. Since the two objects take 30 seconds to pass each other, 30 seconds multiplied by the rate of increase gives you 30(a-b)

Because they are moving parallel to each other, the distance the two objects must travel to pass each other is 300. Therefore,

30(a-b) = 300


A group of 6 students decided to make [i]study groups[/i] and [i]service activity groups[/i] according to the following principle:

Each group must have exactly 3 members. For any pair of students, there are same number of study groups and service activity groups that both of the students are members.

Supposing there are at least one group and no three students belong to the same study group and service activity group, find the minimum number of groups.

I think the answer is 8 groups.

What about if there were $n$ students? (Let's assume that everybody has to be in at least one group) ;)

Clearly the number of groups is 2k: k study groups and k service activity groups.
Let k=3. $G_1, G_2, G_3 ...$ - study groups, $J_1, J_2, J_3 ...$ - service activity groups.
WLOG $G_1=\{1;2;3\}$. Then $J_1=\{1;2;x\} J_2=\{1;3;y\} J_3=\{2;3;z\}$ and it's all J-groups
If x,y and z are different then k>3. (Any two of pairs $\{1;x\}, \{3;y\}, \{2;z\}$ cann't be in one G-group, so it's at least 4 G-groups)
If x=y and x≠z then  k>3. ($G_2=\{1;x;v\} G_3=\{1;x;u\}$ and $u,v$≠$z$)
Finally if x=y=z it's also no solutions.
8 groups are possibly:
$G_1=\{1;2;3\}, G_2=\{1;4;5\}, G_3=\{2;4;6\}, G_4=\{3;5;6\}, J_1=\{4;5;6\}, J_2=\{1;2;4\}, J_3=\{1;3;5\}, J_4=\{2;3;6\}$

The answer is $\boxed{8}.$ Let the students be labelled $1, 2, 3, 4, 5, 6.$ Then, taking the groups $(1, 2, 3), (3, 4, 5), (5, 6, 1), (2, 4, 6)$ to be study groups and $(2, 3, 4), (4, 5, 6), (6, 1, 2), (1, 3, 5)$ to be service activity groups works.

Let's show that less than $8$ groups is not possible. Observe firstly that there must be the same name of groups. Clearly $2$ groups is impossible. It's also easy to see that $4$ groups is impossible. Let's now show that $6$ is impossible, i.e. when there are $3$ of each type of group (study or service activity).

If there is no pair of students who are in at least two study groups together, then the logic is rather easy. Indeed, two study groups must share a triangle, from which we can deduce that the other study group has student not in the first two, together with one student from each of the first two study groups. Then there clearly is no way to partition these people into service activities.

Otherwise, if there is a pair of students who are in three study groups together, the logic is also easy.

Finally, suppose there is a pair of students $(x, y)$ which are in exactly two study groups together. Then, the other study group can contain at most one of these two students. Hence, it's clear that any partition of these edges into service activity groups must have the same two triangle (which contain $(x, y)$). This is a contradiction.

$\square$


If you take the AMC 10/12 at a college you can still qualify for AIME right?  Is it any different than taking it at a high school?

Yes, you can still qualify for the AIME as long as you take it on the correct day. It is the same as if you had taken it at a high school.

Colleges and Universities General Instructions and Information
Page, 4, Item 4 d.
http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/Colleges/2010-College_Booklet.pdf

Steve Dunbar
MAA Director, American Mathematics Competitions


How do you evaluate the following summations?

1. $ \displaystyle\sum_{k\equal{}1}^{n}k^2\binom{n}{k}$

2. $ \displaystyle\sum_{k\equal{}1}^{n}(\minus{}1)^{k\plus{}1}k^2\binom{n}{k}$

hello, for the first sum we can prove by induction that
$ \sum_{k\equal{}1}^{n}k^2\binom{n}{k}\equal{}2^{n\minus{}1}n\plus{}2^{n\minus{}2}n(n\minus{}1)$.
Sonnhard.

[quote="FantasyLover"]How do you evaluate the following summations?

1. $ \displaystyle\sum_{k \equal{} 1}^{n}k^2\binom{n}{k}$

2. $ \displaystyle\sum_{k \equal{} 1}^{n}( \minus{} 1)^{k \plus{} 1}k^2\binom{n}{k}$[/quote]

[b]Solution of 1[/b]

consider $ \ (1\plus{}x)^n\equal{}\sum_{k \equal{} 0}^{n}\binom{n}{k}x^k$

and follow the following steps:

Step 1: Differentiate both side with respect to $ \ x$

Step 2: Multiply both side by $ \ x$

Step 3: Differentiate both side with respect to $ \ x$

Step 4: Put $ \ x\equal{}1$ on both side and there goes your problem.

[b]Solution of 2[/b]

Follow the steps of first solution and in the final step 4:Put $ \ x\equal{} \minus{}1$ on both side and there goes your problem.

A generalization of problem 2

Prove that $ \ \displaystyle\sum_{k \equal{} 0}^{n}( \minus{} 1)^{k}f(k)\binom{n}{k}\equal{}0$ where $ \ f(k)$ is a polynomial of degree $ \ < n .$


limit
real analysis
real analysis solved

Let a1 > 0 be a given real number. Define the sequence {an} recursively as follows :

an+1 = an + an-2

for n = 1, 2, 3, ... . Prove that

n\lim\infty ann-1/3

exists and find it.

We will find \lim a n 3/n which is the same as \lim a n+1 3-a n 3 by Cesaro-Stolz. But a n+1 3-a n 3 =(a n+1- a n)( a n+1 2 +a n+1 a n +a n 2 )=(a n+1 /a n) 2 +a n+1 /a n+1, so we only have to find the limit of a n+1 /a n, which is 1 because a n->\infty.

hi pcalin,  you can have a better approximation

an = (3n)1/3( 1 + ln(n)/(9n) +  \varepsilon  n)

with  \lim  \varepsilon n = 0


I wanted to know what you people do for entertainment, excluding math and other academic excercises of course. Maybe you can give me some fun suggesstions?

Music, movie, and videogame (w/ some espn)

I usually play video games, DDR or watch some Anime.

Video games.  Specifically, Resident Evil 4 and FIFA 2006 (PS2).  So addictive...
I also play the violin some...  I watch tv sometimes (only when there's good stuff on).

[quote="hello"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]

I go out and eat...that's fun.
Unfortunately, I don't live in the city like some of my friends...I live in Hastings on Hudson, a little sleepy town in Westchester.  So even when I have time to go places, it's not that easy.  Occasionally, I go watch an awaited movie with friends or invite them over and go to the park.  But it's not that often...

During the week, I'm generally too busy for much entertainment.  But school is fun, so meh.

I go for a walk....
listen to or play music...
meet friends....

 :)

[quote="hello"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]

Oh. Me and my friends go to the mall to hang out every once in a while. Or the arcade. =P

[quote="hello"]But people! Do you go to any places? Like places outside of your home, anywhere? I dont know where? Anything except video games and tv?[/quote]

Well, I'm not really able to do a lot of these things, but you could go to concerts, movies, malls, restaurants, plays, and lots of other stuff.

I like to hang out with my friends most of the time.  Otherwise, I swim, and I fence.

i drink, smoke hang out with palz, listen to music play rpg's :D  :D

I generally go out and play basketball.......... really don't have a habit of hanging around anywhere though........ :D

I read, write nonsense, watch UEFA Champion's League, and play PC strategy games

[quote="hello"]Any one of you by any chance every played paintball? I am interested in playing that so I am curious. So you people rarely do non studying things, right?[/quote]

Wow, what a generalization.

I'm ALWAYS doing non-studying things.  Juggling clubs and hobbies for years.  Not to mention a social life too.

Never tried paintballing, but I'm told its a lot of fun.  I go laser-tagging more often though, its closer to home for me.

Paintballing is loads of fun, it doesn't really hurt that much either.

Err.........what's paintball ?? I really haven't quite heard of it before...  :oops:

I read sci-fi and fantasy, I juggle, I play video games, I go on AOPS, and I IM people.
That's all I can think of right now, I will probably come up with some more things later.

Well, you could go camping with friends, or go backpacking, or canoeing, or kayaking, or go biking.

[quote="hello"]Ok, juggling. But do you people do things outdoors?[/quote]

Outside?!?  But that's where the (gulp) Sun lives!
(Just kidding.  Although I am currently sitting in my room with the lights off and the shades closed in the middle of the day... :ninja: )

Yes, in reality I do do things outside- cross-country skiing in the winter (which is about 7 months of the year) and rowing or soccer the other five months.

i do baseball, soccer, and tennis, and play a lot of basketball at home

Mr Potter, where do you do kayaking and canoeing, since you live in Pennsylvania. Do you go to the poconos? I am interested in going there, so I am trying to find out.

Uh, I live in Ohio.  But I was just listing ideas for what you could do, I've only gone conoeing once, and that was in Canada, kind of near Toronto.

I play alot of basketball. I also generally go chill with a couple friends in the evening after I'm completely exhausted from doing math in the mornings (this is more applicable to spring break right now, hah).

Yea, video games too

[quote="hello"]Ok, juggling. But do you people do things outdoors?[/quote]

Lol, when I said juggling clubs and hobbies, i meant that I've been balancing a huge number of clubs and committments.  I just realized that sounded funny.  No, I don't juggle, lol.

OK, any more suggesstions here? For those that tried paintball, can anyone explain in more detail?

[quote="bubka"]Must be ballpainting.

Or if you are really imaginative, you really can play around balls using paint brushes ;)[/quote]

L. O. L.
My gosh, that got me laughing, no offense. You're kidding? Or are you serious?

Paintballing is when you go to these places where you can play. You put on like protective gear and get a paintball gun, and buy however many paintballs (balls filled with paint) you want. Then you go in the field or arena or whatever, and like... shoot people.
It's fun. I've only done it once, but it. is. FUN.
Laser tag is fun too.
Both Paintball and laser tag are I guess "easier" for smart people, not sure exactly why.
No stereotyping intended.

Nowadays I have little time, so I play basketball whenever my neighbor's friends are over.

For entertainment I have a very peculiar activity [hide="my peculiar activity"][hide="my very peculiar activity"][hide="my very very peculiar activity"]Post mesages on an odd website called: [url=http://www.artofproblemsolving.com/Forum/index.php?f=139]a very peculliar website[/url][/hide][/hide][/hide]

No but seriously my outdoor hobies are playing Table tennis (I play almost everyday), tennis, and basketball

I've done paintballing several times.  Essentially like a battle, but using paintballs.  Quite fun.  The key is not to hide...I used to do so and shoot people from afar if I could when I was young, being scared  :D
Now, I just run through.  Really, getting shot might hurt a bit, but with all of the adrenaline pumping, the pain is unnoticeable.

i spend most of my time playing video games and playing tennis.

i play tennis, soccer, ride bikes, and try to play on my pc as much as i can.

Music, movie, computer games and graphics, reading, playing quitar, surfing (internet), skiing

[quote="shadysaysurspammed"]Watch T.V(Friends,Simpsons,soccer,tennis,nickelodeon-specificallyKenan n Kel,jimmyneutron,spongebob)I'm really into T.V

I'm not a bookaholic

AoPS(though now i'm trying to come on AoPS for doing math :roll: )

Play Tennis,soccer,table tennis,etc(I'm the best outdoor sportsman you can ever get)

I like to Just sit back and $Imagine$[/quote]

I LUV the simpsons! Remember the time when maggie shot Mr. Burns?


Hello,

I have some questions that the answers have eluded me so far.  Let's say an intelligent person studies on his/her own at home, yet he/she does not do school work due to boredom, et cetera, and thus gets fairly low marks.  If this person wants to continue his studies or research, what should this person do without being able to go to university?  Are there any facilities that this person can make use of?  Thanks in advance.

Masoud Zargar

I can't imagine that "this person" has low enough marks to be unable to go to any university.  Also, somewhat high test scores ought to help.

It doesn't matter if you get bored or not, you have to finish school anyway.......

Well, school is the prison of the small men ;)

As I know there are exceptions possible (here, don't know systems in other countries).

See, this is what I don't get. I was bored at my old school- beyond bored, my lowest grade was in PE (and PE is always a giveaway grade). So I [i]changed schools[/i]. I took the opportunities available to me, and don't tell me there are no opportunities available to you. Off the top of my head, I can name three top public high schools in my state (admission based on talent, not geography), and I live in a state infamous for its education funding. I highly doubt that colleges will accept boredom as an excuse. They expect you to find a way to not be bored.

when i read the first message in this topic, the first thing that came to my mind was my own status at school.  :roll: just like u, i have an interest in math but get bad grades at school. in fact, my lowest grade was in math. :P  :blush:

Those who knew me can probably attest to my TERRIBLE grades during HS (and college for that matter).

Having gone through the experience, I highly suggest you take your GPA far more seriously than you have.

Bleumoose is definitely right about colleges not accepting boredom as an excuse.  However, colleges are willing to make exceptions if you show that you are very good at what you are passionate about.  Consider doing USACO if you are somewhat interested in CS, etc.

My Suggestion is to suround your self with subjects you like then the ones you dont like dosent seem as bad (if you have more that you like then dislike) for example next year im taking 4 math classes Physics 2 and Economics.

that happened to me...

i got good grades my freshman year, but each year my grades basically went down a letter grade.

so right now i'm going to a community college that's non-competitve, meaning there was no admission process, when i applied they weren't full so i got in.  right now i'm trying to get good grades and i'll transfer.

I teach, and am an academic advisor, at a university where admissions is not highly selective, and is fairly automatic based on test scores and GPA. I've seen combinations like a 1510 SAT (the old version) with a 2.7 high school GPA, and I've seen combinations like a 960 SAT with a 4.0 GPA.

The students with the high grades and the low test scores do better in our university than do the students with the low grades and the high test scores.

If I see that combination - high test scores with low grades - I see two possibilities:

1. The person is a misunderstood genius, whose brilliance cannot be captured within the narrow confines of high school, or

2. The person is a flake and a lazy bum.

I'd estimate that there are at least 99 flakes for every misunderstood genius. The bad grades are a bad indicator. And a flake might be able to pass some courses without working at them just on native intelligence - but sooner or later, there will come a course that he can't pass without some serious work.

But life does come with second chances, and I have seen ex-flakes who discovered both their true interest and some useful self-discipline while in college.

Academics mean a lot to students these days though
Now, GPA is just as important as SAT or ACT scores.
-sigh-
that's a bad thing for me

[quote="Aryth"]Now, GPA is just as important as SAT or ACT scores.[/quote]

You say that as if there was a time when the opposite was true.  (In fact, grades are probably more important to colleges than any single standardized test score, as they should be.)


Among the points corresponding to number $1,2,...,2n$ on the real line, $n$ are colored in blue and $n$ in red. Let $a_1,a_2,...,a_n$ be the blue points and $b_1,b_2,...,b_n$ be the red points. Prove that the sum $\mid a_1-b_1\mid+...+\mid a_n-b_n\mid$ does not depend on coloring , and compute its value. :roll:

This is called Proizvolov's identity, I believe. You forgot to mention that $a_1,\ldots,a_n$ are the numbers of the first set [b]in increasing order[/b], while $b_1,\ldots,b_n$ are those in the second set [b]in decreasing order[/b] (or vice-versa, of course; the idea is that the sequences are ordered in opposite directions).

Just observe that in the pairs $(a_1,b_1),\ldots,(a_n,b_n)$, each numebr $\le n$ is paired up with a number $\ge n+1$, so the sum is, in fact, $(n+1)+\ldots+2n-(1+2+\ldots+n)=n^2$.

Maybe this observaton isn't that trivial, but I don't have the time to go into details right now, sorry :).


function
inequalities unsolved
inequalities

this is bulgaria 95 ,i think.and it uses keeping all other values constant except one the function will be linear so minima will be reached when all variables take the boundary values (either of them)


Consider the circumference sorrounding the isoceles triangle ABC  (with AC=BC). In the arc BC opossite from A we choose a point D. Let E be a point in AD such that CE  and  AD  are perpendicular. prove that AE=BD+DE.

 it has a lot of diferent solutions

By Ptolemy's theorem, we have $(DA-DB)\cdot AC=DC\cdot AB$. We have to show that $DA-DB=2DE$, so we have to prove that $\frac{CD}{CB}=\frac{CE}{CM}$, where $M$ is the midpoint of $BC$, which is clear since $CAM,CDE$ are similar.

Is this not the Archimedes Midpoint Theorem? http://mathworld.wolfram.com/ArchimedesMidpointTheorem.html


How many positive three-digit integers less than 400 satisfy each of the following characteristics simultaneously?

-Each of the three digits is prime.
-The sum of the three digits is prime.
-The number is prime.

Also 2 is not a three-digit number... :)

I think the 2 was his answer to your question ;)

Oh I see...

Hint: [hide]consider the three things given in this order:

1.) Each of the three digits is prime.

2.) The number is prime.

3.) The sum of the three digits is prime.[/hide]

I'm afraid there's far more solutions than that...unless you do this systematically and organize your work, sooner or later you'll lose count since there's so many...

exactly why i did not attempt this problem

Think about what the ones digit[b] can't[/b] be. That should reduce the possibilities significantly. Then I guess it's just brute force/trial and error...if anyone have any elegant way of attacking this problem it would be really helpful since it's really annoying by trial and error, especially to check and see if the number is a prime or not...yeck! :lol:

OoOoOoOo....
I was only using 2's and 3's.... you can use the other ones... oops.

I found that it was simpler to check for the sum of the digits being prime first.  So that gives you that the sum of the digits is 19, 17, 13, 11, or 7.  (It's bounded because each digit is between 2 and 7.)  This gives us the following possibilities:
19 = 7+7+5
17 = 7+7+3 = 7+5+5
13 = 7+3+3 = 5+5+3
11 = 7+2+2 = 5+3+3
7 = 3+2+2

So, checking these out, we have to try
577  757  377  737
773  557  337  373
733  553  227  533
353  223

And you only have to check divisibility by primes up to 29 at most for any of them . . . I mean, it's tedious, but I don't see there being an easier way.

The original problem wants all of those primes LESS than 400. furthermore, 377,737,553,and 533 are all composite.

So that shortens it down to 337 373 227 353 223.
Note, that if there was no 400 bound, this would be tedious, if not for a very slick way: [hide]Write a program in Java and let the computer calculate it! :rotfl:[/hide]

Well, logically speaking, the 3 digits can either be 2,3,5, or 7, and the last digit cannot be 2 or 5, or otherwise, the number itself wouldn't be prime. This should narrow down some cases...


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

Thoughts? My own view is essentially that of Tegmark's - I hadn't even thought deeply on this before and more or less implicitly assumed such a statement.

hey also guys did you know godel had a completeness theorem

that was kind of irrelevant oh well I just felt like sharing

guys if you want to discuss Tibet please make another topic

no, it had a much more germane title initially. the main thrust is the original post -- I think it's an interesting thing which I haven't thought about much.

Why should math not apply and be effective in the sciences? This goes back to the whole "math is pure" thing, but math is so abstract and based on the simplest axioms that it would only make sense for it to be true and effective in the sciences. The only parts of math "made up" are those basic axioms which are obviously based on intuition and what we see around us, so math would naturally describe it.

see that's what I implicitly assumed for quite a while. but if you step back and think about it, it seems that your argument really only applies insofar as explaining why math is true with respect to the world around us. it doesn't go further and explain why math is so perfect at explaining almost every single thing in the world around us. (kind of like showing something is necessary but not sufficient). by the way, that same wikipedia article has a link further down on the surprising ineffectiveness of math in biology, the cognitive sciences and so forth.

Okay, but if we wonder why math can explain almost everything around us, then we could ask what else there would be to explain it. I don't know. Nor do I know why math would be ineffective at biology and the cognitive sciences. Possibly those are simply far more complicated? eg. biology is made of cells which has tons of different parts made of molecules working together in ridiculous ways. Perhaps they would be explained at math if we could find a way to examine those sciences so simply? But then we would be back down to, like examining how individual molecules interact, and we get into a different science, so I now I'm not precisely sure what I think about it.

Well, isn't physics made upon axioms itself? I have not studied quantum theory, so I am not sure of this statement, but one can easily imagine a world without gravity. It would probably be more like a sea of materials, but more importantly, it seems possible that an absense of an axiom in physics is somewhat plausible, so it is not mathematical related.

I dont' think I know what I am talkign abotu at all, so I will read more on this later :)

Well, if there was no gravity, then we would develop a mathematical system to describe physics that doesn't involve math. It would be different, but still described by math.

Anyway, what I was really trying to say in my last post is this, and I don't know if it makes sense or if I know what I'm talking about either, but:

The field of biology is complicated. Again, the cell has a lot of different parts working together. Therefore, it is not easy at all to come up with mathematical rules to govern the biological world. Take this analogy: the primes. The primes are very complicated, and we need very very advanced math to describe their distribution well, (Riemann Hypothesis level math, for instance). However, despite how complex it is, we can still do it. So, maybe biology can be described by math, it is just so complicated that it seems on the surface that math cannot describe it [i]effectively[/i].

[quote="fedja"]What does Tibet have to do with it? Is it just a trick to attract more attention to the post?  :spam:[/quote]

The O.P. seems to believe that rules of netiquette don't apply to him.  :spam:, indeed.

[quote="JBL"]The O.P. seems to believe that rules of netiquette don't apply to him. [/quote]

hah, this is another topic for another time. clearly not spam, anyway (dudes totally don't know the definition of spam anymore).

@hance see, it seems very easy to convince ourselves that math should indeed be able to accurately model anything it wants to. like, this line of thought makes perfect sense to me as well. that's why I was kinda surprised to read the wikipedia article (and then some of the papers) in the first place -- people were investigating what seemed to me to be a tautology. so I'm trying to step back and say, well, is there any sense in which this line of thought (math describes everything) might fail? well, is there? or are people just hatin' on math for no reason? I'm not sure (this isn't a topic where I have a specific point I'm arguing, more a topic where I'm like hey this is kind of weird guys)

@anax blehh sorry didn't really understand your point there

[quote="MysticTerminator"]http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Fictionalism[/quote]

Whoa what

I really don't understand that. It's quite difficult to comment on because I don't entirely understand how is "science without mathematics" works. But to me, it looks like he designed his physical system without mathematics, and then said that we can use math even though math is just fiction or something -- what?

I mean what is up with saying "mathematics is a reliable process whose physical applications are all true, even though its own statements are false"?

tjhance: fictionalism is essentially an anti-platonist interpretation of mathematics.  that is, fictionalism claims that mathematical objects are disembodied abstractions completely disconnected from the space and time, and thus, statements referring to them cannot possibly be true.  to a fictionalist, the statement "14 is divisible by 2" is false for the same reason that the statement "santa claus is jolly" is false.  santa claus does not exist, and hence, ascribing characteristics to santa claus is absurd.  similarly, "14" and "2" do not actually exist, so statements that try to say something about them are necessarily false.  (side note: is it necessarily the case that the fictionalist approach simply inverts the convention we use for describing vacuous truths?)  fictionalism, however, concedes that mathematics may be a useful fiction that can help point us towards physical reality, but it can never, in and of itself, contain physical truth.

EDIT:  stanford encyclopedia of philosophy strikes again:  http://plato.stanford.edu/entries/fictionalism-mathematics/

Well okay, that makes a bit more sense. But what exactly is the difference between an object being "abstract" (as in Platonism) and an object begin "fictitious"? Is there a difference between looking at the properties of a "real" (what does this mean?) abstract object and looking at the abstract object which does not exist except in a fictitious system which still is able to describe the physical universe? (I don't know what I'm arguing for here; I just don't understand what the point is or what this means entirely.)

Can I make an analogy without people going crazy? It is just too perfect with some of the recent threads.

Mathematical fictionalism says that math is like religion. It may be useful, but this says nothing about whether what it uses is real or not. In fact, since they are made up, we can almost with certainty say that what is described in them don't exist.

As for what the difference between a "mathematical form" or whatever in Platonistic terms and a "fictitious" object, my guess would be that the philosophical purpose is to eliminate some philosophical problems. Platonists need a way for us to come in contact with or come to know the object that actually exists. These objects also have to exist somewhere. Fictionalists don't have to describe what that process is or where they exist, because they say they don't exist.

Mathematically the difference is probably very subtle. It is probably only in attitude. Platonists' thinking is probably more along the lines of looking for an objective truth and fictionalists are probably more likely to manipulate and change the rules of the game. I don't know, I'm just making stuff up now so I'll stop.


Two consecutive sides of a paralellagram have lengths 13 and 6. Find the length of the longest diagonal.

#1: It's "parallelogram."
#2: Do you mean the longest possible diagonal? As there is not enough information given to determine a unique parallelogram.

Yeah, I agree, and if you specified some ugly angles it would turn into a trig problem.

The longest possible diagonal would be some number infinitesimally smaller than 19.

sorry. It was the longest possible diagonal, and the diagonals were integers. I've already solved the problem.

[hide="solution"]The diagonal is less than 13+6=19, and greater than rt(169+36)=rt205. Therefore, the diagonal is either 15, 16, 17, or 18. We can draw a regular parallelogram ABCD, label AB and CD 13, and BC and DA 6. We also draw height BE, and label it h. Make sure that BE is on the outside of the parallelogram. Now draw EC, and label it x. We now know that

$x^{2}+h^{2}=36$

$169+26x+x^{2}+h^{2}=long_{diagonal}^{2}$

$205+26x=long_{diagonal}^{2}$

We now draw heeight AF, label it h, and label FD x. Now

$169-2x+x^{2}+h^{2}=short_{diagonal}^{2}$

$205-26x=short_{diagonal}^{2}$

We need to find the long diagonal such that the shortest diagonal is also an integer. The only number that works is 17. Therefore, 17 is the longest diagonal that can be achieved.[/hide]


calculus
integration
logarithms
trigonometry
function
calculus computations

This should be moved to Calculus Computation Tutrials.

Japanese Communities Moderator

kunny

is this what you mean:

$ \int^1_0 \ln (\sqrt {1\plus{}x} \plus{} \sqrt{1\minus{}x})$ $ dx$

using integration by parts, with $ u$ $ \equal{}$ $ \ln (\sqrt {1\plus{}x} \plus{} \sqrt{1\minus{}x})$ and $ dv$ $ \equal{}$ $ 1$,

$ du$ $ \equal{}$ $ \frac {\sqrt {1\minus{}x}\minus{}\sqrt {1\plus{}x}}{\sqrt {1\plus{}x}\plus{}\sqrt {1\minus{}x}}$  $ \frac{1}{2 \sqrt{1\minus{}x^2}}$

= $ \frac {2 \minus{} 2\sqrt {1\minus{}x^2}}{4x\sqrt{1\minus{}x^2}}$ (multiplying up and down with $ \sqrt {1\plus{}x}\plus{}\sqrt {1\minus{}x}$ )

= $ \frac {1 \minus{} \sqrt{1\minus{}x^2}}{2x\sqrt{1\minus{}x^2}}$

the original integral is now

$ x \ln (\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}) \minus{} \int x \frac {1 \minus{} \sqrt{1\minus{}x^2}}{2x\sqrt{1\minus{}x^2}} dx$

= $ x \ln (\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}) \minus{} \int \frac {1 \minus{} \sqrt{1\minus{}x^2}}{2\sqrt{1\minus{}x^2}} dx$

= $ x \ln (\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}) \minus{} \frac {1}{2} \int \frac {1}{\sqrt {1\minus{}x^2}} \minus{} 1 dx$

for $ \int \frac {1}{\sqrt {1\minus{}x^2}} dx$, substitute $ u  \equal{}  \sin \theta$
$ du \equal{} \cos \theta d\theta$

$ \int \frac {1}{\sqrt {1\minus{}x^2}} dx$ = $ \int d\theta$

                                                    = $ \theta$

                                                    = $ \arcsin x$

so the original integral is

$ \int \ln (\sqrt {1\plus{}x} \plus{} \sqrt{1\minus{}x})$ $ dx$ = $ x \ln (\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}) \minus{} \frac {\arcsin x}{2} \plus{} \frac {x}{2}$ + $ C$

now just substitute the limits. you'll get:

$ \int^1_0 \ln (\sqrt {1\plus{}x} \plus{} \sqrt{1\minus{}x})$ $ dx$ $ \equal{}$ $ \frac {1}{2} \ln 2 \minus{} \frac {\pi}{4} \plus{} \frac {1}{2}$

[quote="anonym115"]i need the integral (and some explanation) of 1n((1+x)^(1/2) + (1-x)^(1/2)) from 0 to 1.[/quote]

If you want $ \int^1_0 \ln (\sqrt {1\plus{}x} \plus{} \sqrt{1\minus{}x})$
ans is$ \frac{(\minus{}2 \plus{} \Pi\plus{} ln 4)}{4}$

Isn't this subject the same as
[b]integral of a certain natural log expression[/b]?

Let $ x \equal{} \sin \theta$,

$ \int_0^1 \ln (\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x})dx \equal{} \int _0^{\frac {\pi}{4}} \ln (\sqrt {1 \plus{} \sin 2\theta} \plus{} \sqrt {1 \minus{} \sin 2\theta})\cdot 2\cos 2\theta \ d\theta$

$ \equal{} 2\int_0^{\frac {\pi}{4}} \cos 2\theta \cdot \ln (|\sin \theta \plus{} \cos \theta| \plus{} |\sin \theta \minus{} \cos \theta|)d\theta$

$ \equal{} 2\int_0^{\frac {\pi}{4}} \cos 2\theta \ln (\sin \theta \plus{} \cos \theta \plus{} \cos \theta \minus{} \sin \theta)d\theta$

$ \equal{} \int_0^{\frac {\pi}{4}} 2\cos 2\theta \cdot \ln (2\cos \theta)d\theta$

$ \equal{} \left[\sin 2\theta \cdot \ln (2\cos \theta)\right]_0^{\frac {\pi}{4}} \plus{} \int_0^{\frac {\pi}{4}} \sin 2\theta \cdot \frac {\sin \theta}{\cos \theta}d\theta$

$ \equal{} \ln \sqrt {2} \plus{} 2\int_0^{\frac {\pi}{4}} \sin ^ 2 \theta d\theta$

$ \equal{} \ln \sqrt {2} \plus{} \int_0^{\frac {\pi}{4}} (1 \minus{} \cos 2\theta)d\theta$

$ \equal{} \ln \sqrt {2} \plus{} \left[\theta \minus{} \frac {1}{2}\sin 2\theta\right]_0^{\frac {\pi}{4}}$

$ \equal{} \boxed{\frac {1}{2}\ln 2 \plus{} \frac {\pi}{4} \minus{} \frac {1}{2}}$


algorithm
invariant
IMO Shortlist
combinatorics
chip-firing

A finite number of coins are placed on an infinite row of squares. A sequence of moves is performed as follows: at each stage a square containing more than one coin is chosen. Two coins are taken from this square; one of them is placed on the square immediately to the left while the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one coin on each square. Given some initial configuration, show that any legal sequence of moves will terminate after the same number of steps and with the same final configuration.

[quote="bboypa"]There exists $ n \in \mathbb{N}_0$ and $ (a_1,a_2,\ldots,a_n) \in \mathbb{N}^n$ such that $ \min\{a_1,a_n\} > 0$ and such that if $ a_1$ represents the (initial) number of coins in the square $ x$ then $ a_i$ represents number of coins in the square $ x \plus{} i \minus{} 1$. Define also $ y: \equal{} \sum_{i \equal{} 1}^n{a_i}$. Obviously all coins stay always in squares $ \ge x \minus{} y$ and always in squares $ \le x \plus{} n \minus{} 1 \plus{} y$, and we can wlog assume $ x \minus{} y$ the is the square number 0, so all moves is done in a finite number of [i]squares[/i]. So for each time there exist a $ (n \plus{} 2y \plus{} 2)$-upla of non negative integer $ (a_0,a_1,a_2,\ldots,a_{n \plus{} 2y \plus{} 1})$ with sum $ y$ such that $ a_i$ represents number of coins in the square $ i$. Now the number $ \sum_{i \equal{} 0}^{n \plus{} 2y \plus{} 1}{i^2a_i}$ increases at each step by 2, and it is always less of $ y(n \plus{} 2l \plus{} 1)^2$: it shows that the algorithm finishes in a finite number of [i]steps[/i]. It remains only to show that we'll have always the same number of moves and the final configuration is invariant by the algorithm chosen. Denote with $ h_i$ the number of moves done at the end of the game in the square $ i$. Then the final number of coins in the square $ i$ is $ a_i \plus{} h_{i \plus{} 1} \minus{} 2h_i \plus{} h_{i \minus{} 1} \in \{0,1\}$. Assume by contradiction that there exists two distinct sequence $ \{h_i\}$ and $ \{H_i\}$ that satisfies the previous relation. Then we can choose $ \{a_i\}_{0 \le i \le n \plus{} 2y \plus{} 1}$ s.t. $ S: \equal{} \min\{\sum{h_i},\sum{H_i}\}$ reaches minimum. Since the initial configuration is the same by assumption, then there exist $ 0 < j < n \plus{} 2y \minus{} 1$ s.t. $ a_j > 1$, but now the $ n \plus{} 2y \plus{} 2$ upla of non negative integers $ (a_0,a_1,\ldots,a_{j \minus{} 1} \plus{} 1,a_j \minus{} 2,a_{j \plus{} 1} \plus{} 1,\ldots,a_{n \plus{} 2y \minus{} 1})$ reaches a new minimum in S, contradiction.[/quote]

I do not understand the lasst step. How is this a contradiction? More over, even if it is, then why does this imply that they must end in the same configuration?

[hide="English mode"]Note that moves are commutative- that is, the final configuration depends only on which moves we make, not on what order we make them in.

Furthermore, any square with at least two coins on it must be moved at some point.

Look at all the moves legal from the starting position- all of them must be made at some point. No move we make now can block off a different move in the future, so we might as well make every initially legal move simultaneously.

But of course we can continue with this logic, making every legal move simultaneously from every subsequent position; so we will be done if we can show the process must end at some point.

No move can create a gap between two piles of coins wider than any gap that already exists, so the breadth of the group of coins is limited, and therefore there are a finite number of positions the coins can be in. In addition, each move strictly increases the total rotational inertia of the coins around their center of mass, so no position can be repeated- the process must terminate, and we are done.[/hide]

[quote="lokitos"]Look at all the moves legal from the starting position- all of them must be made at some point. No move we make now can block off a different move in the future, so we might as well make every initially legal move simultaneously.[/quote]
For someone like me with terrible intuition, here's an easier way to see this (basically induction): suppose that we have two sequences of moves $(r_1,\ldots,r_m)$ and $s=(s_1,\ldots,s_n)$. Then initially, square $r_1$ must have at least $2$ coins, so there must exist $1\le i\le n$ such that $s_i=r_1$. WLOG $s_1=r_1$. (If not, then letting $i_0>0$ denote the smallest such $i$, it's easy to verify that the sequence $s'=(s_{i_0}=r_1,s_1,\ldots,s_{i_0-1},s_{i_0+1},\ldots,s_n)$ is equivalent to the sequence $s$.) Now we can just operate on $r_1$ and then induct to show that $r$ is a permutation of $s$.

It seems that I don't understand any of the solutions posted here... But combined together, they make sense.

[quote="bboypa"]There exists $ n \in \mathbb{N}_0$ and $ (a_1,a_2,\ldots,a_n) \in \mathbb{N}^n$ such that $ \min\{a_1,a_n\} > 0$ and such that if $ a_1$ represents the (initial) number of coins in the square $ x$ then $ a_i$ represents number of coins in the square $ x \plus{} i \minus{} 1$. Define also $ y: \equal{} \sum_{i \equal{} 1}^n{a_i}$. Obviously all coins stay always in squares $ \ge x \minus{} y$ and always in squares $ \le x \plus{} n \minus{} 1 \plus{} y$, and we can wlog assume $ x \minus{} y$ the is the square number 0, so all moves is done in a finite number of [i]squares[/i].[/quote]

Sorry to say, but I don't understand any of this. What is $n$? What are the $a_i$? Why are all moves done in a finite number of squares?

[quote="lokitos"]Note that moves are commutative- that is, the final configuration depends only on which moves we make, not on what order we make them in.[/quote]

Why? (I am aware that this is true, but I only know how to prove it using the stabilization property, i. e., the fact that final configurations always exist. How do you show this at this point, way before you have proven the stabilization property?)

[quote="lokitos"]No move can create a gap between two piles of coins wider than any gap that already exists,[/quote]

Okay, I finally understood what you mean here with the help of [url=http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=17461]Naphthalin[/url]. The claim is that the "maximum gap" (i. e., the highest distance $j-i$ between two distinct integers $i$ and $j$ such that $i<j$ and such that there are coins at the places $i$ and $j$ but no coins anywhere inbetween) can only decrease but never increase by a move. This is only true when this "maximum gap" is $>0$. Fortunately, this is okay because all we want is that this "maximum gap" is bounded a-priori.

We are now discussing a more complex version of this problem at https://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=428298 . My post there (after some minor modifications) also gives a solution of this problem.

[quote="darij grinberg"]It seems that I don't understand any of the solutions posted here... But combined together, they make sense.

Sorry to say, but I don't understand any of this. What is $n$? What are the $a_i$? Why are all moves done in a finite number of squares?[/quote]

$a_1>0$ is the number of coins in the first [i]non empty[/i] square, called $x$; $a_i \ge 0$ is the number of coins in the square $x+i-1$, for all $i=1,2,\ldots,n-1$; $a_n>0$ is the number of coins in the last[i] non empty[/i] square, placed in $x+n-1$. why finite number of squares? Just think about the best and the worst case..

Oh, I see. You're talking about the [b]initial[/b] configuration. I still don't see why this:

[quote="bboypa"]Obviously all coins stay always in squares $ \ge x \minus{} y$ and always in squares $ \le x \plus{} n \minus{} 1 \plus{} y$[/quote]

is "obvious". (I agree that this is true and easy to show after the problem is solved...)

[quote="bboypa"]Then we can choose $ \{a_i\}_{0 \le i \le n \plus{} 2y \plus{} 1}$ s.t. $ S: \equal{} \min\{\sum{h_i},\sum{H_i}\}$ reaches minimum. Since the initial configuration is the same by assumption, then there exist $ 0 < j < n \plus{} 2y \minus{} 1$ s.t. $ a_j > 1$, but now the $ n \plus{} 2y \plus{} 2$ upla of non negative integers $ (a_0,a_1,\ldots,a_{j \minus{} 1} \plus{} 1,a_j \minus{} 2,a_{j \plus{} 1} \plus{} 1,\ldots,a_{n \plus{} 2y \minus{} 1})$ reaches a new minimum in S, contradiction.[/quote]

We need to prove: If $\sum {{H_i}}  = 0$,there not exist no zero sequence $\left\{ {{h_i}} \right\}$ satisfies ${a_i} + {h_{i + 1}} - 2{h_i} + {h_{i - 1}} \in \left\{ {0,1} \right\}$.

A terribly long, but rather straightforward approach may be found [url=http://artofproblemsolving.com/community/c6h1458641p9083581]here[/url].

I have also outlined a solution of the $2$-dimensional analogue of this exercise (i.e., coins are placed on the integer lattice $\mathbb{Z}^2$ rather than on $\mathbb{Z}$, and accordingly at each step we take $4$ coins from a point and distribute them among its $3$ neighboring squares) in [url=http://www.cip.ifi.lmu.de/~grinberg/t/17s/hw5s.pdf]Math 5707 Spring 2017 homework set #5 exercise 5[/url]. There are a bunch of illustrations in a recent popular article: [url=http://nautil.us/issue/23/dominoes/the-amazing-autotuning-sandpile]Jordan Ellenberg, *The Amazing, Autotuning Sandpile*, Nautilus[/url].

Solution from [i]Twitch Solves ISL[/i]:

[b]First part[/b]: We start by proving the procedure always terminates. Index the squares by ${\mathbb Z}$.

[b][color=red]Claim:[/color][/b]  	For any starting configuration, 	there exists an interval $[A,B]$ such that 	no coin may ever exit $B$.
[i]Proof.[/i]  	Let $n$ be the largest index of a square with any coins. 	Note that for any square $m > n$, 	the following property is true: 	if $m$ ever gains a coin, 	then forever after, either $m$ or $m+1$ has a coin. 	This follows since any move affecting either $m$ or $m+1$ 	will add a coin to at least one of them.
Thus, we make take $B = m + 2c$ where $c$ is the number of coins. 	The choice of $A$ is similar. $\blacksquare$
Now note that the sum of squares of indices of coins increases each step. This shows that configurations may never repeat; but there are finitely many configurations, ensuring termination.

[b]Second part[/b]: Suppose $S = (x_1, \dots, x_n)$ is a valid sequence of moves that leads to an end state. We perform the following procedure for $i = 1, \dots, n$. [list] 	[*]Before the $i$'th move $s_i$, 	look at the leftmost square which has more than one coin. 	It must be operated on eventually, 	since that is the only way it can lose coins. 	Let $x_j$ be the first such move on this square.
[*]Rearrange the sequence so that this move $x_j$ comes next instead. 	That is, apply the change 	\[ (x_i, x_{i+1}, \dots, x_{j-1}, x_j) 		\longmapsto \left( x_j, x_i, x_{i+1}, \dots, x_{j-1} \right). \] 	Note that validity of the whole operation is preserved. [/list] In this way, any valid sequence can be rearranged to a certain [i]canonical sequence[/i] where one always operates on the leftmost possible square. This implies that the lengths of all valid sequences are the same (actually, they are permutations of each other), and also that the ending states match.

First we claim that there exists an interval $[A,B]$ which the coins can never leave, say there doesn't exist such $B$, now if there doesn't exist such $B$, we can reach any square on the right hand side, but that's a contradiction because like consider the rightmost square in the initial configuration, note that if you start from moving $k$ coins towards the right, the number of coins moving towards the right decreases each step, it'll never be $k$ again (due to symmetry of distribution), atmost it can be of the form $k/2 + k/8 ... k/2^{n}$ but that is never equal to $k$ for some finite $k$, so eventually at some point we'll not be able to move on the right hand side. Similar argument holds for the left hand side. Now note that the sum of the square of the coin's indices increase each step so the process terminates. Now no matter how you move, you'll always have the same permutation of moves, so we end up in the same final configuration again.


This question has multiple answers, I'm relatively certain.

[hide="One Solution"]$ a_1 \equal{} a_2 \equal{} ... \equal{} a_8 \equal{} 1$, and $ a_9 \equal{} a_{10} \equal{} a_{11} \equal{} a_{12} \equal{} 2$.[/hide]

I don't understand. Help me, please!!!!!!!!!!!!!


search
function
number theory proposed
number theory

since $ x \minus{} y|x^2 \plus{} y^2 \implies x \minus{} y |2xy \implies (x \minus{} y)k \equal{} 2xy \implies \frac {1}{2y} \equal{} \frac {1}{2x} \plus{} \frac {1}{k}$ so there exists integers like $ a,b,d$ with $ \gcd(a,b) \equal{} 1$ and $ 2y \equal{} dab$ and $ 2x \equal{} da(a \plus{} b)$
now just use these equalities and the fact that $ 1997$ is a prime!

[quote="srinath.r"]Sorry if the following problem has been posted before ,but search function didnt help me .

Find all positive integers $ x,y$ such that $ x^{2} \plus{} y^{2} \equal{} 1997(x \minus{} y)$[/quote]

Let $ x\equal{}au$ and $ y\equal{}av$ with $ \gcd(u,v)\equal{}1$

The equation may be written $ a(u\minus{}v)^2\plus{}2auv\equal{}1997(u\minus{}v)$

$ u\ne v$ and so $ u\minus{}v|2auv$ and so $ u\minus{}v|2a$ and let $ 2a\equal{}k(u\minus{}v)$. We get $ a(u\minus{}v)\equal{}1997\minus{}kuv$ and also $ k(u\minus{}v)^2\equal{}3994\minus{}2kuv$ and so $ k|3994$

Let then $ 3994\equal{}kw$ and we get $ (u\minus{}v)^2\equal{}w\minus{}2uv$ and so $ w\equal{}u^2\plus{}v^2$, hence the solutions :

$ (k,w)\equal{}(1,3994)$ $ \implies$ $ 3994\equal{}u^2\plus{}v^2$ $ \implies$ $ (u,v)\in\{(5,63),(63,5)\}$ and so $ (x,y)\equal{}(1827,145)$
$ (k,w)\equal{}(2,1997)$ $ \implies$ $ 1997\equal{}u^2\plus{}v^2$ $ \implies$ $ (u,v)\in\{(29,34),(34,29)\}$ and so $ (x,y)\equal{}(170,145)$
$ (k,w)\equal{}(1997,2)$ $ \implies$ $ 2\equal{}u^2\plus{}v^2$ $ \implies$ $ (u,v)\in\{(1,1)\}$ and so no positive integer solution
$ (k,w)\equal{}(3994,1)$ $ \implies$ $ 1\equal{}u^2\plus{}v^2$ $ \implies$ $ (u,v)\in\{(1,0),(0,1)\}$ and so no positive integer solution

Hence the only two solutions $ (1827,145)$ and $ (170,145)$

sorry dear mathlinkers i can't find the year but u can see it [url=http://www.mathlinks.ro/viewtopic.php?search_id=1326770934&t=239405]here[/url]!

Yes shoki ,it is bulgaria 1998...I tried to find it using search function but couldnt find it . here is my solution ,
Given ,$ x^{2} \minus{} y^{2} \equal{} 1997x \minus{} 1997y \implies x(x \minus{} 1997) \equal{} y( \minus{} y \minus{} 1997)$ .
It is clearly evident that $ x \neq y$ and leave out the trivial cases where $ x \minus{} 1997 \equal{} 0$ or $ \minus{}y \minus{} 1997 \equal{} 0$ which are not possible .

So we have  $ \frac{x}{y} \equal{} \frac{1997 \plus{} y}{1997 \minus{} x}$

$ \implies \frac{x \plus{} y}{x \minus{} y} \equal{} \frac{3994 \minus{} (x \plus{} y)}{x \plus{} y} \implies ( x \plus{} y)^{2} \equal{}
( x \minus{} y)(3994 \minus{} ( x \minus{} y))$ .

Let $ x \minus{} y \equal{} a$ and $ 3994 \minus{} (x \minus{} y) \equal{} b$ ;$ a \plus{} b \equal{} 3994$ .
Let the $ \gcd(a , b) \equal{} d$ ; $ a \equal{}da'$ and $ b \equal{} db'$ where $ a'$ and $ b'$ are coprime .

So we have $ d ( a' \plus{} b' ) \equal{} 3994 \implies d \mid 3994$.
Since $ 1997$ is a prime  , $ d \equal{}1$ or $ d \equal{} 2$ or $ d \equal{} 1997$ or $ d \equal{} 3994$ ,it is easy to see that $ d \neq 1997 ,3994$ .
If $ d \equal{} 2$
$ a' \plus{} b' \equal{} 1997$ and $ x \minus{} y \equal{} 2a'$ and $ 3994 \minus{} (x \minus{} y) \equal{} 2b'$
Since $ (x \minus{} y)(3994 \minus{} (x \minus{} y) \equal{} (x \plus{} y)^{2}  \implies  4a'b' \equal{} (x \plus{} y)^{2}$ ,we easily observe that $ a'b'$ is a perfect square , but since $ \gcd(a' ,b') \equal{} 1  \implies a' \equal{} m^{2}$ and $ b \equal{} n^{2}$ where $ m ,n \in \mathbb{N}$

We get that $ m^{2} \plus{} n^{2} \equal{} 1997$ ,solving this diophantine (it is easy :D ) we get that the non trivial case to be $ m \equal{}29$ and $ n \equal{} 34$

$ \implies 2a' \equal{} a \equal{} x \minus{} y \equal{} 2 * 29^{2}$  ,  $ 3994 \minus{} (x \minus{} y) \equal{} 2 * 34^{2}$ ,using these two we get

$ x \plus{} y \equal{} 2*34*29$ ,from these conditions we can obtain that $ (x ,y ) \equal{} (1827 ,145)$ .

[b]If[/b] $ d \equal{} 1$ , we have $ a \plus{} b \equal{} 3994$ ,repeating the whole story again we can obtain that,
$ a \equal{} m^{2}$ and $ b \equal{} n^{2}$ and solving this diophantine , $ m^{2} \plus{} n^{2} \equal{} 3994$ by doing some work with $ \mod 7$ ,we get the only non trivial solution which is likely to be possible is $ m \equal{} 5$ and $ n \equal{} 63$ . Thus ,using this we can find $ x , y$ with $ m , n$ and we finally get
$ x \equal{} 170$ and $ y \equal{} 145$
Thus the only solutions $ ( x ,y ) \equal{} (1827 ,145 )$ or $ (x ,y ) \equal{} (170 ,145)$


Let a, b, c be positive real numbers. Prove that
\frac {a^2 + a - c^2}{b + c} + \frac {b^2 + 4b - a^2}{c + a} + \frac{c^2 + 9c - b^2}{a + b}\ge\ 4

hello, do you meant this
Let $ a, b, c$ be positive real numbers. Prove that
$ \frac {a^2 \plus{} a \minus{} c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 4b \minus{} a^2}{c \plus{} a} \plus{} \frac{c^2 \plus{} 9c \minus{} b^2}{a \plus{} b}\ge\ 4$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you meant this
Let $ a, b, c$ be positive real numbers. Prove that
$ \frac {a^2 \plus{} a \minus{} c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 4b \minus{} a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 9c \minus{} b^2}{a \plus{} b}\ge\ 4$?
Sonnhard.[/quote]

We have $ \frac {a^2 \minus{} c^2}{b \plus{} c} \plus{} \frac {b^2 \minus{} a^2}{a \plus{} c} \plus{} \frac {c^2 \minus{} b^2}{a \plus{} b} \equal{} \frac {a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq 0$

$ \frac {a}{b \plus{} c} \plus{} \frac {4b}{a \plus{} c} \plus{} \frac {9c}{a \plus{} b} \geq \frac {(a \plus{} 2b \plus{} 3c)^2}{2(ab \plus{} bc \plus{} ca)}$
But
$ \frac {(a \plus{} 2b \plus{} 3c)^2}{2(ab \plus{} bc \plus{} ca)} \minus{} 4 \equal{} \frac {(a \minus{} 2b \minus{} c)^2 \plus{} 8c^2}{2(ab \plus{} bc \plus{} ca)} \geq 0$


[hide]Position the mother, then the father has to sit across from her. The other six seats are optional. So it's $6!=720$ ways.[/hide]

[hide]After seating mother and father, there are 6!=720 remaining ways to seat the children.[/hide]

[quote="1234567890"]Al, Bob, Carl, David, Earl, and Frank sit down to eat dinner with their mother and father on a circular table. How many distinct ways are there to seat the six children and their parents provided that the mother and father sit across from one another?[/quote]
[hide="unsure"]We can use the mother and father as fixed reference points. There are $\binom{6}{3}=20$ ways to pick three people for the top above them. There are 6*6=36 ways to order the people once the divisions have been assigned. So there are 720 ways, unless you can flip the people, (I'm pretty sure this is overcounted) so I think the answer is $\frac{720}{2}=360$[/hide]


I will start a tournament but instead of being countdown to determine the winner like all the other tournaments are, I will make this tournament normal games.

You will be put into a group and play 3 games of 10 questions 22 seconds and timed. The winners (maybe more depending on the number of groups) of each group will get to play one final game of 20 questions 11 seconds and ranking to determine the overall winner.

Sign up now. I will use lowest rating to determine seeding.  :lol:

mewto55555(1)
ernie(1056)
distracted523(1172)
RomanianGenius(1198)
tinytim
(^_^)
thdanh90
hurdler
math154
ra5249(joined even though you don't want too)

ummmm...this is uncomfortable...because of glitch day, my lowest rating is 1...

i will join and 1056 (the day i bombed), but if u want my real lowest rating- its 1113.

Don't miss me.              :blush:  :blush:

I'll join.

Now I wish I hadn't kept my lowest rating so high.....

There sure have been quite a few tourneys started over the last few days.

Please tell me your lowest ratings if you have not yet.

Sorry, forgot to mention my lowest is 1190.

I'll join.

Oops, my lowest rating is 1000... :P

THDANH90 vs. (^_^), a.k.a. Johny Fasee Waisee
90-75
153-63

He got about 3 blanks in the 2nd games; and this results wouldn't interpret anything about his excellent mind.

group 1
ra5249 vs BOGTRO

group 1337
dannyhamtx vs. tinytim

Advanced
mewto55555(1)
RomanianGenius(2)
Yongyi781(1)

Finals
thdanh90

Needs to get lucky to advance
ernie
distracted523
hurdler

Out
(^_^)

[quote="budi713"]Out
(^_^)[/quote]

Oh, that's sad...and a bit unfair. If you're going to have some consolation advancers, why not make it double-elimination and get it over with?

um i did really understand what you were saying

but this tournament is special since i made if so unfair  :D

BOGTRO DEFEATS ra5246 BY FORFEIT

From: ra5249
To: BOGTRO
Posted: Today, at 8:15 am
Subject: Re: FTW
um just tell budi that i withdraw and that you win by withdrawal because i'm at camp and i can't play

group 1337
dannyhamtx vs. tinytim

Advanced
mewto55555(1)
RomanianGenius(2)
Yongyi781(1)
BOGTRO(1)
hurdler(2)

Finals
thdanh90

[quote="vallon22"][quote="ernie"][quote="ash-Pokemon"]lowest rating is 1088 (tanked)[/quote]

THE ASHTER?  :rotfl:  :rotfl:  :rotfl:[/quote]

ernie, don't u have [i]anything[/i] better to do than respond 2 every single post there is??[/quote]

actually, no. :rotfl:

jk

i was just expressing my laughter on the [b]ashter[/b]

u hav 2 admit, it is pretty funny, but so is "ernie"...

[quote]i was just expressing my laughter on the ashter

u hav 2 admit, it is pretty funny, but so is "ernie"...[/quote]

i completely agree  :D  :rotfl:


What is the smallest integer that can be expressed as the sum of two different perfect squares in two different ways?

I answered 25, 0 + 25 & 16 + 9.  The given answer was 65, 64 +1, & 49 + 16.

Should one consider 0 as a perfect square?

It is true that zero considered as a perfect square and it is...
But in MATHCOUNTS question, they don't include the zero in the answer for perfect square question, from what I know so...
So, that's why they have that answer in the question..

[quote="Ravi B"]Yes, 0 is a perfect square.  So your answer is correct.  What contest did this question come from?[/quote]

I remember seeing this on warmup7

Question is from a 1993-1994 Handbook Warm-up worksheet.

So, I am a little confused.  0 is to be considered a perfect square except in Mathcounts questions?

Perhaps the issue here that influenced Mathcounts scoring is that zero is the additive identity element, so that counting zero as fair game for SUMMING perfect squares leaves no distinction between the smallest number that is the sum of perfect squares in one way and the smallest number that is a sum in two ways.

I think another issue that is in view is that 0 times ANYTHING is 0, so while one dictionary I consulted listed 1 as a perfect square, none that that I found with a list gave 0 as an example in defining what a perfect square is.

Here's a Mathcounts forum thread that raises but doesn't definitively answer the question:

http://mathcounts.org/meeting/Upper_Forum_DB/Queries/Active_Forum.taf?function=CombinedDetail&ParentID=6185

(After edit:) See also the thread on the Getting Started forum about the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]definition of a perfect square[/url]. The short answer is that zero is NOT a perfect square, because it is not a figurate number at all.

I believe the Mathcounts definition of perfect square is "square of a positive integer," not "square of an integer."

I remeber 1 team round we did and we got it wrong cuz we didn't use 0 as a perfect square.

Check for what the handbook says, and then protest it at the contest if something like this ever happens.

Is 0 an even number? Is -2 an even number? Is -1 a factor of 5?

O is a definitely even number.. It is considered as even number because numbers are in pattern of odd, even, odd, even, etc..
So, 1 is odd and 0 is before 1, meaning it's even..
And I heard theone853's team got wrong because they didn't use the zero as a perfect squre.. Can you post the problem so we can look at?

On the MathCounts web site, they have the following "toolbox": [url]http://www.mathcounts.org/Program/toolbox.pdf[/url].  There it says that the perfect squares are 1, 4, 9, etc.  In other words, they don't include 0.  That's not how I would have defined the term, but I guess if you are taking a MathCounts competition, then you should follow their definitions.

theone were u on our team when we did that or were you on a different team? so is 0 a perfect square or not?

but wut about that weird team problem where theone got wrong cause he forgot 0 as a perfect square?

We can't answer directly without seeing the problem..
Problem must mention some special words in it.. Because as I said, zero doesn't count on usual perfect square question on Mathcounts..

This is always an interesting sort of topic.  Personally, I am of the opinion that more inclusive definitions are usually better.

One of the basic properties of mathematics that students must master is that of generalization.  If you create too many special cases where generalization is not allowed (simply due to definition), a student may become reluctant to generalize.

Is a square a special rectangle?  Is a rhombus a special parallelogram?  Is a parallelogram a special trapezoid?  Is an equilateral triangle a special isoceles triangle?  Different books and different mathematicians give different answers.

As I said before, I tend to like the more inclusive definitions as long as properties that apply to the specific case also apply to the general case.

To Ravi B:  This is the first year I recall seeing MATHCOUNTS provide a glossary of mathematical terms.  If MATHCOUNTS has published a definition, then you should probably assume the question-writers will be using the official MATHCOUNTS definitions.  However, Section III of the new MATHCOUNTS Toolbox does not define "Perfect Square".  It simply lists some sample values.  Should we assume 676 is not a perfect square because the table ends with 25^2 = 625?  Of course not!  So I'm not sure you should conclude that MATHCOUNTS has defined "perfect square" to exclude zero.

To tokenadult:  I believe most sources that give a written definition say that a perfect square is the square of an "integer".  That would seem to include 0.  I don't believe I've ever seen a written definition say a perfect square is the square of a "counting number".  On the other hand, most sources that list perfect squares begin with 1.

There is also the possible distinction between a "perfect square" and a "square number".  I don't think anyone would argue that the first 4 "triangular numbers" were 1, 3, 6, and 10.  Or that the third triangular number is 6.  But if you asked what is the zeroeth triangular number, I think most mathematicians would respond "it's zero, of course".  A few might claim it is undefined.  So is zero a triangular number?  Is zero a square number?  Is zero a perfect square?

The best answer is that it depends on the context.

A good question-writer will avoid the ambiguities.  He won't refer to a quadrilateral as a "rectangle" if it is also a square.  If he uses the term "perfect square", he will either use it in a way that it must be positive or he will add the qualifier "non-zero" or "positive" or "including zero" to the problem-statement, or he will define the term.

A good question-taker will consider the possible ambiguities.  He will consider the possibility that the "rectangle" refered to in the problem *might* actually be a square.   He will consider that the "perfect square" *might* be zero.  He will consider that sqrt(17) *might* be negative.  If he finds these ambiguities exist, the good question-taker will reexamine the problem to see if there is an extra constraint the eliminates the ambiguity.  If one case is trivial and one case is non-trivial, he will make note of the trivial case, but solve the non-trivial case.

If ambiguities remain that cannot be resolved, a good math teacher will appreciate that the student noticed the defect in the problem.  In a competition, a student should note the ambiguity on his test and bring the ambiguity to the attention of his or her coach.

Just my opinions.  :-)

I agree with your opinion in general, although I would like to nit-pick some of the particulars.  If one were to list the perfect squares, starting with 1 and continuing to 625, in increasing order, it is quite reasonable to assume that the definition being used does not include 0, since the list could easily have been extended backwards to 0 but could not have been extended back any further.  Thus, the choice to start the list with the number 1 seems quite signficant in this case.  Also, non-mathematical definitions of mathematical terms are consistantly awful.  With all respect to Webster's dictionary and tokenadult and anyone else who refered to such a reference, I would never turn to that kind of source for this kind of question, because the people who write them are not being held to a mathematical level of precision.

From a mathematical standpoint, none of this really matters.  Any of the definitions you listed are completely irrelevant to the mathematical truths associated with them.  It is nice to have standard, comprehensible definitions, and it is usually nice to have them include the widest reasonable array of things possible.  But the issue here isn't really from a mathematical viewpoint; it's from the perspective of math competitions.  In competitions, it matters very much what the rules are, because even if you can solve the question when 0 is a perfect square and when 0 isn't, you only get to put down one answer.  Very careful problem writers [i]should[/i] always define ambiguous terms in their questions, but they don't.  So the default contest rules are quite significant, even given their total ambiguity.

The problem goes something like this:

1 is less than or equal to x is less than 100.

how many x are there that can be expressed as a62+b^2, where a and b integers?

But that question never uses the term "perfect square".  It just says b2 where b is an integer.  Because 0 is an integer, 02 is okay for this problem.

Well, it says to express something as a^2 + b^2 where a and b are integers.. so a and b can be {.. -3, -2, -1, 0, 1, 2, 3, ..}.
Now if it said to express it as the sum of two perfect squares, then we may have a problem. but it doesn't.  this is just what Ravi B said but I don't know how else to explain it.. which part confused you?

Well this is the problem i had trouble with:

How many numbers less than 100 can be expressed as the sum of two squares.

Are you sure you're quoting the question [i]exactly[/i] as it was written?  Because if the question is the way Starry wrote it, there isn't any problem at all -- it'd be a reading issue, not a definition issue.

And as much as I enjoy getting credit for such things, I think you (TripleM) meant to say Ravi B., not JBL, right?

[quote="rcv"]. . . .

To tokenadult:  I believe most sources that give a written definition say that a perfect square is the square of an "integer".  That would seem to include 0.  I don't believe I've ever seen a written definition say a perfect square is the square of a "counting number".  On the other hand, most sources that list perfect squares begin with 1.
. . . .
Just my opinions.  :-)[/quote]

I asked about what the mathematical sources say in the thread on the Getting Started Forum "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]How is 'perfect square' defined in books?[/url]" And after seeing the answer posted to that thread, I have no doubt at all that 0 is NOT a perfect square, and has never been regarded as such by the mathematicians who investigate number theory professionally (who would be the arbiters of usage on such a question).

It appears that this thread developed from a competition problem that was misremembered, or related to another person without sufficient detail, or something like that. Getting the right answer to a mathematical problem always involves, first of all, figuring out what the problem is about. The problem in this thread, it turns out, is not defining the term "perfect square" (which is the proper "problem" posed by the thread I posted) but finding a correct answer given a specified set of constraints, in which it appears the term "perfect square" didn't appear at all (if other participants here are relating the story correctly).

[quote="JBL"]I agree with your opinion in general, although I would like to nit-pick some of the particulars. . . . With all respect to Webster's dictionary and tokenadult and anyone else who refered to such a reference, I would never turn to that kind of source for this kind of question, because the people who write them are not being held to a mathematical level of precision.[/quote]

Now it's my turn to nitpick.  :)  Where did you ever get the idea I would turn to a general English dictionary, or to any nonmathematical source, for the definition of a mathematical term? I notice the absence of quoted text from any of my posts in your discussion excerpted above. I asked, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]in the thread I posted[/url], for definitions from mathematically acceptable sources (and offered some from sources I have at hand) and it's clear to me now that the mathematicians who work with the issue day by day DO NOT call zero a perfect square (for some of the same reasons that they don't call one a prime number).

This, of course, is all in the spirit of "I agree with your opinion in general, although I would like to nit-pick some of the particulars."

[quote="tokenadult"]
I asked about what the mathematical sources say in the thread on the Getting Started Forum "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11100]How is 'perfect square' defined in books?[/url]" And after seeing the answer posted to that thread, I have no doubt at all that 0 is NOT a perfect square, and has never been regarded as such by the mathematicians who investigate number theory professionally (who would be the arbiters of usage on such a question).
[/quote]

The famous theorem that says that a number can be written as the sum of two squares only if any prime factor that has the form 4k-1 has an even multiplicity.

In this case, 0 is a square. Of course it doesn't say perfect square, but really this is pretty similar to this problem..

Just for clarification purposes, the question which originated the thread was verbatim from the 1993-94 Mathcounts handbook, warm-up 18, #3.  It is stated early on in the thread.

Another question brought up was theone's team test question, that we aren't sure of the exact wording.

I now imagine me taking the state test on Saturday in great fear that every question will involve determining whether or not I should consider 0 as a square or a perfect square!  :lol:

Assume that 0 is not a perfect square.  If you get the problem wrong because of that, challenge their answer by showing them the MathCounts Toolbox.

[url=https://en.wikipedia.org/wiki/Square_number]According to Wikipedia[/url] zero is a perfect square


Find all integers $ x$ such that $ x(x\plus{}1)(x\plus{}7)(x\plus{}8)$ is a perfect square
It's a nice problem ...hope you enjoy it!

Daniel


Minimize (a+b)(c+b) subject to abc(a+b+c)=1.

For $ a>0,\ b>0,\ c>0$, $ (a\plus{}b)(b\plus{}c)\equal{}ca\plus{}\frac{1}{ca}\geq 2$ by AM-GM.

I didn't assume a, b, c > 0.  I expanded (a+b)(c+b) and then completed the square to get  a minimum of -((a-c)^2)/4 for b = -(a+c)/2 but am having problems relating this to the condition abc(a+b+c)= 1.


A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of
the vertices of the square lie on the hypotenuse and the other two vertices of the
square lie on the legs of the triangle, what is the length of a side of the square?
Express your answer as a common fraction.

EDIT: I found a mistake in my solution.  Sorry!

However:

[hide="HINT"]
Similar triangles, similar triangles, similar triangles!!
[/hide]

[hide="Another hint"]
Which similar triangle ratios do you want?  What are we trying to solve for? [/hide]

Oh wow!! I never noticed those triangles were similar!!! :rotfl:

yes but how would you relate the side lengths to set up a proportion?

OK I figured it out now.

I labeled the parts of the hypotenuse a,b,c with b being side of square, so a+b+c=17

a/b=15/8
c/b=8/15

a=15b/8
c=8b/15

so 15b/8+b+8b/15=17
409b/120=17
b=2040/409?


Four distinct 3-digit integers have the same hundredth digit and 3 of the numbers are divisors of the sum of the 4 numbers. Find all possible such 4 numbers.

[hide]400a+10b+10c+10d+10e+f+g+h+j is the total sum.
Without loss of generality, let's assume 100a+10b+f, 100a+10c+g, 100a+10d+h are the three numbers that divide the total sum.  By setting up actual divison, we can cancel out the 400a.  This gives us a 4.  Of course, there's a bunch of remainder terms.  10(b+c+d+e) could possibly add up to 360.  f+g+h+j could add up to 36 but in order for that to happen, b doesn't equal c doesn't equal d doesn't equal e and for 360, similar requirements are placed on f+g+h+j.  So the maximum is 360+9+8+7+6 or 390.  The minimum would have 10(b+c+d+e)=0 and have f+g+h+j=0+1+2+3=6.  So the range is between 400a+6 to 400a+390.  We can prove that for a>2 the quotients are 3 and 4 only.  We first prove that a quotient of 2 is impossible.  200a+{0,2,...198}=400a+{6,7,...390}.  200a+{6,7,...390}={0,2,...198} which is obviously not true since a is a digit.  Now, we prove that 5 and numbers greater than 5 will fail.  We pick the smallest a that's greater than 2; this is 3.  300+400(3)=1500 which equals 1500/300=5.  But I can't have three numbers in the 300s average together to equal 400.  The best I can do with those 3 numbers is 399+398+397.  If I was to increase the 300, that would be increasing my lowest number that's increasing my sum by 1.  But for any fraction p/q greater than one, (p+1)/(q+1)<p/q.  Since adding 1 to the lowest number adds 1 to the total, the fractions created by those addtions is less than the original which is less than 5.  Therefore, there's no way it will equal five.  Anything greater than 5 will fail if 5 the fails.  Similarly, anything greater than 300 will also fail.  Now, since there's only 2 possible quotients, but three possible distinct divisors, we have a contradiction.  Now, let's show that a=2 has only 3,4,5 as possible quotients.  We already shown 2 is impossible and 6 is impossible because 6*200/4=300.  In other words, in order for the minimum 200s number to satisfy 6 quotient, it needs to have the 4 numbers average to 300.  That is impossible.  Now, we find that there is no 200s solution.  The lowest possible multiple of 5 is 200*5 or 1000.  The highest multiple of 3 is 299*3=897 so there's no way this can ever match.  So now we go find the 100s solution.  The quotients are 3,4,5,6.  3,4,5 does work in terms of the limit probelm that ruined 200s solutions.  So does 4,5,6 by those terms.  Let's investigate these sequences.  lcm(3,4,5) is 60.  So one of the numbers is divisible by 60/3=20, one divisible by 60/4=15, and one divisible by 60/5=12.  100+199+198+197=694 which is the maximum fraction possible.  100*5=500 is the minimum this number can be so the lowest number can be as high as 690/5=138.  But this number is also a multiple of 12 so the lowest number is 108, 120, or 132.  (Of course, 199,198,197 aren't the real other numbers.)  Now, we multiply each of these three by five and divide them by 3 or 4 and see what happens.  108,136,180,116 is one solution.  All solutions at or above 120 fail because they cause the term divided by 3 to be at 200 or higher.  Now we try 4,5,6 quotients.  The lcm of this is 4*3*5 or 60 so the only difference will be the factors 60/4=15 60/5=12 60/6=10.  Using the previous maximum of 694, we deduce that the minimum this time is 100*6=600.  So we can have from 100 up to 690/6 or 115.  The only two numbers in that set that's divisible by 10 is 100 and 110.  Trying both numbers, we get failures because one number is over 200.  So the only solution is 108,136,180,116.  I think my proof is too long.  Any suggestions?[/hide]

I haven't solved the problem yet, but maybe we should reduce the possible hundreds digits first...

Actually, I made a mistake on my proof.
[hide]540/4=135 not 136 so the number not divisible is 117.  Also, the first 4 sentences of my proof should be removed.  It's not needed.[/hide]


projective geometry
geometry proposed
geometry

Very nice problem,April, and very good application of harmonic division...
[img]http://img98.imageshack.us/img98/9493/aprilnr0.png[/img]
My solution is not the best,but it used only harmonic division theorems. :)
Solution:
Let $ O$ be the intersection of $ AD$ and $ BC$ and $ Z$ be the intersection of tangent lines to $ \omega$ at points $ A,B$.Denote
$ BY\cap AX\equal{}P$.Denote $ PO\cap\omega\equal{}Q,R$(look at the image).Then the division $ (P,Q,O,R)$ is harmonic,hence $ BQAR$ is a harmonic
quadrilateral(Pencil $ Y(P,Q,O,R)$ .Thus $ Z,P.R$ are collinear.Now the problem became obvious,indeed,Let $ T\in ZO\cap AB$,hence if $ F\in CD\cap AB$ then the division $ (B,T,A,F)$ is harmonic,similarly if $ F'\in XY\cap BA$,then $ (B,T,A,F')\equal{}\minus{}1$,thus $ F\equal{}F'$ and we are done.
Thank you.

I'll explan my proof in details,i would like to proof here one part of my solution,the fact $ (P,Q,O,R) \equal{} \minus{} 1$.
Proof:
I think it is well-known that $ O$ lies on the polar of $ P$ w.r.t $ \omega$.But if someone really need,i could post a proof of this well-known fact.
Lemma:
Let $ l$ be an arbitrary line through $ C$.Denote $ P,Q\in\zeta\cap l$.Let $ R\in l$.
Then $ (C,Q,R,P) \equal{} \minus{} 1\Longleftrightarrow R$ lies on the polar $ AB$  of $ C$ w.r.t $ \zeta$.
Proof:
1.Let $ R\in AB$.Then $ PBQA$ is harmonic quadrilateral,thus consider the pencil $ B(PBQA)\cap l$ and we get the result.
2.Let $ A'\in\zeta\cap BR$.Then $ PBQA'$ is harmonic quadrilateral thus $ A' \equal{} A$.
Lemma is proved.
Here is an image to my lemma.
[img]http://img178.imageshack.us/img178/6304/april1rq4.png[/img]

[quote="April"][color=darkred]Let $ A$ and $ B$ be two points lie on the given circle $ \omega$. Let $ C$ and $ D$ be two arbitrary points lie on the tangent lines at $ A$ and $ B$ of the circle $ \omega$, respectively. Denote by $ X$, $ Y$ the second intersections of $ AD$, $ BC$ with $ \omega$. Prove that $ AB$, $ CD$ and $ XY$ are concurrent.[/color][/quote]
[color=darkblue][b]Proof[/b] ([u]without harmonic division[/u]). Denote the intersections $ \|\begin{array}{c} S\in AD\cap BC\ ,\ U\in AC\cap BD \\
 \\
T_1\in AB\cap CD\ ,\ T_2\in XY\cap CD\end{array}\ .$

$ \|\begin{array}{ccc} \triangle CAY\sim\triangle CBA & \implies & \frac {YC}{AC} = \frac {AY}{AB} \\
 \\
\triangle DBX\sim\triangle DAB & \implies & \frac {BD}{XD} = \frac {AB}{BX} \\
 \\
\triangle ASY\sim\triangle BSX & \implies & \frac {SX}{SY} = \frac {XB}{YA}\end{array}\|$ $ \bigodot\implies$ $ \boxed {\frac {YC}{XD}\cdot\frac {SX}{SY} = \frac {AC}{BD}}\ \ (*)\ .$

Apply the [b]Menelaus[/b]' theorem to the transversals :

$ \|\begin{array}{cccc} \overline {T_1AB}\mathrm {\ in\ }\triangle UCD\ : & \frac {T_1C}{T_1D}\cdot\frac {BD}{BU}\cdot\frac {AU}{AC} = 1 & \implies & \frac {T_1C}{T_1D} = \frac {AC}{BD} \\
 \\
\overline {T_2CD}\mathrm {\ in\ }\triangle SCD\ : & \frac {T_2C}{T_2D}\cdot\frac {XD}{XS} \cdot\frac {YS}{YC} = 1 & \implies & \frac {T_2C}{T_2D} = \frac {SX}{SY}\cdot\frac {YC}{XD}\end{array}\ .$

From the relation $ (*)$ obtain $ \frac {T_1C}{T_1D} = \frac {T_2C}{T_2D}$ $ \implies$ $ T_1\equiv T_2\equiv T$, i.e. $ T\in AB\cap CD\cap XY\ .$[/color]

Thank you very much dear [b]Erken[/b] and [b]Virgil Nicula[/b] for your solutions. I liked [b]Virgil's solution[/b] very much, because nowaday we use harmonic so much and forgot other skills.

And sorry dear [b]Erken[/b], could you show me what is the relation between your Lemma and the property $ (P,Q,O,R)$ is harmonic? Or I misunderstood something?  :blush:
Thank you very much. Best regards.

April.

Do you know that in configuration of my first post $ O$ lies on the polar of $ P$ w.r.t $ \omega$.
If you know,then [hide="read it"]From lemma we got that $ (P,Q,O,R)\equal{}\minus{}1$[/hide]
If don't know i could post a proof to this fact...

Oh, I'm sorry dear [b]Erken[/b]. It's a wandering moment  :blush: . So if you know that $ O$ lies on the polar of $ P$ relative to the circle $ \omega$, you can easily to prove that $ O$, $ P$ and $ Z$ are collinear as follows (Please see the [b]Erken's image[/b]).

 Denote by $ F$ the intersection of $ XY$ with $ AB$. We have $ OP$ is the polar of $ F$ relative to the circle $ \omega$. On the other hand, $ AB$ is the polar of $ Z$ relative to $ \omega$, so we have $ Z$ is also lies on the polar of $ F$ w.r.t. $ \omega$. Hence we conclude that $ O$, $ P$ and $ Z$ are collinear. Now, applying [b]Desargue's theorem[/b] to two triangles $ AXC$ and $ BYD$, we have $ P \equal{} AX\cap BY$, $ O \equal{} XC\cap YD$ and $ Z \equal{} CA\cap BD$ are collinear, so we have $ AB$, $ CD$ and $ XY$ are concurrent as desired.

[quote="April"]Denote by $ F$ the intersection of $ XY$ with $ AB$. We have $ OP$ is the polar of $ F$ relative to the circle $ \omega$. On the other hand, $ AB$ is the polar of $ Z$ relative to $ \omega$, so we have $ Z$ is also lies on the polar of $ F$ w.r.t. $ \omega$. Hence we conclude that $ O$, $ P$ and $ Z$ are collinear. Now, applying [b]Desargue's theorem[/b] to two triangles $ AXC$ and $ BYD$, we have $ P \equal{} AX\cap BY$, $ O \equal{} XC\cap YD$ and $ Z \equal{} CA\cap BD$ are collinear, so we have $ AB$, $ CD$ and $ XY$ are concurrent as desired.[/quote]
Applying the [b][size=100]Pascal's theorem[/size][/b], we can also show the collinearity of the points $ Z,$ $ P,$ $ O,$ as the intersection points of the opposite sidelines of the degenereded non convex inscribed hexagon $ AA'YBB'X,$ where $ A'\equiv A$ and $ B'\equiv B$ $ ($ $ Z\equiv AA'\cap BB',$ $ P\equiv AX\cap BY,$ $ O\equiv A'Y\cap B'X$ $ ).$

And then, from the perspective triangles $ ACX$ and $ BDY$, ...

Kostas Vittas.


logarithms
modular arithmetic
number theory proposed
number theory
Analytic Number Theory

Prove that $n!+1$ is not of the form $x^{8}$ for some integer $x$. What about $x^{4}$?

First: just try to prove there are finitely many solutions for $x^{8}+1=n!$, there is a huge number of cases to verify, but we don't care. Next, I have no idea for $x^{4}+1$. The nice fact about $x^{8}+1=n!$ is that it implies the estimation $\frac{3}{4n}{\ln n!}\leq \sum_{p=1\pmod 4, p\leq n}{\frac{\ln p}{p-1}}$, which is not possible for large $n$ (actually, by retaking the estimations in Dirichlet's theorem, one can find explicitly a constant starting with which it's no longer true). Sorry if the hint was too generous.

I got another one:
\[\sum_{p\equiv 3,\, p\leq n}\frac{\ln p}{p-1}\leq \frac{1}{4n}\ln n!.\]
But it is not rigorous, since I used $[n/p]\approx n/p$.

I'm curious about how harazi got $\frac{3}{4n}{\ln n!}\leq \sum_{p=1\pmod 4, p\leq n}{\frac{\ln p}{p-1}}$.
Can anyone explain about this?

@above
First, note that $O\left(\sum_{p, p\leq n}\frac{\ln p}{p-1}\right)=O\left(\frac{\ln n!}{n}\right)$.
$$\prod_{p\equiv 3(mod 4),\, p\leq n}p^{v_p(n!)}\mid x^2-1 \implies\prod_{p\equiv 3(mod 4),\, p\leq n}p^{v_p(n!)}\leq x^2-1<(n!)^{\frac{1}{4}} \implies$$ $$\sum_{p\equiv 3(mod 4),\, p\leq n}\left({\frac{n}{p-1}-\frac{\ln n}{\ln p}}\right)\ln p<\sum_{p\equiv 3(mod 4),\, p\leq n}{\frac{n-s_p(n)}{p-1}\ln p<\frac{\ln n!}{4}} \implies$$ $$\sum_{p\equiv 3(mod 4),\, p\leq n}\frac{\ln p}{p-1}\leq \frac{1}{4n}\ln n!+2$$.
Therefore, $$\frac{3}{4n}\ln n!-2\leq\sum_{p\equiv 1(mod 4),\, p\leq n}\frac{\ln p}{p-1} $$
Now, we can use stirling approximation and use induction on n(if $n$ is true, than $n+4$ is true). But at here, we can get contradiction.


Reverse the two digits of my age, divide by three, add 20, and
the result is my age. How many years old am I?

it's 48
The message is too long. Please make the message smaller before submitting.

[hide=Solution]"My" age can be written as $10a+b$. Flipping gives $10b+a$. Dividing by three gives $\dfrac{10b+a}{3}$. Adding 20 gives $\dfrac{10b+a+60}{3}$. Since that is "my" age, we get \begin{align*}
\dfrac{10b+a+60}{3}&=10a+b\\
10b+a+60&=30a+3b\\
7b+60&=29a
\end{align*} We see that the only pair $(a,b)$ that satisfies (since $a$ and $b$ are single digit integers) is $(a,b)=(4,8)$. So "I" am $\boxed{48}$.[/hide]


HOW TO REPRESENT NORTH CAROLINA AT ARML:

You must first take a regional "Comprehensive" test. These tests vary widely in difficulty from region to region, but they generally have AMC-easy AIME level problems. Score in the top 15% in your region, and you advance to the state level Comprehensive test. The top 30 or so who take that test are invited to ARML, but the place you get doesn't determine the team (A or B) you are assigned to. That is determined on the bus to Penn State, by a Bus Test of 16 problems of individual ARML level problems.

It is possible (but fairly unlikely) to get on the bus and take the bus test without making it through the Comprehensive tests (this is the technique I used last year, when I bombed the regional Comprehensive test; I was invited to the bus because I made USAMO).

And, of course, on the way back from Penn State, the North Carolina participants stop by Hershey Park for a few hours to ride some roller coasters (Storm Runner: highly recommended, if the wait isn't terrible).

As for South Carolina:

[quote="joml88"][size=150][b]South Carolina[/b][/size]

Usually we have a first round of all state in the fall.  Those who qualify go to the second round test at Irmo.  Then, the coaches determine who made the all state team.  Usually about 60 people are named to all state.  Everyone on the all state team are invited to go to ARML.  Usually we get around 30 to go (33 this year).

The process might change a little this year but I think it should stay roughly the same.  We are planning on holding some practices during the year so contact Patrick Rybarczyk by emailing either prybarczyk@rcsd1.org or ronskimomo@hotmail.com as he is the head coordinator for all of this.

If you can't get a hold of him give me a PM and I might be able to help.  Here is this year's all state website: http://155.225.48.46/statemat/stuf2005/teampage.htm .[/quote]

Source: [url=http://www.artofproblemsolving.com/Forum/topic-40434.html]this thread[/url].

[quote="dts"]And, of course, on the way back from Penn State, the North Carolina participants stop by Hershey Park for a few hours to ride some roller coasters (Storm Runner: highly recommended, if the wait isn't terrible).[/quote]

that sounds like fun. maybe we can get the SC team to do that next year. or maybe Carowinds since we stop at rock hill.


Please use English in name of this topic! Locked!