[quote="sandra90"]hi.my name is sandra and i'm 14(almost 15). i live in romania and i just love math although sometimes i feel i can't solve any problem and then i wanna cry.i know that to be good at math means a lot of work and patience but i just can't wait anymore. i'm in love with mathematics.i don't know if you can help me...but please try... :?[/quote]hi there littl' co-national fellow :) what can we help you with, cuz it doesn't sound very clear from what you said here ...

[quote="white_horse_king88"]I think it's an example of the - I really want to understand the solutions to every problem I see, but looking at how much work there is to be done, I really want to know if there is a shortcut to understanding the solutions to everyproblem I see. I used to feel that way, but I think that if you truly love math, doing problems and working at what you love is important.

It's like if you are on a date <Warning: Lousy analogy>, for the first time with a person you've loved for a long time. And you just love them so much that you just want to be in a relationship regardless. Sometimes, things will not work out because you then won't have the underlying foundations. Just like math - taking it one step at a time is crucial. I don't feel bad about not being able to solve USAMO problems because I know I'm not at that level yet. It's nothing to feel worried about, but instead, I work on AIME, and when I know I've mastered that, I proceed. I love math too, and that's why I want to understand it as thoroughly as I can - each level of each topic.

It's like you' re <Warning: ANOTHER Lousy analogy> trying to learn to run, but you can't walk. It's just silly that way. In fact, you will feel more proud of being able to run after you've learned to walk because you know you A) Haven't missed anyting inbetween B) Feel that you've earned it.  ;)

Good luck! Don't stop loving math.[/quote] i know you're wright but there is just one probleme:i don't have so much free time for solving problemes. i have to learn all kinds of stupide things i just don't  have the time i want. it's silly that way...i won't stop loving math because you can't stop loving something or someone that doesn't reply you...you know wath i mean...i love math because it gives me self-confindence...

p.s.:i will allways love math

its better to do lots of little bits - and do the problems yourself! work on suitable problems every week, perhaps set aside 3 hrs on the weekend to work on the problems. soon will get better (=

[quote="sandra90"]
I know you're wright but there is just one probleme: [b]i don't have so much free time for solving problemes[/b]. i have to learn all kinds of stupide things i just don't  have the time i want. it's silly that way...i won't stop loving math because you can't stop loving something or someone that doesn't reply you...you know wath i mean...i love math because it gives me self-confindence...
p.s.:i will allways love math[/quote]

Sandra, I don't want to be mean, but people use to say that the succesful very busy people is the people that usually have more time to do whatever they want. The secret is to find the way of organize your shores efficiently. That does not mean just to do an schedule, but to stick with it.

Unfortunatelly that is something that (a) you can learn, (b) you can't learn it in just a few days, it takes, in the better case, several months, and potentially a lot of [b]Try-Error[/b] attempts, but, ask anyone who have achived something most of the people consider hard (to be on their IMO national team, for example), it require talent, yes, but mainly hard and efficiently applied work.

On the other hand, well, we can give you some advises, but I believe what most of the people can do here is to clarify some of the doubts you may have about some specific problems. By the way, in that sence, I believe the Romanian Forum (in your own language) is quite active!

Best regards,

[quote="sandra90"]i don't have so much free time for solving problemes.[/quote]
Don't wait for free time to come around.... it never will.
Possibly the thought that "I don't [i]have to[/i] do math problems" might be the time-thief here...
Do what you don't have to first, because things you [i]have to[/i], you'll do no matter what. (This works for me, at least.)
<WARNING: bad analogy> In other words, play before you work. Working first will rob you of energy... so that after working, you just don't have the time and energy to play. But if you play before working, you're going to work anyway because you have to.. and who knows, you might even gain new energy to work from playing!

Well, I hope this helped...  Have fun solving problems!

[quote="kueh"][quote="sky9073"]Do what you don't have to first, because things you [i]have to[/i], you'll do no matter what. (This works for me, at least.) [/quote]

Weellll i did maths first and now i'm behind on my school work argh![/quote]
But assuming you're a good student, you'll catch up on that work after the math. That's my point. :)

A good idea is to find a relatively challenging set of problems, and take as much time to solve them as you need (even months), because once [b]you[/b] solve it, you will understand it.  Upon looking at a problem it may look daunting or seeminlgy impossible, but if you give yourself time, you can usually make progress.

[quote="sky9073"]In other words, play before you work. Working first will rob you of energy... so that after working, you just don't have the time and energy to play. But if you play before working, you're going to work anyway because you have to.. and who knows, you might even gain new energy to work from playing!
[/quote]

Hi,

Sky9073, I would like to know, if you don't matter the question, how old are you? I mean, by the way the parragraph is redacted, it seems like you are around 15 or so (just a guess), nevertheless, at least for the population I believe we have here, it is somehow a wise advise. I would guess that the 99% of the highschool and collegue students in this forum are within the top 5% or 10% of their respective institutions, therefore probably they will catch up, then, may be it is better to do first something they like.

Nevertheless, do not abuse, find a point of equilibrium. Remember that you can do math for days and days and suddenly realize that you have two days to do five long projects, and that is unhealthy too (Specially for your grade report)

Best,

I would call this...

scarymathy symptom....

sky9073 is right

do whatever she tells you to do...

[quote="djimenez"]
Sky9073, I would like to know, if you don't matter the question, how old are you? I mean, by the way the parragraph is redacted, it seems like you are around 15 or so (just a guess), [/quote]
Right on the money. I'm 15.

[quote="shinwoo"]sky9073 is right

do whatever she tells you to do...[/quote]
Huh?

wow I never thought about work that way. It seems so counter-intuitive, but it makes sense.

I think I will do nothing until I really need to :D  Of course, 'needing' to do stuff is relative. You have to weigh priorities.

well having to do a 2000-word literature essay in a day is no joke. but i did it  :D

I wonder if i'm similar to anyone else here, because i will literally rip a book in half in frustration from problems (I've done it before).


vector
geometry
complex analysis
complex analysis unsolved

Complex Mapping

Find the image of the following region

[img]http://i452.photobucket.com/albums/qq244/mingcai6172/1.png[/img]

under the action of $ f(z)\equal{}z^5$.

I am horrible with mappings. I wrote $ z^5\equal{}r^5 \cdot e^{5i\theta_0}$, and tried points in both regions, and then I ended up with the whole complex plane as the output and I don't think this is right. How do I find the image of this region? Thanks for the help.

What is the angle between those two lines? Find the angle between the vectors $ (2,1)$ and $ (1,3).$ That angle is $ \frac {\pi}4.$

Now: what does the mapping $ z\mapsto z^n$ do to a wedge between two rays from the origin? It maps it into a new wedge whose angular opening is $ n$ times the angular opening of the original wedge. So $ z\mapsto z^5$ maps a wedge of opening $ \frac {\pi}4$ out into a "wedge" with an angular opening of $ \frac {5\pi}4,$ which is more than half the circle. And in this case, we also have the opposite wedge, which gives of all the points opposite the $ \frac {5\pi}4$ "wedge". That's enough to account for the entire plane, with some areas of overlap.


find the value of

\[ \sum\limits_{m \equal{} 1}^\infty  {\sum\limits_{n \equal{} 1}^\infty  {\frac{{{m^2}n}}
{{{3^m}\left( {n{3^m} \plus{} m{3^n}} \right)}}} }\]

i really enjoyed solving this sum. i will post my solution in a day or two. i hope to find a solution simpler than mine.

[hide="Answer To Verify"]
Answer is $ {9 \over 32}$
[/hide]

I am really surprised that no one was able to answer this question

I have lost patience and hence am posting the solution. in my opinion, this was not really that difficult. maybe a little arduous but not that difficult.  i was actually a bit disappointed that no one could solve this/ or didnt have the time to post the solution anyway, here is my solution. i have put it in the PDF file because it is slightly long and looks better in the pdf.

note that in the pdf you will find a part which says

$ \left[{2,\infty}\right] \cap \text{ (a square indicating unidentified character)}$

this unidentified character is $ {\Bbb N}$, the set of natural numbers

apart from a few other obvious typo's i think it is pretty self explanatory and the file is perfectly safe too.

After you find that symmetry, it is really simple.
We have
\[ 2S = \sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\frac {mn}{3^{m + n}} } = \left ( \sum_{m = 1}^\infty {\frac {m}{3^m}} \right ) \left ( \sum_{n = 1}^\infty {\frac {n}{3^n}} \right )}\]
Now
\[ S' = \sum_{m = 1}^\infty {\frac {m}{3^m}} \text{ and }\frac {S'}{3} = \sum_{m = 1}^\infty {\frac {m}{3^{m + 1}}} \iff \frac {2S'}{3} = \sum_{m = 1}^\infty {\frac {1}{3^m}} = \frac {1}{2} \iff S' = \frac {3}{4}\]
So, $ 2S = \frac {9}{16} \iff \boxed{S = \frac {9}{32}}$

I dont know how i missed that. i guess i must have been thinking too hard.
This always happens to me!! you think too hard and miss the simple things. :wallbash_red:

any way, thanks a lot moonmathpi496


Find a sequence whose limit points are $1,1/2,1/3,..,1/n,..$ & so on.By limit pt of a sequence we mean the limit of its subsequence.
The difficulty of the prob lies on the fact that this sequence has infinitely many limit points.

$0$ must also be one of your limit points.

Just take a sequence of points $\mathcal S_1$ in $(1,2)$ converging towards $1$, a sequence of points $\mathcal S_2$ in $(\frac 12,1)$ converging towards $\frac 12$, and, in general, a sequence of points $\mathcal S_n$ in $(\frac 1n,\frac 1{n-1})$ converging towards $\frac 1n$, for all $n$. Next, in order to get your sequence, simply take an enumeration of the countably many points in the union $\bigcup_{n\ge 1}\mathcal S_n$.

Such a sequence must also have 0 as a "limit" point. (A better name might be "cluster point" for the limits of subsequences.) The set of cluster points must be closed.

For a sequence having cluster points $\frac1n$ and also $0,$ here's an example:

$1,\frac12,1,\frac12,\frac13,1,\frac12,\frac13,\frac14,1,\frac12,\frac13,\frac14,\frac15, 1,\dots$


For $ 1\leq b\leq a\leq b^2,\ 1\leq c\leq a\leq c^2,$ prove that $ 4a\geq (a\minus{}bc\plus{}2)(b\plus{}c).$

$ (a \minus{} bc \plus{} 2)(b \plus{} c) \equal{} ab \plus{} ac \minus{} cb^{2} \minus{} bc^{2} \plus{} 2b \plus{} 2c\leq ab \plus{} ac \minus{} ca \minus{} ba \plus{} 2a \plus{} 2a \equal{} 4a$

$ a \equal{} b \equal{} c \equal{} 1$

That's right. :)
Here is my solution.

$ (b\minus{}1)(c^2\minus{}a)\plus{}(b^2\minus{}a)(c\minus{}1)\plus{}a\minus{}b\plus{}a\minus{}c\plus{}b^2\minus{}b\plus{}c^2\minus{}c\geq 0$

$ \Longleftrightarrow 4a\minus{}2b\minus{}2c\plus{}bc^2\plus{}b^2c\minus{}ab\minus{}ca\geq 0$

$ \Longleftrightarrow bc(b\plus{}c)\plus{}4a\geq (a\plus{}2)(b\plus{}c)$

$ \Longleftrightarrow 4a\geq (a\minus{}bc\plus{}2)(b\plus{}c).$

The equality holds when $ a\equal{}b\equal{}c\equal{}1$, which is trivial, sorry. :oops:


I want to do some investigations about the periodicity of Fibonacci numbers mod m .For me it is kind of open problem ,so I kind of wanted a calculator or a program which will do some computations for me so that I can get some sort of pattern and evolve some conjecture. For example with help I got a C program ,like the program asked me for two inputs ,say k and m ,If I give the value of k as some 5 and value of m as 11 ,the program should give me the remainders left by the first 5(=k) fibonacci numbers when divided by 11(=m),suppose I give the value of k as 10 amd m as 23 , then the program should give me the remainders of first 10(=k) fibonacci numbers when divided by 23 . But the  C program turned to be a flop because after some values it started giving wrong remainders of the corresponding fibonacci numbers and the entire effort was a failure  :(
Can anyone tell me any such similar required calculator or a computer program which can overcome this problem and give me the remainders of first k fibonacci numbers when divided by m correctly ??

what seems to be the problem?
the logic works, so there must be something wrong with the program
you can simply take the last two remainders, add them, and take remainder again
something like:
int main()
{
int k,m;
scanf(%d%d",&k,&m);
int a=0,b=1,c=1;
for(int i=0;i<k;i++)
{
  a=b;
  b=c;
  c=(a+b)%m;
  printf(i!=0?" %d":"%d",a);
}
printf("\n");
return 0;
}
should work.

Now, what could the wrong remainders be coming from? Did you, perhaps, calculate the full Fibonacci numbers and then do the division? That gives you trouble when you hit integer overflow, since that subtracts powers of 2 from everything, and those powers of $ 2$ can't be divided evenly by the numbers you're using. However big your integers are, you will reach this point eventually; even a 32-bit integer (maximum value ~2 billion) is overflowed after about 45 steps in the Fibonacci sequence.

The fix for that is to do everything in modular arithmetic, reducing at you go. If you never work with integers larger than twice $ m$, you never have to worry about overflow.

I didnt calculate the fibonacci numbers directly ,the program will internally calculate it and will just display me the remainders
Here is the program ,can you tell what mistake is made .?

 #include <stdio.h>
 #include <math.h>
 int main()
 {
 int k,m;
 printf("enter the value of k");
 scanf("%d",&k);
 printf("enter the value of m");
 scanf("%d",&m);
 int value;
 unsigned int i=0,j=0,sum=1;
 for(int num=1;num<=k;num++)
 {
 value = sum%m;
 printf("%d,",value);
 i=j;
 j=sum;
 sum=i+j;
 }
 }

@lynbrook
Can you verify your program and again and give a detailed one ,so that I can copy paste it  :D

[quote="srinath.r"]I didnt calculate the fibonacci numbers directly ,the program will internally calculate it and will just display me the remainders [/quote]  "Calculate the Fibonacci numbers directly" means "have the program compute the Fibonacci numbers and reduce them at the end."  The alternative is what the two other posters suggested.  You should read and think about and understand what they've written and then you should have no trouble implementing the necessary minor changes to your program yourself.

Sorry JBL
The problem is that I dont know any good C programming(0.1 knowledge0) to understand programs,my friend made me one and I just copy pasted it in the C program .
I am sorry
@jmerry
Are you telling that I should take the two fibonacci numbers and use the program to do modular arithmetic and display the final remainder.?

An unsigned int is virtually always 32 bits, not 16.

Yes,I wanted their remainders mod m so obviously I will get a wrong value of the fibonacci number themselves are wrong . Is there a possibilty to fix or do you know of any such calculator or any other program?

I found your error. Watch the value of sum while you run your program. You should find that it overflows. What should you do to rectify that?

Yongyi is (for some reason) saying exactly the same thing that everyone else has said.  Listen, the C program you're using isn't very complicated -- I don't know how to use C, but I understand exactly how it works just from reading it.  Lynbrook has given you another similar program.  Go spend an hour or two looking at the two programs, maybe using judicious use of a search engine to find some relevant online resources or look up words people have used that you aren't familiar with, and actually [b]do a bit of thinking and work[/b].  Then, when you're done, you'll have a working program and you won't have lowered everyone's opinion of you by repeatedly trying to find someone to do your work for you without seeming to spend even a moderate amount of thought on your own problem.

[quote="srinath.r"]sorry but what do you mean by overflows ? This is what happened when I ran the program  .
I gave the m=23,k=48 .
The 48th fibonacci number is actually supposed to be congruent to 0 mod 23 and the program should show me the 0  in the sequence . But it is showing it as 11  :([/quote]

What I mean is look at the actual value of the variable "sum" in your code. You are forgetting to do something with it in your loop, and as a result, it is growing without bound, and when it reaches 48 the limits of the "int" type will cause it to wrap around.

[hide="Specifically, you need to..."]
Replace the line:

[code]sum=i+j[/code]

with:

[code]sum=(i+j)%m[/code][/hide]

One other comment: for a quick and dirty exploration involving not-too-big numbers like this, there's a lot to be said for using a spreadsheet (such as Excel) rather than a procedural programming language. It's quick, and you see things laid out in front of you.

With that said, it would be perfectly possible to make the exact same mistake in Excel, and have the exact same fix. Note: look up the syntax of the Excel =MOD(m,n) function.

I'm 99% convinced that jmerry used Excel to create the table in post #7 above. In fact, I just replicated his table exactly, probably the same way he did. I'll note that he was using 16 bit arithmetic (or, rather simulating 16 bit arithmetic) to make a pedagogical point; if you use 32 bits, the same thing happens only it takes longer and the numbers are bigger (and it would have been harder to simulate).

Yes ,it is working correctly ,thanks to all of you  :lol:  :10:
I hope it will continue working

On the 16-bit/32-bit issue: when I had some C++ in high school, the standard was 16-bit. Now, there are some versions of C/C++ that have 32-bit numbers standard.

Making the same mistake in Excel leads to a different failure mode, because all numbers are floating-point numbers in Excel. There, the reduction mod m starts giving errors as soon as the numbers are too big for last-digit precision.


A 12-hour clock shows $ 6: 00$.
  [b]a)[/b] After how many minutes the hour and minute hand will make for first time a right angle ?
 [b] b)[/b] How many times untill 12:00 will the hour and minute hand make a right angle ?
 [b] c)[/b]How many times from 0:00 untill 12:00 will the hour and minute hand make a right angle ?

[quote="mszew"][hide="hint"]Count the amount of right angles in a 6/12hr-period and use that the interval between them is always the same.[/hide][/quote]

 I'd like to see the simplest explanation on these questions, avoiding terms of physics as angular velocity ect.

There two types of right angles: type I with the hour before the minute hand (like at 3:00) and
type II with the minute hour before the hour hand (like at 9:00)

Type I: 3:00,
between 4:05 and 4:10,
5:10-5:15,
6:15-6:20,
7:20-7:25,
8:25-8:30,
9:30-9:35,
10:35-10:40,
11:40-11:45,
0:45-0:50,
1:50-1:55.
In total $ 11$ in  $ 12$ hour period then $ 1$ every $ \frac{12 \times 60}{11}\equal{}65\plus{}\frac{5}{11}$ minutes
Type II: idem but beginning with 9:00.
$ a) 3 (65\plus{}\frac{5}{11})\minus{}180\equal{}16\plus{}\frac{4}{11}$ minutes
$ b) 6$ (type I) $ \plus{}3$ (type II) $ \equal{}9$
$ c) 2 \times 11\equal{}22$

Thanks for answering !
 To a) I have found the same answer. To b) the official answer is 11.Question c) was posed by me and I have the same answer with you deviding 12 with 180/11.But I think that this action is not allowed.

 Babis


Prove that $ \forall n \in N,n>2$,$ n!$ is not a perfect square.

Or, better, find as many different proofs of it as you can.

Here is one for start: by the Bertrand postulate, there is a prime $ p$ with $ \frac n2<p\le n$. So, $ p|n!$ but $ p^2\!{\not\hspace{\minus{}0.6pt}|}\, n!$.

Another solution : No sequence of more than 1 consecutive integers can be a power greater than 1 (erdos-selfridge theorem).

And some other demonstrations [url=http://projectpen.files.wordpress.com/2009/06/pen-a9-a37-o511.pdf]Here[/url]


[url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_3]clickie[url]

So,whattheequationi'mgettingforthatproblemis$ \frac {x \plus{} 10y}{29} \equal{} y$wherexisthelastdigitandyistherestofthenumber.But,whenisolveitiget$ 19y \equal{} x$.Howcanthisbepossiblesincexhastobeonedigit?whereamigoingwrong?

$ x$ is the [i]leftmost[/i] digit, not the last one.

If you don't want to do it the rigorous way, here's a [hide="cheap solution"] Heh, so, since this is the AIME, it'll have to be a three digit answer, since 1 and 2 digits obviously don't work.

So you have $ 100a\plus{}10b\plus{}c\equal{}290b\plus{}29c$, or $ 100a\equal{}280b\plus{}28c$. 7 must divide a, so a must be 7. We now have $ 100\equal{}40b\plus{}4c$, or $ 25\equal{}10b\plus{}c$. You can clearly see that 5 divides c, and c=0 doesn't work. Hence, c=5. Now, b=2, and our number is 725.[/hide]


Let $ a_1,a_2,...,a_n$ be positive real numbers, $ n,k\in N,\ n,k\geq1$, such that $ a^k_1\plus{}a^k_2\plus{}\cdots\plus{}a^k_n\leq n$.
Prove that the following inequality holds
$ \frac{1}{a^{k\plus{}1}_1\plus{}k}\plus{}\frac{1}{a^{k\plus{}1}_2\plus{}k}\plus{}\cdots\plus{}\frac{1}{a^{k\plus{}1}_n\plus{}k}\geq\frac{n}{k\plus{}1}$

[quote="marin.bancos"]Let $ a_1,a_2,...,a_n$ be positive real numbers, $ n,k\in N,\ n,k\geq1$, such that $ a^k_1 \plus{} a^k_2 \plus{} \cdots \plus{} a^k_n\leq n$.
Prove that the following inequality holds
$ \frac {1}{a^{k \plus{} 1}_1 \plus{} k} \plus{} \frac {1}{a^{k \plus{} 1}_2 \plus{} k} \plus{} \cdots \plus{} \frac {1}{a^{k \plus{} 1}_n \plus{} k}\geq\frac {n}{k \plus{} 1}$[/quote]

The inequality is equivalent to :

$ \sum_{i \equal{} 1}^n\frac {k \plus{} a_i^{k \plus{} 1} \minus{} a_i^{k \plus{} 1}}{k \plus{} a_i^{k \plus{} 1}}\ge \frac {kn}{k \plus{} 1}$

Which can be written as :

$ \sum_{i \equal{} 1}^n\left(1 \minus{} \frac {a_i^{k \plus{} 1}}{a_i^{k \plus{} 1} \plus{} k}\right) \ge \frac {kn}{k \plus{} 1}$

From Am-Gm inequality we have :

$ a_i^{k \plus{} 1} \plus{} k \equal{} a_i^{k \plus{} 1} \plus{} 1 \plus{} 1.... \plus{} 1 \ge (k \plus{} 1).a_i$

Thus :

$ \sum_{i \equal{} 1}^n\left(1 \minus{} \frac {a_i^{k \plus{} 1}}{a_i^{k \plus{} 1} \plus{} k}\right) \equal{} n \minus{} \sum_{i \equal{} 1}^n\frac {a_i^{k \plus{} 1}}{a_i^{k \plus{} 1} \plus{} k} \ge n \minus{} \sum_{i \equal{} 1}^n \frac {a_i^k}{k \plus{} 1}$

Thus it's sufficent to Chek that :

$ n \minus{} \sum_{i \equal{} 1}^n \frac {a_i^k}{k \plus{} 1}\ge \frac {kn}{k \plus{} 1}$

Which is equivalent to

$ a^k_1 \plus{} a^k_2 \plus{} \cdots \plus{} a^k_n\leq n$

and we are done  :P

Well done, [b]Abdek[/b]! Thanks!
Just a little remark related to this problem, but not only. Obviously, a particular case of a generalized inequality could be considered much more difficult.
It's not very easy to prove the inequality when the values $ n$ and $ k$ are not stressed very clear. On the other hand, they can be "mixed". But a particular case can be proved using many methods.

[u]For example[/u]
$ (n \equal{} 4,k \equal{} 1)$

Let $ a,b,c,d$ be positive real numbers, such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that
$ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \plus{} \frac {1}{d^2 \plus{} 1}\geq2$

[u]A new example[/u]
$ (n \equal{} 3,k \equal{} 2)$

Let $ a,b,c$ be positive real numbers, such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ \frac {1}{a^3 \plus{} 2} \plus{} \frac {1}{b^3 \plus{} 2} \plus{} \frac {1}{c^3 \plus{} 2}\geq1$

[u]See also[/u] http://www.mathlinks.ro/viewtopic.php?t=322362 .

It looks nice, isn't it? And familiar!  They can be considered as independent problems, but they are particular cases for this inequality. For me, this kind of generalized inequalities is, indeed, a valuable one. What do you think? :)

Yes, very nice problems. But constraints will be inequalities ($ \le$), not equalities as per your general statement :
$ a\plus{}b\plus{}c\plus{}d \le4$
$ a^{2}\plus{}b^{2}\plus{}c^{2}\le 3$ etc.

Dear [b]Shanku[/b],
I've used the inequality sign ($ \leq$) in my initial statement (for the additional condition) [b]because it works[/b], and it is "larger" than a simple equality. Therefore, each equality is just a "particular" constraint of the initial additional condition. I know very well that many proposed inequalities (see the examples) have only equalities as constraints. All these equalities represent the "borders" for the allowed sets of positive real values. I do hope that you understood my explanations.


At a school there are 800 students and 800 lockers, numbered 1 through 800.  Suppose the first student opens each locker.  The second student then closes every second locker.  The third student then comes along and decides to change the state of every third locker(that is, closes the ones that are open and opens the one that are closed). The fourth student then changes the state of every forth locker.  This process continues until the 800th student changes the state of the 800th locker.  Which lockers are open????

I suppose the state of the $n$th locker somehow depends on the number of divisors of $n$.

I've seen high school contest problems based on this. It isn't calculus, and that means it doesn't belong in this forum.

No, you don't really have to know the number of divisors, just some partial information about that number.

How can any of the lockers remain open? If there are 8000 students, we get at least one student to close one locker...

maybe there are 4000 people but each use 2 lockers... or more

Woah that still doesn't make sense.. i dont get it

Maybe a little help in understanding the problem itself needed:
The first student opens each locker, that is
no. 1,2,3, ... , 800
The second changes the condition of
no. 2, 4, 6, ... , 800
The third changes
no. 3, 6, 9, ... , 798
etc.
A hint: note that for a locker to be opened you need it to be changed odd number of times.

[hide="hint"]
Pretend for a minute that we only have 30 lockers and 30 students. Which ones end up open and which ones end up closed?
[hide="big hint"]
Think about how many times each locker gets passed. What does this have to do with the locker number?
[hide="bigger hint"]
Which lockers get passed an odd number of times and which lockers get passed an even number of times? Why?

Make a list of the number of divisors that each number has. Find a pattern. Now prove it.
[/hide][/hide][/hide]

A good starting point:
[hide="Hint 1"] Locker $n$ cannot be changed after student $n$ has gone. [hide="Hint 2"]Try "brute forcing" the first 30 lockers. Remember, to have the students "stop" at 30 though. A pattern should emerge.[/hide][/hide]

...or the "real" solution
[hide="Hint 1"]Think about when a locker's state is changed. For example, which students open and close locker #24?[hide="Hint 2"]If the lockers start closed, think about its final state after $n$ changes of state. After 1 change of state is it open or closed? 2? 3? 4?[hide="MEGA MEGA HINT (basically gives it away)"]If a locker changes state for every factor it has, and is only open if it has been changed an odd number of times, only lockers with an odd number of factors will be open. Now, which numbers have an odd number of factors again?[/hide][/hide][/hide]

Oh ok I get the problem now. The students are not necessarily just closing lockers. They can open lockers too!

So, to solve this, we see that all prime numbered lockers remain closed as well as locker #1.

Also, maybe writing the lockers in binary can help? The first student opens all lockers. The second will open all lockers ending in 1. Or wait...no binary does not work here since we are not skipping lockers...

Will using modulos work? Just putting some thoughts out there.

[quote="perfect_radio"][quote="boxedexe"]Just consider the number of divisors of a special type of number that I shall leave it to the reader to find.

Masoud Zargar[/quote]
What special type are you talking about? Don't you have to consider them all (and then use the well-known formula for computing the numbers of divisors of a given $n$)?[/quote]

I was referring to a property, not the actual values of the divisors.  Also, the solution using only one very basic tool, nothing too special.

[hide="answer"]The state of the locker is equal to the number of divisors of the locker number (mod 2). If the number of divisors is even, the locker will be closed. If the number of divisors is odd, the locker will open. The only numbers with an odd number of divisors are perfect squares. [/hide]

This definetly doesn't belong here. I saw this problem for the first time in 7th grade.


ΑΣΚΗΣΗ

 Ένα τρίγωνο ABC με $ AB<AC$  είναι εγγεγραμμένο σε κύκλο  .Έστω $ (e)$ η εφαπτομένη του κύκλου στο $ A$ .  Ο κύκλος με κέντρο $ A$ και ακτίνα $ AB$ τέμνει την $ (e)$ στα σημεία $ E,Z$ και την πλευρά $ AC$ στο $ D$. Να αποδείξετε ότι η μία από τις ευθείες $ DE ,DZ$  διέρχεται από το έγκεντρο και η άλλη από το παράκεντρο  που αντιστοιχεί στην γωνία  $ A$ .

[quote="stergiu"]ΑΣΚΗΣΗ

 Ένα τρίγωνο ABC με $ AB < AC$  είναι εγγεγραμμένο σε κύκλο  .Έστω $ (e)$ η εφαπτομένη του κύκλου στο $ A$ .  Ο κύκλος με κέντρο $ A$ και ακτίνα $ AB$ τέμνει την $ (e)$ στα σημεία $ E,Z$ και την πλευρά $ AC$ στο $ D$. Να αποδείξετε ότι η μία από τις ευθείες $ DE ,DZ$  διέρχεται από το έγκεντρο και η άλλη από το παράκεντρο  που αντιστοιχεί στην γωνία  $ A$ .[/quote]

just angle chasing...

Πράγματι δεν θέλει κάτι παραπάνω  :)

Εν συντομία η λύση μου . Έστω ότι η διχοτόμος της γωνίας Α τέμνει την ΕD στο I και την DZ sto M.
Το ΑΕD είναι ισοσκελές και εύκολα βρίσκουμε $ \angle{ADE}\equal{}\angle{ABC}$ άρα το τετράπλευρο BIDC είναι εγγράψιμο.
Τώρα έχουμε $ \angle{ACB}\equal{}\angle{EAB}\equal{}2\angle{EBD}\equal{}2\angle{ICB}$ άρα $ \angle{ICB}\equal{}\frac{C}{2}$ άρα το Ι είναι το έγκεντρο του τριγώνου ABC .

Αν η ΑΙ τέμνει τον περ. κύκλο στο Ν τότε το Ν είναι το κέντρο του κύκλου του εγγραψίμου  BIDC άρα ΝΙ=ΝD και
$ \angle{IDM}\equal{}90$ άρα ΝΙ=ΝΜ άρα το Μ είναι το παράκεντρο .  :)

Παιδιά ,  πολύ όμορφα :)  !

  Στο ίδιο μήκος κύματος είναι και η επίσημη λύση (του βιβλίου δηλαδή).΄

 Μπάμπης


determine the maximum of 36x6y4-x8y4-x6y6

for x,y real numbers.

Source: variation on a problem from Math Horizons, Vol. 10, No. 4, Apr 2003

This is the same as [hide]computing the maximum of x6y4*(36 - x2 - y2).  For the maximum we would like this to be positive, and since all powers of x and y are even we can just look for a maximum among those points in the first quadrant at a distance no more than 6 from the center of the circle.

Let's look for a maximum along a circular arc of radius r, r < 6:  thus, x2 + y2 = r2.  Then we need to maximize x6y4.  We can substitute in for y, so we want to maximize x6(r2 - x2)2.  Thus, we wish to find the maximum of (x10 - 2r2x8 + r4x6).  Setting a derivative equal to zero, we get x = 0, r, -r, or x2 = 3r2/5.  The first three results are clearly not what we want, so our maximum definitely occurs at x2 = 3r2/5, y2 = 2r2/5.  We can now plug these values into the original equation to look for r.  We get (108r10/3125*(36 - r2)) as the value of our original expression.  Again setting a derivative equal to zero, we get r2 = 30 or r = 0, so x = 3 :rt2: and y = 2 :rt3: and the maximum value for the expression is 5038848.[/hide]

good job, but there is also a really slick non-calculus way to calculate the maximum, anybody care to find it?

I figured there would be, but it was easy enough to deal with one variable at a time, and I didn't see a nice way of simplifying those expressions ... except that there has to be something with setting z = 6, or something like that -- using the two powers of 6 to homogonize it.

Non-calculus approach: [hide]Rewrite x6y4(36 - x2 - y2) as 108(x2/3)(x2/3)(x2/3)(y2/2)(y2/2)(36 - x2 - y2).  Now apply AM-GM.[/hide]

thats how i did it too, the original question was involved the more general a2x6y4-x8y4-x6y6
I just picked a=6 so that the max would be an integer.


Michael walks at the rate of $ 5$ feet per second on a long straight path. Trash pails are located every $ 200$ feet along the path. A garbage truck travels at $ 10$ feet per second in the same direction as Michael and stops for $ 30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

The answer is $ 5$. Is there any non-bruteforce way to do this?

I just got some graph paper, drew both paths as distance as a function of time, and counted how many times they intersected.  It's a lot easier than trying to solve equations or something like that.

That would take a lot of time on an actual AMC test though. Is there any easier way?

Actually, all you will have to do is graph y=5x and y=10x (with plateaus every 200ft.). It would'nt take [i]tooooo[/i] much time...

Please search for the problem first:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1050534#1050534

If you have any questions, please revive the old thread.


Euler
geometry
incenter
search
circumcircle
geometric transformation
reflection

Let $I$ be incenter of $\triangle ABC$.

a) Prove that Euler lines of triangles $\triangle ABC, \triangle BCI, \triangle CAI, \triangle ABI$ concur.

b) Prove that the isogonal conjugate of this point of concurrency (wrt $\triangle ABC$) is orthocenter of $\triangle DEF$ where $D,E,F$ are points where incircle touches $\triangle ABC$.

For part a.
This proof is in the [b]Mathematics and Youth[/b] Magazine 11/2007 by [b]Le Duc Thinh[/b].
Let $O$ be the circumcircle of triangle $ABC, G$ be the centroid of triangle $ABC, M$ be the midpoint of $BC, G'$ be the centroid of triangle $BIC. AI$ intersects $BC$ at $D$, intersects $(O)$ at $J$. $(I)$ touches $BC$ at $K, JG'$ intersects $OG$ at $S$, intersects $AM$ at $E$.
It's easy to see that $J$ is the circumcenter of triangle $BIC$. Then $J'G$ is the Euler line of triangle $BIC$.
Applying Menelaus's theorem for triangle $GOM$ with line $(S,E,J)$ we have:
$\frac{SG}{SO}.\frac{JO}{JM}.\frac{EM}{EG}=1.$
Then $\frac{SG}{SO}=\frac{JM}{JO}.\frac{EG}{EM}=\frac{JM}{R}.\frac{EG}{EM}$
Applying Menelaus's theorem for triangle $IAM$ with line $(J,G',E)$ we have :
$\frac{JI}{JA}.\frac{EA}{EM}.\frac{G'M}{G'I}=1.$
Since $G'$ is the centroid of triangle $BIC$ we get:
 $\frac{EA}{EM}=2.\frac{JA}{JI}=2.\frac{JA}{JB}=2.\frac{JB}{JD}=2.\frac{JI}{JD}.$
So $\frac{EG}{EM}=\frac{GM}{EM}-1=\frac{1}{3}.\frac{AM}{EM}-1$
$=\frac{1}{3}.(\frac{EA}{EM}+1)-1=\frac{1}{3}.\frac{EA}{EM}-\frac{2}{3}$
$=\frac{2}{3}.\frac{JI}{JD}-\frac{2}{3}=\frac{2}{3}.\frac{ID}{JD}=\frac{2}{3}.\frac{IK}{JM}=\frac{2}{3}.\frac{r}{JM}$
Therefore $\frac{SG}{SO}=\frac{JM}{r}.\frac{EG}{EM}=\frac{JM}{R}.\frac{2}{3}.\frac{r}{JM}=\frac{2}{3}.\frac{r}{R}$
Similarly, the Euler line of $CIA, AIB$ intersect $OG$ at $S$. We are done.

Another synthetic proof. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=278562
In this proof, you note that $BO_b, CO_c, IO$ concur at $I$ then 3 Euler lines of triangles $AIB, AIC$ and $ABC$ are concurrent.

A beautiful solution in fact  but my question is who the hell could have thought of such an elmentary solution man......  :)

Dear Mathlinkers,
I think that a more simple proof is possible. But all the proof are cery interesting because it shows some particular aspect of the general problem.
Sincerely
Jean-Louis

Dear Mathlinkers,
I have found a nice proof of a) (next on my web site)
but can some give a hint of b) or a reference?
Sincerely
Jean-Louis

Here is a property of Schiffler point which is just a different form of the previous property I posted.
Schiffler point of a triangle $ ABC $ is the Kosnita point of $ A'B'C' $, where $ A',B',C' $ are the midpoints of the arcs $ BC,CA,AB $ on $ \odot ABC $.
My proof for Schiffler point reveals this directly.

[quote="RSM"]Schiffler point of a triangle $ ABC $ is the Kosnita point of $ A'B'C', $ where $ A',B',C' $ are the midpoints of the arcs $ BC,$ $CA,$ $AB $ on $ \odot ABC.$[/quote]
Let $I$ and $I_a,I_b,I_c$ be the incenter and three excenters of $\triangle ABC$ against $A,B,C.$ Clearly, the reflection of $I$ about the midpoint $M_a$ of $BC$ is the orthocenter $H_1$ of $\triangle I_aBC.$ Thus,  the reflection of the centroid of $\triangle IBC$ about $M_a$ is the centroid $G_1$ of $\triangle BCH_1$ $\Longrightarrow$ $I_aG_1$ the image of the Euler line  of $\triangle IBC$ under the homothety $(I,2).$

Let $G_a$ be the centroid of $\triangle II_bI_c.$ Then $\triangle I_aI_bI_c \cup G_a \sim \triangle I_aBC \cup G_1$ $\Longrightarrow$ $\angle I_bI_aG_a=\angle BI_aG_1$ $\Longrightarrow$ $I_aG_a,$ $I_aG_1$ are isogonal lines WRT $\angle I_bI_aI_c.$ $I_aG_a$ passes through the complement of the complement of $I$ WRT $\triangle I_aI_bI_c,$ which is the 9-point center $O$ of $\triangle I_aI_bI_c.$ Hence, $I_aG_1$ passes through the isogonal conjugate of $O$ WRT $\triangle I_aI_bI_c,$ i.e. the Kosnita point $K^*$ of $\triangle I_aI_bI_c$ $\Longrightarrow$ Euler line of $\triangle IBC$ passes through the midpoint of $IK^*,$ i.e. the Kosnita point $K'$ of $\triangle A'B'C'.$ Likewise, Euler lines of $\triangle ICA,$ $\triangle IAB$ pass through $K'.$

Dear Mathlinkers,
an article concerning the “Schiffler point and a result by L. and T. Emelyanov’’ has been put on my website
with a new approach.
http://perso.orange.fr/jl.ayme    vol. 9
You can use Google translator
Sincerely
Jean-Louis

[quote="Bismarck"]b) Prove that the isogonal conjugate of this point of concurrency (wrt $\triangle ABC$) is orthocenter of $\triangle DEF$ where $D,E,F$ are points where incircle touches $\triangle ABC.$[/quote]
Rename $X_0,Y_0,Z_0$ the tangency points of the incircle $(I,r)$ with $BC,AC,AB.$ A-excircle $(I_a,r_a)$ touches $BC,CA,AB$ at $X_1,Y_1,Z_1.$ $G$ is the centroid and $(O,R)$ is the circumcircle. $M$ and $D$ are the midpoints of $\overline{BC}$ and the arc $BC$ of $(O).$ $OI_a$ cuts $BC$ at $A_1$ and $AA_1$ cuts $OG,OD$ at $S,U.$ By Menelaus' theorem for $\triangle ODI_a$ cut by $\overline{AUA_1},$ we get

$\frac{UD}{UO}= \frac{AD}{AI_a} \cdot \frac{I_aA_1}{A_1O}=\frac{AD}{AI_a} \cdot \frac{r_a}{OM} \Longrightarrow$

$\frac{UM}{OU}=\frac{1}{R} \left [ \left( \frac{UD}{OU} +1 \right)OM-R \right]=\frac{1}{R} \left [ \left(  \frac{AD}{AI_a} \cdot \frac{r_a}{OM}+1 \right)OM-R \right]$

Let $P$ be the projection of $D$ on $AB.$ Then $\frac{AD}{AI_a}=\frac{DP}{r_a}$ $\Longrightarrow$

$\frac{UM}{UO}=\frac{1}{R} \left [ \left(  \frac{DP}{OM}+1 \right)OM-R \right] =\frac{DP+OM-R}{R}=\frac{DP-DM}{R}.$

But $DP=\tfrac{1}{2}(r+r_a)$ and $DM=\tfrac{1}{2}(r_a-r)$ $\Longrightarrow$ $DP-DM=r.$

$\Longrightarrow$ $\frac{_{UM}}{^{UO}}=\frac{_r}{^R}.$ Hence, by Menelaus' theorem for $\triangle OGM$ cut by $\overline{ASU},$ we obtain

$\frac{SG}{SO}=\frac{AG}{AM} \cdot \frac{UM}{OU}=\frac{2}{3} \cdot \frac{r}{R} \Longrightarrow$ $S$ is Schiffler point of $\triangle ABC.$

Let $E$ be the reflection of $X_1$ about $Y_1Z_1.$ Using the result of the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=495103]V midpoint[/url] for $I_a$ and its pedal triangle $\triangle X_1Y_1Z_1,$ we get that $ASA_1$ goes through $E.$ If $T$ is the orthocenter of $\triangle X_0Y_0Z_0,$ we have then

$\angle TY_0Z_0=\angle IY_0X_0=\angle ICB=\angle CX_1Y_1=\angle X_1Z_1Y_1=\angle EZ_1Y_1.$

Likewise, $\angle TZ_0Y_0=\angle EY_1Z_1.$ Thus, $AZ_0TY_0 \sim AY_1EZ_1$ $\Longrightarrow$ $\angle Z_0AT=\angle Y_1AE,$ i.e. $AT$ and $AE \equiv AS$ are isogonals WRT $\angle BAC.$ Similarly, $BT,BS$ are isogonals WRT $\angle CBA$ $\Longrightarrow$ $S,T$ are isogonal conjugates WRT $\triangle ABC.$

[quote="Bismarck"]
b) Prove that the isogonal conjugate of this point of concurrency (wrt $\triangle ABC$) is orthocenter of $\triangle DEF$ where $D,E,F$ are points where incircle touches $\triangle ABC$.[/quote]
My solution [b]:[/b]

Let $ H $ be the orthocenter of $ \triangle DEF $ .
Let $ I_a, I_b, I_c $ be three excenters of $ \triangle ABC $ .
Let $ O, G, $ be the circumcenter, centroid of $ \triangle ABC $, respectively .
Let $ X, Y, Z $ be the tangent point of $ (I_a) $ with $ BC, CA, AB $, respectively .
Let $ X' $ be the reflection of $ X $ in $ YZ $ and $ M $ be the midpoint of $ BC $ .
Let $ V=ID \cap EF, V'=I_aX \cap YZ, P=XX' \cap YZ, N=DH \cap EF $ .
Let $ K, T, S $ be the intersection of $ AX' $ and $ I_aX, OM, OG $, respectively .

It's well known that $ A, V, V', M $ are collinear ,
so $ \triangle DNV $ and $ \triangle XPV' $ are homothetic with center $ M $ ,
hence combine with $ DN \parallel XP $ we get $ P, M, N $ are collinear and $ XP=DN $ .

Easy to see $ \triangle DEF \cup N \sim \triangle I_aI_bI_c \cup A $ . [b]...[/b] $ (\star) $

From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=469501]here[/url] we get $ Q=AX' \cap BC \in OI_a $ ,
so from $ (\star ) $ we get $ \frac{OT}{TM}=\frac{I_aK}{KX}=\frac{I_aA}{2DN}=\frac{R}{r} $ ,
hence from Menelaus theorem (for $ \triangle OGM $ and $ \overline{AST} $ ) we get $ \frac{GS}{SO}=\frac{2r}{3R} $ .
i.e. $ S $ is the Schiffler point of $ \triangle ABC $

Since $ AEHF \sim AZX'Y $ ,
so we get $ \{ AS, AH \} $ are isogonal conjugate of $ \angle BAC $ .
Similarly, $ \{ BS, BH \}, \{ CS, CH \} $ are isogonal conjugate of $ \angle CBA, \angle ACB $, respectively ,
so we conclude that the orthocenter $ H $ of $ \triangle DEF $ is the isogonal conjugate of $ S $ WRT $ \triangle ABC $ .

Q.E.D


Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991) +1$ can be factored as a product $(x+b)(x+c)$ where $b,c$ are integers.

We need x^2 + (1991+a)x + 1999a + 1 to have integer roots.  Thus the discriminant $(1991-a)^2 - 4$ needs to be a perfect square (say k^2).  So we need a pythagorean triple k^2 + 2^2 = (1991-a)^2.  Can't happen.

Yeah, but you made a mistake. The discriminant is (1991 + a)^2 - 4...[/quote]

$(x+a)(x+1991)+1 = x^{2}+(1991+a)x+1991a+1$

$x = \frac{-(1991+a) \pm \sqrt{(1991+a)^{2}-4(1991a+1)}}{2}$

$x = \frac{-(1991+a)\pm\sqrt{(1991^{2}+2\cdot1991\cdot a-4\cdot 1991\cdot a+a^{2}-4}}{2}$

$x = \frac{-(1991+a) \pm \sqrt{(1991-a)^{2}-4}}{2}$

For x to be an integer, $\sqrt{(1991-a)^{2}-4}$ must be an integer too. This can only happen when the two squares differ by 4 which only happens at 0,4. Thus, $(1991-a)^{2}= 4$. That means that $1991-a = \pm2$ $a = 1989, 1993$. So, we check that and we see that the numerator will indeed be even (as also  required for x to be an integer) and so the two solutions are $\boxed{1989, 1993}$.

Here's a solution without the quadratic formula:

[hide="Solution"]We have $b+c=a+1991$ and $bc=1991a+1$.  Let $b=a+k$, where $k$ is an integer.  Then $c=1991-k$ from the first equation.  Also, we have $(a+k)(1991-k)=1991a+1$ from the second equation.  Expanding and simplifying, we obtain $(1991-a)k-k^{2}=1$.  Then $k(1991-a-k)=1$.  Therefore, $k=\pm 1$.  From $k=1$ we get $a=1989$, and from $k=-1$, we get $a=1993$.  These are the only solutions.[/hide]

Expanding this equation, we see that:
$$x^2 + (1991 + a)x + (1991a + 1) = 0$$
We want the roots of this equation to be integers. This implies that the discriminant must be a perfect square. The discrimant is just:
$$b^2 - 4ac = (1991 - a)^2 - 4$$
For this to be a perfect square, value must be equal to $0$.
$$\implies (1991 - a)^2 = 4$$
$1991 - a$ can be equal to $2$ and $-2$. So $a$ can equal to $\boxed{1993, 1989}$.


Let triangle ABC. M is the midpoint of BC. The bisector of angle A cuts BC at D. The perpendicular line to AD at D meets AB,AM at E,F respectively. The perpendicular line to AB at E meets AD at K. Prove that FK is perpendicular to BC.

It sufficient to prove that $ \frac {FD}{DK} \equal{} tan\frac {|\gamma \minus{} \beta|}{2}\Leftrightarrow \frac {FD}{DA} \equal{} tan^2\frac {\alpha}{2}.tan\frac {|\gamma \minus{} \beta|}{2}\Leftrightarrow \frac {FD}{DA} \equal{}$$ tan\frac {\alpha}{2}.cot\frac {\gamma \plus{} \beta}{2}.tan\frac {|\gamma \minus{} \beta|}{2} \equal{} \frac {|sin\gamma \minus{} sin\beta|}{sin\gamma \plus{} sin\beta}tan\frac {\alpha}{2} \equal{} \frac {|c \minus{} b|}{c \plus{} b}\frac {\sqrt {(p \minus{} b)(p \minus{} c)}}{\sqrt {p(p \minus{} a)}}$.
Since $ MD \equal{} \frac {a(|c \minus{} b|)}{2(b \plus{} c)}$ we have:
$ \frac {DA}{FD} \equal{} cot\angle MAD \equal{} \frac {MA^2 \plus{} DA^2 \minus{} MD^2}{4S_{MDA}} \equal{}$ $ \equal{} \frac {\frac {b^2}{2} \plus{} \frac {c^2}{2} \minus{} \frac {a^2}{4} \plus{} bc \minus{} \frac {bca^2}{(b \plus{} c)^2} \minus{} \frac {a^2(c \minus{} b)^2}{4(b \plus{} c)^2}}{\frac {4|c \minus{} b|S_{ABC}}{2(b \plus{} c)}} \equal{} \frac {p(p \minus{} a)(b \plus{} c)}{|c \minus{} b|S_{ABC}})$ $ \equal{} \frac {(b \plus{} c)\sqrt {p(p \minus{} a)}}{|c \minus{} b|\sqrt {(p \minus{} b)(p \minus{} c)}}$
DONE!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finitely generated group and let $H$ be a subgroup of $G$ of finite index. Prove that there is a subgroup $K$ of $H$ of finite index in $G$ and which is characteristic in $G$ (i.e. it's fixed by all automorphisms of $G$).

(K should be required of non-trivial)
Here is a counter-example:let G=G_1*G_2 ,with G_1,G_2 isomorphic to Z_p.
 H=G_1*e_2, e_2 is the unit of G_2.

First of all, no, it is not necessary to require $K$ to be non-trivial. If the trivial subgroup of $G$ has finite index in $G$, then $G$ is finite and everything is just fine. Second of all, if by $*$ you mean free product, then in your example $H$ does not have finite index in $G$, so the hypotheses of the problem are not satisfied.

I meant direct product.
Thank you for correction.

But then your $G$ is finite, and when $G$ is finite the problem is trivial: just take $K$ to be the trivial subgroup. The interesting case is when $G$ is infinite.

Can we proof that $H$ has only finitely many conjugates (by outer isomorphisms)?

Because then $\bigcap_{\phi \in \mathrm{Aut}(G)}H^\phi$ has finite index in $G$ (because $[G: A%Error. "cupB" is a bad command.
] \leq [G: A]\cdot[G: B]$) and is characteristic.

[quote="gebra"]Can we proof that H has only finitely many conjugates (by outer isomorphisms)? [/quote]
Yes.
Consider the sets $aHb$, for all $a$ and $b$ in $G$. These are the conjugates of $H$ and their cosets.
If $a'=ax$ and $b'=yb$ with $x,y\in H$, $a'Hb'=axHyb=aHb$. The set $aHb$ depends only on which left coset $b$ is in and which right coset $a$ is in. If $H$ has index $n$ in $G$, there are therefore at most $n^{2}$ of these sets. These come in groups of $n$ cosets of each conjugate, so there are at most $n$ conjugates of $H$ in $G$.

With a bit more work, we can show that the intersection has index at most $n!$.

[Edit: Oops, that only applies to inner automorphisms. It suffices to show the existence of a normal subgroup, but not necessarily a characteristic subgroup]
See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=43239]here[/url] for an application of the same idea.

The first step is to show that there is a subgroup $N<H$ with $N$ normal in $G$ and of finite index. This is done in jmerry's post.

Let $n=[G: N].$ Any normal subgroup of index $n$ is the kernel of a homomorphism $G\longrightarrow S_{n}.$ To see this, let $M$ be one such and then consider the homomorphism gotten by the action of $G$ on the cosets $\{aM\}.$ Because $G$ is finitely generated there are only finitely many homomorphisms from $G$ to $S_{n}$, and so there are only finitely many normal subgroups of index $n,$ say $N_{1},N_{2},\ldots, N_{k}.$ Then $K=N_{1}\cap\cdots\cap N_{k}$ is a characteristic subgroup of finite index: characteristic because every automorphism permutes the $N_{i}$'s and finite index because $K$ is the intersection of subgroups of finite index.


What is the greatest common divisor of 45 and 75?

Easy. Prime factorization of 45=5x3x3. 75=5x3x3. Find what is in common. 5x3=15

[hide="Solution"]

Note that you might have made a typo and said ${75 = 5 \cdot 3 \cdot 3}$

${75}$ is equal to ${5 \cdot 5 \cdot 3}$

This is just to clarify for anybody reading.

Bu you are correct. The greatest common factor is, in fact $\boxed{15}$

[/hide]

lol you guys both made typos:

75=5*5*3
45=3*3*5

We see the 5*3 in common and so $\boxed{15}$ is the greatest common divisor.

Wow nice catch
I will change this immediately.


function
vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Let T be the function
T: V x V* ---> hom(V,V)

Show that A in hom(V,V) can be written as T(x_1)+...+T(x_r) for various x_i in V x V* if and only if the rank of A (ie the dimension of the image of A) is less than or equal to r.

Is this just saying that any function A in hom(V,V) is a linear combination of T(x_i) in hom(V,V)?

And I don't understand what the rank has to do with it.

Matrices are easier to understand than abstract tensor products, so let's go there. Let $ V\equal{}F^n$ for some field $ F$ (we can look at the infinite-dimensional case later), so $ \text{Hom}(V,V)$ is the space of $ n\times n$ matrices over $ F$. Identifying the dual of $ V$ with the space of row vectors, the map $ T$ is $ T(v,w^T)\equal{}vw^T$.

Now, there are many different definitions of rank. You're most familiar with row rank and column rank (the dimension of row and column spaces) which are equal by standard linear algebra theorems. This problem introduces a new definition: the tensor rank. The tensor rank of a matrix $ M$ is equal to the minimum number of rank one matrices $ M_i\equal{}v_iw_i^T$ such that $ \sum_i M_i\equal{}M$. You want to show that the tensor rank is the same as each other version of rank.

What tools can we apply? First, note that multiplying on either the left or the right by an invertible matrix doesn't change anything: if $ M\equal{}\sum_i v_iw_i^T$, $ AM\equal{}\sum_i (Av_i)w_i^T$ and $ MB\equal{}\sum_i v_i(w_i^TB)$. That lets you do row and column operations freely without changing the tensor rank.

There are two things to prove:
- If $ M$ has rank $ r$, it can be written as a sum of $ r$ matrices of the form $ vw^T$.
- If $ M\equal{}\sum_{i\equal{}1}^r v_iw_i^T$, it has rank $ \le r$.
Can you do either of these?

..nope.  I also don't understand when you say

[quote="jmerry"]The tensor rank of a matrix $ M$ is equal to the minimum number of rank one matrices $ M_i \equal{} v_iw_i^T$ such that $ \sum_i M_i \equal{} M$.  You want to show that the tensor rank is the same as each other version of rank.[/quote]

It's a definition. There are many such sums; we want one with a minimal number of terms.

Oh, and do try the subproblems. The tools and theorems of basic linear algebra are all you need.

Why is T(v,w)=vw^T?

And I meant the part where you say " You want to show that the tensor rank is the same as each other version of rank."

What is "each other rank"?

[quote="rawkerrxx"]Let T be the function
T: V x V* ---> hom(V,V)[/quote]Just to know what function you mean ... can you give a rigorous definiton of $ T$? There might be plenty of them, but I assume that you mean a very very special one. ;)


Assume that the polynomial $\left(x+1\right)^n-1$ is divisible by some polynomial $P\left(x\right)=x^k+c_{k-1}x^{k-1}+c_{k-2}x^{k-2}+...+c_1x+c_0$, whose degree $k$ is even and whose coefficients $c_{k-1}$, $c_{k-2}$, ..., $c_1$, $c_0$ all are odd integers. Show that $k+1\mid n$.

It reminds me of the lemma that P(x) has no rational root. Any else idea?

[quote="silouan"]A solution can be found here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103844

BTW, Darij do you have another solution ?[/quote]

Yes, I'd like to know an elementary solution.

who have got an elementary solution for this problem?

Thanks to our administrators' wonderful idea to delete the source references of the problems, I don't even know anymore where I have the problem from. Probably it was a contest where I posted all problems. As for this one, I never cared for it and I don't know a solution. I don't understand Rust's solution either.

Official solution:
We have $(x+1)^n-1=P(x)Q(x)$. Call two polynomials $f(x)$ and $g(x)$ [i]similar[/i] and denote it $f(x) \equiv g(x)$ if coefficients at the same powers in $f(x)$ and $g(x)$ have the same parity. So we have $(x+1)^n-1 \equiv (x^k+x^{k-1}+...+1)Q(x)  (1)$. Replace $x$ with $1/x$ and multiply it by $x^{n}$ we have $(x+1)^n-x^n \equiv (x^k+x^{k-1}+...+1)x^{n-k}Q(1/x)  (2)$, where $x^{n-k}Q(1/x)$ is some polynomial with power is lesser or equal $n-k$. Subtract (2) from (1) and get $x^n-1 \equiv (x^k+...+1)R(x)$, for some polynomial $R(x)$. Suppose that $k+1$ didn't divide $n$, so $n=q(k+1)+r, 0<r<k+1$. We have polynomial $x^n-x^r=x^r(x^{q(k+1)}-1)$ is divided by $x^{k+1}-1$, so $x^r-1=(x^n-1)-(x^n-x^r) \equiv (x^k+...+1)R_{1}(x)$ for some polynomial $R_{1}(x)$. But it isn't possible becausef $r \leq k$ and polynomials $x^r-1$ and $x^k+...+1$ are not [i]similar[/i].

Suppose $(x+1)^n - 1 = a(x) b(x)$, where $a$ has degree $k$ and has all coefficients odd. Consider this equation in $\overline{F_2}$, the algebraic closure of the finite field of order 2. Here, $a(x) \equiv x^k + x^{k-1} + \cdots + 1 = \frac{x^{k+1}-1}{x-1}$. Let $\omega$ be a root of $\Phi_{k+1}(x)$ in $\overline{F_2}$. Clearly, $\omega^{k+1} = 1$. Because $\Phi_{k+1}(x)$ is cyclotomic, either the order of $\omega$ is $k+1$ or $k+1$ is even. But $k+1$ is odd, so the order of $\omega$ is $k+1$.

Since $x^k + x^{k-1} + \cdots + 1$ divides $(x+1)^n - 1$ in $\overline{F_2}$, $x=\omega$ and $x=\omega^{-1}$ both satisfy $(x+1)^n = 1$. Thus, $(\omega+1)^n = 1$ and
\[1 = \left(\frac{1}{\omega} + 1\right)^n = \left(\frac{\omega+1}{\omega}\right)^n = \frac{1}{\omega^n} \implies \omega^n = 1. \]
Since the order of $\omega$ is $k+1$, we have $k+1|n$, as desired.

[hide=Minor Rant] When you realize that the solution posted above is exactly what you did :wallbash:, posting anyways :P. [/hide]

[hide=Solution] Consider this divisibility in $F_2[X]$. Let $\bar{F_2}$ denote the algebraic closure of this field. We work with this. Clearly, $x^k+c_{k-1}x^{k-1}+\dots+c_1x+c_0=x^k+\dots+x+1$ in this system. The condition translates into the existence of a polynomial $Q(x)$ such that \begin{align*} (x+1)^n-1=\frac{x^{k+1}-1}{x-1}\cdot Q(x) \end{align*} holds in our field. Now, we consider the $(k+1)^{\text{th}}$ cyclotomic polynomial $\phi_{k+1}(X)$ and take one of its roots, $\omega$ in this algebraic closure. We know for a fact that $\phi_{k+1}(X) \mid \frac{X^{k+1}-1}{X-1}$ and we have the new relation \begin{align*} (x+1)^n-1=\phi_{k+1}(x)\cdot Q_1(x) \end{align*} for some polynomial $Q_1$. Substituting $\omega$ and $\frac{1}{\omega}$ in this relation gives \begin{align*} (\omega+1)^n=\left(\frac{1}{\omega}+1\right)^n=1=\frac{(\omega+1)^n}{\omega^n} \end{align*} which translates into $\omega^n=1$. However, the order of $\omega$ in this field, being a root of $\phi_{k+1}(X)$ is $k+1$. We conclude from the above reasoning that $k+1 \mid n$.  [/hide]

Alex Remorov in his handout gives a hint to use Eisenstein's Criterion. How can this be applied here?


did you guys meet the kids from dodds europe aka connor silveria, ryan boyland, bjorn....., and other asian?
they were from england, germany, korea, turkey
did you meet them and what place did u rank at nats?

What does DoDD stand for?

I was in Korea 2 years ago (6th grade) and I had NEVER heard of MATHCOUNTS until I came to US.

I assume he means the people from the State Dept and Defense Dept?
They're allowed to participate if their parents work for the gov't overseas I believe, something like that..., so they're from all over
I wasn't at nats this year, but I do remember meeting those people last 3 years, they had some cool pins :D

Yuppers, DoDD (Department of Defense Dependents) sends a team. They tend to be in the bottom 10 teams though :(.

dodds mean department of defense schools.


An acute triangle $ ABC$ is inscribed in a circle. The tangents at $ B$ and $ A$ meet the tangent at $ C$ at $ T$ and $ U$ respectively.

Lines $ AT$ and $ BC$ meet at $ P$ and $ Q$ is the midpoint of $ AP$.Lines $ BU$ and $ AC$ meet at $ R$ and $ S$ is the nidpoint of $ BR$.

Prove that $ \angle ABQ \equal{} \angle BAS$

Babis

The same infamous problem has been posted before several times. For instance, see the solutions offered in the topics:

[url]http://www.artofproblemsolving.com/viewtopic.php?t=19806[/url]
[url]http://www.artofproblemsolving.com/viewtopic.php?t=207440[/url]
[url]http://www.artofproblemsolving.com/viewtopic.php?f=46&t=275380[/url]

Surprisingly, it can be done by direct long boring calculation, $AT$ and $BU$ are symmedians of $\triangle ABC$, then take ${X}=BQ\cap AC$, ${Y}=AS\cap BC$, ${Z}=AS\cap BQ$, apply Manelaos few times and get $AZ=BZ$.

Best regards,
sunken rock


inequalities
number theory unsolved
number theory

Let $ p_1,p_2,\ldots,p_k $ be the distinct prime divisors of $ n $ and let $ a_n=\frac {1}{p_1}+\frac {1}{p_2}+\cdots+\frac {1}{p_k} $ for $ n\geq 2 $. Show that for every positive integer $ N\geq 2 $ the following inequality holds:  $ \sum_{k=2}^{N} a_2a_3 \cdots a_k <1 $

[i]Laurentiu Panaitopol[/i]

[b]Lemma 1.[/b]$n\geq 6$Let $p_{1}<\dots<p_{k}$ be all primes not greater than $n$
then $\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}<\frac{n-1}{2n}$
[b]proof[/b]
:
\[\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}=\frac{1}{4}+\frac{1}{9}+\sum_{i=3}^{k}\frac{1}{p_{i}^{2}}< \frac{1}{4}+\frac{1}{9}+\sum_{l=5}^{n}\frac{1}{l^{2}}=\frac{1}{4}+\frac{1}{9}+\frac12\sum_{l=5}^{n}{(\frac{1}{l-1}-\frac{1}{l+1})}=\frac14+\frac19+\frac18-\frac{1}{n+1}<\frac{1}{2}-\frac{1}{2n}=\frac{n-1}{2n}\]
[b]Lemma 2[/b]:$a_{2}\dots a_{n}<\frac{1}{2^{n-1}}$
[b]Proof[/b]:for n<$a_{2}\dots a_{n}\leq (\frac{a_{2}+a_{3}+\dots+a_{n}}{n-1})^{n-1}$
but $a_{2}+\dots+a_{n}\leq \sum_{i=1}^{k}{[\frac{n}{p_{i}}]\frac{1}{p_{i}}}\leq n\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}\leq \frac{n-1}{2}$ (because of Lemma 1)thus
$a_{2}\dots a_{n}<\frac{1}{2^{n-1}}\leq \frac{1}{2^{n-1}}$ with Lemma 1 and
2 we can easily prove the problem.
I hope I am correct.
[b]I hope there is another proof.Anyone has another proof?Could somebody post it?[/b]
Is my solution correct?Please tell me.

[quote="Hawk Tiger"]
$\frac14+\frac19+\frac18-\frac{1}{n+1}$
[/quote]
This is not true... it has to be $\frac14+\frac19+\frac18-\frac{1}{2n+2}$ and then the inequality from Lemma 1 is true for $n>35,$ not for $n>5.$

The following equality holds: $\sum_{k=2}^na_k=\sum_{k=2}^n(\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_k})=\sum_{p\leq n, p prime}\frac{1}{p}\left[\frac{n}{p}\right]$. The folowing inequalities are obvious for $n>0$: \[\sum_{p\leq n, p \text{prime}}\frac{1}{p} \left[\frac{n}{p}\right] \leq \sum_{p\leq n, \text{prime}}\frac{1}{p} \frac{n}{p}=n\sum_{p\leq n, \text{prime}} \frac{1}{p^2} \leq n(\frac14 +\sum_{k=1}^{\inf}\frac{1}{(2k+1)^2})< n(\frac14 +\frac14 \sum_{k=1}^{\inf}\frac{1}{k(k+1)})=\frac{n}{2}.\] Therefore  $\sum_{k=2}^na_k<\frac{n}{2}$. By AM-GM $a_2a_3\cdots a_n<(\frac{a_2+\cdots+a_n}{n-1})^{n-1}<\frac{1}{2^{n-1}}(1+\frac{1}{n-1})^{n-1}<\frac{3}{2^n-1}$. Add this inequalities and obtain: $\sum_{n=2}^{\inf}a_2 \cdots a_n <\frac12+\frac16+\frac{1}{12}+\frac{1}{60}+3(\frac{1}{2^5}+\frac{1}{2^6}+\cdots)=\frac{229}{240}<1$.


calculus
derivative
function
integration
real analysis
real analysis unsolved

[quote="xirti"]Prove that there not exist functions for which $(F(x))'=f(x)$ and $\left(e^{F(x)}\right)'=e^{f(x)}$.[/quote]This is an application of Jensen's inequality.

If there were such functions, say on the interval $[0,1]$, with $F(0)=0$ wlog, we would have
$e^{F(1)} = \int_0^1 e^{f(x)} dx = e^{F(1)} \int_0^1 e^{f(x)-F(1)} dx$.
Now $e^t \ge 1+t$, with equality only at $t=0$. Thus
$e^{F(1)} \ge e^{F(1)} \int_0^1 \left(1 + f(x)-F(1) \right) dx = e^{F(1)} \left(1 + F(1) - F(1) \right) = e^{F(1)}$.
This equation can only hold if $f(x) - F(1) = 0$ for a.e. $x$. In that case, $f(x)$ and therefore $e^{f(x)} > 0$ are constant. Then $F(x)$ must be linear or itself constant. But then $(e^{F(x)})'$ is either not constant or equal to zero, which is a contradiction.


Euler
limit
logarithms
function
real analysis
real analysis unsolved

Let $ q\geq 1$ be a natural number and let $ \gamma_{q}$ be the constants defined by

$ \gamma_{q}\equal{}\lim\limits_{k\rightarrow \infty}\left(\frac{1}{2q\minus{}1}\plus{}\frac{1}{3q\minus{}1}\plus{}\cdots\plus{}\frac{1}{kq\minus{}1}\minus{}\frac{\ln k}{q}\right)$.

Note that $ \gamma_{1}\equal{}\gamma$, $ \gamma_{2}\equal{}\frac{\gamma}{2}\minus{}1\plus{}\ln 2$, so a natural question is whether these constants can be calculated exactly.

You can use the identities $ \minus{}z^{n}\log(1\minus{}z^n)\equal{}\sum_{k\ge 1}\frac 1kz^{(k\plus{}1)n}$ for $ z\equal{}e^{2\pi i/q}$, $ n\equal{}1,2,\dots,q\minus{}1$ to express your sum as a combination of $ \gamma$ and a few "elementary" numbers.

[quote="didilica"]Let $ q\geq 1$ be a natural number and let $ \gamma_{q}$ be the constants defined by

$ \gamma_{q} \equal{} \lim\limits_{k\rightarrow \infty}\left(\frac {1}{2q \minus{} 1} \plus{} \frac {1}{3q \minus{} 1} \plus{} \cdots \plus{} \frac {1}{kq \minus{} 1} \minus{} \frac {\ln k}{q}\right)$.

Note that $ \gamma_{1} \equal{} \gamma$, $ \gamma_{2} \equal{} \frac {\gamma}{2} \minus{} 1 \plus{} \ln 2$, so a natural question is whether these constants can be calculated exactly.[/quote]
I will consider case when $ q > 1$
consider
$ \Gamma(z) \equal{} \lim _{n \to \infty} \frac {n!n^{z}}{z(z \plus{} 1)...(z \plus{} n)}$
then:
$ z(\frac {d}{dz}\ln \Gamma(z)) \equal{} \lim_{n \to \infty} \left( z\ln n \minus{} \sum_{j \equal{} 0}^{n}\frac {1}{1 \plus{} jz^{ \minus{} 1}}\right)$
let's say :
$ (\frac {d}{dz}\ln \Gamma(z)) \equal{} \psi(z)$ (Digamma function)
then :
$ \minus{} z\psi( \minus{} z) \equal{} \lim_{n \to \infty}\left( \minus{} z\ln n \minus{} 1 \minus{} \frac {1}{1 \minus{} z^{ \minus{} 1}} \plus{} \sum_{j \equal{} 2}^{n}\frac {1}{jz^{ \minus{} 1} \minus{} 1}\right)$
sabstitute $ z^{ \minus{} 1} \equal{} q$ thus we have :
 $ \minus{} \frac {1}{q}\psi( \minus{} \frac {1}{q}) \plus{} 1 \plus{} \frac {1}{1 \minus{} q} \equal{} \lim _{n \to \infty} \left( \sum_{j \equal{} 2}^{n}\frac {1}{jq \minus{} 1} \minus{} \frac {\ln n}{q} \right) \equal{} your limit$
hence to calculate $ \gamma_{q}$ it's enough to calculate $ \psi( \minus{} \frac {1}{q})$

[quote="fedja"]You can use the identities $ \minus{} z^{n}\log(1 \minus{} z^n) \equal{} \sum_{k\ge 1}\frac 1kz^{(k \plus{} 1)n}$ for $ z \equal{} e^{2\pi i/q}$, $ n \equal{} 1,2,\dots,q \minus{} 1$ to express your sum as a combination of $ \gamma$ and a few "elementary" numbers.[/quote]
I don't understand well  :(  : How can I express sum by $ \gamma$ and "elementary" numbers? what do you mean :"elementary" numbers and why do you name  so?

Write $ \sum_{n\equal{}0}^{q\minus{}1}z^n\sum_{k\equal{}1}^N\frac {z^{kn}}k$ in 2 ways: in the first case, sum over $ n$ first, in the second case, sum over $ k$ first and write the sums with precision $ o(1)$ (you'll get $ q\minus{}1$ partial sums converging to various logarithms and one that is $ \log N\plus{}\gamma$). Now just look at the resulting identity. By "elementary" numbers I mean here something like $ \log\sin(1/2q)$, i.e., the numbers that can be obtained from rationals applying elementary functions to them.

Thank you for the comments to both of you. It seems that the computations of the constants $ \gamma_{q}$, as I expected, is highly a nontrivial task.


1. if i play a game that has 19 pennies with John, and players on any turn may take 1,2,3, or 4 pennies, how many pennies should I take first (i will start first) to guarentee that I win.  The player who takes the last penny wins.

BOth players dont cheat and they know what they are doing (they are always playing to win and know the logical process to win at any moment)

2. How many pennies should I take first if the player who takes the last penny loses (again I will start-still 19 pennies).

im just making these questions up so i dont know the "official answer" but I think what I am doing is right...

I think im just going to create a lot of questions because i need to study for the AMC 10B.

Hint, consider the number of coins $ \pmod{5}$.

[hide="1"]
You can't control how many pennies John takes from the pile.  You can, however, control how many total pennies that you and John together remove from the pile in two moves.  For example, if John takes 1, 2, 3 or 4 pennies, you can remove 4, 3, 2 or 1 pennies respectively so that after two moves, the total number of coins removed is 5.

This is a good way to guarantee that you take the last coin.

Like azjps said, 19 mod 5 = 4.  Therefore you must remove 4 coins at first.
[/hide]

Number 2 is the same, you just want to leave the opponent with 1 coin


calculus articles vector algebra polynomial integration trigonometry

[quote="Chinaboy"]Well... My parents had a conversation with the principle about our school's lousy calculus policy (must take AB before BC)

Then... he said:
"It's actually good for you because if you take BC next year, then you'll have nothing your senior year. Colleges don't like it when you don't take math every year in HS."

Is this true?[/quote]

It is true that colleges want you to take math every year, but it doesn't mean you should restrict your abilites. If anything, you could go to a local university for a course beyond Calculus BC; that would be even better.

I don't see the point in taking both AB and BC, but school policy is school policy.

I'd recommend for you to check out this article

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php

It may not apply to you, but it is a good read.

Excuse my ignorace, please, but what is Calculus AB and Calculus BC? (I have never been envolved with the organization of HS classes on the US)

Best,

[url=http://www.collegeboard.com/student/testing/ap/sub_calab.html?calcab]Here[/url] is a description of AP Calc AB.

[url=http://www.collegeboard.com/student/testing/ap/sub_calbc.html?calcbc]Here[/url] is a description of AP Calc BC.

I believe I read on a college board handout before that AB is about 3/4 of the first year of college calculus and BC is about the full year of a college calculus class.  So BC has everything AB has and more.  If you take the BC test they give you an AB subscore.  Some of the topics that BC has that AB doesn't are parametric, polar, vectors, a bunch about sequences and series, Taylor polynomials (with Maclaurin), L'Hopital's rule, integration by parts and trig substitutions.  I may have left some things out but those are pretty much the additional topics.

But will it interfere with your chances of getting into a very good college (like Ivy League)?

For example... will let's say, Harvard, look and see that "Bob" did not take a math course during 11th grade (10th BC and 12th Stat). Will this be a bad impact on the admission decision?

In another scenario, what if they see the same thing in scenario I, but also see that Bob has participated in math contests like AMC/AIME and eh, Math League. What will the college think this time?

Like I said before, it is better to take a math course every year. However, if you can provide some evidence that you were still actively engaged in math (such as self-studying multivariate/linear algebra or taking a course from a university) I think you would be fine.

You need to keep busy advancing your ability in the subjects you are most passionate about. The important question to ask is NOT "what impresses colleges?" but rather "How can I get to be better at something until I am really good at it?" There are lots of ways to take courses more advanced than the meager offerings of your high school, for example,

1) distance learning courses (EPGY is an outstanding example)

2) summer programs (e.g. MathCamp, Ross/PROMYS)

3) enrolling at a local college (most community colleges teach up to at least multivariable calculus)

4) "pure" self-study (find a library, and find some books, and ask some knowledgeable adult questions as you read the books and work the problems in them)

and there are other possibilities besides those I have mentioned. Challenge yourself--that's the main thing.

Good luck!

Thank you for your advice, tokenadult!

I do have one question: What are distance learning courses?

Distance learning courses are courses in which any "teacher" you deal with is a long way away. The example I gave above of distance learning courses, the Education Program for Gifted Youth (EPGY)

http://epgy.stanford.edu

consists mostly of courses that are offered by computer disk (CD-ROM) and email, although some of the courses are fully online, just like the AoPS online courses. (Duh! AoPS online courses are the example of distance learning courses most likely to be familiar to people on this forum.) My son has taken EPGY courses in math (various levels), English grammar, and computer programming. I hope to get him into the EPGY physics courses soon. One great thing about the EPGY courses is that although their list price looks high, it is less than going to a private prep school, and EPGY financial aid (= discount from list price for price-sensitive customers) is available to families with middle-class incomes.

Best wishes in finding more math to learn. By the way, I notice, um, you are quite interested in things Chinese. Do you speak the language regularly? I know Miths has gone over to China for summer programs to keep up his language skills. I'm hoping I can get the kids over to Taiwan sooner rather than later to see the other grandparents, see the Taiwanese cousins, and get back in the habit of speaking Mandarin that they once had.

Yes, although I speak Cantonese, not Mandarin, at home, but I did learn Mandarin in China until 2nd grade, and has sparsely learned it over the years. I love going back to China... relatives, great food... and mahjong... hehe. Hope you and the kids have fun!

Thank you for the information! I think I'm going to talk about it with my parents. Thanks, again!


For non-negative $ a,$ $ b$ and $ c$ prove that
\[ \left(a^2\plus{}\frac{3}{4}\right)\left(b^2\plus{}\frac{3}{4}\right)\left(c^2\plus{}\frac{3}{4}\right)\geq\sqrt{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\]


abstract algebra
category theory
superior algebra
superior algebra unsolved

[quote="MeKnowsNothing"]Let $ A$ be a commutative ring with unit. Let $ 0\to M^\prime\to M\to M^{\prime \prime}\to 0$ be an exact sequence of $ A$-modules. Let us suppose that $ M^{\prime\prime}$ is flat. Prove that the sequence $ 0\to M^\prime\otimes N\to M\otimes N\to M^{\prime\prime}\otimes N\to 0$ is also exact for any $ A$-module $ N$.

My problem is that I don't see why $ M^\prime\otimes N\to M\otimes N$ is injective.[/quote]

Have you heard of the Tor functor?  The kernel of M′⊗N → M⊗N is a homomorphic image of Tor(M″,N)=0.

A proof without using Tor is (one direction of) theorem 4.58, page 154 of Lam's Lectures on Modules and Rings.

Exact sequences 0→M′→M→M″→0 that remain exact after tensoring with any module N are called "pure".  One definition of flat is that M″ is flat iff every exact sequence 0→M′→M→M″→0 is pure.

[quote="Soarer"]@JackSchmidt,

Actually what other definition of flatness are there? When I first learnt it, being an exact functor is the definition, and the vanishing Tor being an equivalent condition.[/quote]

The definition with pure exact does not view M″ as a functor.  The functor is -⊗N for arbitrary modules N.  It is required to leave all short exact sequences ending with M″ exact.

The more common definition views M″ as the functor M″⊗- and requires that it leave all exact sequences 0→N′→N→N″→0 exact.

That these two are equivalent is proven directly in Lam, but of course it follows from the fact that Tor is a balanced bi-functor ("the zig-zag lemma" being another name for this).

There are lots of other definitions of flatness that have little to do with either of these.  Three of my favorites are:

* Flats are the "completion" of the frees: flats are exactly the direct limits of the frees.

* Flat and purity (and pure injective) can be defined in a logic context as natural "model theoretic" properties, where solving equations can be done using "finitary methods".  Pure injectives are called "algebraically compact" in this context.

* Flats are the dual of injective under Hom_Z(-,Q) [where it helps if the ring is coherent to make the duality completely symmetric].

Lam covers the first two definitions in a section each, I believe.  Lam covers part of the duality, but it is covered more thoroughly in Enochs&Jenda's Relative Homological Algebra.

[quote="JackSchmidt"]
The more common definition views M″ as the functor M″⊗- and requires that it leave all exact sequences 0→N′→N→N″→0 exact.
[/quote]

I meant this when I said exact functor. Thanks for your info! :lol:


[quote="nguoikhonkho"]let a1,a2,a3,a4 are positive integer numbers such that
i) a1+a2+a3+a4 are divisible a(i) for every i=1,2,3,4
ii) a1+a2+a3+a4 > 2ai  for every i=1,2,3,4
prove that: exist a(m)=a(n) with m,n in{1,2,3,4}[/quote]
Let $S=a_{1}+a_{2}+...+a_{n}$ (n=4). From (1) we have
$a_{i}=\frac{S}{k_{i}},k_{i}\in N$, from (2) $k_{i}>2$. We have \[\sum_{i}\frac{1}{k_{i}}=1.\] If $a_{i}$ distinct, then $k_{i}$ distinct and
$\sum_{i}\frac{1}{k_{i}}\le \frac{1}{3}+\frac 14+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}<1$ give contradition.


[quote]n is powerful if for any prime divisor p of n, p^{2} also divides n.
Prove that there exist k consecutive numbers that are not powerful.[/quote]
When you say 'any prime divisor of p,' do you mean all prime divisors of p, or just at least 1?

Among any four consecutive integers, one is divisible by 4, so k can be at most 3.  However, there are arbitrary long blocks of consecutive integers that are powerful.

[quote="manuel"]$n$ is [i]powerful[/i] if for any prime divisor $p$ of $n$, $p^{2}$ also divides $n$.
Prove that there exist $k$ consecutive numbers that are not powerful.[/quote]

I think he means for all primes $p$ that divide $n$.


Imagine a circle with a area of pi.  You are to put the largest possible square inside this circle.

What is the area of this square?

BONUS-Inside the same square you are to put the largest possible circle inside it.  What is the area of this new smaller circle?

Note-this problem may be hard if you are not familiar with special triangles (geometry/Alg 2).  You by no means NEED it.  By reasoning you should be able to get it.

Hint (for those who need it...  but try it first)
[hide]Of course draw a picture.  Think about this large circle.  What is the radius?  What about the diameter?  Now try and find the diagnal of the square.[/hide]

Still stuck?
[hide]You will have to use the pathagorean theorem (a^2+b^2=c^2 where c is the hypoteneuse and a and b are legs of a right triangle)  Apply this to find out the sides of the square (remember both sides will be equal)[/hide]

Try not to use the hints... draw a picture a work it out.

Big hint:

[hide]
 :sqrt: 2 will pop up somewhere if you did it right.
[/hide]

[hide="It took me a while"]the area of this circle is 3.14or a pi

then the radious has to be 1

and the diameter is 2, so the diagonal of this square has to be 2

by using pathagorean theorem 4 = 2A^2

[/hide]

[quote="somedude"]BONUS-Inside the same square you are to put the largest possible circle inside it.  What is the area of this new smaller circle?
[/quote]

[hide="Solution"]We already know that the length of a side of the square is sqrt(2). The largest circle is one that touches all sides, so it's radius would be half the length of a side, which equals sqrt(2)/2. Then the area would be pi/2.

A comment that might be a little above this level: Hmmm... Could it truly be a coincidence that the area of the smaller circle is half that of the larger?[/hide]


MATHCOUNTS
geometry
rectangle
analytic geometry
3D geometry
tetrahedron
probability

Seeing how there's a Getting Started problem marathon, Mathcounts marathon, Intermediate marathon, and even a Chemistry marathon, I decided to start a PUZZLE MARATHON.

Same rules as in the Getting Started Marathon, which you can find [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33505]here.[/url]

I'll start off with an easy logic problem:

[size=200]Problem 1[/size]

A Number Box is a 3x3 arrangement with a different number 1-9 in each cell. For example, a Number Box could be:

[code]1  2  6
9  4  8
7  5  3[/code]

Given the clues below, can you determine the composition of Number Box 1?

1. The number in the center of the top row minus the number in the center of the bottom row equals 5.
2. The three numbers in the rightmost column of the box add to 8.
3. The numbers in the four corners of Number Box 1 sum to 23.
4. The number in the left cell of the top row minus the number in the right cell of the bottom row equals 2.
5. The number in the center cell of the box isn't 6.

[b]Source: [/b] All Star Puzzles

I tried it using maths, and it led to a contradiction. I assume the problem is at my end? :?

[quote="seamusoboyle"] I assume the problem is at my end? :?[/quote]

Yep :)

There is a unique solution.

I think this may be right...
[hide]
7___8___2
6___4___1
9___3___5[/hide]
I'll post a problem if this is an acceptable solution

[quote="dangvh"][size=150][b]Problem 2[/b][/size]
Given 4 vertical lines: | | | |

Add 5 more straight lines to make 10

note: you'll have to think outside the box to get this one
note: + counts as 2 lines... not four[/quote]

[hide]
Spell it out. TEN - one line for the T, three for the E, one for the N.
[/hide]

[size=150][b]Problem 3[/b][/size]

Without lifting your pencil, connect these nine dots:

[code]. . .
. . .
. . .[/code]

name them as this:

123
456
789

first, go through 1,5,9. Then, go up to go through 6 and 3, but extend farther outside the box. Then the third line goes through 2 and 4 in a diagonal fashion, stretching all the way to align with the bottom row, then connect 7 and 8 to finish off.
[size=150]
[b]Problem 4[/b][/size]

Can you cover the Chessboard with diagonally opposite squares taken off, with 31 2x1 dominoes? Give your reasoning.

[hide="Problem 5?"]In an 8*8 chessboard opposite squares have the same color. If you remove two opposite squares, instead of having 32 white squares and 32 black sqrares the chessboard has 32 of one color and 30 of the other.

A 1*2 rectangle, by definition, has to 1/2 of it on a white square and 1/2 of it on a black square. However, there are more of one color that of the other. Therefore, it is impossible to completely cover the board with 1*2 rectangles. QED.[/hide]

Just wondering, does 5 come after 3?  :P

[size=200]Problem 5 - the REAL Problem 5[/size]

Here's an oldie but goodie:

You are standing in a cell with two guards and two doors. One door leads to freedom while the other leads to death, but you don't know which door is which. One guard always tells the truth while the other always lies. Asking only one question, how can you figure out which door to go through?

[quote="NoSoupForYou"]Just wondering, does 5 come after 3?  :P[/quote]

sorry.. I will change it right now.

[quote="sheepwarrior"]hey, paladin i dont get ur TEN answer.. can you explain again lol  :blush:
and the solution i got is pretty cool (in my opinion)
1+1

=

10 :P im sure you know what i did[/quote]

it said to make 10 out of 9 lines. So thinking outside of box, paladin actally "wrote out" 10 using 9 lines. The English spelling of  10 is TEN, which requires 9 straight lines to be written.

i have a analitical solution:
assume that  left point from the bottom is of coordinates (0,0)
move your pencil like so : (3,0),(0,0),(0,4),(4,0),(0,0),(3,3)
:D
thus this concludes a very hard problem...... :blush:

[quote="mathnerd314"]Ugh. There are [b]exactly[/b] three turtles in a line...

[quote="nebula42"]Uh... is it something ridiculous like the third turtle is lying?[/quote]

That's right!!! :harhar:[/quote]

That isn't cool :rotfl:

[quote="jmerry"][quote="OHO"]Problem:
A team of 25 people attends a game show...

What is the maximum number of people who certainly survive? And what's their plan?[/quote]
I know this as the "Judgment of the Council of Sages", from a Russian contest. They meet the night before to choose their strategy and the order they will stand in line.

My version of the [hide="solution"]During the meeting, the sages develop the ideal strategy to save all but one. After heated discussion, they come to the only reasonable choice of order: seniority.
The junior sage goes to bed and stews: he's been snubbed by the senior sages at every opportunity, and new they leave him to possibly die?
On the day of judgment, the junior sage chooses to betray the others and give the opposite of the agreed answer. Everyone dies except the junior sage, who was right about his own hat by pure chance.[/hide][/quote][hide]I can't understand why all the others die. :ewpu: When the first died, the rest should know.???[/hide]

The executions don't happen until all answers have been given. It just makes more sense that way, with only the answers passed forward.

[quote="z00ey"]well, here's a funny one:

if Mommy is 21 years older than her Kid on this very day, and in 6 years she'd be 5 times older than the Kid, the question is:

[i]where's Daddy now?[/i][/quote]

:spam: :mad:

But solving the question as is:

[hide]Let x be the kid's age.
(x+21)+6=5(x+6)
x+27=5x+30
This is impossible.[/hide]

[quote="mathnerd314"]But solving the question as is:

[hide]Let x be the kid's age.
(x+21)+6=5(x+6)
x+27=5x+30
This is impossible.[/hide][/quote]

why would it be impossible? it's an easy equation... find x and answer the question above. use your imagination, not only calculation

Puzzle:

[code]

HOROBOD

[/code]

What famous word, expression, or name is depicted in the code box above?

[quote]It is known that:
Pure mathematicians tell the truth about their beliefs.
Applied mathematicians lie about their beliefs.
Sane mathematician’s beliefs are correct.
Insane mathematician’s beliefs are incorrect. [/quote]
Knowing this, you meet a mathematician at some insignificant location.

You: "What is 2+2?"
Mathematician: "4"
You are immediately able to deduce that he is either a [b]sane pure mathematician[/b] or an [b]insane applied mathematician[/b].

What is one question that you can ask in order to distinguish him between the two?


Is it possible to put resources, e.g. images, which are uploaded as part of the thread to include within the hide-tag?

I have seen hidden images at G&FF. But those were added using the "Image" tags, so I think you can't hide other attachments.

[quote="herefishyfishy1"]You can hide URLs, also, so as an alternative, you can provide a link to the attachment.[/quote]

I know I could upload it to the ML Galleries, imageshack.us etc. but this only a work-around.


inequalities
LaTeX
triangle inequality
absolute value
inequalities unsolved

Let $a,b,c,d$ be positive real numbers.

Prove

$\sqrt{(a+c)^2 + (b+d)^2} + 2 \frac{|ad-bc|}{\sqrt{(a+c)^2 + (b+d)^2}} \geq \sqrt{(a+b)^2 + (c+d)^2}$

The inequality
$\sqrt{(a+c)^2 + (b+d)^2} + 2 \frac{|ad-bc|}{\sqrt{(a+c)^2 + (b+d)^2}} \geq \sqrt{(a+b)^2 + (c+d)^2}$ does not hold.

The origin inequality is equivalent to
$(a+c)^2 + (b+d)^2+2|ad-bc| \geq \sqrt{(a+b)^2 + (c+d)^2} \sqrt{(a+c)^2 + (b+d)^2}$.

A counterexample is $a=\frac{1}{264}$, $b=\frac{1}{76}$, $c=\frac{1}{36}$, $d=1$.

Are you sure?

It's from Belarus 2002. I don't think a competition problem will be wrong.

I've checked my source again and i've checked that i didn't make any typo.

You can input

"eval((a+c)^2+(b+d)^2+2*abs(a*d-b*c)-sqrt(((a+c)^2+(b+d)^2)*((a+b)^2+(c+d)^2)),

[a=1/264,b=1/76,c=1/36,d=1]);"

 in mathematical software Maple 8. Then you will find the truth.

[quote="Detective"]Let $a,b,c,d$ be positive real numbers.

Prove

$\sqrt{(a+c)^{2}+(b+d)^{2}}+2 \frac{|ad-bc|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}\geq \sqrt{(a+b)^{2}+(c+d)^{2}}$[/quote]

The text of the problem is

$\sqrt{(a+c)^{2}+(b+d)^{2}}+2 \frac{|ad-bc|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}\geq \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\geq \sqrt{(a+c)^{2}+(b+d)^{2}}$

See Hojoo Lee - Inequalities through problems, pag.7, n.41.
Can someone prove this ?

[quote="Leon"]
The text of the problem is

$\sqrt{(a+c)^{2}+(b+d)^{2}}+2 \frac{|ad-bc|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}\geq \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\geq \sqrt{(a+c)^{2}+(b+d)^{2}}$

See Hojoo Lee - Inequalities through problems, pag.7, n.41.
Can someone prove this ?[/quote]

$A=(0,0), B=(a,b), C=(c,d)$, and let $H$ be the foot of the perpendicular from $B$ to $AC$. Then it is equivalent to $AC+2BH \geq AB+BC \geq AC$, which is trivial by triangle inequality.

As $a,b,c,d$ are positive, $B,C$ lie on the first quadrant. So $A< \frac{\pi|}{2}$.

Um... Can u please explain me ???  :huh:
I don't get it, Sung...

[quote="BoesFX"]....
So, $AC+BD \leq BD+DC+2AD = BC+2AD$

The inequality is proven true.

$QED$[/quote]

I thought u'd made a typo...
That must be $BC \leq AC+AB \leq BD+DC+2AD = BC+2AD$
Btw, thx for the explanation 'cause I'd already felt something wrong with the variable, too....  :roll:

[quote="libra_gold"]Then what about $C>\frac{\pi}{2}$?  :huh:[/quote]

i join your problem. i also want to ask it.

[quote="ghjk"]i think the problem leon is extremely easy!I proved it in a few seconds!This is my solution:
Using Mincopxki's inequality We'll have:
square root(a+c)^2+(b+d)^2>=square root (a+b)^2+(c+d)^2
Then we'll have a solution!(because the 2(ad-bc)/square root(a+c)^2+(b+d)^2>=0!)
PS:I think they! don't have absolute value ad-bc!I don't know how to use Latex!Can anyone helps me?thanks
[/img][/quote]

 :huh:  If $a=b=d=1$ and $c=2$, then $2\frac{ab-cd}{\sqrt{(a+c)^{2}+(b+d)^{2}}}<0$
And another counter example for $\sqrt{(a+c)^{2}+(b+d)^{2}}\geq \sqrt{(a+b)^{2}+(c+d)^{2}}$ ($a=1,b=2,c=3,d=4$)

@anthony19891019 : I think the solution that I have modified answer your question ($C>\frac{\pi}{2}$ is impossible)  :wink:

[quote="BoesFX"]Interesting Idea :10:  :clap2:  :thumbup:

But, I think the variables are not right

So, I try to fix the solution.

Let $A(0,0), B(a,b), C(-c,-d)$ dan let $D$ be the foot of perpendicular from $A$ to $BC$

$AB=\sqrt{a^{2}+b^{2}}$, $AC=\sqrt{c^{2}+d^{2}}$

$BC=\sqrt{(a+c)^{2}+(b+d)^{2}}$

$[ABC]=\frac{|ad-bc|}{2}$ (By Shoelace formula)
$AD=\frac{2[ABC]}{BC}= \frac{|ad-bc|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}$

So the inequality becomes: $BC+2AD \geq AB+AC \geq BC$

$\angle A$ is obtuse, since $A(0,0)$, $B$ is in quadran I, and $C$ is in quadran III
Since $\angle A$ is obtuse, $BD+DC$ must be $BC$

By triangle inequality, $AB+AC \geq BC$, $BD+AD \geq AB$, and $DC+AD \geq AC$

So, $AB+AC \leq BD+DC+2AD = BC+2AD$

The inequality is proven true.

$QED$[/quote]

Yesterday I printed off Hojoo Lee's inequalities through problems... working through the problems, I was only able to do about 3 (out of 150ish) but I was skimming... this is one of those three I could do :P (and I ahd the same solution as you...)

[quote] I was only able to do about 3 (out of 150ish) but I was skimming...[/quote]
Lol, I know the feeling, although I only really solved Q22 and then used it to solve two more directly :).

oh nice, do they really build off each other like that?
lol now i'll have to solve Q22 cuz I actually know someone who did it

\[\sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2} \leq \sqrt {(a\plus{}c)^2\plus{}(b\plus{}d)^2} \plus{} \frac {2|ad \minus{} bc|}{\sqrt {a^2 \plus{} c^2} \plus{} \sqrt {b^2 \plus{} d^2} }.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=326139&hilit=belarus+2002


For which $n$ is $2^{n}+105$ a perfect square?

This is the first problem from the first round of the 57th Polish MO.

[quote="Gyan"][quote="Farenhajt"]$2^{n}+105=m^{2}$. <etc> ...........$n\in\{4,8\}$[/quote]
How about n=6? ....   $13$ may be unlucky,  but it's not fair to ignore $13^{2}=169=2^{6}+105$ :)[/quote]

Oh dear, I can't subtract $21-5$... What am I doing here?!

Thanks, Gyan. Edited.

[quote="neelnal"][quote="Farenhajt"]$2^{n}+105=m^{2}$. Looking it modulo 3, we conclude that $n$ is even.
[/quote]

I'm not understanding.[/quote]

All squares are $0, 1\mod{3}$.  Then $2^{n}+105 \equiv 0, 1\mod{3}\implies 2^{n}\equiv 0, 1\mod{3}\implies 2|n$ (just take powers of $2$ mod $3$ and you'll see...)

To add to above,
$2^{n}$ is never zero  mod 3 . ( 3 can not divide a power of 2)
$2^{n}\equiv (-1)^{n}\bmod 3$ is  $1$ only when n is  even.
So $n$ has to be even, if RHS is a perfect square.

[quote="neelnal"][quote="K81o7"]
All squares are $0, 1\mod{3}$.  Then $2^{n}+105 \equiv 0, 1\mod{3}\implies 2^{n}\equiv 0, 1\mod{3}\implies 2|n$ (just take powers of $2$ mod $3$ and you'll see...)[/quote]

All squares are $0, 1\mod{3}$.  How do you know? Can I see a proof? Thanks.[/quote]

Legendre: $(\frac{2}{3})=-1$

[quote="neelnal"]
All squares are $0, 1\mod{3}$.  How do you know? Can I see a proof? Thanks.[/quote]
Basically  just try, any number can be written as $3k+r$ (You divide by 3 and take the remainder) where $r=0,1,2$
Now the square is $9k^{2}+6kr+r^{2}$, and since first two terms are divisible by 0, you are left with only  $r^{2}$ , and which can only be (0,$1^{2}= 1$ or $2^{2}= 4 \equiv 1 \bmod 3$ .. IOW the square is either 0 or +1 mod 3.

Similary, another nice result (try to prove it) to know is that square of an odd number is always 1 mod 8.

Thanks everyone so much. I actually learned what a modulo is today in application after years of wondering what "mod" is lol.


[quote="Jars"]How would you compute this problem?

If $x+y = 6$ and $x^{2}y+y^{2}x = 60$, find $x^{3}+y^{3}$.

[size=75][color=darkred]Added LaTeX.
-nebula42[/color][/size][/quote]
The easiest way to do this problem is to notice that \[(x+y)^{3}=x^{3}+y^{3}+3x^{2}y+3y^{2}x\] or \[6^{3}=(x^{3}+y^{3})+3(60)\] so \[x^{3}+y^{3}=216-180=\boxed{36}\]

[hide]Yes.

Also, if you can't notice that, tehn divide the two equations to get $xy=10$.

Then $(x+y)^{3}-3xy(x+y)=x^{3}+y^{3}$.

So $x^{3}+y^{3}=216-180=36$

$36$[/hide]


geometry
3D geometry
octahedron
counting
distinguishability
rotation

I don't know how to do this problem:

Eight congruent equilateral triangles each of a different color are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look identical to the other.)

[hide]I would say $ \frac{8!\cdot4!}{2\cdot 4\cdot\ 4!}=\frac{8!}{8}=7!=5040$, but I'm not sure.[/hide]

There are $ 8!$ ways to color the octahedron. However, since the octahedron has 6 vertices, if you take any one of them to be the "top", you can rotate it 4 times, which makes 4 different coloring but same octahedron. Therefore, for each coloring, there are $ 6\cdot4$ indistinguishable octahedrons.

$ \frac{8!}{6\cdot4}=1680$

I agree with Allan. My method was different and incorrect.

yes, allan's answer was correct, and I get it now. thanks!