Find $max c$, where $c=const$ such that the sequence {$a_n$}$_1^\infty$ $a_n=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}+\frac{1}{n+c}$ is monotonely decreasing.

[hide]I don`t see the sense of that problem, but.. [/hide]

Monotonically decresaing means:

$n < m \rightarrow a_n > a_m$

So set $m=n+1$, do the maths, and i got that:

$c < \sqrt{2}-1$

Im not sure, but this means a max dosnt exist... you can put c as close as you want to it, but never equal to it, so you can always get a higher c.

(PS not to sure about mods liken homework posted!!!)


geometry
3D geometry
prism
sphere
AMC
AIME
rectangle

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units.  Given that the volume of this set is $(m + n \pi)/p$, where $m$, $n$, and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p$.

[quote="Treething"]I think you forgot to subtract 6. 487 sounds more correct.[/quote]Hmm.. i think 487 sounds correct. Treething is reminding you to subtract the volume inside the rectangular parallelepiped that is not within one unit of the sides. You can visualize it in 2-d by thinking about the area in a 3 x 4 rectangle that is at least 1 unit from the sides. What he did was $(3-1-1)(4-1-1)(5-1-1)=6$

Note that the question asks for the volume "[b]inside[/b] or within one unit of a rectangular parallelepiped" (emphasis mine)  :wink:

505 is the correct answer (checked on AMC website).

[quote="E^(pi*i)=-1"]Note that the question asks for the volume "[b]inside[/b] or within one unit of a rectangular parallelepiped"[/quote]oh oops.. :blush:
[quote="Ubemaya"]Even if you did have to subtract 6, since when did 505 - 6 = 487?[/quote]$60+94+12\pi+\frac{4}{3}\pi-6=\frac{462-18+40\pi}{3}=\frac{444+40\pi}{3}$ which adds up to 487.

[quote="BarbieRocks"][quote="Ubemaya"]Even if you did have to subtract 6, since when did 505 - 6 = 487?[/quote]
subtracting six is the same as subtracting 18 from the answer, because there's a denominator of 3. :wink:[/quote]This has been responded by minsoens 3 years ago...making your comment [u]superfluous[/u].

[quote="mathemonster"]problems like this are impossible to visualize for me :( any suggestions?[/quote]It isn't impossible but it takes practice. Draw more 3D figures, study them, rotate them, do more problems related to solid geometry. The point is to develop your mind to "visualize" objects from familiar ones and make more complex shapes. Even watch more 3D movies (if that helps  :P ).

[quote=chesspro][hide="Answer"]The box's volume is 60.

The points external to the box within one unit can be broken up into three types of pieces: rectangular prisms, quarter cylinders, and eighth-spheres. The prisms can be found by 2(3*4 + 4*5 + 5*3) = 94. There will be a total of 12 quarter cylinders, each with radius 1. Four will have a height of 3, four will have a height of 4, and four will have a height of 5. We can thus combine them, and we have 3 cylinders of radius 1 and heights 3, 4, and 5. Their volume is $\pi(3 + 4 + 5) = 12\pi$. For the eighth-spheres, we have a total of 8, so we can combine them to form a regular sphere with radius 1. Its area is $\frac{4}{3} \pi$. Adding these volumes, we have

$60 + 94 + 12\pi + \frac{4}{3}\pi=\frac{462 + 40\pi}{3}$

Therefore the answer is 505.[/hide]

I haven't done an AIME problem in a while, so I might have messed up somewhere.[/quote]

Hello, in this solution does anyone know where he gets the quarter cylinders from? (I'm not great at visualization and trying to understand a good way to do this problem). Thank you in advance.

I assume its from the points that are on the edge but are not vertices, as the vertices are the eighth spheres and not part of the cylinder


Could anyone tell me how exactly one derives (rigorously, if possible) from Maxwell's equations (any subset of the latter) Coloumb's law - or point me to an appropriate source. Also, in  the case of e.g. electrostatics, does the the priniciple of superposition of electric fields as caused by different charges (i.e. the elctric field at a point in space is the vector sum of the electric fields caused by individual charges) follow from Maxwell's equations (any subset of them)? And to finish it off (O.K. I'm obviously asking too many questions, I hope you lot can answer at least some of them :roll: ), what exactly does it mean that light is an electromagnetic wave - that is, (precisely) which conditions must the E and H vectors in space satify for us to be in a position to claim that "there is" light in this space, or to be more illustrative, when does an electromagnetic field have the "god-given" right to say: "Let there be light!"?
Thanks for any answers and thoughts on the subject.  :lol:

I don't know about Coulomb's law, but the principle of superposition of electric fields follow from the same principle of forces. I.e. (I hope you can follow):

$F=\sum_i F_i = \sum_i K\frac{qq_i}{r_i^2} = qK\sum_i \frac{q_i}{r_i^2}=q\sum_i E_i$

$F$ being the net force on a charge, $F_i,q_i,r_i$ being the appropriate things for each individual charge that affects the "test" charge. Since we desire $F=qE$ it follows that

$E=\sum_i E_i$

And that's superposition for ya. (Replace everything with vector equivalences for a stringent proof.)

--

Electromagnetic waves [b]in vacuum[/b] satisfy the [i]wave equations[/i]:

$\nabla^2 \vec{E} = \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t}$
$\nabla^2 \vec{B} = \frac{1}{c^2} \frac{\partial \vec{B}}{\partial t}$

This follows from Maxwell's equations ($c$ being the speed of light). I don't know of any such equations for electromagnetic radiation through matter.


[quote="nayel"]i searched but couldn't find this problem in the forum. so i posted it.

Find all primes $p,q$ such that $p+q=(p-q)^{3}$.[/quote]

maybe i'm wrong...i'm not so good :blush:  but this is what i've found:

if $p+q=(p-q)^{3}$  then $(p-q) | (p+q)$
$\frac{p+q}{p-q}= \frac{p-q+q+q}{p-q}=1+\frac{2q}{p-q}$
then $(p-q)| 2q$   =>     $2q=np-nq \Rightarrow p= q+\frac{2q}{n}$
either $n=1 \Rightarrow p=3q$  but both $p$ and $q$ are primes
or      $n=2 \Rightarrow p=2q$  but both $p$ and $q$ are primes
or      $n=q \Rightarrow p=q+2 (1)$
or      $n=2q \Rightarrow p=q+1 (2)$
if $(2)$then $2q+1=1$ but $q$ is a prime
if $(1)$then $2q+2= (q+2-q)^{3}$
       $2q+2=8$
       $q=3 ; p=5$

The only solution is $(5,3)$

EDIT: I added the case n=2q

nice solution! :D

i have the following one:
if $p,q\ge 5$, we check the four cases: (congruences are taken $\bmod 6$ if otherwise not stated)
1. $p\equiv q\equiv 1$, then $LHS\equiv 2$ while $RHS\equiv 0$
2. $p\equiv 1, q\equiv-1$ then $LHS\equiv 0, RHS\equiv 2$
3. $p\equiv-1, q\equiv 1$ then $LHS\equiv-2, RHS\equiv0$
4. $p\equiv-1, q\equiv-1$ then $LHS\equiv 0, RHS\equiv-2$

so both $p$ and $q$ cannot be >3.

if $p=3$ we get $q<3$ so nothing. similarly $p=2$ implies nothing.
if $q=3$ we have after expanding, $p(p^{2}-9p+26)=30$ so $p|30$ and from here $p=5$ because $p=3$ isn't a solution.
if $q=2$, $p(p^{2}-6p+11)=10\Rightarrow p|10\Rightarrow$ no solution. :wink:

[quote="nayel"][quote="EUCLA"]Could you please explain what do LHS and RHS mean...i haven't found it  :blush:
thanks[/quote]

LHS=Left Hand Side (of the equation)
RHS=Right Hand Side (of the equation) :)[/quote]
thanks, now i've understood your solution  :lol:


logarithms
quadratics
algebra
quadratic formula

[quote="sharkman"]Solve each log equation given.

(1) log_5 (x^2 + x + 4) = 2

(2) log_3 (3^x) = -1[/quote] Use the log rule that $ \log_x (a) \equal{} b$ can be written as $ x^b \equal{} a$, to get a nice expression. (Eg. $ \log_5 (x) \equal{}2$ can be written as $ x\equal{} 5^2$.)

 Simplify then solve for x.

In particular, the exact solutions to number 1 are $ \frac{\minus{}1 \pm \sqrt{85}}{2}$.

In number 2, if you use the same binom did in his 3rd step on his first step, you immediately get x=-1

[quote="binomial_4eva"]1)
$ \log_5(x^2 + x + 4) = 2\implies5^2 = x^2 + x + 4\implies x^2 + x - 21 = 0$
using the quadratic formula $ x_1 = 4.11$ and $ x_2 = - 5.11$ is the solutions

2)
$ \log_3(3^x) = - 1\implies3^{ - 1} = 3^x\implies\frac {1}{3} = 3^x$
$ \ln{\frac {1}{3} = \ln{3^x}\implies x = \frac {\ln{\frac {1}{3}}}{\ln{3^x}} = - 1\implies x = - 1}$[/quote] Alternatively recall that $ \log_a a = 1$.

[hide="Hence,"] $ \log_3(3^x) = - 1 \Leftrightarrow x \log_3 3 = - 1.$

I.e. $ x = - 1$.[/hide]


Does anyone have any non-proof AIME-like probs which I can try for these? The AoPS book is kinda short on them.
Also, has Menalaus/Ceva appeared in previous AIMEs?

Thanks.

[quote="joml88"]Menelaus/Ceva [i]very[/i] rarely comes up on non-Olympiad, non-proof type problems.  It wont show up nor ever has shown up on AIME (unless there's an alternate way using Ceva/Menelaus).

Nonetheless, here is one:

Given triangle ABC, with AB = 12 and CA = 15. The median from B, the altitude from C, and the angle bisector from A are all concurrent inside the triangle. Compute the length of BC.[/quote]

I thought that was an old aime problem?

Using angle bisector, we know that the parts of $BC$ are $12x$ and $15x$ which are separated by the angle bisector, for some common ratio $x$, so using Ceva, we have $12x\cdot\frac{15}{2}\cdot a=15x\cdot\frac{15}{2}\cdot b \Longrightarrow a=\frac{5}{4}b$, where $a$ and $b$ are the parts of $AB$ which are seperated by the altitude.

From using Pythagorus, we have $15^2-a^2=(27x)^2-b^2$, but since $a+b=12$, so $b=\frac{16}{3}$, we can simplify this to $(27x)^2=15^2-(\frac{20}{3})^2+(\frac{16}{3})^2$, and someone can solve this...


Ребята  помогите мне  две последние задачи!
 РАННЕЕ СПАСИБО!!!
http://rapidshare.com/files/145515024/IMG_3961.JPG.html

[quote="Erken"]Out of curiosity,where do you study?
quite clear that in Russia,but where exactly?due to the problems it is your first year?[/quote]
Да. Это д.з
Я  в Томске и  это  задачи для первого курса.
  Я  уже  решил все  но не очень красиво.
И  хочу  все вместе  решат.


I've been a TA for 5 schools in the Westchester Area (30 miles north from NYC). Each year some way or the other I'm stuck dealing with the graduation gown ordering. Each of the schools use a rep that comes in an measures all of the students. I've had many problems with this as for some odd reason when it comes graduation time we are left with a few students that don't have a graduation gown.

Has anyone thought of ordering directly from a company and placing the order yourself?

[url=http://www.graduationsource.com]Graduation Source[/url]

The most appropriate forum for this question is probably [url=http://www.artofproblemsolving.com/Forum/index.php?f=138]the Round Table[/url].

I'm not sure how it works in your area, but at my high school the seniors are all given order forms on a certain day after school and told to turn them in a week or so afterward.  The students without graduation gowns are just the students that didn't order them, which is more or less their problem.


A ball is elevated off the ground.  A person is trying to get the ball knocked down onto the ground.  After two minutes of failed attempts, the ball is finally on the ground.  However, once the ball hits the floor, he regrets having tried to knock it down in the first place.  What is going on?

Feel free to ask questions.

Erm... The ball is a bowling ball. He is trying to get it down from a high shelf to bowl. But, once he gets it down, it hits his toe, making him regret getting the ball down.

Second guess: The ball is an egg of a bird. The guy is hungry and tries to get it from the nest. But, he accidentally hits the egg over without catching it and it splatters everywhere, leaving him with no food.

No to both of those.  There was virtually no chance of him getting hurt in the process.

Again, feel free to ask questions.

did knocking the ball down do any damage to anyone or the environment in which he was standing (ex. break a floor)?

Did knocking the ball down do any damage to anyone or the environment in which he was standing? [b]No[/b]

[hide="Hint"]The person was nowhere near the ball.  But he was trying to get it knocked down... how could this be?[/hide]

[hide]The ball was stuck on top of something, and the person threw something at the ball to knock it down. When he finally knocked the ball down, the other thing got stuck up there?[/hide]

The ball was stuck on top of something, and the person threw something at the ball to knock it down. When he finally knocked the ball down, the other thing got stuck up there? [b]No[/b]

How was the ball elevated?

...Or are we only allowed yes or no questions?
In that event, what the ball elevated with a string?

Please ask yes/no questions only.  However, The_Scintillator, you could find out how the ball was elevated by asking more questions.

In that event, what the ball elevated with a string? [b]No[/b]
Is the ball spherical? [b]Yes[/b]

Question time! :D

Was he playing Cricket? Or some odd subsort? With a ball instead of stumps....

Was he playing Bocce? (I don't know much about that game, though.)

Was the ball elevated on a solid surface?

Was the ball elevated on a liquid surface?

Was the ball elevated onto a shelf?

Was the ball elevated onto a pole?

Was the ball involved in some sort of sport?

Was the ball and scenario insolved in some sort of sport?

Was the ball setting off a chain reaction?

Does the person now suffer from psychological/emotion trauma due to the effects of the ball? (Sounds kind of silly, but would rule out some stuff.)

Does it matter who elevated the ball?

Does it matter what sort of space/dimension/universe he was in?

Was the diameter of the ball $ \ge 5cm$ ? $ \ge 10cm$ ? $ \ge 20cm$ ? $ \ge 50cm$ ? $ \ge 1m$ ? $ \ge 10m$ ? $ \ge 1km$ ?
Was the ball red? green? blue? yellow? cyan? magenta? white? black? transparent?
Was the ball shattered after it fell down?
Was something destroyed after it fell down?
Did the incident cause any loss to the person?

Was he playing Cricket? Or some odd subsort? With a ball instead of stumps.... [b]No[/b]

Was he playing Bocce? (I don't know much about that game, though.) [b]No[/b]

Was the ball elevated on a solid surface? [b]Yes, if by that you mean it was supported by something solid[/b]

Was the ball elevated on a liquid surface? [b]No[/b]

Was the ball elevated onto a shelf? [b]Yes[/b]

Was the ball elevated onto a pole? [b]Yes, but not an upright pole[/b]

Was the ball involved in some sort of sport? [b]Outside this scenario, probably not[/b]

Was the ball and scenario insolved in some sort of sport? [b]Some consider it a sport, others don't[/b]

Was the ball setting off a chain reaction? [b]No[/b]

Does the person now suffer from psychological/emotion trauma due to the effects of the ball? (Sounds kind of silly, but would rule out some stuff.) [b]No[/b]

Does it matter who elevated the ball? [b]No[/b]

Does it matter what sort of space/dimension/universe he was in? [b]No[/b]

Was the diameter of the ball...? [b]Based on official measurements, the diameter is about 1 meter (officially 40 inches)[/b]

Was the ball red? Green? Blue? ... [b]It was either red or blue.  Theoretically, the ball could've been another color.  However, in the scenario given, the balls are either red or blue.[/b]

Was the ball shattered after it fell down? [b]No[/b]

Was something destroyed after it fell down? [b]No[/b]

Did the incident cause any loss to the person? [b]Yes, but not physically[/b]

Did the incident cause an emotional loss? Any mental loss...?

Was the object on the pole AND the shelf? At the same time?

Was the object elevated onto a sideways pole?

Is the sport known in the U.S.?

Does the sport merit a page in Wikipedia?

What sport is this, anyways? (Not an official question.)
Grarg. I don't know.

[hide="Is the first letter of the first word of the name of the game:"]
a?
b?
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[/hide]

[hide="Is the second letter of the first word of the name game:"]
a?
b?
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[/hide]

[hide="Is the third letter of the first word of the name of the game:"]
a?
b?
c?
d?
e?
f?
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[/hide]

How many words are in the name of the game?

[quote]How many words are in the name of the game?[/quote]

PSST!! Not a yes or no question..

nice approach, though.

Cheap shot.  But whatever. :P

The first three letters of the name of the game: [b]FIR[/b]

Is the game directed towards...
adults (20+)?
seniors (65+)?
teenagers (13-19)?
children (12 and under)?

To win, do you try to score as many points as possible?

Is the ball put into motion by anything other than a human? (such as a club or whatnot)

Is the game directed towards...
adults (20+)? [b]No[/b]
seniors (65+)? [b]No[/b]
teenagers (13-19)? [b]Yes[/b]
children (12 and under)? [b]No[/b]
To win, do you try to score as many points as possible? [b]Yes[/b]
Is the ball put into motion by anything other than a human? [b]Yes[/b]

Does the name contain "ball"?
Is it called "fireball"?
Does the sport involve the hand?

Does the name contain "ball"? [b]No[/b]
Is it called "fireball"? [b]No[/b]
Does the sport involve the hand? [b]Yes, but they don't touch the ball[/b]

[hide="Hint"]The organization that created this game also created another game called "Triple Play."  This game also has its own Wikipedia article.[/hide]

Now it seems too easy

Is it FIRST Overdrive?

Is it FIRST Overdrive? [b]Yes[/b]

I just gave it away, didn't I?

are there usually mmore than 5 participants in this game?

does playing the game involve more than the ball, the pole, and the participant?

are there usually more than 10 balls?

I'm officially calling this solved.

[hide="Explanation"][url=http://www.youtube.com/watch?v=yViRWaSODIc]General overview of game[/url]
He was trying to knock the ball down in order to move it around the track, earning points.  However, he got it down before the end of the game, so he lost an opportunity for a 12-point bonus in the end.  In addition, he lost the opportunity to go around the track, which would still earn points.[/hide]

I tried to offer a hint with the question I asked.
[quote="krsattack"]Does this sport require [b]human[/b] (physical) strength?[/quote]

Of course, then lingomaniac88 changed my question (BTW, did you do that on purpose?):
[quote="lingomaniac88"]Does this sport require physical strength? [b]No[/b][/quote]

I was hoping that someone would catch my wording...

Well, by that, krsattack, I assumed that you explicitly meant physical strength.

Of course, it does require mental strength, i.e., strategy.  "Should I try to grab and hurdle (throw over the overpass) the ball, or should I just herd it under the finish line?"  Efficiency vs. quantity of points.  For those who have no idea what I'm talking about, watch this:

[youtube]yViRWaSODIc[/youtube]

Wow... the "40 inches to be exact" and the "red or blue. in this situation" should have immediately tipped me off...


Consider a string of $n$ 7's, $7777\cdots77$, into which $+$ signs are inserted to produce an arithmetic expression. For example, $7+77+777+7+7=875$ could be obtained from eight 7's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value 7000?

[quote="musicheck"](Sorry, haven't installed latex yet)

[hide]Looking at the number 7000, we obviously see the maximum number of sevens: a string of one thousand sevens.  Then, we see that the minimum is 28 sevens: 777 X 9 + 1 =7000.  The next step is to see by what interval the value of N increases.  Since 777 is 3 sevens, and 77*10+7 is 21 sevens, we can convert a 777 into 77's and 7's and add 18 to the value of N.  Since we have 9 777's to work with, this gives us 28,46,64,82,100,118,136,154,172,190 as values for N.  Since 77 can be converted into 7*11, we can add 9 to N by converting 77 into 7's.  Our N=190, which has 0 777's 90 77's and 10 7's.  We therefore can add 9 to N 90 times by doing this.  All values of N not covered by this can be dealt with with the N=46 (8 777's 10 77's 2 7's) up to 190.  We also have our 28 term of 9 777's and 1 7.  (1000-46)/9=106, but this does not cover N=46 or N=28.  So we have a grand total of 108 possible values. [/hide][/quote]

You DONT have to have [i]anything[/i] installed to use $\LaTeX$ on the forums.  Where does this myth come from  :?

To use Latex you only need to write dollar signs $.
If you want to write x^2 just write $x^2$.

wait...dondigo, how did you post those dollar signs without having them turn on LaTeX?

Check the box that says "Disable BBCode in this post" when you are posting (it's below the message body in the options section).

[quote="musicheck"]
[hide="fixed up latex for him..."]Looking at the number $7000$, we obviously see the maximum number of $7's$: a string of $1000 \ 7's$.  Then, we see that the minimum is $28 \ 7's: \ 777*9 + 1 =7000$.  The next step is to see by what interval the value of $N$ increases.  Since $777$ is $3 \ 7's, \ 77*10+7 is 21 \ 7's$, we can convert a $777$ into $77's$ and $7's$ and add $18$ to the value of $N$.  Since we have $9 \ 777's$ to work with, this gives us $28,46,64,82,100,118,136,154,172,190 (=28+18n | 1\leq n\leq 9)$ as values for $N$.  Since $77$ can be converted into $7*11$, we can add $9$ to $N$ by converting $77$ into $7's$.  Our $N=190$, which has $0 \ 777's \ 90 \ 77's \ 10 7's$.  We therefore can add $9$ to $N \ 90$ times by doing this.  All values of $N$ not covered by this can be dealt with with the $N=46 \ (8 \ 777's \ 10 \ 77's \ 2 \ 7's)$ up to $190$.  We also have our $28$ term of $9 \ 777's$ and $1 \ 7$.  $(1000-46)/9=106$, but this does not cover $N=46, N=28$.  So we have a grand total of $\boxed{108}$ possible values. [/hide][/quote]

By the way, are you sure that [hide="answer"]108[/hide] is the answer?
Here is how I get my answer:
[hide="??"]Divide through by $7$ in the question to rewrite it as a string of $1's$ adding to $1000$. So we have $a+11b+111c=1000$. Notice we can rewrite variables to get $a+d+e=1000$, which normally has $\binom{1002}{2}$ solution in positive integers. However, we can only rewrite when $d$ and $e$ are divisible by $11$ and $111$ respectively. This happens $\frac{1}{11*111}$ of the time, so we find our answer by $\lfloor \frac{\binom{1002}{2}}{11*111} \rfloor = \boxed{136}$. Since the numbers are so big, even using a ballpark figure still gives 136...[/hide]

[hide]Yes, $108$ is the answer.  (I checked the official answers on the AMC page.)  Basically, $n$ can be any number equivalent to $1\pmod{9}$ between $28$ and $1000$ inclusive, except $37$.  I got $109$ because I didn't get rid of $37$.  :lol:[/hide]

hm aren't there a lot more than $ 108$ solutions to $ a\plus{}11b\plus{}111c\equal{}1000$, or am I overlooking something in the problem?

Yes, but many of them give the same total number of $ 7$'s.

Here is a fairly quick method for the beginning:
[hide]
We have

$ a\plus{}11b\plus{}111c\equal{}1000$

and

$ a\plus{}2b\plus{}3c\equal{}n$.

So subtract the second from the first to get

$ 9b\plus{}108c\equal{}1000\minus{}n$.

And now it's obvious that $ 9|1000\minus{}n\Rightarrow n\equiv 1\pmod 9$.
[/hide]


function
geometry
3D geometry
inequalities

Determine all the functions $f : Z \rightarrow Z$ such that
$f(x + 1959) \leq f(x) + 1959$  and  $f(x + 2002) \geq f(x) + 2002$.

Determine the smallest base $b > 1$ for which there exist a perfect cube $xyxy$ in base $b$

Problem 2:  [hide]Let's rewrite this algebriacally.  xb^3+xb+yb^2+y=z^3 where x and y are less than b.  Now, let's factor this equation.  xb(b^2+1)+y(b^2+1) or (xb+y)(b^2+1).  This means our cubic has to be divisible by the base squared plus one.  Now, we test our possibilities.  2 yields 5 which fails because the smallest cube that's divisible by 5 is 125 which is too big.  3 yields 10 which fails because the smallest cube that's divisible by 10 is 1000 which is too big.  4 yields 17 which fails since the smallest cube that's divisible by 17 is 17^3, too big.  Similar results come in for 5(26) and 6(37).  But when we try 7, we get 50 which is divides 1000 or 10^3.  Dividing 50 into 1000 yields 20 so we have 7x+y=20 where x and y are under 7.  (2,6) is a solution.  Just to verify, 2(343)+6(49)+2(7)+6=686+294+14+6=980+20=1000.  So seven is the smallest mod that works.[/hide]

Let's assume f(x) is linear for now.  We'll expand the case later.  y=a(x+1959+b) greaterthanorequalto y=a(x+b)+1959.  We cancel out the x and b terms to yeild 1959a islessthanorequalto 1959, so a islessthanorequalto 1.  But when we apply this function to 2002, after simplifing, we get 2002a isgreaterthanorequalto 2002, so a isgreaterthanorequalto 1.  This pretty much shows that a=1.  So all linear equations that come in the form y=x+b will work.  Now, we expand to quadratics.  a(x+1959)^2+b(x+1959)+c islessthanorequalto ax^2+bx+c+1959 so we cancel out the bx, c, ax^2 terms.  We get 1959(2)(ax)+(1959)^2(a)+1959(b) islessthanorequalto 1959 so we divide by 1959.  We get 2ax+1959a+b islessthanorequalto 1.  Applying the same process to the second inequality, we get 2ax+2002a+b isgreaterthanorequalto 1.  Obviously, a isn't negative.  But now I'm stuck.  Any ideas?


[quote="manjil"]I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!

So if anyone is from the NE then please give a brief introduction here![/quote]

Surprising it seems I am the only one!

[quote="chemrock"]Probably u can spread it that side.[/quote]

Yes why not? I am trying, but the people here are very lazy! :rotfl:  :P

You've just got another mathlinker from the north- east.....................................hi this is supranta here.

why are you guys dividing india into parts
? :huh:  :rotfl:

Hey man, I am not dividing India into parts......I was just curious!

dont take my post seriously (i dont know how you do it   :wink:  :rotfl: )

[quote="nuclear_alchemist"]Lol, he was joking! One of the purposes this site performs is to bring the best minds from all over India together. You'll meet people from all parts of the country here. Why look for people in your region when you can meet them everyday in any case?  :rotfl:[/quote]

(and in the US too)  :P

[quote="Rajiv"][quote="manjil"]I was just curious how many Indian Mathlinkers are from the North East and Assam in particular!

So if anyone is from the NE then please give a brief introduction here![/quote]
mathlinkers..........!?
The site artofproblemsolving too supports the same forums.[/quote]

Sure that is implied!

Anyone.I know one or 2 from tripura who visit here regularly

well thats too distant

never knew Aops is so widespread :)

There's me ,murgi,muks,akashnil,bengal(NRI)
That's all I can remember! :D

and Keshav. Also, I lived in Kolkata for 6 years :P

i am from tripura

[hide]but i live in delhi now, in Vmc reg batch[/hide]

I am in for jee 2011,
 in reg-4.

My name is Shankhadip Majumder.


geometry
3D geometry
tetrahedron
sphere
circumcircle
calculus
integration

What is the 4d volume of a 4d tetrahedron with side length one? Show work...

The tessaract is the wrong name for it, but I can't change that

(and if there are two solids here then list them both... I dont know the answer im just curious)

I always thought a tesseract was the 4d equivalent of a 3d cube. (Which would imply that it has volume 1 unit4)

Yes, I also thought that a tesseract was a hyper-cube.

Also, there are 6 platonic solids in 4 dimensional space -- I was under the impression that there were [i]two[/i] simplexes, and thus you're asking a question with a reference to two different shapes.  (I do know there are 6 platonic solids in 4-D . . . I don't know if the second bit is correct.)

im sorry i accendately wrote the wrong thing... Ill change it

Okay, so I looked it up on Mathworld and it didn't tell me anything drastically interesting, except that there is only 1 real equivalent of the tetrahedron in 4-space.

Now, I'm going to outline how you would find its volume, and maybe do it some other time.

Solution by analogy --
to get from a line segment (a 1-simplex) to an equilateral triangle (a 2-simplex), we go to the centroid and draw a perpendicular and take a point on it such that all three segments we end up with are equal.
To get from the equilateral triangle to the tetrahedron (3-simplex), we similarly go to the centroid, draw a perpendicular to the plane, and pick the point on it such that all 6 segments are equal.
To get from there to the 4-simplex, we have to draw a perpendicular to the space we are in, and go a distance such that all the segments connecting it to the vertices of the tetrahedron are equal in length to the side of the tetrahedron.

Now, I can do this -- but I'm not going to do it now.  See if you can figure it out.  (Note that finding the length you need to do doesn't solve the problem entirely, but it's about 1/2 of the way there.  You're also going to need to solve for the area of the tetrahedron and its height in order to get anywhere.)

There is one unique 4-simplex (or hyper-tetrahedron, or pentatope as they are called) in four dimensions. As somebody already pointed out, the tessaract is actually another name for the hypercube.

This is a very cool problem though. There are lots of interesting ways to solve it. I worked on this quite a bit when I was in high school, as well as finding the volumes of hyper-spheres. They are fun problems to try. It leads to an interesting study of the sphere packing problem in higher dimensions.

Has anyone made any progress with this problem yet?

This was very much fun.  I claim that the volume of an n-dimensional simplex of side s is given by
Vn = sn:sqrt:(n + 1)/(n! * 2n/2).

To do this, I need the following fact:  the circumradius of an n-dimensional simplex is (n - 1)/n times the height of the same simplex.  I'm sure that this can be verified with an integral or a volume argument identical to that which can be used to prove that the medians trisect each other at the centroid.  Anyhow, let me take it as a given.  Then we have that if hn is the height of an n-dimensional simplex,
hn = s2 - ((n - 1)/n * hn - 1)2
which is just an application of the Pythagorean Theorem.

Then, inductively we can show that in general,
hn = s:sqrt:(n*(n + 1)/2)/n.

Now, Vn = Vn - 1*hn/n.  This definitely can be verified by integral, I think :).  Anyhow, if we continuously apply that identity to itself, and then make the substitution for the hi's, you get exactly the formula I wrote above.

Yep that's right. Now a further project you could try, is to find the volume of an n-sphere. Then figure out what is the ratio of the volume of an n-sphere inscribed in an n-simplex to the volume of the simplex. And what is the ratio of the total volume of "packed" n-spheres inscribed in an n-simplex.

And finally what happens to this ratio as the number of inscribed spheres tends to infinity? The ratio turns out to be very unexpected as n varies.

[quote="ComplexZeta"]What do you think about this interesting fact? The ``volume'' of an n-sphere of radius 1 is maximized when n=5.[/quote]

Do you know how to show that? And in what dimension is the "n-tetrahedron" maximized?

Well, for s = 1, we can see that in will start getting smaller pretty quickly (since both of the pieces of the denominator get big quite fast, while the numerator doesn't) and quickly checking the first few cases shows that the volume is minimized at n = 1 (or 0, I suppose).  However, if we take bigger values of s, we should be able to make it be just about any dimension we want.  Letting s = :rt2: makes it uniquely n = 1.  Letting s be bigger should definitely make it become higher dimensions, until the n! and :rt2:n catch up.  I don't really want to calculate higher values, though, because :rt2: was just so nice and I'm very lazy.  Perhaps when I'm back at school with Mathematica . . .


geometry MATHCOUNTS function trigonometry AMC USA(J)MO USAMO

Sorry if this is going to violate the "no whining" rule. I just had to get it out.

Recently, my math teacher had criticized me again. For what seemed to be the "1 followed by an excessive amount of zeros" times. My math--(middle school Algebra and Geometry) teacher had criticized me for thinking that I was a braggart, an egomaniac, and a show-off. (My math teacher plays favorites, they're the MC coach--and only favors those who are good at MATHCOUNTS, and I'm not good at MC at all.) Putting that aside, though, we got the AMC 10 and 12 scores. (Yes, this is a bit dated, and I apologize.) I scored dismally. I don't mean 144 and didn't get a perfect score--I mean outright failed. I was unable to function at all that day, and scored, well, well below 80 on both.

Why did this matter? Well, at my school, like many other public schools throughout the nation, many kids pretended to be stupid. I, myself, and not a victim and hope to never succumb to this [i]disease[/i]. I didn't hide the fact that math was something interesting for me. I never hid the fact that it is truly an art and is breathtakingly beautiful. However, I [i]did[/i] hide the fact that I was more advanced than others. I didn't feel I needed to rub it in other peoples' faces, but I did want them to appreciate the beauty of it. This is where the problem began.

One day, I let it accidentally slip when we were deriving the Law of Cosines that "oblique triangles were frustrating...*mumble**mumble**mumble*...". Right triangles can be so much easier to work with, on occasion. I didn't think anyone had heard, except for my friend sitting next to me, who then promptly asked, "What are oblique triangles?". And me being me, I answered that "They, in short, are triangles that are....erm....not right." That's probably not the exact definition, as I was frazzled by the fact that I had lost control and let a little bit of vocabulary slip. That usually didn't happen. However, I shook it off and proceeded.

I let little bits of vocabulary slip-but didn't feel this was a huge deal. So I knew a little bit of advanced vocabulary. Whoop-e-dee-doo-dah-day. My teacher, however, did not quite feel the same way. My teacher remembered all these incidents and strung them together in her mind. My teacher then proceeded to nail me about everything. Not only did she target me for one little thing I'd say, for example, "Could you please keep it down over there?", and she'd get all mad and have her tirade about how they were discussing mathematics and how it was perfectly fine--even if I had just voiced everyone else's opinion. I knew I wasn't the only one who was about to say it. I just happened to be the only that did. She would always jump on me about things like that, and really chastise me in front of everyone else. Most people just looked the other way. This is middle school, remember?

Now let's bring this information all back into the (somewhat) present. The day the scores came. My teacher was as shameless as usual, and let me the scores--but not before "inconspicuously" flaunting everyone else's for a few moments before pointing me to my scores. Needless to say, they were at the very bottom. That's when the lecture started. She managed to impact me in such a brutal way in such few sentences that I was partially traumatized. She talked about how, and this is a direct quote--"...you really don't know all those things..." I kept on my serious veneer the entire time I was in there. Just because she wanted the satisfaction of seeing me break doesn't mean I was going to give it to her. Another quote was:
Me: "But I never said I knew all those things."
Teacher: "But you always used advanced vocabulary and other things...."
Me: (Trying not to break....)

And in case you were wondering, I did break but managed to keep myself together. The way I've grown up--I have excellent self-control over my emotions. I shed a few tears, and no one even noticed. I felt much better after my friends tried to cheer me up when i traumatized and wasn't talking or really, just showing any emotion whatsoever.

I'm sorry if depresses anyone. But I really felt I needed to get it out--just to type it. I'll probably take it off after a little while and I've cooled off. Thanks. :)

But, um, out of curiosity (not to meantion blatent disregard for other's feelings), has anyone had someone do this to them? I know this has happened to me a lot, but has anyone else ever been subject to anything like this--at all? Just a simple yes or or not or maybe will suffice, I don't need stories that will make me use up an entire box of tissues--but I'd like to hear them anyways, even if we run out of tissues. :P

[quote="The_Scintillator"]I'm sorry if depresses anyone. But I really felt I needed to get it out--just to type it. I'll probably take it off after a little while and I've cooled off. Thanks. :)[/quote]

This is why we have blogs.  :)

And to your problem, the advanced vocabulary is simply a better way to get your point across. The next time that incident happened where your teacher criticizes you for, simply explain that it's a better way to communicate (not to show off), and the other possibility is to have you speak in redudancy (using words like "thingamajig, whachacallit, blah"). Granted you might get detention for being smart with the teacher, but this is middle school (like it matters?), and the teacher needs to be taught that gifted math kids should not be limited. Using the term "oblique" triangle is not a crime.

However, I do think that you should tone your vocab down a bit. If I can picture myself sitting in a regular math class with no "AoPS" math in my mind, I do find that people using advanced terminology is quite self-centering. I think it's best if you're modest, and trudge along math class like everyone else. I do this frequently and as I'm getting by in math, my peers still see me as a relatively advanced math student. The respect is there, but the effort in math class diminishes.

Don't worry about it. You're going to end up with bad [math] teachers in high school. It's important that you pass the class (with an A) and not make too many enemies. Being a good student with a "secret AoPS math life" is the modest way to get by.

I truly feel sorry for you. I do not know what grade you are in, possibly middle school, and I never thought that the teachers would be involved in your life as much as you described. I already feel this in elementary school, when less-gifted students in another class get a higher grade than me when I score [i]slightly[/i] less on a harder, more advanced assessment. As you said, they "nailed" me and I haven't emotionally developed a rock-hard emotionless feeling to last for quite a while.

Of course, there isn't a lot of fancy vocabulary that is difficult to master in elementary school. Nonetheless, I have a slight ability to have a good sense on what things should be. I'm involved with every club after school, which are difficult (especially Black Saga!), there are many "Bees" (a.k.a. Spelling Bee, Geography Bee), classwork assignments, homework, [b]science fair*[/b] and socializing with friends and classmates.

*About science fair. Last year, it was tragic. I wasn't really counting on winning; after all, you don't win anything in fourth grade. But to see such...such insolent people winning and gloating here and there, I was completely shattered. My emotions got carried away, and the bathroom became my permanent home (for two minutes). So, I thought, "Why not get help from my brother and his friends? That'll help me win!" and indeed, I got that help from them. I was so sure, because after attending science fair night, the competition was blown away with my board.

During the announcements to see who would receive Top 10 spots, I was pretty happy for my friends, but still waiting for my name to be called. After every person, my hands were shaking and I was growing nervous. Why wasn't my name being called for the Top 10? I wasn't  [i]too[/i] arrogant, was I? The judges can't hold anything against you, right? I'm pretty sure they can't, but they did give some of the kids who needed help learning English and can't get along very well with others the Top 10 spot. Of course, I was outraged and shocked.

From there on out I was emotionless, stony, hard, cold, like a robot. No feelings. No heart. I had no choice anymore. If life was that unfair, think of the [b] real world [/b]. But after consulting with friends, resuming normal life on the cybernetwork and just chilling around made me feel better.

There are things you want, but things you can't get. Teachers will try to challenge you and think of you so highly that their standards are practically in the sky and you have to die to reach them. It's sad to think of the changes I've seen come and gone in the same school for my whole life, me starting at the same school in Pre-K and my brother starting in the same school since Kindergarten (he's now in ninth grade).

[quote="7h3.D3m0n.117"]Being a good student with a "secret AoPS math life" is the modest way to get by.[/quote]

This is good advice. Letting the world (or at least the school) know of your AoPS life may not be the best decision you make... but regarding the vocab, I think it's more of the teacher's fault than yours. Sure, advanced vocab may make other students feel bad but from your post, it doesn't seem like you are getting carried away with it.

@D3m0n Shad0w: You probably lost points for using extensive vocabulary in your research report (you used my college biology textbook, remember?) so they probably thought you cheated.

@Buzzer11: Yeah telling your high school you're a math nerd won't help you  :P .

Aye, getting carried away [b]will and always[/b] lead to arrogance. You could slip up, because arrogance is the banana peel or the bar of soap just sitting there. You could avoid it by previewing the other chapters, brushing up on some old units, or even moving on to another subject. Teachers do socialize, don't they? They're bound to know you're doing excellently in say, math, but failing in Biology.

@7h3.D3m0n.117: I did not cheat, just used you as a "resource" that others wouldn't think or have the heart to use. One girl who got 3rd place used her mom, who was a real scientist.

I  :( feel :| comforted :)!
He. Yeah, I modified that from someone's sig...

I guess I can just live my double life happily and avoid arrogance.
Thanks for all the advice. :D

:dry: And I'm not failing any of my classes...

I think I need a blog.

And *small chuckle*....a kid in elementary school? using a college textbook for a science fair, by then, i think they would think that you've been plaigerizing. I know that happened to me before. Apparently, when you don't do a baking soda volcano for a science project, questions arise.  :ninja: :lol:

@illusiat
Hmm. I wonder how the "1st smartest person" feels at your school. Then I wonder how all the other, i.e., "3rd smartest person" and below feel. Ouch.  :o

How would you be able to judge, anyways? :huh:

...

This will take time, I was much the same in middle school...but you have to be comfortable with your abilities and love what you're doing for the sake of doing it, not for the sake of getting any kind of recognition (this isn't a direct response to the OP, but rather to some other things that were said).  I remember in my freshman year, I made it a big goal to qualify for the USAMO, and almost judged my worth by my competition scores.  By the time I had gotten to junior year, on the other hand, I was more comfortable with my mathematical abilities, and took competitions more as an opportunity for fun rather than a way to measure myself.

Nowadays, although it's generally accepted that I'm far away the best (if you can define "best") math student at my school, I don't try to measure mathematical ability.  At that point, it's not so much about competing with the next person and trying to score the highest.  Instead of being "best" or "second best" or whatever, you all love the subject and pursue it together.  These fields become even more team-oriented as you move up into college and beyond (I don't know if you're looking that far).  You'll probably encounter some sort of camp or research program in your high school years: I got some experience with the team nature of math at PROMYS.

So although it's great to pursue things you love, you don't necessarily have to be the best in every competition, and you shouldn't feel too bad if competitions don't go your way.  If you do badly, the outcome is that your abilities haven't depreciated, you've learned something interesting, and all you've lost is a chance to add another entry to your list of achievements you use to qualify yourself.  Also I should say some of the kids at my school don't do very well on competitions, but are extremely smart.  I share math problems with them all the time, and they know me not just as a person who loves and is good at math, but also as a helpful person who is open-minded and willing to share knowledge in mathematics.

I'm not saying competition is bad: it's fun, and drives you to do well.  But don't take it so seriously that you alienate your competition: remember, they're your peers, and it becomes much less pressured and more fun when it's a friendly competition rather than a fight to the death.

As far as dealing with teachers goes...I can guess a lot of your peers say that you are smarter than the teacher.  The teacher probably [b]does not[/b] like this.  Instead, take a position as the open-minded student, only seeking to learn more.  I'm not saying to suck up to the teacher, but just swallow your ego and take the stance of "I'm just trying to learn from you"...from the position you're in now, it may help to approach the teacher, apologize (even though you aren't at fault in any way, the teacher will feel better and will be more forgiving), and try to find out what the she wants.  If you're in class throwing terms around (even if it's unintentional), it disrupts the teacher's plan and makes it appear to the rest of the class that you know more than the teacher.  When you're talking, students will tend to focus their attention on you rather than the teacher because you are known to be smart, and the teacher knows this.  I'm sure it's just a misunderstanding and the teacher feels the class is unstable with someone like you around (this is bound to happen in a structured sort of class in middle school level): it's not like she randomly wants you to suffer.

When you get to high school (and even more in college, I'm sure), you'll get teachers who are more involved in their subjects.

[quote="The_Scintillator"]
Putting that aside, though, we got the AMC 10 and 12 scores. (Yes, this is a bit dated, and I apologize.) I scored dismally. I don't mean 144 and didn't get a perfect score--I mean outright failed. I was unable to function at all that day, and scored, well, well below 80 on both.
[/quote]

I'm not trying to be harsh or anything. But are you sure you're not rationalizing?

[quote="The_Scintillator"]And *small chuckle*....a kid in elementary school? using a college textbook for a science fair, by then, i think they would think that you've been plaigerizing. I know that happened to me before. Apparently, when you don't do a baking soda volcano for a science project, questions arise.  :ninja: :lol:[/quote]

Well, the college textbook was more reliable to use. And, a college textbook wouldn't be as straightforward as what I put in my research report. Even my teacher said it was ok, but I guess it wasn't.

One more thing: Don't worry about showing your teachers how smart you are. I have found that a student's brightness shines through without the need for help from the student him or herself. AMC scores, for one. Even if your scores this year weren't that great, in your opinion, you'll get more chances. These are the things that show your teachers, as well as your peers, how intellegent you are, not the vocabulary you use in class or how much you talk about higher levels of mathematics.

Still feeling comforted. :)
But yeah, I guess I've only got a quarter left with that teacher, and she can only continue to influence me for a few more years....she knows people at the high school I'll be attending.
I'm hopeful that she won't know people at the college I'm attending. Hopefully.
I'm also not particularly worried about showing my teachers how smart I am. I don't ask for that kind of attention--it drives me insane. I don't want recognition. I mostly just want teachers to leave me alone. I just want to do what I do. And if failure comes along with it, if recognition comes along with it, so be it.
Eh. I'm rambling. Forgive me.

@SimonM
I might be rationalizing about my scores--I knew most of the things on there, but applying them was more difficult. [i]I[/i] think it was due to the fact that I was hyper about my birthday, the four gifts I had just received, and the amount of candy I had ingested--all without regard for the fact I was taking the AMC 10 and 12 that day, since I had forgotten. I almost didn't go. Ah, well. I've pretty much gotten over my scores by now. :D

@K81o7
I've tried the approach with the whole "I just want to learn from you". I honestly do. I want to learn from them, anything and everything they can teach me. It doesn't work. :(
But the advice is much appreciated. Thank you.

@buzzer11
I should hope I get more chances. Thanks. :)

@D3m0n Shad0w
Heh. Enough said. :lol:

Confront the teacher. Tell her that she made you feel bad, and that you feel she purposely tried to break your spirit. Tell her how immature it is of a teacher to feel the need to bring down a student with sarcasm and malicious comments. And if she feels there is something wrong with your attitude, she has handled it very poorly if she almost made you cry. She should and probably will feel ashamed.

Well, I don't know how old you are, so I guess it can be really hard to confront her since you're not an adult, but at least it's what I'd like to do

[quote="The_Scintillator"]I'm also not particularly worried about showing my teachers how smart I am. I don't ask for that kind of attention--it drives me insane. I don't want recognition. I mostly just want teachers to leave me alone. I just want to do what I do. And if failure comes along with it, if recognition comes along with it, so be it. [/quote]

There could be full potential underneath your skin. Having teachers recognize a talent (I don't know your age, so if your in high school this could very well not work) that you don't usually show will allow them to involve you with clubs, extra classes, and private tutors. It could work well with college.

A bit on advanced vocabulary, though: math isn't about vocabulary, and sometimes people can get carried away with using very specific terms that most other people don't understand. I wouldn't call it "your fault" if you do that, but you should be aware that using such vocabulary can sometimes be more of a hindrance than an aid to communication, even when talking to other people who are good at math. I for one didn't know what an oblique triangle was.

That doesn't like a very nice teacher. I'm in middle school too, and my geometry teacher also scolded me when I overly expressed my knowledge of math, but in private, and acknowledged that I was indeed a "bright student", so I didn't feel depressed afterwards. A teacher should never try to humiliate a student, especially when he or she does not fully understand why a student acts that way (I think your teacher hasn't interacted with too many advanced learners).
     I think it's a bit sad that mathematically gifted students are often overlooked, whereas those best in other subjects, especially sports, can easily show off their talent without being criticized. I have also been thought as a braggart before, but mostly by other students rather than teachers (most of my teachers were pretty easygoing). If I really needed to show off something, I would scribble it on paper, show it to the person and quickly cross it away.

In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are. I learned this in, what, 9th grade? Now I do this on a subconscious level, and life is much easier.

[quote="serialk11r"][quote="1=2"][quote="Lazarus"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]

At least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]
Dude wtf that appears in english class...[/quote]

I mean on a daily basis.

I don't think this should be a question of whether advanced vocabulary is allowed or not. In my opinion, high level vocabulary can be used, but with a limit in mind. If you start stringing your sentences with mathematics jargon, you probably went too far. :)

wow.. i really need to get something out right now...
it relates to the topic..

One day in Geometry, we were talking about right triangles and solving for angles. I raised my hand because I knew a much simpler way that leads for less error. My teacher, who knew that I was studying PreCalc at home said, "Now you're not relating to the Law of Sines or anything like that. right?"
That just blew me off. All because I knew some more than the class should know, she jumped to the conclusion of thinking that I was going to show off again. I maintained my composure and said, "No." and the stupid thing is that the problem didn't even need to use the freakin law of sines.

Also!

my teacher is one of the most unprepared teacher of all times. She loses her transparencies, she grades papers so slowly (and her excuse is that she gives out partial credit, but the partial credit she gives out is like half a point off for every single question).
Also, we were doing standardized test preparation. she knew she didn't have to go over it, so we just did really quick problems that the whole class solved in 2 minutes. Guess who gets all three questions wrong. my teacher. She claims that she has an off day. By now, i'm really sick and tired of her..
Oh. and I taught point symmetry to the class. She had NO FREAKING IDEA WHAT THE HECK POINT SYMMETRY WAS!!

My math teacher claims to like me. So far only two math teachers i've had at school liked me. the rest hated me for being accelerated.
What's the big deal?
shouldn't they be happy that they have bright students in their class?

Sure, with genius comes arrogance, but it's at least controlled.
I could doze off in class if I wanted to, but the kid behind me keeps asking me questions, and i'm right next to the overhead cart, which means that she could catch me doing anything.

@Scintillator, I am very sorry if I am in a better situation than you are, but yes, We all feel that way...

My favorite subject right now is math.
My least favorite subject right now is history.
My favorite teacher is my US History teacher.
My least favorite teacher is my math teacher.

I'm so glad that I have AoPS to resort to.

[quote="1=2"][quote="Lazarus"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]

At least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]

...

>_>

[quote="1=2"][quote="Lazarus"]In life, you have to dumb down your vocabulary, mannerisms, logic, everything, when you're around people who aren't as interested in pointlessly technical things as you are.[/quote]

At least until you get a job that let's you have an advanced vocabulary. Once you get a really good job, you can say words like rapport.[/quote]
QFT at lazarus's reply.
Seriously, ?????????

Shouldn't this be in a blog though?
Meh, the worst thing a teacher who hated me did was give me a C as a grade.


*)Lemma. Among any $ 2n \minus{} 1$ integers one choose $ n$ integers whose sum is divisible by $ n$.
Proof. If lemma holds for $ m,n$ then lemma holds for $ mn$.
So we have prove for $ n$ is prime. Let $ 2n \minus{} 1$ integers are $ a_1,a_2, . . . ,a_{2n \minus{} 1}$.
Indeed, assume any $ n$ integers whose sum isn't divisible by $ n$, then apply theorem Fermat:
$ \sum_{1\leq i_1 < i_2 < . . . < i_n\leq 2n \minus{} 1}(a_{i_1} \plus{} a_{i_2} \plus{} . . . \plus{} a_{i_n})^{n \minus{} 1}\equiv (^{2n \minus{} 1}_n)$(mod n).
Impossible, hand-left is divisible by $ n$, hand-right isn't divisible by $ n$.
*)We have $ 2n \minus{} 2$ integer:
$ (0,0, . . . ,0,1,1, . . . ,1)$ satisfy no $ n$ integer whose sum is divisible by $ n$.
*)So $ n$ smallest in problem is $ 35$.

Hello, I have two questions about your proof...
Why is the following true?[quote="Thjch Ph4 Trjnh"]If lemma holds for $ m,n$ then lemma holds for $ mn$.[/quote]
Also why is the left hand side divisible by $ n$ if $ n$ is prime?[quote="Thjch Ph4 Trjnh"]$ \sum_{1\leq i_1 < i_2 < . . . < i_n\leq 2n \minus{} 1}(a_{i_1} \plus{} a_{i_2} \plus{} . . . \plus{} a_{i_n})^{n \minus{} 1}\equiv \binom{2n \minus{} 1}{n}\pmod{n}$[/quote]
Thank you,

[quote="moldovan"]Determine the smallest natural number $ n$ such that among any $ n$ integers one can choose $ 18$ integers whose sum is divisible by $ 18$.[/quote]

sorry, but i could not understand your question.
according to the statement of the question, i am allowed to choose any $ n$ naturals out of which i choose some $ 18$ numbers whose sum is divisible $ 18$.
if so then $ n$ can be $ 18$ itself as i can choose $ 18$ multiples of $ 18$ and their sum is obviously divisible by $ 18$.

there is something missing..(either in the question or my comprehension)..please clear it

[quote]Hello, I have two questions about your proof...
Why is the following true?[quote="Thjch Ph4 Trjnh"]If lemma holds for $ m,n$ then lemma holds for $ mn$.[/quote]
[/quote]
Assume lemma holds for $ m,n$. Let $ 2mn\minus{}1$ integers are:$ a_1,a_2, . . .,a_{2mn\minus{}1}$.
Among $ a_1,a_2, . . .,a_{2n\minus{}1}$ has $ n$ number whose sum is divisible by $ n$, assume is $ a_1,a_2, . . . ,a_n$.
Among $ a_{n\plus{}1},a_{n\plus{}2},. . .,a_{3n\minus{}1}$ ...
....
So have $ (2m\minus{}1)$ $ n\minus{}$tuples $ (a_{i_1},a_{i_2},. . .,a_{i_n})$.
Donote $ b_i\equal{}\frac{a_{i_1}\plus{}a_{i_2}\plus{}. . .\plus{}a_{i_n}}{n}$ so $ b_i\in Z$.
We obtain $ (2m\minus{}1)$ $ n\minus{}$tuples $ b_i$.
So have $ m$ numbers of these whose sum is divisible by m.
Completes.

Hi, can someone post a clearer proof of the generalization? Thanks :)

Hi, the generalization actually cannot be proven with elementary means I believe; see [url=http://en.wikipedia.org/wiki/Zero-sum_problem]here[/url].

As mavropnevma sir says
[quote="mavropnevma"]Erdös-Ginsburg-Ziv theorem: [color=#0000FF]Out of $2n-1$ integers there exist $n$ with sum divisible by $n$. [/color]
the general proof is elementary, but kind of complicated;[/quote]

in this thread British MO 2000  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=592776

This is a very similar but much simpler question

[quote=Thjch Ph4 Trjnh]
$ \sum_{1\leq i_1 < i_2 < . . . < i_n\leq 2n \minus{} 1}(a_{i_1} \plus{} a_{i_2} \plus{} . . . \plus{} a_{i_n})^{n \minus{} 1}\equiv (^{2n \minus{} 1}_n)$(mod n).
.[/quote]
How do your prove that the LHS is divisible by n?

Actually it has been a long time since the discussion, but anyway.
So the problem can be generalized for every k positive integer and here is why:
[b][i][size=150][size=110]Claim1[/size][/size][/i]:[/b]  It works for any $p \in P$.
Indeed, it is straightforward that the answer is at least $2p-1$, otherwise there is a tuple $(0,0,0,0...0,1,1,1...1)$, where $0$s and $1$s appear exactly $p-1$ times, the sum is never divisible by $p$.
So it is left to prove that it is sufficient. For the sake of contrary let $a_1, a_2, ..., a_{2p-1}$ be the numbers, so that the $(a_1+a_2+...+a_p)$ is not divisible by $p$ for all permutations of $a_i$. There are exactly $\binom{2p-1}{p}$ ways to pick $a_i,$ it is clear that $\binom{2p-1}{p} \not\vdots p$.
Let's count the coefficients of $a_1a_2...a_j$ in $\sum{(a_1+a_2+...+a_p)}^{p-1}$ . From the one hand, by Fermat's, $LHS$ is just $1+1+...+1$  $\binom{2p-1}{p}$ times, meaning that $LHS \not\vdots p$. From the other, each term $a_1^{e_1}*...*a_j^{e_j}$ for $e_1+...+e_j=p-1$ appears once for $(a_1+...+a_j+...)^p$, so there are $\binom{2p-1-j}{p-j}$ appearances of $a_1^{e_1}*...*a_j^{e_j}$, which is divisible by $p$, the contradiction.
[b][i][size=150][size=110]Claim2[/size][/size][/i]:[/b] if it works for $k_1, k_2$, then it works for $k_1k_2$
Let's pick $2k_1-1$ among $2k_1k_2-1$, by hypothesis, there are $k_1$ numbers, whose sum $\vdots k_1$. Return the rest $k_1-1$ numbers and do this operation again and again until we get $2k_2-1$ groups where each group's sum $\vdots k_1$. let $b_i=s_i/k_1$, then $ b_i \in Z$. For $b_1, b_2, ..., b_{2k_2-1}$ we have $k_2$ numbers whose sum $\vdots k_2$, so then the union of all these consists of $k1*k2$ terms, the sum is divisible by $k_1*k_2$ and we are done
Answer: $2*18-1=35$


A substance contains $40 \% C$, $53.33 \% O$, $6.66 \% H$. $13.39mg$ of this substance has volume $10mL$ in normal conditions.

Find the formula of the substance.

I get some "leftovers" when computing. Am I doing something wrong?

I'm rly not sure but I got $CHO_2$ just assuming it's out of 100 g......or is it suppose to be out of 13.39 mg?.....I dont' know how the mg and vol. fit in....though

I'd focus on the weight ratios.  Since we are looking for a formula, we don't care how much is really there.  Just divide the wt% by mole weight and end up with 40% C/ 12 = 3.33; 53.33% O/16 = 3.33; 6.66% H/1 = 6.66.

This leads me to the mole ratio of C:O:H of 1:1:2, which leads me to the formula of CH2O.

Now, why would we get volume and mass information?  They must want to give us an ability to calculate the molecular weight (from mass and number of moles from PV=nRT).

I'll leave the rest of this to you.

PV=nRT?  We're probably not working with gases, as the stuff is more dense than water.

I'm not sure where we can get the molecular mass.

It's probably some sort of monosaccharide, as they have the $CH_2O$ empirical formula.  I don't know which one in particular; it could be glucose, galactose, fructose, ribose...

I guess I could look for a table of densities.

I was also thinking something like glucose or fructose. I don't know if you can really find the molecular formula from the density alone (without a table of densities) for solids and liquids.

If water has a density of 1 g / 1 mL, and the lowest density liquid has a density of 0.5 g/1 mL, then the compound must be a gas if 0.013 g occupies 10 mL.  Formaldehyde is the only possibility.


Find, step by step, second derivative of function:

$ f(x)\equal{}x^{4}(1\minus{}x)$

Please refer to the second announcement in this sub-forum.

[hide]$ f(x)\equal{}x^4\minus{}x^5$

$ f'(x)\equal{}4x^3\minus{}5x^4$

$ f''(x)\equal{}12x^2\minus{}20x^3$[/hide]


A half-cylinder of glass has radius $R$ and index of refraction $n$. A ray of light lies in
the plane of the figure as shown, and it is incident on the plane face at an angle $\theta$. It goes
on to hit the curved surface; for what range of incident positions on the flat surface will the
light be able to leave the cylinder when it first hits the curved surface? Let the position
that it hits the flat surface be $x$ as measured from the centerline of the flat. For what
combination of $n$ and $\theta$ will this range extend over the whole flat surface $(-R<x<R)$?

${\text{DFA}=\tan^{-1}\left(\frac{x \cos (\varphi )}{\sqrt{r^{2}-x^{2}\cos^{2}(\varphi )}}\right)}$ - You can ask for it on a geometryforum

We dont need to do that, but. just to show
[quote]${n_{1}\sin (\theta )=n_{2}\sin (\varphi )}$
${n_{2}\sin (\text{DFA})=n_{1}\sin (\psi )}$

after Simplify
${\sin (\theta ) n_{1}=\sin (\varphi ) n_{2}}$
${r \sin (\psi ) n_{1}=x \cos (\varphi ) n_{2}}$[/quote]

${r^{2}-x^{2}\cos^{2}(\varphi )\geq 0\Rightarrow x^{2}\leq \left(\frac{r}{\cos (\varphi )}\right)^{2}}$

the argument is The angle DFA need to exist.


[quote="yildobekir"]If a square ABCD and has side a and P a point in ABCD square,$ PA \equal{} x,PB \equal{} y,PC \equal{} z$.PROVE THAT;
$ (a^{2} \plus{} y^{2} \minus{} z^{2})^{2} \plus{} a^{2} \plus{} y^{2} \minus{} x^{2} \equal{} (2ay)^{2}$[/quote]
[color=blue][size=150][b]hi
I think u mean
(a^2+y^2 -z^2)^2 +(a^2+y^2 -x^2)^2 = (2 a y )^2
If that I will post acomplet solution
mrmohamed[/b][/size][/color]

[color=red]Hi all
here is my idea[/color]

[url=http://www.servimg.com/image_preview.php?i=457&u=12899977][img]http://i19.servimg.com/u/f19/12/89/99/77/almonk11.gif[/img][/url]
with my best wishes
mrmohamed


number theory
least common multiple
floor function
inequalities
number theory solved

I found this problem in a Russian book Selected problems from AMM.
There isn't solution in that book (the author claims that it is very hard
problem), but it seems to me that the statement doesn't hold. It goes like this

"Let $a_1<a_2<\dots<a_n\leq2n$ be a sequence of positive integers. Then

\[\min[a_i,a_j]<6(\lfloor n/2\rfloor+1),\]

where $\, [a_i,a_j]\, $ denotes the least common multiple of $a_i$ and $a_j$. This is
the best possible estimate."

This is even better than the best possible estimate since for $n=4$ we can
take numbers 5,6,7,8 and $\min[a_i,a_j]=24>6\cdot3$. However, one can prove
that for $n>4$ the statement almost holds, then we have

\[\min[a_i,a_j]\leq6(\lfloor n/2\rfloor+1),\]
 (I replaced $<$ with $\leq$) and the equality occurs for numbers $n+1,n+2,\dots,2n$ (it is not so hard to prove).
On the site 20000 problems under the see I found the same problem, with the
same formulation. I wonder if I am misunderstanding something here. Does
anyone have access to AMM? This was problem AMM3834 by Paul Erdos.

As you notice, the strict inequality is incorrect, and has to replaced by a large one, and only holds for $n \neq 4$.
Since $n=1,2,3$ can be checked by hand, let's assume that $n \geq 5$.

Now, here is the published answer from C. R. Phelps :

For any $n$, consider first the sequence $a_i = n+i$.
The first even number in this sequenec is $2( [ \frac n 2 ] + 1)$ and the number $3( [ \frac n 2 ] + 1)$ is also in the sequence, so min$ lcm(a_i , a_j) \leq 6( [ \frac n 2 ] + 1)$ for this special sequence.

For any other sequence $a_1 < a_2 < \cdots < a_{2n} \leq 2n$, for any $a_i$ there exists an integer $k_i \geq 1$ such that $n < k_ia_i \leq 2n$. If $k_ia_i = k_ja_j$ for some distinct $i,j$ then $lcm (a_i,a_j)$ divides $k_ia_i$ and therefore is not greater than $2n$.
Otherwise the $k_ia_i$ are all distinct ($i=1, \cdots, n$) and form exactly the special sequence in some order, so that the above inequality holds.

Moreover, the bound $6( [ \frac n 2 ] + 1)$ is best possible since it is the minimum for the special sequence for each $n$ : Suppose that for some $n$, for the special sequence $lcm(a_i,a_j) =M < 6( [ \frac n 2 ] + 1)$ where $a_i < a_j$. Then $a_j \geq n+3$ so that $a_j > 2( [ \frac n 2 ] + 1)$ with $M = ma_j$ for some integer $m$. Hence $m=2$, and $M=ka_i = 2a_j$, so that $k \geq 3$.
If $k=3$ then $a_i < 2( [ \frac n 2 ] + 1)$ with $a_i$ even, a contradiction.
If $k \geq 4$ thn $a_i < \frac 3 2( [ \frac n 2 ] + 1)$ so that $a_i < n+1$ also impossible. Hence $M < 6( [ \frac n 2 ] + 1)$ is impossible.

Pierre.

sbos, does that site with 20000 problems work? I've been trying to browse it for months, getting only stuff like "layout not found" (whatever that means). It's a great resource, but I thought it had too many problems to be useful. Could you give me the address? (maybe ou're talking about some other site?)

Now I have troubles with that site too. I found the above problem long time ago when the site worked normally.
Thank you pbornsztein!


calculus
integration
number theory
Diophantine equation
superior algebra
superior algebra unsolved

Let d be a square free integer.

Prove that if (x^2)-d(y^2)=-1 has an integral solution,
then (x^2)-d(y^2)=1 also has an integral solution.

Think about $ (x \plus{} y\sqrt {d})^2$.

$ (x\plus{}y\sqrt{d})(x\minus{}y\sqrt{d})\equal{}\minus{}1$.  Square both sides.  See what you get.

Ok. I squared it and it simplified to the second equation!

So does this simply mean that any integral solution to equation 1 is also an integral solution to equation 2?

Not quite.

[b]Definition:[/b]  Given a number $ z \equal{} x \plus{} y \sqrt {d}$ where $ x, y$ are integers, let $ \bar{z} \equal{} x \minus{} y \sqrt {d}$ be its [i]conjugate.[/i]

[b]Definition:[/b]  Given $ z \equal{} x \plus{} y \sqrt {d}$, let $ N(z) \equal{} z \cdot \bar{z} \equal{} x^2 \minus{} dy^2$ be its [i]norm.[/i]

[b]Lemma:[/b]  $ N(zw) \equal{} N(z) N(w)$.

The first Diophantine equation means $ z \equal{} x \plus{} y \sqrt {d}$ has norm $ \minus{} 1$.  That means $ z^2 \equal{} (x^2 \plus{} dy^2) \plus{} (2xy) \sqrt {d}$ has norm $ 1$, so $ (x^2 \plus{} dy^2, 2xy)$ is a solution to the second Diophantine equation.


The integer sequence $ a_i$ is defined by $ a_0 \equal{} m, a_1 \equal{} n, a_2 \equal{} 2n\minus{}m\plus{}2, a_{i\plus{}3} \equal{}3(a_{i\plus{}2} \minus{} a_{i\plus{}1})\plus{} a_i$. It contains arbitrarily long sequences consecutive terms which are squares. Show that every term is a square.

It is easy to show, that $ a_i\equal{}i^2\plus{}(n\minus{}m\minus{}1)i\plus{}m$.


MATHCOUNTS
AMC
AMC 8
AMC 10
AMC 12
USA(J)MO
USAMO

Greeting,

Maybe in next week or before, I'll post thirty questions for beginners to do. Answers must be send to me by P.M. with explanation, meaning if you only put the answer, you won't get a credit at all.  Questions will cover from Mathcounts to maximum of AHSME.  No questions will be exactly same with any of past years' competition so, there is no way you're going to cheat.  Problems, however, will be similar.

For 25 questions, it's like this:

5 MC
5 AMC8
5 AMC10
5 AMC12
5 AHSME

Total point is 100, meanig correct question worth 4 points each. Unanswered question is given 1.5 points while incorrect answers are given no point at all.  Plus, any answes with no expalnation is absolutely disqualified so, make you sure you do right.
Test will be on for exactly 3 days, and all answers must be submitted before I draw up the questions.  After two or three days, there will be rank of people from first to last. Although it's for beginners, intermediater and advancer may join too.  Unlike Rep123max's getting started contest, it's actually a competition in here except there isn't going to be prize (sorry!).

Silverfalcon

[i]Edited: I erased last part about next level competition[/i]

Are calculators allowed?  And, I assume that it goes without saying that work needs to be shown?

Calculators are perfectly allowed. For showing work, calculations doesn't have to be shown like this:

[tex]5x = 20[/tex]
[tex]x   = 4[/tex]

But what I meant by explanation is that whenever somebody sees your answer, they must say, "oh, that's the way you do it"

So, try to put some words that match with your answers to get full credit.

According to what I know, CMO is Canadian Mathematics Olympiad. I think it's in level of AMC10.

CMO is far harder than AMC10. It's the Canadian equivalent of the USAMO, but the questions are easier.

Unlike having all questions at once, I'll post 5 questions from each level.

Number one from MC level:

1. 6372 has 24 factors. What is the sum of the largest and smallest factor?

Number two from AMC8 level:

2. Chord with length of 140cm is drawn inside the circle, and it's 48cm away from the center.  Then what is the area of circle? Write your answer in terms of [size=150]:pi:.[/size]

Number three from AMC10 level:

3. [tex](x^2+z^3)^4)^5)^6)^7[/tex] has degree of what?

Number four from AMC12 level:

4. Prove that [tex]x^7-x^6-3x^5+3x^4+3x^3-3x^2-x+1[/tex] has two solutions 1, and -1 by factoring it out.

Number five from AHSME level:

5. [tex]10!9!8!7!6!5!4!3!2!1![/tex] has how many factors?

Please don't discuss them under here. If you think you have the answer, please p.m. to me with them, and they are only accepted by Wednesday, 5/19.

My questions aren't totally for intermediaters.
If you understand the questions and pretty much terms of math, you are able to do most of the questions.
But it's true that most of them are pretty hard for beginners...

anytime on the 19th? So like, could I PM u in the evening, or do you want it before then?

Rep, in my opinion these are easier than your questions. =)

In fact, ones I just posted are the ones that are pretty "easy"

[quote="BHorseMath"]Thats confusing silver falcon.  No one send me answers so everyone turn them in???  Do you want them yet or not?  You are contridicting yourself or else I am very tired and just confusing myself.[/quote]

Pretty sure he meant "no one sen[b]t[/b] me answers yet."  Give him a break ;-)

[quote="BHorseMath"]Thats confusing silver falcon.  No one send me answers so everyone turn them in???  Do you want them yet or not?  You are contridicting yourself or else I am very tired and just confusing myself.[/quote]

Sorry about my bad wording.  But I hope you understand that you hadn't p.m.ed me with your rest of solution in round two since like last two weeks.  If you don't p.m.ed me soon, your score would be 4, point for only question you got right.

Sincerely,

P.S. Thanks probability 1.01

Did I send you a part solution silverfalcon?  if you can remind me what problem I will send the rest.

Finally!!! Last five questions are made:

Number twenty-one from MC level:

In this arithmetic sequence: 1,4,7..... What's the 26th term?

Number twenty-two from AMC8 level:

Word Latex has 120 different combinations.  Word Telax is nth number. Define n.

Number twenty-three from AMC10 level:

How many different values of b's are possible in expression [tex]x^2+bx-15[/tex]to factor? (Trust me, it can be factored).

Number twenty-four from AMC12 level:

For this question, listen very carefully.  Segment AB and AC are tangent to circle Z, and they are same length.  Square of the length of the AB is equal to product of segement AD, which is secant to circle Z and pass point E, whch is the one of outer point of the circle Z.  Chord ED is bisected perpendicularly by diameter KP at J.  If segment AC has length of 8, AE has length of 4, and KJ has length of 16, find the length of JP.

Number twenty-five from AHSME level:

Rationalize the numerator in following expression:

[tex]\displaystyle \frac {\sqrt {x+h} - \sqrt x}{h}[/tex]

Good luck! These five questions are considered hardest five questions of all and I hope you all do pretty well.  Especial luck to our two finales, Chinaboy and Ragingg.....

Silverfalcon, can you check your PM box, cuz i sent you my answers 3 days ago!

[quote="Silverfalcon"]Number twenty-two from AMC8 level:

Word Latex has 120 different combinations.  Word Telax is nth number. Define n.
[/quote]
Please clarify what  you mean by the nth number and the "Word Telax"

[quote="Silverfalcon"]
Number twenty-three from AMC10 level:

How many different values of b's are possible in expression [tex]x^2+bx-15[/tex]to factor? (Trust me, it can be factored).
[/quote]
Do you mean integer values of b? Cuz there are infinite values of b by which you are phrasing it.

For the first question, it's just word "Telax" like "Latex," "Texla," etc... For the second question, it meant integer value to factor the trinomial into two binomials. Here's hint:

[hide]There isn't really infinite number of solutions because remember, in FOIL process, the middle term is sum of in and out like this:

[tex](a+2)(a+3) --> 2a+3a --> 5a[/tex]So, think on how many factors are in the last term..[/hide]

I assume that you mean they are ordered in alphabetical order.  That is probably the part that confused some people.

[quote="Silverfalcon"]Number twenty-four from AMC12 level:

For this question, listen very carefully.  Segment AB and AC are tangent to circle Z, and they are same length.  [b]Square of the length of the AB is equal to product of segement AD, which is secant to circle Z and pass point E, whch is the one of outer point of the circle Z.[/b]  Chord ED is bisected perpendicularly by diameter KP at J.  If segment AC has length of 8, AE has length of 4, and KJ has length of 16, find the length of JP.
[/quote]

You totally lost me there. How can there be the product of a segment and a point.

I know my word is confusing but try this way:

[hide]What are the two other ways of power of a point theorem other than two chords intercepting? Now think about it. K? It should be a big help if you see which one to use[/hide]

Contest has been ended.

Winner: theone833

Answers will be soon posted.

Sincerely,

Silverfalcon

Solutions: Odds first

1. 6372's largest factor is  6372 and smallest is 1. Therefore, it's 6372+1 or 6373.

3. Simply 3*4*5*6*7 = 2520

More will be posted.


algebra
polynomial
inequalities
function
inequalities proposed

If $ a,b,c$ are nonnegative real numbers, then

$ (a)  \ \ \ \  10\sum a^4\plus{}64\sum a^2b^2 \ge 33\sum ab(a^2\plus{}b^2)$;

$ (b)  \ \ \ \  29\sum a^4\plus{}118\sum a^2b^2 \ge 72\sum ab(a^2\plus{}b^2)$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ (a) \ \ \ \ 10\sum a^4 \plus{} 64\sum a^2b^2 \ge 33\sum ab(a^2 \plus{} b^2)$;

[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ $ abc\equal{}w^3$ and $ u^2\equal{}tv^2.$
Hence, $ \sum_{cyc}(10 a^4 \minus{}  33a^3b\minus{}33a^3c\plus{}64 a^2b^2 )\geq0\Leftrightarrow$
$ \Leftrightarrow270u^4\minus{}657u^2v^2\plus{}450v^4\geq55uw^3.$
But $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0$ gives $ w^3\leq3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}.$
Id est, it remains to prove that
$ 270u^4\minus{}657u^2v^2\plus{}450v^4\geq55u\left(3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}\right).$
But $ 270u^4\minus{}657u^2v^2\plus{}450v^4\geq55u\left(3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow190u^4\minus{}411u^2v^2\plus{}225v^4\geq55u\sqrt{(u^2\minus{}v^2)^3}\Leftrightarrow$
$ \Leftrightarrow190t^2\minus{}411t\plus{}225\geq55\sqrt{t(t\minus{}1)^3}\Leftrightarrow$
$ \Leftrightarrow(75t^2\minus{}155t\plus{}81)(21t\minus{}25)^2\geq0,$ which is obviously true.

Let f(a,b,c) = L.H.S - R.H.S
Fix S = a+b+c and T = ab+bc+ca then f(a,b,c) is linear function on P = abc. So, f(a, b, c) achieves its minimum when P max or min. In these cases, two variables are equal. So, it is sufficient to show that
    $ f(a, b, b) \ge 0$
In firts case this leads to $ (5a^2 \minus{} 3ab \plus{} b^2)(a \minus{} 3b)^2 \ge 0$
In second case we come to $ (29a^2 \minus{} 28ab \plus{} 8b^2)(a \minus{} 2b)^2 \ge 0$

By this method, we can prove the generalization:

If $ (2A \plus{} B \minus{} 2C)x^4 \minus{} 2Cx^3 \plus{} 2Bx^2 \minus{} 2Cx \plus{} A \ge 0$ for all x

then $ A(a^4 \plus{} b^4 \plus{} c^4) \plus{} B(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \ge C(ab(a^2 \plus{} b^2) \plus{} bc(c^2 \plus{} a^2) \plus{} ca(c^2 \plus{} a^2))$ for all a,b,c.

Nice, Arqady.
Nice, Namdung.
For the second method, we need to verify also that $ f(0,b,c) \ge 0$ for any nonnegative $ b$ and $ c$. We need to show that $ f(0,b,c)\ge 0$ and $ f(a,1,1) \ge 0$ for all $ a,b,c$.
For instant, the inequality
$ \sum a^4 \plus{} 8\sum a^2b^2 \plus{} 8abc\sum a \ge 5\sum ab(a^2 \plus{} b^2)$
is not true, but the condition $ f(a,b,b)\ge 0$ (or $ f(a,1,1)\ge 0$) is verified.

Thank you Vasc! I forgot the case P = 0!

Note that inequalities of these type are independent in such meaning: no one can be obtained from another one. In other words, we can not say 1) is stronger that 2) or versi versa.

Namdung, your reasoning works for all symmetric homogeneous polynomial  of degree 5 with 3 non-negative variables.

Yes. Indeed.

It is interesting to characterize a simple criterium for A, B, C to have
  $ A(\Sigma a^4) \plus{} B(\Sigma {a^2b^2}) \ge C(\Sigma {ab(a^2 \plus{} b^2})$


Does anyone know how many students there are per room and per apartment?

2 people per room, 4 rooms per apartment.

apparently the group size in each apartment is becoming smaller.  Last year I think there were 8 students, plus a counselor, per apartment, which meant that everyone slept in a double except the counselor, who slept in the common room.

This year, there are six students, so the counselor gets a room too.  Which is good news for me, because I'll be a counselor. =)

I will see some of you there.  I look forward to getting to know all of you!  And if you're unlucky, maybe you'll even be in my apartment/class (I think I'm assisting in computational geo).

[quote="roadnottaken"]Then why are there only three lines for rooming requests?  If we write in more students below the lines, will those requests be honored?[/quote]
so can we write more?

[quote="bpms"][quote="13375P34K43V312"]well i think they already confirmed my room but i wrote down all five people on the three lines, it should be fine[/quote]
:rotfl: :rotfl:[/quote]
i dont get it

I haven't gotten something asking for roommate preferences...

it came with all the information and forms and such.


trigonometry limit integration logarithms

fie f:[0;1] -> [b]R[/b] , definita astfel:
f(x) = arctg(1/x) , daca x apartine (0;1] ; pi/2 , daca x=0

a) Sa se arate ca f e continua  (evident)
b) Sa se arate ca f e derivabila si sa se calculeze f'(x) , pentru orice x din [0;1]
c) sa se calculeze integrala de la 0 la 1 din f(x)dx

[quote="abedhu"]Fie$\ f: [0;1]\rightarrow\mathbb{R}$ , definita astfel:
$\ f(x) =\arctan\left(\frac{1}{x}\right)$ , daca $x\in\ (0,1]$ si $\frac{\pi}{2}$ daca $\ x=0$.

a) Sa se arate ca $\ f$ e continua.
b) Sa se arate ca $\ f$ e derivabila si sa se calculeze $\ f'(x)$ , pentru orice $\ x$ din$\ [0;1]$.
c) Sa se calculeze $\int_{0}^{1}f(x)dx$[/quote]
Mai bine pui problemele pe Avansati Olimpiada, ca aici nu prea te baga nimeni in seama.

Ba eu cred ca aici e locul acestei probleme !

pentru b) aplicam regula lui Hôpital ca sa gasim derivata (daca exista!) in 0
Obtinem f'(0)= -1.
pentru c) ma mai gindesc...sa gasesc o primitiva ... :mad:

Un anume [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=17823]stancioiu sorin[/url] a postat punctul (c) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107358]aici[/url].

$2.\blacktriangleright\ R_{0}(x)\equiv\frac{f(x)-f(0)}{x-0}=-\frac{\frac{\pi}{2}-\arctan\frac{1}{x}}{\tan (\frac{\pi}{2}-\arctan\frac{1}{x})}$ $\Longrightarrow f'(0)\equiv f'_{d}(0)=\lim_{x\searrow 0}R_{0}(x)=-1\Longrightarrow f'(0)=-1\ .$
$f'(x)=\{\begin{array}{ccc}-1 & \mathrm{daca}& x=0\\-\frac{1}{x^{2}+1}& \mathrm{daca}& 0<x\le 1\end{array}\Longrightarrow \boxed{\ f'(x)=-\frac{1}{x^{2}+1}\ ,\ 0\le x\le 1\ }\ .$

$3.\blacktriangleright\ 0<x\le 1\Longrightarrow F=\int\arctan\frac{1}{x}\ dx=x\arctan\frac{1}{x}+\frac{1}{2}\ln\frac{1}{1+x^{2}}+C \Longleftarrow$ $\boxed{\ \begin{array}{ccc}u(x)=\arctan\frac{1}{x}& \Longrightarrow & u'(x)=-\frac{1}{1+x^{2}}\\ v'(x)=1 & \Longrightarrow & v(x)=x\end{array}\ }\ .$
$I=\int_{0}^{1}f(x)\ dx=F(1)-F(0+)$ $\Longrightarrow \boxed{\ \int_{0}^{1}f(x)\ dx=\frac{1}{2}(\frac{\pi}{2}+\ln 2)\ }\ .$


Putnam
integration
trigonometry
college contests

Evaluate

$ \int_{0}^{\pi/2}\eft( \frac{dx}{1+(tan x)^\sqrt{2}}\right)$


Find the trisection points of line joining (-6,2) and (3,8)?
Sorry I have no idea.   :lol:

[hide]
To get from (-6, 2) to (3,8) you would have to move 9 units right and 6 units up in total. < 9,6 >. The trisection points split the whole distance into thirds, so they will be found by moving (9/3)=3 units right and (6/3)= 2 units up twice from the point (-6, 2).

The first trisection point will be (-6[color=red][b]+3[/b][/color], 2[color=red][b]+2[/b][/color]) or (-3, 4)
The second trisection point will be (-6[color=red][b]+3+3[/b][/color], 2[color=red][b]+2+2[/b][/color]) or (0,6)
[/hide]


modular arithmetic
number theory unsolved
number theory

Let $n$ be an integer greater than $6$. Prove that if $n-1$ and $n+1$ are both prime, then $n^{2}(n^{2}+16)$ is divisible by $720$. Is the converse true?

If $n-1$ and $n+1$ are both prime, then $n$ is obviously a multiple of $6$.
Let $n=6p$
$n^2(n^2+16)=(6p)^2[(6p)^2+16]=144p^2(9p^2+4)$.
If $p=5q$, then $n^2(n^2+16)$ is a multiple of $144\times 5=720$.
If $p=5q\pm 1$, then $n=30q\pm 6$. Is is easy to see that either one of $n-1$ or $n+1$ is divisible by $5$.
If $p=5q\pm 2$, then $9p^2+4\equiv 9(4)+4\equiv 0\pmod{5}$, i.e. $n^2(n^2+16)$ is a multiple of $144\times 5=720$.
The converse is not true. For instance, when $n=720$, $n^2(n^2+16)$ is a multiple of $720$. But $721=7\times 103$ is not a prime.

[quote="kennylcc"]Prove that $n$ is a multiple of 6.[/quote]

$n-1\equiv 1(mod 2)$ $\implies n\equiv 0(mod 2)$

Any three consecutive integer has at least one of them is a multiple of three . $n-1$ and $n+1$ are prime , so $n$ must be the one with multiple of three  :D

At Olympiad level would you have to prove that $n \equiv{0} \mod{6}$ or would they take it as "obvious"? I proved it, but my entire solution was two whole sides of A4 (and considering I only started this problem with 20 mins to go, that was bad :)), so it could have used some cutting down.

There is no contradiction $n=4$ since $n>6$...

To Ilthigore's question: you can prove it the following way:
All primes except $2$ and $3$ are $\equiv \pm 1 \mod 6$. Prove: For $\equiv 0,2,4$ they are divisible by $2$ and for $\equiv 3$ they are divisible by $3$.
So when we have two primes ($\neq 2;3$) with difference $2$, then the smaller one ($n-1$) is either $\equiv -1 \mod 6$ and it follows $n \equiv 0 \mod 6$ or it is $\equiv 1 \mod 6$, but then the greater one would be $\equiv 3 \mod 6$ which is impossible.

:blush:  :blush:

Ill read it next time!


In a big city, all the streets have double-hand, but they have only two possible directions (north-south or east-west). During a tour in a car, the same place wasn´t visited more than 1 time, and the tour finished in the start-point. 100 left turns were done. How many right-turns were done, then???? (all posibilities are asked).

I'm guessing the asnswer is either all even integers or all integers divisible by four, I couldn't really understand the question...

[quote="me@home"]I couldn't really understand the question...[/quote]

If it helps for anyone that can translate it better, look for "Nivel Juvenil", problem 4...

http://www.oma.org.ar/enunciados/ptic052.htm

did you get my asnwer to your other question?

I saw it, but I don't know the answer, so I can't tell you if your answer is OK or not...

Does the proof seem okay conceptually?
I was guessing from your other post with that 1997 problem that the two were similar, and from the same test possibly. Is this the case? Also, how come this problem is so easy to crack compared to the 1997 problem?


inequalities
algebra
polynomial
inequalities unsolved

For $a,b,c$ positive reals such that $a+b+c=1$ prove that

$ 5(a^2+b^2+c^2) + 18abc \geq \frac{7}{3}$

This is quite routine.
First we homogenize, making the inequality
\[
5(a^2+b^2+c^2)(a+b+c)+18abc \geq \frac{7}{3}(a+b+c)^3
\]
after expanding it, the ineq becomes
\[
4(a^3+b^3+c^3)+6abc \geq 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)
\]
and this is just the sum of Schurs and AM-GM:
\begin{eqnarray*}
3(a^3+b^3+c^3)+9abc &\geq& 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \\
a^3+b^3+c^3 &\geq& 3abc
\end{eqnarray*}

hi, may i know what is "homogenize"? do you mean making each term into same power?
if so, on the right hand side, why did you choose (a+b+c)^3 instead of (a^3+b^3+c^3) or any other possible choice?
Furthermore, if there's no a+b+c=1 or similar condition, how will you do the homogenization?
Thanks!

[quote="blahblahblah"]Homogenize means to make all terms of the same power.  If all terms are not of the same power, well, then you don't have too many options.  This is useful because AM-GMing, Cauchy, Schur's, etc, applied to terms of degree t will give you more terms of degree t.

a+b+c=1 essentially de-homogenizes the equation, so bill is just reversing that.

On the RHS, he picked (a+b+c)^3 because it is equal to 7, which is convienient, and because it transforms the LHS into a polynomial of degree 3.  After this, all he has left is a big symmetric polynomial of degree 3, and we all know that such inequalities are usually very easy with Schur and weighted AM-GM (and Muirhead if necessary)[/quote]

Thank you!


geometry
circumcircle
trigonometry
angle bisector

In a triangle $ ABC$ with $ AB\equal{}1,\ BC\equal{}\sqrt{7},\ AC\equal{}2$ the interior bisector of $ \angle{CAB}$ intersects $ BC$ at $ D$ and intersects the circumcircle of $ ABC$ again at $ E$. Let $ O$ be the circumcenter of $ BED$.

(1) Find the length of $ DE$.

(2) Find the value of $ \tan \angle{EBO}$.

[hide="number 1"]
Let $ \overline{AD} \equal{} h, \angle CAD\equal{}\angle BAD \equal{} \alpha$

From the cosine law
$ c_1^2 \equal{} a_1 \plus{} h^2 \minus{} 2a_1h\cos{\alpha} \qquad(1)$
$ c_2^2 \equal{} a_1 \plus{} h^2 \minus{} 2a_2h\cos{\alpha}\qquad(2)$

Sub (1) and (2)
$ a_2c_1^2 \minus{} a_2a_1^2 \minus{} a_2h^2 \equal{} a_1c_2^2 \minus{} a_1a_2^2 \minus{}a_2h^2$
$ h^2(a_1\minus{}a_2) \equal{} (a_1c_2^2\minus{}a_2c_1^2) \plus{}a_1a_2(a_2\minus{}a_1)$

Using the angle bisector theorem this simplifies to
$ h^2 \equal{} \minus{}\frac{a_1a_2}{(a_1\plus{}a_1)^2} \plus{} a_1a_2$

With $ a_1 \equal{} 2, a_2 \equal{} \sqrt{7}$
$ h^2 \equal{} \minus{}\frac{2\sqrt{7}}{(2\plus{}\sqrt{7})^2}\plus{}2\sqrt{7} \equal{} \frac{56\minus{}4\sqrt{7}}{9}$

$ \Rightarrow h \equal{} \frac{2\sqrt{14\minus{}\sqrt{7}}}{3}$...

So now we apply the formula $ \overline{AD}\cdot\overline{DE}\equal{}\overline{BD}\cdot\overline{DC}$

$ \overline{DE}\cdot\frac{2\sqrt{14\minus{}\sqrt{7}}}{3}\equal{}\frac{2\sqrt{7}}{(2\plus{}\sqrt{7})^2}$

$ \overline{DE} \equal{} \frac{11\sqrt{7}\minus{}28}{3\sqrt{14\minus{}\sqrt{7}}}$ :)

[/hide]

Yet another mistake, how about now??????

[quote="ocha"]Bah, I always make silly mistakes with your questions.
Here I have put $ \overline{AB} \equal{} 2$ and $ \overline{BC} \equal{} 1$ instead of the other way arround

Is the answer; $ \frac {7}{3}$

EDIT: :rotfl: I made a mistkae in my correction as well. IT'S A CURSE!

Note to self; just work more thoroughly before posting![/quote]

Now You get the answer of Q1.

[hide]Part1. $ \frac {CD}{DB} = \frac {AC}{AB} \Rightarrow CD = \frac {2\sqrt {7}}{3}, \; BD = \frac {\sqrt {7}}{3}$
$ AD^2 = AC.AB - CD.BD = 2 - \frac {14}{9}\Rightarrow AD = \frac {2}{3}$
$ CD\cdot DB = AD.DE \Rightarrow DE = \frac {7}{3}$[/hide]
[hide]2.
$ \cos \angle CAB = \frac {AC^2 + AB^2 - BC^2}{2AC\cdot AB} = - \frac {1}{2} \Rightarrow \angle CAB = 120^\circ \Rightarrow EBC = 60^\circ \; \Rightarrow \Delta EBC$is equilateral
$ \Rightarrow EB = \sqrt {7}$
$ \cos \angle EDB = \frac {ED^2 + BD^2 - EB^2}{2ED\cdot DB} = - \frac {1}{2\sqrt {7}} \;\;\sin \angle EDB = \frac {3\sqrt {3}}{2\sqrt {7}}$
$ EB = 2R\sin \angle D \\
Rightarrow R = \frac {7}{3\sqrt {3}}$ skipping some obvious steps $ \tan \angle EBO = \frac {1}{3\sqrt {3}} = \frac {\sqrt {3}}{9}$[/hide]


function
geometry
geometric transformation
rotation
parameterization
real analysis
calculus

Let $f: \mathbb{R}\to \mathbb{R}$ be a $C^{\infty}$ function. Suppose that $u: \mathbb{R}^{n}\to \mathbb{R}$ is a $C^{2}$ function that satisfies
$\Delta u+f(u)=0$
in the ball $B=B(0;R)$ and $u=0$ on the boundary $\partial B$. Prove that $u$ is radially symetric !

if the solution u is unique, it must be radial, because rotations of u also satisfy the equation.

Is solution unique?

Even for the linear function $f(t)=ct$ there may be a one-parameter family of solutions (eigenfunctions of the Laplacian).

I tried to think this way: if $u$ is a solution, then it is equal to the convolution of $-f(u)$ with Green's function for $B$ (which can be written explicitly, and does have certain symmetry). In other words, $u$ is a fixed point of an explicit nonlinear integral operator. However, this did not lead me anywhere. :(


If $ x$ has to be an integer, well Fermat's Last Theorem, that's how.

Alternatively, show that $ f(x) \equal{} \left ( \frac {3}{5} \right )^x \plus{} \left ( \frac {4}{5} \right )^x$ is strictly decreasing

[quote="williams"]3^x +4^x=5^x.
obviously,x=2,but HOW???[/quote]
[hide="Hint"]
I think calculus will be useful to prove that there is one solution
[/hide]

If we want to use calculus to show strictly decreasing, then
$ f(x)\equal{}e^{x\ln{.6}}\plus{}e^{x\ln{.8}}$
so
$ f'(x)\equal{}\ln{.6}e^{x\ln{.6}}\plus{}\ln{.8}e^{x\ln{.8}}\equal{}\ln\frac{3}{5}\left(\frac{3}{5}\right)^{x}\plus{}\ln\frac{4}{5}\left(\frac{4}{5}\right)^{x}$
Now you can proceed from here.

Why?  It is a basic property of exponential functions that $ b^x$ is strictly decreasing for $ 0 < b < 1$ and strictly increasing for $ b > 1$; see, for example, Rudin's discussion of the construction of the exponential.

[hide]

Taking modulo 4, $ (\minus{}1)^x\equiv 1 \mod4$. So, $ x$ must be even. Let $ x\equal{}2a$. So, now:

$ (3^a)^2\plus{}(4^a)^2\equal{}(5^a)^2$

Since $ (3^a, 4^a, 5^a)$ is a primitive Pythagorean triple,

$ 3^a\equal{}m^2\minus{}n^2$ (1)

$ 4^a\equal{}2mn$ (2)

$ 5^a\equal{}m^2\plus{}n^2$ (3)

From (2), m and n must be powers of 2. Also, one is even and one is odd. The only power of 2 that is odd is 1, and since $ m>n$ due to (1), $ n\equal{}1$. So, $ 3^a\equal{}m^2\minus{}1\equal{}(m\plus{}1)(m\minus{}1)$. Hence, $ m\equal{}2$. $ a\equal{}1$ and thus, $ x\equal{}2$

[/hide]

[quote="lifeisacircle"] $ (3^a, 4^a, 5^a)$ is a primitive Pythagorean triple,

$ 3^a \equal{} m^2 \minus{} n^2$ (1)

$ 4^a \equal{} 2mn$ (2)

$ 5^a \equal{} m^2 \plus{} n^2$ (3)
[/quote]

$ (3^a, 4^a, 5^a)$ is not a primitive pythagorean triple; $ (3a, 4a, 5a)$ is.

I thought that (3,4,5) was the only primitive triple of the form (3a,4a,5a)...

yeah, but i was just pointing out that $ (3^a,4^a,5^a)$ can't be a pythagorean triple unless a=1

Pythagorean triples have nothing to do with this problem except that they inform the only solution; in particular, $ x$ is not required to be an integer.


probability
integration
probability and stats

Two components of a minicomputer have the following joint pdf for their useful lifetimes $ X$ and $ Y$: $ f(x,y) \equal{} \begin{cases} xe^{ \minus{} x(1 \plus{} y)} \ x \geq 0 \ \text{and} \ y \geq 0 \\
0 \ \ \ \ \ \ \ \ \ \   \ \ \text{otherwise} \end{cases}$

(a) What is the probability that the lifetime $ X$ of the first component exceeds $ 3$?

(b) What are the marginal pdf's of $ X$ and $ Y$? Are the two lifetimes independent?

(c) What is the probability that the lifetime of at least one component exceeds $ 3$?

$ \Pr[X > 3] \equal{} \int_{x \equal{} 3}^\infty \int_{y \equal{} 0}^\infty f_{X,Y}(x,y) \, dy \, dx$.

$ f_X(x) \equal{} \int_{y \equal{} 0}^\infty f_{X,Y}(x,y) \, dy$.

$ f_Y(y) \equal{} \int_{x \equal{} 0}^\infty f_{X,Y}(x,y) \, dx$.

X and Y are independent iff $ f_{X,Y}(x,y) \equal{} f_X(x) f_Y(y)$.

$ \Pr[X > 3 \cup Y > 3] \equal{} 1 \minus{} \Pr[X \le 3 \cap Y \le 3] \equal{} 1 \minus{} \int_{x \equal{} 0}^3 \int_{y \equal{} 0}^3 f_{X,Y}(x,y) \, dy \, dx$;

if X and Y are independent, then the above simplifies to

$ \Pr[X > 3 \cup Y > 3] \equal{} 1 \minus{} \Pr[X \le 3] \Pr[Y \le 3] \equal{} 1 \minus{} F_X(3)F_Y(3)$.

(a) and (c) are just integrals. Find the appropriate bounds and compute (Using the complement makes one simpler).

The joint density of a pair of independent random variables is of the form $ f(x)g(y)$. There's no way you're going to get an $ e^{\minus{}xy}$ term out of that.