quadratics
FTW
algebra
special factorizations

In [u][b]every[/b][/u] contest I take, I always make a simple mistake and it costs me a LOT.
For example, in MC chapters, I got 4 questions wrong out of the first ten questions, and i got the rest 20 questions right. When I found out that happened....i was paralyzed...
Also, in MC state, the number of "simple/stupid" decreased, but it didnt decrease completely so I ended up getting 7th Place Top Ten, and I failed in CountoDown because I made simple mistakes.

If I actually didnt make any simple mistakes...I would've made it to nats, and my life would've much better. Also, this was my first and last year for MC.....I feel really bad...

The types of mistakes I make: Addition/Subtraction----MOST OF THE TIME
                                          Reading the question wrong
                                          and some others..

Help anyone?

Other than being organized, do easy problems twice (if it doesn't take long, and absolutely not on a countdown). This can be at the end when you have finished all the problems.

I think the key to not making mistakes is thinking clearly. In order to do that, you have to think calmly yet quickly. Basically, be confident and do not be nervous. Sure, checking your work may help, but sometimes you make the same mistake over again.

Practice hard and feel less stressed out when you're taking the real thing. GL :)

The most important this is that you should [i]know[/i] that you won't make careless mistakes. If you think you will, then you definitely will. You don't want to be nervous either.

Before you do each problem, you can estimate what you think it will be so you won't get it way off. Also, see [hide="Here"]http://www.ehow.com/how_2172879_avoid-stupid-math-errors.html[/hide]

I know exactly what you mean...This is my second year w/ MC and i still make the random silly mistake. I know it sounds cliqued to say like underline the important ideas, but it really works. I once spent a really long time on this problem and ended up putting the total amount, not the amount at the end. Hope it works :) !

To not make that many mistakes concentrate or focus all of it into problem. i am not trying to be funny. it seriously works

This happened to me today(agh), the most useful advice: keep a clear head, don't get distracted, reread the question at least twice if you have the time, and try to double check your answers if you can.

[quote="generalmotors"]
The types of mistakes I make: Addition/Subtraction----MOST OF THE TIME
                                          Reading the question wrong
                                          and some others..

[/quote]
Try doing the problem 2 times to see if you get the same answer, or 2 different ways (if possible). I usually do the "short way" to a problem first, then a "longer way". For example, if I have to solve a quadratic equation (I did this on a test in May), I solved it by completing the square and by using the quadratic equation. Same answer...and I got a 100% on my test. :)

Everyone makes mistakes like that.
You can't completly get rid of them.
No one is perfect.

It cost me a lot last year.
Just keep practicing. I suggest you do countdown problems, which increase your speed AND doing it accurately makes you make less mistakes in sprint/target.
It helped me a lot.

In order for you to not make simple mistakes, you should clear your head.

A strategy I use is to simply put the pencil down and empty your brain by rubbing the temples.

It really helps, you should try it!

After you've done that, you should be able to see the problem before you in a WHOLE different light!!!

on ftw everyone mistypes once in 3 problems.

the way to stop silly mistakes is to think that you have time and just clear your mind.

competition makes you make silly mistakes.

You know how much that sounds like me? I also made stupid mistakes in cd and sprint. And I got 7th. In state.

Here are a few tips of how to stop making stupid mistakes.

1. Check the 1st 10 problems thoroughly! If you can get the rest of the 20 right, then that's pretty good. The first 10 problems aren't that hard so it probably won't take you that long.
2. When you're doing the problem the first time, make a mental note to yourself "Am I doing anything wrong?"
3. Practice Addition/Subtraction.  :wink:
4. Carry on with life. There are much more contests later on in your life. Such as AMCs and a lot more advanced stuff.

I gotta ask you a question. Do you find yourself finishing everything in time? Or you still have a few questions left and you're out of time. Or is it a piece of cake and your done with like 8 minutes left?

Hm....thanks for your help...I'll try those things :)

Also, I have another serious problem...

Remember when I went to state? I got a score of ..38(26 on sprint, and 12 on Target)

And now the state rounds are posted online so I did it today and got a score of 45..

It seemed so easy! But I think this is the reason:

At States: I was totally tense, and scared...Really wanted to get a high score...My heart was pounding.. 1000b/s :maybe:

At Home: I was calm and relaxed...Didnt care about my score..Just doing it for fun :)

And that made a difference of 7 points...

How do I decrease my tenseness when I go to a test? I cant stop myself from being scared, no matter how hard I try..Especially if its some big contest..

Feeling nervous is natural, but there are always ways to calm yourself down. Is there some food that makes you feel better? I remember one author of a chess book said that it always helped him to suck on a ginger root. For me, it's tic-tacs.  :P (Now I can't do math without them.)

Try to be confident - after all, you worked hard to reach this point, so what happens will happen and you can't change it. Take deep breaths. Find someone you know there and crack some jokes, laugh a little, and relax.

Also, never eat potatoes. I swear it's bad luck.

Also, do get rid of nervousness (however you spell it!) by thinking of a happy place such as you eating a ginourmus cookie.

Note that when you re-take a test, you remember some of the answers (or at least some of the methods of the problems you got wrong).  For example, I'm looking at the 2009 State test right now - I have 6 of the last 10 Sprint memorized and 3 Target memorized (and I don't spend time trying to memorize answers).

GeneralMotors, this is (unfortunately) the reality of speed-based math competitions. When you go on to study mathematics at the university level, you will have plenty of time to work on challenging problems (REAL problems). But when you're racing to solve problems like $ 4x^2\plus{}4x\plus{}1\equal{}0$ against 15 other matheletes in an MC Countdown, it is very likely you will make a mistake!

Just try to relax and think calmly.  What I do is I just ask myself, if I miss this problem will the world come to an end?  will I die? and then I realize it just doesn't matter that much and then I do much better.

To get every problem right you have to be calm.

i do much better in countdowns when i dont get pressured so try to stay calm and double check ur work if u have enough time


geometry
rectangle
combinatorics proposed
combinatorics

Let $m$ and $n$ be positive integers. Suppose that some given rectangle (consisting of a number of unit squares) can be tiled, without overlap, by a combination of horizontal $1 \times m$ and vertical $n \times 1$ strips.

Show that it can be tiled using just one of the two types of strips as well.

Pretty well-known idea, isn't it? :)

Partition the rectangle into unit squares, place a $1$ in the top left corner, and then, for each square containing the complex number $t$, place $tz_1$ below it and $tz_2$ to its right, where $z_1,z_2$ are $n$'th and $m$'th primitive roots of unity respectively.

In a rectangle which can be tiled as described the sum of all the entries must be zero, and it's easy to see that this is only possible if either its width is divisible by $m$ or its height is divisible by $n$.


logarithms
function
inequalities
induction
number theory
real analysis
real analysis unsolved

Prove, that for naturals $ n\geq 2,p\geq 3$:
\[ \prod_{k=2}^{n}(1-\frac{1}{k^{p}})>\frac{p}{p+1}.\]

Who can prove??? It`s very interesting... :ninja: ... I don`t know the solution  :maybe:

really interesting :)

it suffices to show
\[ \sum_{k\equal{}2}^{\infty}\log\left(1\minus{}\frac{1}{k^p}\right)\ge\log\frac{p}{p\plus{}1}\]

this series.... isn't it one of known functions in number theory?
if it is, surely there are some lower evaluations for it

I found a solution... Maybe i`t will be interesting for somebody...  :roll:

First we need to prove that for all $ 0 \leq a_i \leq 1$ there is an inequality:
\[ \prod_{i \equal{} 1}^{n}(1 \minus{} a_i) \geq 1 \minus{} \sum_{i \equal{} 1}^{n}a_i
\]
It`s easy to prove by induction:
i) For $ n \equal{} 1$ it`s true;
ii) We expect that for $ n$ it`s true;
iii) For $ n \plus{} 1$ we have:
\[ \prod_{i \equal{} 1}^{n \plus{} 1}(1 \minus{} a_i) \geq (1 \minus{} a_{n \plus{} 1})(1 \minus{} \sum_{i \equal{} 1}^{n}a_i) \equal{} 1 \minus{} \sum_{i \equal{} 1}^{n \plus{} 1}a_i \plus{} a_{n \plus{} 1}\sum_{i \equal{} 1}^{n}a_i \geq 1 \minus{} \sum_{i \equal{} 1}^{n \plus{} 1}a_i
\]
Inequality proved.

Now we have:
\[ \prod_{k \equal{} 2}^{n}{(1 \minus{} \frac {1}{k^p})} \geq 1 \minus{} \sum_{k \equal{} 2}^{n}{\frac {1}{k^p}}
\]
So we have to prove that:
\[ 1 \minus{} \sum_{k \equal{} 2}^{n}{\frac {1}{k^p}} > \frac {p}{p \plus{} 1}
\]
We have:
\[ 1 \plus{} \sum_{k \equal{} 2}^{n}{\frac {1}{k^p}} < 2 \minus{} \frac {p}{p \plus{} 1}
\]
We can put here $ \zeta(x) \equal{} \sum_{k \equal{} 1}^{\infty}{\frac {1}{k^x}}$ and we have to prove that:
\[ \zeta(p) < 1 \plus{} \frac {1}{p \plus{} 1}
\]
Last inequality we can easily prove by integration...
[b]pleurestique[/b] have done it there [url]http://www.mathlinks.ro/viewtopic.php?t=180795[/url] Big thanks to him  :roll:


[quote="rainynightstarz"]if the girl doesn't like u than she doesn't like u[/quote]

Hmm, this statement that rainynightstarz brought up is interesting to examine. It was posted in the "How to Make a Girl Laugh" thread, but since that thread was getting somewhat long, I didn't know how many people still read it. The quoted statement is claiming that no matter how a boy acts, if the girl he is trying to woo just doesn't like him, then there is nothing that the boy can do about it.

Let us consider the scenario in reality: each person is acquainted with many many people of the opposite sex in the entire world; let us call put these people of the opposite sex that a particular person knows in set S. This person is also deeply attracted to a few people of the opposite sex within set S; let us take this to be the subset T. Now rainynightstarz statement asserts that the subset T is taken randomly from set S (since otherwise, the actions of one person would affect whether the person is included is the set T of another person -- a contradiction to the statement). The size of T is, let us suppose, 4 to 5 people (and this is probably stretching it). So the size of T compared to the size of S is very very small. Thus, theoretically, the probability that the set T_A of person A contains person B and that the set T_B of person B contains person A is very very low. However, look at all of the people in this world who have boyfriends/girlfriends/husbands/wives -- in most of these cases, both the male and the female are attracted to each other.

Therefore, how can we explain this paradox: the mutual attraction of many pairs of people in this world contradict the low theoretical probability of that attraction being present (assuming the initially quoted assertion is true)?

[quote]No matter how a boy acts, if the girl he is trying to woo just doesn't like him, then there is nothing that the boy can do about it[/quote]

While a boy may be unable to do anything to change a girl's opinion about him, can't her opinion about him still change as a result of other things?  Say, if the girl goes off to math camp for a summer and realizes that math is loads of fun, then wouldn't her opinion on the boy who owns when it comes to math competitions change?

Also, concerning this quote:

[quote]In most of these cases, both the male and the female are attracted to each other. [/quote]

Is this necessarily true?  A lot of times relationships fail because people no longer like each other all that much - or they realize that they never did in the first place.

I don't see how relationships is anything math related...that may be the problem here Fierytycoon.

[quote="rainynightstarz"]by "if the girl doesn't like u than she doesn't like u" i meant that if the girl doesn't like u, u can't do anything to make her like u. but of course the girl can still change her opinion on u. i mean u can still be her friend, and once she knows u better, maybe she will like u[/quote]

let's hope that the truth...

[quote="Fierytycoon"]
 Now rainynightstarz statement asserts that the subset T is taken randomly from set S (since otherwise, the actions of one person would affect whether the person is included is the set T of another person -- a contradiction to the statement). [/quote]

Besides, nothing asserts that T is taken randomly from S. Just because the actions of one is insignificant doesn't mean it's [i]random[/i].

Okay, inspired by the original post, here is another paradox  -
[quote="sunnydaystarz"]if the man doesn't  pay your college tution, he doesn't pay your college tution. [/quote]

Hmm, this statement that sunnydaystarz brought up is interesting to examine. It was posted in the "How to Make a man   pay your college tution" thread, but since that thread was getting somewhat long, I didn't know how many people still read it. The quoted statement is claiming that no matter how a boy acts, if the man he is trying to  fund his college is not his parent and does not feel like funding it,  then there is nothing that the boy can do about it.

Let us consider the scenario in reality: each person is acquainted with many many people of the older generation  in the entire world; let us call put these people of the older generation  that a particular person knows in set S. This person is also deeply attracted to a few people of the older generation within set S; let us take this to be the subset of "parents"  T. Now sunnydaytstarz statement asserts that the subset T is taken randomly from set S (since otherwise, the actions of one person would affect whether the person is included is the set T of another person -- a contradiction to the statement). The size of T is, let us suppose, 4 to 5 people (and this is probably stretching it- most authorities say it is 2 and can be even less). So the size of T compared to the size of S is very very small. Thus, theoretically, the probability that the set T_A of person A contains person B and that the set T_B of person B contains person A is very very low. However, look at all of the people in this world who have parents/fathers/mothers/legal guadians/ -- in most of these cases, both the  child and the parents are  a  "family group"

Therefore, how can we explain this paradox: the mutual attraction of many people who form a family in  this world contradict the low theoretical probability of that group bonding being present (assuming the initially quoted assertion is true)?

Okay -  Its a slow day here ... Just  trying to have some fun.. :)

[quote="Gyan"] Now sunnydaytstarz statement asserts that the subset T is taken randomly from set S (since otherwise, the actions of one person would affect whether the person is included is the set T of another person -- a contradiction to the statement).[/quote]

That line of reasoning is incorrect. "No effect" and random are not one and the same.

[quote]That line of reasoning is incorrect.[/quote]
Exactly ! (Not,  that   *that* ws the *only* incorrect resoning there. :) .) but I just cut and pasted the resoning  from  the first post.    :)  Don't take it so seriously..

[quote]In fact, there are very few things in the actual world, not math world, that are truly random.[/quote]

Don't tell that to those quantum physicists.!  There is randomness in every wave function.
In fact  all "things"  are just "probablity  wave functions"  No one really knows if  an electron was really in that spark chamber in my old HighEnergyLab ... No one knows that that if that particular  atom will  decay  today ...I know lot of electrons are jumping from one  energy level to another to give me light so that I can type this post .. but which electron ..its random...

  :)  :)

[quote="Gyan"]There is randomness in every wave function.[/quote]

At least until we get a better theory than Quantum Mechanics gives us ...

:)

HOw on EARTH did this thread get resurrected. *sigh* Okay, as for that college thing...I say look for scholarships.

There is a pattern for everything, but that doesn't mean that humans are able to comprehend the pattern or figure it out...

Even waves... :)


PQRS (anti-clockwise with P at the top left-hand-corner) is a square. T is on QR and U is on RS. Angle QPT = angle UPS = 30 degrees.
Prove that the area of triangle PTU is one third of the area of the square PQRS.

[hide="My Solution"]
Let the side length of the square be $ s$ and let $ A$ be the area of $ \triangle PTU$. It follows that the area of square $ PQRS$ is $ s^2$. $ \triangle PQT \cong \triangle PSU$ by AAS Congruence because  $ \angle QPT \equal{} \angle UPS$,  $ PQ \equal{} PS$, and $ \angle USP \equal{} \angle TPQ$. These triangles are also $ 30 \minus{} 60 \minus{} 90$ right triangles due to the fact that they have a $ 30^\circ$ angle and a right angle. $ PU \equal{} PT$ by CPCTC. Because $ PT$ is the hypotenuse of a $ 30 \minus{} 60 \minus{} 90$ triangle, its length is $ \frac {2s\sqrt {3}}{3}$, which is congruent to the length of $ PU$. Finally, $ \angle TPU \equal{} 30^\circ$ because $ \angle QPS$ is right.

By the trigonometric area formula for a triangle,

$ A \equal{} \frac {1}{2}\left(\frac {2s\sqrt {3}}{3}\right)^2\sin{30^\circ} \equal{} \frac {12s^2}{18}\cdot \frac {1}{2} \equal{} \frac {s^2}{3}$. $ \Box$

[/hide]


analytic geometry
combinatorics proposed
combinatorics

The distance between two cells of an infinite chessboard is defined as the minimum nuber to moves needed for a king for move from one to the other.One the board are chosen three cells on a pairwise distances equal to $ 100$. How many cells are there that are on the distance $ 50$ from each of the three cells?

We'll prove that there is exactly one such point.
Let us choose cell coordinates $ (x,y)$ where $ x,y$ are integers such that one of the chosen cells is $ (0,0)$.
It's easy to notice that for a cell $ (x_0,y_0)$, the set $ S$ of cells with distance $ d$ from it is of the form $ S \equal{} \{(x_0 \plus{} d,y_0 \plus{} k): k\in \{ \minus{} d,\ldots,d\} \} \cup \{(x_0 \minus{} d,y_0 \plus{} k): k\in \{ \minus{} d,\ldots,d\} \} \cup \{(x_0 \plus{} k,y_0 \plus{} d): k\in \{ \minus{} d,\ldots,d\} \} \cup \{(x_0 \plus{} k,y_0 \minus{} d): k\in \{ \minus{} d,\ldots,d\} \}$
In other words, $ S$ consists of 4 blocks forming a square (without the interior of sorts).
WLOG we can assume that the second chosen cell lies on the upmost part of the rightmost face of this figure, that is, it's coordinates are $ (100,k)$ for some $ 0\le k \le 100$, $ k$ being an integer.

Lemma:
Two of these three cells lie either in the same line (horizontal or vertical).

Proof:
If $ k \equal{} 0$ (i.e. the second cell is $ (100,0)$, the lemma is proved already.
If $ k\ne 0$, then there are exactly two cells that can be picked as the third cell (i.e. are in the intersection of two squares with centres in two already chosen cells).
One is $ (0,100)$, the other one is $ (100,k \minus{} 100)$. In the first case, the first and the third cell are on the same vertical line, in the second case, the second and the third cell are on the same vertical line.

Using the lemma above, we can again change the coordinates so that two cells are in $ (0,0)$ and $ (0,100)$.
WLOG we can also assume that the third cell is on the right-upper side of the board, i.e. it has coordinates $ (d,k)$ for some $ 0 \le k\le 100$, $ k$ integer.

The set $ X$ of points with distance 50 from first two cells now form a vertical bar $ (50,n)$, $ \minus{} 50 \le n\le 50$, $ n$ integer.
The set of points with distance 50 from the third cell forms a square-like-figure (without the interior part).  Now, if $ k\ne 0, k\ne 100$ then only the intersection of the bottom-most block (call it $ Y$) with $ X$ is not empty.
Moreover, $ X$ is horizontal, $ Y$ is vertical, so that the intersection consists of exactly one point.

The case $ k \equal{} 0$ or $ k \equal{} 100$ is now also trivial, the point is in the intersection of the vertical bars and two faces of the square, one of them being the bottom-most horizontal one.

best regards,
Wojtek


function
integration
induction
binomial coefficients
calculus
calculus computations

prove that for $ N,M \in \mathbb{N}$ such as $ N>M$ sum
\[ \sum_{k\equal{}0}^{n}(\minus{}1)^{n\minus{}k}C_{n}^{k}\frac{(N\plus{}k)!}{(M\plus{}k)!}\]
is equal to $ n!$ if $ N\minus{}M\equal{}n$ and $ 0$ if $ N\minus{}M<n$

$ C_n^k \equal{} C_k^n \equal{} _nC_k \equal{} \binom{n}{k} \equal{} \frac{n!}{k!(n\minus{}k)!}$.

Ahh, the choose function. It is weird how the $ k$ is written on the top.

the general case:
$ \sum_{k \equal{} 0}^{n}( \minus{} 1)^{n \minus{} k}C_{n}^{k}\frac {(N \plus{} k)!}{(M \plus{} k)!} \equal{} (N \minus{} M)! \binom{N}{n \plus{} M}$

if we have N<M  sum equals 0 for every $ n$
and if you put $ n\equal{}N\minus{}M$ and$ N\geq M$ sum equals to $ (N\minus{}M)!\equal{}n!$(i used generating function)

[quote="JBL"]$ C_n^k \equal{} C_k^n \equal{} _nC_k \equal{} \binom{n}{k} \equal{} \frac {n!}{k!(n \minus{} k)!}$.[/quote]

I don't what to be rude, but the first equality is false! An expression should not be distinguished from another just by the names of it's variables. If you have $ C^x_y$ would that be $ \frac{x!}{y!(x\minus{}y)!}$ or $ \frac{y!}{x!(y\minus{}x)!}$. And how would you know?! Personally, I've always used the binomial version since I thought the version using "C" was somewhat unambiguous. Seems my thinking wasn't far off..

I was giving five commonly-used ways of writing the same expression.  I agree that the fourth is preferable, in large part because of confusion between the first and second.  (You meant "ambiguous" where you wrote "unambiguous," though :).)  It also generalizes better to multinomial coefficients, $ q$-binomial coefficients, etc.

Uh.. Yes.. Wrong sentence syntax.. But the message was delivered !

[quote="sumita"]...
(i used generating function)[/quote]
and what generating function have  you used?

I guess what he used is $ f(z) = \sum_{m=0}^{\infty} \left[ \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \binom{l+m+k}{m} \right] z^m$.

If we can show that $ \sum_{m=0}^{\infty} \binom{l+m}{l+n} z^m = \frac{z^n}{(1-z)^{l+n+1}}$, then we obtain the desired result by interchaging two summations in $ f(z)$.

[hide="And here is my solution"]Note that $ \frac{1}{2\pi i} \int_{|z-z_0|=\epsilon} \frac{z^{m}}{(z-z_0)^{n+1}} \, dz =\binom{m}{n} z_0^{m-n}$ for $ \epsilon > 0$. This gives

\begin{eqnarray*}
\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \frac{(l+m+k)!}{(l+k)!}
& = & \frac{m!}{2\pi i} \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \int_{|z-1|=\epsilon} \frac{z^{l+m+k}}{(z-1)^{l+k+1}} \, dz\\
& = & \frac{m!}{2\pi i} \int_{|z-1|=\epsilon}\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k}  \frac{z^{l+m+k}}{(z-1)^{l+k+1}} \, dz\\
& = & \frac{m!}{2\pi i} \int_{|z-1|=\epsilon} \left( \frac{z}{z-1} - 1 \right)^n \frac{z^{l+m}}{(z-1)^{l+1}} \, dz\\
& = & \frac{m!}{2\pi i} \int_{|z-1|=\epsilon} \frac{z^{l+m}}{(z-1)^{l+n+1}} \, dz\\
& = & m! \binom{l+m}{l+n}
\end{eqnarray*}[/hide]

Nise solution sos440  :) .
My way is different.
I use the fact

if we fixed some $ h$ and  denote by
$ df(x) \equal{} f(x\plus{}h)\minus{}f(x)$
and
$ d^{n}f(x) \equal{} d^{n\minus{}1}f(x\plus{}h)\minus{}d^{n\minus{}1}f(x)$
also $ x_{k} \equal{} x_{0}\plus{}kh$
then it's easy to prove by induction that

\[ d^{n}f(x_{0}) \equal{} \sum_{k\equal{}0}^{n}(\minus{}1)^{n\minus{}k}\binom{n}{k}f(x_{k})\]
...


$ 16a_{n+1}=48+ a_{n}^2    ,   4≤a_{n}<12
what is the value of lim {n→\infty} a_n ?$

sorry i'm poor at text message  :blush:

Modified.

Second Condition is ,  4 < = a_{n} <12

Question is   "what is the value of  lim {n→infinite} a_{n} " :blush:

We have $ a_{n\plus{}1}\leq a_n$ because $ 16a_n\minus{}a_n^2\minus{}48\equal{}(a_n\minus{}4)(12\minus{}a_n)\geq 0$. Then $ a_n$ is decreasing and bounded below (by 4), so it converges to some $ L$ (namely, the greatest lower bound of the $ a_n$).

$ 16L\equal{}48\plus{}L^2\Rightarrow (L\minus{}12)(L\minus{}4)\equal{}0$. So either $ L\equal{}12$ or $ L\equal{}4$. But also $ L\leq a_1$. In particular, $ L\neq 12$. We get $ L\equal{}4$.

(Note: If we were just given $ 4\leq a_1<12$, we would prove $ 4\leq a_n<12$ for all $ n$ by induction first.)


Έστω $x,y,z$ θετικοί ακέραιοι ώστε $x^2+y^2+z^2=1993.$ Να αποδείξετε ότι ο αριθμός $x+y+z$ δεν είναι τετράγωνος.

Kalispera se olous, einai i prwti fora pou stelnw sto elliniko forum,

[quote="nickolas"]Έστω $x,y,z$ θετικοί ακέραιοι ώστε $x^2+y^2+z^2=1993$. Να αποδείξετε ότι ο αριθμός $x+y+z$ δεν είναι τετράγωνος.[/quote]

Loipon: As ipothesoume arxika oti o $x+y+z$ einai tetragwnos arithmos.

Afou $x^2+y^2+z^2=1993$, ara apo anisotita C-S exoume $x+y+z \leq \sqrt{3(x^2+y^2+z^2)}$, ap'opou pairnoume

$x+y+z \leq 77 \ \ (1)$.

Ap'tin alli, logw tis tautotitas $(x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+xz) \ \ (2)$  exoume $x+y+z \geq 45 \ \ (3)$.

Apo tis $(1),(3)$ ta monadika tetragwna pou tis ikanopoioun einai $x+y+z=49,64$. Omws i lisi $x+y+z=64$ aporriptetai logw tis tautotitas $(2)$ (artios einai isos me peritto).

H moni periptwsi pou menei na eksetasoume einai ean iparxoun thetikoi akeraioi $x,y,z$ me $x^2+y^2+z^2=1993$ kai $x+y+z=49$ $(\Sigma )$

Alexandros

Kata lathos esteila to sigekrimeno post. Koitakste to parakatw

Kathws me rwtisan 2-3 atoma gia to pws linetai to sistima twn diophantikwn eksiswsewn sto parapanw post kai gia na min tin afisw etsi, stelnw tin lisi. Kalo tha itan na tin deite giati den moiazei me tis sinithismenes tou eidous kai isws tin lisete me kalitero tropo apo auto pou parathetw parakatw. Autos einai kai o logos pou tin kanw "hide". (distixws den vriskw kapoio kalitero tropo gia na apofigw merikes periptwseis (sto telos). Ean vlepete kati pou den vlepw steilte tin lisi an kai idi mou fainetai oti einai ena provlima me ergasiotherapeia xwris ousiwdeis skepseis. Kati san to 4o thema tis persinis Valkaniadas Newn).

Na lithei loipon to sistima $x^2+y^2+z^2 = 1993$ kai $x+y+z=49$.

[hide]Loipon, kathws $x^2+y^2+z^2=1993 \ \ (1)$ kai $x+y+z=49 \ \ (2)$ ara apo tin tautotita $(2)$ sto proigoumeno post, exoume $xy+yz+xz=204 \ \ (3)$. As ipothesoume logw simmetrias oti $x \geq y \geq z$. Tote apo tin $(3)$ exoyme oti $204=xy+yz+xz \geq 3z^2$ ap'opou $z \leq 8 \ \ (4)$.

Ap'tin alli linontas ws pros $y$ tin $(2)$ kai antikathistwntas stin $(1)$, exoume ena triwnimo  kai to opoio gia na exei lisi ws pros $x$ tha prepei i diakrinousa tou na einai $\geq 0$, pragma pou dinei telika oti prepei $-3z^2 + 98z+1585 \geq 0$ ap'opou $z \in [5, 44] \ \ (5)$. Sindiazontas to me tin $(4)$, exoume $z=5,6,7,8$ tis opoies eksetazoume poli eukola memonwmena kai den vgazoume liseis.
[/hide]

Ti lete ki eseis? Exete kati kalitero na proteinete? Mou fanike ligo periergo gia Roumania h sigekrimeni askisi ean den iparxei kaliteros kai sintomoteros tropos lisis.

Alexandros


calculus
integration
analytic geometry
trigonometry
vector
geometry
graphing lines

The coordinate of $ P$ at time $ t$, moving on a plane, is expressed by $ x = f(t) = \cos 2t + t\sin 2t,\ y = g(t) = \sin 2t - t\cos 2t$.

(1) Find the acceleration vector $ \overrightarrow{\alpha}$ of $ P$ at time $ t$ .

(2) Let $ L$ denote the line passing through the point $ P$ for the time $ t%Error. "neqo" is a bad command.
$, which is parallel to the acceleration vector $ \overrightarrow{\alpha}$ at the time. Prove that $ L$ always touches to the unit circle with center the origin, then find the point of tangency $ Q$.

(3) Prove that $ f(t)$ decreases in the interval $ 0\leq t \leqq \frac {\pi}{2}$.

(4) When $ t$ varies in the range $ \frac {\pi}{4}\leq t\leq \frac {\pi}{2}$, find the area $ S$ of the figure formed by moving the line segment $ PQ$.

1) $\vec a=4t(-\sin{2t}\ \hat {x}+\cos{2t}\ \hat {y})$.
2) The line $L$ in question is $Y={1\over \sin{2t}}-{\cos{2t}\over \sin{2t}}\ X.$ The point $Q: (X,Y)=(\cos{2t},\sin{2t})$ is clearly on the line. The line from the center of the unit circle to this point $Q$ has a slope of $\tan{2t}$, and thus this line is perpendicular to $L$.
3) ${{\text d}x/{\text d}t =-\sin{2t}+2t\cos{2t}}$. This is clearly negative in $[{\pi\over 4},{\pi\over 2}].$ It is also negative in $[0,{\pi\over 4}]$ because $\tan\theta\ge \theta$ in  $[0,{\pi\over 2}].$
4) Note that $\int_{\pi/4}^{\pi/2} y\left | {{\text d}x\over {{\text d}t}}\right | {\text d}t={\pi(96+7\pi^2)\over 192}$. Then the area in question is $7\pi^3/192.$


Yes! I made it with 37. Feel free to post your scores and if you made it or not (if you want to).

Is that the exam after OML????

Well I got a 38 on the OML, I misread the first question and gave the greatest instead of the least fraction :(
I also missed that really easy exponent question. All of this and I sat idly for 12 mins after I finished early :(

The OHMIO is the contest after OCTM. OML is totally unrelated I believe.

I didn't make it.

JKLOL.

What is the OHIMO.

[quote="roadnottaken"]I made it with a 32 (and 5 stupid mistakes :blush: ).  Hopefully I'll do better next year.[/quote]

Sorry, it was a 33.  There was one question that I answered but forgot to write down.  Afterwards, I thought that I hadn't gotten anywhere near the answer (the 5 stupid mistakes figure still stands).

As to the other two questions, they were problems involving triangles.  I brute forced one of them after the test was over, and the other one was the angle bisector problem where the sides were simply in equivalent ratios :oops: .

[hide="If I had known that during the test, been more careful on 5 of the other questions, and had a little more time, I would have gotten a 40 :rotfl: ."]Of course, you can make a similar statement about every test you take :roll: .[/hide]

I was disappointed to only get 30...  :( That was five less that I predicted; next time I'll make sure to write down my answers so I can grade myself accurately.

Anyway, I did make it to the next level, so I guess I can live.

Hehe, made OHMIO with a 27, exactly the cutoff score.  It's kind of annoying, because I made completely retarded mistakes on about 8 of them (i.e. forgetting the negative sign on the csc one), but I was going on about 2 hours of sleep, and I still qualified, so I guess it's all good.  I'll just have to pwn it next year...

ehehheehee thinks the purpose of the OCTM is to make people mess up.  People who don't mess up win.  Of course ehehheehee messed up.

I got a 28, which is what I ALWAYS, yes everytime, got on previous tests that I practiced with, but does anyone know where to get previous OHMIO tests?


[hide="easy way out"]
because of the mutual equality in the second equation, the easiest real solution to find is
$x=y=z=\frac{1}{3}$
but I don't think that's all of the solutions
[/hide]

someone else gave me the problem and he said the only solution is x=y=z... but how can one prove it?


Consider sequences of positive real numbers of the form $x,2000,y,\dots$, in which every term after the first is $1$ less than the product of its two immediate neighbors. For how many different values of $x$ does the term $2001$ appear somewhere in the sequence?

\[\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than 4}\]

Can someone please help? I don't get how to do it/

i saw this when you first posted it....is the first tem x and is this an old amc or something?

was it older than 2000?
and, again, is x the first term...


Ν.δ.ο. για καθε μιγαδικο $ z$ ισχυει
$ |z\plus{}10|\plus{}|z\plus{}11|\plus{}|z\plus{}19|\leq|z\plus{}8|\plus{}|z\plus{}12|\plus{}|z\plus{}20|$
Kατα προτιμηση, ας αποδειχθει με γνωσεις Γ' Λυκειου πρωτα και μετα.....

Αν και έμαθα σχετικά πρόσφατα μιγαδικούς, θα την επιχειρήσω.

Λοιπόν: $ \forall \ \ z_1, z_2 \in \mathbb{C}$ ισχύει: $ |z_1 \pm z_2| \leq |z_1| \pm |z_2|$
Είναι:
$ |z \plus{} 10| \plus{} |z \plus{} 11| \plus{} |z \plus{} 19| \equal{}$
$ |z \plus{} 8 \ \ \plus{} 2| \plus{} |z \plus{} 12 \ \ \minus{} 1| \plus{} |z \plus{} 20 \ \ \minus{} 1| \leq |z \plus{} 8| \plus{} 2 \plus{} |z \plus{} 12| \minus{} 1 \plus{} |z \plus{} 20| \minus{} 1 \equal{} |z \plus{} 8| \plus{} |z \plus{} 12| \plus{} |z \plus{} 20|$

Edit: Έχεις δίκιο NickNafplio. Γράψε λάθος :blush:. Είναι:$ |z_1 \pm z_2| \leq |z_1| \plus{} |z_2|$

[quote="Orfikoss"]Αν και έμαθα σχετικά πρόσφατα μιγαδικούς, θα την επιχειρήσω.

Λοιπόν: $ \forall \ \ z_1, z_2 \in \mathbb{C}$ ισχύει: $ |z_1 \pm z_2| \leq |z_1| \pm |z_2|$
Είναι:
$ |z \plus{} 10| \plus{} |z \plus{} 11| \plus{} |z \plus{} 19| \equal{}$
$ |z \plus{} 8 \ \ \plus{} 2| \plus{} |z \plus{} 12 \ \ \minus{} 1| \plus{} |z \plus{} 20 \ \ \minus{} 1| \leq |z \plus{} 8| \plus{} 2 \plus{} |z \plus{} 12| \minus{} 1 \plus{} |z \plus{} 20| \minus{} 1 \equal{} |z \plus{} 8| \plus{} |z \plus{} 12| \plus{} |z \plus{} 20|$[/quote]

Οκ, αφου εκανες edit κανω κι εγω :P

Παντως δεν νομιζω να βγαινει με τριγωνικη ανισοτητα γιατι εχω δοκιμασει διαφορα τρικ και δεν βγαινει κατι. Επισης πηγαινοντας τα ολα στο 1ο μελος με z = x + yi και συναρτηση δεν βρισκεται ευκολα κανενα ακροτατο ισο με 0 ενω προκειπτουν πολυ πολυπλοκες πραξεις. Αρα xp οταν λες γνωσεις Γ Λυκειου μηπως εννοεις τιποτα με ορισματα που ειναι εκτος σχολικης υλης;

oχι εννοω με την υπαρχουσα υλη
η βαλε μια λυση με οτι γνωσεις θελεις. Απλα θελω να δω με ποσους τροπους βγαινει (Εχω δυο ηδη)

Η ιδέα για τη λύση σε hint

[hide]Η ανισότητα είναι ισοδύναμη με $ f(10)\minus{}f(9)\plus{}f(9)\minus{}f(8)\leq f(20)\minus{}f(19)\plus{}f(12)\minus{}f(11)$  με την $ f$ κυρτή άρα την $ f'$ γν αύξουσα :wink: [/hide]

βασικα πρεπει να βγαινει σιγουρα με αναλυτικη μεθοδο... υψωση στο τετραγωνο και τα 2 μελη, αναπτυγματα ταυτοτητων, και μετα πραξεις... αλλα θελει υπερβολικα πολες πραξεις, μολις καλυψα μιση σελιδα τετραδιου μονο για τις πραξεις στο 1ο μελος

[quote="silouan"]Η ιδέα για τη λύση σε hint

[hide]Η ανισότητα είναι ισοδύναμη με $ f(10) \minus{} f(9) \plus{} f(9) \minus{} f(8)\leq f(20) \minus{} f(19) \plus{} f(12) \minus{} f(11)$  με την $ f$ κυρτή άρα την $ f'$ γν αύξουσα :wink: [/hide][/quote]

Α, νομιζω οτι καταλαβα τωρα, εφαρμοζουμε ΘΜΤ για την $ f(x) \equal{} |z \plus{} x|$ και εχουμε οτι υπαρχουν $ j_1, j_2, j_3, j_4$ στα διαστηματα [8, 9], [9, 10], [11, 12], και [19, 20] αντιστοιχα, με $ f'(j_1) \equal{} \frac {f(9) \minus{} f(8)}{9 \minus{} 8}, f'(j_2) \equal{} \frac {f(10) \minus{} f(9)}{10 \minus{} 9}, f'(j_3) \equal{} \frac {f(12) \minus{} f(11)}{12 \minus{} 11},$ και $ f'(j_4) \equal{} \frac {f(20) \minus{} f(19)}{20 \minus{} 19}$

και απο f κυρτη => f' γνησιως αυξουσα, αφου $ j_1 \leq j_3$ και $ J_2 \leq J_4$, εχουμε $ f'(j_1) \leq f'(j_3)$ (1) και $ f'(j_2) \leq f'(j_4)$ (2), και με προσθεση κατα μελη των (1) και (2) προκειπτει το ζητουμενο :wink:

ΥΓ: Πολυ ωραια η σκεψη σου για τη κυρτοτητα της f(x) = |z + x| και τη μονοτονια της f' και το ΘΜΤ, αν και εχω λυσει αρκετες σχολικες ασκησεις με ανισοτητες και ΘΜΤ σε αυτη την ασκηση δεν θα καταφερνα να βρω αυτη τη λυση οση ωρα και αν περνουσε, πραγματικα εντυπωσιαστηκα

[quote="xp"]Ρε μπαγασα Sil δεν ειπαμε οτι διαβαζουμε εκθεσοβιολογια.......και δεν μπαίνουμε στο Forum .Τελικα ειδες βγηκε καλη κατασκευη.....
Nic απ το ομορφο Ναυπλιο (εκανα την θητεια μου στο Μηχανικο) για $ z$ μηδεν η οποιοδηποτε θετικο πραγματικο στην αρχικη για κοιτα τι γινεται...[/quote]

Ναι εχεις δικιο, ολα οσα γραφω (μαζι και η παρατηρηση) ισχυουν για z οχι πραγματικο, για z πραγματικο η συναρτηση f' ειναι σταθερη και οχι γνησιως αυξουσα οποτε εφαρμοζοντας παλι το ΘΜΤ στα ιδια διαστηματα προκειπτει οτι ισχυει η ισοτητα, για z ομως οχι πραγματικο η f' ειναι γνησιως αυξουσα και οπως εγραψα και πιο πανω δεν ισχυει ποτε η ισοτητα... τελος παντων, βγαζω την παρατηρηση :P

$ Bil$ πολυ ωραια λυση .Μπραβο σου .(Επειδη στις Γενικες εξετασεις  γραφεις και εκθεση το μ[b]ι[/b]γαδικοι με [b]ι[/b])
Δεν χρειαζεται σε καθε post  το ονομα μου ... [/b]

Ευχαριστώ για την επισήμανση αλλά δυστυχώς έχω πολύ μεγαλύτερο ορθογραφικό πρόβλημμα γενικότερα.. :P ....
Η άλλη λύση που έχετε που αναφέρατε πιο πάνω (μία σχολικά που έβαλε ο Σιλ και o Νick και μία άλλη με ολυμπιακό πνεύμα)??

H λυση σου δεν διαφερει ιδιαιτερα απ την προηγουμενη και ας φαινεται διαφερετικη .Απλα εσυ κανεις διαδοχικες εφαρμογες της $ (Jensen)    f(\frac{a\plus{}b}{2})\leq\frac{f(a)\plus{}f(b)}{2}$ ενω η προηγουμενη βασιζεται στην γενικοτερη  αποδειξη των ανισοτητων $ Jensen$ με Θ.Μ.Τ.στην κλαση των παραγωγισιμων συναρτησεων
Αλλος ενας τροπος  στοιχειωδης ειναι στην ανισωση αφου προσθεσουμε το $ |z\plus{}9|$ και στα δυο ,μετα ανα δυο με τετραγωνισμο ευκολα.....

[quote="wallace"]Αρχικά ονομάζω z+8=z'
Γράφω την τριγ. ανισ. στη μορφή: από $ |a \plus{} b|\leq|a| \plus{} |b|$ σε $ |z' \plus{} \frac {a \plus{} b}{2}|\leq|\frac {z' \plus{} a}{2}| \plus{} |\frac {z' \plus{} b}{2}| (1)$
Έτσι η δικιά μας από την $ (1)$ γίνεται:
$ |z' \plus{} 2| \plus{} |z' \plus{} 3| \plus{} |z' \plus{} 11|\leq\frac {|z'| \plus{} |z' \plus{} 4|}{2} \plus{} \frac {|z'| \plus{} |z' \plus{} 6|}{2} \plus{} \frac {|z' \plus{} 10| \plus{} |z' \plus{} 12|}{2}\leq\frac {|z'| \plus{} |z' \plus{} 4|}{2} \plus{} \frac {|z'| \plus{} \frac {|z' \plus{} 4|}{2} \plus{} \frac {|z' \plus{} 8|}{2}}{2} \plus{} \frac {\frac {|z' \plus{} 8|}{2} \plus{} \frac {|z' \plus{} 12|}{2} \plus{} |z' \plus{} 12|}{2} \equal{} |z'| \plus{} \frac {3|z' \plus{} 4|}{4} \plus{} \frac {|z' \plus{} 8|}{2} \plus{} \frac {3|z' \plus{} 12|}{4}\leq|z'| \plus{} \frac {3|z' \plus{} 4|}{4} \plus{} \frac {\frac {|z' \plus{} 4|}{2} \plus{} \frac {|z' \plus{} 12|}{2}}{2} \plus{} \frac {3|z' \plus{} 12|}{4} \equal{} |z'| \plus{} |z' \plus{} 4| \plus{} |z' \plus{} 12|$

Κύριε Πατήλα και οποιοσδήποτε άλλος επειδή δν τους ξέρω καλά τους μηγαδικούς ενημερώστε με για τυχόν λάθη......Sorry και για τυχόν typo...Το πρώτο μου $ LaTeX$ βλέπετε...[/quote]
Η λύση είναι αξιοθαύμαστη λόγω [b]γεωμετρικής ερμηνείας[/b]. Στηρίζεται στην ανίσωση $ 2\mu_a \le b\plus{}c$ (Α λυκείου) και στην αναπαράσταση των μιγαδικών με διανύσματα. Το πρώτο μέλος αποτελείται από διαμέσους ενώ στο δεύτερο αξιοποιούνται οι διάμεσοι με κατάλληλες πλευρές ""τριγώνων" ώστε τελικά να καταλήξουμε στις επιθυμητές πλευρές


let $ x_1,x_2,...,x_n$  are natural and different.
prove(or disprove):$ T: (x_1,...,x_n) \rightarrow (\frac{x_1\plus{}x_2}{2},\frac{x_2\plus{}x_3}{2},...,\frac{x_n\plus{}x_1}{2})$ is periodic

[quote="stvs_f"]let $ x_1,x_2,...,x_n$  are natural and different.
prove(or disprove):$ T: (x_1,...,x_n) \rightarrow (\frac {x_1 \plus{} x_2}{2},\frac {x_2 \plus{} x_3}{2},...,\frac {x_n \plus{} x_1}{2})$ is periodic[/quote]

What is the meaning of "periodic" for a function from $ \mathbb N^n\to\mathbb N^n$ ?

I took it to mean:  For any starting point $ \mathbf{x} \equal{} (x_1,\dots,x_n)$, if we do $ \mathbf{x}$, $ T(\mathbf{x})$, $ T(T(\mathbf{x}))$, and so on, then eventually we return to the starting point $ \mathbf{x}$.

But clearly that's false: $ (1,3) \mapsto (2,2) \mapsto (2,2) \mapsto \dots$ never returns to $ (1,3)$.

Nor is it necessarily eventually periodic.  $ (1,2,3) \mapsto (1.5, 2.5, 2) \mapsto (2, 2.25, 1.75) \mapsto \dots$ converges toward $ (2,2,2)$ but never repeats.

So I don't know what the OP intended.


Suppose that $ P(x)$ has exactly $ n$ distinct roots: $ x_1, x_2, \dots, x_n$, where $ P(x)$ is a polynomial of the $ n$-th degree. Let $ k$ be an arbitrary positive integer, such that $ k < n-1$. Prove that the following identity holds:
\[ \sum_{i=1}^{n}\frac{x_i^k}{P'(x_i)}=0
\]

$ \frac{1}{P(x)} \equal{} \sum_{k\equal{}1}^{n} \frac{1}{P'(x_i)(x \minus{} x_i)}$ by l'Hopital's rule, so

$ \frac{x^{n\minus{}1}}{x^n P \left( \frac{1}{x} \right)} \equal{} \sum_{k\equal{}1}^{n} \frac{1}{P'(x_i) (1 \minus{} x_i x)}$

and the rest follows by comparing the coefficient of $ x^k$ on both sides.


[quote="10000th User"][quote="kimby_102"][quote="pianoforte"]Find the sum of the roots $\tan^2{x}-9\tan{x}+1=0$ that are between $x=0$ and $x=2\pi$ radians.

$\text{(A)} \ \frac{\pi}{2} \qquad \text{(B)} \ \pi \qquad \text{(C)} \ \frac{3\pi}{2} \qquad \text{(D)} \ 3\pi \qquad \text{(E)} \ 4\pi$[/quote]
Using the quadratic equation,  we get $\tan x = \frac{9\pm\sqrt{77}}{2}$. However, reject $\frac{9+\sqrt{77}}{2}$, since it's bigger than 1.
Then..... I can't simplify anymore.. Are u sure about the pbm.?[/quote]Why are you rejecting $\frac{9+\sqrt{77}}{2}$? $\tan x$ can be larger than 1... There is nothing wrong about the problem.[/quote]
What a mistake!! Sorry :blush:

note that we can say of r is a solution, then 1/r is a solution...what does this correspond to in the sum of the x values?

[quote="pianoforte"]Find the sum of the roots $\tan^2{x}-9\tan{x}+1=0$ that are between $x=0$ and $x=2\pi$ radians.

$\text{(A)} \ \frac{\pi}{2} \qquad \text{(B)} \ \pi \qquad \text{(C)} \ \frac{3\pi}{2} \qquad \text{(D)} \ 3\pi \qquad \text{(E)} \ 4\pi$[/quote]

I have a guess...

[hide]The graph of $\tan x$ goes up until it to infinity before it reaches $\frac{\pi}{2}$ (don't know how to explain that) and then starts from negative infinity after $\frac{\pi}{2}$ and goes to back up to $\frac{3\pi}{2}$ amd then goes from the negative to 0 as it approaches $2\pi$. (If you don't understand, that's ok).

So, we have already found out that what $\tan x$ is.  One solution is before $x$ reaches $\frac{\pi}{2}$ and one is after it, but since it is symmetrical, the sum of those two solutions is $\pi$.

Doing the same thing for the other period, we get the sum of those solutions is $3\pi$.

So I think the sum is $4\pi$ if I didn't make any mistakes.

Obviously, if it isn't symmetric, then my answer is wrong :( [/hide]

Sorry for the bad explanation.  I can't really do it w/o a picture.

[quote="pianoforte"]Find the sum of the roots $\tan^2{x}-9\tan{x}+1=0$ that are between $x=0$ and $x=2\pi$ radians.
$\text{(A)} \ \frac{\pi}{2} \qquad \text{(B)} \ \pi \qquad \text{(C)} \ \frac{3\pi}{2} \qquad \text{(D)} \ 3\pi \qquad \text{(E)} \ 4\pi$[/quote]

$\tan^{2}x-9\tan x +1=0\iff\sec^{2}x=9\tan x\iff\sin 2x=\frac{2}{9}$

Then the sum of your solutions $x\in\left[0,2\pi\right]$ is

\[ \sum_{x\in\left[0,2\pi\right]} x =\sum_{n=0}^{3}\frac{n\pi +\left(-1\right)^{n}\arcsin\frac{2}{9}}{2}=3\pi \implies\boxed{D}  \]

[quote="matt276eagles"]...
Is this a valid solution?  It seems a little shaky to me.  If $\tan\alpha\tan\beta =1$, can we assume $\alpha+\beta=\frac{\pi}{2},\frac{3\pi}{2}$?[/quote]

$\tan\alpha\tan\beta =1$
$\sin\alpha\sin\beta=\cos\alpha\cos\beta$
$\cos(\alpha+\beta)=0$
$\alpha+\beta=\frac{\pi}{2}+k\pi$ ($k\in\mathbb{Z}$)

[quote="amcavoy"][hide]$\sin 2x=\frac{2}{9}$[/hide]
[/quote]

[hide]Continuing from there, when you graph that you get $\sin x$ with a period of $\pi$ and 4 values of $x$.  The first two sum up to $\frac{\pi}{2}$ because they are symmetrical with respect to $\frac{\pi}{4}$ so you just double that.

The last two sum up to $\frac{5\pi}{2}$ because they are symmetrical with $\frac{5\pi}{4}$ being in the middle so you just double that.

So the answer is $\boxed{3\pi}$.[/hide]

It's good that I realized what I did wrong, and I didn't know the identity that joml gave so that helped me :)

[quote="matt276eagles"][hide]There are two roots for $\tan x$.  Since $\tan x$ has period $\pi$, there are two roots, $r_{1}$ and $r_{2}$ in $[0,\pi]$.  There are two more roots, $r_{1}+\pi$ and $r_{2}+\pi$ in $[\pi,2\pi]$.  Then the sum of the roots is $2(r_{1}+r_{2})+2\pi$.

By Vieta's sums, we have $\tan r_{1}\tan r_{2}=1$.  Then $\tan(r_{1}+r_{2})=\frac{\tan r_{1}+\tan r_{2}}{1-\tan r_{1}\tan r_{2}}=\frac{\tan r_{1}+\tan r_{2}}{0}$, which is undefined.  Therefore, $r_{1}+r_{2}=\frac{\pi}{2},\frac{3\pi}{2}$, and the sum of the roots is $\boxed{3\pi}$ in either case.[/hide]

Is this a valid solution?  It seems a little shaky to me.  If $\tan\alpha\tan\beta =1$, can we assume $\alpha+\beta=\frac{\pi}{2},\frac{3\pi}{2}$?[/quote]

Or...you can arrive at that fact directly from

$\tan(r_{1}+r_{2})=\frac{\tan r_{1}+\tan r_{2}}{1-\tan r_{1}\tan r_{2}}=\frac{\tan r_{1}+\tan r_{2}}{0}$, which is undefined

Tangent is undefined at $\frac{\pi}{2},\frac{3\pi}{2}$, so you don't need to worry about that :).

[quote="amcavoy"]

$\tan^{2}x-9\tan x +1=0\iff\sec^{2}x=9\tan x\iff\sin 2x=\frac{2}{9}$

Then the sum of your solutions $x\in\left[0,2\pi\right]$ is

\[ \sum_{x\in\left[0,2\pi\right]} x =\sum_{n=0}^{3}\frac{n\pi +\left(-1\right)^{n}\arcsin\frac{2}{9}}{2}=3\pi \implies\boxed{D}  \][/quote]

hmm i get the first part... but the whole summation part is confusing me... can someone explain?

Well, since $\sin 2x=\frac{2}{9}$ and $x\in\left[0,2\pi\right]$, the values of $x$ which satisfy this are

$x=\frac{\arcsin\frac{2}{9}}{2}$

$x=\frac{\pi-\arcsin\frac{2}{9}}{2}$

$x=\frac{2\pi+\arcsin\frac{2}{9}}{2}$

$x=\frac{3\pi-\arcsin\frac{2}{9}}{2}$

The problem asks for the sum of these, so add them and you get $3\pi$ (the inverse sines cancel out).

hmm wait are u just subtracting $\arcsin \frac{2}{9}$ from $\frac{\pi}{2}$? and then adding it to $\pi$, and etc?

can someone explain why?


[quote="CircleSquared"][quote="Temperal"]1) Sid votes in an election. Currently, there are $ s$ support votes, $ n$ oppose votes, where $ 0 < s,n < 10$, and no neutrals. A neutral vote is counted as half-support, half-oppose. When Sid votes, what is the probability that the percentage of support votes goes down by more than 5%?

2) Ante votes in the same election after Sid. He, however, opposes. Find the change in the percentage after he votes in terms of the change in the percentage after Sid voted.[/quote]

what does sid vote[/quote]
I think it's any of the 3, oppose, support, or neutral.


parameterization
linear algebra
matrix
quadratics
probability and stats

Can anyone help me to understand how the following parameters would be estimated:

The model,

$ y_1(t) \equal{} a_{11}\cdot x_1(t) \plus{} a_{12}\cdot x_2(t) \plus{} u_1(t)$

$ y_2(t) \equal{} a_{21}\cdot x_1(t) \plus{} a_{22}\cdot x_2(t) \plus{} u_2(t)$

and we know that

$ a_{11} \plus{} a_{12} \equal{} a_{21}$

$ a_{11} \minus{} a_{12} \equal{} a_{22}$

$ u_1(t) \sim N(0,\sigma^2)$

$ u_2(t) \sim N(0,4\sigma^2)$

and $ u_1(t)$ and $ u_2(t)$ are independent

Can anyone help me in understanding how to estimate the parameters $ a_{11}$, $ a_{12}$, $ a_{21}$, $ a_{22}$, and $ \sigma^2$.

When considering how to estimate the parameters, I thought that it looks very familiar to using a statistical test of some sort, like the Ramsey test, Glejser Test, Goldfeld and Quandt test, etc. , but cannot see the direct connection with which statistical test is more appropriate.

Can anyone help me understand this model better by possibly explaining how the parameters would be estimated.

Thanks.

I'm not really an expert, but I would just consider $ Y\equal{}\begin{bmatrix}y_1\\ \frac{y_2}2\end{bmatrix}$, $ X_1\equal{}\begin{bmatrix}x_1\\ \frac{x_1\plus{}x_2}2\end{bmatrix}$, $ X_2\equal{}\begin{bmatrix}x_1\\ \frac{x_1\minus{}x_2}2\end{bmatrix}$, $ U\equal{}\begin{bmatrix}u_1\\ \frac{u_2}2\end{bmatrix}$, which reduces the model to
\[ Y\equal{}a_1 X_1\plus{}a_2 X_2\plus{}U\]
with $ a_1\equal{}a_{11}, a_2\equal{}a_{12}$ in your notation, i.e., to a $ 2$-dimensional linear regression with standard $ 2$-dimensional Gaussian error $ U$ and would use the good old least squares method to estimate $ a_1,a_2$ and $ \sigma^2$. Do you see anything wrong with this approach? (Note that the division by $ 2$ is there to make $ U$ "round", which is always a good idea unless you prefer to minimize some fancy quadratic form instead of the usual sum of squares of distances; that the variables are $ \mathbb R^2$-valued rather than $ \mathbb R$-valued shouldn't matter much either though you'll have to adjust the standard formulae a little bit).

Fedja,

It looks as though you choose to start with putting the model in matrix form; that is one method I didn't consider.

From your explanation it looks like it might work, but I would have to work through it.

The model appears as though you should use a statistical test of some sort, like the Goldfeld and Quandt test, since it is presented in two groups. I believe it is represented most like the Goldfeld and Quandt test.

What I have determined is that in any statistical test used, there is always a transformation of some sort, which ultimately uses the OLS method after the model is transformed in such a way that OLS estimation would work.

[quote="Cezanne123"]
From your explanation it looks like it might work, but I would have to work through it.
[/quote]
Sure. Feel free to ask questions any time (though I may be not always able to answer).
[quote]
What I have determined is that in any statistical test used, there is always a transformation of some sort, which ultimately uses the OLS method after the model is transformed in such a way that OLS estimation would work.[/quote]
That was my impression too when I was learning all that stuff (which was 20 years ago and I cannot say I remember much nowdays :( but, fortunately, there are younger and more knowledgeable people here to help you out :) )


199^13^15^17mod19 ?
answer is 9
pls suggest a short solution


The number 313726685568359708377 is the 11th power of some number n. Find n.

[hide]
$ n^{11}=313726685568359708377$

to figure out the approximate range of the number, we do some rounding

$ 1 \cdot 10^{11}< n^{11}< 1 \cdot 10^{22}$

$ 1 \cdot 10^{1}< n < 1 \cdot 10^{2}$

so n is a 2 digit number

now to get the exact number, we look at some modulo
It is clear that n is odd or $ n \equiv 1 \mod 2$

Looking in mod 3

$ n^{11}\equiv 1 \mod{3}$
$ n \equiv 1 \mod{3}$

combining them by CRT
our answer must be $ 1 \mod{6}$

Looking in mod 5
$ n^{11}\equiv 2 \mod{5}$
$ n \equiv 3 \mod{5}$

combining again by CRT
$ n \equiv 13 \mod{30}$

this leaves us a handful of values
n = 13, 43 or 73

I'm going to show that n < 50 to show that 73 is our answer

$ 50^{11}= \left(\frac{100}{2}\right)^{11}= \frac{10^{22}}{2048}\approx 10^{19}$

since our number is about 300 times bigger than that, n=50 is too small so $ n=73$ is our answer
[/hide]


function
inequalities
number theory unsolved
number theory

Consider the function f defined as:f(n)=n+p,where p is the product of digits of n. Show that for every n (n belongs to N) f :^o: f :^o: f :^o: ...  :^o: f(n) becomes constant after a certain number of compositions.
(i couldn't find a way to write a sequence using emoticons and i used a function ).

The answer doesn't have very much to do with number theory.

Obviously, $f(n)=n$ if and only if the decimal expansion of $n$ has a zero.
If $n<10^k$, $f(n)-n \le 9^k$. Also, if $10^k \le n < 10^k+10^{k-1}$, $f(n)=n$ (since these numbers have a zero in the second place).
Now, if $k \ge 73$, $\left(1+\frac19 \right)^{k-1} \ge \left(1+\frac19 \right)^{72} \ge 1+\frac{72}{9} = 9$ by Bernoulli's inequality, so $\left(\frac{10}{9}\right)^{k-1} \ge 9$ and $10^{k-1} \ge 9^k$. Therefore, if $n < 10^k$ (with $k\ge 73$), $f(n) < 10^k+10^{k-1}$ and if $10^k \le n < 10^k+10^{k-1}$, $f(n) < 10^k+10^{k-1}$.

Therefore, if the first term of the sequence is less than $10^k+10^{k-1}$, for some $k \ge 73$, all terms are less than $10^k+10^{k-1}$. An increasing bounded sequence of integers is eventually constant, and we are done.

(I could have used integers of less than 73 digits there, but there was no need for a more refined inequality)


is everything divided by sqrt(3) or just the 7?

7*x+(7/sqrt(3))=1  then find   x+sqrt(x)=?

$7x + \frac{7}{\sqrt{3}} = 1$.  What is the value of $x + \sqrt{x}$?

Uh, x is negative. Are you expecting people to use complex numbers here?


Bob needs $ (n\plus{}1)$ heads to win the game where $ n$ is the number of tails. What is the probability that Bob will win with exactly 4 heads?

Sorry, I don't have the answer for this problem.

i think you are talking about National's target #8?
that question is almost the same with only 1 thing changed

Is the answer 1/32? From the information given, we know that Bob will flip 4 heads and 3 tails to win and that the last flip is a head and the first is a tail. There are 128 possible outcomes when flipping a coin 7 times. The favorable ones are:

THTHTHH
TTHHTHH
TTHTHHH
TTTHHHH

The reason why there are only 4 favorable outcomes is that he number of heads flipped can never exceed the number of tails flipped until the 7th flip. That is why THTHTHH works while THHTTHH doesn't. In THHTTHH Bob would have won on the 3rd flip, making it an unfavorable outcome. This also makes invalid any sequence beginning with a head.

Thus, there are 4 favorable outcomes out of 128 total possible. So the probability is 1/32. I hope my logic is correct.

I am not sure but I think you forgot

THTTHHH

Yes, he forgot  :)

We have:

TTTHHHH
TTHTHHH
TTHHTHH
THTTHHH
THTHTHH

Answer: 5/128

Is 5/128 correct? Would someone please tell us?


Show (and if you can, find) that there exists exactly one positive integer n such that 2^8 + 2^11 + 2^n is a perfect square.

It's easy to find: n=12. But how to show it is the only value :?:

Simple Unique Factorization.  :wink:

Masoud Zargar

$2^{8}+2^{11}=48^{2}$. So $2^{n}=(k-48)(k+48)$.
So we have that there are $a$ and $b$ such that $k-48=2^{a}$ and $k+48=2^{b}$...
But i don't know how to find $a$ and $b$ quickly to prove that the only answer is $12=a+b=n$

We have that $2^{n}=p^{2}-2^{8}-2^{11}\implies 2^{n}=(p-48)(p+48)$.  Let $n=u+v$.  Then By unique factorization, $2^{u}=p-48$ and $2^{v}=p+48$.  Subtracting the first from the second, we have $2^{u}-2^{v}=96=2^{5}\cdot 3\implies 2^{v}(2^{u-v}-1)=2^{5}\cdot 3$.  By unique factorization, $2^{v}=2^{5}\implies v=5$ and $2^{u-5}-1=3\implies u=7$.  Thus $n=7+5=12$.  Q.E.D.

Masoud Zargar


Let $a$ be an odd number such that $3\leqq a \leqq 9999$.Find all of $a$ such that $10000|a^2-a$.

[hide]
$a^2-a = a(a-1)$. Since $a$ is odd $2^4|(a-1)$ and obviously $5^4$ cannot divide both $a$ and $a-1$ so it divides on of them. But if $5^4$ divides $a-1$, then $10000|(a-1) \Rightarrow a-1 \ge 10000$. So it can't. Then $5^4|a$. So we look for all numbers such that:

$5^4|a$ and $2^4|(a-1)$

Let $a=5^4k$.
We see that $5^4 \equiv 1 \pmod{2^4}$ so $k \equiv 1 \pmod{2^4}$ for $a-1 \equiv 0 \pmod{2^4}$. So we have $k=1$ as our only possibility because if $k=17$, then $a > 10000$. So $a=625$ is the only solution. [/hide]

Nice solution, you can also solve it using 'automorphic' numbers, numbers whos squares end in themselves. The only 3-digit numbers are 376 and 625. (Prove this!) However 376 does not work. It is know there are no four digit automorphic numbers.

[quote="seamusoboyle"]$0$, $1$, and $625$[/quote]

$a \ge 3$ =P.

I read the problem, got carried away, and didnt finish it! :D

[quote]So a=625 is the only solution.[/quote]
$a=9376$  :D  :D  :D
Okay Okay 9376  is NOT odd, but that's not my fault... :)  :)  :)


Sa lahat naman ng filipino jan, na napapadaan lang sa Philippine math forums:

$ **$ [b]Mag-register naman kayo dito kung nangangailangan kayo ng need sa math homeworks, problems, etc.[/b] :wink:
$ **$  [b]Welcome lahat dito...!!  :lol:
$ **$ Mag post naman kayo ng mga math problems nyo......[/b] :D
$ **$[b] Pwede rin kayo mag-discuss dito!![/b]  :roll:
$ **$ [b]Kung may kilala kayong mahilig din sa math, invite nyo din na mag-register na!![/b] :lol:
$ **$ [b]Walang mag spam!![/b] :mad:
$ **$ [b]Open Ang Math Marathons dito[/b] :P

$ **$ [b] Kaya Kung May question kayo suggestion o violent reactions o anuman yan Reply lang kayo!! [/b] :P


geometry
circumcircle
angle bisector
geometry unsolved

[quote="cosinator"]Given a convex quadrilateral $ ABCD$ such that $ OA \equal{} \frac {OB\cdot OD}{OC \plus{} OD}$, where $ O$ is the intersection of the diagonals. The circumcircle of $ ABC$ intersects line $ BD$ at the point $ Q$.  Prove that $ CQ$ is the bisector of $ \angle DCA$.[/quote]
[hide="Solution"]
Let the bisector of $ \angle ACD$ meet $ BD$ at $ P$.  The Angle Bisector Theorem yields that $ OP \equal{} \frac {OC\cdot OD}{OC \plus{} OD}$.  Thus, $ \frac {OP}{OA} \equal{} \frac {OC}{OB}$.  This, combined with the fact that $ \angle AOP \equal{} \angle BOC$ yields that $ \triangle AOP\sim \triangle BOC$, so $ \angle BCA \equal{} \angle BPA$, so $ ABCP$ is cyclic, so $ P$ is the intersection between the circumcircle of $ \triangle ABC$ and $ BD$, which implies that $ P \equal{} Q$, so $ CQ$ bisects $ \angle ACD$. [/hide]

EDIT: Sorry, [b]yetti[/b], my solution does not completely work.  The $ OD$ in the denominator should be a $ CD$.  Therefore, the problem is reduced to showing that $ OD\equal{}CD$.

[quote="The QuattoMaster 6000"]...The Angle Bisector Theorem yields that $ OP \equal{} \frac {OC\cdot OD}{OC \plus{} OD}$...[/quote] What ? :what?:

[b]QuattoMaster[/b], sorry for pulling your leg. The problem is wrong, the condition $ OA \equal{} \frac{OB \cdot OD}{OC \plus{} OD}$ does not make $ CQ$ bisector of the angle $ \angle DCA.$ The correct condition is $ OA \equal{} \frac {OB \cdot OD}{OC \plus{} CD}$ and then your solution is fine. I was about to point out the error, when you posted.

The problem was posted before (with "iff") at
[url]http://www.mathlinks.ro/viewtopic.php?t=203401[/url],
[url]http://www.mathlinks.ro/viewtopic.php?t=198133[/url].


Two concentric circles are drawn such that the inner circle coveres $ 81\%$ of the area of the outer circle. Given that the radius of the outer circle is ten units, how many units are in the radius of the inner circle?

If the radius of the outer circle is $ 10$ units, its area is $ 100\pi$.

The problem gives us that the inner circles area is $ 81$% of that of the outer circle. Thus, the area of the inner circle is $ \frac{81}{100}\pi$.

Now, we work backwards. First, divide by $ \pi$ \[ \frac{\frac{81}{100}\pi}{\pi} \implies \frac{81}{100}\] Next, take the square root of that to get the radius of the inner circle, our answer \[ \sqrt{\frac{81}{100}}\implies \boxed{\frac{9}{10}}\]

Actually, the area of the inner circle is $ 0.81 \times 100\pi\equal{}81 \pi$ :wink:

From there we divide by $ \pi$ and take the square root to get the answer of $ \boxed{9}$.

1ST WE know that the outer circle radius is 10 so the area is 100pi the inner circle is 81/100pi working backward we get 9/10

 :bruce:  :bruce:  :bruce:  :bruce:  READY FOR HAND COMBAT!

Hello, I don't think that answer is right.  We know that the outer circle radius is $10$, so the area is $100\pi$.  The inner circle covers $81\%$ of this area, which is $81\pi$.  So the radius is $\sqrt{81} = \boxed{9}$.

Lol Chuck Norris doesn't have his own smiley but Bruce Lee does :)


[hide="1"]
Sum cyclically the inequality
\[ 11\cdot\frac{a^{4}}{b^{2}}\plus{}2\cdot\frac{b^{4}}{c^{2}}\plus{}\cdot\frac{c^{4}}{a^{2}}\geq 14\cdot\frac{a^{3}}{b}\]
proven by AM-GM. (It's clear that the ratio of $ \frac{b^{4}}{c^{2}}$ to $ \frac{c^{4}}{a^{2}}$ terms must be 2 to 1 to clear the $ c$'s, and then 11 was chosen to satisfy $ 4k\minus{}2\equal{}3(k\plus{}3)$.)
[/hide]

[hide="2"]
In the case $ a\equal{}8,b\equal{}4,c\equal{}1$, the left hand side is
\[ 2^{8}\plus{}2^{8}\plus{}2^{\minus{}6}\]
and the right hand side is
\[ 2^{9}\plus{}2^{3}\plus{}2^{\minus{}2}\]
which is greater.
[/hide]


function
real analysis
real analysis unsolved

let f be continuous on $ \mathbb{R}$. show that both $ f$ and its fourier transform $ \widehat f$ cannot be compactly supported unless f is identically 0.

If $ f$ is continuous and compactly supported then $ \widehat{f}$ is analytic.  But the only analytic compactly supported function is $ f\equiv 0$...

that is also the method my frd suggested...any real variable methods?

The notion of [url=http://en.wikipedia.org/wiki/Analytic_function]analytic functions[/url] makes sense in a real analytic context as well; they are just functions (locally) given by their power series.


problem: a bicycle has a pedal gear of 24 teeths with a speed of 45 revolutions/min and a rear gear of 16 teeths. If the diameter of the tire is 20 inches, what's the speed of the bicyle in mph? The answer is 4.02, but i dont see how...

[hide="Solution"]The revolutions/minute is inversely proportional to the amount of teeth in the gears, so:

$ 24\cdot 45 \equal{} 16x \Rightarrow x \equal{} \frac{135}{2}$ revolutions per minute.

Then, since the diameter of the tire is $ 20$ inches, the circumference is $ 20\pi$ inches, and the speed of the tire is $ \frac{135}{2}\cdot 20\pi$ inches per minute.

Then, using dimensional analysis, we have as our final answer:

$ \frac{135}{2} \cdot 20\pi \cdot 60 \cdot \frac{1}{12} \cdot \frac{1}{5280} \equal{} \frac{225\pi}{176} \approx \boxed{4.016240608}$ miles per hour.[/hide]


I've waited a while and there is no check mark next to round 4 on my profile.  I sent my solutions on monday night.  I sent an email to the solutions email address but my solutions weren't attached because I don't have my laptop with me right now since I'm away from home.  I can get them sent in later today if necessary.  Help. what should I do?

You probably do not need to worry. You might have noticed, above your scores, there is a bolded message that says:

"Most Round 4 Papers will not be marked received until March 24."

So, everything is probably okay (unless there is not a check there by March 25).

What [b]JGeneson[/b] wrote is correct.  The person who processes the USAMTS entries is on vacation :), so no entries will be processed until March 24.

What if it turns out that there is no check next to your name by the 25th? By then, the round would be under active discussion, so would your submission even count?

If you turned it in, there will be a checkmark next to your name by the end of the day on March 25th.


[quote="mathgeniuse^ln(x)"][quote="seamusoboyle"]That looks like some kind of trig identity[/quote]

Well, the first one is cos 45, the second is cos 22.5, the third one is cos (11.25), etc.[/quote]

[hide="So you might want to know the identity..."] $\frac{\sin x}{x}= \cos \frac{x}{2}\cos \frac{x}{4}\cos \frac{x}{8}...$ [/hide]

Is it easy to prove?

Then in that case the answer is 1/2, right.

No, not exactly...

[hide="Read more closely"] Look at the $x$ in the denominator.  In this expression we have $x = \frac{\pi}{2}$, so then

$\frac{\sin x}{x}= \boxed{ \frac{2}{ \pi }}$

According to MathWorld, this identity is due to Vieta. [/hide]

In the Pi book it was noted that this was probably the most beautiful and least efficient Pi calculation algorithm since it takes forever to converge...


Find minimum value of function:
$ f(x) \equal{} 6|x \minus{} 1| \plus{} |3x \minus{} 2| \plus{} 2x$

[quote="uytki"]Find minimum value minimum value of fuction:
$ f(x) \equal{} 6|x \minus{} 1| \plus{} |3x \minus{} 2| \plus{} 2x$[/quote]

Just study the three situations :
1) $ x\le \frac 23$

2) $ \frac 23 < x \le 1$

3) $ 1<x$

[b]CASE 1[/b] $ x < \frac {2}{3}$

$ f_1(x) \equal{} 6(1 \minus{} x) \plus{} (2 \minus{} 3x) \plus{} 2x$

$ \equal{} 8 \minus{} 7x$

[b]CASE 2[/b] $ 1 > x \geq \frac {2}{3}$

$ f_2(x) \equal{} 6(1 \minus{} x) \plus{} (3x \minus{} 2) \plus{} 2x$

$ \equal{} 4 \minus{} x$

[b]CASE 3[/b] $ x \geq 1$

$ f_3(x) \equal{} 6(x \minus{} 1) \plus{} (3x \minus{} 2) \plus{} 2x$

$ \equal{} 11x \minus{} 8$

Now we find the vertices (intersections).

$ f_1 \cap f_2 \equal{} (\frac {2}{3}, \frac {10}{3})$

$ f_2 \cap f_3 \equal{} (1,3)$

Clearly, the minimum is $ \boxed{3}$.

EDIT: WOW FAIL!!!

[quote="modularmarc101"]
Clearly, the minimum is $ \boxed{\frac {10}{3}}$.[/quote]
But  $ \frac{10}{3} > 3$, so the minimum is $ 3$...

hello, if $ x\equal{}1$ we get the global minimum $ 3$.
Sonnhard.


function
limit
integration
real analysis
real analysis solved

$\{\varphi_k\}\subset L^2(E)$ is an orthonormal system, and there is a $\Phi
\in L^2(E)$ s.t. $|\varphi_k| \leq |\Phi|, \mbox{ a.e. } x\in E$. If $\sum\limits_{k=1}^\infty a_k\varphi_k(x)$ converges almost everywhere, show that $a_k\to 0(k%Error. "toinf" is a bad command.
)$.

$\{e_n\}$ is an orthonormal system (I don't know if it is the correct word, so I write the meaning here), which means that it is an orthogonal system and $\|e_n\|=1$

[hide="Hint 1"] Since the series converges almost everywhere, the functions $\psi_k=b_k^2\varphi_k^2$ where $b_k=\min(1,|a_k|)$ tend to $0$ almost everywhere.[/hide]
[hide="Hint 2"] $\psi_k$ are also dominated by $\Phi^2$.[/hide]
[hide="End of solution"] The dominated convergence theorem now yields $
\lim_{k\to\infty}b_k^2=\lim_{k\to\infty}\int_E \psi_k = 0\,.
$
[/hide]


The product of three whole numbers is 36 (two of the numbers used could be the same).  What is the least possible sum of these three numbers?

A.10	B.11	C.12	D.13	E.14

[hide=" click to reveal hidden content"]

I make this $ 10$:

$ 3 \times 3 \times 4 \equal{} 36$ and $ 3\plus{}3\plus{}4\equal{}10$.

[/hide]

[hide]to get the smallest sum we try to find numbers as close to each other as possible that multiply to 36. We find that 3,3,4 work, since 3*3*4=36. Thus, the smallest sum is 3+3+4=10, which is (A).[/hide]

Hi;

[hide]

By examining the divisors of 36 and your choices.

3 * 4 * 3 = 36 and 3 + 3 + 4 = 10[/hide]

3 numbers are 3, 3, and 4. They sum to $ \boxed {10}$


solve in positive integers
x^n+y^n=2^2002

Aaargh...it has been moved into the intermediate topics before I post my solution...Too late, I cannot post it anymore :D

Pierre.

A very nice (and forgotten ?) problem !

Let's consider : x^n + y^n = 2^N

The case n=1 or N=1 is not too exciting and is ignored  ;)

I think I've proved that it only have the solutions x=y=2^k and kn+1=N

Let's suppose n > 1 and N > 1.

A few facts :
. x,y have same parity,
. if x and y are both even dividing by the appropriate power of 2 we can only consider the case x and y both odd (and prove there are no solution in this case)
. if x and y are odd and n is even then then x^n = 1 [4] so there are no solutions.
. if x and y are odd then (x,y) = 1.

So let x, y, n odd such as : x^n + y^n = 2^N

Then x+y | 2^N since n is odd, so x+y = 2^k with 1 < k < N.

x = 2X+1, y = 2Y+1 so X+Y+1 = 2^(k-1)

expand S = (2X+1)^(2k+1) + (2Y+1)^(2k+1) with binomial shows that
S = 2^k * odd_number, which is only posible if k=N, impossible.

Returning to our problem, solutions are (2^n, 2^n, k) with kn=2001, and (2^2002-y,y,1).

Hope it's correct.


If points A, B, C, and D are any 4 points on a circle, and P, Q, R, and S are the midpoints of the arcs AB, BC, CD, and DA respectively, show that PR is perpendicular to QS.

[hide]
The angle between $PR$ and $QS$ is $\frac{arc(PQ)+arc(RS)}{2}$

if we show that $arc(PQ)+arc(RS) = 180$, then it will follow that the chords are perpendicular

since $P,Q,R,S$ are midpoints, arcs $PB, BQ, SD, DR$ can be paired with arcs of equivalent measurements on $PS$ and $QR$

this means that the arcs we are focusing on comprise half of the full circle, or $180$ degrees, proving that the chords are perpendicular
[/hide]


$f: \mathbb R^{n}\longrightarrow\mathbb R^{m}$ is a non-zero linear map. Prove that there is a base $\{v_{1},\dots,v_{n}m\}$ for $\mathbb R^{n}$ that  the set $\{f(v_{1}),\dots,f(v_{n})\}$ is linearly independent, after ommitting Repetitive elements.

Let $k>0$ be the dimension of the range of $f$. We can then write $\mathbb R^{n}=\ker f\oplus V$ with $\dim\ker f=n-k,\dim V=k$. Choose a basis $(e_{i})_{i=1}^{n}$ of $\mathbb R^{n}$ with $(e_{i})_{i=1}^{k}\subset V,\ (e_{j})_{k+1}^{n}\subset\ker f$. Now the basis $(e_{i})_{i=1}^{k}\cup(e_{1}+e_{j})_{j=k+1}^{n}$ satisfies the requirements.


[color=darkblue]Let $ a\not \equal{} b, b\not \equal{} c, c\not \equal{} a$. Prove that: $ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(c \plus{} a)^2}{(c \minus{} a)^2}\ge 2$[/color]

[hide="well, a start..."]
$ 3 \plus{} (\frac {4ab}{(a \minus{} b)^2} \plus{} \frac {4bc}{(b \minus{} c)^2} \plus{} \frac {4ca}{(c \minus{} a)^2}\ge2$

If all three positive or negative, this is trival. 1 non-negative 2 negative is equivalent to 2 non-negative 1 negative, so WLOG, let's take a to be negative. Then we make the variable change a-->-a to get all the variables non-negative, and:

$ ( \minus{} \frac {4ab}{(a \plus{} b)^2} \plus{} \frac {4bc}{(b \minus{} c)^2} \plus{} \minus{} \frac {4ca}{(c \plus{} a)^2}\ge \minus{} 1$

$ \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \ge \frac {4ab}{(a \plus{} b)^2} \plus{} \frac {4ca}{(c \plus{} a)^2}$

$ (b \plus{} c)^2(a \plus{} b)^2(c \plus{} a)^2 \ge 4ab(b \minus{} c)^2(c \plus{} a)^2 \plus{} 4ca(b \minus{} c)^2(a \plus{} b)^2$

Which I don't want to bash out, but I'm sure that Muirhead and Schurs would clean it up.

Lol, can't think this early in the morning.

Edit: we can also assume $ b\ge c$ since it symmetric by b and c.[/hide]

[hide="This is something very familarly to us...."] we call  inequality above is Dao Hai Long' inequality, it has a very simple proof, that is set $ \frac{a\plus{}b}{a\minus{}b}\equal{}x; \frac{b\plus{}c}{b\minus{}c}\equal{}y; \frac{c\plus{}a}{c\minus{}a}\equal{}z$
So the main idea is to change the given inequality into new inequality with the condition, we have the expression;
$ (x\plus{}1)(y\plus{}1)(z\plus{}1)\equal{}(x\minus{}1)(y\minus{}1)(z\minus{}1)$
Extending the above expression, we get the condition $ xy\plus{}yz\plus{}zx\equal{}\minus{}1$
But we change the given inequality into $ x^2\plus{}y^2\plus{}z^2 \geq 2$
It's equivalent to $ (x\plus{}y\plus{}z)^2 \geq 0$ which is true [/hide] :)


How do you change a percent fraction to a fraction or a decimal?

i.e. how many thirty-sixths are in (83 1/3)%

Well percent means "out of one hundred"

That translates to $ x\% \equal{} \frac{x}{100}$

See if you can use this to find what the fraction or decimal would be and

hint: $ 83 \frac{1}{3} \equal{} \frac{250}{3}$, which is much easier to do arithmetic with.

$ 83\frac{1}{3}\equal{}\frac{250}{3}$
Since this is a percent, you can express this as
$ \frac{\frac{250}{3}}{100}$
Multiply both sides by 3, and you get
$ \frac{250}{300}$
Simplifying, this is
$ \frac{5}{6}$

Didn't he ask how many 1/36's there are in 83 1/3%? That means that you would divide 5/6 by 1/36 which would be 30.

If he didn't want the answer to that, then sry ppl =D


maybe it is not the right place that i post this super hard problem but the idea is really nice.
Let $H=(V,E)$ be a graph with maximum degree d,and let $V=V_1\bigcup V_2...V_r$ be partition into r pairwise disjoint sets.suppose each set $V_i$is of cardinality $|V_i|\geq 25d$.then there is an independent set of vertices $W\subseteq V$,that contains at least one vertex from each $V_i$.

It's the degree of a vertex, and it means the number of edges which have that vertex as one of their endpoints. The maximum degree of the graph is the maximal element of the set formed by the degrees of the vertices.

[quote="sam-n"]
Let $H=(V,E)$ be a graph with maximum degree d,and let $V=V_1\bigcup V_2...V_r$ be partition into r pairwise disjoint sets. Suppose each set $V_i$is of cardinality $|V_i|\geq 25d$. Then there is an independent set of vertices $W\subseteq V$,that contains at least one vertex from each $V_i$.[/quote]
I got nowhere with this problem..  :?

Someone willing to try? :)


Dillon has a 8.5 by 11 inch piece of paper. He folds it in half four times. He then uses a hole punch to punch a hole through the folded paper. What percent of the paper is removed if the punch's diameter is 1/2 inch?

There will be 16 layers, so the total area of the holes is $ .25^2(16)\pi$.  The area of the paper is $ 8.5 \times 11$.  Thus the answer is $ \frac{.25^2(16)\pi}{93.5} \implies \frac{\pi}{93.5} \equal{} \boxed{\sim 3}$ percent.

Good job.  :lol:

Extension: What is the maximum percentage of paper that can be punched away with one punch using this method, assuming a piece of paper can be folded infinitely many times, and that Dillon can only fold in half? Also, after each fold, the paper retains a thickness of zero.

I don't know what you mean y thickness0, but I think it is 100% as you can fold it so the hole is bigger the the piece of paper.  (of course the hole puncher has to be like really strong to go through like 10,00 papers)

thickness zero means that even if you fold it many times, the paper will still be flat. No thickness. zero.

well, let me clarify. The hole must fit in the paper. otherwise its not really a "punch" through the paper.

We fold it many times until we end up with sides .5625 and .53125.  The area is then ~.298828125 and the area of the punch $ .0625\pi$.  The answer is then $ .0625\pi \div .298828125 \equal{} \sim 66$ %

[quote="buzzer11"]How did you get .5625?[/quote]

$ 1.125 \div 2 \equal{} .5625$


[quote="eminem"]umm, sorry if this is stupid

IN TRIANGLE ABC

if $cos^2A + cos^2B + cos^2C = 1$

prove triangle is right angled[/quote]

$cos^2A + cos^2B + cos^2C = 1$

$\Rightarrow {\rm cos}^{\rm 2} {\rm A   =  sin}^{\rm 2} {\rm C  -  cos}^{\rm 2} {\rm B}$

$\Rightarrow {\rm cos}^{\rm 2} {\rm A   =   - cos(B + C) cos(B - C)}$

(in the triangle ABC, cos(B+C)= - cosA)

$\Rightarrow {\rm cosA\{ cosA - cos(B - C)\}  = 0}$

$\Rightarrow {\rm cosA\{  - cos(B + C) - cos(B - C)\}  = 0}$

$\Rightarrow 2{\rm cosAcosBcosC = 0}$

Hence any one of the angles is right angle.

[quote="Rushil"][quote="shyong"]why is it $sin^2C-cos^2B$ = $-cos(B+C)cos(B-C)$ ? :?[/quote]

Better expand and see. Open the RHS using the formula for $ \cos{A+B} $[/quote]

Oh ok , I just notice that I forget to use $cos^2+sin^2=1$ to further expand it  :P

By the way in any triangle $ABC$ we have \[\cos^2{\alpha} + \cos^2{\beta} + \cos^2{\gamma} + 2\cos{\alpha}\cos{\beta}\cos{\gamma} = 1,\] I think you should memorize that identity.


"Determine the last two digits of $ 3^3$, where there are 2007 threes."
OK, note that there is an error in the LaTeX code that I used.  It's supposed to have 3 to the third to the third to the third to the third to the third... and on.  Sorry that I don't know how to display that with LaTeX.

This isn't an actual MathCounts problem, but it is very similar to the MathCounts number theory problems.  Can anyone help?

so find a pattern.
FInd last two digits.
First it is just
With one 3- 3=03
With two threes- 3^3=27
With three threes- 27^3=83
With four threes- 83^3=87
With five threes- 87^3=03

SO it repeats with every 4 threes. SInce there are 2007, and 2007 mod4 is 3, the answer is when you have three threes, or 83.

I understand what you mean, Ihatepie, but I would've thought of what you thought of if I could.  I think you believe that there are parentheses.  If there is just a tower of threes as I described, then you solve the highest part first.  In other words, let's say the tower has:

One three.  Then $ 3^3 \equal{} 27.$
Two threes.  Then 3^27=a big number.

I can't even describe three threes because the number is WAY too big.  Do you understand that you did the reverse process?

That's why this problem is so hard.  BTW, the answer is 87.

You can't use a calculator, and you have about 3 minutes per problem.  Even if you could use a calculator, it wouldn't help much; the number is way too big.  I own Mathematica, which is a semi-expensive extremely powerful mathematics program.  It can do 1000000000! in less than 2 minutes (at least on my fast computer).  However, it can't do this problem.  I tried it.

This problem isn't that hard, necessarily, just sort of tedious. List the last two digits of powers of 3.

3^1=03
3^2=09
3^3=27
3^4=81
...43
29
87
61
83
49
47
41
23
69
07
21
63
89
67
01
03--yay finally!!! It cycles in groups of 20.

so now applied to the larger problem. again, we look for patterns.

3^3=27
3^3^3=3^27=87 (because it cycles once through the pattern of 3's, and then goes 7 more to 87)
3^3^3^3=3^3^27...hm...this looks tough. But think about it. We don't need the VALUE of 3^27. Just it's value (mod 20). So pattern time again!
3^1=3 mod 20
...9 mod 20
7 mod 20
1 mod 30
3 mod 30...and guess what? It cycles!!! (groups of 4)
So 3^27 would be 3 mod 4. So it would be 7 mod 20. So again, it would be 87 as the last two digits.
Then for 3^3^3^3^3=3^3^3^27
We do the same thing again. 3^27=3 mod 4---7 mod 20.
Then again, pattern "3^7"=87 as last two digits. But this time, it could be 3^3^___87. Luckily, it cycles in 20s. And divisibility by 20 is easy. The ___ part is n x 100! So it would cancel out. So again, 7 mod 20 etc.
"3^7" pattern=87 as last two digits.
 :D
See a pattern?
27-87-87-87-87...87...87...
answer:$ \boxed{87}$
QED pwn3d!!! :D

It is known that most numbers, including 3, end in 01 when to the 20th power, 40th power, etc. If we take the last 3 threes, we have have 3^(3^3). This simplifies to $ 3^{27}$. Thus, $ last2digits[3^{27}] \equal{} last2digits [(3^{20})(3^7)] \equal{} last 2 digits [(01)(3^7)] \equal{} last2digits[2187] \equal{} 87$ ($ 3^7$ can be done without a calculator.) Note that 100 is divisible by 20, so we only need to focus on the last two digits. So $ 3^{2187}$ leaves the same last two digits as $ 3^{87} \equal{} (3^{80})(3^7)$. $ 3^{80}$ ends in 01, so that leaves $ 3^7$, which = 2187, so the last digits are again, 87! This means the pattern keeps on going. If the number of threes > 2, then the last two digits are 87. This means using 2007 threes, the answer is $ [87]$. :D

OK. That's interesting. I'll mem that. But if you didn't know, I think my method would work best.

[quote="l Bob l"]It is known that most numbers, including 3, end in 01 when to the 20th power[/quote]

We can be more specific:  any number not divisible by $ 2$ or $ 5$ has this property.

That definition is a lot better.

And for 2, the only difference is that after 02, it starts with 04 and cycles in 20s.

Using Euler's Formula (which is a generalization of FLT), we find that \[ 3^{40}\equiv 1 \text{ (mod 100)}\](I know that $ 3^{20}\equiv 1$ (mod 100), though.  You take this result and show that it is impossible for $ 3^{20}$ to be congruent to 49, 51, or 99 in mod 100, so thus it must be congruent to 1 in mod 100) So if we can find $ 3^{3^{3^{...}}}$ in mod 40 (that's 99 three's, btw), then we're basically done.  There doesn't seem to be any easy way to do this though.  If we use the formula again, this reduces to finding $ 3^{3^{3^{...}}}$ in mod $ \phi(40)=16$, with 98 threes.  And we do it again: $ 3^{3^{3^{...}}}$ with 97 threes in mod 8.  Now, it's pretty easy.  Since $ 3^2\equiv 1\text{ (mod 8)}$, $ 3^{3^{3\&{...}}}$ with 97 threes is 3 in mod 8.  Now, $ 3^{3}\equiv 11\text{ (mod 16)}$, and then $ 3^{11}\equiv 27\text{ (mod 40)}$, and finally \[ 3^{27}\equiv 3^7\equiv 87\text{ (mod 100)}\]Final answer: $ \boxed{87}$ (hmm, that actually was pretty interesting)

[quote="alanchou"](I know that $ 3^{20}\equiv 1$ (mod 100), though.  You take this result and show that it is impossible for $ 3^{20}$ to be congruent to 49, 51, or 99 in mod 100, so thus it must be congruent to 1 in mod 100) [/quote]

Or... $ 3^{20} \equiv 1 \bmod 100$ follows directly by [url=http://mathworld.wolfram.com/CarmichaelsTheorem.html]Carmichael's Theorem[/url].  :)


geometry
rectangle
combinatorics unsolved
combinatorics

There are $ ab$ points arranged in a grid $ a$ x $ b$, each colored in one of the $ k$ colors. It is known that every four points which form a rectangle (as vertices) with sides parallel with the axis, are not all of the same color. For a fixed $ k$, determine the greatest possible $ ab$.

There is no upper bound for $ ab,$ just consider $ a \equal{} 2$ and color one of the rows in one color and the other row in another color.

Some similar problem can be very interesting, though; because the related problem with $ a \equal{} b \equal{} n$ is I think already known, what about the following one:

[i]Given that the grid of points $ \{1, \ldots, n\} \times \{1, \ldots, n\}$ is colored in $ k$ colors in a way that any [color=darkred]square[/color] with vertices on the grid and sides parallel to the axis does not have all vertices in the same color. Find the maximum of $ n$ as a variable of $ k$.[/i]

Using the [url=http://en.wikipedia.org/wiki/Lovasz_local_lemma]Local Lemma[/url] we can find that $ n_\text{max} > \tfrac{k^3}{4e}.$ What about finding the exact maximum?


quadratics
induction
number theory unsolved
number theory

Not really a problem exactly, but how would you prove that for odd primes, a is a square mod p if and only if a is a square mod p^e?

Are there any elementary solutions ?  :maybe:

Rust: really _very_ helpfull answer... ;)

RDeepMath91: Is there any nonelementary solution¿ "Elementary" is a term without proper objective meaning. And Hensels lemma is surely elementary if correctly stated.

Induction on $e$ produces a trivial solution...


Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1$ and
\[ f \left( x \plus{} \frac {13}{42} \right) \plus{} f(x) \equal{} f \left( x \plus{} \frac {1}{6} \right) \plus{} f \left( x \plus{} \frac {1}{7} \right).
\]
Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x \plus{} c) \equal{} f(x)$ for all $ x \in \mathbb{R}$)


Let $ z \in \mathbb{C}^*$ so that $ \left| z^3\plus{}\frac{1}{z^3} \right| \le 2.$ Prove that:

\[ \left| z\plus{} \frac{1}{z} \right| \le 2\]

[quote="moldovan"]Let $ z \in \mathbb{C}^*$ so that $ \left| z^3 \plus{} \frac {1}{z^3} \right| \le 2.$ Prove that:
\[ \left| z \plus{} \frac {1}{z} \right| \le 2\]
[/quote]

We suppose that ( i ) : $ \left| z \plus{} \frac {1}{z} \right| > 2$.
Thus, (ii) : $ \left| z \plus{} \frac {1}{z} \right|^2 \minus{} 3 > 1$.

Multiplying (i) by (ii), we get  $ \left| z \plus{} \frac {1}{z} \right|^3 \minus{} 3\left| z \plus{} \frac {1}{z} \right| > 2$. But $ \left| z^3 \plus{} \frac {1}{z^3} \right| > \left| z \plus{} \frac {1}{z} \right|^3 \minus{} 3\left| z \plus{} \frac {1}{z} \right|$. So $ \left| z^3 \plus{} \frac {1}{z^3} \right| > 2$, which is in contradiction with the condition $ \left| z^3 \plus{} \frac {1}{z^3} \right| \le 2$. So what we supposed is wrong.
The conclusion follows immediately.


function
algebra
polynomial
geometry
rectangle
3D geometry
prism

Hey guys, I have this worksheet i have to do with exponential functions and three term polynomials. I dont really understand it no matter how much i tried. So could you guys help?

These instructions come from the worksheet, you dont have to do it
Instructions-
Take a large piece of thin cardboard and cut 20 by 30

From each cornor cut off a square. Size is your decision, just make consistent

Fold up the "Flaps" and use tape.

1. Math teacher, Mrs. X wants to incorporate math in real world. "Each of you cut off squares in different side lengths, I want you to model the volume of your own box in terms of x, which represents the length of the side of the square. Then use that formula to find the volume. You must give the formula in simplified form i.e. without parentheses, an equation s V=(three term polynomial)

2. Three kids,  used different values. The product of the side lengths are 120 and the sum of the side lengths is 15. Assuming they used consecutive integers, Calculate the volume of the box using the formula from part 1. List them in order.

I found out for part 2, the numbers were 4, 5, 6. I just needed help in finding the formula.

Thank you for all help given.

I can't quite understand the first set of instructions.  But the volume of a box is given by $ V\equal{}l\cdot w\cdot h$, where $ l$ is the length of the box, $ w$ is its width, and $ h$ is its height.  So in the second part, the volume of the box is just the length times the width times the height, or $ 4\times 5\times 6\equal{}120$ (doesn't really ratter which number is which).

So you see, you have this 20 by 30 rectangle  piece of cardboard. You snip of squares out of each corner. So it would become a fat cross, or a multiplication sign if you wish. I needed a formula on how to calculate the volume of the "Cross"

This is an easy way:

1) Find the area of the original figure and let it be $ a$.
2) Find the area of a square you cut out.  Let that be $ b$.
3) Plug each number into this equation: $ a \minus{} 4b$.  I basically found the area of the big thing and subtracted the area of the tiny things.

I get what you mean, but how would I make that into a three term polynomial...

Isn't it simply $ 6,000 \minus{} 4x^2$ where $ x$ is the length of the side of the square?

Actually Goldenboy, you dont end with a rectangular prism, its more of a really fat cross. So i dont get what you mean.

YOu fold up those pieces that are sticking out to make a rectangular prism.

If you wanted to know the area of the cross after snipping, you would have 5 rectangles with areas ((20-2x)(30-2x)),
2*((20-2x)x), and 2*((30-2x)x) which all simplifies to
(600 - 100x + 4x^2) + 2(20x - 2x^2) + 2(30x - 2x^2). If you simplify even further, you get 600 - 4x^2. But that's pretttty obvious since you only subtracted 4 x-by-x squares from 600. :roll:  :lol:

Well its a rectangular prism without a top. Once again i need a three term polynomial

[quote="goldenboy1.618"]So if you start with a 20-by-30 rectangle and cut off an x-by-x square from each corner, you are left with a rectangular prism with dimensions (20-2x) (30-2x) (x). So you distribute every factor and you get 600x-(40x^2)-(60x^2)+(4x^3) which simplifies to [b]4x^3 -100x^2 + 600x[/b].
For example, if you cut off a 5-by-5 square, you get (20-10) (30-10) (5) which simplifies to 10*20*5 =1000.[/quote]

$ \boxed{4x^3 \minus{} 100x^2 \plus{} 600x}$

Oh thanks, I thought the ^ was another term.  SO is the "x" the length of what you cut off?

Yeah, $ x$ is the side length of the squares you cut off.

Could you please explain the following sentences? I dont get it
So if you start with a 20-by-30 rectangle and cut off an x-by-x square from each corner, you are left with a rectangular prism with dimensions (20-2x) (30-2x) (x). So you distribute every factor and you get 600x-(40x^2)-(60x^2)+(4x^3) which simplifies to 4x^3 -100x^2 + 600x.
For example, if you cut off a 5-by-5 square, you get (20-10) (30-10) (5) which simplifies to 10*20*5 =1000.

Try doing this visually and make a model.  The volume of a rectangular prism is $ lwh$ (you should be able to tell what each variable means).  The length of the prism is less $ 30 \minus{} 2x$ because there are two squares.  The width is $ 20 \minus{} 2x$ because there are another two squares.  The height is $ x$.  Then you have:

\[ (30 \minus{} 2x)(20 \minus{} 2x)(x) \implies \boxed{4x^3 \minus{} 100x^2 \plus{} 600x}
\]

Oh ok, i get it now! Thank you 007math. Apparently you have solved all my other math dillemas! KUDOS TO YOU


let a,b,c is non-negative such that $ a\plus{}b\plus{}c\equal{}2$, no two of which are zero. prove that:
     $ \frac{ab}{c\plus{}\sqrt{c^2\plus{}ab}}\plus{}\frac{bc}{a\plus{}\sqrt{a^2\plus{}bc}}\plus{}\frac{ac}{b\plus{}\sqrt{b^2\plus{}ac}} \leq 1$

[quote="can_hang2007"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a \plus{} b \plus{} c \equal{} 2$, no two of which are zero. prove that:
     $ \frac {ab}{c \plus{} \sqrt {c^2 \plus{} ab}} \plus{} \frac {bc}{a \plus{} \sqrt {a^2 \plus{} bc}} \plus{} \frac {ac}{b \plus{} \sqrt {b^2 \plus{} ac}} \leq 1$[/quote]
It is very old and must be due to Hungkhtn's. You have transformed his inequality from:
$ \sum \sqrt {a^2 \plus{} bc} \le \frac {3}{2}(a \plus{} b \plus{} c)$
to this form. I think this way to create inequality is not so good.
See here: http://www.mathlinks.ro/viewtopic.php?t=197672[/quote]
yes.it is very easy

[quote="can_hang2007"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a \plus{} b \plus{} c \equal{} 2$, no two of which are zero. prove that:
     $ \frac {ab}{c \plus{} \sqrt {c^2 \plus{} ab}} \plus{} \frac {bc}{a \plus{} \sqrt {a^2 \plus{} bc}} \plus{} \frac {ac}{b \plus{} \sqrt {b^2 \plus{} ac}} \leq 1$[/quote]
It is very old and must be due to Hungkhtn's. You have transformed his inequality from:
$ \sum \sqrt {a^2 \plus{} bc} \le \frac {3}{2}(a \plus{} b \plus{} c)$
to this form. I think this way to create inequality is not so good.
See here: http://www.mathlinks.ro/viewtopic.php?t=197672[/quote]
you say is correct. it is an interesting result of hungkhtn.I am wrote it under form is ditto.but I have found a better result is:
let a,b,c are non-negative such that a+b+c=2.prove that:
$ \frac {a^2b^2}{4c^2 \plus{} ab} \plus{} \frac {b^2c^2}{4a^2 \plus{} bc} \plus{} \frac {c^2a^2}{4b^2 \plus{} ac} \leq 1$
can you please prove it?

[quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a + b + c = 2$, no two of which are zero. prove that:
     $ \frac {ab}{c + \sqrt {c^2 + ab}} + \frac {bc}{a + \sqrt {a^2 + bc}} + \frac {ac}{b + \sqrt {b^2 + ac}} \leq 1$[/quote]
It is very old and must be due to Hungkhtn's. You have transformed his inequality from:
$ \sum \sqrt {a^2 + bc} \le \frac {3}{2}(a + b + c)$
to this form. I think this way to create inequality is not so good.
See here: http://www.mathlinks.ro/viewtopic.php?t=197672[/quote]
you say is correct. it is an interesting result of hungkhtn.I am wrote it under form is ditto.but I have found a better result is:
let a,b,c are non-negative such that a+b+c=2.prove that:
$ \frac {a^2b^2}{4c^2 + ab} + \frac {b^2c^2}{4a^2 + bc} + \frac {c^2a^2}{4b^2 + ac} \leq 1$
can you please prove it?[/quote][quote="can_hang2007"]
Hello, I am sorry but I think you result is not better than Hungkhtn's one. Because under the condition $ a,b,c \ge 0$ and $ a + b + c = 2,$ it is clear that

$ \frac {a^2b^2}{4c^2 + ab} \le \frac {ab}{c + \sqrt {c^2 + ab}}.$ ;)[/quote]
why?
we have:
$ (\sum{\frac{ab}{c + \sqrt {c^2 + ab}})^2 \le 3\sum{\frac{a^2b^2}{(c+\sqrt{c^2+ab})^2} \le 3\sum{\frac{a^2b^2}{4c^2+ab}}}}$

[quote="tuantam1lan"][/quote][quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a + b + c = 2$, no two of which are zero. prove that:
     $ \frac {ab}{c + \sqrt {c^2 + ab}} + \frac {bc}{a + \sqrt {a^2 + bc}} + \frac {ac}{b + \sqrt {b^2 + ac}} \leq 1$[/quote]
It is very old and must be due to Hungkhtn's. You have transformed his inequality from:
$ \sum \sqrt {a^2 + bc} \le \frac {3}{2}(a + b + c)$
to this form. I think this way to create inequality is not so good.
See here: http://www.mathlinks.ro/viewtopic.php?t=197672[/quote]
you say is correct. it is an interesting result of hungkhtn.I am wrote it under form is ditto.but I have found a better result is:
let a,b,c are non-negative such that a+b+c=2.prove that:
$ \frac {a^2b^2}{4c^2 + ab} + \frac {b^2c^2}{4a^2 + bc} + \frac {c^2a^2}{4b^2 + ac} \leq 1$
can you please prove it?[/quote][quote="can_hang2007"]
Hello, I am sorry but I think you result is not better than Hungkhtn's one. Because under the condition $ a,b,c \ge 0$ and $ a + b + c = 2,$ it is clear that

$ \frac {a^2b^2}{4c^2 + ab} \le \frac {ab}{c + \sqrt {c^2 + ab}}.$ ;)[/quote][quote="tuantam1lan"]
why?
we have:
$ (\sum{\frac {ab}{c + \sqrt {c^2 + ab}})^2 \le 3\sum{\frac {a^2b^2}{(c + \sqrt {c^2 + ab})^2} \le 3\sum{\frac {a^2b^2}{4c^2 + ab}}}}$[/quote]
Hello, you are right with this Cauchy Schwarz step. But I think, following this step, we must prove that
$ \sum \frac{a^2b^2}{ab+4c^2} \le \frac{1}{3},$ which is not true.
However, the inequality
$ \sum \frac{a^2b^2}{ab+4c^2} \le 1$ as you stated is very nice and I like it. :)
Please see the attachment below for my proof. :)

Anyway, I am sorry for the previous mistake.  :oops:

[quote="tuantam1lan"][/quote][quote="can_hang2007"][/quote][quote="tuantam1lan"][quote="can_hang2007"][quote="tuantam1lan"]let a,b,c is non-negative such that $ a + b + c = 2$, no two of which are zero. prove that:
     $ \frac {ab}{c + \sqrt {c^2 + ab}} + \frac {bc}{a + \sqrt {a^2 + bc}} + \frac {ac}{b + \sqrt {b^2 + ac}} \leq 1$[/quote]
It is very old and must be due to Hungkhtn's. You have transformed his inequality from:
$ \sum \sqrt {a^2 + bc} \le \frac {3}{2}(a + b + c)$
to this form. I think this way to create inequality is not so good.
See here: http://www.mathlinks.ro/viewtopic.php?t=197672[/quote]
you say is correct. it is an interesting result of hungkhtn.I am wrote it under form is ditto.but I have found a better result is:
let a,b,c are non-negative such that a+b+c=2.prove that:
$ \frac {a^2b^2}{4c^2 + ab} + \frac {b^2c^2}{4a^2 + bc} + \frac {c^2a^2}{4b^2 + ac} \leq 1$
can you please prove it?[/quote][quote="can_hang2007"]
Hello, I am sorry but I think you result is not better than Hungkhtn's one. Because under the condition $ a,b,c \ge 0$ and $ a + b + c = 2,$ it is clear that

$ \frac {a^2b^2}{4c^2 + ab} \le \frac {ab}{c + \sqrt {c^2 + ab}}.$ ;)[/quote][quote="tuantam1lan"]
why?
we have:
$ (\sum{\frac {ab}{c + \sqrt {c^2 + ab}})^2 \le 3\sum{\frac {a^2b^2}{(c + \sqrt {c^2 + ab})^2} \le 3\sum{\frac {a^2b^2}{4c^2 + ab}}}}$[/quote]
sory can_hang2007,i was wrong when use cauchuy-schawrs
 i readed your proof!
my proof is:
assume $ \sum\frac {a^2b^2}{kc^2 + ab} \le 1$ ($ \frac {1}{3} \le k \le 4$)
we have:
$ \frac {a^2b^2}{kc^2 + ab} = ab(1 - \frac {kc^2}{kc^2 + ab})$
inequality is equivalent:
$ ab + bc + ca \le 1 + kabc\sum{\frac {c}{kc^2 + ab}}$
apply cauchuy-schawrs:
$ \sum{\frac {c}{kc^2 + ab} \ge \frac {(\sum{a})^2}{k\sum{a^3} + 3abc} = \frac {4}{k\sum{a^3} + 3abc}}$
we must prove that:
$ \sum{ab} \le 1 + \frac {4kabc}{k\sum{a^3} + 3abc}$
let $ q = \sum{ab}$,$ r = abc$
we have:$ q \le 1 + \frac {4kr}{k(8 - 6q + 3r) + 3r}$
<=> $ r(3 + 7k - 3kq - 3q) + 2k(q - 1)(3q - 4) \ge 0$(1)
 because $ k \ge \frac {1}{3}$=> $ 3 + 7k - 3kq - 3q \ge 3 + 7k - 4k - 4 = 3k - 1 \ge 0$
apply schur third degree:$ r \ge \frac {p(4q - p^2)}{9} = \frac {8(q - 1)}{9}$
(1)<=> $ (q - 1)(\frac {8}{9}(3 + 7k - 3kq - 3q) + 2k(3q - 4))) \ge 0$
   <=> $ (q - 1)(15kq - 12q + 12 - 8k) \ge 0$(2)
choose $ k = \frac {24}{23}$(such that$ \frac {1}{3} \le k \le 4$)=>(2)<=> $ \frac {84}{23}(q - 1)^2 \ge 0$
equality when $ a = b$,$ c = 0$

hmmmm.... quite easy.....
$ \sum_{cyc}\frac {ab}{c \plus{} \sqrt {c^2 \plus{} ab}}.\frac {c \minus{} \sqrt {c^2 \plus{} ab}}{c \minus{} \sqrt {c^2 \plus{} ab}} \equal{} \sum_{cyc}\frac {ab(c \minus{} \sqrt {c^2 \plus{} ab})}{ \minus{} ab} \equal{} \sum_{cyc}\sqrt {c^2 \plus{} ab} \minus{} c$
so $ \sum_{cyc}\sqrt {c^2 \plus{} ab}\le3$ $ \rightarrow$ $ 2\sum_{cyc}\sqrt {c^2 \plus{} ab}\le3(a\plus{}b\plus{}c)$
it's known that it's cyclic so we can assume $ a\le{b}\le{c}$ ,then by Cauchy we have $ 2\sqrt{b^2\plus{}ca}\plus{}2\sqrt{c^2\plus{}ab}\le 2\sqrt{2(b^2\plus{}c^2\plus{}ca\plus{}ab)}$
but we have $ 2\sqrt{a^2\plus{}bc} \le 2a\plus{}c$
so we just need tu proof $ 2\sqrt{2(b^2\plus{}c^2\plus{}ca\plus{}ab)}\le a\plus{}3b\plus{}2c$ , square both side we get $ (a\minus{}b\minus{}2c)^2\plus{}8c(b\minus{}c)\ge0$ so, equality holds if $ (a,b,c)\equal{}(1,1,0)$ and their permutations $ Q.E.D$

[quote="matrix41"]
it's known that it's cyclic so we can assume $ a\le{b}\le{c}$[/quote]Wrong! You assume that WLOG (with loss of generality). :D  WOLOG (without loss of generality), you may assume that $ a$ is the minimum, but you cannot assume that $ c$ is the maximum at the same time. You must consider both cases.

Oh. I didn't read the problem carefully. After I read the word "cyclic", I simply claim that the assumption is wrong. :P

ow sorry
wrong typing, hehehe.... it should be symmetric, I already change it....


is there a combinatorial proof / interpretation of the fact that -
$ 3f_n \equal{} f_{n\plus{}2} \plus{} f_{n\minus{}2}$

can someone please reply ..
how do you prove that it is a 1-3 correspondence between - ${ f_n}$ and $ f_{n-2}+f_{n+2}$?? :(

There is a combinatorial proof in "Proofs that Really Count," by Benjamin and Quinn, page 6.  You can find it on Google Books.

thanks!!!! it is just the book i am looking for!! the book is very nice thank you again..


I don't know if this one is right or not :
         Prove that there can't exist a, m,k which are positive integers such that [a^(k*m)]- 1=(p_1^a_1).....(p_n^a_n) and (a^k) - 1=(p_1^b_1)....(p_n^b_n) where a_1,b_1,..... are positive integers,p_1,... are primes.
      Please help me!!!!!! :(

[quote="Treegoner"]Prove that there can't exist a, m,k which are positive integers such that [a^(k*m)]- 1=(p_1^a_1).....(p_n^a_n) and (a^k) - 1=(p_1^b_1)....(p_n^b_n) where a_1,b_1,..... are positive integers,p_1,... are primes.
Please help me!!!!!!

Hi,your question is really a problem IMO Shortlisted 1997:Find all triples of positive integers b,m,n such that b>1,m>n and the set of prime divisors of $b^m-1$ is the same as the set of prime divisors of $b^n-1$.
Answer:b=2 k -1;m=2;n=1 where k is any integer greater than 1.
Maybe,you will get more confidence in this problem and I hope you can find out its solution!It's a beautiful and useful problem. :P


geometry
geometric transformation
rotation
analytic geometry
linear algebra
matrix
trigonometry

how can i show that every rotation in the four-dimensional space $ \mathbb{R}^4$ can be resolved into six rotations, in the planes, given a coordinate system $ (x,y,z,ct)$, $ xy, zy, xz, tx, ty, tz$, and that the most general form for a rotation in the $ tx$ plane is given by $ x \equal{} x'\cosh \psi \plus{} ct' \sinh\psi, ct \equal{} x'\sinh\psi \plus{} ct' \cosh\psi$?

The standard meaning of "rotation" in $ \mathbb{R}^n$ is that the matrix in question belongs to $ SO(n).$ That is, $ A^TA\equal{}I$ and $ \det A\equal{}1.$ The set of all rotations - that is, $ SO(n)$ - is a compact subset of the $ n\times n$ matrices. A rotation preserves the Euclidean norm of a matrix: that is, $ \|Ax\|\equal{}\|x\|,$ where $ \|x\|\equal{}\sqrt{x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2}.$

Those hyperbolic cosines and hyperbolic sines in what you wrote tell us that you [i]don't[/i] mean this sense of rotation; if we let $ \psi$ range over all of $ \mathbb{R},$ the result will be a non-compact set of matrices.

I'm assuming that what you mean by "rotation" is a matrix $ A$ such that $ m(Av)\equal{}m(v)$ where $ m(v)\equal{}x^2\plus{}y^2\plus{}z^2\minus{}c^2t^2.$

Is that what you're talking about?

yes, i'm assuming we're working in minkowsi space, where, as you told, $ \|x\|$ is substituted by $ m(x)$


when can two different quadrilaterals have the products of all their sides the same??

I don't know if there would be any relationship between the two, but an easy example is if you take a square with a side of 2, and then replace two of the adjacent sides with segments of length 1 and 4.  The product is still the same (2*2*2*2=2*2*4*1=16), but you have 2 distinctly different quadrilaterals.

are you sure about this problems statement? :)

Hi all,

     If you connect the same sides in different order the quad will be different,
     but product will be the same, I guess.

     Pestich.

    P.S.  Was it from Polish olympiad?

Take a square with sides 1, 1, 1,  and 1. Then take a kite with sides 0.5, 0.5, 2, and 2.