A={x l x=$m^{2}-n^{2}$, m, n $\in Z$}

Prove:

1. If s, t$\in$A, then st$\in$A

2. If s, t,$\in$A, t doesn't equal 0, then $\frac{s}{t}=p^{2}-q^{2}$, where p and q are rational.

[hide="Determinant method"]

1.  If $s,t\in\mathcal{S}$, then $s = \left|\begin{array}{cc}a & b \\ b & a\end{array}\right| = a^{2}-b^{2}$ and $t = \left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = c^{2}-d^{2}$ for some integers $a,b,c,d$. Therefore, \[\begin{aligned}st & = \left|\begin{array}{cc}a & b \\ b & a\end{array}\right|\cdot\left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = \det\left(\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]\cdot\left[\begin{array}{cc}c & d \\ d & c\end{array}\right]\right) = \det\left(\left[\begin{array}{cc}ac+bd & ad+bc \\ bc+ad & ac+bd\end{array}\right]\right) \\ & = (ac+bd)^{2}-(ad+bc)^{2}\ \in\ \mathcal{S}.\end{aligned}\] 2.  Clearly, it's enough to prove that if $t\in\mathcal{S}$, then $\frac{1}{t}= a^{2}-b^{2}$ for some $a,b\in\mathbb{Q}$. Let $t = \left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = c^{2}-d^{2}$ for some integers $c,d$. Then we have to prove that \[\exists a,b\in\mathbb{Q}: 1 = (a^{2}-b^{2})t = \left|\begin{array}{cc}a & b \\ a & b\end{array}\right|\cdot\left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = \left|\begin{array}{cc}ac+bd & ad+bc \\ bc+ad & ac+bd\end{array}\right|.\] So choose $a,b$ such that $ac+bd=1$ and $ad+bc=0$, i.e. $\left(a,b\right) = \left(\frac{c}{c^{2}-d^{2}},\frac{d}{d^{2}-c^{2}}\right)$.
[/hide]

[quote="computer_nuke"][quote="Kurt Gödel"][hide="Determinant method"]

1.  If $s,t\in\mathcal{S}$, then $s = \left|\begin{array}{cc}a & b \\ b & a\end{array}\right| = a^{2}-b^{2}$ and $t = \left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = c^{2}-d^{2}$ for some integers $a,b,c,d$. Therefore, \[\begin{aligned}st & = \left|\begin{array}{cc}a & b \\ b & a \end{array}\right|\cdot\left|\begin{array}{cc}c & d \\ d & c \end{array}\right| = \det\left(\left[\begin{array}{cc}a & b \\ b & a \end{array}\right]\cdot\left[\begin{array}{cc}c & d \\ d & c \end{array}\right]\right) = \det\left(\left[\begin{array}{cc}ac+bd & ad+bc \\ bc+ad & ac+bd \end{array}\right]\right) \\ & = (ac+bd)^{2}-(ad+bc)^{2}\ \in\ \mathcal{S}.\end{aligned}\] 2.  Clearly, it's enough to prove that if $t\in\mathcal{S}$, then $\frac{1}{t}= a^{2}-b^{2}$ for some $a,b\in\mathbb{Q}$. Let $t = \left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = c^{2}-d^{2}$ for some integers $c,d$. Then we have to prove that \[\exists a,b\in\mathbb{Q}: 1 = (a^{2}-b^{2})t = \left|\begin{array}{cc}a & b \\ a & b \end{array}\right|\cdot\left|\begin{array}{cc}c & d \\ d & c \end{array}\right| = \left|\begin{array}{cc}ac+bd & ad+bc \\ bc+ad & ac+bd \end{array}\right|.\] So choose $a,b$ such that $ac+bd=1$ and $ad+bc=0$, i.e. $\left(a,b\right) = \left(\frac{c}{c^{2}-d^{2}},\frac{d}{d^{2}-c^{2}}\right)$.
[/hide][/quote]

For the second part, why is it suffice to just prove that if $t\in\mathcal{S}$, then $\frac{1}{t}= a^{2}-b^{2}$? And what is the set %S%?[/quote]

Euhm, I meant $A$ of course.  :lol:

Hmm, you're right, it's non-trivial...I think. But let $\mathcal{U}= \{x\in\mathbb{Q}\ |\ \exists p,q\in\mathbb{Q}: x = p^{2}-q^{2}\}$. Clearly, both proofs still hold, i.e. $s,t\in\mathcal{U}\Rightarrow st\in\mathcal{U}$ (Theorem 1) and $t\in\mathcal{S}\Rightarrow\ \frac{1}{t}\in\mathcal{U}$ (Theorem 2). But $\mathcal{S}\subset\mathcal{U}$, so if $s\in\mathcal{S}$, then $s\in\mathcal{U}$, so $s,t\in\mathcal{S}$ implies that $s,\frac{1}{t}\in\mathcal{U}$ by theorem 2 and thus $\frac{s}{t}\in\mathcal{U}$ by theorem 1.

Correct me if I'm wrong  :)

[quote="Kurt Gödel"][hide="Determinant method"]
2.  Clearly, it's enough to prove that if $t\in\mathcal{S}$, then $\frac{1}{t}= a^{2}-b^{2}$ for some $a,b\in\mathbb{Q}$. Let $t = \left|\begin{array}{cc}c & d \\ d & c\end{array}\right| = c^{2}-d^{2}$ for some integers $c,d$. Then we have to prove that \[\exists a,b\in\mathbb{Q}: 1 = (a^{2}-b^{2})t = \left|\begin{array}{cc}a & b \\ a & b \end{array}\right|\cdot\left|\begin{array}{cc}c & d \\ d & c \end{array}\right| = \left|\begin{array}{cc}ac+bd & ad+bc \\ bc+ad & ac+bd \end{array}\right|.\] So choose $a,b$ such that $ac+bd=1$ and $ad+bc=0$, i.e. $\left(a,b\right) = \left(\frac{c}{c^{2}-d^{2}},\frac{d}{d^{2}-c^{2}}\right)$.
[/hide][/quote]

Whoa your solution is really long for the second one.
The one the book gave was
[hide]$\frac{s}{t}=\frac{st(\text{since we have already proved st}\in A)}{t^{2}}=\frac{m^{2}-n^{2}}{t^{2}}=(\frac{m}{t})^{2}-(\frac{n}{t})^{2}$[/hide]

[hide="1"]
Hmm, after some exploration I think I came up with a theorem...

An integer can be expressed in the form $m^{2}-n^{2}$ iff it is the product of two even or two odd integers. (ie. $\equiv 1,0 \pmod 4$)

Proof:
Given two odd integers $2x+1$ and $2y+1$,
$(2x+1)(2y+1)=((x+y+1)-(x-y))((x+y+1)+(x-y))=(x+y+1)^{2}-(x-y)^{2}$
Do similar for two even integers.
For the "only if" part, consider the parity of $(m+n)$ and $(m-n)$, the two values always have the same parity.

So for part 1, $s$ and $t$ $\equiv 0,1 \pmod 4$, therefore $st \equiv 0,1 \pmod 4$, and $st \in A$
[/hide]

[quote="eryaman"] it is the product of two even or two odd integers. (ie. $\equiv 1,0 \pmod 4$)[/quote]
Numbers in the form $m^{2}-n^{2}$ can also be $-1\mod4$.

[hide="A useful identity"] $(a^{2}-b^{2})(c^{2}-d^{2}) = (ac+bd)^{2}-(ad+bc)^{2}$[/hide]
The problem is nearly trivial by the determinant solution, though.

Edit:  Eryaman, your theorem is a nice idea, but it needs a little work.

[hide="Salvage"] [b]Theorem:[/b]  An integer is expressible as the difference of two squares iff it is congruent to $0, 1, 3 \bmod 4$.

Proof:  $m^{2}= 1+3+5+...+(2m-1)$
$n^{2}= 1+3+5+...+(2n-1)$

$m^{2}-n^{2}= (2n+1)+...+(2m-1)$

It's immediately obvious that we get every odd integer this way.  In addition, we get every even integer expressible as the sum of two consecutive odd numbers, one of which is congruent to $1 \bmod 4$ and the other of which is congruent to $3 \bmod 4$.  Hence, the sum is congruent to $0 \bmod 4$.  QED. [/hide]
Edit 2:  [b]Follow-up:[/b]

Count the number of ways an arbitrary integer $a$ is expressible as a difference of two squares.


trigonometry
Pythagorean Theorem
geometry

The navigator of a ship at sea has the harbor in sight at which the ship is to be docked. She spots a lighthouse that she knows is 1 mile up the coast from the mouth of the harbor and she measures the angle between the line-of-sight observations of the harbor and the lighthouse to be 20 degrees. With the ship heading directly toward the harbor, she repeats this measurement after 5 minutes of traveling at 12 miles per hour. If the new angle is 30 degrees, how far is the ship from the harbor?

The answer is 1.88 miles.

How do I find it?

After 5' the ships traveled 1 mile
By law of sine, we have the distance between the ship at that time and the lighthouse spot is d=$ \frac{\sin20}{\sin10}$
So the distance between the ship and the coast is $ d\cos30$
By Pythagorean theorem, we have the distance between the harbor and the coast is $ \sqrt{1-(d\sin30)^{2}}$
So the answer is $ d\cos30+\sqrt{1-(d\sin30)^{2}}\approx1.88$


Assume that x,y,z are non-negative real numbers satisfy the condition:
[tex]x+y+z=3[/tex]
Find the maximum value of m satisfying the inequality:
[tex] x^{m}y+y^{m}z+z^{m}x \le 3 [/tex]

It is similar to this problem.[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2549]www.mathlinks.ro/Forum/viewtopic.php?t=2549[/url]
but not same competely.

first use (2,1,0) to see
$2^{m} \leq 3$
so m 0 or 1.

to m=0 nothing to prove.
to m=1

$xy+yz+zx= \frac{(x+y+z)^2 -(x^2 +y^2 +z^2)}{2}$

Using $QM-AM$ and $x+y+z=3$
$(\frac{1}{3}(x^2 +y^2 +z^2))^{\frac{1}{2}} \geq \frac{1}{3}(x+y+z)$

$x^2 +y^2 +z^2 \geq 3$

$xy+yz+zx \leq \frac{9-3}{2}$

$xy+yz+zx \leq 3$

so m=1 is also a solution.


[quote="Vasc"]Let $ a,b,c$ be positive numbers such that $ abc=2$. Prove that

(a) $ (a+b+c-3)(6-ab-bc-ca)\le 1$;

(b) $ a+b+c-3 \ge \sqrt\frac{a^{2}+b^{2}+c^{2}-3}{3}$.[/quote]

Very simple in appearance, but hard to find an elementary solution. Mixing variable method kills then, also EV method can directly show that it suffices to consider $ a=b$. But I need to think more... about a less complicated solution...  :lol:

[quote="hungkhtn"] Mixing variable method kills they. :lol:[/quote]
I don't think so. :D

Can you tell me what is mixing method and EV method????????? :blush: Thanks a lot.

I agree with Vasc's opinion  :D Maybe Mixing variable method cann't kills them.
And they are really hard problems, the equality occurs if and only if $ a=2,b=c=1$

I don't agree ! My solution is the following (mix some variables to $ 1$)

Assume that $ a,b$ is chosen for which $ (a-1)(b-1)\ge 0$ (we can always do so), then $ a+b\le 1+ab$.

Denote $ F(x)=(c+x-3)(6-ab-cx)$. We have to prove that $ F(a+b)\le 1$. Indeed, $ F(1+ab)=(c+ab-2)(4-ab-c)=1-(c+ab-3)^{2},$ therefore

\[ F(a+b)-1=F(a+b)-F(1+ab)+F(1+ab)-1\]
\[ =(a+b-1-ab)(6-ab-c(c-3)-c(a+b)-c(1+ab))+F(1+ab)-1\]
\[ =(a-1)(b-1)(ab+bc+ca+(c-1)^{2}-5)-(c+ab-3)^{2}.\]

Notice that

\[ (c+ab-3)^{2}=(c+\frac{2}{c}-3)^{2}=(c-1)^{2}(1-ab)^{2}\ge (c-1)^{2}(1-a^{2})(1-b^{2}).\]

So, it suffices to consider that
\[ (c-1)^{2}(1+a)(1+b)\ge ab+bc+ca+(c-1)^{2}-5,\]
or
\[ (c-1)^{2}(ab+a+b)+5\ge ab+bc+ca.\]
If above one is true, we are done. otherwise, we refer that $ ab+bc+ca\ge 5$. Moreover,
\[ a+b+c \le 1+ab+c=4+\frac{(c-1)(c-2)}{c}\]
If $ (c-1)(c-2)\le 0$ then $ a+b+c\le 4$ and we are done. If $ c\ge 2$ then
\[ 1-(a+b+c-3)(6-ab-b-ca) \ge 1-(c+ab-2)(1-(c-1)^{2}(a+b+ab))\]
\[ =(c-1)^{2}(a+b+ab)(c+ab-3)-(c+ab-3) =(c-1)^{2}(a+b+ab)(c+ab-3)-(c-1)(1-ab)\ge 0.\]
It suffices to consider the last case $ c\le 1$. In this case, $ a,b\ge 1$. We will show that this case can not happen. Indeed,
\[ (c-1)^{2}(ab+a+b)+5\le ab+bc+ca \Rightarrow c(c-2)(ab+a+b)+5\le (c-1)(a+b)\]
\[ \Rightarrow c(2-c)(1+2ab)+(c-1)(1+ab) \ge 5\]
\[ \Rightarrow (2-c)(c+2)+c+1-ab\le 5.\]
This is wrong since $ ab\ge 1\ge c$.

The problem is proved completely.

Nice observation and solution :lol:

And I remember that there is a result (somehow basic to me, however I don't know how to prove it easily) that when $ a+b+c$ and $ abc$ is fixed, then $ ab+bc+ca$ is minimized when $ a\ge b=c$. If so, this problem seems solvable by direct computation.

Ingenious solution, Hungkhtn, but sophisticated enough. :o
We can generalized these inequalities as follows.

Let $ a_{1},a_{2},...,a_{n}$ be positive numbers such that $ a_{1}a_{2}...a_{n}=2$. Prove that

(a) $ (a_{1}+a_{2}+...+a_{n}-n)(n-\frac 1{a_{1}}-\frac 1{a_{2}}-...-\frac 1{a_{n}})\le \frac 1{2}$;

(b) $ a_{1}+a_{2}+...+a_{n}-n \ge \sqrt\frac{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n}{3}$.

[quote="hungkhtn"]I don't agree ! My solution is the following (mix some variables to $ 1$)

Assume that $ a,b$ is chosen for which $ (a-1)(b-1)\ge 0$ (we can always do so), then $ a+b\le 1+ab$.

Denote $ F(x)=(c+x-3)(6-ab-cx)$. We have to prove that $ F(a+b)\le 1$. Indeed, $ F(1+ab)=(c+ab-2)(4-ab-c)=1-(c+ab-3)^{2},$ therefore
\[ F(a+b)-1=F(a+b)-F(1+ab)+F(1+ab)-1 \]

\[ =(a+b-1-ab)(6-ab-c(c-3)-c(a+b)-c(1+ab))+F(1+ab)-1 \]

\[ =(a-1)(b-1)(ab+bc+ca+(c-1)^{2}-5)-(c+ab-3)^{2}. \]
Notice that
\[ (c+ab-3)^{2}=(c+\frac{2}{c}-3)^{2}=(c-1)^{2}(1-ab)^{2}\ge (c-1)^{2}(1-a^{2})(1-b^{2}). \]
So, it suffices to consider that
\[ (c-1)^{2}(1+a)(1+b)\ge ab+bc+ca+(c-1)^{2}-5, \]
or
\[ (c-1)^{2}(ab+a+b)+5\ge ab+bc+ca. \]
If above one is true, we are done. otherwise, we refer that $ ab+bc+ca\ge 5$. Moreover,
\[ a+b+c \le 1+ab+c=4+\frac{(c-1)(c-2)}{c}\]
If $ (c-1)(c-2)\le 0$ then $ a+b+c\le 4$ and we are done. If $ c\ge 2$ then
\[ 1-(a+b+c-3)(6-ab-b-ca) \ge 1-(c+ab-2)(1-(c-1)^{2}(a+b+ab)) \]

\[ =(c-1)^{2}(a+b+ab)(c+ab-3)-(c+ab-3) =(c-1)^{2}(a+b+ab)(c+ab-3)-(c-1)(1-ab)\ge 0. \]
It suffices to consider the last case $ c\le 1$. In this case, $ a,b\ge 1$. We will show that this case can not happen. Indeed,
\[ (c-1)^{2}(ab+a+b)+5\le ab+bc+ca \Rightarrow c(c-2)(ab+a+b)+5\le (c-1)(a+b) \]

\[ \Rightarrow c(2-c)(1+2ab)+(c-1)(1+ab) \ge 5 \]

\[ \Rightarrow (2-c)(c+2)+c+1-ab\le 5. \]
This is wrong since $ ab\ge 1\ge c$.

The problem is proved completely.[/quote]

This is quite similar with my solution. However, I think this is only proof of one part, if I'm not wrong...

Yes, exactly, this is only solution to part (a). This solution is quite sophisticated, but elementary.

For the general form, of course, EV is what we think first (for me, nSMV theorem is also useful to prove it easily), but this way may be not elementary. Personally, I think the elementary for general form, if exists, will be too complicated.


[quote="xxxyyyy"]Find consecutive four positive integers that are divisible by $ 13$, $ 15$, $ 17$ and $ 19$ respectively.[/quote]

[hide="brutal force.........."]

Let the smallest number be $ 13x$ , where $ x$ is positive integer

Then $ 13x\plus{}1 \equiv 0 \pmod{15}$
$ \Leftrightarrow x\equal{}15k\plus{}8$ for some integer $ k$

$ 13x\plus{}2\equal{}195k\plus{}106 \equiv 8k\plus{}4 \equiv 0 \pmod{17}$
$ \Leftrightarrow k\equal{}17y\plus{}8$ for some integer $ y$

$ 13x\plus{}3\equal{}195k\plus{}107\equal{}3315y\plus{}1667 \equiv 9y\minus{}5 \equiv 0 \pmod{19}$
$ \Leftrightarrow y\equal{}19z\plus{}9$ for some integer $ z$

So $ 959685z\plus{}31499, 959685z\plus{}31500, 959685z\plus{}31501, 959685z\plus{}31502$ ,where $ z$ is any non-negative integers, are the possible answers.

[/hide]


Post place this year at nats or at state and what state

10 in NH this year, aiming for top 4 in NH next

Except, Phil, technically you won't be back next year. :P

Unless of course you decide to visit again.

5th in Nats, Indiana.
Even though I'm also in 8th grade, I will be coming back next year as a guest with my brother, and, possibly, an assistant coach for Indiana. :)

But I doubt that's what you had in mind when you said coming back. :P

I want to be an assistant coach too, and Pardha too

As assistant coach, a slim chance, but possible.

1st in Chapter
1st in State
16th in Nation
1st of all Penguin!s

I'm in 8th Grade now.
So I won't be back as a competitor.  :(
oh, well. It was fun while it lasted.
Only 5 more days of school!

I'm in 8th grade, but I might still be helping out.  Our school never sent a student to nationals before, so maybe it will give our struggling program a boost.
3rd in chapter
3rd in state (unless you go by national scores, where I'm second in state)
105th in nation

2nd at State (Ind)
30th at Nats (Ind)
7th Grade
1st in Team (State)
1st in Team  (Nats)
Our team consisted of 6,7,8,8 graders

[quote="Xantos C. Guin"]Only 5 more days of school![/quote]

WHAAA??? Our school still has over a month...

You're lucky.  Our schedules are all skewed.  We go back to school in August, and now they're trying crunch summer vacation and turn Thanksgiving holidays into a week long.  What a waste!  They look at Japan and say, "Hmm.  They do well, and they have year-round school.  Let's have year-round school here!" except what they don't realize is that Japan does well not because they spread the same number of days around more (like they're trying to do here), but they simply have MORE DAYS OF SCHOOL.

Spreading the same number of days???
I thought WE were the ones that were speading the days around!
Lets see, fall vacation, winter vacation, holiday vacation (meaning christmas, new year vacation) spring vacation, summer vacation, and all those other holidays.
Why don't we do it the Chinese way? One big summer, one big Winter, and some scattered 3-days weekenders here or there? They still come out to be ~180 days, but the 2 big vacations wcan be 2-3 months long!
Ah well, we shouldn't be complaning. ther's gotta be SOME reason why it's been this way for 200+ years...

[quote="dwx314"]
Ah well, we shouldn't be complaning. ther's gotta be SOME reason why it's been this way for 200+ years...[/quote]

Because it's america. ;)

[quote="uberl33tmage"]1st chapter, 4th state, 17th national. AND YES I SHALL RETURN![/quote]

5th at chapter
2nd at state
41st national

and I'm not coming back...

Individual
1st place chapters, 2nd place CD
11th place states, 9-16th place CD
Team:
1st place chapters (entire team in CD)
1st place states (2 members in CD)
7th grade

[quote="uberl33tmage"]1st chapter, 4th state, 17th national. AND YES I SHALL RETURN![/quote]

Kevin? uberl33tmage? aznness? 2 accounts? Whaaa...?

1st in chapter, 39th at state (stupid target round)
But, I'll probably stay and help coach the team next year, since I am in eighth grade.

[quote="Phelpedo"]David and Pardha and Bob Fisher coaching the Terre Haute team?[/quote]

yes, mr. fischer agreed to letting me and david coach his team next year.

:huh: I think he was joking... coz there can only be one assistant coach, if you want to come up stage.....

4th chapter (1st CD)
5th state (2nd CD)
8th grade :rotfl:
155th nats

[quote="bigHC"]i dont care about coming on stage, i just want to help out the team[/quote]

That's the spirit!  ;)

1st chapter
1st state (2nd before CD)
91st nationals
Maine
7th grade

Since people are now posting their chapter and state ranks:

1st chapter (don't remember score)
2nd state (37) - 3rd in unofficial bracket countdown (won in consolation)
58th nationals (24) - in a tie of which one person of nine got on the list, darnit
Ohio
8th grade

School Rank:
1st at chapter
1st at state (3 in CD, and 2 edit: went to nats)

School rank:
1st in chapter
1st in state(4 in CD, except one was an indivdiual, 2 went to nats)

School:
5th Chapter
6th State
1 countdown both times.

3rd at chapter

2nd at state(alabama)...countdown round was thrown out because they accidently mixed up my score and placed me at 8th where i lost to the 9th place person..so the results were based only on written results

42nd at nats

*cracks knuckles*

Aiming for Countdown next year.

7th grade

2nd in Chapter
1st in State
63rd in Nats

The reason why I won at States but screwed up awfully at Nats is because I can't do the written rounds [b]at all[/b].

2nd at chapter
22 at state
 :(
6th grade


On reading through the posts on this forum  I notice that many students either play this piece or like listening to it .
Can anyone explain why this piece is so popular.?(The composer never published it in his lifetime because of its ressemblances to a work of  another composer.)What are the qualities we find in Chopin"s music? Is this work representative of these qualities?

Another work mentioned several times is the Mozart Symphony No 40.What state  was the composer trying to cnovey in this work?Is it radiant with joy as Wagner said , or something quite different?
I for one am unable to fathom this remark by Wagner.I find this piece too distressing for words.

I'm not a musician, but I can see that a lot of chopin's music came from polish folk songs, which is different from many other composers.

its fast, its running like nobody's business, and the consecutive notes are so near each other.. my friends classify fantasie as a "ëtude" rather than a music piece.. haha..

I completely agree with Arne. I played the Fantasie-Impromptu ages ago and it's nothing special. It's not hard at all. The Chopin Sonatas and Ballades are his best works in my opinion, especially Sonata No.3 (sighs). And I think Ballade No.1 is the cornerstone to all piano music.

[quote="Arne"]I really don't think the fantaisie impromptu is so special compared to, for example, the ballades, scherzos and sonatas (2 and 3). Of course it's not too hard so that's probably why many people play it, and it's easy to listen to. Personally, I love playing the second scherzo...[/quote]

yesterday i heard a person play the second scherzo. I really like it, but the way this person played it, it was a complete disaster.  :P

When I started playing it, I was also messing up. It's quite a nasty piece...

My piano teacher just gave it to me a week ago, i agree its overplayed.

I think the popularity of a piece widely rely on how it was introduced to the people and if people associate something else with it.

For example: Fur Elise (spelling?) is so overplayed because of the many movies that use it, now its even in DDR.

[quote="Arne"]I really don't think the fantaisie impromptu is so special compared to, for example, the ballades, scherzos and sonatas (2 and 3). Of course it's not too hard so that's probably why many people play it, and it's easy to listen to.[/quote]

I heard a master class by Lydia Artymiw tonight, and she commented on one student who played the fantasy impromptu. It is a piece that's rather forgiving for the student musician: it can sound good before it is really played well. Listening to tonight's master class made me appreciate what is hard about that piece if one is to play it as Chopin wrote it.

Does anybody know where you can download sheet music of the Fantasie Impromptu, for free?

[quote="pianoforte"]try the Sheet Music Archive.

[url=http://sheetmusicarchive.net/index.cfm]sheetmusicarchive.net/index.cfm[/url][/quote]

Thanks!

I don't find the Fantasie-Impromptu any specialer or better than most of Chopin's works either. And yes, I do like the Ballades, Scherzos, and Sonatas better than the dreaded Fantasie-Impromptu.

Why is it so popular? It's pretty easy, it sounds impressive, and you don't need much brains to play it.
 :P

Ballade No. 3 is truly a challenging piece for me. Argh. And I have to perform it in three days.

Talk about it, my piano teacher gave me 2 totally new pieces and told me to go to an international comp.


geometry
parallelogram
trigonometry
3D geometry
AMC

5. In circle O, chord AB has length 9, chord CD has length 6, AB and CD meet at E, and AB is perpendicular to CD.  Tangents to the circle are drawn at A and D.  The tangents meet at a point P, with angle
APD = 20 degrees.  Find the measure of minor arc AD.

6. Compute the largest prime factor of   3^14 + 3^13 - 12.

7. Find all ordered pairs of real numbers, (x, y), such that
(x - 1/x)(y - 1/y) = 4 and (x - 1/y)(y - 1/x) = 25/6.

8.  In parallelogram ABCD, internal angle bisectors are drawn from each of the angles.  If AB = 5, BC = 4, and the sine angle ABC is s, compute the area of the region bounded by the four angle bisectors.

Wow, this was really interesting!  So long as the side lengths differ by 1, the area will be

[hide]s/2[/hide]

Yeah -- I was playing around with parallelograms and I just stumbled across the formula for that question.  I thought it was really neat.

How about you just make the problems out of 7 like most comps where you need to give solutions.. then you just say, yep you're right, 1/7 :)

Good job.  You can also factor (3^6 + 1) as (3^2 + 1)(3^4 - 3^2 + 1), the factorization for a sum of cubes.

4/7.  Your answer doesn't check with the original equations.  You found x/y = 2/3 or 3/2, but this doesn't imply that the only solutions are (3, 2) or (2, 3).

I was too lazy to just use algebra so I was using guess and check.  Thanks Joel!  :-).

-------------guess and check really really sucks  :? ------------------------

No one has presented a solution to 5 or 8 :(

we have 90 = (AD + CB)/2 so AD + CB = 180, and consequently, so does AC + DB. Then we have 20 = (CA + CB + BD - AD)/2, or 40 = 180 + CB - AD, which simplifies to CB + 140 = AD. Substituing in 180 - AD for CB, we get that 2AD = 180 + 140. DIviding by two, we have 90 + 70 = 160 as ARC AD.


algebra
polynomial
linear algebra
matrix
quadratics
group theory
abstract algebra

Let $ M$ be the set of all non-singular matrices of size $ 3\times3$ over $ \mathbb{Z}_2.$ Find the smallest positive integer $ n$ such that $ A^n \equal{} I$ for all $ A\in M.$
(A. Bondarenko)

In other words, we want to find the smallest positive integer $ n$ such that every characteristic polynomial of a matrix in $ M$ divides $ x^n \minus{} 1$.  Since the maximal degree of such a polynomial is $ 3$, and since the only cyclotomic polynomials of degree less than or equal to $ 3$ are $ \Phi_2, \Phi_3, \Phi_4, \Phi_6$, we conclude that if $ n$ exists it must divide $ 12$.  It remains to find examples of matrices with characteristic polynomials divisible by $ \Phi_4, \Phi_6$, which is computation.

The group has order $ 168$; there are sevens in there. The choice of field matters.

The answer should be $ 84$, since there are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21942]no elements of order $ 8$[/url].

My apologies; I don't have much experience in how characteristic polynomials behave over finite fields  :oops:

Since there are elements of order 3, 4, and 7, it suffices to show that $ A^{84} \equal{} I$ for all $ A\in M$.

Consider the characteristic polynomial $ p$ of $ A$.  Since the product of all irreducible polynomials of degree 1 or 3 is $ x^8 \minus{} x$, (basic finite fields result) if $ p$ is irreducible then $ A$ has order 7.  The product of all irreducibles of degree 1 or 2 is $ x^4 \minus{} x$, so if $ p$ is the product of one linear and one quadratic irreducible, $ A$ has order 3.

So, we must have $ p \equal{} (x \plus{} 1)^3$.  But this divides $ x^4 \minus{} 1$, so $ A$ has order 1, 2, or 4, and we are done.

Somehow I'm surprised to deduce that there are no elements of order, say, 6.

In general, the exponent of a finite group is the product of the maximum orders of elements in the Sylow subgroups for each prime. This doesn't tell us the extra information (elements of composite order) that LydianRain just found.

[quote="jmerry"]In general, the exponent of a finite group is the product of the maximum orders of elements in the Sylow subgroups for each prime.[/quote]

Interesting... do you know where I can find a proof of that?

Over $ \mathbb{Q}$ this is easier, because each $ \Phi_n$ is irreducible.  I believe $ GL(r,\mathbb{Q})$ has an element of order $ n \equal{} \prod p^\alpha$ if and only if $ r \geq \sum \tilde{\varphi}(p^\alpha)$, where $ \tilde\varphi(n) \equal{} \varphi(n)$, except for $ \tilde\varphi(2) \equal{} 0$.  (We can take $ A$ to be block diagonal, with each block the companion matrix of $ \Phi_{p^\alpha}$.  If $ 2\parallel{}n$, we can skip a block, and just multiply by -1.)

Over $ \mathbb{F}_p$ it's a different story.  The multiplicative group of $ \mathbb{F}_q$ is cyclic, so it contains a primitive $ n$-th root of unity iff $ n|q \minus{} 1$.  So if $ p\nmid n$, then the degree of a root of $ \Phi_n$ over $ \mathbb{F}_p$ is the smallest $ d$ such that $ n|p^d \minus{} 1$, i.e., the order of $ p$ mod $ n$.  And if $ n \equal{} p^\alpha m$, with $ p\nmid m$, then $ \Phi_n(x)\equiv\Phi_m(x)^{\varphi(p^\alpha)} \mod p$.

Recall that the product of all the irreducible (mod $ p$) polynomials of degree dividing $ d$ is $ x^{p^d} \minus{} x$.  So each such polynomial is either $ x$ or divides some $ \Phi_n$ with $ n|p^d \minus{} 1$.

Taking $ p \equal{} 2$, the only irreducible polynomials of degree $ d \equal{} 1,2,3$ are $ x$, $ \Phi_1$, $ \Phi_3$, and the two factors of $ \Phi_7$, say $ \varphi_7$ and $ \phi_7$.  So the minimal polynomial of $ A$ must be some product of $ \Phi_1,\Phi_3,\varphi_7,\phi_7$, with total degree no more than 3.  So there aren't that many cases to consider:
$ \Phi_1$: order 1
$ \Phi_1^2$: order 2
$ \Phi_1^3$: order 4
$ \Phi_3,\Phi_1\Phi_3$: order 3
$ \varphi_7,\phi_7$: order 7.

So the only possible orders in $ GL(3,\mathbb{F}_2)$ are 1,2,3,4,7.  (Compare 1,2,3,4,6 in $ GL(3,\mathbb{Q})$, same as for $ GL(2,\mathbb{Q})$.)  Moving up to 4 by 4 matrices, we have 1,2,3,4,5,6,7,15 in $ GL(4,\mathbb{F}_2)$.  (Compare to 1,2,3,4,5,6,8,10,12 in $ GL(4,\mathbb{Q})$.)

Is there a simple formula for the minimal $ r\equal{}r(n)$ for which $ GL(r,\mathbb{F}_p)$ contains a matrix of order $ n$?  If $ d(n)$ denotes the degree of an irreducible factor of $ \Phi_n$, we have over $ \mathbb{F}_p$,

$ d_p(p^\alpha) \equal{} 1$, $ d_p(q^{\beta \plus{} 1}) \equal{} q d_p(q^\beta)$, $ q\neq p$, $ \beta \gg 1$,
$ d_p(mn) \equal{} lcm(d_p(m),d_p(n))$,

whereas over $ \mathbb{Q}$, $ d_0(n) \equal{} \varphi(n)$, so of course

$ d_0(q^{\alpha \plus{} 1}) \equal{} q d_0(q^\alpha)$, $ \alpha\geq1$,
$ d_0(mn) \equal{} d_0(m)d_0(n)$.

For example, over $ \mathbb{Q}$, to get a matrix of order 15, we have $ r_0(15)\equal{}d_0(3) \plus{} d_0(5) < d_0(15)$, so it's more 'efficient' to take the companion matrices for $ \Phi_3$ and $ \Phi_5$, while over $ \mathbb{F}_2$, $ d_2(3) \plus{} d_2(5) > d_2(15)\equal{}r_2(15)$, so it's better to just take a factor of $ \Phi_{15}$.

For $ p\equal{}2$, the minimal $ r(n)$ goes 1,2,2,3,4,4,3,5,....  Surprisingly, this isn't in Sloane.

[quote="jmerry"]It's elementary. If an element has order $ p_1^{e_1}\cdots p_r^{e^r}$, some power of it has order $ p_1^{e_1}$. You can't beat the power of $ p$ you get in the Sylow subgroup.[/quote]

Touché.


function
trigonometry
inequalities unsolved
inequalities

Find the maximum and the minimum of the function: \[y=\sin^{3}x+2\sqrt{2-\sin^{2}x}-2\]

[quote="nobody1"]Find the maximum and the minimum of the function:
\[y=\sin^{3}x+2\sqrt{2-\sin^{2}x}-2 \]
[/quote]

The minimum is when $\sin x =-1$

$y=-1+2\sqrt{2-1}-2 =-1+2-2 =-1$

Any bode can find the max value of it for me?

$f(x) = sin^{3}x+2 \sqrt{2-sin^{2}x}-2$
$f'(x) = sin x \cdot cos x \cdot (3sinx+\frac{2}{\sqrt{2-sin^{2}x}})$

To find the extreme point, just let $f'(x) = 0$, we will have $sinx = 0,1,-1, \frac{\sqrt{30}-\sqrt{5}}{6}, \frac{\sqrt{5}-\sqrt{30}}{6}$.
You can check one by one and you can get the maximum value of $f(x)$..


If $ A \in M_n{\mathbb{(R)}}$  is a matrix such that : $ Tr(A\minus{}A^T)^{2n}\equal{}0$ for some integer $ n>0$ .Show that : $ A\equal{}A^T$

We assume $ Tr((A\minus{}A^T)^{2p})\equal{}0$.
$ H\equal{}A\minus{}A^T$ is a skew real matrix, then there exist positive real numbers $ a_1,\cdots,a_k$ s.t. $ spectrum(H)\equal{}\{0_1,\cdots,0_{n\minus{}2k},\pm{i.a_1},\cdots,\pm{i.a_k}\}$.
$ Tr(H^{2p})\equal{}(\minus{}1)^p2\sum_{j\equal{}1}^ka_j^{2p}\equal{}0$. Thus for any $ j$, $ a_j\equal{}0$ and $ H$ is a nilpotent matrix. Moreover $ H$ is diagonalizable over $ \mathbb{C}$ (because it is a normal matrix) and $ H\equal{}0$.


A tunnel is being digged from the north pole through the south pole of the earth . An object of mass $m$ is released from rest into the tunnel from the north pole . By assuming that the earth is a uniform sphere with radius $R$ and doesnt rotate , and let its gravity acceleration be $g$ .

(i) Find an expression for the tme taken for the object to come out from the south pole in terms of $g$ and $R$ .

(ii)If the object is released at the height of $H$ ( where $H$<<$R$) into the tunnel , prove that the velocity of the object when it reaches the center of the earth is given by

$v=\sqrt{gR}(1+\frac{H}{R})$

Can someone help me ?

The acceleration at distance $H=\alpha R$ ($\alpha<1$) from the center of Earth will be $\alpha g$.

i) The expression for the time taken for the object to travel from the north pole to the south pole in terms of  $R$ and $g$ is $\large T=\pi\sqrt{\frac{R}{g}}$

bus , I still dont understand how that helps ?

phucnv87 , can you show how ?

As Bus said, the gravitational force increases linearly with distance from the center of the earth, right?  Well, if so... doesn't that remind you of another law?

Specifically, Hooke's Law? $F=kx$ is, in this case, $F=\alpha g$
And you know how to work with springs...

The acceleration at the surface of the Earth is $\large g=\frac{GM}{R^2}=\frac{4}{3}\pi\rho R$
The acceleration at distance $x$ from the centre of the Earth is $\large g_x=\frac{GM_x}{x^2}=\frac{4}{3}\pi\rho x$
So, $\large g_x=\frac{g}{R}x$
The Newton's second law equation in this case is
$\large -F_x=-mg_x=-m\frac{g}{R}x=mx''$
And $\large x''=-\frac{g}{R}x$
Consider $\large \omega=\sqrt{\frac{g}{R}}$
We have the answer $\large \Delta t=\frac{T}{2}=\pi\sqrt{\frac{R}{g}}$

It is interesting that the answer does not change if the tunnel was not entirely vercital. If you for example dug a straight tunnel from London to New York and let a train "fall" down the tunnel frictionlessly, it would take it the same time to get to the other end. Taking this to extremes, a ball set on a flat surface not following the curvature of earth, oscillates naturally with this frequency:)

[color=blue]Really? It's also a problem in my test[/color]


For $ k \in N*$ and $ p > 3$ be a prime,let $ A \equal{} \frac {1}{1^k} \plus{} \frac {1}{2^k} \plus{} ... \plus{} \frac {1}{(p \minus{} 1)^k}$.Prove that if $ A \equal{} \frac {a}{b}$,where $ a,b \in N$ and $ gcd(a,b) \equal{} 1$,then $ p|a$.

If $ k\mid p\minus{}1$ it is trivial since in $ \mathbb{Z}/p\mathbb{Z}$ we have $ S_k: \equal{}\sum^*{i^{\minus{}k}}\equal{}\sum^*{1}\equal{}0$ (where * denotes that the sum is extended over all positive integer $ i<p$). Otherwise, let $ g$ a [i]primitive[/i] root mod p, $ \displaystyle \sum^*{i^{\minus{}k}}\equal{}\sum^*{g^{\minus{}ki}}$ $ \equal{} g^{\minus{}k}\sum_{0 \le i \le p\minus{}2}{(g^{\minus{}k})^i}\equal{}g^{\minus{}k}\frac{g^{\minus{}k(p\minus{}1)}\minus{}1}{g^{\minus{}k}\minus{}1}\equal{}0$.


Uh...all flights into boston from where we are have been cancelled for...freezing rain?

Anybody else have this issue?  (I heard TJ is too?)

YEah, same here.  I don't understand why they would just reject us like that.

You guys can fly into like NYC and take the Amtrak. That's a little bit of a stretch though.

It's not exactly freezing rain as of now, it's legit snow.

Yes, TJ will not be attending due to the questionable intelligence of our superintendent. (i.e. I see absolutely no ice on the roads)

Questionable to be sure.  The airlines told us freezing rain yet all weather reports simply say snow, light or heavy.  Even more odd is that out of the five flights from Detroit DTW to Boston (BOS?) only ours was cancelled -_-''...

any more ideas?  (Amtrak would take kinda long).

Fly into MHT and take a taxi? Probably expensive and that airport might have flights canceled too but you could try...

[quote="Phelpedo"]To be fair, the snow is rather heavy at this point. Missing HMMT is not the end of the world.[/quote]
Says you.
For some of us HMMT is the world.

Wait, WHAT?  I'm supposed to fly to Philadelphia on Wednesday.  Should I be scared?   :(  :(   I'm afraid of this cold frozen stuff, but, I mean, the flight stuff... :maybe:

[quote="Twiz"]I left my house in Georgia today at about 7:00 am today to fly to Boston. We had a connection in Virginia, and when we got there, we were unable to go on to Boston, so we spent the rest of the day trying to get back home. Finally, more than 15 hours after leaving my house, I am back precisely where I started. :mad: I hate to complain, but there's not much else to do after spending 15 hours on planes and in airports.[/quote]
same here from san fran to chicago and back
man i am pissed

Yeah, my plane from Baltimore to Boston yesterday was canceled.  Luckily, I was able to get Amtrak tickets.  After a 2 hour delay because they had to replace the engine (!) and a 7.5 hour train ride, I finally arrived in Boston around 12:30 AM.  Frustrating, to say the least, but at least I did end up getting to Boston.


$ \sum\frac{c^2\plus{}c\minus{}b}{a\plus{}b\plus{}2c}\leq\frac{1}{2}(\sum\frac{a^2}{b\plus{}c})$

$ a,b,c>0$

Using Cauchy chwarz inequality we have $ \frac {c^2}{a + b + 2c}\le \frac {1}{4}(\frac {c^2}{a + c} + \frac {c^2}{b + c})$

Hence it remains to prove that

$ \frac {1}{2}\sum_{cyc}(\frac {a^2}{b + c})\ge \sum_{cyc}(\frac {(a + c) - (a + b)}{(a + c) + (c + b)}) + \sum_{cyc}\frac {1}{4}(\frac {c^2}{a + c} + \frac {c^2}{b + c})$
$ \Leftrightarrow \frac {1}{2}\sum_{cyc}(\frac {1}{b + c})(\frac {a^2}{2} - \frac {b^2}{2} + \frac {a^2}{2} - \frac {c^2}{2})\ge \sum_{cyc}(\frac {(a + c) - (a + b)}{(a + c) + (c + b)})$
$ \Leftrightarrow \frac {1}{2}\sum_{cyc}(\frac {1}{b + c})(\frac {((a + c) - (b + c))(a + b)}{2} + \frac {((a + b) - (b + c))(a + c)}{2})\ge \sum_{cyc}(\frac {(a + c) - (a + b)}{(a + c) + (c + b)})$
Now let $ x = a + b$ and $ y = b + c$ and $ z = c + a$

The inequality becomes as

$ \sum_{cyc}(\frac {xz}{2y} - \frac {(x + z)}{4}) \ge \sum_{cyc}(\frac {z - x}{y + z})$
$ \Leftrightarrow \sum_{cyc}(\frac {xz}{2y} + \frac {x + y}{z + y})\ge 3 + \frac {x + y + z}{2}$

Which is obviously true !!  :D


linear algebra
matrix
absolute value
linear algebra unsolved

Prove that for an arbitrary [i]fixed[/i] matrix $A$, all sufficiently "small" matrices (i.e. the absolute value of its elemets is smaller than some $\epsilon$) $X$ satisfy $\displaystyle \textrm{rank} \left( A+ X \right) \geq \textrm{rank} \left( A \right)$.

what if I take X=-A and A is a "small matrix" with a single non zero element?? sorry if i didn't understand the problem correctly , but it seems false

I added now the word "fixed".

$A$ is a fixed matrix. For $X$ small enough, (we can find a matrix "smaller" than $-A$ and than $A$) $A+X$ has rank $\geq 1$. (in your case) Do you understand it now?

yeah...makes sense....

FORZA STEAUA!!!! :mad:

Write $A = PDQ$, $P,Q$ are invertible matrices and $D$ is a diagonal matrix of the form $I_k \oplus O$. Multiplication of invertible matrices preserves rank.
Thus,
$rank(A+X) = rank(D+PXQ)$ and $rank(A) = rank(D) = k$
For a sufficiently small matrix, $PXQ$ is also small, and we claim that adding $PXQ$ won't affect the rank of the first k rows of D. It suffices to prove the required result for invertible matrices to show this.

Note that $(I+H)^{-1} = I - H + H^2 +o(|H|^2)$ so when $|H|$ is small enough we see that $I+H$ is invertible thus the rank is unchanged.

you can also see that this way: If $rank(A)=r$ then there exists a $r \times r$ submatrix of $A$ that is invertible. And everyone know that the set of invertible matrices is open (determinant is continuous). So the conclusion follows because when you add a small matrix, then this submatrix stay invertible.


Every point of a plane is colored red or blue, not all with the same color. Can this be done in such a way that, on every circumference of radius $1$,
(a) there is exactly one blue point
(b) there are exactly two blue points

For i), no: suppose it were possible.  Draw a circle of radius 1, pick the blue point on it, and call this point $O$.  Every unit circle which passes through $O$ has all other points red, and every point at most distance 2 from $O$ is on such a circle, so in particular the unit circle centered at $O$ contains no blue points, contradiction.

For b), locate the blue points on // lines y=2n (n is an int)


combinatorics solved
combinatorics
graph theory

Suppose that each of the n guests at a party acquaited with exactly 8 other guests. Furthermore, suppose that each pair of guests who are acquainted with each other have four acquaintances in common at the party, and each pair of guests who are not acquainted have only two acquaintances in common. What are the possible values of n ?

Is, by any chance, n=21? Here's what I did. Each n is a common acquaintance (I'll replace this nasty word with AC) of 8C2 pairs of people. Then 8C2*n=4n*4+(nC2-4n)*2 because there are 8n/2=4n segments in the graph (2 people are connected if they know each-other). If we solve this we get n=21 (I think). Is it right?

No such n exists. First you can reduce it to the case n = 21 as you did it probably by counting the number of ordered triples (u,v,w), where u knows v and w but v does not know w.  But checking the case n = 21 yields a contradiction.

Oh, sorry abt that! I found that n must be 21, but I forgot to check if it was valid, since the way you get to n=21 doesn't guarantee that the soln is good.

See ya! :D


MATHCOUNTS
number theory
prime factorization

[hide="The relevant theorem, with two proofs"] [b]Theorem:[/b]  Let the prime factorization of some integer $n = \prod_{i=1}^{k}p_{i}^{e_{i}}$.  Denote $\sigma(n)$ to be the sum of the factors of $n$.  Then

$\sigma(n) = \prod_{i=1}^{k}\left( 1+p_{i}+p_{i}^{2}+...+p_{i}^{e_{i}}\right)$

There are two ways to prove this.  One way is simply to expand the above expression.  You see that you get every possible number of the form $\prod_{i=1}^{k}p_{i}^{d_{i}}$ where $d_{i}= 0, 1, 2, 3, ... e_{i}$; these numbers are precisely the divisors of $n$.

The other way is by two lemmas.

[b]Lemma:[/b]  Given $n = p^{k}$ for some prime $p$, we have $\sigma(n) = 1+p+p^{2}+...+p^{k}$ (this is obvious).

[b]Lemma:[/b]  $gcd(a, b) = 1 \implies \sigma(ab) = \sigma(a) \sigma(b)$ (every factor of $ab$ is a product of a factor of $a$ and a factor of $b$) [/hide]

[quote="samath"]$225=3^{2}5^{2}$

The sum of the factors is $(1+3+9)(1+5+25)=13\times 31=403$ (can you figure out why?)[/quote]

[hide]Because if you distribute, you get every combination of factors possible.[/hide]

check the mathcounts toolbox:
Finding the Sum of the Factors of a Number:
Example: What is the sum of the factors of 10,500?
(From the prime factorization 22 ´ 31 ´ 53 ´ 71, we know 10,500 has 3 ´ 2 ´ 4 ´ 2 = 48 factors.)
The sum of these 48 factors can be calculated from the prime factorization, too:
(20 + 21 + 22)(30 + 31)(50 + 51 + 52 + 53)(70 + 71) = 7 ´ 4 ´ 156 ´ 8 = 34,944.

[quote="k8reindeer"]I think you're missing quite a few exponents there. it took me a while to figure out what you were trying to say.[/quote]
sorry, i just cut and paste from the mathcounts toolbox

[quote="jli"][quote="mr. math"]is there any quick way to do this problem?

what is the sum of the positive integer factors of 225?[/quote]

what about the sum of the factors of 1000 that are not multiples of 4?[/quote]

[hide]the factorization of 1000 is $2^{3}5^{3}$.  however, we can only use one of the 2's, or else it will be a multiple of four.  So, the answer is $(1+2)(1+5+25+125)=\boxed{468}$[/hide]


analytic geometry
rotation
geometry
geometric transformation
linear algebra
matrix
reflection

Here's a problem I just made up.

[b]PROBLEM:[/b] An equilateral triangle $ABC$ has its rightmost vertex $B$ 2 units up from the x-axis.  A point P is inside triangle ABC, where $PA=5$, $PB=12$, and $PC=13$.  What are the coordinates of this equilateral traingle rotated 60 degrees to the right around its leftmost vertex $A$ having coordinates $A(0,0)$?

NOTE: There may be more than one solution.

Enjoy! :)

Masoud Zargar

well, use the method in your previous problem, rotate...etc. to determine the side length, then find point B (where it has a y coordinate of 2) intersect that with a circle radius of the triangle (there would be 2 answers, but we take only the rightmost one)

then find C using the intersection of 2 circles with radius of the triangle, or you could use the rotation matrix rotating B 60 ccw, reflect C over AB to D, the new coordinates for A',B',C' are the points A,D,B

[quote="shobber"]
Since we have:
$\sqrt{\frac{a}{b+c}} \geq \frac{2a}{a+b+c}$
Add them together.[/quote]

You can also normalize, and let $a+b+c=1$ :D

[quote="shobber"]$a=x^2$, $b=y^2$, $c=z^2$, then it becomes:
\[ \sum \sqrt{\frac{a}{b+c}} \geq 2  \]
Since we have:
\[ \sqrt{\frac{a}{b+c}} \geq \frac{2a}{a+b+c}  \]
Add them togather.[/quote]
wow!!
thanks
another one:
$a,b,c>0$
$a+b+c>abc$
prove that:
$a^2+b^2+c^2>abc$

Firstly, we have the very weak inequality:
$(a^2+b^2+c^2)^2>abc(a+b+c)$
But since $abc(a+b+c)>(abc)^2$, we now get
$(a^2+b^2+c^2)^2>(abc)^2$
$\Rightarrow a^2+b^2+c^2>abc$

P.S. Please post separate problems in separate topics next time. ;)

Also, search before posting. The inequality $\frac{x}{\sqrt{y^2+z^2}}+\frac{y}{\sqrt{x^2+z^2}}+\frac{z}{\sqrt{x^2+y^2}}>2$ has been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=50390 , and, in the equivalent version $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$, at http://www.mathlinks.ro/Forum/viewtopic.php?t=48825 , http://www.mathlinks.ro/Forum/viewtopic.php?t=48311 (stonger version), http://www.mathlinks.ro/Forum/viewtopic.php?t=46428 , http://www.mathlinks.ro/Forum/viewtopic.php?t=27481 and probably elsewhere, too. The inequality $a^2+b^2+c^2>abc$ was given as problem 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=66657 and in some threads before.

  Darij


Find a set of $ n$ events which are pairwise independent, however, any set of $ k\ge 3$ events are NOT independent.

[quote="JBL"]Please don't give your threads such useless titles.

Draw $ n$ lines on the plane in general position, and color their intersections with two colors independently and uniformly.  Your events are that all the intersection points on a given line are colored the same.  I imagine this problem has appeared before.[/quote]

Thank, and sorry, I edited title.


function
inequalities
inequalities proposed

[quote="silouan"]Let $x,y,z>0$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}$

$5(xy+yz+zx)-12(x+y+z)+12\geq 0$[/quote]
The following inequality is stronger:
Let $x,y,z>0$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}.$
Prove that
$3(xy+yz+zx)-8(x+y+z)+12\geq 0.$

Put $a=\frac{2}{x}, b=\frac{2}{y}, c=\frac{2}{z}$, then $a+b+c=3$. After that, we can mix $f(a,b,c) \to f(a,\frac{b+c}{2},\frac{b+c}{2})$


geometry
incenter
circumcircle
geometry theorems

I think that it is well known.
I'm posting this because I think that it is a useful tool:

$\triangle ABC$, $(I,r)$ incircle, $w=(O,R)$ circumcircle
$AI \cap w =\{A,N\}$

$\Rightarrow NB=NC=NI$

Here is my aproach:

[hide="Proof 1"]Obviously, $N$ is the midpoint of an arc $BC$. So $NB=NC$.

From the triangle $\triangle ANB$ we have $NB = 2Rsin\frac{A}{2}$

The (negative) power of $I$ w.r.t. $w$ is $2rR$ (easy proof using $OI^2=R^2-2rR$)

Hence $AI \cdot NI = 2rR\Rightarrow NI = \frac{2rR}{AI} = \frac{2rR sin \frac{A}{2}}{AI sin \frac{A}{2}}= \frac{2rR sin \frac{A}{2}}{r}\Rightarrow$

$NI = 2Rsin\frac{A}{2}$

Hence $NI=NA=NB$[/hide]

[hide="Proof 2"]$\widehat{BIC}=90^0+\frac{A}{2}$

$ABCN$ cyclic $\Rightarrow \widehat{BNC}=180^0-A$

So, the obtuse angle $BNC$ is $180^0+A \Rightarrow$

$\widehat{BIC} = \frac{1}{2} \check{BNC} \Rightarrow$

$I$ lies on the circle $(N,NA)$, on the arc $BC$ which contains the interior points of $\triangle ABC$[/hide]

I would like to see many proofs for this, if there are  :)

I agree with you pontios.It's a really useful tool.This my solution(I solve it 4 years ago):
It's well-known that $\widehat{AIC}=90^o+\frac{\widehat{ACB}}{2}\longrightarrow \widehat{BIN}=90^o-\frac{\widehat{ACB}}{2}$ [color=red](1)[/color]
Otherwise,$\widehat{IBN}=\widehat{IBC}+\widehat{CBN}=\frac{\widehat{ABC}}{2}+\frac{\widehat{BAC}}{2}=90^o-\frac{\widehat{ACB}}{2}$[color=red](2)[/color]
From [color=red](1)[/color] and [color=red](2)[/color] we have $\widehat{IBN}=\widehat{BIN}\longrightarrow NB=NI$

$NB=NC=NI=NI_a$, where $I_a$ is the exincentre w.r.t. the angle $A$.


If $A,B\in\mathcal l M_n(\Re)$ such that $A^2+B^2=\pi(AB-BA)$ then $\det(A^2+B^2)=0$.

Are we allowed to use the fact that $pi$ is transcendental? I think it is a little bit too much just to solve the problem... ;)

i don`t know anyother way to solve it without using the fact that $\pi$ is $transcendental$.


linear algebra
matrix
Ring Theory
superior algebra
superior algebra unsolved

1. Find all ideals in $ H$, the ring of Hamilton quaternion.

2. It is easy to poitn out that $ F\equal{}\{\left(\begin{array}{cc}a & b \\\minus{}b&a\end{array}\right)  : a, b \in \mathbb{R}\}$ is a field under the usual matrix operation. Is $ F$ isomorphic to any well-known field?

3. For a fixed prime $ p$, let $ \mathbb{Z}_p\equal{}\{\frac{r}{s}\in\mathbb{Q}: (r,s)\equal{}1,p\nmid s\}$. Prove that $ \mathbb{Z}_p$ is a subring of $ \mathbb{Q}$ (I met confusing on proving product of two elements in $ \mathbb{Z}_p$ belong to $ \mathbb{Z}_p$)

1.  The quaternions are a division ring, so have no nontrivial ideals.

2.  The complex numbers.  Identify the generic matrix with $ a \minus{} bi$.

3.  The product of two fractions in lowest terms is not necessarily a fraction in lowest terms, but that doesn't matter.  The second condition is the important one.

[quote="t0rajir0u"]
[quote="kidmath"]3. For a fixed prime $ p$, let $ \mathbb{Z}_p \equal{} \{\frac {r}{s}\in\mathbb{Q}: (r,s) \equal{} 1,p\nmid s\}$. Prove that $ \mathbb{Z}_p$ is a subring of $ \mathbb{Q}$ (I met confusing on proving product of two elements in $ \mathbb{Z}_p$ belong to $ \mathbb{Z}_p$)[/quote]

3.  The product of two fractions in lowest terms is not necessarily a fraction in lowest terms, but that doesn't matter.  The second condition is the important one.[/quote]

Indeed, the construction works over any ring as long as the denominator set forms a monoid,
which always holds true for the complement of a prime ideal. This generalization of the fraction
field construction is known as "localization". It is one of the most basic constructions in algebra.


If the numbers e, pi, pi^2, e^2 and e(pi) are irrational, prove that at most one of the numbers pi + e, pi - e, pi^2 - e^2, pi^2 + e^2 is rational.

[hide="casework"]
Assume to the contrary at least two of them are rational. We split it into cases:

$ \pi\pm e$ and $ \pi^2 \plus{} e^2$: $ \pi^2\pm2\pi e \plus{} e^2 \minus{} (\pi^2 \plus{} e^2) \equal{} 2\pi e$ is irrational, which is a contradiction.
$ \pi \plus{} e$ and $ \pi \minus{} e$: $ \pi \plus{} e \plus{} \pi \minus{} e \equal{} 2\pi$ is rational, contradiction.
$ \pi\pm e$ and $ \pi^2 \minus{} e^2$: $ \frac {\pi^2 \minus{} e^2}{\pi\pm e} \equal{} \pi\mp e$ which is thus rational which is a contradiction by the previous case.
$ \pi^2 \plus{} e^2$ and $ \pi^2 \minus{} e^2$: $ 2\pi^2$ is rational: contradiction.
[/hide]

Thanks. ( I suspected that casework is the only way to do it.)


calculus
integration
trigonometry
function
complex analysis
complex numbers
calculus computations

Need help in the following integral

         $\int_{0}^{2\pi} {e^{\cos \theta}}.{\cos(\sin \theta)} d\theta$

I will be very happy if anybody can send me the codes of writing all the mathematical functions in latex.  :lol:

[quote="Moubinool"]${e^{\cos \theta}}.{\cos(\sin \theta)}$ is the real part of $e^{e^{i\theta}}$[/quote]
Plz solve the integral.
 :)

Use power serie of $e^{z}$ and justify you can permut $\int$ and $\sum$

you'll find $2\pi$

[quote="jmerry"]Or- we can do this the complex analysis way.

We have $e^{\cos\theta}\cdot\cos(\sin\theta)=\text{Re }e^{e^{i\theta}}$. Let $z=e^{i\theta}$, $dz=ie^{i\theta}\,d\theta$ so $\frac{-i}z\, dz=d\theta$.
We have $\int_0^{2\pi}e^{e^{i\theta}}\,d\theta=\int_{|z|=1}\frac{-ie^z}{z}\,dz$. By the residue theorem (there is a simple pole at $z=0$ with residue $-i$), this last integral is $2\pi i\cdot -i=2\pi.$ The real part is $2\pi,$ and we are done.[/quote]

In $\int_{0}^{2\pi} {e^{e^{i\theta}}} {d\theta}$ .
Taking, $z={e^{e^{i\theta}}}$ will cause both the limits of the integral to be same. Then how did you proceed further.
I dont know residue theorem. Can you explain it to me.

Bye :)

That's a complex line integral around a closed loop. You don't know the residue theorem because you don't know complex analysis- just ignore the solution.


Sometimes I get the feeling that life is all a big waiting game. After you are born, you wait until you are old enough to go to school. Although school does teach you what you need to know to do well in life, it seems that it is more of a place for you to stay until you get older and are mature enough to handle a job, relationship, etc. - school seems like a babysitting place. All through life it seems as though you are simply waiting for time to go by.

What do you think?

Seize the day!  Don't wait for things to happen.  Make it happen!

Who says school teaches you what you need to know to do well in life?  AoPS does, but school?  That highly depends upon the school.

Don't let the systems and conventions which are imposed upon you hold you back.  You can learn without school.  You can have a job or a relationship before you ever go to college.

Time goes by too quickly.

We all aspire not to live such boring lives, a few actually try hard not to live our lives as a waiting game, but only a few will succeed in escaping life as a waiting game.

[quote="Singularitarian"]
Who says school teaches you what you need to know to do well in life?  AoPS does,...[/quote]

I'd have to disagree. What does AoPS teach you in life? Problem solving? Sure. But with that definition of "teach you in life", school "teaches" a lot to, like learning that life is not so great after all.

But that's besides the point.

I completely agree that you should seize the day, and take matters into your own hands (that is, If you can handle it). However, there are some things you have to wait for, like you arent going to get a raise the first day on the job, realistically. And if I want something really bad for my birthday? And I have no money now and no way of making money and the only way I am going to get that thing is for my birthday? Sorry. Gotta wait 4 months.

Seek all options! Keep an open mind!

Life is like a rollercoaster, you wait 2 hours for the ride, and some consider the wait totally worth it, and some gets scared and refuse to do it ever again.

It is up to you to manage your time between savoring the present and investing in the future. As a student, that might mean balancing studying with hobbies, sports, friends, and parties. Later, that might mean doing a good job at work from 9 to 5, and then spending quality time with family. It means different things for different people. But it's always a question of priorities and  time management.

Whether or not life is a waiting game is entirely up to you.

I agree with all of the above posts.

Sometimes in school it seems as though we are simply wasting time (some of the activities seem to have no point or educational value). One day we had to dress like people from the time of the American Revolution and play tug-of-war (social studies). This seems to be a waste of time and a hassle (the clothing of people living during the time of the American Revolution or tugs-of-war are not listed anywhere in the Hawaii Content Standards for 8th grade social studies). It wasn't a good or bad experience, but I was thinking "Why are we doing this?" the entire day.

Sometimes I think that it would be possible to learn everything you needed to know from middle school in 1 1/2 years instead of 3. It makes me think that middle school is made 3 years long just so that you can grow up and mature, not just learning.

If anyone here has not seen Randy Pausch's talk on time management, I would recommend you do.
Very inspiring!!
[url]http://video.google.com/videoplay?docid=-5784740380335567758[/url]

[quote]Sometimes I think that it would be possible to learn everything you needed to know from middle school in 1 1/2 years instead of 3. It makes me think that middle school is made 3 years long just so that you can grow up and mature, not just learning.[/quote]

From my experiences at the Ross program, the fastest students are capable of progressing through course material 5 times faster than the slowest students, or more.  And the crazy thing is, the slowest students were very smart, usually among the top few best students at the schools they had come from.

Students should be set free to learn at their own paces.  If they were, it would be common for smart students to finish middle school in 1.5 years instead of 3.  It'd be common for smartest students to progress 5 times faster than slowest students.  And some exceptional students may progress 10 times faster than the worst students.  This would all be a good thing.

@sailingdog: I havent seen that one, but I did see the one about being hisd last lecture. Isnt he supposed to die?

Yeah.

My brother read his book, and told me his reply when someone asked what he wanted his gravestone to say:
"Randy Pausch - died 30 years after diagnosis"
=D


There's a Soccer Tournament.

Let $A$, $B$ and $C$ be the three teams playing this tourney.

For all teams (no matter which teams are playing) we have:

$P(v)=\frac{1}{3}$, where $P(v)$ is the probability of victory;

$P(t)=\frac{1}{3}$, where $P(t)$ is the probability of tie;

$P(d)=\frac{1}{3}$, where $P(d)$ is the probability of defeat.

For each victory one team gets 3 points.
For each tie one team gets 1 point.
For each defeat one team gets zero points.

We know that $A$, $B$ and $C$ will go to play a total of 3 matches.

Find the probability that after those 3 matches, one team will finish the tourney with 6 points, other will finish with 3 points and the third place will finish with zero points.

Come on..

I think this problem is cool :)

But i'm not sure if my solution is good enough  :mad:

And if it's too easy or too hard for this forum i ask a [b]mod[/b] to move it.

Thanks in advance.  ;)

[hide="Answer"]$\displaystyle\frac{2}{9}$[/hide]

i dont know what the point of hiding it is, but...
[hide]
There are three games A vs B(#1), B vs C(#2), A vs C(#3).
The only way the score can be 6-3-0 at the end of the tournament is when one team wins twice, one teams wins once and loses once, and the last team loses twice.
There are 3! ways to assign the teams A, B, C into the scores of 2-0, 1-1, 0-2
There are $3^3 = 27$ ways for the outcomes of all three games
So $\frac{6}{27} = \frac{2}{9}$
i think
[/hide]


geometry
congruent triangles
geometry unsolved

(B.Frenkin) Does a convex quadrilateral without parallel sidelines exist such that it can be divided into four congruent triangles?

What a wonderful example!!! :wink:  :lol:


[quote="math92"]Cybernations is an online simulation game where you get to make your own nation and rule it.  You can set tax rates, a national government, buy technology and infrastructure, and wage war, and loads of other things.

http://www.cybernations.net

My nation name is math92[/quote]

wow nice!

lol is it text based or does it have actual graphics like Age of Empires, etc.?

[quote="daermon"]kewl!  my nation name is daermon
wow math92 u have a reeeeeeeaaaaalllllyyyyy powerful nation! :D[/quote]
yeah watch out or i'll nuke you!!!!!!!! (jk, i wont nuke anybody)

ok i'm going to make an account. How long have you people been playing?

sort of like nationstates but at the same time more complex

[quote="math92"][quote="7h3.D3m0n.117"]lol do u have to log in almost everyday? for example the game Utopia is text based (very good i mite add) except that if u don't log in nearly every 24 hours ur kingdom mates will kill u cuz of inactivity (literally)[/quote]
i do log in everyday.[/quote]
wait a minute can that happen?

you mean your kingdom mates attacking ur province? yeah. unless you are in vacation mode  :lol: . some kingdoms are nice and some arn'et so u have to be active most of the time

i've been doing it for 18 days
yay i just got some technology :P

Some dorkface already took the name mustafa, so I'm going to be mustafa15.

My nation's headquarters is the Wisconsin Badgers football team's field


[quote="negative"]Let $ H$ and $ O$ denote the orthocenter and circumcenter respectively of a triangle $ ABC$.Prove that
$ 3R \minus{} 2OH \leq HA \plus{} HB \plus{} HC \leq 3R \plus{} OH$ where $ R$ is its circumradius[/quote]
Dont help him!!!!!!!!!!!!,here is the newest problem in this issue of MYM in our country,ofcourse the test is running
So ,I think mods should warm him


let a,b be two positive integers such that $\ b^{n}-1|a^{n}-1$ for all $\ n=1,2,3,...$
prove $\ a=b^{k}$ with certain k of integer.

thank you.
I'm very amazed because I'm given the solution very fast.
on VietNam mathematical forum ,the solution come slower.


How do you prove that a number is or is not a perfect square?

couldnt you just put it into prime factorization?

Here's something I wrote recently-

PERFECT SQUARE SIEVE

To test if a number is a perfect square we make use of the fact that it must produce a quadratic residue for every prime. If it does not do this for a single prime it cannot be square.

Take a list of, say, 50 small primes. Call these P1, P2, P3, ...P50.

Assume n is not a perfect square.

Let n = q (mod p) for any prime. The chances that q is a quadratic residue of p is 1/2, since there are (p + 1)/2 quadratic residues (including zero) and (p - 1)/2 nonquadratic residues.

Let n = q (mod P1). The chances of q being a quadratic residue of P1 are 1/2.

Let n = q2 (mod P2). The chances of q2 being a quadratic residue of P2 are 1/2.

1/2 x 1/2 = 1/4.

Let n = q3 (mod P3). The chances of q3 being a quadratic residue of P3 are 1/2.

1/2 x 1/2 x 1/2 = 1/8.

Continue in this way to P50. The chances of a non-square n filtering through the entire sieve is only 1 in 2^50.

To proove that the chances that q is a quadratic residue is 1/2, consider p consecutive integers for any prime, p. These will be reduce to 0,1,2,...p - 1 modulo p (not necessarily in that order). There will be one integer congruent to zero and (p - 1)/2 congruent to q such that q is a quadratic residue and (p - 1)/2 congruent to r such that r is a nonquadratic residue. So the chances are 1/2 for any given integer.

If n = qi = 0 (mod Pi), Pi|n and it must be that n = 0 (mod Pi^2) if n is a square because each prime divisor of n must occur to an even power in its decomposition into factors. If this is not the case, n cannot be square and it can be rejected on this count alone.

The chance of a nonsquare integer getting past say, P7 is only 1 in 2^7 < 1%.
I recently tried all integers up to 100 million using a 50 prime sieve and no nonsquare got through.

For large numbers you can increase the prime base to 1000 in which case the chances are 1 in 2^1000 or about 1 in 10^301.

I think the three most widely used methods are
1) try to show that $n^2<N<(n+1)^2$;
2) try to prove that $p\mid N$ and $p^2\nmid N$ for some prime $p$;
3) check remiders modulo 3, 4, 8, etc.

[quote="jimhu"]Hi,

Think of it this way:

Every expression can be a perfect square if you plug in the right number  ;)[/quote]

No:
\[
2\lfloor x\rfloor^2, \mbox{ with }\lfloor x\rfloor \neq 0
\]

I believe he meant every number x is the "perfect square" of the number $\sqrt{x}$.

[quote="chaose"]I think I worded my question wrong. It should be how do you test if an expression is always a perfect square or never a perfect square?
For example: 4K^2 + 4K + 1 is a perfect square, but 4K^2+4K may or may not be a perfect square.[/quote]

Well if you can factor it into a square, it is obviously one: $4K^2+4K+1 = (2K+1)^2$. But if you can't, I think there's always some number for which it isn't a perfect square (except for maybe strange functions that we don't really care about).

[quote="Danbert"][quote="nr1337"][quote="Treething"]Well, 3x + 2 is [b]never[/b] a square number, and it's an algebraic function...[/quote]

x = 2/3.[/quote]

I think that he meant with x an integer.[/quote]

Then he meant

$3\lfloor x\rfloor + 2$

Or
     _    _
  3|   x   | + 2. Yeah, that's what I meant :P .

Was that an attempt to write $3 \lfloor x \rfloor + 2$?


Ross Mathematics Program
summer program
PROMYS
AMC
AIME
Honors Summer Math Camp
number theory

This forum is for discussion about [url=http://www.txstate.edu/mathworks/camps/hsmc.html]Honors Summer Math Camp[/url].  Run by Texas Mathworks, this camp is in the same vein as the ROSS program at Ohio State, and the PROMYS program at Boston University.

Set in the heart of Texas at Texas State University - San Marcos, Honors Summer Math Camp is an intensive six-week experience for students of all backgrounds interested in doing math at a high level.

Teachers, students, parents, prospective applicants, etc., are all encouraged to post questions regarding the Camp!

Mathworks Honors Summer Math Camp applications are now online!
[url]http://www.txstate.edu/mathworks/applications.html[/url]

1. Fill out student application
2. Have a teacher complete a recommendation form
3. Obtain official school transcript
4. Mail the above to Mathworks

Application deadline is 4/30, and admissions are considered on a rolling basis.

@ anyone who has gone there:
Is it fun? :)

and i think its sort of stupid that they dont have a qualifying test.

Yes, it's quite a bit of fun.

First year students take Number Theory, Mathematica Programming, and Problem Sessions.
Every week there is at least one guest speaker, and weekends are filled with trips.
You work on the daily problem sets in groups of 3-4, allowing for great discussions.

You can always ask a teacher from years past.  It doesn't have to be your current math teacher.
All three parts are weighed pretty equally, 1) application form, 2) teacher rec, 3) essay.

Make the essay as formal as you want.  But we're expecting more than just a couple of sentences.  I'd recommend one page typed.

[quote="veezbo"]Not to be...well, I don't have a word to describe this but:

How important are the teacher recommendations in the application package? Unfortunately, my teacher and I don't exactly share the best relations.

Also, how formal would you recommend the essays be? I haven't really written any before since this is my first time applying to summer camps.[/quote]

Tell us a little about yourself, how you think the Math Camp can benefit you, and what your goals are.
Just don't type in IM talk, haha.

[quote="veezbo"]Oops, I guess I should have asked, how informal can the essays be  :blush:

Because my essay right now is really informal, almost conversation-like (without the slang)[/quote]

It's not directly comparable.  We don't focus on competition math.  We emphasize proofs, mathematical reasoning, and working in groups.  First year student's main course is Number Theory, where you work with Diophantine Equations, prove Fermat's Little Theorem, etc.

[quote="AdithyaGanesh"]Forgive me if it's on the website, but what is the difficulty of this camp?  Could it be compared with any of the AMCs, AIMEs, etc.?[/quote]

[quote="veezbo"]How important are the teacher recommendations in the application package? Unfortunately, my teacher and I don't exactly share the best relations.
[/quote]
 :roll: funny, I have the same problem  :blush:

[quote="pumpkinpi"]How's the food during the summer? (Is there a variety of choices? Is the food just plain dull? etc.)

Did the attendees mostly come from Texas in the years before? (Obviously, I can't ask about this year.) Not that that's a bad thing, of course.[/quote]

About 1/3 of the students are from states [b]other[/b] than Texas.  This varies year to year depending on how many non-Texans find out about the Honors Summer Math Camp.

Room and board is included in the cost of the Camp.  The cafeteria is "buffet" style where you eat what you want.  If you want to bring extra pocket money, there are also various places off-campus you can grab a bite.

[url=http://www.txstate.edu/mathworks/applications/HSMC-application.html]Apply today![/url]

"Im looking for a math camp over the summer does anyone know any other good ones that suit my age im 11 and in the 6th grade?"

Texas Mathworks has a 2-week residential program for talented middle school students.
http://www.txstate.edu/mathworks/camps/jsmc/residential.html

Hi I am found out about this camp but I don't know if I am ready for level 5.
I am a seventh grader
SAT math:570
I am doing alg 2 and a little bit of geo on my own. What exactly do level 5 people learn at camp?
I am talking about the junior camp.

anyone got their acceptance/rejection letters yet?

If accepted: can you post your stats (like cc except that cc doesn't have a thread for thsmc)?

[quote="kewi"][quote="ahsiau"]First year student's main course is Number Theory, where you work with Diophantine Equations, prove Fermat's Little Theorem, etc.
[/quote]
But what if already know how to prove Fermat's Little Theorem, the Sophie Germain Identity and all that stuff?
Would you still recommend Honors Summer Math Camp in this case?
:)[/quote]

The sophie germain identity isn't that important in number theory, it just lets you create problems that are tricky to solve without it...

AND

don't underestimate the usefulness of mathematica programming, I am learning it now, and it is crazy how much cool stuff you can do with mathematica!

I am not underestimating the usefulness of mathematica programming :). I would love to know how to use it ^.^

FYI, here's how our Honors Summer Math Camp (HSMC) students fared in the Siemens Competition this year, [url]http://www.txstate.edu/mathworks/about/news/regional_siemens_competition_2010.html[/url]

And here is a historical listing of all HSMC team projects recognized by the Siemens Competition, [url]http://www.txstate.edu/mathworks/camps/hsmc/Siemens_Competition_Results.html[/url]

I should note that returning students (students attending for the second or third summers) to the HSMC have the opportunity to work on original mathematics research mentored by university faculty.

[quote="ilikecake"]How much does it cost?[/quote]

The full tuition cost is $2,700.  We do offer need-based scholarships - be sure to fill out the Financial Aid section of the application.

thanks, I'll be sure to think carefully about applying!


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
geometry
AMC 10

As many people know (or at least I hope), I'm conducting math discussion through AIM.  The thing is, however, we don't have enough people now.  So, if you're interested, please tell me or Julie.  I want this in kinda easy level but it would be great if we can get some REALLY high mathy people who could teach us.

Thanks!

I'm interested.  My AIM s/n is BHorseMath.

I'd love to come...u kno my sn already...

What's your (silverfalcon) AIM? Who's Julie? and could everybody put their AIM on this forum. It's been so long I've used AIM, I have no clue who's who on AIM. Why do you need all this? Just read books and fill your life with AP classes!

My AIM is fantasychrono.

I'll be glad to answer any questions people might have (that I can answer, of course). :)

my AIM is caocoprez. i'm online a lot, but usually pretty busy. IM me and i might be able to help

Hi, I am Julie.  (My s/n is js1527k.)

Yes, I suggested to Silverfalcon that the reason nobody wants to come is b/c the problems are too easy, but then again you don't want to be too difficult or then the younger people will not want to come.

Anyway everyone should come, but watch out b/c I will ask lots of questions if I don't get it! =)

My sn is joml88 (who would have guessed?).  Or I might be on my older sn BaseballJoe2002  It depends.

No kidding no hard ones...If theres another one like the last one I went to then I'm not coming anymore.  All that one served to do was confuse me!

when's the next one??? do they happen regularly (ie: every wed night at 7 est or something)???

Not to be a jerk or anything, but I will not want to come if its too easy.

My advice is that challenge is good.  VERY GOOD.  If you don't understand something then ask questions.  If you are not challenged then you will not have enthusiasm for something, aka boring.

The next discussion:

6:00 EST Friday 5/7/04

However, since we're getting new people, the upcoming is going to be just chat for deciding specific level of discussion.  It's quite useless to have discussion without knowing specific level of each people who's involve in it.

Not to be rude, Julie, your question is quite a lot challenging to myself too.  It seems quite simple factoring but not all people overcome the high level of factoring in their school math yet.. So..

Well it's really hard to appease everyone's level of math.  So we will just have to work something out.  Maybe do some easy and some a little bit harder.

[quote="joml88"]  Maybe do some easy and some a little bit harder.[/quote]

For example?

Well I don't have any specific examples but I was thinking maybe do some mathcounts level and then maybe the level of the first couple of aime problems as the highest.  I know that aime is intimidating but the first few questions are normally attainable to people on mathcounts.  That is they can understand most of the concepts but wouldn't get them on their own otherwise(this is just on the average there are exceptions that go to USAMO and stuff).

Ya, good idea joml, I agree, some easy, some hard.

Silver, we can do math after we talk about how much we know....right?  =)

I don't know a lot of math right off the bat, workign on that though, but I do learn really quickly. Something about younger brains generally takes them a lot longer to absorb something. So I'd be interested in going to either

A.)One with older people, preferably not everyone like Nukular, etc. who knows way more than me, unless you all want to tutor me lol.

B.) Or one with mostly older people, and hte younger ones who are know a lot already, or are supergeniuses.

Anythings going on like that?

my aim is scm200x, but I'm rarely on. So just post whatever is going on in the near future, or plans for something here, or PM me. Then we can talk on AIM. BTW AIM sucks! You always have to use msconfig to keep it from starting up. Anybody know of a permanent way from keeping programs from starting up, even if you reopen them?

[quote="scm007"] Something about younger brains generally takes them a lot longer to absorb something. [/quote]

Actually I think that it is  quite the opposite.  Normally when someone gets older they can't learn as well.

[quote="scm007"]Anybody know of a permanent way from keeping programs from starting up, even if you reopen them?[/quote]

Look in the AIM preferences -- or, better yet, notice that when you have AIM open, before you've signed on, there are two buttons you can press.  One of them saves your password, the other is "auto-login."  If you have auto-login selected, it will start automatically when your computer starts.

I'll be sure to try that out JBL. And joml, I see where you are coming from in cases like languages, but many times, unless the person is a prodigy, it takes a lot longer for a young person to learn something than one who is older. When I first started playing piano when I was little I was horrible for a very long time, like 5 years. Recently I started playing guitar and now I am very good and have only been playing for around 5 months. I'm much better than many younger people who have been playing for numerous years.

I agree with joml88.  You see, there was research that stated that as people get older, their brain cells began to die, resulting hard to comprehend the datas than when they're young...
What you said is not totally wrong, however, because that if you have past knowledge, and you're learning something that's related to it, it's easier to learn.. So, it's quite easy to do Algebra II than Algebra I since you had basics in Algebra I.

For the doing problems, I'm not sure whether we'll do it or not.. I know it's hard to get the level of each person but at least, it's worth thing to do and I want to spend some time on it..

I'm not disagreeing that people learn concepts better at younger ages, but when you are learning a new idea that pertains to what you already know, older people learn it faster. I imagine that everybody that posts here had an early start in mathematics. Whether this means visualizing objects in space, counting, etc. we probably all have this comprehension of math. But then something like learning about power series, or somethign that younger people usually do not learn, would be grasped easier by an older person. That is all that I'm saying. I'm very good at math, and I just want to learn it at a quicker rate.

That we need to have a math chat for older people and smarter younger people.

It sounds good, however, in order to do that we need more people such like qualifiers of USAMO.  Plus, I'm not technically sure about younger people in what age? 6th grade? The important thing is that we have to have more people and more experienced instructor on this since it's not like MathJams where administrators take charges. I myself take charge on it but since I don't know any math further than algebra II, I usually face some problems from geometry, which I haven't cover.

It is nice you point out these things while most people think discucssion is just for competition.  I personally think it should help on learning, not just test prep..

I agree. I mean if a bunch of people at my relative knowledge level would be willing to get together and discuss some problems, ideas, and techniques that would be great. I find it much more stimulating, motivating, and enjoyable to learn as a group. Sometimes it is really hard to concentrate for hours on end reading and studying, but not so difficult if you have other people to work with.

Yes.  Like if it's just REALLY smart people with just smart people mixed together and solving difficult problems, just smart people would get lost while REALLY smart people will answer everything..
But the problem is, how am I going to group the people?

I'm mathy, but I'm not very good (114 on this years AMC10a, blech.)

AIM sn = applepuppy88

114???I would kill for that score!  I can't even do the AMC 8! hmph.

[quote="Silverfalcon"]I agree with joml88.  You see, there was research that stated that as people get older, their brain cells began to die, resulting hard to comprehend the datas than when they're young...
What you said is not totally wrong, however, because that if you have past knowledge, and you're learning something that's related to it, it's easier to learn.. So, it's quite easy to do Algebra II than Algebra I since you had basics in Algebra I.
[/quote]

Agreed :). Experience makes up for a very big part of problem solving, and learning.

Freshman year, I took the USAMO and gave a blank stare for 9 hours, and said "how am I ever going to be able to do these? They're crazy!" After about two years of work and experience, I'm doing very well. But it takes work, motivation, and experience to develop skills.

Think of everything you learn without prior experience when you're younger: language, basic counting and math skills, reading, etc. Without patterns to put new information into, these things seem almost miraculous. But you learn more easily at a young age (I think younger than age 6-7, especially), so its not that big a problem. When you're older, you have the experience to categorize new things in terms of things you already know.

Which is another reason why people shouldn't get discouraged over AMC/AIME/USAMO. Put more time into it (quite possibly a lot of time, especially once you're olympiad level) and you'll do better. Just have fun with it. That's the point.

Good saying..

Hear that everyone? Don't panic!


Whenever I try  to put something in quotation marks, the left quotation mark is backwards:

$
\mbox{\"Hello!\"}
$

Why?

$\mbox{``Hello!\"}$

Use `` for left quotation.


Find all ordered pairs $(m,n)$ of integers such that $mn-2(m-n)^2 = 2005$.

m*n - 2*(m-n)^2 = 2*(n-m)*(m-n)

2*(n-m)*(m-n) = 2005

(n-m)*(m-n) = 2005 / 2

2005 / 2 is not integer =>
 There does not exists any integers m,n for this problem.

LHS = -2m^2 + 5nb -2n^2 = (2n-m)(2m-n) = 2005 = 5*401

So we have 2n-m = 5
2m-n = 401
4m-2n = 802
3m = 807
[b]
m = 269
n = 137
[/b]

(or vice versa)

EDIT: Whoops, missed a  spot!

We can also have 2m-n = 1 and 2n-m = 2005, for which we get  m=669 and n=1337.  n=1337 and m=669 is also a solution.


linear algebra
matrix
linear algebra unsolved

Say that a square matrix $ Z$ is metric if it is symmetric, its diagonal entries are all 1, its off-diagonal entries are all in $ [0,1)$, and $ Z_{ij} Z_{jk} \le Z_{ik}$ for all $ i,j,k$.  Prove, or disprove, that every metric matrix is invertible.  (Note the j's are not summed over).

Source: T. Leinster

[quote="alekk"]you should maybe define what is a metric matrix..[/quote]
Umm... he did¿  :wink:


To reignite this forum and really get it going, I propose a fun game: Biology Quiz Bowl. Each person will answer the question that the poster above them has proposed, then they will submit a question of their own. Let's get started. For example:

What is the term that describes the segment of non-coding DNA where RNA polymerase binds to the genome to initiate transcription?

Promotor.

The transcribed RNA goes to this part of the cell where it is translated by ribosomal RNA.

Ribosomes translate at the rough ER or in the cytoplasm, just depends.

Ehhhh, a little bit like chemistry but still has biology applications:

For a spontaneous, exothermic reaction, what is the sign on $\Delta{G}$?

Negative.

What is cork cambium composed of?

cork cambium is ring of dividing cells (if i remember correctly) and produces parenchyma cells on its inner face and then on its outer face produces cells that will develop into the outer cork of the plant.

Easy one: The Calvin Cycle takes place where in the chloroplasts? (Be specific)

Light reactions occur on the thylakoid membrane and the Calvin cycle takes place in the stroma.

Actually I was thinking of cork cells in the periderm, which consist of suberin.  As for the cork cambium, I don't know exactly what it's made from.  It originates from the cortex.

Anyways, here's a somewhat easy one: what is the refractory period and when does it occur?

you could be a little more specific. the refractory period refers to two things that i know of:

a period of time after an action potential where a neuron will not generate another action potential.

also in the sexual response cycle, the period after the male orgasm where orgasm cannot be acheived.

Haha, I was not aware of the second.  I never read the chapters on the reproductive cycle so I wouldn't know.

Anyways, this has become kind of pointless.  If only more people participated.

Yeah I know, not too many people visit the biology forum. It's okay, I'm sure more will as AoPS continues to grow.

What is the function of the MPF, and what is it composed of? How does the concentration of one of its components affect the progression of the cell cycle?

MPF- mitosis promoting factor. A function which stimulates the cell division processes of meiosis and mitosis.

MPF is composed of cyclin and kinase.

As the cyclin rhytmically changes, the MPF increases like this:
[img]http://home.earthlink.net/~dayvdanls/MPF.GIF[/img]
IMG SOURCE: http://home.earthlink.net/~dayvdanls/MPF.GIF

My biology question for the quiz bowl:
Detail the symptoms of anarchic hand or Alien hand syndrome.

Alien hand syndrome occurs often when people have had their brain surgically separated, or sometimes if they have epilepsy. This syndrome is caused by damage to the motor reflex area of the brain. It also can be caused or occurs with infection, or stroke. Alien hand syndrome is when a person, who is the owner of the hand, feels that the hand is another living thing and "has a mind of its own." In many cases, sufferers of alien hand syndrome refer to the hand as "it" or "he". And just like its name suggests, actions involving the hand often become involuntary or uncontrollable.

Here's an easy one:
what is the primary function of the neurolemmocyte? (schwann cell)

Are any of you guys doing Science Bowl? A big part of science bowl is biology.

[quote="rituraj007"]Schwann cell increases the speed of conduction in the neuron...is that right?? :oops:[/quote]
Yes, a Schwann cell is a glial cell that serves to myelinate PNS axons (oligodendrocytes in CNS). Myelinated axons can conduct potentials at much faster speeds than can unmyelinated axons.

Next question (a number-crunching one):
What is the net osmotic pressure in atm on a semipermeable membrane (only permeable to water) separating one solution of 0.23 M aluminum chloride and one solution of 0.30 M sucrose at 20 degrees Celsius?


I said I would like to raise discussion of the important issue of how to improve primary education (especially in the United States--but I hope the international participants in this forum will bring their valuable perspective to the discussion). Improvement of education for persons of all ethnic groups and all levels of ability is an issue that has long been close to my heart, particularly because I had several teachers in my school days who encouraged my interest in that issue. I have been actively engaged in discussion, research, and advocacy related to that issue for, well, at least twice as long as most participants on this forum have been alive.

I'll open with one very concrete suggestion (which is perhaps not the most important or controversial of the suggestions I would make) just to open the discussion. If I had sufficient personal influence on the United States educational system to make this happen, I would set up incentives so that all schools would make available to teachers and learners the excellent mathematics instruction materials developed in Singapore, the Primary Mathematics series for elementary grades, the New Elementary Mathematics series for early secondary grades, and the various upper-secondary series used in the best schools in that country. That, I think, would provide a substantial improvement in mathematics education for most learners in the United States. For teacher professional development of teachers, I would also make sure all teacher candidates have available [url=http://www.amazon.com/exec/obidos/ASIN/0805829091/learninfreed]Knowing and Teaching Elementary Mathematics[/url]by Liping Ma (which is already required reading in many of the best teacher-training programs in mathematics) and the forthcoming book [url=http://www.ams.org/amsmtgs/2078_abstracts/993-97-876.pdf]Elementary Mathematics for Teachers[/url] by Thomas H. Parker and Scott J. Baldridge or the already published books Mathematics for Elementary Teachers [url=http://www.amazon.com/exec/obidos/tg/detail/-/0321129806/learninfreed]volume 1[/url] and [url=http://www.amazon.com/exec/obidos/tg/detail/-/0321149149/learninfreed]volume 2[/url] by Sybilla Beckman, and that they have all seen the [url=http://math.berkeley.edu/~wu]interesting writings of Professor Hung-hsi Wu[/url].

There is a lot more I could advocate, and probably will, as this thread develops, if anyone else is interested. What constructive suggestions do you have?

(I apologize in advance for the length and grammer of this post.. it was kind of a free form ramble).

This is a complex and daunting issue. First let me just say this, I don't think that education is that bad in ALL parts of the country. Some school and districts (in all kinds of environments), actually do very well. On the other hand, some schools and districts are lagging way behind. Furthermore, the issues differ from State to State and even between urban and rural environments. People commonly think the biggest problems are in urban cities, when in fact some of the absolute worst cases are in poor (southern) rural areas.

I think that changing certain academic curricula is a good step to improving education. Remember though, that each curriculum is for the most part, set at the local (school district and state) level. And along with any change in curriculum comes a huge amount of bureaucracy. Supplying new textbooks en mass for a population of students is expensive - even for just for one subject. Then teachers have to be trained to be able to teach this new curriculum, which is also expensive.

Another problem, is that in many of the worst schools and districts you have teachers teaching subjects (especially math) for which they were not trained, and honestly, know just barely enough to get by themselves. This is particularly true in my state, where some schools are so desparate for math teachers they will take just about anyone off the street who can add. The only other alternative would be to crowd the students into classes with the few people qualified to teach.

As it is now, different districts and schools use a variety of books and curricula and techniques for math education. Some are decent and some are not. But the most important factor is always good, motivated, and knowledgable teachers who take a personal interest in their students. Without that, the curriculum you install won't be effectively taught.

The biggest changes that need to be made though are political. Starting with the school board systems. School boards should not be elected, they should be appointed. By whom? I don't know. But elected school boards often have people who either a) don't know that much about education, or b) are only beholden to the most powerful and wealthy (i.e. voting segment) of the population in each district. So they make decisions based on what's good for some people, but not everyone.

That's why I tell everyone, get involved in local politics!! If you don't know who's on your school board or who your city or county councilpeople are, you need to! These are the people who have the most control over your day to day lives - even more so than some of the bigger offices.

The single most important thing to a school or districts' success though is good leadership. It's amazing what a motivated and good principal can do for a school. They are the field generals - the captains of the ship. It takes extraordinary and very dedicated people in those positions to turn poor schools around. Unfortunately, there are more schools in need of them than there are people like that available and willing to do the job. Part of that has to do with money (why work in education when you can take your management skills somewhere else and make more money?); and part of it is social stigma. Its often, unfortunately, assumed that people who do that kind of work are doing it because they can't do anything else.

My immediate solution to the problem is to first reprioritize expenses (people assume that it's always a lack of money, when in some cases it's just a misappropriation of money).I'd actually be interested to see how much we spend per student on education compared with some other western nations, because I have a feeling we probably spend a lot more with less results.

Another possible solution might be to model education slightly on business principles. For example, within schools and districts they shouldn't have to pay all teachers the same amount. They should be able to recruite star teachers, and pay them according to what they think they're worth. The inherent danger in doing this... is that it could further the problem of inequity in education. The wealthy schools would being able to outbid poorer schools for star teachers. That already more or less happens across district lines now. But maybe it would raise the quality of teaching and education across the board.

Which leads to the most difficult question about whether we are willing to sacrifice quality of education in some places for equality of education everywhere.

Here's my impractical and rather extreme suggestion.  Make the starting salary for teachers $80,000.  Honestly, I think that would basically do the trick -- I go to Harvard, which has no end of students who would be qualified to teach in almost any field, and a reasonable number of people who I think might even make good teachers (not just knowledgable or smart teachers).  The problem is, they can all make 5 times as much money not being teachers.  By increacing teacher rewards, you will increace the number of people (and particularly the number of qualified people) who will want to be teachers, which should allow such nice things as reduced class size (assuming there is enough physical classroom space available, which is another (unrelated) problem that goes essentially unaddressed), which has proven to be one of the most effective ways of increasing achievement.

And, if anyone is planning on suggesting more testing and higher standards as a good fix here, I recommend you go read the recent New York Times articles on the "Texas Miracle" as realized in Houston.


我们刚刚学了这篇课文.语文老师徐稚引发了我们的思考:文中感情按照 乐------>痛------>悲的情绪发展. 作者提到, 即使有很快乐的事,也是暂时的.虽然可以给人一时的愉悦,但时光在不知不觉中流逝.人不可能一辈子有美好的回忆,他终究会因为霜鬓而叹息岁月的残酷.等待人的,只有死亡,而且不可避免!人生苦短,岂不痛哉!

    那么人应当怎样活着呢? 如果怀着"一生死,齐彭殇"的看法,未免太悲观.若云:"生死亦大矣.",最后又不免感叹.这样,人又应该怎样活着呢??

附:
兰亭集序（晋）王羲之
　　永和九年，岁在癸丑，暮春之初，会于会稽山阴之兰亭，修禊事也。群贤毕至，少长咸集。此地有崇山峻岭，茂林修竹；又有清流激湍，映带左右，引以为流觞曲水，列坐其次。虽无丝竹管弦之盛，一觞一咏，亦足以畅叙幽情。

　　是日也，天朗气清，惠风和畅，仰观宇宙之大，俯察品类之盛，所以游目骋怀，足以极视听之娱，信可乐也。

　　夫人之相与，俯仰一世，或取诸怀抱，晤言一室之内；或因寄所托，放浪形骸之外。虽取舍万殊，静躁不同，当其欣于所遇，暂得于己，快然自足，不知老之将至。及其所之既倦，情随事迁，感慨系之矣。向之所欣，俯仰之间，已为陈迹，犹不能不以之兴怀。况修短随化，终期于尽。古人云：“死生亦大矣。”岂不痛哉！

　　每览昔人兴感之由，若合一契，未尝不临文嗟悼，不能喻之于怀。固知一死生为虚诞，齐彭殇为妄作。后之视今，亦犹今之视昔。悲夫！故列叙时人，录其所述，虽世殊事异，所以兴怀，其致一也。后之览者，亦将有感于斯文。

这篇课文是高一语文课本的课文，我记得那时老师死活都要我们全背过，说什么能提高修养，实在不解呀~

我语文太差劲了~


modular arithmetic
trigonometry
algebra unsolved
algebra

Express

    nC3 + nC7 + nC11 + ...

in simplest form.

give a minor hint.

[quote="debanik2"]Express

    nC3 + nC7 + nC11 + ...
in simplest form.
[/quote]

$ (1\plus{}z)^n\equal{}\sum_{k\equal{}0}^n\binom nkz^k$
$ (1\minus{}z)^n\equal{}\sum_{k\equal{}0}^n\binom nk(\minus{}z)^k$.

So $ f(z)\equal{}\sum_{k\equal{}0,\text{ odd }k}^n\binom nkz^k\equal{}\frac{(1\plus{}z)^n\minus{}(1\minus{}z)^n}2$

And the required sum is $ S\equal{}\sum_{k\equal{}0,k\equal{}3\pmod 4}^n\binom nk\equal{}\frac{if(1)\minus{}f(i)}{2i}$ $ \equal{}\frac{i2^n\minus{}(1\plus{}i)^n\plus{}(1\minus{}i)^n}{4i}$

So $ S\equal{}\boxed{\sum_{k\equal{}0,k\equal{}3\pmod 4}^n\binom nk\equal{}2^{n\minus{}2}\minus{}2^{\frac n2\minus{}1}\sin(\frac{n\pi}4)}$


Here is a funny little one,Its very basic :D :

Let $a,b,c,d,e \ge 0$ then prove

$a^2+b^2+c^2+d^2+e^2 \ge ab+ac+ad+ae$

This inequality is true for all a,b,c,d,e  :)


Who know American math forum and England Math forum?

[url]http://mymathforum.com[/url]
[url]http://www.artofproblemsolving.com[/url]
[url]http://forum.awesomemath.org[/url]
[url]http://www.kalva.demon.co.uk/index.html[/url]
@ chien than :spam: :spam:

[quote]@ chien than :spam: :spam:[/quote]
I don't think so.This is my question


Prove if $n \in Z$

$2n^3 + 3n^2 + n \equiv 0(\bmod 6)$

[hide]
first factor out a n from the equation to get
$n(2n^2+3n+1)$
Then factor the inside polynomial, so then you have
$n(2n+1)(n+1)$
well, you need to prove that this is divisible by 6
if you divide the above polynomial by 6 you have
$\frac{n(2n+1)(n+1)}{6}$
this is the formula for the sum of the first n squares
therefore, it must be an integer, meaning that
$n(2n+1)(n+1)\equiv0(\bmod6)$
[/hide]

Creative solution, ampis!

[hide]
Factor into $n(n+1)(2n+1)$.  Either $n$ or $n+1$ must be even, so for it to be divisible by 6, one of the terms must be divisible by 3.  The only way $n$ or $n+1$ isn't divisible by 3 is if $n\equiv 1 \pmod 3$ and $n+1 \equiv 2 \pmod 3$.  However, in this case, $2n+1 \equiv 3 \equiv 0 \pmod 3$, so one of the three terms is always divisible by 3, and thus the expression is always divisible by 6.
[/hide]

[u]Classic.[/u]

$p\in N,\ p>1;\ P\in Z[X],\ \left(\forall\right) k\in \overline {0,p-1},\ p \mid P(k)\Longrightarrow \left(\forall\right) z\in Z,\ p \mid P(z).$

[u]Indication.[/u]$z=qp+r,\ 0\le r<p\Longrightarrow p\mid P(z)-P(r).$

[u]The proposed problem.[/u] $P(z)\equiv 2z^3+3z^2+z.$

$P(0)=0,\ P(1)=6,\ P(2)=30\Longrightarrow$

$\left(\forall)\right) z\in Z,\ 2 \mid P(z),\ 3 \mid P(z)\Longrightarrow \left(\forall)\right) z\in Z,\ 6 \mid P(z).$


Find All The Continue Functions f, with:
[img]http://www.gnux.be/latex/data/b0ee356738932f39da9b3e300b656feb.png[/img]
Continue means :  [img]http://www.gnux.be/latex/data/fcb31215a3b9e6fd55989e8c33e8483a.png[/img]
And Thanks  :D

[hide="hint"]For all $a\in\mathbb{R}$, fixed it. Assume $(x_{n})$ is the sequence defined by $x_{1}=a,x_{n+1}=x_{n}^{2}+\frac{1}{4}$. We have $f(a)=f(u_{n})(n=1,2,...)$. Now is $(u_{n})$ conveger?[/hide]

[quote="N.T.TUAN"]Now is $(u_{n})$ conveger?[/hide][/quote]
What is $u_{n}$?  :P
Anyway it is always increasing and for $-{1\over 2}\leq a\leq{1\over 2}$ it converges to ${1\over 2}$.
For $a\geq{1\over 2}$ the sequence $v_{1}=a,v_{n+1}=\sqrt{v_{n}-{1\over 4}}$ is decreasing and converges to ${1\over 2}$ with $f(a)=f(v_{n})(n=1,2,\ldots)$.
Also $f(a)=f(a^{2}+{1\over 4})=f((-a)^{2}+{1\over 4})=f(-a)$ for all $a\in\mathbb{R}$.
This gives $f(x)\equiv f({1\over 2})$ constant.

[quote="arkhammedos"]Bulgaria 96/97 :wink:[/quote]
Hello !  :lol:
Its Morrocco 2007 4th Test, right??


simplify:
4x^4 - 17X^2 + 4 = 0

Thanks for help.


superior algebra
superior algebra unsolved

What's the quickest way to solve x^3 = 38 (mod 195)?

I see that 195 = 3*5*13
I think I should use the Chinese Remainder Theorem.
I find that
in mod 3, the solution is 2,
in mod 5, the solution is 2,
in mod 13, the solutions are 4,10,12.
So there should be 3 solutions to the given problem. However, I don't know how to proceed from here. The cubic part and the fact that I can break it up into 3 primes is a little different than the other ones in the text.

Thanks for any help.

You need to combine your solutions by a second application of CRT.  The first solution $ x_1$ satisfies

$ x_1 \equiv 2 \bmod 3$
$ x_1 \equiv 3 \bmod 5$
$ x_1 \equiv 4 \bmod 13$.

The solutions $ x_2, x_3$ satisfy the first two congruences as well but the last is replaced with $ 10$ and $ 12$ respectively.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ ABC$ be triangle. $ P,Q$ are isogonal conjugate points. Show $ AP \equal{} AQ$ iff $ B,P,Q,C$ are cyclic.

We draw the circumcircle $ (K),$ of the triangle $ \bigtriangleup BPQ$ and we denote as $ D,$ $ C',$ its intersection points from the sidelines $ AB,$ $ AC$ of $ \bigtriangleup ABC,$ respectively.

It is enough to prove that $ C'\equiv C.$

Because of the center $ K$ of $ (K),$ lies on the midperpendicular of the segment $ PQ$ and also because of $ \angle DAP \equal{} \angle C'AQ,$ we have that the point $ C',$ is the reflection of $ D,$ with respect to the line segment $ AK$ and so, we have $ DC'\parallel PQ$ $ \Longrightarrow$ $ PD \equal{} QC'$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ \angle DBP \equal{} \angle C'BQ$ $ ,(2)$

From $ (2)$ and because of $ \angle DBP\equiv \angle ABP \equal{} \angle CBQ,$ we have that $ \angle C'BQ \equal{} \angle CBQ$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ BC'\equiv BC$ $ \Longrightarrow$ $ C'\equiv C$ and the proof is completed.

Kostas Vittas.

PS. By a similar way, we can prove that $ AP \equal{} AQ,$ when the points $ B,$ $ P,$ $ Q,$ $ C,$ are concyclic.


A sequence is defined as follows:

$a_1=1$, $a_n=\frac{5a_{n-1}-13}{3a_{n-1}-7}$

Find $a_{50}$

[hide]By simple trial and error, we find that $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 1$, and so the sequence will repeat every three terms.  Since $50 \equiv 2 \pmod 3$, $a_{50} = a_2 = 2$.[/hide]

Can you prove your conjecture??? I'm not getting it..... :blush:  :oops:  :(


logarithms
calculus
calculus computations

[quote="kihe_freety5"]let a sequence be determine that :
  ${ \{x_1}$=0;${ \{x_2}$=2;  ${ \{x_n + 2}$=${ \{2^{ - x_n}}$+$ \frac {1}{2}$ ;
prove that sequence has limit n  $ \n\to\infty$ infinity determine that![/quote]
What does it mean ? Does it mean  $ x_{n+2}= 2^{-x_n}+\frac 12$  :?:
 :cool:

$ f(x)\equal{}2^{\minus{}x}\plus{}\frac 12\Longrightarrow \forall _{x\geq 0},|f'(x)|\leq \ln (2) < 1$ . The attractive fixed point of $ f$ is $ 1$, , so the given sequence converges to $ 1$ .
 :cool:


Hello   :)
Let [img]http://www.gnux.be/latex/data/887fb68a10cbd4369b27c90bee0334d8.png[/img]
Prove : [img]http://www.gnux.be/latex/data/93478bce6df1899362867113dffb02d4.png[/img]
And Thanks  :wink:


There are already a lot of topics about NBA. Use the [url=http://www.mathlinks.ro/Forum/search.php]search[/url] function.