modular arithmetic
invariant
combinatorics unsolved
combinatorics

let $p\geq  5$ be a prime.Prove that for any partition of the set {$1,2,...p-1$} into three subsets there exist numbers $x,y,z$, each belong to a distinct subset,such that $x+y=z modp$

what? did you mean p=5? what does $p\rightarrow5$ mean?

Assume the converse, i.e., there exists a partition $A\cup B\cup C = \{ 1,\dots,p-1 \}$ such that there are no numbers $x,y,z$ belonging to distinct sets with $x+y=z\pmod p$. All operations below are considered modulo $p$.

Without loss of generality we can assume that $0<|A|\leq |B|\leq |C|$ and that $1\in A$ (otherwise take any $a\in A$ and mutiply all numbers in $A,B,C$ by $a^{-1}\bmod p$).

Since $1\in A$, for any $x\in B$ the number $x\pm 1$ must not belong to $C$. In other words, if we put all numbers $1,\dots,p-1$ in a row, neighbors of every element of $B$ belong to either $A$ or $B$, and neighbors of every elements of $C$ belong to either $A$ or $C$.

It immediately follows that $p-1\in A$. Indeed, if, say, $p-1\in B$ then take a pair of elements $x,x+1$ with one element from $A$ and the other from $C$ (such a pair always exists) and construct a forbidden solution $(x+1)+(p-1)=x$.

We will show that there exist pairs $x,\ x+1\in B$ and $y,\ y+1\in C$.

Suppose that there is no pair $x,\ x+1\in B$. Between any two elements of $B$ in a row $1,\dots,p-1$ there exists an element of $A$. Since there are also elements of $C$ somewhere in this row and since $|A|\leq |B|$, we must have $|A|=|B|$ and the following row configuration \[ 1=b_1, a_1, b_2, c_2, \dots, b_i, a_i, c_1, c_2, \dots, c_{|C|}, a_{i+1}, b_{i+1}, a_{i+2}, \dots, a_{|A|}, b_{|B|}=p-1  \] implying $p-1\in B$. The obtained contradiction proves that there is a pair $x,\ x+1\in B$. Similarly, there is a pair $y,\ y+1\in C$.

Let the chains of elements $x,x+1,\dots,x+l_B-1\in B$ and $y,y+1,\dots,y+l_C-1\in C$ of length $l_B, l_C\geq 2$ be the longest chains in $B$ and $C$ respectively.
Consider pairwise sums of their elements. They form a chain  $x+y, x+y+1, \dots, x+y+l_B+l_C-2$ of the total length $l_B+l_C-1>\max\{l_B,l_C\}$. Neither element of this chain can belong to $A$ (e.g., if $x+y+2\in A$ then $(x+1)+(y+1)=x+y+2$ forms a forbidden solution). And neither element can be zero (if $x+y+k=0$ for $k>0$ then $x+y+k-1=p-1\in A$, and if $x+y=0$ then $x+y+1=1\in A$). Therefore, all these elements must belong to the same set, either $B$ or $C$, meanning that we found even a longer chain there. The contradiction proves that our original assumption is wrong, and there no such partition exists.

Ciao.

My solution is a bit different.

[size=75][b]Lemma:[/b] [i]Suppose we have such a partition. Let $\scriptstyle k$ be an integer between $\scriptstyle 2$ and $\scriptstyle p-1$. Suppose $\scriptstyle A$ contains $\scriptstyle 1, 2, \dots, k-1$ and $\scriptstyle B$ contains $\scriptstyle k$. Then $\scriptstyle A$ is strictly larger than $\scriptstyle C$.[/i]

[b]Proof:[/b] I will prove a stronger result: $\scriptstyle card(A) \geqslant (k-1) (card(C)+1)$. I call an integer value [i]valid[/i] if it is less than $\scriptstyle p$.

I will prove that if $\scriptstyle x \in C$ and the block of $\scriptstyle k-1$ integers preceding $\scriptstyle x$, namely $\scriptstyle x-1, x-2, \dots, x-k+1$ aren't in $\scriptstyle C$, then they must be in $\scriptstyle A$ and, if valid, also the following block ($\scriptstyle x+1, x+2, \dots, x+k-1)$ is in $\scriptstyle A$.

In fact: we know that every element in the block preceding $\scriptstyle x$ is not in $\scriptstyle C$. It cannot be in $\scriptstyle B$ as well, otherwise $\scriptstyle i, x-i, x$ would be in different subset. Now take a valid value $\scriptstyle x+i$. It cannot lie neither in C ($\scriptstyle x-(k-i),k,x+i$), nor in B ($\scriptstyle i, x+i, x$). So it is in $\scriptstyle A$.

Now, let $\scriptstyle x$ be minimal in $\scriptstyle C$. $\scriptstyle x > k$, by hypothesis. So the preceding block is valid and, by minimality, not in $\scriptstyle C$. I apply the property just proved and find that the following block is in $\scriptstyle A$. If not all the elements of the following block are valid, then $\scriptstyle C$ cannot contain other elements and I have finished. Otherwise, I restart with the next minimimal element in $\scriptstyle C$ (if it exists): the preceding block cannot be in $\scriptstyle C$ by minimality, so it must be in $\scriptstyle A$, etc...

After this process $\scriptstyle A$ must contain: the first block $\scriptstyle 1, \dots, k-1$ + a preceding block of $\scriptstyle k-1$ elements for each element in $\scriptstyle C$, and we get $\scriptstyle card(A) \geqslant (k-1) (card(C)+1)$. []

[b]Remark: [/b] The relation $\scriptstyle x+y=z \pmod p$ is invariant under multiplication by a nonzero constant, so we can multiply every element in the partition by a nonzero constant and obtain another partition with the same properties. In particular, I can suppose that $\scriptstyle 1$ is in the smallest subset in the partition.

Now we are there. Suppose $\scriptstyle A$ is the smallest subset in the partition and $\scriptstyle 1 \in A$. Let $\scriptstyle k$ be the smallest number not in $\scriptstyle A$ (it obviously exists). Apply the Lemma and get a contradiction by finding another subset strictly smaller than $\scriptstyle A$. [][/size]

M.


Find all pairs (m,n) of integers satisfying the following two conditions:

a) two of the digits of m are the same as the corresponding digits of n, while the other two digits of m are both 1 less than the corresponding digits of n;

b) both m and n are four-digit squares

[hide]Equation to solve is something like :

A^2 + 10^u + 10^v = B^2 + 1111 with u !=v in 1..4

or

(A+B)*(A-B) in 1100,1010,1001,110,101,11 (*)

A+B and A-B have same parity so we can eliminate even numbers in (*) not divisible by 4 (i.e. 1010 and 110).

33 < A,B < 100 so we can eliminate 11.

We are left with 101 prime, 1100=2511 and 1001=7*11*13.

With A+B > 66, we can still reduce the cases.

For example, A+B=7*13 and A-B=11 gives 2601=51 and 1600=40

And so on ...

I find :

2025, 1024

2601, 1600

2601, 2500

3600, 2500

5625, 4624

[/hide]


War Nerd. A war hobbyist who post interesting, insightful and sometimes funny articles about war all around the world.

http://www.exile.ru/archive/by_author/gary_brecher.html

http://www.launch.yahoo.com

i like free music.

I have a bunch to post.

http://developer.apple.com/
http://dhost.info/mathproject (my site, not much on it yet)
http://s3.invisionfree.com/dnas_message_board
http://www.cocoadevcentral.com/

Amendment:

http://www.flashplayer.com/animation/glassofwater.html

This is one of my favorites, I'm not totally sure as to whether it's my all-time favorite.  This man writes humor columns that are absolutely hilarious.

[url]http://www.geoffmiller.net[/url]

http://www.albinoblacksheep.com   I thought I lost tetris forever when I got rid of my old computer.  But ha.  Um, their tetris doesn't stop at level 10, and you can't get negative points!  (I think after like 32,000 pts, it starts going negative...but I never got to zero. -13, something... I think).  Haha, it's got a fun game "Defend Your Castle" where you kill the little stick-figures. :-D  And a video, "Cat!  I'm a kitty-cat!  And I dance dance dance, and I dance, dance dance."  It's very interesting.

[quote="SingTheSorrow"]www.albinoblacksheep.com   I thought I lost tetris forever when I got rid of my old computer.  But ha.  Um, their tetris doesn't stop at level 10, and you can't get negative points!  (I think after like 32,000 pts, it starts going negative...but I never got to zero. -13, something... I think).  Haha, it's got a fun game "Defend Your Castle" where you kill the little stick-figures. :-D  And a video, "Cat!  I'm a kitty-cat!  And I dance dance dance, and I dance, dance dance."  It's very interesting.[/quote]

i've played defend your castle. that games pretty funny. also, the "end of the world" cartoon is funny as well.

[quote="yif man12"]i've played defend your castle. that games pretty funny. also, the "end of the world" cartoon is funny as well.[/quote]
Defend Your Castle loses all point when you get to a high enoug hlevel that no stick figures every even get to your castle because of the archers.

http://www.isketch.net

Fun game for those with excellent verbal skills. I'm still trying to find a math section there...

Unbelievably funny:

http://www.jibjab.com

Check out the "This Land is Your Land" animation.

The site was created by two brothers. It got 5 million hits in the past few weeks simply by word-of-mouth about this particular animation. They were featured on "The Today Show" and were on Nightly News this evening. It is a brilliant animation!!!!

The site I visit most, after AoPS, is

http://news.google.com/

which is a habit I developed during the SARS epidemic.

I visit this site just almost as often as aops:

http://www.google.com/

http://www.www.com
surprisingly (to me at least), a real website

and also, my favorite game on the web, city jumper.
http://www.nationlocation.com

[url=http://www.google.com/]Google[/url]
[url=http://forums.gentoo.org/]Gentoo Forums[/url]
Not so much these days, but.. [url=http://forum.defcon.org/]Defcon Forums[/url]
[url=http://en.wikipedia.org/]Wikipedia[/url]

The AoPS Forum server seems slow.. Is it always like this, or is it because of heavy load or something?

[url=http://www.mathlinks.ro]www.xxx.com[/url]

no, just kidding! :D

[quote="RobinHood3000"]Ahem, you might want to change that link.  You can make it display a link different from the site to which it actually leads, you know...[/url][/quote]

...or just delete the message entirely.

[quote="beta"][url]http://www.shuku.net/dblx/html/34/689-2-0.html[/url]
I am bored out of my mind that I am reading Jinyong wuxia novels(Chinese) now online.[/quote]

hehe i just got back from China and I've watched She Diao Yin Xuong Zhuan and Qi Tian Zhu Long Ji  they were awsome!  I also watched half of the first series in Jinyong (forgot what its called) but the actors wasnt as good so i stopped

Yeah, I watched Xiao Ao Jiang Hu(TVB) and Tian Long Ba Bu(TVB) in China. They were pretty good(although the fightings were really fake, too much dynamite). I really want to watch Xiao Ao Jiang Hu(CCTV) now, Xiao Ao Jiang Hu is definitely my favorite Jinyong novel.

what about Anyone here enjoy learning about our White House?   :twisted:

[i]edited by a mod[/i]

This is ridiculous, guys. WarpedKlown and Peter VDD's posts should be deleted immediately.

We need a moderator's attention to this thread, pronto.

I enjoy reading [url]http://www.cockeyed.com[/url], which sounds similar to the links Peter and Yannik posted, but is completely different.  It's the personal website of a self-acclaimed funny guy named Rob Cockerham who conducts pseudoscientific experiments to determine how much is inside things like toothpaste and sharpie markers.  (This comprises the 'How Much Is Inside?' portion of his website.)  He also makes costumes (California and Papparazzi are two), pulls pranks (for example, he created a fake McDonalds' ad for apple pie that featured the star of American Pie, then put it in several McDonalds' drive-thrus), and sells silly things (spring shoes, diamonds*, etc.) on eBay.

I also like [url=http://en.wikipedia.org]Wikipedia[/url] and [url=http://gmail.com]Gmail[/url].

ehh... I dunno about favorite or not.. but I visit these the most besides AoPS:

[url=http://www.google.com]Google[/url]
[url=http://www.xanga.com]Xanga[/url]

[quote="WarpedKlown1335"]Oops I meant http://www.whitehouse.org

I clicked on the link and eh...yeah.  Sorry.[/quote]
err, why dont you make an edit to your original post? that could save the mods some trouble

I thought I [b]DID[/b] make my link point to another site :? :?

Anyway sorry for that...

If you review my post now I think you will get the fun of it though :D

[quote="Peter VDD"][url=www.mathlinks.ro]www.xxx.com[url]

no, just kidding! :D[/quote]

Peter is just tricking us

[quote="Ragingg"]www.fark.com

Great news site.[/quote]

Mm, I'd say there's more than just news there.

https://www.youtube.com/channel/UCFod3BWeZwhg2W1kBi15-pA
here's mine :)


[quote="penny77114"]Prove there exists an N such that if you have n>N points in general position, two-colored in the plane, there exists a monochromatic empty 4-gon.[/quote]what is exactly empty 4-gon?


linear algebra
matrix
algebra
polynomial
system of equations
algebra proposed

Solve the system of equations:
$ \frac{x}{a\minus{}30}\plus{}\frac{y}{a\minus{}4}\plus{}\frac{z}{a\minus{}14}\plus{}\frac{t}{a\minus{}10}\equal{}1$
$ \frac{x}{b\minus{}30}\plus{}\frac{y}{b\minus{}4}\plus{}\frac{z}{b\minus{}14}\plus{}\frac{t}{b\minus{}10}\equal{}1$
$ \frac{x}{c\minus{}30}\plus{}\frac{y}{c\minus{}4}\plus{}\frac{z}{c\minus{}14}\plus{}\frac{t}{c\minus{}10}\equal{}1$
$ \frac{x}{d\minus{}30}\plus{}\frac{y}{d\minus{}4}\plus{}\frac{z}{d\minus{}14}\plus{}\frac{t}{d\minus{}10}\equal{}1$
where $ a;b;c;d$ are distinct numbers,none of which belong to the set $ \{4;10;14;30\}$

[hide]
\[ \begin{pmatrix} 1 & \frac {a \minus{} 30}{a \minus{} 4} & \frac {a \minus{} 30}{a \minus{} 14} & \frac {a \minus{} 30}{a \minus{} 10} & a \minus{} 30 \\
1 & \frac {b \minus{} 30}{b \minus{} 4} & \cdots \\
\vdots & & \ddots \\
\end{pmatrix}
\]
Since $ \frac {b \minus{} r}{b \minus{} s} \minus{} \frac {a \minus{} r}{a \minus{} s} \equal{} \frac {(r \minus{} s)(b \minus{} a)}{(b \minus{} s)(a \minus{} s)}$, we have (after canceling out $ b \minus{} a$ from and multiplying by $ (b \minus{} 4)(a \minus{} 4)$ row 2)
\[ \begin{pmatrix} 1 & \frac {a \minus{} 30}{a \minus{} 4} & \frac {a \minus{} 30}{a \minus{} 14} & \frac {a \minus{} 30}{a \minus{} 10} & a \minus{} 30 \\
0 & 26 & \frac {16(b \minus{} 4)(a \minus{} 4)}{(b \minus{} 14)(a \minus{} 14)} & \frac {20(b \minus{} 4)(a \minus{} 4)}{(b \minus{} 10)(a \minus{} 10)} & (b \minus{} 4)(a \minus{} 4) \\
0 & \frac {26}{(c \minus{} 4)(a \minus{} 4)} & \cdots \\
\vdots & & \ddots \\
\end{pmatrix}
\]
You can continue on in this way to get the answer. I, however, need to go to sleep or else my parents will shut off the computer and I'll lose all my work. So, g'night!

Oh, and it would be really nice if there's an quicker/easier way to get to the answer. Without a calculator, that is.
[/hide]

[hide="there is an easier way"]
The polynomial
\[ (u\minus{}4)(u\minus{}10)(u\minus{}14)(u\minus{}30)\minus{}x(u\minus{}4)(u\minus{}10)(u\minus{}14)\]
\[ \minus{}y(u\minus{}10)(u\minus{}14)(u\minus{}30)\minus{}z(u\minus{}4)(u\minus{}10)(u\minus{}30)\minus{}t(u\minus{}4)(u\minus{}14)(u\minus{}30)]\]
is a polynomial in $ u$ of degree 4, leading coefficient 1, and roots $ a,b,c,d$. So the polynomial must be $ (u\minus{}a)(u\minus{}b)(u\minus{}c)(u\minus{}d)$. Plugging in 30, 4, 14, and 10 for $ u$ gives
\[ x\equal{}\minus{}\frac{(30\minus{}a)(30\minus{}b)(30\minus{}c)(30\minus{}d)}{8320}\]
\[ y\equal{}\frac{(4\minus{}a)(4\minus{}b)(4\minus{}c)(4\minus{}d)}{1560}\]
\[ z\equal{}\frac{(14\minus{}a)(14\minus{}b)(14\minus{}c)(14\minus{}d)}{640}\]
\[ t\equal{}\minus{}\frac{(10\minus{}a)(10\minus{}b)(10\minus{}c)(10\minus{}d)}{480}\]
[/hide]

[quote="scorpius119"]
The polynomial
\[ (u \minus{} 4)(u \minus{} 10)(u \minus{} 14)(u \minus{} 30) \minus{} x(u \minus{} 4)(u \minus{} 10)(u \minus{} 14)
\]

\[ \minus{} y(u \minus{} 10)(u \minus{} 14)(u \minus{} 30) \minus{} z(u \minus{} 4)(u \minus{} 10)(u \minus{} 30) \minus{} t(u \minus{} 4)(u \minus{} 14)(u \minus{} 30)]
\]
is a polynomial in $ u$ of degree 4, leading coefficient 1, and roots $ a,b,c,d$. So the polynomial must be $ (u \minus{} a)(u \minus{} b)(u \minus{} c)(u \minus{} d)$. Plugging in 30, 4, 14, and 10 for $ u$ gives
\[ x \equal{} \minus{} \frac {(30 \minus{} a)(30 \minus{} b)(30 \minus{} c)(30 \minus{} d)}{8320}
\]

\[ y \equal{} \frac {(4 \minus{} a)(4 \minus{} b)(4 \minus{} c)(4 \minus{} d)}{1560}
\]

\[ z \equal{} \frac {(14 \minus{} a)(14 \minus{} b)(14 \minus{} c)(14 \minus{} d)}{640}
\]

\[ t \equal{} \minus{} \frac {(10 \minus{} a)(10 \minus{} b)(10 \minus{} c)(10 \minus{} d)}{480}
\]
[/quote]

How do you know that the roots of the polynomial are $ a, b, c, d$?  Also, where do the denominators of your solutions come from?


AMC
AIME
geometry
analytic geometry
trigonometry
angle bisector
Pythagorean Theorem

It seems very do-able, but its #14 on a very hard mock aime.

In triangle $ ABC$, $ AB\equal{}308$ and $ AC\equal{}35$.  Given that $ AD$, $ BE$, and $ CF$ intersect at point $ P$ and are an angle bisector, median, and altitude of the triangle, respectively. compute the length of $ BC$.

It can be done very quickly if you know
Ceva's Theorem

[quote="unimpossible"]It seems very do-able, but its #14 on a very hard mock aime.

In triangle $ ABC$, $ AB \equal{} 308$ and $ AC \equal{} 35$.  Given that $ AD$, $ BE$, and $ CF$ intersect at point $ P$ and are an angle bisector, median, and altitude of the triangle, respectively. compute the length of $ BC$.[/quote]

[hide]
Ceva's obviously.

[/hide]

[hide="Solution, its not that bad"]
By [[Ceva's Theorem]],
\[ \frac {AE}{EC} \cdot \frac {CD}{DB} \cdot \frac {BF}{FA} \equal{} 1
\]
Since $ E$ is the midpoint of $ AC$, $ AE \equal{} EC$. By the Angle-Bisector Theorem, $ \frac {CD}{DB} \equal{} \frac {CA}{AB} \equal{} \frac {35}{308} \equal{} \frac {5}{44}$. Thus this reduces to
\[ \frac {BF}{FA} \equal{} \frac {44}{5}
\]
Also $ BF \plus{} FA \equal{} 308 \Longrightarrow FA \equal{} \frac {220}{7}, BF \equal{} \frac {44^2}{7}$. By the Pythagorean Theorem on $ \triangle AFC$, $ FC^2 \equal{} 35^2 \minus{} \left(\frac {220}{7}\right)^2 \equal{} \frac {11625}{49}$, and again on $ \triangle CFB$ gives $ BC^2 \equal{} FC^2 \plus{} BF^2 \equal{} \frac {11625}{49} \plus{} \left(\frac {44^2}{7}\right)^2 \equal{} 277^2$, and so $ BC \equal{} \boxed{277}$.

I actually solved a mock aime geometry problem .. yay.  :lol: And post conflict twice.
[/hide]

I just cranked the whole thing out using a coordinate system. It's not as ugly as one would imagine, though it is nowhere near as nice as ceva's

[hide]
Let $ A$ be the origin and $ AB$ be the positive x axis.

Let $ \theta$ be $ \angle CAB$

This gives equations for the three lines

CF: $ x \equal{} 35 \cos\theta$

AD: $ y \equal{} (\tan\frac{\theta}{2}) x$

BE: $ y \equal{} \frac{\frac{35}{2}\sin\theta}{\frac{35}{2}\cos\theta \minus{} 308}(x\minus{}308)$

The first two equations can be subbed into the third one to solve for $ \cos\theta$. After a trig identity for the $ \tan\frac{\theta}{2} \equal{} \frac{\sin\theta}{1\plus{}\cos\theta}$ things drop out nicely. It turns out to be linear in $ \cos\theta$. Then you can use the law of cosines to find $ CB$ fairly easily.
[/hide]


probability
inequalities
analytic geometry

[hide="not sure..."]is it $\frac{1}{3}$?[/hide]

[quote="rnwang2"]Let $A(0,0)$, $B(1,0)$, $C(\frac{89}{400},0)$, and $D(x,0)$ be $4$ points on a graph with $x$ a random value such that $0<x<1$. Find the probability that all $3$ of the following conditions are true: $AD+BD>CD$, $AD+CD>BD$, $CD+BD>AD$. Express your answer as a common fraction reduced to lowest terms.[/quote]
[hide] The first condition is always true.
$x+\left|x-\frac{89}{400}\right|>1-x\iff x>\frac{163}{400}$

$\left|x-\frac{89}{400}\right|+1-x>x\iff x<\frac{311}{400}$
So, $\frac{311}{400}-\frac{163}{400}=\boxed{\frac{37}{100}}$.  :maybe: [/hide]

[hide]
Summing the inequalities, we have

$2AD+2BD+2CD=2(AD+BD+CD)>AD+BD+CD$

which will always be true given $AD+BD+CD>0$. Thus the probability is 1.[/hide]

[quote="SM4RT"]it cant be 1 clearly lotrgreengrapes7926 solution shows values for which the inequality holds true and its not all fo them..[/quote]

That's weird.. what's wrong with my solution then?

well i think its cuz ur assuming the three inequalities hold true no matter what..

Dude... you can't add up inequalities out of the blue  :wink:

[hide]
Alright, first of all we can forget about the second coordinate because its always zero.

The first inequality is always true because either $AD>CD$ or $BD>CD$ is true if the points are unique, and if $D=C$ then it is obviously true. Valid D: $[0,1]$

For the second, $C<D<B$. We therefore have:
$D+D-89/400>1-D \to D> 163/400$
Valid D: (163/400,1]

For the third, again, $C<D<B$. Thus, $CD+DB$ is a constant $311/400$, and we have $311/400>D$
Valid D: $[89/400,311/400)$

The valid interval is the intersection of these, or $(163/400,311/400)$.

The probability we want is $\frac{311/400-163/400}{1-0}=\frac{148}{400}=\boxed{\frac{37}{100}}$
[/hide]


[quote="metdzoi"]If  abc = 1 and a,b,c >o .Prove :

 $ \sqrt {\frac {a}{a \plus{} b \plus{} 7}} \plus{} \sqrt {\frac {b}{b \plus{} c \plus{} 7 }} \plus{} \sqrt {\frac {c}{c \plus{} a \plus{} 7 }} \geq 1$[/quote]
Let $ S \equal{} \sqrt {\frac {a}{a \plus{} b \plus{} 7}} \plus{} \sqrt {\frac {b}{b \plus{} c \plus{} 7 }} \plus{} \sqrt {\frac {c}{c \plus{} a \plus{} 7 }} \geq 1$
By ineq Holder's
$ S^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} 7(a \plus{} b \plus{} c)) \geq (a \plus{} b \plus{} c)^3$
We must prove that :$ (a \plus{} b \plus{} c)^3 \geq (a \plus{} b \plus{} c)^2 \minus{} ab \minus{} bc \minus{} ca \plus{} 7(a \plus{} b \plus{} c)$(*)
 Let
$ a \plus{} b \plus{} c \equal{} p \geq 3$and$ ab \plus{} bc \plus{} ca \equal{} q$
we have:$ q^2 \equal{} (ab \plus{} bc \plus{} ca)^2 \geq 3abc(a \plus{} b \plus{} c) \equal{} 3p$
(*) becomes :$ p^3 \minus{} p^2 \minus{} 7p \plus{} ab \plus{} bc \plus{} ca \geq 0$
Let$ p \equal{} t^2, t \geq \sqrt {3}$
we have :$ t(t \minus{} \sqrt {3})(t^4 \plus{} \sqrt {3}t^3 \plus{} 2t^2 \plus{} 2\sqrt {3}t \minus{} 1)\geq 0$
( obviously  right )($ S^2 \geq 1 \iff S\geq 1$)
we have done completly.


The notation $ [x]$ stands for the greatest integer that is less than or equal to $ x$. Calculate $ [\minus{}1.2]$.

The greatest integer less than or equal to -1.2 is -2.


A regular polygon with 2n + 1 sides is inscribed in a circle. Three of the polygon’s vertices are
selected at random and the triangle formed by connecting these vertices is drawn. Prove that
the probability that the center of the circle lies inside the triangle is n+1
2(2n−1) .

Shakthi S,

Please do not post problems here that are not from Alcumus. If they are, use Gamebot instead by clicking the "Discuss this problem on the forum" link.


inequalities logarithms calculus derivative integration geometry trapezoid

Prove that if $ 0 < a < b$ then $ \ln \frac {b^2}{a^2} < \frac {b}{a} \minus{} \frac {a}{b}$

If you multiply each of $ a$ and $ b$ by the same positive number, both sides of the inequality remain unchanged. Hence, WLOG, we may assume that $ a\equal{}1$ and we are trying to prove that $ 2\ln b<b\minus{}\frac1b$ for $ b>1.$

Let $ f(x)\equal{}x\minus{}\frac1x\minus{}2\ln x.$ We're trying to prove that $ f(x)>0$ for $ x>1.$

Note that $ f(1)\equal{}0.$ Now take the derivative:

$ f'(x)\equal{}1\plus{}\frac1{x^2}\minus{}\frac2x\equal{}\frac{x^2\plus{}1\minus{}2x}{x^2}\equal{}\frac{(x\minus{}1)^2}{x^2}>0$ for $ x>1.$

Hence $ f$ is strictly increasing on $ [1,\infty),$ hence $ f(x)>0$ for $ x>1.$

If $ 0 < a < b$ then we can just observe the equation:

$ \ln x^2 < x \minus{} \frac{1}{x}$ for $ x > 1$

Now, we can see that for $ x \equal{} 1$ we have:

$ \ln x^2 \equal{} x \minus{} \frac{1}{x}$

so we only need to prove

$ \frac{2}{x} < 1 \plus{} \frac{1}{x^2}$

$ 0 < 1 \minus{} \frac{2}{x} \plus{} \frac{1}{x^2}$

$ 0 < \left(1 \minus{} \frac{1}{x}\right)^2$

and this is true..

Let $ s\equal{}\ln\frac ba$. We want to show that $ 2s\le e^s\minus{}e^{\minus{}s}$.
There are many ways to do this. The most obvious to me is the series expansion $ e^s\minus{}e^{\minus{}s}\equal{}2s\plus{}\frac{2s^3}{6}\plus{}\frac{2s^5}{5!}\plus{}\cdots$, in which all terms are nonnegative.

(Wow, two posts while I was composing this. It's different, so it goes up anyway.)

$ \ln \frac {b^2}{a^2} < \frac {b}{a} \minus{} \frac {a}{b}\ (0 < a < b)$

$ \Longleftrightarrow \int_a^b \frac {1}{x}\ dx < \frac {1}{2}\left(\frac {1}{a} \plus{} \frac {1}{b}\right)(b \minus{} a)\ (0 < a < b)$

Considering the areas of two figures for $ y \equal{} \frac {1}{x}\ (x > 0)$ in using convexity of the graph gives the proof. :lol:

[quote="kunny"]Considering the areas of two figures for $ y \equal{} \frac {1}{x}\ (x > 0)$ in using convexity of the graph gives the proof. :lol:[/quote]

Nice solution dear Kunny  :lol: I'm very interested in undestanding what I quoted. and how can I prove it, sorry if that is elementary, but i dont see it clearly  :blush:

Thanks in advance

[quote="Tiger100"]

Nice solution dear Kunny  :lol: I'm very interested in undestanding what I quoted. and how can I prove it, sorry if that is elementary, but i dont see it clearly  :blush:

Thanks in advance[/quote]

$ \frac 12 \left(\frac 1a \plus{} \frac 1b\right)(b \minus{} a)$ is the area of the trapezoid for the function $ 1/x$ with bases at $ a$ and $ b$. The top boundary of the trapezoid is a secant line of the graph. The graph of $ 1/x$ is below the secant line, thus the area under the function is less than the area of the trapezoid.

[quote="Kent Merryfield"]If you multiply each of $ a$ and $ b$ by the same positive number, both sides of the inequality remain unchanged. Hence, WLOG, we may assume that $ a \equal{} 1$ and we are trying to prove that $ 2\ln b < b \minus{} \frac1b$ for $ b > 1.$[/quote]

Can you explain it a bit? Thanks.


Interesting story for Careers in Mathematics

http://www.usatoday.com/money/perfi/funds/2006-05-26-hedge-funds-usat_x.htm

I want to quote the last paragraph of the article:

[quote]The managers didn't just line their own pockets. "I suspect they have made more money for their clients than themselves," says Charles Davidson, hedge fund director at Standard & Poor's. [/quote]
What ever Charles Davidson said is mathematically incorrect. Simon, the #1 on the list (with earning 1.5 billion in one year), who charges 5% of the asset plus 44% of the earning simply can't make more for his client than for himself.

One thing always bothers me is that the whole trading/ transections do not enrich the society (no real product/goods/true service/new value generated for the benefits of the society).

[quote="qvc122"]
One thing always bothers me is that the whole trading/ transections do not enrich the society (no real product/goods/true service/new value generated for the benefits of the society).[/quote]

While part of the reason I left trading was that I didn't feel day-to-day like I was generating benefits for society, I don't think it's true that in general trading (or even I in my old job) does/did not generate benefits for society.  In addition to tangible goods such as the financial instruments that have made housing loans and student loans possible, there are intangible benefits.  Even if the traders do nothing more than provide liquidity (i.e. the ability to buy and sell assets), that is a service in and of itself, as, for example, it makes these loans more accessible and cheaper.

Certainly the great mob of traders narrow the price between bid and ask at the very least.  For example, (this is a very simplistic example) imagine a world with no trading-for-profit's-sake traders.  The price of, say, corn in such a market might be bid at 72 and offered at 75 (meaning you could sell corn at 72 and buy it at 75).  Now, let's add on the traders.  Suppose we add enough speculating traders to this market to triple the trading volume.  What's likely to happen is that the bid-ask spread will narrow, say to 73-74.  Now, Farmer Bob gets a buck more when he sells his corn, and the feed store gets it a buck cheaper (a portion of these savings are then ultimately passed to the consumer).  So, Farmer Bob benefits, and so do consumers.

Indeed, this is a simplistic example, but market liquidity alone does have social value.  Another is market efficiency.  Suppose Government A is very stupid, and doesn't allow its banks to sell bonds, only to buy them.  However, the banks need to sell bonds to hedge their interest rate exposure.  So, they instead sell bond futures, which have a price based on the price of bonds.  What ultimately happens because the banks are so big is that the bond futures gradually get pressed down relative to the price of the bonds.  In a world with no profit-driven traders, this might result in the banks  (and those people with their money in the banks) getting hosed, as the price is driven ever cheaper.  However, in come the arbitrageurs.  They buy the future when it gets too cheap and sell short the bonds and wait for the prices to converge.  They'll make some money due to this stupid government policy, but the banks will be able to sell at a higher price.  Again, this is a very simplistic example and glosses over a lot of details, but providing efficiency to the market does have societal good.

All this said, I rarely went home jacked about making Mexican markets more efficient when I was trading.  Yet my trading did have that effect.  I do find my current work more rewarding in a societal sense, but I do think there is some tough-for-me-to-feel-while-you're-doing-it value to trading solely for profit.

rrusczyk, I have great respect for you for the fact that you left trading to start AoPS. If I was in your shoes, I probably would not make the same choice.

I do agree that trading has values, representing service and providing liquidity, but not for the outrageous fees imposed. In other words, those outrageous fees cannot be supported by the service and the liquidity generated.

When you think about investors of hedge funds, I could careless for the wealthy ones, since in most of the situations, even they are robbed by fund managers, they can sustain the loss and will not become poor. However, most of the hedge funds investors are institutional investors, the second layer of managers, many of them, managing retirement funds for routine people, who did not have slightest idea that some of their retirement is in hedge funds. The ordinary people are risking their retirement capitals for whatever the remaining profit left for them after Simon takes that huge fees (in his specific situation, close to 60% of total profit) and after the fees for the second layer of managers (the institutional managers). Simon risks nothing, even when there is no profit, he is still entitled for that 5% asset. The only risk he has is if his hedge fund doesn’t do well, he has to close his fund. Then he will just start again with a hedge fund in different name. When you think about 8000 hedge funds in the market, 1000-2000 will go under each year, it is quite disturbing. The managers never need a long term goal, since they calculated and take fees quarterly, why should they even bother with long term. If it doesn’t work, just re-start another hedge fund.

I can only hope that ordinary people do not further lose their purchasing power when they retire after this redistribution of money between fund managers and the investing public. The truth is the investing public, the routine people really do not have money to invest. They truly treat the portion of their pay check taken into their 401k as another tax. Hopefully, those whose hands are entrusted with public retirement money have conscience.

I think you're aiming at a different issue here than the societal good of trading, and it's one that I do agree with to some extent.  For example, investing in large mutual funds with high fees is typically a disastrous way to invest money, as compared with, say, finding an index fund with fees that are as low as possible.  (In fact, for long-term stock investing, it's tough to beat a strategy of 'choose broad-based funds with the lowest fees.)  However, there are cases in which large funds can offer opportunities not available to small investors, such as access to certain markets.

That said, I tend to agree that money managers are overpaid in the same sense that rock stars are.  Very few rock stars are notably better (if at all) than the many would-be Madonnas who don't make it, just as I think few money managers are better than the zillions of mid-level investment bankers who don't make it to the top of the heap.  But they won a competition that pays the winners very well, indeed, because money management is also like music in that it is somewhat of a herd business.  If you can somehow define yourself as the front of a herd (often a marketing feat, or the result of luck, or in some rare cases, skill), then you win.

One thing you touch on that is a real problem in the trading world is that traders have downside protection - the worse that can happen (provided they don't do anything illegal) is that they can lose their job.  But, finding another one is typically no problem.  So they have an incentive to swing for the fences.  Good traders keep this attitude in check, but it's hard.

As for people's retirement money ending up in the hands of hedge fund managers, I think the people (and their companies that set up the 401k plans) have the primary responsibility to protect against this - they are not forced to give their money to mutual funds or hedge funds, etc.

I should also note that most fund managers are heavily invested in their own funds.  Moreover, a well-designed company has most of its traders invested in the firm.  For example, at Shaw, a portion of my bonus each year was withheld and paid out over the next number of years as deferred compensation, and that money grew or shrank with the return of the firm.  So, when we lost a lot of money, it cost me personally.  This goes all the way up the food chain.  As I understand it, LTCM's implosion in 1998 followed a couple years of the the firm's returnign capital to investors so that the principals would have a greater share of their own money earning their high (until 1998) returns.  I'm sure these people were diversified, but nonetheless, it's not the case that they have nothing to lose.

Perhaps the issue of fund manager pay is really a result of wealth increasing faster than the number of vehicles through which to invest them.  If the number of funds quadruples (as is happening), one would expect a fundamental change in pay structures.  But the pay to traders will always be higher than some deem just - like it or not, providing services to wealthy people or institutions will almost always pay more than providing them to poor people.

Some of the top people at LTCM were not diversified at all and basically lost everything.  One guy divorced his wife when he got suddenly rich, married a one-generation-younger woman, then lost the money in the collapse.


In a right triangle, ratio of the hypotenuse over perimeter of the triangle determines an interval on real numbers. Find the midpoint of this interval?

$\textbf{(A)}\  \frac{2\sqrt{2} \plus{}1}{4} \qquad\textbf{(B)}\  \frac{\sqrt{2} \plus{}1}{2} \qquad\textbf{(C)}\  \frac{2\sqrt{2} \minus{}1}{4} \\ \qquad\textbf{(D)}\  \sqrt{2} \minus{}1 \qquad\textbf{(E)}\  \frac{\sqrt{2} \minus{}1}{2}$

[hide]$ \varrho={c\over c(1+\sin\alpha+\cos\alpha)}={1\over{1+\sqrt{2}\sin\left(\alpha+{\pi\over 4}\right)}}$

For an acute angle $ \alpha$ we have $ {\sqrt{2}\over 2}<\sin\left(\alpha+{\pi\over 4}\right)\leqslant 1$, hence $ \varrho$ is between ${ {1\over {1+\sqrt{2}\cdot 1}}=\sqrt{2}-1}$ and $ {1\over {1+\sqrt{2}\cdot{\sqrt{2}\over 2}}}={1\over 2}$, so the desired midpoint is $ {1\over 2}\left(\sqrt{2}-1+{1\over 2}\right)={2\sqrt{2}-1\over 4}$, and the answer is $ D$.[/hide]


MATHCOUNTS
AMC
AIME
geometry
probability
arithmetic sequence
number theory

2. For each positive integer k, let S(k) denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. For example S(3) is the sequence 1, 4, 7, . . . . For how many values of k does S(k) contain the term 2005?

(#) Means subscript.
I thought this was a national sprint level problem, so take your shots guys (and girls). The answer has to be an integer from 000 to 999. Don't peek at the AMC forum, for the answer, and hide your solutions as usual. I'll post my solution friday night.

Ha!  That's exactly what I told someone! "#2 on the AIME was a mathcounts-level problem!"

[hide="Click for a hint."]
What happens if you subtract 1 from each term?
[/hide]

Maybe even State Target!

Sorry, I already read the AIME topics, but it has one fun solution!

[hide]Here is a interesting fact, hope it isn't giving it away... For a composite number with prime factors $a^x b^y c^z \cdots $, there are exactly $(x+1)(y+1)(z+1)\cdots $ divisors. VERY USEFUL! Use whenever you can get a problem to be a 'how many divisors does this number have' one. For example, we can take a  easy case, like 12. The prime factorization is $2^2 3^1$, so by the formula, there are $(3)(2)$ divisors. We list them, and get :
1
2
3
4
6
12
So indeed it works.[/hide]

Billy

Billy, can you hide that as a hint, bc it really does give away the entire question. That was the exact same way I did it.

[hide]12. if k= 1,2,3,4,6,12,167,334,501,668,1002,2004. Subtract 1 from the number you're looking for (in this case it's 2004) and find the factors.[/hide]

Now biffandoc, can you do it without going case by case? Try Solafidefarms' property.

[hide]Way I did this was that i said xy - x + 1 = 2005 where x is the common difference. We're trying to find all the different values of x so I said that x = 2004/(y-1). So all you have to do is find the number of factors in 2004 which is 12 because it it is 2^2 * 3 * 167. I'm not sure if I'm right, but I think the answer is 12.[/hide]

Yes. That method is also correct. There are two un-mentioned methods left, one which is the brute force method and the other, which I did, which shortcutted the brute force. I think tarquin may have used the formula, but I'm not sure, b/c his post doesn't say. Your equation makes this solution "elegant", as MCrawford says, but it may be unecessary if you are doing Mathcounts. After a few more solutions are posted, I'll post my way, which you guys pretty much have done. Here's another real easy problem from the AIMEs this year: It helps to have geometry for this problem, but there is another way without any geometry knowledge. I've seen four general solutions to this problem, so let's see how you all do:

Six congruent circles form a ring with each circle externally tangent to the two circles adjacent to it. All six circles ar einternally tangent to a circle C with radius 30. Let K be the area of the region inside C and ouside all of the six circles in the ring. Find bKc.

[hide]All of the little circles must have a radius of 10 so its just 900pi - 600pi.[/hide]

So that everyone can see, can you explain your method tarquin? You're absolutely right...

Heh, those two accounted for 1/3 of the AIME problems I solved...so I'll keep quiet.

All right, hockey.
[hide]
We can see that there are, 6 circles inside the big circle. Knowing this, we know that if we were to draw a two radii tangent to the same small circle, it would form a 60 degree angle. Now we also know that if that is the case, we can circumscribe an equilateral triangle about one of the small circles, using the radius of the large circle as the height. Now the radius of the small circle can be found by using the formula rs=A where r is the radius of the circle, s is semiperimeter, and A is the area of the triangle, or we can simply take 1/3 of the height of the triangle because it is equilateral. Since the height of the triangle is 30, the radius of the small triangles must be 10. Therefore the area inside the large circle but outside the small circles is 30^2*pi - 10^2*6*pi which is approximately 942.
[/hide]

[quote="hockeyadi23"]Funny.. I think I had no clue how to solve the quartic equation but I guessed 45 and I think I got it right! Probability of that: 1/1000!!![/quote]
you could complete the square of a square

[quote="hockeyadi23"]Funny.. I think I had no clue how to solve the quartic equation but I guessed 45 and I think I got it right! Probability of that: 1/1000!!![/quote]
The [i]same thing[/i] happened to me!!!!! We must be hardwired to detect answers automatically, huh?  :D


You can improve your programming skills in much the same way that you can improve your mathematical problem solving skills: by competing in programming contests. Some popular contests are [url=http://www.uwp.edu/sws/usaco/]USACO[/url], [url=http://www.topcoder.com/]Topcoder[/url], the [url=http://cm2prod.baylor.edu/login.jsf]ICPC[/url], and many more. These contests vary in level between high school, collegiate, and open-to-anyone. Of course, you should first learn a programming level relatively well (I recommend C++), but there are tons of resources available for that (I can list some if you want).

Yeah, But I am in the final year of school right now, I am pretty good at C, and not so proficient at C++.

Do list the resources that you had mentioned. Thanks!

C is usually enough to solve the programming contest type problems. C++ is more useful for maintaining large, complex software projects that multiple people, possible in different places, are working on over long periods of time. But some C++ features like the std::string and some STL containers and algorithms are still very convenient and can save you a lot of headaches due to things like out-of-bounds memory access.

Maintaining large software projects with complicated interdependencies between classes is a whole other realm of programming and requires skills that might be considered perpendicular to the skills you need to do well in programming contests. If you are interested in acquiring these skills, you would want to learn about [url=http://en.wikipedia.org/wiki/Design_pattern_(computer_science)]design patterns[/url] and [url=http://en.wikipedia.org/wiki/Software_framework]frameworks[/url], and also look into C++ class libraries like [url=http://www.boost.org/]Boost[/url] and [url=http://www.sgi.com/tech/stl/table_of_contents.html]STL[/url]

But, in my opinion, programming contests are where the most fun programming is. :lol: Even if you are not a student, you can still, for example, try old contest problems. You can follow some of the links I gave you to get to databases of old contest problems. They also usually provide lots of testcases with those problems, so you can check you're solution and make sure it works correctly. Furthermore, you can compete in contests even if you are not associated with any school. USACO allows literally anyone to register for free and compete in their contests. Although USACO makes all non-high-schoolers compete as "observers", you still get a score and get to see you're ranking against everyone else. I believe that Topcoder also allows anyone to register, and winning their contests actually pays you money, although I'm not sure whether there is an entrance fee or not.

The point is that you should just dive in and try to program something and see if you like it. If not, try to program something else. The old contest problems, especially the USACO training website, provide tons of good material to hone your programming skills.

As for the C++ resources, my favorite online one is this:

http://www.cplusplus.com/doc/tutorial/

and my favorite printed one is this:

http://www.amazon.com/Primer-4th-Stanley-B-Lippman/dp/0201721481/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1239114164&sr=8-1

[quote="Gen8"]Hi all, I am quite new to this this particular "forum" of AOPS. I would like to ask a question.

How can I improve my coding skills to a great/dramatic extent? How can I become an expert programmer?[/quote]
Try to write unit tests first before you finish your implementation. Use a testing harness like JUnit can help. And practice this religiously. Become proficient at debugging.

[b] @ Fibonacci Fan and @ Polishcalculator [/b] : I have finished my under-graduation in computer science, and I don't intend to participate in IOI ( and I'm not eligible too).

Another query- When beginning to learn a new language say 'Python' do I immediately begin thinking of problems and start developing the code for it, or do I first 'master' the syntax and semantics of the language (I mean 'Python' here)?

[It doesn't necessarily mean 'Python' but any new programming language]

What do you people do if you were in a situation as mine? (I think you people are very good at coding).

If you feel pretty confident about your knowledge in C, I'd say find problems and solve them.  You'll learn syntax tricks through errors and thus, they'll be reinforced in your mind the next time you run into them.  Since you should already know most programming techniques and capabilities, it shouldn't be too hard to find an equivalent across languages.

If you're not too sure about your knowledge in C, I'd say find a book about whatever language and restart from scratch.  This way, you'll learn what the language is capable of first, so you don't write overly complex looparound solutions.

C is really good language I'll say. Its fast, compiled and has lots of support.


In square $ ABCD$, $ AD$ is 4 centimeters, and $ M$ is the midpoint of $ CD$. What is the ratio of $ OC$ to $ OA$? Express your answer as a common fraction.

[asy]size (3cm,3cm);

pair A,B,C,D,M;

D=(0,0);
C=(1,0);
B=(1,1);
A=(0,1);

draw(A--B--C--D--A);

M=(1/2)*D+(1/2)*C;

draw(B--M);

draw(A--C);

label("A",A,NW);
label("B",B,NE);
label("C",C,SE);
label("D",D,SW);
label("O",(0.5,0.3));
label("M",M,S);[/asy]

ABO and CMO are similar with $ \frac{AB}{CM}\equal{}2$. So $ \frac{OC}{OA}\equal{}\frac{1}{2}$?

This is not :spam:.  I'm just clarifying that Bugi's answer is correct, because Bugi put a ? in the end of his/her statement.

Is there a way to use mass points to solve this? I believe there is, I am having trouble though...

But why would you want to use mass points to solve a similarity problem? :)

Why does this post keep getting revived? (I guess this is sort of a revive, but better than 3 years... lol)

Is there a way to use mass points for this problem?

[quote=blep]What is mass points anyway?[/quote]

[url]https://www.artofproblemsolving.com/wiki/index.php?title=Mass_points[/url]

Mass points could solve this, but it is easier to use similar triangles to solve this one.

[quote=Dark_Lord]isnt 1/2 supposed to be oc/ca tho? by similar triangles[/quote]

no

wow this post was made a year after i was born.... anyway thanks for the mass points and similarity help, I couldn't find a good article explaining mass points.


email search AMC USA(J)MO USAMO inequalities

I suppose this could go in the "Other Contests" forum as well, but I figured it would be most appropriate here.

The first problem (mine!) has been posted in the Mandelbrot Problem Corner ([url]http://www.mandelbrot.org[/url]).  As of right now, there is a typo in the problem ('>' appears when ':ge:' should), but this should change soon.  (As you probably know, problems are often rephrased and typeset when posted at the Problem Corner, and this typo apparently snuck in during the process.)

Please, do [b][i]NOT[/i][/b] post any solutions, hints, questions, etc. regarding this or any problem on the Mandelbrot Problem Corner here at AoPS.  If you have a question about the problem, feel free to PM or email me (please make the subject clear).  Ditto with solutions (if you are a high school student and want your name on the Problem Corner, you will need to email me with your name and school as per the instructions on the site).

Again, do [b][i]NOT[/i][/b] post anything regarding the problem on AoPS.
Thanks.

Apparently fairyofwind is correct.  I wasn't aware of the fact that this problem had appeared before when I submitted it to the problem corner, but that's no excuse for not doing anything about it now.   I've suggested to the guy at Mandelbrot to take the problem down, but we'll just have to wait and see what happens.

I need an easy way to determine if a problem has appeared before...  I seem to run into this trouble way too much.

Wait, but if the problem is so famous, how come the people at Mandelbrot didn't notice this fact before posting the problem?

But yeah, a simple google search resulted in solutions to this problem. They should really take it down.

The ``well-known'' problem involves only 2s rather than arbitrary integers, as in Setofallsets's problem.

But a simple generalization would yield a solution to Setofallsets's problem, right?

Probably. I have neither looked at the solution nor attempted the problem in question.

i believe the older problem using "2" is usamo 1991/3.

Oh, yeah, that looks right. But I have also googled up sections of Setofallset's problem and found problems using just a constant (and not just '2').

The problem has been removed from the Problem Corner.

I'd like to thank fairyofwind for making me aware that the problem had appeared previously.

I suppose this all goes to show that some "well-known" problems are not exactly well-known to everyone...

Well, if it's on the USAMO, many people will have seen it, but most people (even those who take the USAMO) haven't gone through all the USAMOs. More often we work on problems from just a few different years.

I think it is worth mentioning that a new first problem is up in the Mandelbrot Problem Corner.  A fun question, if I do say so myself.  An inequality problem.  Anyhow, go give it a try.
http://www.mandelbrot.org/dialogue/corner.html


i see that u boys have formed quite a gang around urselves...

hi everybody
i am new to this forum and definitely have a lot to learn from u all (all academic only) ....

to b straight in approach,  i aim the iit(s)

i am just a novice in the field of maths ....so help me out

thats it...

let us roll the ball and start playing the mathematical game ... :lol:

hello nakul. why dont u start by trying some of the problems in the iit jee thread?

[quote="Cyber-Shreyas"]Welcome Nakul. You seem to pretty poetic. You can learn poetry from shruthibaskaran and Soumyashant Nayak. ;)

Anyway which class are you in?[/quote]

And also from Ritu :D :D

see i m the worst poet if i write real poetry
but anyways,thanks 4 compliment!!!!! :D  :rotfl:  :D

thanks for the gr8 responses....

simply awesome  :roll:

 well to continue the poetic trend , read below

[hide]iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit

heard the tune.....  :rotfl: [/hide]

Hi dude..
where ayoo from? there s nothing like we ve formed a gang among ourselves. i am new here also, but jus started talkin freely to many ppl on this forum. we ve lotta fun. so cm on the jee thread. post problems, solutions. we can all gain a lot frm each other. and considering dat there are sm gods like pardesi, soumya madness chemrock varun shreyas etc etc (sorry if ive missed out sm1's name) who actively participate, and other gods who arent as active here but busily slogging,
our innovation in mathematics and other sciences too can expand.  :D
so welcome.  :)

[quote="pardesi"][quote="NAKUL07"]
[hide]iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit iit

heard the tune.....  :rotfl: [/hide][/quote]

ok what's that :rotfl:  :rotfl:  :P[/quote]

International Institute of Typewriting.

For admission to which you have to write CRAP, Common Research Admission Programme.

or InDIAN INSTITUTE OF TRAINING
 :D  :rotfl:  :P

[quote="NAKUL07"]gr8 ....

ok to start it, which resource is suitable for last minute revision for iit

[hide]people actively participate and do mention for PCM[/hide][/quote]

Start a new thread if you want to shift the topic focus. :)

one more idea--INDIAN INSTITUTE of TIMEWASTING!!! :D  :D


Euler
function
number theory
relatively prime
superior algebra
superior algebra unsolved

let $ \phi$ be the Euler phi function
(i) If $ p$ is a prime number, and $ r$ an integer $ \geq 1$, show that $ \phi(p^r)\equal{}(p\minus{}1)p^{r\minus{}1}$

(ii) Prove that if $ m,n$ are positive relatively prime integers, then $ \phi(mn)\equal{}\phi(m)\phi(n)$

Use the definition and the Chinese remainder theorem.


I'm not really a fan of arithmatic problems...

so I'll just say the answer

[hide="no looky"]1[/hide]

lol let me do the longer way!

(5-square root of 5^2-3^2)^2
(5-square root of 25-9)^2
(5-square root of 16)^2
(5-4)^2
-1^2=-1*-1=1
so the answer is one
1[/hide]

I thought that one of the purposes of this website was to learn to do things more efficiently, not try and do the longest way. . . ;)

I agree that it is importatnt to learn the long way, and you should learn that before the short way.  So maybe I was wrong. . .This forum's about [u][b]learning[/b][/u] math.


geometry
circumcircle
conics
induction
geometry proposed

Given a cyclic $2n$-gon with a fixed circumcircle s.t. $2n-1$ of its sides pass through $2n-1$ fixed point lying on a line $\ell$, show that the $2n$’th side also passes through a fixed point on $\ell$.

[try it with a conic instead of a circle :)]

I don't understand this problem. Can you explain it? Thank you!

I'm not Grobber, but let me try:

Let $n$ be a positive integer, $k$ a circle and $\ell$ a line. Consider $2n-1$ points $P_{1}$, $P_{2}$, $P_{3}$, ..., $P_{2n-1}$ on the line $\ell$. Prove that there exists a point $P_{n}$ on the line $\ell$ such that for every $2n$-gon $A_{1}A_{2}...A_{2n}$ inscribed into the circle $k$ satisfying that $P_{1}\in A_{1}A_{2}$, $P_{2}\in A_{2}A_{3}$, ..., $P_{2n-1}\in A_{2n-1}A_{2n}$, we also have $P_{2n}\in A_{2n}A_{1}$.

Nice problem, by the way, though I don't see an elementary enough for the IMO proof...

  darij

I think I have it.

I will prove it for $n=2$, then the rest will easly folow with induction.

Let $A,B,C$ be fixed points on line $\ell$ and let $X,Y,Z,W$ be fixed points on conic so that $A,X,Y$ are colinear, $B,Y,Z$ are colinear and $C,Z,W$ are collinear.

We must prove that map $X\mapsto W$ is involution on conic with center on line $\ell$.

Now let $X`,Y`,Z`,W`$ be variable points on conic so $A,X`,Y`$ are colinear, $B,Y`,Z`$ are colinear and $C,Z`,W`$ are collinear.

By Pascal theorem for hexagon $ZYXZ`Y`X`$ points $A,B$ and $P=Z`X\cap ZX`$ are collinear.
Again by Pascal theorem for $ZWXZ`W`X`$ points $P,C$ and $D=XW\cap X`W`$ are collinear.
Since $XW$ is fixed and $PC=\ell$ is fixed, so is point $D$ wich proves our statement.

I` l leave to you induction step.

You can prove by induction. By the way, this was discovered a long time ago by Poncelet.


Prove the inequality:

$ (a^3 \minus{} a \plus{} 2)^2 > 4a^2(a^2 \plus{} 1)(a \minus{} 2)$

I'm sure this has an easy solution, just can't find it.

I expanded and rearranged to get $ a^6\minus{}4a^5\plus{}6a^4\plus{}9a^2\minus{}4a\plus{}4\equal{}a^4(a\minus{}2)^2\plus{}2a^4\plus{}8a^2\plus{}(a\minus{}2)^2>0$. Yay...

I thought there was a solution without expanding it.
Thank you once again, I'll post better problems when I work on this list tomorrow, sorry about that   :(


Evaluate  $\lim_{x\to2}\frac{\sqrt[4]{x-1}-\sqrt[3]{3-x}}{\sqrt[3]{x-1}-\sqrt{x-1}}$.
without hopita rules :rotfl:

Isn't that exactly L'Hospital's Rule?

He is looking for a factorization of $\sqrt[4]{x-1}-\sqrt[3]{3-x}$ involving the quantity $\sqrt[3]{x-1}-\sqrt{x-1}$, which is just algebra

EDIT: didn't see post

dont use hospital?

i didnt see that

but why this!?

i cannot use plus and minus operator too? to become more hard?

Are you making a joke about the $\pm$? I am a little confused...

The point of not using L'Hospital's is that, first of all, I believe this is a non calculus form. Second of all, they are looking for a nicer solution (this would be like a challenge problem in a middle to high school level test)

Did you know the solution before posting the problem?

yes because it s my solution and where is the prob :wink:


[quote="gemath"]Given $n$ point $A_{1},A_{2},...A_{n}$ lie on a circle ,find place of these point for the sum $A_{1}A_{2}+A_{2}A_{3}+...+A_{n-1}A_{n}$ is maximum[/quote]
we must chose these points such that $A_{i}A_{i+1}=R$ for $i=1,2,...n-1$
$R$ is the radius of circle
the maximum is $nR$

Maybe you are wrong, the answer is $A_{1},...,A_{n}$ must be a vertex of star-regular polygon(note that not a regular polygon). The following figure is for $n=5$ a star.


[hide="Hint"]Prove that the both sides plus $n$ are equal to the number of the integral points satisfying the conditions $x^{y}\leq n,x\leq n,y\leq n$.[/hide]


function Euler trigonometry MATHCOUNTS calculus integration

I know that $\zeta(2)=\displaystyle\frac{\pi^2}{6}$, but can someone show me the proof? I haven't been able to find it anywhere.  Also, I know there is a closed form for all $\zeta(2k)$, where k is an integer, but I don't know the exact form.  Thirdly, is there a closed form for $\zeta(2k+1)$?

Thanks

There are 14 proofs here

[url]http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf[/url]

There is a closed form in terms of bernoulli numbers; depending on how you define them, we get

$\zeta(2k)=\frac {1}{2}(2\pi)^{2k}\frac {B_k}{(2k)!}$

where $B_k$ is the $k$th bernoulli number.

Thanks $Blah^3$  (Actually Proof 1  I had seen only recently, and Proof 2 was again nice)

One old way (attributed to  Euler ) goes like:

Use   $\frac {\sin x} x = 1 - \frac {x^2}  6 +  \frac {x^4} {5!} \dots $   as a function of $y=x^2$
The roots of y are  $\pi^2,(2\pi)^2,(3\pi)^2 \dots $
 so sum of the reciporcal of roots is qual to the cofficient of  (-) $y$ hence
$\frac 1 6 = (\frac 1 {\pi^2})(\frac 1 1 + \frac 1 {2^2} + \frac 1 {3^2} \dots$

(For   more rigour see the  proof 10 in the above reference)

[quote="gopherhole112"]Thirdly, is there a closed form for $\zeta(2k+1)$?[/quote]
There is none known till know, even not if they are irrational/transcendental (except for some like $\zeta (3)$, where at least the irrationality was proven already).

Also http://www.mathlinks.ro/Forum/viewtopic.php?t=52359 could be of interest.

FWIW -  Although the  some of the proofs are  a  little involved, to prove that these serieses do  converge,  (and get very accurate values of other $\zeta s$ 's too)  is  easy .. and can  to be done  at the Mathcounts level -

for example Noticing for $n>1$
  $\frac 1 {n^2} >  \frac 1 {n(n+1)} = \frac  1  n -  \frac 1 {n+1} $
 and                          $\frac 1 {n^2} < \frac 1 {n-1} - \frac 1  n $

One can see that the  $(3/2)<\zeta (2) < 2$

And If one sums up  a few terms (say five)   by hand, and for the  rest  apply Euler-Maclaurin type  formula to  correct the  error due to  using an  integral instead of summation (Integral of  (1/x^2) is easy)  you get something like:

=  sum of (five terms) + $\frac 1 6 + \frac 1 2 \frac 1 {6^2} - \frac 1 6 \frac 1 {6^3} + \dots $
and result is prob. correct for  4-5 significant digits.


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And since there are so many Christians (5/7 thus far) I'd like to add: "What does it mean to you to be a Christian?" Or even, "What does it mean to you to be _____?" Discuss your beliefs, and your lifestyle/morals, and how your religious beliefs affect that, if they do at all.

[quote="stupidkid"]Well to be a Christian you simply have to accept Christ as your savior and that it is only through him that your sins are forgiven which allows you to go to heaven after death.[/quote]

I'm a Christian and proud of it, but I believe there is more to being a Christian than what was mentioned above. As Christians, we also have the responsibility to live the Christian lifestlye. That is more than simply going to church on Sundays, or maybe even once during the week. Living the Christian lifestyle includes so many more things.

I can't remember the reference, but it says somewhere that, basically, we are God's temple, and when we ask him into our lives, he becomes a part of us. Now, since God is living inside of us, we need to take care of our bodies, and not put things into them that will harm us. Things such as drugs, alcohol, etc. It is also our responsibility, as Christians, to witness to others around us and to set a good example. Even if it means being the one at the party that is "no fun" because you won't drink the beer, it's our responsibility to God and to ourselves. As for the witnessing, we need to put aside our fears of being turned down, because it is bound to happen. Now, I'm not saying to go out in the middle of the hallway at school and scream that Jesus is the Lord and Savior, however, if you feel that that's what God has called you to do, then so be it. But, witnessing in the small ways works too. Just smiling at someone in the hallway goes a loooong way. As Christians, we need to BE Christians 7 days a week, not just on Sundays and that possible weekday. So I guess I agree with 'stupidkid' in what was said, but I believe that being a Christian goes much much deeper.

[quote="stupidkid"]Well to be a Christian you simply have to accept Christ as your savior and that it is only through him that your sins are forgiven which allows you to go to heaven after death.[/quote]

Nicely put, but its not that easy to accept Christ sometimes.  It can be a struggle sometimes.

Also, there's different viewpoints within Christanity, for instance, some Christians say that it is not a religion, but a relationship.  Not about going to church all the time, saying prayers all the time, but just about accepting Jesus as a friend to guide you.

[quote="henryyangrui"]Is it :( to be atheistic?[/quote]
I don't think there is anything wrong with atheism, it's just what you believe and no one can do anything about it.

According to [url=http://www.religioustolerance.org/]Religious Tolerance[/url],

[quote="Religious Tolerance"]Agnosticism is a concept, not a religion. It is a belief related to the existence or non-existence of God.

An agnostic is a person who feels that God's existence can neither be proved nor disproved, on the basis of current evidence. Agnostics note that some theologians and philosophers have tried to to prove, for millennia, that God exists. Others have attempted to prove that God does not exist. Agnostics feel that neither side has convincingly succeeded at their task.

Are they Theists? No, because Agnostics do not believe in a God, or a Goddess, or in multiple Gods, or multiple Goddesses or in a pantheon of Gods and Goddesses.[/quote]

I consider myself agnostic. I do not believe that one or more gods or goddesses exist, nevertheless, I consider just as fanatist to completely deny as completely believe the possibility of the existence of such thing.

I consider that if a fair god exist, it is not going to care if you believed on it or not, it is going to judge you according to how you lived your life. According to what I consider to live a good life, well, I may be too liberal for many people, because I consider that as long as you follow the most basic ethic rules that have been present in most of the civilization since the begining of the human civilizations, you should be OK, and those are simple:
[list]You don't kill unless it is a self preservation situation.
You don't steal from others.
You don't force anyone to do something they don't want, unless you honestly believe that it is for their own good
As long as you can, you help others
You try to be tolerant with the rest of the people[/list]
Not that hard. And those are not exactly moral concepts, just a simple way to live your life.

i am a christian, and like math fiend am also proud of it.  to be a christian it is more than just accepting christ. soooo much more.  we as christians are supposed to spend our lives glorifying god and leading others to christ. we are supposed to demonstrate many characteristics such as love, joy, peace, patience, gentleness, perseverence, and self-control.  just because we are christians, does not mean we are perfect.  in fact we are still sinners like everyone else except there is one difference, we have been forgiven.  math fiend and i go to school together and we hang out. we went to a bible study last wednesday morning and the girl said that once we ask for forgiveness, god completely forgets what we have done. he doesn't care anymore. we are to be as white as snow.  a lot of people have the misconception that christians are too good for them. we are just like everyone else in the terms that we live on the same planet, but in the bible we are told to be set apart from everyone else. we are supposed to be a light to those in the darkness. the world should see a difference between those who claim to be christians and those who really are. where i come from there are so many people that go to church and claim to be christians, but if you would just listen to the words that come from their mouths and the stuff that they are suggesting by wearing some of the clothes that they wear, you would be like yeah, i really want to be like them.  most people are repelled by how those pretending to be christians act, so they never get the chance to see what christ can really do in someones life. you dont have to be a specific person to receive christ. he accepts you the way that you are. he doesn't care what you look like or what you have done in the past. he loves you anyways!! he loved you even before you were ever born. he wants you to be with him. he wants you to love him and spend time with him. he wants you to trust him and let him guide your life!! accept him!!

[quote="Scrambled"]isnt buddhism technically a concept too?[/quote]

Why should it be?

[quote="stupidkid"]I was wondering, for all those people that picked other, what is your religion?[/quote]

May be some wicca practicant. I have been reading about it, and it is amaizingly interesting, and just very recently accepted in the US as a religion.

greek/roman gods though since i still live with my parents, i dont sacrifice my meals yet (and neither did the greek children, i am led to believe, and they ate some of the stuff they sacrifieced because the gods would enjoy the scent, so its just like a barbeque)
not very widely practiced

just kidding for those who didnt realize that yet

[quote="stupidkid"]I was wondering, for all those people that picked other, what is your religion?[/quote]

I think as a Chinese, you should know there is a native religion called "Dao Jiao"(道教）,  it is similar to the Shintoism of Japan.

[quote="ZetaX"]Instead of quoting everything to write one line you could have quoted something form the bible instead, proving your claim...

Well, how old are you (plural you)¿ For some of you, I've the impression that you believe and defend your believes without any reason.[/quote]
The Dead Sea Scroll claims it.

I don't see any reason that we should give out our ages to you, ZetaX.
Also you are considering only your own impression, not others.
Just because you have an impression about other people and you can tell about them, doesn't mean that we should give out our ages to you.
Not only that, your demand does not even relate to what we were talking about...
But if you really want to know, I'm 14.

[quote="ZetaX"]Instead of quoting everything to write one line you could have quoted something form the bible instead, proving your claim...

Well, how old are you (plural you)¿ For some of you, I've the impression that you believe and defend your believes without any reason.[/quote]
I don't need to prove any claims. What/which claim are you talking about?

Why would I defend my believes without any reason? Throughout the past few years, I went through the trouble of almost visibly proving my believes to myself, and that is enough reason to defend them.

By the way, a large finite number is nowhere near infinity.

Agnostic

(strong agnosticism )

 :lol:

my eyes are hurting from reading all the posts

[quote="stupidkid"]From what I have noticed, there seems to be many religions among AoPSers, what religion are you?[/quote]

Wow, I leave for two years, and you're discussing religion?  FINE!

I was raised Catholic, went to Catholic school, and then became agnostic.  Things calling themselves religions, or called by others religions, that are actually philosophies became of interest to me.  Now, I'd say I see religion as an interesting study in anthropology:  what people will believe is interesting, although the beliefs themselves are usually not at all interesting.  One of my favorite books of the past few years was Christopher Hitchens' "Religion Poisons Everything".  I believe that many people who have or had great minds have had their growth stunted by religion, when their personal interests or conclusions were contradicted in the religion that they were either forced to be a part of, or failed to see their way out of.  I consider religions to be "mind traps".  I am proud of anyone who finds their way clear of such mind traps, but this doesn't mean I judge people who are still in one.  Essentially, a religion is a paradigm of thought.

I'm very glad that we exist in a time where religions are tolerated, but in which many people are free of them, and being free of them doesn't mean you get burned alive, stoned to death, etc.  My view is that people are in search of answers to "big questions".  Sometimes, when you're desperate, any answer is better than no answer.  When you're not desperate, though, and you have time to consider and think, as well as look, I think it becomes apparent that most religions end up in a infinite regress, with God as "first cause", or Prime Mover Unmoved.  This is unsatisfactory to an honest mind:  it leaves you wondering "Okay, but how did God get there".  The Big Bang, of course, runs into this also.  I think it's much more intellectually honest to simply say "I don't know how the ball got started rolling, all I know is that there is a ball, and it is rolling".  I think it's good to cultivate a sense of wonder and awe, a willingness to tolerate a question without an answer, rather than suffering incomplete or illogical answers.  Maybe one day we'll know how and why the universe started, how consciousness came to exist incarnated in physical bodies, and if it's possible for our sense of identity to persist beyond the point of the death of the body.  Until that time, willingness to accept stupid, ignorant, or incomplete answers "on faith" will probably continue to lead to wars, dissolution of families, hatred and prejudice.  Wouldn't it be better to simply abolish belief systems based on antiquated and false information, and perceive the universe as it is, explore it for it's information content, and live happy, healthy lives?

My policy about religion is that 99.9% of the time it is a set belief, and discussing it is fruitless, or becomes illogical.

e.g.  Those who believe in <insert religion here>  will use the <Insert holy book here> to prove Christianity.

Which is like:  Find the exact value of $ \phi$.  Well, if we accept that $ \phi$=3, then $ \phi$=3.  This assumes that you accept $ \phi$=3 already.

Religion brings war.  Crusades.  Fights over the holy land.

We all need to believe, and I am not talking about a specific religion, but these are effects of religion.

hmm I think religion is a byproduct of man's free thinking, therefore redundant.

haha this is an essay I wrote for school on this topic, for a class of 4/5 Christians :D

[url]http://www.fictionpress.com/s/2440934/1/Creation_vs_Evolution_Faith_vs_Reason[/url]

pretty much sums up what I believe, though I admit some of the research may be faulty.

who else believes that [i]cogito ergo sum[/i]? :)

I am an atheist. I don't believe in god.

1st of all I can't believe in something written in any book (every religion is based on some book) without any evidence. For me it's just like believing Peter Pan actually flew around.

2nd which is closely connected to 1st, there are no physical evidences any god exists. A book is not an evidence as people write books.

3rd: If there is a god why is there so much evil in this world? If there ever was a god it must have abandoned us long time ago.

Money is the modern god (Balzac). - Not my belief but people act that way.

It seems like any sightings of god or whoever could be explained by mental illness/hallucination...
or they could be real.
The universe is definitely not infinite, but space definitely extends infinitely beyond the universe. and if there were other universes, shouldn't there be infinitely many? which would mean somewhere out there is an exact copy of me who is typing these very words...

And then I get to thinking about why [i]anything[/i] exists, why there is [i]space[/i] in the first place...

That's when I start to feel like taking a nap.  =D

As for my religion, I don't feel like discussing it, except that I definitely don't believe everything any religion has to explain.

athiests ftw?
Meh, my mom's Christian and is trying to convert me.
:|

I would escape from my home if my mom would try to convert me.

[quote="Identity"]I think I'm part agnostic part apatheist.[/quote]
I think tht's called being a non-believer who acknowledges a possibility of something else with their mouth but not with their soul..

[quote="henryyangrui"]Is it :( to be atheistic?[/quote]
No, and those who look down on any religion(including atheism) can't really say anything on their own.  I believe that tolerance of people, regardless of religion, beliefs, culture, etc., is the hallmark of what makes this world great.  I don't believe there's a moral repungance to atheism.  We all have our own opinions, and I think we should all respect that.  :D


find all values of a,b for which a+b=ab and a,b are even and a>b

[quote="mathsrox"]find all values of a,b for which a+b=ab and a,b are even[/quote]

$a,b$ are positive integer numbers ?

[quote="mathsrox"] a+b=ab [/quote]

From $a+b=ab$ we have $(a-1)(b-1)=1$ therefore $a-1=b-1=1$,... :roll:


Einai profanos xazo na rotisoume ean mporoume na grapsoume to [0,1] os enosi anoixton diastimaton !
Alla, mporoume na grapsoume to (0,1) os arithmisimi enosi kleiston diastimaton xoris na temnontai ana dio ?

Stergios

Σου αρέσει πολύ το Cantor, ε? ;)

Cheerio,

Durandal 1707

Den katalaba to sxolio sou.
Isos exeis kapoia lisi diaforetiki apo autin pou exo kata nou, opote tha xeromoun na tin akouso.

Stergios

Ok, πες μου αν κάνω κάπου λάθος (probability 1 ;) )

Let $ J_0 \equal{} (0,1)$.

Set $ I_1 \equal{} [1/3,2/3]$ (middle third) and remove it from $ J_0$ to obtain $ J_1 \equal{} (0/3,1/3)\cup (2/3,3/3)$.

Let $ I_2 \equal{} [1/9, 2/9]$ be the middle third of the first segment of $ J_1$, and $ I_3 \equal{} [7/9,8/9]$ the middle third of the second segment. Remove these parts to obtain $ J_2 \equal{} \ldots$ and proceed ad infinitum...

I think this should do the trick: $ \displaystyle\cap_{k \equal{} 1}^\infty J_k \equal{} \varnothing \implies \cup_{j \equal{} 1}^\infty I_j \equal{} (0,1)$ and, clearly, $ I_j\cap I_l \equal{} \varnothing$.

Δουλεύει ή πάλι λέω βλακείες?

Cheerio,

Durandal 1707

Edit: Λέω βλακείες  :mad:

Askisi : Na deikseis oti to $ (0,1)\minus{} \cup_{j\equal{}1}^\infty I_j$ einai iperarithmisimo. :wink:

Stergios

[quote="Anto"]Askisi : Na deikseis oti to $ (0,1) \minus{} \cup_{j \equal{} 1}^\infty I_j$ einai iperarithmisimo. :wink:

Stergios[/quote]

Gaaaah, πόσο #(*@(*κας μπορώ να γίνω?! Φυσικά και αφήνω το Cantor απ'έξω, μ'αρέσει που από αυτό ξεκίνησα κιόλας! :(

Καλά, παραιτούμαι... έχω max λίγα λεπτά την ημέρα να ασχοληθώ με ότι γίνεται posted και μπαίνω μονίμως στο λούκι να γράφω χωρίς να έχω σκεφτεί καθαρά. Τι να πώ πια...

Sorry,

Durandal 1707

Ok, αφού έφαγα και άλλη μία ώρα στην παγίδα του Anto, να μία διαφορετική απάντηση (είμαι σίγουρος ότι θα γράφεται και πιο κομψά, αλλά τέτοια ώρα τέτοια λόγια...). Anto, hope this works (fingers crossed   :) ):

Έστω ότι το $ U \equal{} (0,1)$ γράφεται στη μορφή $ \cup_1^\infty I_k$ όπου $ I_k$ κλειστά διαστήματα με $ I_j\cap I_k \equal{} \varnothing$ (θα χρειαστεί σχεδιάκι μάλλον παρακάτω...).

Παίρνουμε το κλειστό $ I_1$ και θεωρούμε (μη κενό!) κλειστό $ J_1$ που να τέμνει (γνήσια) το $ I_2$ αλλά να είναι ξένο ως προς το $ I_1$ (αυτό είναι δυνατό επειδή $ I_1\cap I_2 \equal{} \varnothing$).

Αν το $ I_3$ τέμνει το $ J_1$, θεωρούμε (μη κενό) κλειστό $ J_2 \subset J_1$ που να περιέχει (γνήσια) την τομή $ I_3\cap J_1$ και να είναι ξένο ως προς το $ I_2$.

Από την άλλη, αν το $ I_3$ είναι ξένο ως προς το $ J_1$ θέτουμε $ J_2 \equal{} J_1$ και προχωράμε με την αρίθμηση των $ I$ έως ότου πέσουμε στην παραπάνω περίπτωση.

Then, proceed ad nauseum...

(Βασικά, αυτό που θέλουμε να αποφύγουμε είναι η περίπτωση να διαλέξουμε $ J_n \equal{} I_{n \plus{} 1}$ επειδή αυτό θα μας σταματήσει την κατασκευή...)

Κατ'αυτόν τον τρόπο, αποκτούμε μία κιβωτισμένη (υπ)ακολουθία μη κενών κλειστών διαστημάτων $ J_n$ που θα έχει μη κενή τομή $ \varnothing \neq \cap_1^\infty J_n \subset (0,1)$. Όμως, εκ κατασκευής, το σημείο που θα ανήκει σε όλα τα $ J_k$ αναγκάστικα δε θα ανήκει σε κανένα $ I_k$, και καταλήγουμε σε άτοπο.

Ελπίζω να μη χάνω κάτι αυτήν τη φορά γιατί δεν έχω άλλο χρόνο... :(

Cheerio,

Durandal 1707

[quote="Anto"]Eisai se kalo dromo  :)

Exo omos tin entiposi oti i kataskeui sou den einai kala orismeni. p.x Ean J3 den temnei to I1 tote J2=J1 kai mporeis na sinexeis na kataskeuaseis Ji's pou omos temnoun to I2 ? I mipos den katalaba kati ?

Stergios[/quote]

Δεν είναι πρόβλημα αυτό... τελικά, κάποιο $ I_{k\plus{}2}$ θα τμήσει το $ J_k$ (εξ' υποθέσεως) και τότε το $ J_k$ θα μικρύνει... δεν είναι τόσο κομψό όσο το να έχουμε πχ $ J_k\cap I_k \equal{} \varnothing$, αλλά μου φαίνεται ότι εξακολουθεί να δουλεύει...

Cheerio,

Durandal 1707


algebra
polynomial
linear algebra
matrix
geometry
3D geometry
limit

1. Let $ A$ and $ B$ are two $ n*n$ matrixes. Prove that
\[ det(AB \minus{} xE) \equal{} det(BA \minus{} xE).
\]
(in other words, two characteristic polynomials of $ AB$ and $ BA$ are the same.

2. Assume that $ x_1,x_2,...,x_n$ are eigenvalues of $ A$, then $ f(x_1),f(x_2),...,f(x_n)$ are eigenvalues of $ f(A)$, where $ f$ is a polynomials.

Problem 2 is pretty easy, but I just wonder about the order of each eigenvalues?

The algebraic multiplicities do exactly what you would expect; the algebraic multiplicity of $ c$ is the sum of the algebraic multiplicities of $ r$, for all $ r$ with $ f(r)\equal{}c$ in some extension of the field in which the characteristic polynomial splits. Geometric multiplicities don't behave so nicely.

To prove this, pass to an extension in which the polynomial splits, and look at Jordan form.

An example, to clarify some of what's going on: $ A\equal{}\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$, over $ \mathbb{R}$. The eigenvalues are the cube roots of $ 1$, each with multiplicity 1. Over $ \mathbb{R}$, there's only one cube root of $ 1$, while there are three in $ \mathbb{C}$. We have $ A^3\equal{}I$, with eigenvalue $ 1$ of multiplicity three.

I am not sure, perhaps my idea is similar to jmerry. I think if we consider the order of roots of characteristic polynomials of matrix A (means, the order of eigenvalues), then we have to consider the matrix in algebraic-closed field (in particular, $ C$). In this case, I proved (2) but with a pretty long proof. I have to consider the subspace $ ker(f\minus{}x*id_C)$ for each eigenvalue $ x$.

Maybe I use a wrong jargon. "Order" of eigenvalue should be called "multiplicities".

How about (1)? It is not obvious.

Consider C = [B, -xI; I, 0], D = [A, -xI, I, -B], and use $ det CD \equal{} det DC$

Another solution is using limit. Since the problem is clearly true if detA or detB is non-zero.

If $ detA$ is non-zero, problem is obvious. So assume that $ detA\equal{}0$. Consider $ A_n$ as a sequence of matrix such that $ \parallel{}A_n\minus{}A\parallel{}\to 0$, then
\[ det(AB\minus{}xI)\equal{}\lim_{n\to \infty} det(A_nB\minus{}xI)\equal{}\lim_{n\to \infty} det(BA_n\minus{}xI)\equal{}det(BA\minus{}xI).\]
This true for all $ x$, so we have done.

Dear Soarer, could you write your solution more clearly? thank you.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=116836]here[/url], and follow the links.

$ C \equal{} \begin{bmatrix}B & \minus{} xI \\
I & 0\end{bmatrix}$, $ D \equal{} \begin{bmatrix}A & \minus{} xI \\
I & \minus{} B\end{bmatrix}$
Then $ CD \equal{} \begin{bmatrix}BA \minus{} xI & 0 \\
A & \minus{} xI\end{bmatrix}$,$ DC \equal{} \begin{bmatrix}AB \minus{} xI & \minus{} xAI \\
0 & \minus{} xI\end{bmatrix}$
So $ det CD \equal{} det DC$ gives us what we need, except the trivial $ x \equal{} 0$ case.


function
real analysis
real analysis solved

harazi please check your syntax. in this form the char. equation of the sequence would be 2r+2 = 1 + 1 / 2(r+r+1) which obivously doesn't admit 1 as a solution.

Also 2r= 1 + 1 / 2(2r+1) so there is a problem with the problem :)

This is just a partial solution (I think the problem is as I wrote it in my last post).

From the properties of the functions 2^x and x+1 we find that x_(n+2) is between 1 and [x_(n+1)+x_n]/2, so
|x_(n+2)-1| <= 1/2*|x_(n+1)-1+x_n-1| <= (|x_(n+1)-1|+|x_n-1|)/2. Let a_n = |x_n-1|. We have found that 0 <= a_(n+2) <= (a_(n+1)+a_n)/2. A sequence having this property is convergent. My solution is only partial because I'm not proving this here (I have proved it before, but I have to remember the proof). Let l be the limit of a_n. It's now easy to see from the properties of the sequence x_n that this limit must be 0.

I know it's not really a solution, but I'll be back... :D

Ok, so the problem is this: a sequence (a_n)n>=1 of non-negative numbers has the property a_(n+2) <= (a_(n+1)+a_n)/2. Prove that a_n is convergent.

Let b_n = max(a_n,a_(n+1)). We can see that a_(n+2) <= b_n, and since a_(n+1) <= b_n we have 0<=b_(n+1)<=b_n, so b_n is ecreasing and convergent to l>=0. This shows that b_n>=l for all n. Let  \varepsilon>0 . It's obvious that there are only finitely many terms of the sequence a_n >= l+  \varepsilon . All we need to prove now is that there are also finitely many terms <= l- \varepsilon.

Let's assume there are infinitely many such terms. If a_n<l then a_(n+1)>=l because b_n>=l. For n large enough such that a_n <= l- \varepsilon a_(n+1) must be < l+ \varepsilon/2 because a_(n+1)=b_n and b_n ->l. This, in turn, means that a_(n+2) <= (a_(n+1)+a_n)/2 < l- \varepsilon/2 <l, so a_(n+3) <= (a_(n+2)+a_(n+1))/2 < l, so b_(n+2) <l, but this is a contradiction because b_n>=l for all l, which means that there are only finitely many n's for which a_n<=l- \varepsilon for any  \varepsilon>0.

I think this ends it. Sorry it's long and ugly.  :D


Well the first one is:

3/4x+5/2y=11
1/16x-3/4y=-1

The answer is (8,2).

I think that first I use LCD on both, which is LCD=4 for the first one and LCD=16 for second one then multiply but 6 for first one and 5 for sencond one to get rid of the y part but I am not getting 8.

HELP PLEASE!!! and show all work ( I need that...)

We have:

$ \frac{3}{4}x \plus{} \frac{5}{2}y \equal{} 11$

$ \frac{1}{16}x \minus{} \frac{3}{4}y \equal{} \minus{}1$

Fractions are very unwieldy, so let's eliminate them. Multiply the first equation by 4, and the second one by 16.

$ 4 \left ( \frac{3}{4}x \plus{} \frac{5}{2}y \equal{} 11 \right )$

becomes

$ 3x \plus{} 10y \equal{} 44$

$ 16 \left ( \frac{1}{16}x \minus{} \frac{3}{4}y \equal{} \minus{}1 \right )$

becomes

$ x \minus{} 12y \equal{} \minus{}16$

So now we have the equations:

$ 3x \plus{} 10y \equal{} 44$

$ x \minus{} 12y \equal{} \minus{}16$

Take (2) and rearrange it like this:

$ x \equal{} \minus{}16 \plus{} 12y$

Substitute this value into the first equation and proceed that way.

Please ask if you have any questions. (I didn't solve the whole thing since it appears to be homework. :) )

After canceling out the fractions you should get:
\begin{align*} 3x + 10y & = 44 \\
x - 12y & = 16 \end{align*}
Multiply the second equation by $ - 3$ what happens?

Edit: Oh, don't be thrown by the matrices if you have never seen them before. They are used here just to make things a little bit cleaner (think of them as the coefficients neatly organized).

$ 3/4x \plus{} 5/2y \equal{} 11$
$ 1/16x \minus{} 3/4y \equal{} \minus{} 1$

become

$ 3x \plus{} 10y \equal{} 44$
$ x \minus{} 12y \equal{} \minus{} 16$

by multiplying the top equation by 4 and the bottom equation by 16.
This is represented by this augmented matrix:

$ \left[ \begin{array}{cc} 3 & 10 \\
1 & \minus{} 12 \end{array} \right] \left[ \begin{array}{c} 44 \\
\minus{} 16 \end{array} \right]$

Using Gaussian elimination, we can solve for one of our variables. For example, we can multiply the bottom row by -3:

$ \left[ \begin{array}{cc} 3 & 10 \\
\minus{} 3 & 36 \end{array} \right] \left[ \begin{array}{c} 44 \\
48 \end{array} \right]$

Then add the rows together and replace the bottom row:

$ \left[ \begin{array}{cc} 3 & 10 \\
0 & 46 \end{array} \right] \left[ \begin{array}{c} 44 \\
92 \end{array} \right]$

Putting this back into equation form, we have:

$ 3x \plus{} 10y \equal{} 44$
$ 46y \equal{} 92$

We can easily solve for $ y$ by dividing both sides by 46, and we get $ {y \equal{} 2}$.
Substituting this into the top equation:

$ 3x \plus{} 10(2) \equal{} 44$
$ 3x \equal{} 24$
$ x \equal{} 8$

So our solution is $ \boxed{(8,2)}$

[quote="senpai"]Ah, so your saying get rid of the x's not the y's. I see now.
Thanks Isabella I get substation just not elimination. Hee~[/quote]

Elimination is actually very simple. I'll show a separate example, and then you try your problem. :)

Say we have:

$ 3x \minus{} 2y \equal{} 17$

and

$ 5x \plus{} 2y \equal{} 39$

If we add these two equations together, we see that the $ 2y$ terms cancel out.

$ 8x \equal{} 56$

Then you can solve for $ x$ and substitute that value into either equation, thus finding the value of $ y$.

But what if we have something like:

$ 3x \minus{} 2y \equal{} 17$

and

$ 4x \plus{} 4y \equal{} 36$

Well, multiply the second equation by 2.

$ 2(3x \minus{} 2y) \equal{} 2(17)$

$ 6x \minus{} 4y \equal{} 34$

Now we can add our equations.

$ 10x \equal{} 70$

And proceed like that.

Hope this helps!  :)

[quote="gauss202"][quote="grn_trtle"]Edit: Oh, don't be thrown by the matrices if you have never seen them before. They are used here just to make things a little bit cleaner (think of them as the coefficients neatly organized).

Using Gaussian elimination, we can solve for one of our variables.[/quote]

If someone has never seen matrices before, why would you think they would know what Gaussian elimination is? :huh:

You have to know your audience, and be reasonable in your expectations of what they know when you give an explanation. If someone is asking for help with regular elimination, it probably isn't very helpful to pull out Matrices and Gaussian Elimination just yet.[/quote]

I didn't know about matrices when I learned about Gaussian elimination, but whatever :)
Doesn't regular elimination work essentially the same way? (you line up equations, multiply if necessary, add them to cancel out variables?)

Regular elimination, I find, is much easier to understand. senpai is new to systems of equations, so Gaussian Elimination is probably a bit hard to understand. I still have trouble with it, myself. :)

See [url=http://en.wikipedia.org/wiki/Gaussian_elimination]Gaussian Elimination[/url]
By the way, for the matrices:
$ \left[ \begin{array}{cc} 3 & 10 \\
1 & - 12 \end{array} \right] \left[ \begin{array}{c} 44 \\
- 16 \end{array} \right]$
means
\begin{align*}3x+10y&=44\\x-12y&=-16\end{align*}


analytic geometry
Pythagorean Theorem
geometry

Find the distance between the points $ (2,2)$ and $ (\minus{}1,\minus{}1)$.

We can use the distance formula or the Pythagorean Theorem (which are pretty much the same thing). I will use the Pythagorean Theorem.

We have a triangle with legs 3 and 3, and we wish to find its hypotenuse. Using the Pythagorean Theorem:

$ x \equal{} \sqrt {3^2 \plus{} 3^2}$

$ x \equal{} \sqrt {2(3^2)}$

$ \boxed{x \equal{} 3\sqrt {2}}$

EDIT: beaten.

[hide="A little shortcut that I found for distance problems, very quick."]We can translate the points such that one of them is the origin. This means we can add and subtract coordinates as long as we do it do both, that way it wont affect the distance between them. In this case, we get

$ (2,2),(\minus{}1,\minus{}1) \rightarrow (2\plus{}1,2\plus{}1),(\minus{}1\plus{}1,\minus{}1\plus{}1) \rightarrow
(3,3),(0,0)$

Now all you have to do is pythagorean theorem. Once you get a hang of it, you can go really fast.

And in this case, this is just a $ 45\minus{}45\minus{}90$ triangle so the distance is $ \boxed{3\sqrt{2}}$.[/hide]

[quote="justinthepark"]If you graph this you get an isosceles right triangle with sides of 3, so the hypotenuse, or distance is obviously $ 3SQRT{2}$[/quote]
You don't exactly have time to graph it during FTW.


I still don't get QED..
how do people use it???
 :?:
can someone give me an example?

QED is an acronym for Quad Erat Demonstradum, which directly translates to "That which was to be demonstrated," in Latin. It means what it sounds, you use it when you have demonstrated something that you wanted to demonstrate, that is, you have finished a proof of a claim.

Some people use it at the end of their solutions to numerical answer problems as well, such as "what is the measure of this angle?"

"blahblahblah, thus the measure of this angle is blahblahblah, QED,"

but usually it is used in the following way, "prove that A, B, and C are collinear."

"blahblahblah, thus they are concurent, QED."

You just put it at the end of a proof. Any proof. QED (or [b]Q[/b]uod [b]E[/b]rat [b]D[/b]emonstrandum) means "That which was to be demonstrated."

[quote="i_like_pie"]@ 1337 - Collinear $ \ne$ concurrent, but everything else is correct.[/quote]

Many collinearity problems can be turned into problems of concurrency and vice versa.  For example, 1337 may have been talking about pole-polar transformations in projective geometry.

I think we get the point about this QED, wiki it for extra info!  (lock)