Two cards are randomly selected from a standard 52 card deck. What is the probability of getting 2 hearts or 2 numbers less than 4 (aces count as 1)? Note: A result of 1 heart and 1 number less than 4 is not a favorable outcome.

Probability that you get two hearts:

$ \frac{13}{52} \times \frac{12}{51}$

Probability that you get two numbers less than $ 4$:

$ \frac{12}{52} \times \frac{11}{51}$

Adding yields $ \frac{13 \times 12 \plus{} 12 \times 11}{52 \times 51}\equal{}\frac{288}{2652}\equal{}\boxed{\frac{24}{221}}$.

don't you have to subtract when it includes both?

Yes. There are $ 13$ Hearts, $ 12$ cards less than $ 4$, and $ 3$ cards that are both a Heart and less than $ 4$, so there are $ 13\plus{}12\minus{}3\equal{}22$ cards that satisfy the given condition. Thus, the probability is $ \frac{13\cdot12\plus{}12\cdot11\minus{}3\cdot2}{52\cdot51}\equal{}\boxed{\frac{47}{442}}$

Bah, fail.

Probability that you get both:

$ \frac{3}{52} \times \frac{2}{51}$

Subtract:

$ \frac{288}{2652}\minus{}\frac{6}{2652}\equal{}\frac{282}{2652}\equal{}\boxed{\frac{47}{442}}$.

EDIT: beaten.


trigonometry
geometry
circumcircle
geometry unsolved

Let $D$ be a point on the base side $BC$ of an isosceles triangle $ABC$. Point $F$ in triangle $ABC$ is situated on the arc of the circle passing through points $A$, $D$ and $C$. The circle passing through points $B$, $D$ and $F$ intersects side $AB$ at point $E$. Prove that $CD\cdot EF + DF\cdot AE = BD\cdot AF$.

We can see that $ACDF$ is similar to $DBEF$.
We are trying to prove $CD*EF + DF*AE =BD*AF$.
Since $AE = AC-EB$,  We can re-write the equation as $CD*EF+DF*AC-DF*EB=BD*AF$.
Again, we can re-write this as $CD*EF-DF*EB=BD*AF-DF*AC$.
$\frac{CD}{EB}=\frac{DF}{EF}$. Due to similar quadrilaterals.
So, $CD*EF-DF*EB=0$.
Also, $\frac{BD}{AC}=\frac{DF}{AF}$.
Thus, $BD*AF-DF*AC=0$
So, $CD*EF-DF*EB=BD*AF-DF*AC=0$.
Thus, the original statement is proven true.

[quote="Jimmy"]We can see that $ACDF$ is similar to $DBEF$.[/quote]

Why?

  darij

$\angle EBD = \angle ACD$ since $\Delta ABC$ is isosceles.
$FD = DF$ reflexive property.
$\angle EFD + \angle EBD = \angle ACD + \angle DFA = 180^{\circ}$.  Since both quads are cyclic.
So $\angle EFD = \angle DFA$.
Also, $\angle FDC + \angle FAC = \angle BEF + \angle BDF=180^{\circ}$.
I'm not sure if this is enough to prove similarity, but if we draw the diagonals of each quad, then since they are cyclic they will for for pairs of equal angles.  We could then prove similar triangles.

Hi, jimmy, i don't think you are correct. My drawing in gsp shows that they are not similar..

I don't know why, but I thought on it rather long. Solution is very simple. Here it goes:
We have to prove that $CD\cdot{EF}+DF\cdot{AE}=BD\cdot{AF}$. Let $X=DF\cap{AB}, M=AF\cap{BD}$.It is easy to see that $FX$ is the bisector of $\angle{EFA}$, thus $\frac{EX}{AX}=\frac{EF}{AF}\Rightarrow\frac{EX}{AE}=\frac{EF}{EF+AF}\Rightarrow{AE=\frac{(EF+AF)EX}{EF}}$   (*).
On the other hand $\angle{EFX}=\angle{DFM}, \angle{XEF}=\angle{FDM}$, hence triangles $XEF$ and $MFD$ are similar, from this we obtain: $EX=EF\cdot\frac{MD}{FD}$. Putting the last relation in (*) we get: $AE=\frac{(EF+AF)\cdot{MD}}{FD}$, now put this in the original statement to have :$CD\cdot{EF}+DF\cdot{AE}=CD\cdot{EF}+MD(EF+AF)=BD\cdot{AF}$ or $EF\cdot{CM}=AF\cdot{BM}$ or $\frac{EF}{AF}=\frac{BM}{CM}$. From triangles $ABM$ and $AMC$  we have $\frac{BM}{CM}=\frac{AB\sin{EAF}}{AC\sin{AEF}}=\frac{EF}{AF}$.
QED
Isn't it nice? :)
[i]Edited by Myth: just added a figure.[/i]

[quote="mecrazywong"]but anyway, Myth, this olympiad took place in 2004, not 2005 ;)[/quote]
Ops, I don't know at what stage this mistake appeared.

[quote="Myth"][quote="mecrazywong"]Let $D$ be a point on side $BC$ of isosceles triangle $ABC$. Point $F$ in triangle $ABC$ is situated on the arc of the circle passing through points $A$, $D$ and $C$. The circle passing through points $B$, $D$ and $F$ intersects side $AB$ at point $E$. Prove that $CD\cdot EF + DF\cdot AE = BD\cdot AF$ iff $AB=AC$ :) [/quote]
Do you mean "Let $D$ be a point on side $BC$ of triangle $ABC$." (without isosceles ;) )?[/quote]
Yes, sorry for my stupid mistake :blush:
I'll correct it.

[quote="mecrazywong"]A more challenging question:
Let $D$ be a point on side $BC$ of an arbitrary triangle $ABC$. Point $F$ in triangle $ABC$ is situated on the arc of the circle passing through points $A$, $D$ and $C$. The circle passing through points $B$, $D$ and $F$ intersects side $AB$ at point $E$. Prove that $CD\cdot EF + DF\cdot AE = BD\cdot AF$ iff $AB=AC$ :)

The official solution (which I read from a book and is very nice) works for this case as well. I also have a different trigo proof of it (but not troublesome).[/quote]
Can you post the official solution?

Sure.

Let the extension of AF meet the circumcircle of $\triangle BDF$ at K. Therefore $\triangle AEF\sim\triangle AKB$. By some substitutions, we know that $CD\cdot EF+DF\cdot AE=BD\cdot AF\Leftrightarrow CD\cdot BK+DF\cdot AK=BD\cdot AB$. Note that $\angle KBD=\angle KFD=\angle C\Leftrightarrow BK//AC\Leftrightarrow [ABC]=[AKC]\Leftrightarrow [ABD]=[DCK]+[ADK]\Leftrightarrow CD\cdot BK\sin C+DF\cdot AK\sin C=BD\cdot AB\sin B.$

Thanks mecrazywong. Excellent solution!!!!! :)


Consider $ n,z \in \mathbb{N},(n,z)\equal{}1$. Prove that at least one of the numbers $ 1,1\plus{}z,1\plus{}z\plus{}z^2,...,1\plus{}z\plus{}...\plus{}z^{n\minus{}l}$ is divisible by $ n$.

[quote="moldovan"]Consider $ n,z \in \mathbb{N},(n,z) \equal{} 1$. Prove that at least one of the numbers $ 1,1 \plus{} z,1 \plus{} z \plus{} z^2,...,1 \plus{} z \plus{} ... \plus{} z^{n \minus{} 1}$ is divisible by $ n$.[/quote]

[hide="hint"]
[color=blue]
Consider residue class of $ \ n,\ \  \{0,\ 1,\ 2,\ \dots,n\minus{}1\}$
[/color]
[/hide]

[hide="complete solution"]
[color=blue]
For $ \ z\equal{}1\ ,1 \plus{} z \plus{} ... \plus{} z^{n \minus{} l}\equal{} n,$ divisible by $ \ n.$

Let $ \ z\ne 1.$

Let  $ \ a_r\equal{}1 \plus{} z \plus{} ... \plus{} z^{r \minus{} l}\equal{}\frac {z^r\minus{}1}{z\minus{}1}$ for $ \ r\equal{}1,\ 2,\ 3,\dots,n$

As there are n numbers in $ \ \ \{1,1 \plus{} z,1 \plus{} z \plus{} z^2,...,1 \plus{} z \plus{} ... \plus{} z^{n \minus{} 1}\}$, either all belongs to different residue class or atleast two belongs to same class.

If all belongs to different residue class,then one of them is divisible by $ \ n$  and we are done with the problem.

Let us consider no of them is divisible by $ \ n\implies$ atleast two of them say $ \ a_i$ and $ \ a_j\ ,(i < j,\ \ j\minus{}i\equal{}k\le n\minus{}1,\ k\in\mathbb N)$ belongs to same class.

$ \implies n|(a_j\minus{}a_i)$

$ \implies n|\left(\frac {z^j\minus{}1}{z\minus{}1}\minus{}\frac {z^i\minus{}1}{z\minus{}1}\right)$

$ \implies n|\left(\frac {z^j\minus{}z^i}{z\minus{}1}\right)$

$ \implies n|z^i\left(\frac {z^{k}\minus{}1}{z\minus{}1}\right)$

$ \implies n|\left(\frac {z^k\minus{}1}{z\minus{}1}\right)\   ($ As $ \ n$ and $ \ z$ are co-prime $ \implies n$ and $ \ z^i$ are co-prime)

$ \implies n|a_k$ where $ \ 1\le k\le n\minus{}1$

Hence at least one of the numbers $ 1,1 \plus{} z,1 \plus{} z \plus{} z^2,...,1 \plus{} z \plus{} ... \plus{} z^{n \minus{} 1}$ is divisible by $ n$.
[/color]
[/hide]


greatest common divisor
number theory
least common multiple
MATHCOUNTS
prime factorization

7. The least common multiple of two numbers is $ 2^3\cdot 3^4\cdot 5\cdot 7$ The greastest common divisor of the same two numbers is $ 2\cdot3\cdot5$ One of the numbers is 210. What is the other?

[hide="Answer"]3240[/hide]

[hide] Okay, you are given the prime factorization of the union and intersection (using a loose definition of both terms) and you are even given one number.

First, prime factorize 210:

$ 3\cdot 7\cdot5 \cdot2$

Now, GCF is the common factors, so the other number  must also have $ 2, 3$ and $ 5$, but it does not have $ 7$. Now, the LCM is the highest power of each occurring number so this means that the other number has:

$ 2^3 \cdot 3^4$ in its prime factorization because it occurs in the LCM and is not accounted for in $ 210$. This means we have our number, it is:

$ 2^3 \cdot 3^4 \cdot 5 \equal{} \boxed{3240}$

[/hide]

Use the fact that the $ LCM(m,n)*GCD(m,n)\equal{}m*n$. Always true for natural numbers $ m,n$.  Proof is not needed since its mathcounts =)


Given four positive numbers a, b, c and d, prove that

(a+b)(b+c)(c+d)(a+d) is at least equal to 16abcd.

$ a \plus{} b\geq2\sqrt ab$ then multiple these inequalities

Nice one! My work is longer, 'cause i disregarded other facts.

[hide]First, choose an arbitrary arrangement of numbers in terms of value. Choose a be the smallest, b, c , and d as the largest. Also, these numbers may or may not be distinct. Therefore, there must be some distance between these points. So, let k1, k2, k3 (not necessarily distinct) be the distances between these points, respectively, and call the distance between the least and the greatest number k.

Replacing the numbers gives

[a+(a+k1)][b+(b+k2)][c+(c+k3)][(d-k)+d]

(2a+k1)(2b+k2)(2c+k3)(2d-k)

Looking at the expression, We can get its least possible value if we make these k's sufficiently small. So, if all k's are zero, we have them all equal. But for the sake of having them in the general form, we have its least value to be 16abcd.

Just treat a=b=c=d the special case, when k approaches a very small positive number

correct me if i'm wrong[/hide]


Notice how by grouping each pair of fractions, they are reciprocals of each other. Therefore, the product is 1^5 = 1.


conics
ratio
geometry
geometric transformation
reflection
projective geometry
geometry solved

let (C) be a circle .AB and AC are tangents to (C) in points B and C.M is the mid point of arc BC.and N is the mid

point of chord BC.an arbitrary line goes through M,andintersects AB,AC,(C) in points P,Q,R respectively.

PC and QB intersect each other in point S.prove that:

                                                                            R,S,N are colinear.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=28195 .

  darij

Well, although there are several solution, I'l post mine:

Map $BQ \mapsto CP$ from pencil through $B$ to pencil through $C$ is projective, since map $Q\mapsto P$ from line $AC$ to line $AB$ is perspective wrt $M$. So locus of point $S$ is some conic $\mathcal C$.

Since $C\in \mathcal C$, map $S\mapsto P$ is projective(fom conic $\mathcal C$ to line $AB$). Also map $P\mapsto R$ is projective (from $AB$ to circle). There for, map
$S \mapsto R$, wich is projective, induce projective map $NS \mapsto NR$ of pencil
through $N$ on itself.

Now this map is identity, since it has three fixed points (it is not hard to find them),
so points $N, S, R$ are collinear.

P.S. How do we define cross-ratio for 4 tangents (on fixed circle)?

[quote="Virgil Nicula"]I note $C=C(O,R)$, the reflection $L$ of the point $M$ w.r.t. the center $O$, the intersections $U\in BC\cap AS,\ V\in PQ\cap AS,\ T\in BC\cap PQ,\ U'\in BC\cap RL,\ N'\in RS\cap BC$ and $h.p.$ ([i]harmonic pencil[/i]), $h.d.$ ([i]harmonic division[/i]). The line $BC$ is the polar line of the point (pole) $A$ w.r.t.the circle $C$. Particularly, the division $(ALNM)$ is harmonically.

$\blacksquare\ 1^{\circ}.\ (PVQT)-h.d.\Longrightarrow A(PVQT)-h.p.\Longrightarrow$
$A(BUCT)-h.p.\Longrightarrow R(BUCT)-h.p\Longrightarrow (BUCT)-h.d.$
$RT\perp RL,\ \widehat {BRL}\equiv \widehat {CRL}\Longrightarrow R(BU'CT)-h.p.\Longrightarrow (BU'CT)-h.d.$
$(BUCT),\ (BU'CT)-h.d.\Longrightarrow U\equiv U'\Longrightarrow \overline {\underline {\left| U\in RL\right| }}.$

$\blacksquare\ 2^{\circ}.\ (BUCT)-h.d.\Longrightarrow P(BUCT)-h.p.\Longrightarrow$
$P(AUSV)-h.p.\Longrightarrow R(AUSV)-h.p.\Longrightarrow$
$R(ALN'M)-h.p.\Longrightarrow (ALN'M)-h.d.\Longrightarrow$
$(ALNM),\ (ALN'M)-h.d.\Longrightarrow N\equiv N'\Longrightarrow \overline {\underline {\left| N\in RS\right| }}.$[/quote]

thanks ... ;)


April Olymon  did anyone get the answer for the functional equation?

Yes, it was a very contrived problem.

Hint: make a term on the L.H.S. equal the R.H.S. what happens?

The solution can be found here http://www.math.utoronto.ca/barbeau/olymon92.pdf (my solution too :) ).


Hi,
Recently, I have been thinking about a math problem that might have a solution, well..almost all mathematical problems have solutions, but they may be undiscovered.  The problem is: what is x when (3^x)+(2^x)=35; x=3, but I am looking for a solution for this problem, not just an answer. Thanks for you time.
Best Regards,
Masoud Zargar

Yeah, $x=3$ works by inspection.  Then note that $f(x) = 2^x+3^x$ is increasing, so 3 is the only solution.  I don't really think there's a way to "solve" it per se, but $x=3$ is quite easy to find.

Thanks for your reply, but I was hoping for a solution to these kind of problems, even if they are hard to figure out from the top of someone's mind. Hope this continues on. Does anyone have any ideas of how this could be solved because, I'm out of them.

there is of course a general solution that gives you solutions with as much accuracy as you desire.
for example here, define 2^x=y and take log 3 in base 2 to be n.
then y^n +y =35 which is a ploynomial problem. (of course n is not integer)
Anyways that as much as you can get out of this equation.
Algorithm. Take a rational as close to n as you wish, say p/q, and then solve y^p = (35-y)^q.
-Ali


calculus integration MATHCOUNTS function algebra polynomial derivative

Um...if you want an advanced way...you can use something like Taylor Expansion or something but I think this probably needs to be moved to a more advanced forum.

You know Taylor expansions ? and youre particpating in mathcouts holy s***!
well you can do it without Taylor expansions..... you don't need integration either, although i tried and got the right answer :D

Did some googling, it looks like calculus and it looks hard. I don't think Mathcount people are supposed to know this, are they? I haven't seen a problem like that on in a competition. (unless i am googling wrong).

1/3 of the people in Mathcounts don't make it to nats, more like 1/500 :wink:

edit:
[quote="Yongyi781"]What do you mean, a MATHCOUNTSer is not supposed to know Taylor Expansions? I think at least one third of the MATHCOUNTS participants know those.  :| [/quote] I referred to this one, and then you posted something else. Oh, well, I see what you mean.

What do you mean? :huh: I said 1/3 of the people WHO MAKE NATIONALS... :|

The general idea is that you can approximate any function (more often non-polynomial, such as $ \ln(x)$ or $ \sin(x)$ or $ \sqrt [3]x$) with a polynomial. The higher the degree of the polynomial, the more accurate the approximation. The notion of a [i]Taylor[/i] series is that the polynomial has infinite degree, and as such, is an [i]exact[/i] approximation of the function. The actual derivation of such a polynomial is calculated by taking successive derivatives of the function at 0 (or some other constant) and using them to create the coefficients of $ x^n$.

... As this is a [b]Classroom Math[/b] forum, I will not go into detail here; more information can be found [url=http://en.wikipedia.org/wiki/Taylor_series]here[/url].

The reason he was talking about using a Taylor expansion to approximate the cube root of 2 is that as $ \sqrt [3]x$ is not a polynomial function, one can use Taylor expansions to approximate it.

taylors expansion  :huh:

never heard of it.....

seriously...stick with guess and check....

No, seriously, guess and check is bad... using a calculator is SO much better! :D

Oh, and if you didn't know, $ 126^3 \equal{} 2000376$, so that tells you that $ 1.26$ is a REALLY good approximation of $ \sqrt[3] 2$.

[quote="Yongyi781"]No, seriously, guess and check is bad... using a calculator is SO much better! :D

Oh, and if you didn't know, $ 126^3 \equal{} 2000376$, so that tells you that $ 1.26$ is a REALLY good approximation of $ \sqrt [3] 2$.[/quote]

i doubt anyone know $ 126^3$ by heart....and calculator is bad, you need to go to
http://www.artofproblemsolving.com and ask this question on the mathcounts forum, you are sure to receive soem very helpful answers.

Umm... this IS the MathCounts forum!! I'm a little worried about you now. :o

[quote="Yongyi781"]You're right. Taylor expansions are not MATHCOUNTS material. However, given that MATHCOUNTS people (the ones who make Nationals) are REALLY good at math (like exceptional), I'd say that a good number of them know advanced topics like this.[/quote]

being smart does NOT mean that you know advanced math. Especially calculus at a middle-school level.

[quote]Causation does not imply correlation.[/quote]

I'm not saying that [i]because[/i] you go to MATHCOUNTS, you know advanced math. I'm just saying that they [i]happen[/i] to know this stuff. :)

I highly doubt that someone at mathcounts would know taylor series because its just too freakin' advanced! And to have any sufficient use of it you would have to do some MAJOR calculus

HAHA! I broke Yongyi's pattern of every other post  :rotfl:

ok sorry for the spam

I've never heard of Taylor expansions before... well, that's not saying much because everyone knows, especially Yongyi, that I'm stupid....

Poincare, apparently you and Yongyi know about the taylor series.

*Sigh*

[quote="Poincare"]
What is the cuberoot(2) to the nearest hundreth?

thanks,
Poincare
[/quote]

I think we have already answered this. :D [/quote]


MIT
college
algorithm
ARML
geometry
3D geometry
AMC

Just wondering, is everyone only interested in mathematics or do some of you do other Olympiads as well?

Also, does your country stop you participating in more than one?

I think the poll will only let you choose one... so choose the one you do best in and comment below if you do more. :)

I've been in Olympiad in Informatics for 3 years... I was in the training team of Hong Kong, never been to IOI or National OI, but have been to SEARCC contest (Southeast Asian level). There're several people in both Mathematics and Informatics Olympiad training, although none have been to both of IOI and IMO. There were a good chance to have someone in both teams, I was even officially announced to be a member of IOI 2003 Hong Kong team... for two days, until they found a bug in the marking system and I had been given 10 marks I shouldn't have... And then I come down to the 5th place, orginally assigned for a place in our National OI team but I was two months too old... That left me with nothing. Still having someone in both teams in really possible for 2004 too, there's a IMO 2003 team member that came 6th place in the same OI selection test and he's likely to be in both next year... On the other olympiads, Hong Kong hasn't been participating in, although we will send a team for Physics in 2004

And I was very disappointed when I know I'm in Maths department in university so I can't go to ACM... (A contest similar to OI but for university students)

dude I'm in maths department and I CAN go to the ACM ... in fact I went to the first selection test for the team, but afterwards I got carried away with other stuff ... but I know the guys that went 4th place at the finals (world ACM final) from Bucharest University in 1999 they were all at maths department ... so ...

unfortunately in Romania it was only possible to go to one National Olympiad, as all of them are taking place in the same spring break. Untill 2 years ago, only the NOI was separated from all others, and the usual combination was that there were some maths people who went to the NOI and got some prizes. The only one I know of being in the training squads in the same time was [url=http://www.mathlinks.ro/phpBB/profile.php?mode=viewprofile&u=100]Pliviu[/url] in 2000. He didn't qualify for neither of them, missing 1 place in IMO Team and a couple places in IOI team. Next year we both went to IMO.

hey ! There's also the national geographic world championship, which used to be named igo, but the US olympic comitee decided that "olympic" could not be used as it wsa a trademark... :(  so in 2001 they changed the name...
Last year (it's hels every 2 years), at Tampa, Fl, the rankings were :
1. USA (97/28)
2. Germany (87/25)
3. France (85/18)  :D (I was in the team)
4. Romania and Russia (84)
...
(score for the tests/score for the final)
__
and France doesn't participate to the ipho, the ibo, the iao,... :(

i have been to chemistry and physics at the national level but never managed to make it to the team :(  :(  :(
and i were to sit in the bio olympiad i wud score a big 0. :D

Something else:literature (national level multiple times,best place:4th,worst:13th)

The only time I was interested in something else was in the 8'th grade (I'm in the 12'th now), when I qualified for the final round of the literature olympiad. I had also managed to go to the preparation camp for the Junior Balkan ( never went there, though :)), and when I came back from there I had lost interest in literature, so I didn't want to go to the final round anymore!

That's strange, huh? I guess I could say I was completely changed after contact with the reall beautiful maths problems. I just couldn't imagine myself going to the literature contest.. I still like reading, though :).

I was choosed to NOI(National Olympiad of Infomatics) twice, but there always have a time coincidence, and Obviously I'll choose CMO first....
So I've never been to other national level olympiads.

Lots of incredible folks have gathered here  :D

How about this guy?
---------------------------
Quote:
--------------------------------------------------------------------------------

 Hello everyone. I'm Reid Barton, 17, from Arlington, MA. This is my
second year at camp. I was on the US team last year and won gold medals at
the BOI, CEOI, and IOI.

I started programming in 6502 assembly language, Basic, and Logo sometime
before kindergarten, when I began programming in Pascal. I switched to C
in third or fourth grade. Since then I have learned C++, Perl, and Lisp.

At the end of eighth grade I started working part-time at the MIT
Laboratory for Computer Science in the Supercomputing Technologies Group,
with Professor Charles Leiserson, the L in Introduction to Algorithms by
CLR. I worked on the Cilkchess parallel chess program, one of the
applications developed in Cilk, a C-based language for programming parallel
supercomputers.

Since August 1999, when Charles took a two-year leave from MIT, I've been
working part-time at Akamai Technologies (http://www.akamai.com), a company
which speeds up internet content delivery. I'm a developer in the
Performance Analysis Group, where recently I've been learning more than I
ever wanted to know about how streaming media works. :)

In addition to computer science, I am also interested in math. I have
competed in various math competitions including the IMO, where I won gold
medals the past three years. Other interests of mine include playing the
piano (12 years) and cello (10 years) and learning foreign languages.

I look forward to seeing all of you at camp. In the meantime, I'll see
some of you at ACSL and ARML (at Penn State). You can talk to me on ICQ
(#9959478) or AIM (MOPReid, for historical reasons).

Regards,
Reid

P.S. I'll be joining former cowmates Percy and John at MIT in the fall,
along with Yuran. Tom and Thuc--where will you be going?

--
Regards,
siuman

well I haven't seen Reid on MathLinks (yet :P) but I've met him at MOP in 2001. Also, I have seen him at ACM in Pragues this spring (he was a one-man-team for MIT :D)

He's like the man with the most gold medals ever (I've counted at least 9 of them! 4 in IMO, 2 in IOI, 1 in BIO, 1 in CEIO and 1 in Balkan Informatics Olympiad ! yes the US team was invited in 2000 to participate, and all the Romanian guys were angry cuz one of the gold medals got stolen :D)

That quote was taken from

[url]http://oldweb.uwp.edu/academic/mathematics/usaco/2001/camp/barton.htm[/url]

What was your impression about Reid,when you met him,Valentin?

since i did not make it to the IMO =), i'm tryimg to get into the chemistry olympiad team... chances are not bad now, just 6 ppl left.

Whom exactly would I make proud by doing what?

Mathematicians by stopping my prostitution to chemistry? *confused*  :lol:

Well, there might have been some suspect ice cubes around, who knows...  :maybe:

This other Olympiad remind me of the Programming Section
Will it appear anytime soon?
I think even though it is not directly related to math, it is quite interesting for many of us.

[quote="Valentin Vornicu"]I think he means the German people (which lately, considering some soccer games they had with Romania ;);) cannot say they have anything left to be proud of :P:P)

just kiddin' of course ...[/quote]

I mean the mathlinks community.I am proud of any member who gets good results at any subject

well yeah I knew that :D ;) ... I was just kiddin' about the soccer thing. I'm sure that Germany will make it to the finals of the EuroCup 2006, as for Romania, we are staying home :(:(

Haha, WHOA that kid has issues... Okay, what were their parents thinking?!  "Oh honey, we just shot you taking your first step... I guess now we should start teaching you how to go to the bathroom by yourself, and after that, it is Q Basic!"  Haha, I'm just jealous I didn't have parents like that.  :P

Haha, btw, I learned Basic as an Enrichment program in [b]6th grade[/b]!

I met Reid in '01, when we were playing glock with the americans... he was a nice guy :-) My 2 cents... I'm just trying not to study..

I had to choose between IMO and IPhO (I won both olympiads) but I had to choose and I chose for IMO :)

[quote="liyi"]why time coincidence?
NOI is in summer while CMO is in winter.[/quote]

yes ,NOI is between July and August ,and every team must have a girl ! :lol:

[quote="luqash"]if anyone should be interested: I'm going to go to IChO as a member of the german team :-)[/quote]

Well ICHO takes place in Germany right now: http://www.icho.de/

This year (8th grade) I'm trying as hard as I can to get good scores in the USAMO (for the IMO) and the USACO (for the IOI).

Hopefuly I can get to the final levels this year, but that's a really big goal, since this is my first year for both of them.

Anders Kaseorg may not be too sad though he did not qualify for IMO but he is on the IOI team ( http://oldweb.uwp.edu/academic/mathematics/usaco/home/ ) for Greece this year.  :)

This topic is older than some AoPS users...

This topic is 12 years old.... only a bit younger than the entire AoPS site itself! :P
Also, I think the poll at the top is the most unpopular poll ever... 3 votes in 12 years?

[quote=Xinyan]I don't see why it's in FF. And we should probably stop reviving it.[/quote]

This is very likely the first topic in FF, maybe they didn't understand.

Still, its cool that such an old post still exists!

This is the first topic in FF: http://www.artofproblemsolving.com/community/q1h41p127

[quote=azmath333]
Also, I think the poll at the top is the most unpopular poll ever... 3 votes in 12 years?[/quote]

Correction*

8 votes

40 replies

3500 Views

I wouldn't call that unpopular, 3500 views is a lot.

[quote=azmath333]This topic is 12 years old.... only a bit younger than the entire AoPS site itself! :P[/quote]

Wow


I typed \mathbb{Z} for the integer symbol into my latext file but my output still generates a normal looking Z except that it seems to be in italics.

Instead of$\mathbb{Z}$, I am getting something like [i]Z[/i]

Could there be a package i'm missing or something?

You need amssymb. In fact I would put something like \usepackage{amsmath,amssymb,amsthm} in every document then you will have lots of symbols and facilities always available.


Hey,

can anyone help me on the problem:

Find all integers k such that if x,y,n are positive integers with n=(x^2+y^2+kxy)/(xy+1)>k, then n is actually a perfect square.

Its is true for all non-positive integers k, its also true for k=1,3,4,5,15 etc, not to mention their exists a test by which you can 'test' individual values of n like n=13, much like Eristothines's sieve for primes, but how do you attack in its general form.

-BOmB

[quote]Source is unavailable. [/quote]

What does it mean???

[quote]

can anyone help me on the problem: [/quote]

Thus it is a unsolved problem?

Pierre.

[quote]Quote
Source is unavailable.

What does it mean???

Quote

can anyone help me on the problem:

Thus it is a unsolved problem?

[/quote]

You are probably right in your assumption pbornsztein but I'm not really sure what he means by it either.

Sorry it looks weird. I should of previewed. Just look up down one. It's pierre's exact message.


There's been speculation as to who's gonna run for president in 2008. Just a few of the names that have come up alot:

Hillary Clinton (D-NY)
Rudy Giuliani (R-NY)
Barack Obama (D-IL)
John Kerry (D-MA)
Mitt Romney (R-MA)
Ralph Nader (I)

Who else have you heard? And who do you think will end up winning?

OBAMA!!!!

Actually he might do better to wait a bit longer till he's more experienced and established.

[quote="junggi"][hide="and how can you forget"]GW BUSH THE GREAT?</sarcasm>[/hide][/quote]

Actually, there's still Jeb.  I dunno...I don't keep up with politics much.  Does anyone know if it's feasible for him to run (i.e. would he consider it, could he pull enough votes, etc)?  The part about being related to a rather unpopular president might not be going too much in his favor though..

What about Al Gore?

And yes, McCain is a likely candidate.

[quote="13375P34K43V312"]Gore...eh the last time he ran he lost because Nader took away 4% of his votes...he probably would have won Florida[/quote]
I was thousand of miles away... but I never got why people were all talking about the popular vote thing and whether or not Gore did have more votes if you simply added them all up.
That may be so, but how could that have become an issue, there's a difference between questioning a system and questioning the way it was executed :ninja:  :maybe: ?

Personally, I don't think a female or non-white candidate has a chance. Maybe after the last generation passes by and we hit 2050 that we might actually see a female or non-white president.

Olympia Snowe should run, I'd totally vote for her.

[quote="Treething"]Oprah should run.[/quote]
:rotfl: :rotfl: :rotfl:

Sorry, I couldn't resist.

[quote="mustafa"]If Oprah ran, she would do pretty well :rotfl:[/quote]

Yeah, and Arnold Schwarzenagger would probably do pretty well too. Shows you where our culture's values are  :!:

Isn't John Edwards going to run?

ha! oprah give me a break.

anyways, i think we need better crackdown on scandalized officials. by that i mean boot them out of office.

MA is about to have some history, the two gubenetorial frontrunners are Deval Patrick and Kerry Healey. Deval Patrick is African-American and Kerry Healey is female.

I would like to see Hilary Clinton win. :)

[quote="Treething"]Oprah should run.[/quote]

Better than Duhbya.

Or maybe we could start a third.....no one needs fucia to start funcia's sentence because funcia was being sarcastic

I would like to see Indiana Jones or Me or Artemis Fowl or Some Math Prof.... win. It is all the same with the state of our government. Did anyone yet realise that d and r are the same thing? what would happen? would are government fall apart?

Yes doen't he have two young adultish aged girls???

RUDY GULIANI FOR PRESIDENT! He deserves to win. Look what he did to get New York back on its feet after the Sepember 11th attacks.

AND KERRY HEALEY FOR GOVERNOR!


[hide="Solution"]
$ f$ is constant function!  :maybe:

[/hide]

[quote="sinajackson"]
$ f$ is constant function!  :maybe:
[/quote]

Not so simple, a lot of non constant functions are solution too :

Example : Let $ g(x)$ be any function from $ [0,\frac{1}{2}]$ in $ \{-1,+1\}$, then $ f(x)=g(|\{x\}-\frac{1}{2}|)$ is solution (where $ \{x\}$ is fractional part of $ x$).

[quote="maradona"]Determine all functions R-> R such that :

f(x+f(y) ) = f( x.f(y) )[/quote]

Another example : $ f(x)=a+b\chi_{%Error. "mathBB" is a bad command.
{Q}}(x)$ where $ a$ and $ b$ are rational numbers and $ \chi_{%Error. "mathBB" is a bad command.
{Q}}(x)=1$ if $ x\in\mathbb{Q}$ and $ \chi_{%Error. "mathBB" is a bad command.
{Q}}(x)=0$ if $ x\notin\mathbb{Q}$.

In fact, it is easy to build solutions if $ |\{f(x),x\in\mathbb{R}\}|$ is finite (and even countable, I think).

An example of countable image :
If it exists $ n\in%Error. "mathBB" is a bad command.
{N}$, $ p\in%Error. "mathBB" is a bad command.
{Q}$ and $ q\in%Error. "mathBB" is a bad command.
{Q}$ such that $ x=p2^{\frac{1}{n}}+q$, then $ f(x)=n$ else $ f(x)=-1$

So there are a lot of solutions and it seems difficult to give a general expression for them.


The avarage barrel of crude oil costs 60 dollars. A barrel of crude oil
contains 0.85g/cm3. Calculate the total revenue from oil deposits.

[quote="CircleSquared"]How many barrels of oil are to be sold?
And what is the volume of a barrel?

 :maybe: [/hide][/quote]

One barrel is 55 gallons, but that doesn't help unless sharkman makes the problem more clear.

I think it is trying to say something like crude oil is such and such percent stuff we want & we sell the good stuff at 60 dollars a barrel.  We sell $x$ barrels of the good stuff.  Compute the total revenue.


[url=http://www.ocf.berkeley.edu/~wwu/riddles/medium.shtml#oneRealSolution]Source[/url]
Mathematics departments at some south-western universities received Mr. H.N.’s sly letters asking for the one real solution $ x$ of the following two equations:\[ 18 \equal{}\frac{(1\plus{}x)^{18}}{x}\]\[ 17 \equal{}\frac{(1\plus{}x)^{17}}{x}\]Professor A.S. at one of those departments sent Mr. H.N. the following brief solution:
$ \frac{18}{17}\equal{}\frac{\frac{(1\plus{}x)18}{x}}{\frac{(1\plus{}x)17}{x}}\equal{} 1\plus{}x$, so $ x \equal{}\frac{1}{17}$

Are there any other real solutions x? Why?

[hide="does it remind anyone of a particular"]inequality? :D [/hide]

I might be miscalculating, but when I plugged in $ x\equal{}\frac{1}{17}$, it doesnt work.

$ 18\equal{}\frac{(1\plus{}x)^{18}}{x}$
$ 18x\equal{}(1\plus{}x)^{18}$
$ \frac{18}{17}\equal{}(\frac{18}{17})^{18}$
which isnt true...

The same goes for the other equation, $ 1\equal{}(\frac{18}{17})^{17}$, which is also not true.

Is his method flawed, or did I miscalculated?

Looks flawed. There is a loss of information in combining the two equations. The numerator and denominator of his result can equal a lot of things besides $ 18$ and $ 17$, and their quotient can still be $ 18/17$.

It is pretty clear though that $ 1\plus{}x \equal{} 18/17$, if anything. So I'm seeing no solutions... not surprising since it's a 1-variable system with two equations.

[quote="MellowMelon"]There is a loss of information in combining the two equations. [/quote]

Well, sure, but the real issue is that the two equations might contradict each other.  Anyway, for a resolution to this problem I would suggest looking under the hide tags ;)

Darn. Well that nicely trivializes everything.

[hide="If you have not seen it"]
$ (1\plus{}x)^{n}\geq 1\plus{}nx$ (Bernoulli's)
$ 18 \equal{}\frac{(1\plus{}x)^{18}}{x}$ but $ 18x < 1\plus{}18x\leq (1\plus{}x)^{18}$. So $ 1\plus{}x$ has to be negative for anything to work, but that's impossible because then the LHS is less than 0 and the RHS is bigger than 0. The first equation can't be satisfied. I think the second one has some negative root.
[/hide]

Is it possible that the "one real solution" is individual to each question?  :maybe: If so, then we are looking for [i]two[/i] different solutions to [i]two[/i] different problems. :ninja:

The discussion in the problem implies a system, which is strange.


[quote="zxzzh"]In any triangle, prove that:
\[ h_a+w_b+m_c\leqslant3(R+r).  \][/quote]
What is \[ w_b \] ?

$w_b$ is bisector of angle $\angle B$, $l_b$ is more familiar than $w_b$.


greatest common divisor
AMC
USA(J)MO
USAMO
trigonometry

Για να μην πάει ούτε μια άσκηση χαμένη  !

 Α! Όσοι στέλνουν λύση να κάνουν quote πάνω στην εκφώνηση και στη συνέχεια να γράφουν και τη λύση. Με αυτόν τον τρόπο  εκφώνηση και λύση είναι μαζί και είναι εύκολη η ανάγνωση της λύσης. Εκτός αυτού , έτσι είναι εύκολο να την πάρουμε και για το βιβλίο.

                  [color=red][b]  Καλή Αρχή !!![/b][/color]

[color=red][b]ΑΣΚΗΣΗ 1[/b][/color]

Let (έστω) $ a, b, c$ are the sides of a triangle ( οι πλευρές ενός τριγώνου), such that(έτσι ώστε) $ a\plus{}b \equal{} 2c$. Prove that the centres of the inscribed and circumscribed circles are concyclic with the midpoints of the sides $ a$ and $ b$.( Να αποδείξετε ότι το έγκεντρο , το περίκεντρο  και τα μέσα των πλευρών a,b  του τριγώνου αυτού είναι ομοκυκλικά).
 ( Από το ποεριοδικό Komal, Ουγγαρία - Μάης 2007)

[b]ΛΥΣΗ[/b] [color=red](Μιχάλης Σάββας)[/color]

[i]Πολύ ωραία άσκηση, κ. Στεργίου.[/i]

Έστω $ D$ το ίχνος της καθέτου από το $ I$ στην $ AB$, $ K$ το μέσο της $ CA$, $ M$ το μέσο του τόξου $ \widehat{BC}$ και $ N$ το αντιδιαμετρικό του ως προς τον περιγεγραμμένο κύκλο του τριγώνου. Θα είναι

$ AD \equal{}\frac{b\plus{}c\minus{}a}{2}\equal{}\frac{a}{2}\equal{} BM$

 και $ \angle DAI \equal{}\frac{A}{2}\equal{}\angle KBM$

Άρα τα ορθογώνια τρίγωνα $ DAI, KBM$ είναι ίσα, συνεπώς $ MK \equal{} ID \equal{} r$. Τώρα από το ορθογώνιο τρίγωνο $ MBN$, αφού $ BK\perp MN$ έχουμε :

 $ BM^{2}\equal{} MK*MN \equal{} 2Rr$, άρα:

 $ IA^{2}\equal{} BM^{2}\equal{} 2Rr$. Επομένως στο τρίγωνο  $ OIA$ ισχύει  :

$ OA^{2}\equal{} R^{2}\equal{} R^{2}\minus{}2Rr\plus{}2Rr \equal{} OI^{2}\plus{}IA^{2}$,οπότε $ \angle OIA \equal{} 90$.

Δηλαδή το $ I$ ανήκει στον κύκλο με διάμετρο $ OA$ που είναι ο περιγεγραμμένος κύκλος του τριγώνου που σχηματίζεται από το $ O$ και τα μέσα των $ b,c$.

[quote="Protonios"][i]Θα παρακαλέσω όποιος μπορέσει να βάλει ένα σχήμα.[/i]

[color=red][b]ΑΣΚΗΣΗ 2 [/b][/color]

Θεωρούμε ένα τραπέζιο [b]ΑΒΓΔ [/b]με [b]ΑΒ//ΓΔ [/b]και [b]ΑΒ>ΓΔ[/b]. Έστω [b]Κ [/b]ένα σημείο του [b]ΑΒ[/b] και [b]Λ[/b] ένα σημείο του [b]ΓΔ[/b] τέτοια , ώστε [b]ΑΚ/ΚΒ=ΔΛ/ΛΓ[/b]. Αν είναι [b]Ρ, Σ[/b] δύο σημεία του [b]ΚΛ[/b] τέτοια , ώστε [b]<ΑΡΒ=<ΒΓΔ [/b]και [b]<ΓΣΔ=<ΑΒΓ[/b], να αποδειχθεί ότι τα [b]Ρ, Σ, Β, Γ[/b] είναι ομοκυλικά.

[color=red]Moder.[/color]

 [i]Είναι άσκηση SL 2006 και προτρέπω να στείλλετε όλες τις λύσεις αναλυτικά για να τις συμπεριλάβουμε στο βιβλίο.[/i][/quote]

Ευχαριστώ κύριε Βήττα για το σχήμα. Αναζήτησα και έχω υπ' όψιν μου δύο λύσεις. Μία κομψότερη και πιο ανεβασμένη :wink:  και μία που μοιάζει λίγο πολύ με τη δική μου απλούστερη προσέγγιση :blush: . Δεν τις βάζω ακόμη μήπως βρεθούν κι άλλες. :P

Από τα $ A,B$ φέρνουμε παράλληλες στις $ DS,CS$ που τέμνονται στο σημείο $ Q$ της $ MK$. Τότε $ \angle AQB\equal{}\angle DSC\equal{}\angle ABC\equal{}180\minus{}\angle BCD\equal{}180\minus{}\angle APB$, άρα το τεράπλευρο $ APBQ$ είναι εγγράψιμο, οπότε $ \angle MSC\equal{}\angle DSC\minus{}\angle DSM\equal{}\angle ABC\minus{}\angle AQM\equal{}\angle ABC\minus{}\angle ABP\equal{}\angle PBC$, που είναι το ζητούμενο.

ΑΣΚΗΣΗ 3 (παιχνιδάκι για σας αλλά είναι ωραία)

Θεωρούμε ένα ισοσκελές τραπέζιο $ ABCD$ με $ AB$ παράλληλο στη $ CD$. Ο εγγεγραμμένος του $ BCD$ τέμνει τη $ CD$ στο $ E$. Έστω σημείο $ F$ στην εσωτερική διχοτόμο της γωνίας $ <DAC$ τέτοιο ώστε $ EF$ κάθετη στη $ CD$. Αν ο περιγεγραμμένος του $ ACF$ τέμνει τη $ CD$ και στο $ G$, να αποδείξετε ότι το $ AFG$ είναι ισοσκελές.

Ωραία άσκηση Κώστα :) αν και στην αρχή να πω την αλήθεια μου φαινόταν πιο δύσκολη από όσο πράγματι είναι.

Εστω $ I$ το κέντρο του εγγεγραμμένου στο $ BDC$ και $ O$ το σημείο τομής της $ BI$, με την $ AF$.
Τότε κατ' αρχάς $ AOBC$ εγγράψιμο αφού $ \angle OAC \equal{} OBC$. Κι αφού ο περιγεγγραμμένος του ΑΒC διέρχεται από το $ D$ το Ο βρίσκεται στον περιγεγγραμμένο του $ BCD$. Αρα $ OI \equal{} OC$ (1)

Επίσης έχουμε έυκολα (αφού $ FE$ κάθετη στην $ AB$) ότι $ \angle OIF \equal{} 90\minus{}IBA \equal{} 90\minus{}FAB \equal{} OFI$.
Αρα $ OI \equal{} OF$ (2)

Απο 1,2 παίρνουμε $ OC \equal{} OF$ κι οπότε $ \angle OCF \equal{} OFC$ .
Απτο εγγράψιμο $ ACOD$ έχουμε $ \angle OCD \equal{} OAD \equal{} OAC$

Τώρα εύκολα έχουμε απτο εγγράψιμο $ AGCF$ παίρνουμε $ \angle GAF \equal{} GCF \equal{} OCF\plus{}DCO \equal{} OFC\plus{}OAC \equal{} AGC\plus{}CGF \equal{} AGF$ κι άρα $ AF \equal{} GF$

Πολύ ωραία λύση mim :wink: . Να μια ακόμα προσέγγιση.

Γνωρίζουμε ότι το σημείο επαφής του εγγεγραμμένου κύκλου με μια πλευρά ενός τριγώνου είναι συμμετρικό ως προς το μέσο της πλευράς με το σημείο επαφής του αντίστοιχου παρεγγεγραμμένου κύκλου με την πλευρά. Άρα λόγω συμμετρίας του ισοσκελούς τραπεζίου ως προς τη μεσοκάθετο των παράλληλων πλευρών του το $ F$ είναι το παράκεντρο του τριγώνου $ \triangle ADC$, άρα $ \angle AGF \equal{} 180\minus{}\angle ACF \equal{} 90\minus{}\frac{\angle ACD}{2}$ και $ \angle GAF \equal{}\angle DCF \equal{} 90\minus{}\frac{\angle ACD}{2}\equal{}\angle AGF$.

Δημήτρη πολύ ωραία η λύση σου :) . Δεν έχω κάποια άλλη υπ' όψιν μου διότι η δική μου ήταν η ίδια ακριβώς με του Μιχάλη. Δεν ήταν άσχημη η άσκηση γιατί ήταν ένας έξυπνος τρόπος χρήσης του σημείου επαφής του εγγεγραμμένου κύκλου που μεταφερόταν στο άλλο τρίγωνο σαν σημείο επαφής του παραγεγραμμένου. Νομίζω ήταν από USAMO ή κάτι αμερικάνικο εν πάση περιπτώσει. :)

ΑΣΚΗΣΗ 4:

Δίνεται τρίγωνο $ ABC$ και $ D$ το μέσο της $ BC$. Το $ M$ είναι σημείο της $ BC$ τέτοιο ώστε $ <BAM\equal{}<DAC$. Αν $ L$ είναι το σημείο τομής του περιγεγραμμένου του $ CAM$ με την $ AB$ και $ K$ το σημείο τομής του περιγεγραμμένου του $ BAM$ με την $ AC$ να δείξετε ότι $ KL//BC$.

Ωραία ασκησούλα . :)
[color=red]ΛΥΣΗ ΑΣΚΗΣΗΣ 4[/color]
Αφού $ \angle{BAM} \equal{} \angle{DAC}$ έχουμε ότι η $ AM$ είναι η συμμετρική της $ AD$ ως προς τη διχοτόμο . Άρα ΑΜ είναι συμμετροδιάμεσος . Από το θεώρημα Steiner έχουμε ότι $ \frac {BM}{MC} \equal{} \frac {AB^{2}}{AC^{2}}$  (1)
Τώρα από το θεώρημα της δύναμης σημείου ως προς κύκλο έχουμε

$ BL\cdot BA \equal{} BM\cdot BC$ ή $ \frac {BL}{BA} \equal{} \frac {BM\cdot BC}{BA^{2}}$ (2) και όμοια

$ \frac {CK}{CA} \equal{} \frac {CM\cdot CB}{CA^{2}}$ (3)

Tα δεύτερα μέλη των (2) , (3) είναι ίσα λόγω της (1) άρα $ \frac {BL}{BA} \equal{} \frac{CK}{CA}$ άρα $ KL\ //\ BC$
***
Δείτε επίσης ότι οι $ BK,\ CL,\ AD$ συντρέχουν . :wink:

Ωραία λύση Σιλουανέ :wink:

Παραθέτω και εγώ μια λύση πολύ πιο άκομψη βέβαια :oops:  με αντιστροφή (την δίνω περιληπτικά γιατί είναι μεγάλη...)

Λοιπόν κάνουμε αντιστροφή με κέντρο Α και μια ακτίνα $ r$. Τότε κατ' αρχάς $ B'M'K'$ συνευθειακά, $ C'M'L'$ συνευθειακά και $ AB'M'C'$ ομοκυκλικά.

Εύκολα βγαίνει τότε απ' τον τύπο $ \frac{BM}{MC}\equal{} (\frac{AB}{AC})^{2}$ ότι $ \frac{B'M'}{C'M'}\equal{}\frac{AC'}{AB'}$ ( θεωρώ γνωστό τον τύπο $ BM \equal{}\frac{r^{2}}{AB'*AM'}*B'M'$).
Από αυτό προκύπτει (απ' το νόμο συνημιτόνων π.χ.) κι απ' το γεγονός ότι $ AB'M'C'$ ότι η $ AM'$ τέμνει την $ B'C'$ στο μέσο της, έστω $ P$.
Τότε $ B'C'$ παράλληλο στο $ K'L'$ αφού $ AP, L'C', K'B'$ συντρέχουν στο $ M'$.

Συγχωρέστε μου που στη λύση μου παραλείπω αρκετές πράξεις :blush:

Σόρρυ για την αργοπορία... Επιτρέψτε την ταπεινή μου λύση (καλά εσείς τα ξέρετε κυριολεκτικά όλα :) ). Πού είσαι ρε Δημήτρη φέτος να με διδάξεις λίγο γεωμετρία... :)

Θα δείξουμε ότι $ CK/LB\equal{}AC/AB$ (ταπεινός πλην διαχρονικός Θαλής)
Παρατηρούμε την ομοιότητα των $ ABC$, $ MKC$ διότι έχουν την κοινή $ C$ και $ <CMK\equal{}180\minus{}<BMK\equal{}<KAB$ (εγγράψιμο AKMB).
Ομοίως όμοια και τα $ ABC$, $ MBL$. Άρα καταλήγουμε ότι όμοια είναι τα $ MKC$, $ MBL$.
Διαδοχικά λοιπόν $ CK/LB\equal{}KM/BM\equal{}(AMsinKAM/sinAKM)/(AMsinMAB/sinMBA)\equal{}sinKAM/sinMAB$ (διότι από το ίδιο εγγράψιμο $ <AKM\equal{}180\minus{}<MBA$
'Ομως $ sinKAM/sinMAB\equal{}sinDAB/sinDAC\equal{}(BDsinBDA/AB)/(CDsinCDA/AC)\equal{}AC/AB$

Εστω τρίγωνο $ ABC$ του οποίου ο εγγεγραμμένος κύκλος τέμνει τις πλευρές $ BC, AC, AB$ στα σημεία $ D, E, F$ αντίστοιχα. Αν $ X$ εσωτερικό σημείο του τριγώνου τέτοιο ώστε ο εγγεγραμμένος κύκλος του $ XBC$ τέμνει τις $ BC, XC, XB$ στα σημεία $ D, Y, Z$ αντίστοιχα να αποδείξετε ότι το τετράπλευρο $ EFZY$ είναι εγγράψιμο.

Βλέπω σε παίδεψε πολύ... :D  Πάντως αρχικά τον trig Ceva είχα υπ' όψιν για το πρώτο μισό αλλά δε σκέφτηκα ότι το ίδιο γίνεται και στο δεύτερο. :(  Είχα βρει μια άλλη πιο κατασκευαστική λύση για το δεύτερο μισό που πιάνει πολύ χώρο και γι' αυτό φαντάστηκα ότι θα ήταν δυσκολότερη η άσκηση...  :maybe:

[b][size=100][color=DarkBlue]ΠΡΟΤΑΣΗ. - Δίνεται ισοσκελές τρίγωνο $ \bigtriangleup ABC$ με $ AB \equal{} AC,$ εγγεγραμμένο σε κύκλο $ (O)$ και έστω $ A',$ το αντιδιαμετρικό σημείο του $ A.$ Η εφαπτομένη του $ (O)$ στο σημείο $ C,$ τέμνει την ευθεία $ AB$ στο σημείο έστω $ P.$ Αποδείξτε ότι η ευθεία $ PA',$ τριχοτομεί την πλευρά $ BC.$[/color][/size][/b]

Κώστας Βήττας.

[b][size=100][color=DarkBlue]PROBLEM. - An isosceles triangle $ \bigtriangleup ABC$ with $ AB \equal{} AC$ is given and let $ A'$ be, the antidiametric (= diametric opposite) point of $ A.$ The tangent line of $ (O),$ at vertex $ C,$ intersects the sideline $ AB,$ at one point so be it $ P.$ Prove that the line segment $ PA',$ trisects the side-segment $ BC$ of $ \bigtriangleup ABC.$[/color][/size][/b]

Kostas Vittas.

Πάλι trigo και ποιος τον ακούει τον Μιχάλη  :surrender:

Θετουμε $ \angle{CPD} \equal{} P_1$ και $ \angle{DPB} \equal{} P_2$
Πάντως είναι κάτι πολύ σύντομο . Από trig Ceva στο PBC για το σημείο Α' έχουμε :

$ \frac {\sin P_1}{\sin P_2}\cdot\frac {\sin 90}{\sin\frac {A}{2}}\cdot\frac {\sin\frac {A}{2}}{\sin\frac {A}{2}} \equal{} 1$

άρα $ \frac {\sin P_1}{\sin P_2} \equal{} \sin\frac {A}{2}$  (1)

Όμως από το νόμο ημιτόνων στα PDC και PDB έχουμε :
$ \frac {\sin P_1}{\sin P_2} \equal{} \frac {CD\sin A}{BD\cos\frac {A}{2}}$   (2)

(1)=(2) και  $ \sin A \equal{} 2\sin\frac {A}{2}\cos\frac {A}{2}$  άρα BD=2CD                 QED
 :D

Παρατηρήσεις : Αν η BA' τέμνει την CP στο Κ τότε τα K,D,A συνευθειακά .
Αν S το σημείο τομής της OC με την ΑΚ και Ν το μέσο της AC , τότε P,S, N συνευθειακά  :)

[quote="silouan"] Παρατηρήσεις : Αν η BA' τέμνει την CP στο Κ τότε τα K,D,A συνευθειακά .
Αν S το σημείο τομής της OC με την ΑΚ και Ν το μέσο της AC , τότε P,S, N συνευθειακά  :)[/quote]
Σιλουανέ σε ευχαριστώ για τη σύντομη λύση σου και κυρίως για τα δύο νέα (σε μένα τουλάχιστον) συμπεράσματα που ανέφερες.

Νομίζω ότι μπορούμε να τα αποδείξουμε, όχι δύσκολα, με αρμονικά συζυγή και βασισμένοι σ' αυτά, να αποδείξουμε ότι τα σημεία $ B,$ $ D,$ είναι αρμονικά συζυγή των $ M,$ $ C,$ όπου $ M,$ είναι το μέσον του $ BC$ και από όπου προκύπτει ότι $ DC \equal{} 2DM$ $ \Longrightarrow$ $ BD \equal{} 2DC,$ δηλαδή το ζητούμενο.

Αυτή είναι μία άλλη συνθετική απόδειξη από αυτήν που είχα υπόψη μου όταν έβαλα το πρόβλημα εδώ, ως οριακή περίπτωση του προβλήματος [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=181210]A line through a fixed point[/url] $ ($ όταν οι ισογώνιες ευθείες δια της κορυφής $ B$ του $ \bigtriangleup ABC,$ ταυτίζονται με τις πλευρές του $ AB,$ $ BC$ $ ).$

Θα βάλω αργότερα περισσότερες λεπτομέρειες, εκτός και αν με προλάβει κάποιος άλλος, κάτι που εύχομαι να συμβεί. :)

Κώστας Βήττας.

[quote="vittasko"][b][size=100][color=DarkBlue]ΠΡΟΤΑΣΗ. - Δίνεται ισοσκελές τρίγωνο $ \bigtriangleup ABC$ με $ AB \equal{} AC,$ εγγεγραμμένο σε κύκλο $ (O)$ και έστω $ A',$ το αντιδιαμετρικό σημείο του $ A.$ Η εφαπτομένη του $ (O)$ στο σημείο $ C,$ τέμνει την ευθεία $ AB$ στο σημείο έστω $ P.$ Αποδείξτε ότι η ευθεία $ PA',$ τριχοτομεί την πλευρά $ BC.$[/color][/size][/b]

Κώστας Βήττας.
[/quote]

[i][color=blue]Βάζω ένα πρόβλημα από BMO 1985 (με κάπως διαφορετική διατύπωση βέβαια απ' το πρωτότυπο) που μοιάζει στο σχήμα με αυτό το πρόβλημα:[/color][/i]

[color=red][b]ΑΣΚΗΣΗ 99[/b][/color]

Δίνεται ισοσκελές τρίγωνο $ \bigtriangleup ABC$ με $ AB \equal{} AC,$ εγγεγραμμένο σε κύκλο $ (O)$ και έστω $ A',$ το αντιδιαμετρικό σημείο του $ A.$ Εστω επίσης $ D$ το σημείο στην $ BC$ ώστε $ 3CD \equal{} BC$.Προεκτείνουμε την $ AC$ κατά ίσο τμήμα $ AC \equal{} CE$. Να αποδείξετε ότι $ BE\perp{A'D}$.

[color=orange][i]Επαναλαμβάνω ότι το πρωτότυπο πρόβλημα είχε αρκετά διαφορετική διατύπωση απλά το έγραψα έτσι για να φανεί η ομοιότητα του με αυτό το πρόβλημα (αφού αναφερόμαστε και πάλι στην $ PA'$ :wink: )[/i][/color]

[color=blue][i]Καλό  είναι να διατηρήσουμε την αρίθμηση των ασκήσεων για να κάνουμε εύκολα αναφορές.Και να γίνει ένα λαθάκι , δεν πειράζει. Το διορθώνουμε.
 Οι λύσεις είναι επίσης πρακτικό να γίνονται με Quote.[/i][/color]

 [color=red][b]ΑΣΚΗΣΗ 100[/b][/color]

 Στο παρακάτω σχήμα οι $ MKL, EMZ$ είναι παράλληλες προς τις πλευρές του ορθογωνίου $ ABCD$.Να αποδειχθεί ότι
 $ EK \perp LZ$

[b]ΣΧΟΛΙΟ[/b]

 Έχω μια διανυσματική λύση και χθες έκανα και μια συνθετική. Θέλω να τη ρίξετε μια ματιά για καμιά πιο σύντομη λύση.Φυσικά πρόκειται για μέτριας δυσκολίας άσκηση έως και εύκολη για επίπεδο διαγωνισμών.Επειδή κάπου χρησιμοποιώ  μετρικές σχέσεις[hide]δύναμη σημείου[/hide] , δεν την βάζω στους μικρούς , αν και στη Balk. Junior αυτό θεωρείται γνωστό.

[quote="stergiu"] [color=red][b]ΑΣΚΗΣΗ 100[/b][/color]
 Στο παρακάτω σχήμα οι $ MKL, EMZ$ είναι παράλληλες προς τις πλευρές του ορθογωνίου $ ABCD$.Να αποδειχθεί ότι
 $ EK \perp LZ$
[/quote]

Όταν έπρεπε να τον χρησιμοποιήσω δεν το έκανα (βλέπε ΙΜΟ 2007 pr 2) αλλά τώρα δίνει μια εξαιρετική λύση .
Από το θέρωμα Simpson για το τρίγωνο BDC παίρνουμε ότι αν $ S$ το σημείο τομής των $ ZL,\ DB$ τότε $ MS\bot DB$
Από το θεώρημα Simpson για το τρίγωνο ADB παίρνουμε ότι τα σημεία E,K,S είναι συνευθειακά .
Τώρα το MSLD είναι εγγράψιμο άρα $ \angle{DSL}\equal{}\angle{DML}\equal{}\angle{MDE}\equal{}\angle{MSE}$  αφού το EMSD εγγράψιμο.

άρα $ \angle{DSL}\plus{}\angle{ESD}\equal{}\angle{MSE}\plus{}\angle{ESD}$ άρα $ \angle{ESL}\equal{}90$    qed  :D

[quote="silouan"][quote="stergiu"] [color=red][b]ΑΣΚΗΣΗ 100[/b][/color]
 Στο παρακάτω σχήμα οι $ MKL, EMZ$ είναι παράλληλες προς τις πλευρές του ορθογωνίου $ ABCD$.Να αποδειχθεί ότι
 $ EK \perp LZ$
[/quote]

..................................................................................
Από το θέρωμα Simpson για το τρίγωνο BDC παίρνουμε ότι αν $ S$ το σημείο τομής των $ ZL,\ DB$ τότε $ MS\bot DB$
Από το θεώρημα Simpson για το τρίγωνο ADB παίρνουμε ότι τα σημεία E,K,S είναι συνευθειακά .
..........................................................................................................[/quote]
Υπέροχα !

 Βέβαια , από τις δύο πρώτες γραμμές  προκύπτει ότι τα σημεία $ E,M, S,L ,D$  είναι ομοκυκλικά , διότι τα $ AMLD ,MSLD$ είναι προφανώς εγγράψιμα. Άρα :

 $ \angle ESL \equal{} \angle EML \equal{} 90$
 και τελειώνεις !

[b] ΣΧΟΛΙΑ[/b]

 [b] α)[/b] Αυτή τη λύση δεν την γνώριζα  και έτσι άξιζε τον κόπο που  έβαλα την άσκηση στο ... μικροσκόπιό σας.
 [b] β)[/b] Νομίζω ότι η λύση που έκανες πρέπει να είναι και ο τρόπος που δημιουργήθηκε η άσκηση. Γι αυτό έχει ιδιαίτερη σημασία.
[b]γ) [/b]Μακάρι τέτοιες εμπνεύσεις να σας έρχονται και την ώρα των διαγωνισμών !

  Μπάμπης

[quote="stergiu"]
 [color=red][b]ΑΣΚΗΣΗ 100[/b][/color]

 Στο παρακάτω σχήμα οι $ MKL, EMZ$ είναι παράλληλες προς τις πλευρές του ορθογωνίου $ ABCD$.Να αποδειχθεί ότι
 $ EK \perp LZ$

[b]ΣΧΟΛΙΟ[/b]

Επειδή κάπου χρησιμοποιώ  μετρικές σχέσεις [i]δύναμη σημείου[/i], δεν την βάζω στους μικρούς , αν και στη Balk. Junior αυτό θεωρείται γνωστό.[/quote]

Βάζω και εγώ μια λύση σε αυτό το ωραίο πρόβλημα αν και πολύ φοβάμαι ότι η λύση μου συμπίπτει με τη δικιά σας κ. Στεργίου αφού χρησιμοποιώ κι εγώ δύναμη σημείου. Πάντως είναι αρκετά σύντομη :)

Εστω ότι η $ MLK$ τέμνει ξανά τον κύκλο στο $ P$. Τότε εύκολα αποδεικνύεται ότι $ MK\equal{}LP\Longleftrightarrow KP\equal{}ML$.
Από δύναμη σημείου του $ K$ έχουμε:
$ KM\cdot{KP}\equal{}KA\cdot{KB}\Longleftrightarrow KM\cdot{LM}\equal{}ME\cdot{MZ}$
Ara $ \frac{KM}{ME}\equal{}\frac{LM}{MZ}\Longleftrightarrow\tan{MLZ}\equal{}\tan{MEK}\Longleftrightarrow\angle{MLZ}\equal{}\angle{MEK}$

Aπό αυτό το σημείο το ζητούμενο είναι προφανές :)

[quote="bilot4"][quote="stergiu"]
 [color=red][b]ΑΣΚΗΣΗ 100[/b][/color]

 Στο παρακάτω σχήμα οι $ MKL, EMZ$ είναι παράλληλες προς τις πλευρές του ορθογωνίου $ ABCD$.Να αποδειχθεί ότι
 $ EK \perp LZ$

[b]ΣΧΟΛΙΟ[/b]

Επειδή κάπου χρησιμοποιώ  μετρικές σχέσεις [i]δύναμη σημείου[/i], δεν την βάζω στους μικρούς , αν και στη Balk. Junior αυτό θεωρείται γνωστό.[/quote]

Βάζω και εγώ μια λύση σε αυτό το ωραίο πρόβλημα αν και πολύ φοβάμαι ότι η λύση μου συμπίπτει με τη δικιά σας κ. Στεργίου αφού χρησιμοποιώ κι εγώ δύναμη σημείου. Πάντως είναι αρκετά σύντομη :)

Εστω ότι η $ MLK$ τέμνει ξανά τον κύκλο στο $ P$. Τότε εύκολα αποδεικνύεται ότι $ MK \equal{} LP\Longleftrightarrow KP \equal{} ML$.
Από δύναμη σημείου του $ K$ έχουμε:
$ KM\cdot{KP} \equal{} KA\cdot{KB}\Longleftrightarrow KM\cdot{LM} \equal{} ME\cdot{MZ}$
Ara $ \frac {KM}{ME} \equal{} \frac {LM}{MZ}\Longleftrightarrow\tan{MLZ} \equal{} \tan{MEK}\Longleftrightarrow\angle{MLZ} \equal{} \angle{MEK}$

Aπό αυτό το σημείο το ζητούμενο είναι προφανές :)[/quote]

  ΔΗΜΗΤΡΗ , έπεσες μέσα ! ΑΥτή είναι και η δική μου τριγωνομετρική λύση.
 Φυσικά με τον ίδιο τρόπο μπορείς να χρησιμοποιήσεις και τη μέθοδο καθετότας με τη διαφορά τετραγώνων και με λίγο Πυθαγόρειο φτάνεις πάλι στη σχέση που απέδειξες με τη δύναμη.
  Καλή Κυριακή!
 Μπάμπης

[color=red] (edited-mod)[/color]

[quote="bilot4"][quote="stergiu"]
  [color=red]ΣΤΕΛΙΟ[/color] , έπεσες μέσα ! ΑΥτή είναι και η δική μου τριγωνομετρική λύση.
 Φυσικά με τον ίδιο τρόπο μπορείς να χρησιμοποιήσεις και τη μέθοδο καθετότας με τη διαφορά τετραγώνων και με λίγο Πυθαγόρειο φτάνεις πάλι στη σχέση που απέδειξες με τη δύναμη.
  Καλή Κυριακή!
 Μπάμπης[/quote]

κ. Στεργίου ήθελα να σας το πω κι από τα personal messages που μου στέλνατε αλλά νόμιζα ότι ο Στέλιος (mostel) το είχε ξεκαθαρίσει.
Εγώ είμαι ο Δημήτρης Παπαδημητρίου :!:
 :P
Καλή Κυριακή και σε εσάς![/quote]

  Δημήτρη , καλή χρονιά  !!!

[color=red][b]Ου γαρ έρχεται μόνον !!!![/b][/color]

 Ποιος ξέρει τι μου ήρθε και σας ταύτισα.Τώρα καταλαβαίνω και ένα μήνυμα του Σιλουανού που μου έγραψε ότι αντί του Στέλιου έγραψα κάποτε σε σένα. Κοίταξα στο αρχείο , αλλά πάλι βλέποντας Bilot νόμισα ότι έγραψα  στο Στέλιο.
 Τέλος πάντων , από αυτό μου το μπέρδεμα , μάλλον προκύπτει ότι σας αγαπώ όλους το ίδιο  και δεν έχει σημασία σε ποιον γράφω , αλλά το ότι γράφω.Κάτι είναι και αυτό !
 Συγχωρέστε με και οι δύο . Αν ξανασυμβεί , προωθείστε το μήνυμα ή γράψτε μου.

                              Ευχαριστώ - Μπάμπης

[i][color=blue]( Αφού σου έδωσα και δεύτερο όνομα  :lol:  πρέπει ως νουνός να σου κάνω και ένα πρωτοχρονιάτικο δώρα! Έτσι είναι. Διάλεξε λοιπόν ένα από τα βιβλία μου που δεν έχεις και γράψε μου σε PM για να στο στείλλω δωρεάν. Μη διστάσεις και κυρίως μη ντραπείς! Θα το χαρώ πολύ . Θα έχουμε και κάτι να θυμόμαστε και να γελάμε !Ουδέν  κακόν αμιγές καλού !)[/color][/i]

[color=red][b]ΑΣΚΗΣΗ 101[/b][/color]

[quote="stergiu"]Στο παρακάτω σχήμα $ K$ είναι το περίκεντρο του τριγώνου $ ABC$ , $ O$ είναι το περίκεντρο τυχαίου κύκλου που περνάει από τα σημεία $ B,C$ και  $ L$ είναι το περίκεντρο του τριγώνου $ ADE$.Να αποδείξετε ότι
  [color=red]α)[/color] $ AK \parallel OL$  και στη συνέχεια ότι
  [color=red]β) [/color] $ ON \perp AN$, όπου $ N$  είναι το δεύτερο σημείο τομής των κύκλων $ K$ και  $ L$

[b] ΣΧΟΛΙΟ[/b]

[i] Στους παλαιότερους το δεύτερο ερώτημα κάτι θα θυμίσει.[/i][/quote]

[i][b]Μήπως ξέρει κάποιος να μου πει πώς θα βάλω τις εικόνες στην κατάλληλη  σειρά, διότι δεν το έχω ξαναεπιχειρήσει ; [/b][/i]

[quote="bilot4"]Για το πρώτο ερώτημα έχουμε:
$ DE\perp{LO}$ (αφού $ LO$ η μεσοκάθετος του $ DE$). Aρα αρκεί $ DE\perp{AK}$.

Πράγματι ξέρουμε ότι $ \angle{BAK} \equal{} 90 \minus{} \angle{C} \equal{} 90 \minus{} \angle{EDA}$ αφού το $ EDBC$ έιναι εγγράψιμο.

Αρα $ \angle{BAK} \plus{} \angle{EDA} \equal{} 90$ και τελειώσαμε :)

Το δεύτερο ερώτημα δεν το έχω σκεφτεί ακόμα :blush:[/quote]

[quote="silouan"]Όλες οι λύσεις που ήξερα μέχρι τώρα γγια την άσκηση είχαν κάτι μετρικό μέσα τους , και έίπα να βρω μια καθαρα γεωμετρική απόδειξη .

Έχουμε ότι $ \angle{DAL} \equal{} 90 \minus{} \angle{AED} \equal{} 90 \minus{} \angle{ABC}$ άρα $ AL\bot BC$
Επομένως $ AL//KO$ και $ AK//LO$ (από το α) ερώτημα) άρα $ AKOL$ παραλληλόγραμμο και έστω $ P$ το σημείο τομής των διαγωνίων του .Έστω ότι η $ KP$ τέμνει την $ AN$ στο $ S$ . Όμως η KL είναι μεσοκάθετη της ΑΝ άρα το S είναι το μέσον του ΑΝ.
Άρα η ΡS που συνδέει τα μέσα των πλευρών στο τρίγωνο $ ANO$ είναι κάθετη στην ΑΝ άρα το ίδιο θα συμβαίνει και με την ΟΝ .

Η λύση είναι αφιερωμένη στον κ .Κώστα Βήττα  :)[/quote]

[color=red][b]ΑΣΚΗΣΗ 102[/b][/color]

[i]Μια ωραία γεωμετρική ανισότητα από BMO [/i]

Εστω τρίγωνο $ ABC$ και $ D, E, F$ σημεία στις πλευρές $ BC, AC, AB$ αντίστοιχα τέτοια ώστε το $ AFDE$ να είναι εγγράψιμο. Να αποδείξετε ότι:

$ 4\frac {(DEF)}{(ABC)}\leq (\frac {EF}{AD})^2$, όπου με $ (XYZ)$ συμβολίζουμε το εμβαδόν του $ XYZ$.

Βάζω το σχήμα . Αρχικά τραβάω την παράλληλη CK ως προς την ΑΒ . Ο λόγος των εμβαδών των τριγώνων ABC και ACK είναι $ \dfrac{(ACK)}{(ABC)}=\dfrac{DK}{AD}$ .Επίσης το $ \triangle{(ACK)}$ είναι όμοιο με το $ \triangle{(DEF)}$  άρα ο λόγος των εμβαδών τους είναι $ \dfrac{(DEF)}{(ACK)}={(\dfrac{EF}{AK})^2}$ . Άρα από αυτα τα δύο συνεπάγεται ότι
$ \dfrac{(DEF)}{(ABC)}={(\dfrac{EF}{AK})^2*\dfrac{DK}{AD}}$αντικαθιστώντας στην ανίσωση έχουμε  $ 4[{(\dfrac{EF}{AK})^2*\dfrac{DK}{AD}]\leq{(\dfrac{EF}{AD})}^2}$ $ \Rightarrow 4\dfrac{DK}{AK^2}\leq\dfrac{1}{AD}$$ \Rightarrow AK^2\geq4DK*AD$ που προφανώς ισχύει .(χρησιμοποιώντας την ανισότητα $ {(x+y)}^2\geq4xy$ καθώς το ΑΚ=DK+AD) So QED :D


search
linear algebra
linear algebra unsolved

a flag of a($ dim(E)\equal{}n$) vectorial space $ E$ is a decomposition of this space in to $ n\plus{}1$ verctoriel spaces $ F_i$ such that $ F_0\equal{}\{0\},F_{n\plus{}1}\equal{}E ,\forall i\in[1,n] F_iCF_{i\plus{}1}$ and $ dim(F_i)\equal{}i$
let $ D$ and $ D'$ be two flag of $ E$,prove that there existe a base $ (e_1,...,e_n)$ adapted two both of the flags

[quote="drapt@"]... prove that there existe a base $ (e_1,...,e_n)$ [b]adapted two both of the flags[/b][/quote]
I don't understand the question :(. Could you make it clearer?

take a look at:
http://www.mathlinks.ro/viewtopic.php?search_id=413923847&t=6329
for the more accurate statement.


algebra
polynomial
vector
algorithm
Ring Theory
superior algebra
superior algebra unsolved

Hi all, I have trouble understanding polynomial algebra.

Let F[x] in Z_3[x] be the set of all polynomials modulo 1+x^2 where Z_3[x] is the set of all polynomials with coefficients in Z_3. Addition and multiplication are defined in the usual way but modulo 1+x^2 and the arithmetic of the coefficients are in Z_3. Compute the addition table.

I don't know what members are in F[x]. Are they 1,x,1+x and x^2, that is, permutations of polynomials of degree no higher than 2? I'm trying to draw similarities between polynomial modulo and integer modulo, where I know for the latter, members of Z_p are {0,1,...,p-1}.

Also, after I know what the members are, do I fill in the table by considering all possible "combinations", i.e. 1+x is obtained from the 1-row and x-column of the table, divide that "combination" by 1+x^2, keeping coefficients in Z_3 then take the remainder as the result?

Thank you.

Regards,
Rayne

Can you tell us what is means "modulo" $1+x^{2}$, I never seen that term before.

If we're dealing with only integers, then modulo 4 would be divide any integer by 4 and take the remainder. For example,
2 modulo 4 = 2
4 modulo 4 = 0
6 modulo 4 = 2

So I guess that modulo 1+x^2 is divide any polynomial by 1 + x^2 and take the remainder.

Maybe you mean,

$\mathbb{Z}_{3}[x]/<1+x^{2}>$? No?

Cuz, $1+x^{2}$ is irreducible in $\mathbb{Z}_{3}[x]$.
And therefore $<1+x^{2}>$ is an ideal for $\mathbb{Z}_{2}[x]$, which implys the factor ring is well-defined.

If then it has 9 elements and is a field.

(I edited my post, dont worry I'm not stealing what you wrote JMerry).

The element of this quotient ring are of the form $p+I$, where $p$ is an element of the larger polynomial ring and $I$ is the ideal generated by $1+x^{2}$. The set of all polynomials of degree $\le 1$ is a set of representatives for these classes, so the quotient is a 2-dimensional vector space over $F$. The quotient has zero-divisors if $1+x^{2}$ factors, and is a field if $1+x^{2}$ is irreducible.
Since $1+x^{2}$ is irreducible over $F_{3}$, the quotient is a field with 9 elements.

If you're thinking in the analogy to integers, "less than" is "lower degree than". The connection is the division algorithm; we can always find a quotient which makes the remainder "smaller" than the dividend.


algebra
polynomial
calculus
derivative
integration
calculus computations

Show that the polynomials
\[p_{n}(t): =\sqrt{\frac{2n+1}{2}}\frac{1}{2^{n}n!}\frac{d^{n}}{dt^{n}}\left ((t^{2}-1)^{n}\right ) \]
forms an orthonormal sequence in the space $L^{2}([-1,1])$.

General hints: treat the orthogonality separately from the normalization. All of those constants before the derivative are about the normalization; ignore then when you're working on the orthogonality.

As for proving the orthogonality: rather than trying to prove that $P_{n}$ is orthogonal to $P_{m},$ it's easier to simply prove that $P_{n}$ is orthogonal to any polynomial of degree less than $n.$ That proof calls for an integration by parts.


Some interesting effect occurs if you iterate some TeX-commands. For example, ["\bar\mathbb{Q}"] results in [$ \bar\mathbb{Q}$] instead of [$ \bar{\mathbb{Q}}$] (note the [] to mark the point where the code is; you can even copy it).
Whereas it isn't dangerous (as long as no other forum get's knowledge about our stealth-TeXnologies as they then may want to attack us), it may be somewhat disturbing if code isn't visible.

[quote="ZetaX"]For example, ["\bar\mathbb{Q}"] results in [$ \bar\mathbb{Q}$] instead of [$ \bar{\mathbb{Q}}$] (note the [] to mark the point where the code is; you can even copy it).[/quote]That's because there's an error in the LaTeX code.  \bar{\mathbb{Q}} $ \bar{\mathbb{Q}}$ is the correct version.
If you use \bar\mathbb{Q} in a LaTeX document, you will see the error it gives you. LaTeX tries to decide what is meant and ignores the bit it doesn't understand.


Λυπάμαι που δεν μπορώ να συνεισφέρω καθόλου τελευταία, έχει πλακώσει πολλή δουλειά. Όμως σας παραθέτω ένα interesting προβληματάκι που πέτυχα την περασμένη βδομάδα:

Prove or disprove: Υπάρχει κάποιο ανοιχτό υποσύνολο $ \mathcal{V} \subseteq \mathbb{R}^{n\plus{}1}$, $ n \geq 1$, με την παρακάτω ιδιότητα;

Για κάθε $ \mathcal{U} \subseteq \mathbb{R}^n$ ανοιχτό, υπάρχει  κάποιο $ x \in \mathbb{R}$ τέτοιο ώστε $ \mathcal{U} \equal{} \{ y \in \mathbb{R}^n \; : \; (x,y) \in \mathcal{V} \}$. (Δηλαδή, για κάθε $ \mathcal{U} \subseteq \mathbb{R}^n$ ανοιχτό, να υπάρχει κάποια οριζόντια φέτα του $ \mathcal{V}$ που να είναι αντιγραφή του $ \mathcal{U}$ και μόνο αυτού.)

Καλή συνέχεια παίδες.

Anyone? Τόσο uninteresting (ή πασίγνωστο) :huh:

Gaaah, όχι, εγώ δεν το ξέρω... όταν το είχες βάλει δοκίμασα να δω μήπως είχα καμιά ιδέα να ξεπετιέται άμεσα, δε μου ήρθε κάτι και μετά το ξέχασα.

Εμένα σίγουρα μ'ενδιαφέρει το πρόβλημα (είναι πολύ όμορφο) αλλά (να'ναι καλά το defense μου) δε βλέπω να ασχολούμαι μαζί του any time soon (παλιά έσωζε την κατάσταση η αναμονή στο μετρό, τώρα όμως με το αμάξι συνέχεια... :( ).

Anyway, kill it if you want, και η λύση επίσης μ'ενδιαφέρει σίγουρα... δε μου λες όμως, δεν το αφήνεις κανά διήμερο ακόμα μήπως και?

Cheerio,

Durandal 1707

Ενδιαφέρον είναι. Το είχα σκεφτεί χωρίς όμως να βρω τίποτα. Έχεις κανένα hint να μας δώσεις;

[hide="Hint"]Can be done :!:, even with $ x \in [0,1)$, i.e. $ \mathcal{V} \subset [0,1) \times \mathbb{R}^n$[/hide]

Μια προσέγγιση την οποία είχα σκεφτεί πριν από καιρό αλλά κατάφερα να πείσω τον εαυτό μου ότι δεν δούλευε. Μετά το hint κατάφερα να με πείσω πως δουλεύει. Για να δούμε...

Ας σκεφτούμε την περίπτωση ν=1 αν και η γενική περίπτωση είναι παρόμοια: Υπάρχουν ένα αριθμήσιμο σύνολο Α ανοικτών συνόλων του R ώστε κάθε άλλο ανοικτό σύνολο μπορεί να γραφτεί σαν ένωση κάποιων εξ' αυτών των συνόλων. (Π.χ. μπορούμε να πάρουμε όλα τις ανοικτές μπάλες με ρητό κέντρο και ρητή ακτίνα.). Αριθμούμε τα στοιχεία του Α (γίνεται και χωρίς επιλογή), έστω $ A \equal{} \{A_1,A_2,\ldots\}$ Μετά για κάθε $ x \in [0,1]$ γράφουμε το $ x \equal{} 0.x_1x_2\ldots$ στο διαδυκό σύστημα και θέτουμε $ B_x \equal{} \bigcup\limits_{\{i: x_i \equal{} 1\}}A_i$. Αν ο $ x$ μπορεί να γραφτεί με δύο τρόπους τότε χρησιμοποιούμε άπειρα μηδενικά αντί άπειρους άσσους. Παίρνουμε τώρα $ V \equal{} \cup_x \{x\} \times B_x$ και αρκεί να δείξουμε ότι το $ V$ είναι ανοικτό. Έστω $ (x,y)$ ένα σημείο του $ V$. Ας υποθέσουμε πρώτα ότι το ανάπτυγμα του $ x$ δεν τελειώνει σε άπειρα μηδενικά. Τότε υπάρχει $ i$ ώστε $ x_i \equal{} 1$ και $ y \in A_i$. Παίρνουμε $ j > i$ με $ x_j \equal{} 0$ και $ k > j$ με $ x_k \equal{} 1$. Τότε για κάθε $ z \in (x\minus{}2^{\minus{}k},x\plus{}2^{\minus{}k})$ θα έχουμε $ z_i \equal{} 1$ και άρα η μπάλα με κέντρο το $ (x,y)$ και ακτίνα $ 2^{\minus{}k}$ θα περιέχεται στο $ V$. Ας υποθέσουμε τώρα ότι το ανάπτυγμα του $ x$ τελειώνει σε άπειρα μηδενικά. (Εδώ είναι το σημείο που είχα πειστεί πως δεν μπορούσε να δουλέψει) Έστω ότι $ x_r \equal{} 0$ για κάθε $ r \geqslant j$. Yπάρχει $ i$ ώστε $ x_i \equal{} 1$ και $ y \in A_i$. Τότε για κάθε $ z \in [x,x\plus{}2^{\minus{}j})$ θα έχουμε $ z_i \equal{} 1$ και άρα $ A_i \subseteq B_z$. Παίρνουμε τώρα $ k > j$ ώστε $ i \in A_k \subseteq A_j$. Τότε για κάθε $ z \in (x\minus{}2^{k\plus{}1},x]$ θα έχουμε $ z_k \equal{} 1$ και άρα $ A_k \subseteq B_z$. Άρα η μπάλα με κέντρο το $ (x,y)$ και ακτίνα $ 2^{\minus{}k}$ θα περιέχεται στο $ V$.

Και το πρόβλημα πολύ όμορφο και η λύση του Δημήτρη εξίσου όμορφη. Θα ήθελα να προσθέσω απλά μια μικρή απλοποίηση στην ιδέα του Δημήτρη. Έστω
\[ X \equal{} \{0,1\}^\mathbb{N}\]
Εργαζόμαστε στο χώρο   $ X\times\mathbb{R}$ . Αν $ \{U_k\}$ μία αρίθμηση των ανοικτων διαστημάτων του$ \mathbb{R}$ με ρητά ακρα τότε το σύνολο:
\[ B \equal{} \{((a_n),x): x\in\cup_{n: a_n \equal{} 1}A_n\}\]
είναι ανοικτό όπως έπεται πολύ εύκολα από τον ορισμό της τοπολογίας του $ X$.  Το $ X$ όμως είναι ομοιομορφικό με το σύνολο Cantor $ C$. Συνεπώς το $ B$ είναι ομοιομορφικό με ένα ανοικτό υποσύνολο του $ C\times \mathbb{R}$, το οποίο πρέπει να είναι η τομή ενός ανοικτού του $ \mathbb{R}^2$ με το $ C\times\mathbb{R}$,  και τελειώσαμε.

Όπως και με την αρχική απόδειξη του Δημήτρη, η απόδειξη αυτή γενικεύται τετριμμένα για το $ \mathbb{R}^n$.


geometry
3D geometry
arithmetic sequence
number theory unsolved
number theory

[quote="knol"]Find an infinite nonconstant arithmetic progression of positive integers such that each term is not a sum of two perfect cubes. Determine if an infinite of such progressions exist.[/quote]

Since no sum of two perfect cubes is $ 3$ or $ 4$ modulo $ 7$, any arithmetic progression such that $ 3\plus{}(7a)n$ or $ 4\plus{}(7a)n$ fits the requirements.


I have KazaaLite, and I heard they now charge money for downloading...
so I figure thats why theres not as many people as there used to be

Anyone know of any good P2P programs??
One without SPYWARE and viruses (or is it virii)?

AoPS - MathLinks does not encourage the use of such means of downloading files. This topic is closed, and it's not deleted because I ask all of you to not talk about P2P, filesharing, warez or other kind of such activities on this forum. Not respecting this amendment may lead to your banning of the forums.


what exact types of music do you guys like? any lifehouse fans? or are you more into... what exactly?

I like classical music although I can listen to a lot of other types of music.  I mainly like concertos, sonatas (including other works such as noctures, waltzes, etc...), and symphonies especially by Beethoven, Chopin, Mozart, and some others.

-interesting_move

I like Rock (yeah, I know...weird!).  My two favorite bands are Counting Crows and Dave Matthews Band.  Any other DMB fans out there?

Oh, and old stuff like The Beatles are good, too.  <Hums "Carry that Weight" to self>

Classical, trance and progressive house.

I only listen to classical, and it's usually cello music.

Yes, I do play the cello. I play the piano also, but I prefer listening to cello music most of the time.

wow classical for everybody huh? anybody go off into celtic... i mean, its still instumental...

dmb is cool yes.

Cello, that's cool.  Do you like Yo-Yo Ma? I think he's awesome.

-interesting_move

Yes, he is definitely among the best cellists. It would be awesome to be that good.

Hi from summer camp, finally got some comp access. I listen to classical on the radio every now and then, but I also like some rock, for intance  Radiohead. At CTY, my previous summer camp, we had our own canon with a bunch of awesome songs, for instance:

It's the End of the World as we Know It by REM
Istanbul by They Might be Giants
Blister in the Sun by the Violent Femmes
American Pie by Don McLean
Stairway to Heaven by Led Zeppelin
Birdhouse in Your Soul by They Might Be Giants
Like a Prayer by Madonna
Brown Eyed Girl by Van Morrison
Nightswimming by REM
Time Warp from Rocky Horror Picture Show
Forever Young by Alphaville
and so on and so forth.

-Isaac

[color=white]Yeah, andy17null...I went to that camp about two years ago.  Those songs are so familiar...yet they were so long ago...sniff  :cry: [/color]

I like Toby Mac's music. It's really in it's own class.

[quote="andy17null"]American Pie by Don McLean...[/quote]

Ah, oldies.

[quote="andy17null"]Stairway to Heaven by Led Zeppelin[/quote]

Mm.. classic.  I love stairway.

[quote="andy17null"]Brown Eyed Girl by Van Morrison[/quote]

And more oldies...

You're making me nostalgic too.  Although I'd have to be 4 times my age to actually remember those when they first came out...still, I like oldies.  Anyway, there's another ramble set down on paper.

Shaddup  :D, they may be oldies but they're still good. And they're not all I like...

 :oops:  :oops:  :oops:

Does anybody like [i]any[/i] Eminem? The only song that I think is any good is Sing for the Moment and even that is pushing the limits of "good." As said in a much earlier post, I also like Final Fantasy music.

I like FF music, classical makes me wish I could dance except I can't and soft rock/pop is okay. Maybe I should put it this way - I listen to anything that isn't 'Matrix Reloaded' type. Some oldies are really good...it really depends on the individual songs. Oh - anyone like Kitaro?

[quote="andy17null"]Shaddup  :D, they may be oldies but they're still good. And they're not all I like... [/quote]

On the contrary... I like Oldies too.  Ever heard "Baker Street"...Gerry Rafferty...old but good... yeah.  'nother ramble.

eminem... i like parts of that song from 8 mile. its (surprisingly) a little positive... at the beginning. and he doesnt cuzz at the beginning much either. yet another surprising thing for him. i wish you could get the beat without his voice.

I know what you mean. Try searching for "lose yourself.mid", might find something.

-Isaac

[color=white]Yeah, try that.  What's the "Matrix Reloaded" type music?  I've never seen it...but my grandpa found this case full of mini discs that I can't play, but one of them is labeled "The Matrix Soundtrack".  If only I had a MD player...anyway, could I download it somewhere?[/color]

You could hear some of it on the official website => http://whatisthematrix.warnerbros.com/ Go to the 'soundtrack' section under Reloaded. I don't call it music, actually. More like noise... *shudders*

rippledance wrote:Classical makes me wish I could dance except I can't and soft rock/pop is okay.

I thought you said you really liked to dance.  Confusion, confusion.  Which oldies in particular?  I'm trying to think of any song that will get the "Jiffy Lube" jingle out of my head.

[hide]</kidding>  I don't even remember that jingle.[/hide]

daaang...lotsa classical music fans out there. I mean, I love classical too (beethoven, bach, chopin, mozart, schubert, etc), but I'm also crazy about every other genre. Don't tell me there aren't any alternative rock fans out there. Thursday kicks some major ass, and if haven't ever heard of them...then...BAH to you! BAH! Then there's pop, hip hop, and the norm. Opera's pretty darn good too. Oh and soundtrack music. And of course asian pop (Jay Chou esp. I'm gonna marry him!)

Alright, enough said. Punk/emo and Jay Chou lovers out there...HOLLA BACK!

hmm...I dont listen to classical...
But I do like pop/rock/alternative.
AZN pop is pretty good too :-) (Jay Chou, SHE, etc...)


Find all integral triples $ (a,b,c)$ if the roots of $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ are $ a, b,$ and $ c$.

[hide="references"]\[ a\plus{}b\plus{}c\equal{}\minus{}a\] \[ ab\plus{}bc\plus{}ac\equal{}b\] \[ abc\equal{}\minus{}c\][/hide]
Assume $ c\neq 0$, $ a,b,c\in \mathbb{Z}$
It follows that $ ab\equal{}\minus{}1$. However, when both sides of Eq.2 are multiplied by $ a$ and we sub in $ ab\equal{}\minus{}1$, the result is \[ c(a\plus{}1)\equal{}1\text{    ,    } a,c\in\mathbb{Z}\] This is a contradiction, and so it must be that $ c\equal{}0$. Application of this quickly results in $ (a,b,c)\equal{}\{(1,\minus{}2,0),(0,0,0)\}$.
That is to say, the only two monic cubic polynomials with its integer coefficients as its roots are $ x^3$ and $ x^3\plus{}x^2\minus{}2x$.


Plot the three vertices as coordinates and use the distance formula

[quote="archimedes1"]Hint
[hide]Use 30-60-90 triangle ratios[/hide][/quote]

[hide]Ya basically using that just drop prependiculars from the intersection points to the x-axis. Now you know by SAS that the triangle are congruent and using that you know that the small triangles created are 30-60-90s.

So

$x\sqrt{3}=\frac{x^{2}}{2}$
$\sqrt{3}=\frac{x}{2}$
$2\sqrt{3}=x$

Therefore one side is $4\sqrt{3}$[/hide]

[quote="sonny"][quote="archimedes1"]Hint
[hide]Use 30-60-90 triangle ratios[/hide][/quote]

[hide]Ya basically using that just drop prependiculars from the intersection points to the x-axis. Now you know by SAS that the triangle are congruent and using that you know that the small triangles created are 30-60-90s.

So

$x\sqrt{3}=-\frac{x^{2}}{2}$
$\sqrt{3}=\frac{x}{2}$
$2\sqrt{3}=x$

Therefore one side is $4\sqrt{3}$[/hide][/quote]
Or you could just halve the triangle with the y-axis..


AMC
AIME
geometry
AMC 10
AIME I
\/closed

Hello,
I have signed up for WOOT, but now I am having doubts if I will have trouble understanding the material.
Last year, I received a 139.5 on the AMC 10 and a 6 on the AIME I (this is my first time taking the AMC).  I have worked in-depth through Intro Geometry, Volume 1, and Inter Algebra and I'm currently working on Inter C+P, Acops, and Volume 2.  Since I am taking a tough schedule in school, I have decided if I don't take WOOT this year (incoming 10th grader), I will not take WOOT as a junior.  Should I take WOOT, or should I take AIME A and B classes?  Thanks.

You'll get a lot more out of WOOT (provided you take the time to work on it).


abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a group is which $ a^2\equal{}e$, the indentity for all $ a\in G$, show that $ G$ is Abelian.

Since $ a^{2}\equal{}a\circ a\equal{}e$ for all $ a\in G$, each $ a$ serves as its own inverse, so $ a\equal{}a^{\minus{}1}$.  Then $ a\circ b\circ a^{\minus{}1}\circ b^{\minus{}1}\equal{}a\circ b\circ a\circ b\equal{}(a\circ b)^{2}\equal{}e$, which means that $ G$ is abelian.

$ abab\equal{}(ab)^2\equal{}e\equal{}a^2b^2\equal{}aabb$. Multiply by $ a^\minus{}1$ from the left and by $ b^{\minus{}1}$ from the right to get $ ba\equal{}ab$.


Let [x] be the floor function. Find the largest prime p such that[(n^2)/3] = p for some integer n.

Break it up into Floor[3*(n/3)^2]

Does that help?

Hmmm..not really.  :(

This problem is pretty weird. I mean, the floor function is pretty weird itself, but combine that with primes, and..whew!

Ok, well think about it

IN order for Floor[3*(n/2)^2] to be a prime, what must be true of (n/2)^2?

Well, it can't be an integer (and thus 2 cannot divide n) since then the whole expression would be divisible by 3 and n/2.

Um, sorry a typo on my part.
I meant (n/3)^2

if the expression itself is a prime, then there is only one possible value for n:1.

Now, try what happens when the entire expression isn't a prime.
But u kno the difference between that number and a prime must be less than 1.

Now, prove that n can only equal 4.

[quote]if the expression itself is a prime, then there is only one possible value for n:1. [/quote]

Huh? Wouldn't the expression be (1^2)/3 = 1/3?

[quote]Now, try what happens when the entire expression isn't a prime. But u kno the difference between that number and a prime must be less than 1. [/quote]

The entire expression being 3*(n/2)^2, I assume. Hmm, ok, but I don't get "u kno the difference between that number and a prime must be less than 1".

OK, first i meant n=3. Sorry, i meant (n/3)^2 is equal to one, which makes n=3.

Second, the expression is 3*(n/3)^2, not 3*(n/2)^2

Third, since we know that if 3*(n/3)^2 is an integer, the only possible floor of that is 3. Now, this means we need to explore when 3*(n/3)^2 isn't an integer, but it needs to be prime. Which means that p<3*(n/3)^2<p+1. Do you see why?

Because if 3*(n/3)^2 is more than one away from a prime, the floor of it won't be a prime!!!

Ok, I get it now. Sorry, I didn't realize that you were assuming the entire expression was a prime. Anyways, I get why p<3*(n/3)^2<p+1 must be true now, but I am stuck there.

n is an integer. We are looking at the floor of (n^2)/3. Obviously, some integer can be one of three different residue classes mod 3. If n^2 is congruent to 0 mod 3, then the floor of (n^2)/3 is simply equal to (n^2)/3, and n^2=3p. If it is congruent to 1, the floor is (n^2 - 1)/3, hence n^2 - 1=3p. If it is congruent to 2, the floor is (n^2 - 2)/3.

We know that if p > 3, n^2 is not 3p. It is also easy to verify that a perfect square cannot be congruent to 2 mod 3. So n^2 - 1 = 3p. n^2 - 1 can be factored as (n-1)(n+1). Since p is prime, p | n - 1 or p | n + 1, likewise for 3. So 3 = n - 1, n = 4, and p = 5.

teratomato: ok, I understand your solution, except I am not sure of the proof that "It is also easy to verify that a perfect square cannot be congruent to 2 mod 3." Is there a quick one that you could post?

Try it out fiery:

n^2=n*n

Now, n can only be congruent to 0,1,2 (mod3)

Now, if n=0(mod3), then n^2=0(mod3)
if n=1(mod3), then n^2=1(mod3)
if n=2(mod3), then n^2=1(mod 3)

So n^2 cannot be congruent to 2(mod3)

Ah, ok. I should really brush up on my number theory. Thanks.

No prob

Elementary number theory is bout all the number theory i can do...

Here's the full solution if you want it

[hide]

We already know that n is either 1 or 0 mod 3. If n is 0, then the only number it can be is 3 ( do you see why?) So either p = 3 or there's a different answer.

If there is a different answer, n:^2: can be expressed as 3p+1

n:^2:-1 = 3p

(n+1)(n-1) = 3p

We're working with integers so it's pretty simple.

n-1 = 3

n+1 = 5

And therefore p =5

Aww man what a beautiful problem.

[/hide]


Let $\frac{1}{9}\leq x\leq 1$ find the minimum and the maximum of the expression
$A=(1-\frac{1}{9x})^{2}+(1-x)^{2}$

Expanding, collecting like terms:
$x^{2}-2x+2-\frac{9}{2x}+\frac{1}{81x^{2}}$
Take the derivative:
$2x+frac{2}{9x^{2}}+\frac{2}{81x^{3}}-2$
Find the roots of the derivative.
The minimums over the interval given are A = $\frac{7}{9}$
The maximum over the interval give is A = $\frac{8}{9}$


inequalities
function
inequalities unsolved

Let $ a$ be a positive real number, show that
\[ (a^4\plus{}a^2\plus{}1)(a^4\plus{}a\plus{}1) \ge \frac{3}{4}a^2(a\plus{}2)^2\]

[quote="Rijul saini"]Let $ a$ be a positive real number, show that
\[ (a^4 \plus{} a^2 \plus{} 1)(a^4 \plus{} a \plus{} 1) \ge \frac {3}{4}a^2(a \plus{} 2)^2\]
[/quote]

$ (a^4 \plus{} a^2 \plus{} 1)(a^4 \plus{} a \plus{} 1) \ge \frac {3}{4}a^2(a \plus{} 2)^2$

$ \Longleftrightarrow a^8 \plus{} a^5 \plus{} \frac {5}{4}a^4 \plus{} a^6 \minus{} 2a^3 \minus{} 2a^2 \plus{} a \plus{} 1\geq a^5 \plus{} a^4 \minus{} 2a^3 \minus{} 2a^2 \plus{} a \plus{} 1\equal{}(a\minus{}1)^2(a\plus{}1)^3\geq 0$

[quote="dgreenb801"]I don't think that's right because equality doesn't hold in the original inequality when $ a \equal{} 1$.[/quote]
Actually, I had the same apprehensions...
But I guess he cut out some terms...so the inequality he uses must be strict...
But still I haven't yet checked if his expansion is true in the first place...

just expansion and use of AM-GM gives that LHS is exactly bigger. the RHS seems very small.

@rijul, the ques is perfectly right!!!

its quite easy as well, expanding the inequality and a little manipulationg gives us:-
(a^2-1)^2 + (a^4+a^2/2)^2 + a(a^2-1)^2 >=0

which is obviously true as a is positive[/code]

the problem is that in test almost EVER1 thinks cauchy..AM-GM..power mean,..tchebychef...rearrangement etc etc..NOBODY thinks about the good old
x^2 > 0 ... :p ....thats y i'm sure almost every1 tried using the above mentioned inequalities..not the easy old showing expression as sum of squares...BTW can any1 show (this is a kinda request to arvind/akashnil) whether the mentioned function in the ineq(let it be f(a) ) is convex or concave in the interval (0,1/2)

[hide="Shorter Way"]

$ (a^4 \plus{} a^2 \plus{} 1)(a^4 \plus{} a \plus{} 1) \equal{} (a^2 \minus{} a \plus{} 1)(a^2 \plus{} a \plus{} 1)(a^4 \plus{} a \plus{} 1)$.

Now, $ a^2\minus{}a\plus{}1 \equal{} \left (a\minus{}\frac{1}{2} \right)^2 \plus{} \frac{3}{4} \ge \frac{3}{4}$.

Therefore,

$ (a^4 \plus{} a^2 \plus{} 1)(a^4 \plus{} a \plus{} 1) \equal{} (a^2 \minus{} a \plus{} 1)(a^2 \plus{} a \plus{} 1)(a^4 \plus{} a \plus{} 1) \ge \frac{3}{4} (a^2 \plus{} a \plus{} 1)(1 \plus{} a \plus{} a^4)$.

Now, by Cauchy-Scwarz,

$ (a^2 \plus{} a \plus{} 1)(1 \plus{} a \plus{} a^4)  \ge (a\plus{}a\plus{}a^2)^2 \equal{} a^2(a\plus{}2)^2$,

which completes the proof.

[hide="Note"]
It is the original solution provided. My solution involved brute force followed by SOS.  :what?:
[/hide]
[/hide]

[quote="sahilsharma94"]what is SOS?[/quote]

Sum Of Squares which sometimes Saves Our Souls!


If $ 0\rightarrow A\rightarrow A\oplus B\rightarrow B\rightarrow 0$ is any exact sequence of finite abelian groups then the induced sequence $ 0\rightarrow Hom(B,A)\rightarrow Hom(B,A\oplus B)\rightarrow Hom(B,B)\rightarrow 0$ is exact.

$ Hom$ is left exact so one only needs to prove that the sequence is exact at $ Hom(B,B)$.

Take a look at the sequence $ 0\to B\to A\oplus B\to A\to 0$ induced by the above splitting.

It is not given that the first sequence is split exact. It just happens that the middle term is $ A\oplus B$. Otherwise we would not need the groups to be finite.


calculus
integration
limit
logarithms
calculus computations

Find $ \lim_{a\rightarrow\infty}\int_{a}^{a\plus{}1}\frac{x}{x\plus{}\ln x}\ dx$.

[hide="A possible solution"]Let $ f(x) \equal{}\frac{x}{x\plus{}\ln x}$. For large $ x$, $ \frac{d}{dx}\ln f(x) \equal{}\frac{\ln x\minus{}1}{x (x\plus{}\ln x)}> 0$ and $ f$ is increasing.

Then $ f(a)\leq\int_{a}^{a\plus{}1}f(x)\; dx\leq f(a\plus{}1)$ and by squeezing theorem, this limit converges to $ 1$.
[/hide]

[hide]$ f(x) \equal{}\frac{x}{x\plus{}\ln x}\equal{} 1\minus{}\frac{\ln x}{x\plus{}\ln x}$
$ \lim_{x\to\infty}f(x) \equal{} 1$

Hence, $ \lim_{a\to\infty}\int^{a\plus{}1}_{a}\frac{x}{x\plus{}\ln x}dx \equal{}\lim_{a\to\infty}\int_{a}^{a\plus{}1}1 dx \equal{} 1$[/hide]


Euler
number theory proposed
number theory

Hi

Show that for every integer n>0 we have n|phi(2^n-1)

Bye

I'm pretty sure it was posted before. Try searching for it.

The result remains true even if we replace 2 by any integer $a \geq 2$ :

Since $a$ is coprime with $a^n - 1$ it is inversible in $E = Z/(a^n - 1)Z$ (and its inverse is clearly $a^{n-1}$).
The order of $a$ in $E$ is $n$ since for each $0 < m < n$ we have $1<a^m < a^n$ so that $a^m \neq 1$ mod[$a^n - 1$].
But, from Fermat-Euler theorem, we know that $a^ { \varphi (a^n - 1)} =  1$ mod[$a^n - 1$], thus $\varphi (a^n - 1)$ is a multiple of $n$.

Pierre.

it is really easy one u can prove that k|phi(a^k-1) use order of numbers it is almost trivial.


find all n nonnegative positive integers such that
            [b] n l 2^(n-1)+1[/b]

[quote="epsilonist"]You can use the extreme principle on $ n$, it works.

Suppose that $ n_{0}$ be the least positive integer $ \geq 2$
...  :)[/quote]
and then what we must do?


trigonometry
function
logarithms
complex numbers

How to solve for $ x$? $ x\equal{}\frac{10^{\minus{}x} \plus{} 10^x}{2}$

this definitely should not be in 'basics'...

[hide="Answer"]

[u]Non-negative x:[/u]

$ \frac {10^{ \minus{} x} \plus{} 10^x}{2} > \frac {10^{x}}{2} \equal{} 5 \cdot 10^{x \minus{} 1} > e \cdot e^{x \minus{} 1} \equal{} e^x \geq x \plus{} 1 > x$

[u]Negative x:[/u]

$ \frac {10^{ \minus{} x} \plus{} 10^x}{2} > 0 > x$

No real solutions.[/hide]

thanks! by the way, do you know how to delete a post? thanks! (i know that i should click on the edit/delete but what next?)

I don't know what to do in MathLinks mode, but in AoPS mode, there is a separate delete button which is visible to the poster of the most recent post in the thread, unless it is the original post. Once someone posts after your post, you can no longer delete that post.

Alternatively, you can ask a moderator to delete it.

Clearly, no real solutions.  But what about imaginaries/complex?  The first thing I noticed is we can rewrite this as $ x\equal{}\cosh(\ln(10)x)$ or if $ x\equal{}iy$ we have $ iy\equal{}\cos(\ln(10)y)$ which is impossible for real $ y$'s.  So the only solutions are complex.

If we let $ x\equal{}a\plus{}bi$ then we have $ a\plus{}bi \equal{} \frac{e^{2ln(10)a}(\cos b \plus{} i \sin b) \plus{} (\cos b \minus{} i \sin b)}{2e^{ln(10)a}}$ hence $ a \equal{} \frac{(e^{2ln(10)a}\plus{}1)\cos b}{2e^{\ln(10)a}}$ and $ b\equal{}\frac{(e^{2ln(10)a}\minus{}1)\sin b}{2e^{\ln(10)a}}$... I'm not sure if these equations are solvable.

There's a solution at $ x \equal{} 0.24814 \plus{} 0.589191i$. I found it using Mathematica. I don't think it's expressible in terms of elementary functions, though.

I made a mistake: I forgot to include the $ \ln 10$ with $ b$ in my algebra.  So the real equations are: $ a \equal{} \frac {(10^{2a} \plus{} 1)\cos (\ln(10) b)}{2\cdot 10^{a}}$ and $ b \equal{} \frac {(10^{2a} \minus{} 1)\sin (\ln(10) b)}{2\cdot 10^{a}}$.

TheAmazingOne, how can you do this in mathematica?  I just got mathematica, and really have no clue how to use it.  I tried the Solve command on the original equation, but it never gave an answer.  Either way, though, the original equation is impossible to solve algebraically.  The only thing I can think of is that the solution must satisfy the series

$ x \equal{} \frac {1}{2}\bigg(\sum_{n \equal{} 0}^{\infty}\frac {(\ln(10)x)^n}{n!} \plus{} \sum_{n \equal{} 0}^{\infty}\frac {( \minus{} \ln(10)x)^n}{n!}\bigg) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {(\ln(10)x)^{2n}}{(2n)!}$

Mathematica can't do it with Solve, so I used FindRoot, which tries to find a numerical answer. If you put in the initial guess as a complex number, it will try to solve it numerically in the complex plane. So the command was

FindRoot[2 x == 10^(-x) + 10^x, {x, 1 + I}]

I believe there are an infinite number of values based on entering different start values (I, 2I, 3I, ...)

Yeah, there does appear to be an infinity of solutions, but Mathematica even told me that there is a solution that has no real part when i gave the start value 5I, but it is obvious that this is not the case, so I don't think its approximations can distinguish between "close to" a "root" that really is not a root and an actual root.


Simplify and express your answer in simplest radical form:
$ \sqrt[5]{32^2 \plus{}16^3\plus{}8^4\plus{}4^5}$

You add up the sums to get $ 10240$.  Factor that into $ 1024 \cdot 10$.  Then, you immediately can come to the answer: $ 4\sqrt[5]{10}$

Which I think you have to type as:

4\sqrt[5]{10}

[quote="ernie"]You add up the sums to get $ 10240$.  Factor that into $ 1024 \cdot 10$.  Then, you immediately can come to the answer: $ 4\sqrt [5]{10}$

Which I think you have to type as:

4\sqrt[5]{10}[/quote]
I thought it was $ 1024\sqrt[5]{10}$

[quote="myyellowducky82"][quote="ernie"]You add up the sums to get $ 10240$.  Factor that into $ 1024 \cdot 10$.  Then, you immediately can come to the answer: $ 4\sqrt [5]{10}$

Which I think you have to type as:

4\sqrt[5]{10}[/quote]
I thought it was $ 1024\sqrt [5]{10}$[/quote]

No, cuz when you take out a number out of a radicand (if that's what you call it, I don't recall), you have to change that number back into its original form.

Thus you would have to find $ \sqrt[5]{1024}$, which happens to be $ 4$.


induction
modular arithmetic
algebra
binomial theorem

$19^{n}=(12+7)^{n}$. Now use Newton's binomial formula ...

[quote="julien_santini"]Now use Newton's binomial formula ...[/quote]

When writing a solution like this, please try to explain any formulas you're alluding to and certainly complete the proof if possible. And this applies to everyone. Thanks.

Generally, $a^{n}\equiv b^{n}\mod m$ iff $a \equiv b \mod m$.

Sketch of Proof: WLOG let $a>b$. Thus, $a=b+km$ for some integer k. When we raise this to the power n using the binomial theorem, all of the terms will contain an $m$. Except one, and that's the $b$.

[quote="julien_santini"]Newton's binomial formula ...[/quote]

Unnecessary.  A simpler lemma suffices:

[b]Lemma:[/b]  $a \equiv b \bmod m, c \equiv d \bmod m \implies ac \equiv bd \bmod m$

Ergo, $a \equiv b \bmod m \implies a^{n}\equiv b^{n}\bmod m$.

[quote="mathnerd314"]Generally, $a^{n}\equiv b^{n}\mod m$ [b]iff[/b] $a \equiv b \mod m$.[/quote]

Quite untrue.  For example, $(-1)^{2}\equiv (1)^{2}\bmod p$ for any integer $p$ but it does not follow $-1 \equiv 1 \bmod p$.

It's an if, not an iff.

$a \equiv b \pmod{m}\rightarrow a^{n}\equiv b^{n}\pmod{m}$.


How does a computer generate random numbers? Can it generate numbers perfectly randomly, or with some flaw?

What about a calculator?

The standard is [url=http://en.wikipedia.org/wiki/Pseudorandom_number_generator]pseudo-random[/url]. For "perfect" randomness, there are ways to use environmental noise. Those can only produce random numbers at a limited rate, so they aren't used for most applications.


The quadratic equation $ x^{2}\plus{}5x\plus{}7\equal{}0$ has roots $ \alpha$ and $ \beta$. Find an equation with roots $ 2 \alpha$ and $ 2 \beta$.

One way to solve this is $ u\equal{}2 \alpha$ or $ u\equal{}2 \beta$ means $ \alpha \equal{} \frac{u}{2}$ or $ \beta \equal{} \frac{u}{2}$, so $ \frac{u}{2}$ satisfies the given equation and sustituting and simplifying gives $ u^{2}\plus{}10u\plus{}28\equal{}0$.

Use the same method to get from the equation $ x^{2}\plus{}4x\plus{}7\equal{}0$ with roots $ \alpha$ and $ \beta$ to an equation with roots $ \alpha \plus{} 2 \beta$ and $ 2 \alpha \plus{} \beta$.

[hide="OR, just use Vietas"]By Vieta's formulas,

$ \alpha \plus{} \beta \equal{} \minus{} 4 \implies (\alpha \plus{} 2 \beta) \plus{} (2\alpha \plus{} \beta) \equal{} \minus{} 12$

$ \alpha \cdot \beta \equal{} 7 \implies (\alpha \plus{} 2 \beta)(2\alpha \plus{} \beta) \equal{}  2(\alpha^2 \plus{} \beta^2) \plus{} 5\alpha \cdot \beta$

$ \equal{} 2(\alpha^2 \plus{} \beta^2) \plus{} 35$

$ \equal{} 35 \minus{} 2[ 4(\alpha \plus{} \beta) \plus{} 2(7)]$

$ \equal{} 39$

Hence, the equation we are looking for is $ \boxed{x^2 \plus{} 12x \plus{} 39 \equal{} 0}$.[/hide]

Is it possible to solve this problem without manually finding the roots or using Vieta's?

[hide="Using Substitution w/ Vieta Simplification"]
We have $ \alpha \plus{} 2\beta \equal{} \beta \minus{} 4$ and $ 2\alpha \plus{} \beta \equal{} \alpha \minus{} 4$ from $ \alpha \plus{} \beta \equal{} \minus{}4$. Substitute $ y \equal{} x \minus{} 4\implies x \equal{} y \plus{} 4$ to get $ y^2 \plus{} 8y \plus{} 16 \plus{} 4y \plus{} 16 \plus{} 7 \equal{} 0\implies \boxed{y^2 \plus{} 12y \plus{} 39 \equal{} 0}$.
[/hide]

Thanks both of you. I did exactly the same as TachyonPulse but was bothered because I'd used the sum of the roots is $ \minus{}4$ as I couldn't see any way of avoiding it.


algebra
polynomial
number theory unsolved
number theory

Determine all positive integers $ m$ satisfying the following property: There exist prime $ p_m$ such that $ n^m\minus{}m$ is not divisible by $ p_m$ for all integers $ n$

Let $ m \equal{} d^j$ with $ j$ as large as possible (i.e. $ i$ is not a perfect power). Let $ i$ be the product of all prime divisors of $ d$.

Now consider the $ i$th cyclotomic polynomial in $ d$. Since all prime divisors are $ \equiv 1( mod$ $ d)$ but not $ \equiv 1(mod$ $ di)$, it has a prime divisor $ q$ such that $ x^d \equiv d(mod$ $ q)$ has no solution.
Hence, if $ x^m \equiv m (mod$ $ q)$ has a solution, $ x^d\equiv d(mod$ $ q)$ must have a solution, which is not possible.

Typing error. I meant every prime divisor is $ \equiv 1(mod$ $ i)$ rather than $ 1(mod$ $ d)$.
I think we have proved a slightly better result than $ x^m\equiv m$. The prime $ q$ is such that for every prime divisor $ q_1$ of $ m$ such that $ q_1^{k_1}|m$, $ q_1^{k_1 \plus{} 1}\not|m$, $ x^{q_1^{k_1}}\equiv m(mod$ $ q)$ has no solution.

However, showing that there are infinitely many $ q\equiv 1(mod$ $ m)$ such that $ x^m\equiv m(mod$ $ q)$ has no solution seems rather difficult.

$ n^m\minus{}m\mod p$ periodic, and had $ \frac{p\minus{}1}{(p\minus{}1,m)}\plus{}1$ means. Obviosly, if $ p|m$ then for $ p|n$ we get $ p|n^m\minus{}m$. Therefore $ p_m\not |m$. If $ d\equal{}(p\minus{}1,m)\equal{}1$, then exist $ n$, suth that $ p|n^m\minus{}m$.
I think for any m>1 exist prime $ p\equal{}1\mod m$, suth that $ p\not |(m^{(p\minus{}1)/m}\minus{}1)$. But I don't prove it. It give $ p_m\equal{}p$.
If $ m\equal{}2\mod 4$, it is easy to prove, that exist prime $ p\equal{}m\plus{}1\mod 2m$, suth that $ p\not |(m^{(p\minus{}1)/2}\minus{}1).$ It give $ p_m\equal{}p$.


search
function
real analysis
real analysis unsolved

Construct a non-empty perfect set in $\mathbb{R}$ consisting only of irrational numbers.

already posted before. please use the search function .

1) covering the rationals by small open intervals, you can get a closed uncountable subset $A\subset \mathbb R\setminus \mathbb Q$
2) Let $B$ be the set of all condensation points of $A$.


Propose a structure, together with a mechanism that explains its formation, for compound C below:

$ Oxazole \plus{} \ce{CH2 \equal{} CHCN}/Et_3N/80^oC \longrightarrow C\,\,(C_6H_4N_2)$

[hide="my guess"]
Is this reaction something like diels alder [4+2] addition?
[/hide]

If that's your guess, why don't you continue and propose a reasonable structure? Why stop in the middle and ask "Is this correct?".


Given $ a, b, c \geq\ 0$. Prove that:
$ \frac{3a^3\plus{}abc}{b^3\plus{}c^3} \plus{} \frac{3b^3\plus{}abc}{c^3\plus{}a^3} \plus{} \frac{3c^3\plus{}abc}{a^3\plus{}b^3} \geq\ 6$

Some friends asked me about this ineq (sorry, I can't find the link). So, I post my proof in this topic  :oops:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3a^3 \plus{} abc}{b^3 \plus{} c^3} \plus{} \frac {3b^3 \plus{} abc}{c^3 \plus{} a^3} \plus{} \frac {3c^3 \plus{} abc}{a^3 \plus{} b^3} \geq\ 6$
[/quote]
After expanding we need to prove that
$ \sum_{cyc}(3a^9 \minus{} 3a^6b^3 \minus{} 3a^6c^3 \plus{} a^7bc \plus{} 3a^4b^4c \minus{} a^3b^3c^3)\geq0,$ which is obvious.

Your solution has a flaw.
In 4th line you swapped an AM GM.
But AM GM works for POSITIVE reals only.
The condition states: $ a, b, c\ge0$
So you should consider the case of $ a, b, c\equal{}0$     which is a tough job!? :huh:

[quote="Potla"]Your solution has a flaw.
In 4th line you swapped an AM GM.
But AM GM works for POSITIVE reals only.
The condition states: $ a, b, c\ge0$
So you should consider the case of $ a, b, c \equal{} 0$     which is a tough job!? :huh:[/quote]
Right!Thank you for reminding :)
but if $ a\equal{}0$ the ineq is trivial

Find the greatest value of k:
$ \sum\frac{ka^3\plus{}abc}{b^3\plus{}c^3} \ge \frac{3(k\plus{}1)}{2}$

[quote="babyloverain"]Find the greatest value of k:
$ \sum\frac {ka^3 \plus{} abc}{b^3 \plus{} c^3} \ge \frac {3(k \plus{} 1)}{2}$[/quote]
$ k \equal{} 3$ is the greatest value, babyloverain. With my ineq, equality occur when $ a \equal{} b \equal{} c$ or $ a \equal{} b$;$ c \equal{} 0$  :wink:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3a^3 \plus{} abc}{b^3 \plus{} c^3} \plus{} \frac {3b^3 \plus{} abc}{c^3 \plus{} a^3} \plus{} \frac {3c^3 \plus{} abc}{a^3 \plus{} b^3} \geq\ 6$

Some friends asked me about this ineq (sorry, I can't find the link). So, I post my proof in this topic  :oops:[/quote]

Can you put the document in PDF?
Dear [b]Nguoivn[/b] the world uses pdf.... thanks


show that at least one subset of four elements from any seven integers has a sum that is divisible by 4. Is there a particularly elegant/short solution without going through every case? is there a generalization?

I believe it is a theorem of Erdos that from any 2n-1 numbers one can pick n whose sum is divisible by n.  Sadly I've yet to hit upon a proof.

Bugzpodder, yes, there is an elegant solution to your problem.  It uses the following easy result:

(*) Given 3 or more integers, 2 of them have a sum that is divisible by 2.

Here is the solution to your problem in spoiler: [hide]Suppose we have 7 integers.  By (*), two of them (a and b) add up to 0 mod 2.  By (*) again, two of the remaining (c and d) add up to 0 mod 2.  By (*) yet again, two of the remaining (e and f) add up to 0 mod 2.  Now apply (*) one last time to (a+b)/2, (c+d)/2, and (e+f)/2.[/hide]

Alison, yes, that generalization is due to Erdos, Ginzburg, and Ziv.  The hard part is to prove it when n is a prime.  I've seen proofs of the prime case using Fermat's Little Theorem.  Once you've proved the prime case, the general result follows pretty easily (just like I derived the case n = 4 from the case n = 2 above).

[quote="Ravi B"]Bugzpodder, yes, there is an elegant solution to your problem.  [/quote]

beautiful... exactly what i was looking for

Okay, I will post a proof for n = 3 below.  Then I will challenge you to generalize the proof to any prime n.

Spoiler: [hide]Let the 5 numbers be a, b, c, d, and e.  Given a triple (say a, b, and c), define its weight to be (a + b + c)2.  Note that if a triple adds to 0 mod 3, then its weight is 0 mod 3.  Otherwise, by Fermat's Little Theorem, its weight is 1 mod 3.

Consider the sum of the weights of all 10 (unordered) triples.  In other words, consider the sum

(a + b + c)2 + (a + b + d)2 + ... + (c + d + e)2.

If we expand all these terms and add them, we get

6(a2 + b2 + ... + e2) + 6(ab + ac + ... + de).

Thus the sum of the weights is 0 mod 3.

But all the weights are either 0 or 1 mod 3.  Thus the number of weights that are 1 mod 3 must be 0 mod 3.  Thus, because there are 10 weights, the number of weights that are 0 mod 3 must be 1 mod 3.  Hence the number of triples that add to 0 mod 3 must be 1 mod 3.  In particular, there is at least one triple that adds to 0 mod 3.

[/hide]