Hi,

Regarding Marius Cavachi and Mihaly Bencze I would like to have some information concerning their job, involvement with mathematical olympiad, journals and books. Thanks.

The first one is a teacher in Constanta, and I guess that either of amfulger or pliviu can tell you more about him. He generally proposed some nice and very hard problems to mathematical olympiads.

The second one is a teacher in Brasov, and is currently involved in Number Theory. He proposed a big number of problems in different mathematical journals, and also he is the chief editor of the Romanian-Hungarian journal Octogon. He is part Hungarian, part Romanian.

Hi pliviu and amfulger,

can you provide with more information please ? :)


geometry
incenter
circumcircle
parallelogram
trapezoid
ratio
geometry proposed

[quote="limes123"]The inscribed circle of triangle $ ABC$ touches the sides $ BC,CA,AB$ in $ D,E,F$ respectively and $ I$ is the incenter of triangle $ ABC$. Let $ L,M,N$ be the orthocenters of triangles $ IBC,ICA,IAB$ respectively. Prove that:
(a) $ MN,NL,LM$ pass through $ D,E,F$ respectively;
(b) if the excircles relative to $ A,B,C$ touch $ BC,CA,AB$ in $ D_1,E_1,F_1$, the lines $ MN,NL,LM$ are right angles to $ AD_1,BE_1,CF_1$ and their respective meetpoints lie on the incircle of triangle $ ABC$.[/quote]

a, Let $ BI\cap (O)\equal{}\{O_b\}, CI\cap (O)\equal{}\{O_c\}$ then $ O_b, O_c$ are circumcenters of triangles $ AIC, AIB$, respectively.
Since $ BN\perp AI, CM\perp AI$ we obtain $ BN//CM$
On the other side, $ BN\equal{}2O_cJ, CM\equal{}2O_bJ$ then $ \frac{BM}{CN}\equal{}\frac{O_cJ}{O_bJ} (1)$
But $ \Delta O_cIO_b\sim \Delta BIC$ thus $ \frac{O_cJ}{O_bJ}\equal{}\frac{BD}{CD} (2)$
From $ (1)$ and $ (2)$ we get $ M,D,N$ are collinear, similar for $ N,L,E$ and $ L,M,F$.
b, Denote $ AD_1\cap (I)\equal{}\{U\}$. From a well-known result: $ U,I,D$ are collinear.
To prove $ MN\perp AD_1$ we will show that $ \angle BND\equal{}\angle IAU$ (because $ O_bO_c//BN$)
Since $ AI\perp BN, BD\perp IU then \angle NBD\equal{}\angle AIU (3)$
$ \frac{BD}{BN}\equal{}\frac{BD}{2O_cJ}\equal{}\frac{ID.\cot B/2}{2IJ.\cot B/2}\equal{}\frac{IU}{AI} (4)$
From $ (3)$ and $ (4)$ we have $ \Delta AIU\sim \Delta NBD$ therefore $ \angle BND\equal{}\angle IAU$.
We claim $ MN\perp AD_1$ at $ T, \angle UTD\equal{}90^o$ so $ T$ lies on $ (I)$.

[quote="shoki"]Added problem:
$ \Delta LMN$ and $ \Delta ABC$ have equal surface.[/quote]
Let $ O_a$ be the circumcenter of triangle $ IBC$.
Since $ \angle LDN\equal{}\angle AD_1B$ so $ S_{ABC}\equal{}S_{MNL}$ iff $ LD.MN\equal{}BC.AD_1 \Leftrightarrow \frac{MN}{BC}\equal{}\frac{AD_1}{LD} (1)$
$ \frac{UI}{AU}\equal{}\frac{BD}{DN}\equal{}\frac{DC}{DM}\equal{}\frac{BD\plus{}DC}{DN\plus{}DM}\equal{}\frac{BC}{MN} (2)$
Let $ V$ be the intersection of $ IO_a$ and $ (O_a), D'_1$ be the projection of $ V$ onto $ BC$.
It's easy to see that $ LBVC$ is a parallelogram thus $ LD\equal{}BD'1$ and $ BD\equal{}CD'_1 \Rightarrow D'_1\equiv D_1$
We get $ \frac{LD}{AD_1}\equal{}\frac{VD_1}{AD_1}\equal{}\frac{UI}{AU} (3)$
From $ (1), (2)$ and $ (3)$ we are done.

[quote="shoki"]Added problem:
$ \Delta LMN$ and $ \Delta ABC$ have equal surface.[/quote]
another solution for this is to use the fact that if $ ABCD$ is a trapezoid with $ AB\parallel{}CD$and $ E$ is the intersection of $ AC$ and $ BD$ then $ S_{\Delta BEC}\equal{}S_{\Delta AED}$.
For a better comprehension of what will be said, see the picture 28 of livetolove.
Note that $ AMBL$, $ BMCN$,and $ ALCN$ are trapezoids it follows that
$ S_{\Delta FBM}\equal{}S_{\Delta FAL}$,$ S_{\Delta BMD}\equal{}S_{\Delta CND}$,$ S_{\Delta ALE}\equal{}S_{\Delta CNE}$.
Let $ Q$ be the intersection of $ AC$ and $ LM$ then it's obvious that $ S_{QENDBF}\equal{}S_{QENDBF}$.
by adding these equalities the result directly follows.
as u saw most of this kind of problem can be solved by the fact mentioned.here is another example that its principle idea is the fact mentioned:(it should be well-known)
Let $ ABC$ be a scalene triangle. A point $ P$ is chosen in it such that $ \angle PBC\equal{}\angle PCA< \angle PAB$.The line $ PB$ intersects the circumcircle of $ ABC$ at $ B$ and $ E$ and the line $ CE$ intersects the circumcircle of $ APE$ at $ E$ and $ F$.Prove that the following ratio is constant while $ P$ varies:$ \frac{S_{APEF}}{S_{\Delta ABP}}$

If I'm not wrong, the problem you mentioned above was solved by [b]plane geometry[/b] in Mathlinks a long time ago.

More generally:
$ \mbox{Let K an arbitrary point of the plane triangle ABC. If }H_a, H_b, H_c \mbox{ is the orthocentres of } \\
\triangle{KBC}, \triangle{KAC}, \triangle{KAB}, \mbox{then }\triangle{H_aH_bH_c} \mbox{ and } \triangle{ABC} \mbox{ have equal surface.}$
([b]Neuberg[/b]; MATHESIS-1881,p.168, problem 70)


Sign up now!!! Once again, I'll need your FTW rating.

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Note: I'm not as sneaky as r15. [hide=" :D "] Which probably means r15 will help me with problems, right? [/hide]

Haha, good luck making this as pwnage as the original LTM (which was pretty good)!

Join: 2053

[quote="isabella2296"]I'll sign up. 1703.[/quote]
Aw, dang, I was going to make you a victim in the story. :D
[quote="FlyAgaric"]Hmmm... Can I sign up if my rating is 1200? Also did you have to be in the 1st LTM?[/quote]
Yes and no.

I sign up.

Too bad the first had to die!

Rating: 1658

The first one died because the players had too much work to do (e.g. finding coordinates to attack, etc.)  Hopefully Dojo won't make the same mistake.

[b]1200[/b] I want to play!

How many spots are left?

I sign up. My rating is somewhere in between 1300 and 1400, i'll see when i get to a comp that actually has a flash plugin

[quote="Dojo"]1. EggyLv.999- 1229 MRU = 229
2. isabella2296- 1703 MRU = 703
3. FlyAgaric- 1200 MRU = 200
4. AIME15- 2053 MRU = 999
5. lotsofmath- 1658 MRU = 658
6. r15s11z55y89w21- 1637 MRU = 637
7. spaceguy524- 1188 MRU = 188
8. louiethechewy123- 1037 MRU = 37
9. Chompy- 1200 MRU = 200
10.
11.
12.
13.[/quote]

can you change that to 1237? my rating dropped by 8 b/c of a memmer

Sign up, I think my rating is somewhere between 1520 and 1580... Haven't played for ageeeees

I think I can still join? If so, I join. my ftw score is 1175. Is it okay if it changes from now?

[quote="EggyLv.999"]can you change that to 1237? my rating dropped by 8 b/c of a memmer[/quote]
No, I cannot. What you sign up as is what you sign up as.

1. EggyLv.999- 1229 MRU = 229
2. isabella2296- 1703 MRU = 703
3. FlyAgaric- 1200 MRU = 200
4. AIME15- 2053 MRU = 999
5. lotsofmath- 1658 MRU = 658
6. r15s11z55y89w21- 1637 MRU = 637
7. spaceguy524- 1188 MRU = 188
8. louiethechewy123- 1037 MRU = 37
9. Chompy- 1200 MRU = 200
10. vahalla- 1218 MRU = 218
11. Bugi- 1520 MRU = 520
12. tennis123- 1175 MRU = 175
13. dragon96 - 1518 MRU = 518

wait, so if your FTW rating is too low, you cannot join? :(\

And uh, you skipped Algerbra2000 and me on DA LIST. :grimace:

EDIT: AGH, I'm not in the game.

EDIT2: oh, so MRU is figured by how much u have over 1000.

EDIT3: PLZ Let me join! You let algebra2000 join :(

[quote="EggyLv.999"]the game has begun on another thread[/quote]

wait...WHAT?!?!?!?!?!

where?


quadratics algebra polynomial Vieta LaTeX MATHCOUNTS

Two students attempted to solve a quadratic equation $x^2 + ax + b = 0$. Although
both students did the work correctly, the first miscopied the middle term and obtained
the solution set ${2, -2}$. The second miscopied the constant term and obtained the
solution set ${-1, -2}$. What is the correct solution set?
(A) ${-4, 1}$ (B) ${-2, -1}$ (C) ${-2, 0}$ (D) ${-1, -1}$ (E) ${-4, 3}$

just use vietas, this problem is so common

Yeah, really...

Btw, in $\LaTeX$ you have to put \{ if you want to actually see the brace.  Otherwise they are just compiled as delimiters.

[hide="Solution"]
When you have a quadratic equation of the form $a_2x^2+a_1x+a_0=0$ the sum of the solutions is $-\frac{a_1}{a_2}$ and the product of the solutions is $\frac{a_0}{a_2}.$  So, even though the first student miscopied the middle term the other two terms still hold and therefore we can glean the product of the solutions: $r_1r_2=2\times -2=-4$.  Likewise, we get the sum of the solutions from the second students info: $r_1+r_2=-3$.  So, from here you can plug $r_1=-3-r_2$ into $r_1r_2=-4$ and solve to find $\{1,-4\}$ is the solution set.[/hide]

Thanks.  I had that much, but I was stupid and didn't use simple algebra to plug in values and solve in the final step.  Ugghhh.

Joe,

In what forum does this type of problem belong?  This is on my level.  I can't find many problems on AoPS on my level; they are all at too high (Getting Started) or too low (MathCounts) of a level.  Should I just start in GS?

This is probably getting started level.  The content in getting started [i]should[/i] be at your level.  But it all depends on who's posting and what's being posted.  I haven't visited Getting Started in a while so I don't know what it's like in there.  But it should be a good place for you.

[quote="mysmartmouth"]You know what Shobhit, either help or don't.  I'm tired of your superior attitude when you yourself got like a 900 on the SAT in 8th grade.[/quote]

geez sorry, I had to go and wanted to give you a hint before i left. And no, it was actually 720

Vieta's is just showing that the sum of the roots is $-\text{middle term}$ and products is $\text{constant term}$ in a quadratic with lead coefficient of $1$. And you know the roots that the accidentally got and know which term they apply to....

Now it's just a matter of putting everything together :)

and knowing you, i don't think Getting Started should be difficult for you ;)

BTW, are you going to Coastal Carolina Sean?

I'll be at CCU for AC Flora unless I get suspended for skipping class.  (To go to MathCounts, wtf).

[quote="furious"][quote="mysmartmouth"]I'll be at CCU for AC Flora....[/quote]
summerville opposes any motion to strenghten ac flora's level 1 team. you should respresent your middle school.  :P[/quote]

Flora has lower division, Summerville has upper.  Who wins BY MORE will determine the outcome....


limit
function
integration
inequalities
real analysis
real analysis unsolved

[quote="harazi"]Suppose that $f_{n}$ is an orthogonal sequence in $L^{2}(0,1)$ such that for all $k$ we have $|f_{k}(x)|\leq M$ ae in $x\in (0,1)$, where $M$ does not depend on anything. Then for almost all $x$ we have $\lim_{n\to \infty}{\frac{1}{n}{(f_{1}(x)+f_{2}(x)+...+f_{n}(x))}}=0$.[/quote]

Let <,> be the scalar product. Then $<\frac{1}{n}(f_{1}+f_{2}+...+f_{n}) , f_{i})> =\frac{1}{n}<f_{i},f_{i}>\le \frac{M^{2}}{n}\forall i\ge 1$ Let $\lim_{n\to \infty}{\frac{1}{n}{(f_{1}(x)+f_{2}(x)+...+f_{n}(x))}}=f$ wich implies that $<f,f_{i}> =0$ so f is orthogonal with $f_{i}, \forall i\ge 1\implies f\equiv 0~allmost~ everywhere$

[quote="harazi"]Sorry, I'm a beginner in this field, but can you tell me why the orthogonal space of the linear space spanned by the $f_{n}$ is formed by functions zero almost everywhere? I mean, we don't assume that $f_{n}$ forms a total family here. Thank you.[/quote]

If ${f_{n}}$ not total we can reconsider the problem in $span\{f_{i}, i\ge 1, f_{i}\ne 0~ allmost~ everywhere\}$ wich is a subspace of $L^{2}(0,1)$ and in this space $f_{i}$ is a total orthogonal system.

[quote="alekk"]cckek, you proved that $\lim_{n \to \infty}\frac{f_{1}+f_{2}+\ldots+f_{n}}{n}= 0$, with convergence in $L^{2}([0;1])$. But this very different from what we are looking for. One wants to prove that for almost every $x \in [0;1]$, $\lim_{n \to \infty}\frac{f_{1}(x)+f_{2}(x)+\ldots+f_{n}(x)}{n}= 0$. (you can find a lot of examples of sequences $F_{n}\in L^{2}([0;1])$  ,even bounded with the supremum norm,  that tends towards 0 in $L^{2}([0;1])$  but such that $\lim_{n \to \infty}F_{n}(x)$ never exists)[/quote]

You're right. My mistake. Sorry

remark that one can easily prove the weaker result that for every $\epsilon > 0$, for almost every $x \in (0;1)$, $\lim_{n \to \infty}\frac{1}{n^{1+\epsilon}}(f_{1}(x)+f_{2}(x)+\ldots+f_{n}(x)) = 0$
To see that, notice that
$\int_{[0;1]}\sum_{n \geq 1}[\frac{1}{n^{1+\epsilon}}(f_{1}(x)+f_{2}(x)+\ldots+f_{n}(x))]^{2}< \infty$

If a sequence converges in $L^{2}$ then it has a subsequence that converges a.e. So we are done because it is easy to se that  $\frac{f_{1}(x)+...+f_{n}(x)}{n}$ is a Cauchy sequence a.e.

Not easy for me.  :maybe:

OK, let $F_{n}=(f_{1}+\dots+f_{n})/n$. We know that $\int_{0}^{1}F_{n}(x)^{2}\,dx\le \frac{M^{2}}{n}$, which by Chebyshev's inequality implies $|\{x\colon |F_{n}(x)|\ge\epsilon\}|\le \frac{M^{2}}{n\epsilon^{2}}$. By the [url=http://en.wikipedia.org/wiki/Borel-Cantelli_lemma]Borel-Cantelli lemma[/url] we have $\lim_{k\to\infty}F_{k^{2}}(x)=0$ a.s., because $\sum_{k=1}^{\infty}|\{x\colon |F_{k^{2}}(x)|\ge\epsilon\}|<\infty$ for any $\epsilon$. For $k^{2}<n<(k+1)^{2}$ we have $|F_{n}(x)|\le (k^{2}|F_{k^{2}}(x)|+2kM)/n$, which implies that $\lim_{n\to\infty}F_{n}(x)=0$ whenever $\lim_{k\to\infty}F_{k^{2}}(x)=0$.


number theory
divisibility tests
prime factorization

What is the sum of the positive whole number divisors of 210?

A good knowledge of divisibility tests is helpful for problems like these.

$ 210$ is divisible by $ 1$, $ 2$, $ 3$, $ 5$, $ 6$, $ 7$, $ 10$, $ 14$, $ 15$, $ 21$, $ 30$, $ 35$, $ 42$, $ 70$, $ 105$, and $ 210$. Note that you only need to check the first half to find the others.

Adding these up, $ 1\plus{}2\plus{}3\plus{}5\plus{}6\plus{}7\plus{}10\plus{}14\plus{}15\plus{}21\plus{}30\plus{}35\plus{}42\plus{}70\plus{}105\plus{}210\equal{}\fbox{576}$, interestingly enough a perfect square.

Actually, we just need to know the prime factorization:

$ 210\equal{}2 \times 3 \times 5 \times 7$, so the sum of its divisors is $ (1\plus{}2)(1\plus{}3)(1\plus{}5)(1\plus{}7)\equal{}\boxed{576}$.

May you please explain the process you used?


[quote="236factorial"]What is the 100th digit in the decimal expansion of $\frac{2}{7}$?[/quote]
[hide]
2/7=0.285714285714285714285714....

1st  2nd   3rd   4th   5th   6th
2     8       5      7      1       4

100/6=36 remainder 4.

so the 100th digit is 7.
[/hide]

[hide]the pattern is 2 8 5 7 1 4, so 100/6=Remainder 4 and the answer is 7 :idea: our answer[/hide]

[hide]
the pattern of the expansion is: .285714285714285714
100 divided by 6 is _ r4 so four digits after the point is [b]7[/b][/hide]


function
algebra
polynomial
limit
real analysis
real analysis unsolved

Let $ c$ be a complex number (or real, if you like). Prove that the differential equation
$ f'(x)\equal{}f(x)^{2}\minus{}x^{2}\minus{}c$
has a rational solution (a quotient of two polynomials, both with complex coefficients(or, again, real, if you like)) iff $ c$ is an odd integer.

I haven't solved this problem yet, but I proved that if $ f(x) =\frac{P(x)}{Q(x)}$ where $ P(x)$ and $ Q(x)$ are monics polynomials and $ P(\pn\sqrt{-c})\neq 0$ then $ c$ is a odd negative integer. In the following suppose one of such $ f$.

$ \left(\frac{P(x)}{Q(x)}\right)' =\left(\frac{P(x)}{Q(x)}\right)^{2}-x^{2}-c$
$ P'(x)Q(x)-Q'(x)P(x) = P^{2}(x)-Q^{2}(x)(x^{2}+c)$ (*)

Let $ n$ the degree of $ P$ and $ m$ the degree of $ Q$, and without lose generality suppose that $ P$ and $ Q$ are relative primes.

Suppose that $ \lim_{x\to\infty}f(x) = L <\infty$ then $ \lim_{x\to\infty}f'(x) =\lim_{x\to\infty}\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\lim_{x\to\infty}\frac{f(x+h)-f(x)}{h}= 0$, this is impossible becouse $ f(x)^{2}-f'(x) = x^{2}+c$ so $ n > m$.

All the roots of $ P$ and $ Q$ have multiplicity 1, because if $ P(x_{0}) = P'(x_{0}) = 0$ implies $ Q^{2}(x_{0})(x^{2}+c) = 0$, which is impossible ($ P$ and $ Q$ are relative primes), an similar for $ Q(x_{0}) = Q'(x_{0}) = 0$, so $ P(x) =\prod_{i = 1}^{n}(x-\alpha_{i})$ and $ Q(x) =\prod_{i = 1}^{m}(x-\beta_{i})$

In (*) LHS has degree $ m+n-1$ (The coeficient of $ x^{n+m-1}$ is $ m-n\neq 0$) but $ m+n-1 < 2n$ so in RHS $ x^{2n}$ must be cancelled with $ x^{2m+2}$, so $ 2n = 2m+2$, $ n = m+1$. But $ 2n-1 = m+n > m+n-1$ so $ x^{2n-1}$ must be cancelled too, that implies that if $ P(x)$ is $ x^{m+1}+ax^{m}+....$ then $ Q(x)$ is $ x^{m}+ax^{m-1}+....$ then dividing $ P(x)$ by $ Q(x)$ gives quotient $ x$ and dividing $ Q(x)(x^{2}+c)$ by $ P(x)$ gives quotient $ x$ too (**).

(*) is equivalent to $ Q(x)\left(P'(x)+Q(x)(x^{2}+c)\right) = P(x)\left(Q'(x)+P(x)\right)$, so there exist polinomials $ r$ and $ s$, with degree 1, such that $ Q(x)r(x) = Q'(x)+P(x)$ and $ P(x)s(x) = P'(x)+Q(x)(x^{2}+c)$. Note that $ \frac{P'(x)}{P(x)}=\sum_{i = 1}^{n}\frac{1}{x-\alpha_{i}}$ and $ \frac{Q'(x)}{Q(x)}=\sum_{i = 1}^{m}\frac{1}{x-\beta_{i}}$. Usin these results and (**) we can find that $ \frac{P(x)}{Q(x)}= x-\sum_{i = 1}^{m}\frac{1}{x-\beta_{i}}$ and $ \frac{Q(x)(x^{2}+c)}{P(x)}= x-\sum_{i = 1}^{n}\frac{1}{x-\alpha_{i}}$, multiplying the last to equations and using $ \frac{x}{x-y}= 1+\frac{y}{x-y}$ it gives $ x^{2}+c = x^{2}-2m-1+\left\{-\sum_{i = 1}^{n}\frac{\alpha_{i}}{x-\alpha_{i}}-\sum_{i = 1}^{m}\frac{\beta_{i}}{x-\beta_{i}}+\sum_{i = 1}^{n}\sum_{i = 1}^{m}\frac{1}{(x-\alpha_{i})(x-\beta_{i})}\right\}$
The term between braces tends to 0 as x tends to $ \infty$ so $ c =-2m-1$

This is a nice problem, but I haven't solve it  :( . Anybody has a diferent aproach?, please excuse my english.

In the previous post taking $ f(x)\equal{}\minus{}\frac{P(x)}{Q(x)}$ where $ P$ and $ Q$ are monics plynomials and $ P(\sqrt{\minus{}c})\neq 0$ we find that $ c\equal{}2m\plus{}1$, so, for necessary condition, we need to prove that if $ f(x)\equal{}k\frac{P(x)}{Q(x)}$ for som complex k and $ f'(x) \equal{} f(x)^{2}\minus{}x^{2}\minus{}c$ then $ k\equal{}\pm 1$ and $ P(\sqrt{\minus{}c})\neq 0$. to give the sufficient cocondition we have to prove that we can find for all $ m$ and $ k\equal{}\pm 1$, $ {\alpha_{i}}$ and $ {\beta_{i}}$ such that $ \minus{}\sum_{i \equal{} 1}^{n}\frac{\alpha_{i}}{x\minus{}\alpha_{i}}\minus{}\sum_{i \equal{} 1}^{m}\frac{\beta_{i}}{x\minus{}\beta_{i}}\plus{}\sum_{i \equal{} 1}^{n}\sum_{i \equal{} 1}^{m}\frac{k}{(x\minus{}\alpha_{i})(x\minus{}\beta_{i})}\equal{}0$


function LaTeX trigonometry algebra domain quadratics polynomial

You need to explain your notation/terminology.
1) What do you mean, "the connection?"  Are you asking for a relationship between the roots that doesn't involve alpha?  If so, do you care what form it takes?

2) What is D of a function?

[hide][quote="Po|de"]2)Can D of the function f(x)=ax^2+bx+c be a)D=2001 b)D=2002 ?[/quote]

Assuming by D you mean the discriminant ($D = b^2 - 4ac$), and a, b and c are integers. then:
$b^2 - 4ac = 2001$ ?
$a = 2$
$b = 45$
$c = 3$

Assume:
$b^2 - 4ac = 2002$ ?
when b is odd, b^2 is also odd and so D cannot be even.
when b is even, is can be written as b=2x. with x an integer:

$(2x)^2 - 4ac = 2002$
$4x^2 - 4ac = 2002$
$4(x^2 - ac) = 2002$

However 4 does not divide 2002 and so x cannot be an integer. so b is neither even or odd (or zero clearly) and this is a contradiction. so D cannot be 2002.[/hide]

:blush: Oops...

Yes, the second one is correct...in the first one, i meant the "relationship", like for example: x1 = x2 - 5, so that there will be no alpha in the equation.

[quote] If so, do you care what form it takes? [/quote]
                    What do you mean with that?

The result is said to be not "pretty".

[quote="Po|de"]1)Determine the connection between the roots of the function f(x) = x^2 - 2xcos :alpha:  + 1 - sin :alpha: , so that the connection will not include  :alpha: .[/quote]
[hide]
$f(x) = x^2 - 2x(cos \theta ) + (1 - sin \theta)$

The roots of $f(x)=0$ are expressed as:

$\frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

Adding the roots cancels the term $\sqrt{b^2 - 4ac}$ and leaves $\frac{-b}{a}$, so the sum of the roots is:

$\sum_{of}^{sum} {roots} = 2cos\theta$ (interestingly independant of c, which makes sense after some thought)

The product of roots is a similar story. The roots being conjugates, you know that a load of junk is going to cancel. Product of roots is given by:

$\sum_{of}^{product} {roots} = \frac {c}{a} = {1 - sin \theta} $

so calling the roots $p$ and $q$, so that $f(x) = (x - p)(x - q)$,

$p + q = 2cos\theta$
$pq = 1 - sin \theta$

*at this point I went out to get a glass of water and came back having forgotten what I set out to do. I hope you enjoyed my nonsense LaTeX story*
 ;)

Oh well, I did warn you  :) [/hide]
Note: Clicking the above link is likely to result in the wasting of approximately 30 seconds of your life.

:(  :(  :(  :(

Oh well........you did warn me  :P

Anyone got any ideas??? I really need this...!!! pleaseeeeeeee

Well, you have sine and cosine of an angle -- how do you think you should elliminate them?

I don't know.... :(
I tried calculating the discriminant and then the roots of the function, and then i could replace co^s 2  with 1-sin 2 , but oops, only for triangle corners...

Maybe I wrote the function wrong.... ;)

No, the function is fine.  As a general proposition, don't think about the roots right now, but say that I gave you f(x) = cos(t) and g(x) = sin(t) (or something) and I asked for you to eliminate t; how would you do it?

what?!?  :blush:  I haven't got a clue...

Do you know a formula with cos and sin it in that simplifies to something very simple?
(you've already mentioned it once).

Then just substitute f(x) and g(x) into it..

...do you mean sin 2  + cos 2  = 1? (because i only wrote this one ;) ) So... f(x) 2 + g(x) 2  = 1...but isn't that only for 0°<:alpha:<90°?

No, it's a defining property of the sine and cosine functions over their entire domain (whether that be the first quadrant, the reals, or the complex numbers).  So, express sine and cosine of alpha in terms of the roots and then plug into that equation.

I don't know what you are trying to tell me  :(

I got this function from my professor, and i'm only second grade (17 y.o.), so i wouldn't be suprissed if you are talking about an equation, that we haven't even mentioned in school. But the prof said it can be done only using the basic knowledge, that we have done (quadratic function, corner functions in a triangle, corner functions up to 180°).

Can you just put the solution. please!!!  :?

1.  Express $\cos \alpha$ in terms of the two roots $p$ and $q$.  (mikewd already provided the expression.)

2.  Express $\sin \alpha$ in terms of $p$ and $q$.  (Again, mikewd already provided the expression.)

3.  Plug both of those expressions into the identity $\cos^2 \alpha + \sin^2 \alpha = 1$.  You will then have an equation that involves only the two roots $p$ and $q$.  In other words, you will be done.

:D
But doesn't  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/16e8e52af60b1a6179558fa25be3bf4b.gif[/img] come out of Pithagoras theorem (and is therefore only valid for sharp corners)?

No.  The identity is true for all angles, not just acute angles.

You are right that, for acute angles, the identity follows directly from Pythagoras.  Nevertheless, the identity is true for all angles.


function
integration
inequalities
IMC
college contests

a) For each $1 < p <\infty$ find a constant $c_{p}< \infty$ for which the following statement holds: If $f : [-1, 1] \rightarrow R$
is a continuously differentiable function satisfying $f(1) > f(-1)$and $|f '(y)| \le 1$ for all $y \in [-1, 1]$, then there
is an$x \in [-1, 1]$ such that $f'(x) > 0$ and $|f(y)-f(x)| \le c_{p}(f'(x))^{\frac{1}{p}}|y-x|$ for all$y \in [-1, 1]$.

b) Does such a constant also exist for $p = 1$?

a) $\frac{|f(y)-f(x)|}{|y-x|} = \frac{1}{|y-x|} \left|\int_{x}^{y} f'(t)\,dt \right|$. This motivates us to use the Hardy–Littlewood maximal function.
Let $M g (x) = \sup_{\{I : x\in I \subset [-1,1]\}}\int_{I} |g(t)| \,dt $ , where $\sup$ is taken over all intervals contained in  $[-1,1] $.

It is well known that $g\to M g$ is $L^{p} [-1,1]$ bounded operator when $p>1$ .  i.e.

$ \|M g\|_{L^{p}[-1,1]} \leq C_{p} \|g\|_{L^{p}[-1,1]}$ , where $C_p$ depends only on $p$.

Now, assume on the contrary a) is not true and let $f_n \in C^{1} [-1,1] , \|f'_{n}\|_{L^{\infty}}\leq 1, f_{n}(-1) <f_{n}(1) \,,n=1,2,\ldots $ such that for each  $n$ and each $x\in[-1,1]$ with $f'_{n}(x) >0$ there exists $y\in [-1,1]$ satisfying:

$\frac{1}{|y-x|} \left|\int_{x}^{y} f'(t)\,dt \right| > n f'(x) ^{1/p}$

Denote $P_n=\{x\in [-1,1] \mid f_n'(x) > 0\} $ . We have $\int_{P_n} |f'(t)| \, dt > 1/2 \cdot \| f'\|_{L^{1}[-1,1]}$ since $f(-1)< f(1)$.

Let $g_n = f'_{n}$. It follows:

$\forall x \in P_n \,:\, ( M g_n (x)) ^{p} > n g(x) \, \Rightarrow $

$ \|M g_{n}\|^{p}_{L^{p}[-1,1]} > n \int_{P_{n}} g_{n}(t) \,dt \geq 1/2\cdot n \|g_{n}\|_{L^{1}[-1,1]}\geq 1/2\cdot n \|g_{n}\|^{p}_{L^{p}[-1,1]} $ , since $|g_n(x)| \leq 1$,

The above inequality contradicts with the boundness in $L^p$ of the Hardy–Littlewood maximal operator.

b) For $p=1$ such a constant does not exist. A counterexample can be given, which also agrees with the fact that Hardy–Littlewood maximal operator is not bounded in $L^1$.


A circular race track, with center $ O$, consists of two concentric lanes. The inner lane has a radius of $ 50$ meters; the outer lane has a radius of $ 52$ meters. The finishing points on both lanes are labeled $ A$ and $ B$ on the inner and outer lanes respectively, with $ A$ and $ B$ lying on the same radius. If the starting point on the inner lane is also at $ A$ (one time around the track only) and $ C$ is the starting point on the outer lane, find $ \angle COB$ to the nearest tenth of a degree so that the race lanes will be of equal length.

The length of the inner track is $ 100\pi$. If $ \angle COB\equal{}x^\circ$, then the length of the outer track is

$ \frac{360\minus{}x}{360}\cdot\pi\cdot104\equal{}100\pi\implies360\minus{}x\equal{}\frac{4500}{13}\implies x\equal{}\frac{180}{13}\approx\boxed{13.8}$.


modular arithmetic
group theory
number theory

[hide] isnt it just 1 because of fermats little theorem?[/hide]

Agreed with nateharman. Pretty simple, you just have to know the theorem.

Basically, anything to the n-1 power is 1 mod n, given that n is a prime number.

Or more generally, if $\gcd\left(a,n\right)=1$, then $a^{\phi\left(n\right)}\equiv 1\pmod{n}$ which directly implies Fermat's Little Theorem since $\phi\left(p\right)=p-1$ for each $p\in\mathbb{P}$.

As for a proof, the only way I know how to do it is by using a little group theory, but I'm sure there are other ways.

Those who used Fermat's Theorem were correct.  It's a very simple problem, but it looks complicated - that's why I like it.

[hide]The answer is 1 because of fermat's theorem.  $n^{x-1}=1 (\mod x)$[/hide]

[quote="seesaw"]Here's a nice problem:

What is the remainder when $191^{192}$ is divided by $193$?[/quote]

[hide]

The problem translates to $191^192\pmod{193}$
Fermat's Little Theorem states that $A^(p-1)=1\pmod{p}$
Since 193 is prime, The answer is: 1.

[/hide]


$\ a,b,c\geq0$  .Prove that
 $\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ca}}+\frac{1}{\sqrt{4c^{2}+ab}}\geq\frac{4}{a+b+c}$
          Good luck  :D


Let $ a,b,c$ be positive real numbers such that $ a \ge b \ge c$. Show that

$ \frac{a^2\minus{}b^2}{c} \plus{}\frac{c^2\minus{}b^2}{a}\plus{}\frac{a^2\minus{}c^2}{b} \ge 3a\minus{}4b\plus{}c$

This is really neat ...  :)
Cf. the attachment

wow that was fantastic.  why does the condition imply those two equations?  and also on the second pdf, why did you let $ t\equal{}0$?  thanks

[quote] why does the condition imply those two equations? and also on the second pdf, why did you let t = 0?

[/quote]

* You mean: ... those two "inequalities"?

* Just have a look at the expression appearing on LHS of  the last inequality.
It consists of nothing but "+"-terms!

* 'cause t = 0 <=> b = c
And it's always a good guess to check the cases of equal variables.
(As you probably have noticed s = 0 wouldn't work.)
And remember: We have to show in some way that x > 4 can not happen in general ...

I hope that helps


$a,b,s$,are three positive number such that $a+b=2s,s \leq  1$
prove or disprove:$[(\frac{n}{a})^k-(\frac{a}{n})^k][(\frac{n}{b})^k-(\frac{b}{n})^k] \geq [(\frac{n}{s})^k-(\frac{s}{n})^k]^2$  for $n\geq 2,k \in R_+$

looks like possible to solve with holders

i am in rush, at school, but will try to explain.

Let x =  (n/a)^k, x' = 1/x, same for y, z; this is for convenience in my writing.

Then $(x - x')(y - y') = (x+x')(y+y') - 2(x'y+xy')$

Also, $(z-z')^2 = (z+z')^2 - 4zz'$.

Now $2zz' \geq x'y + xy'$ is easily proved.  So we just need the other part.

Now by holders, $(x+x')(y+y') \geq (\sqrt{x/y}^k + \sqrt{y/x}^k)^2$.

So we just need $(\sqrt{x/y}^k + \sqrt{y/x}^k) \geq (z + z')$

I think this is doable, but no time to try.  Will return to this question later.

okay, returning to the question

I have messy solution

$f(x) = x + 1/x$.

Expand.  We need $f(\frac{n^2}{ab}) + f(1) \geq f(a/b) + f(\frac{n}{(a+b)/2})$.

Let x = a/n, y = b/n.  Notice these are in between 0 and 1.

We need $f(xy) + 2 \geq f(x/y) + f(\frac{x+y}{2})$.

So it is $\frac{(xy)^2 - x^2 - y^2 + 1}{xy} + 2 \geq \frac{(x+y)^2 + 4}{2(x+y)})$.

or $2(x+y)[(xy)^2 - x^2 - y^2 + 1 + 2xy)] \geq xy[x^2 + y^2 + 2xy + 4]$

or $2(x^3y^2 + x^2y^3 - x^3 - x^2y - y^3 - y^2x + x + y + 2x^2y + 2y^2x) \geq x^3y + xy^3 + 2x^2y^2 + 4xy$

This is a cubic in x, $(2y^2 - y - 1)x^3 + x^2(2y^3  -2y^2 + y) + x(+2y^2 - y^3 - 4y) + (-2y^3 + 2y) = 0$.

At x = 0, it is positive.  At x = 1, it is negative.  So it must have a root somewhere.

Maybe can use mathematica to find a root?

well,thank you $Singular$,but I think I haven't get your meaning now,have you solved it?
indeed it is come from magazine,the orginal problem is :
$a_i$ are positive numbers,such that $\sum_{i=1}^na_i=s,s \leq 1$,
prove that $ \prod_{i=1}^n((\frac{1}{a_i})^k-a_i^k) \geq ((\frac{n}{s})^k-\frac{s}{n})^k)^n$(k is a positive real number)
and it says it haven't been sloved.
if we can prove $s \leq 2$ in my problem,I think we can solve it. :)


pigeonhole principle
number theory unsolved
number theory

Consider a sequence of $ N$ positive integers containing $ n$ distinct integers. If $ N\geq 2^n$, show that there is a consecutive block of integers whose product is a perfect square.

[hide="My idea"]We separate each of these $ n$ distinct integers into $ n$ different pigeonholes. Since $ N>n$ for all $ n>0$, we can safely state that $ N>n$, since if $ n\equal{}0$, then there would be no consecutive integers.

Since $ N>n$, there must be at least 8 integers that must fit into these three pigeonholes, and at least one pigeonhole must have 2 elements. Each element in this pigeonhole is the same so we would have the square of a consecutive block of integers as requested.[/hide]

What's wrong with this?

[quote="#H34N1"]Each element in this pigeonhole is the same so we would have the square of a consecutive block of integers as requested[/quote]

I think you're only mentioning a specific case.  And anyways, I don't understand how you found the consecutive block of integers.

Here is how I would go about it:
[hide="hint"]Let the sequence of integers be $ (a_1,a_2,...,a_N)$.  Let the $ n$ distinct integers be $ q_1,q_2,...,q_n$.  Now, let $ b_k\equal{}a_1a_2\cdots a_k$.  Notice that each $ b_i$ can be represented as $ q_1^{e_1}q_2^{e_2}\cdots q_n^{e_n}$.  Consider the various parities of $ e_1,e_2,...,e_n$.[/hide]

Wait which case did I miss? What I was trying to say is that if there are at least two same elements in a hole, then there is already a perfect square.

The way you suggested is how the book does it.

But how do you know that those integers are consecutive (in the sequence)?

And I thought you were only mentioning a specific case because you used numbers (i.e. 8,3) which I thought only applied to the case $ n\equal{}3$.


analytic geometry
linear algebra
matrix
rotation
geometry
algebra
system of equations

if you know linear algebra, you can calculate the determinant of the matrix that represents the system of linear equations

if det(A) = 0, then at least one pair of equations are linearly dependent upon each other

in this case, A:

1, 1, 1
3, -2, 1
1, 6, 3

det(A) = 1(-6-6) - 1(9-1) + 1(18- -2) = -12 - 8 + 20 = 0

it becomes evident WHY if the determinant is 0, then at least one pair of
equations are linearly dependent on each other once you know the
definition of the determinant for an nxn matrix, but that may be beyond
the scope of the beginner's board.

[quote="VeCTer"]if you know linear algebra....[/quote]

If he's learning Algebra II right now then I think an explaination relying too heavily on Linear Algebra is a bit beyond his level. I think Determinants are taught in Algebra II, but the full implications of what they mean probably isn't.

But to answer your question goldenfire, you're right. It is not just that equation 3 is dependend on equations 1 and 2 - it's also true that equations 1 and 2 are depended on either of the other two.

Graphically, what this means that the three planes don't intersect in just one single point. It could mean that they meet in a line (which is in deed the case here); or it could mean that they don't meet at all (i.e. that they're all parallel), or that they meet in a plane (i.e. that they're all multiples of each other and represent the same plane).

hey vecter, are u the same vecter as from college confidential, if so, just wanted to say hey, it looks like u just joined this thing a few days after i did.

wait, i thought evaluating a system of 3 equations was part of algebra II. I just finished algebra II and we evaluated them with a program that figured out the determinents

[quote="goldenfire6"]my book solved that system in terms of z [tex](-\frac{3}{5}z+\frac{17}{5},-\frac{2}{5}z+\frac{18}{5}, z)[/tex], but for another problem like that the answer key gives the solution in terms of y? Here it is:

[tex]x-4y+z=6[/tex]
[tex]2x+y-3z=1[/tex]
[tex]4x-7y-z=13[/tex]

Combining equations (1)/(2) and (2)/(3) you get the same equation, [tex]9y-5z=-11[/tex].  It doesn't seem to be much different from the first one but the answer key says the answer is [tex](\frac{11y+19}{5},y,\frac{9y+11}{5})[/tex].  Is there a way to tell which variable to solve in terms of or would solving in terms of any variable be correct?[/quote]

Any variable would be correct.

[quote="SingTheSorrow"]In my algebra II class, we learned Cramer's rule...[/quote]

Using Cramer's rule, you get that the denominators of all three variables are 0.

[quote="goldenfire6"]my algebra II book has a problem showing how a system of 3 equations and 3 variables can have an equation dependent on the other 2:

[tex]x+y+z=7[/tex][tab]  (1)
[tex]3x-2y+z=3[/tex][tab] (2)
[tex]x+6y+3z=25[/tex][tab] (3)

Add -3 times equation (1) and add it to equation (2) to get equation (4).
Add -1 times equation (1) and add it to equation (3) to get equation (5)

[tex]-5y-2z=-18[/tex][tab] (4)
[tex]5y+2z=18[/tex][tab]   (5)

Add equation (4) to equation (5):

[tex]0=0[/tex]

The book says that shows that equation (3) is dependent on equations (1) and (2)?  How do you know that its (3) thats dependent on the others and not (2) or (1)?  How would their corresponding planes intersect in a coordinate system?[/quote]

Call your five equations E1 through E5. You derived equation 4 as
E4 = -3E1 + E2. You derived equation 5 as E5 = -E1 + E3. Finally, you
added E4 + E5 to get the all-zero's equation, which in this kind of
calculation acts like the number zero. So E4 + E5 = 0.

Substituting in for E4 and E5 gives (-3E3 + E2) + (-E1 + E3) = 0,
which simplifies to -4E1 + E2 + E3 = 0. Since a linear combination of the
original three equations can give the all-zeros equation, that system of
three equations is dependent. The relationship -4E1 + E2 + E3 = 0 says
that the dependence is reciprocal: any two of the equations E1, E2, and E3
can derive the other. So E3 is dependent on E1 and E2, E2 is dependent
on E1 and E3, and E1 is dependent on E2 and E3.

[quote="goldenfire6"]ugggh I can't get this!

Someone tell me what I did wrong here:

[tex]x-y+3z=-8[/tex]
[tex]2x+3y-z=5[/tex]
[tex]3x+2y+2kz=-3k[/tex]

I experimented with systems that were inconsistent and had infinitely many solutions and made an assumption that:

If [tex]D=0[/tex] and [tex]D_x[/tex],[tex]D_y[/tex], and [tex]D_z[/tex] all equal 0, then the equation has infinitely many solutions, and if [tex]D=0[/tex] and at least one of [tex]D_x[/tex], [tex]D_y[/tex], or [tex]D_z[/tex] doesnt equal zero, then the system in inconsistent and has no solutions.

Using my new assumption I got that if [tex]k=2[/tex], the system has no solutions because if [tex]k=2[/tex] then [tex]D_x=-14[/tex] and [tex]D=0[/tex], and when [tex]k[/tex] was anything else it had only one solution because 2 was the only value that made [tex]D=0[/tex].  But when I did some linear combination I found that if [tex]k=2[/tex], then [tex]x=-\frac{7}{5}[/tex], [tex]y=\frac{21}{10}[/tex], and [tex]z=-\frac{3}{2}[/tex], and if [tex]k=1[/tex] then there were infinitely many solutions?! Help![/quote]

I assume that [tex]D[/tex] is the determinant of the matrix of coefficients,
[tex]D_x[/tex] is the determinant of the matrix of coefficients with the
column of [tex]x[/tex]-coefficients replaced by the constants column,
[tex]D_y[/tex] is the determinant of the matrix of coefficients with the
column of [tex]y[/tex]-coefficients replaced by the constants column, and
[tex]D_z[/tex] is the determinant of the matrix of coefficients with the
column of [tex]z[/tex]-coefficients replaced by the constants column.

In that case, if [tex]k=2[/tex], then [tex]D=10[/tex] not [tex]0[/tex]. You made an arithmetic error.

To simplify the problem, you can take the determinant while leaving
[tex]k[/tex] as an unknown. For example [tex]D = 10k - 10[/tex].

The writer of this homework problem pulled a little stunt that isn't fair to
inflict on a beginner. Taking the system of equations and Guassian reducing
it to upper triangular form (I left E1 alone, replaced E2 by E2 - 2E1, and
replaced E3 by E3 - E1 - E2) gives

[tex]\begin{eqnarray*} x-y+3z &=&-8\\ 5y-7z &=& 21\\ (2k-2)z &=& 3-3k \end{eqnarray*}[/tex]

If [tex]k\ne 1[/tex], then we could divide the last equation by [tex]k-1[/tex] to get
[tex]2z = -3[/tex]. But if [tex]k=1[/tex], then the last equation is the all-zeros equation,
so the original E1, E2, and E3 have E1 + E2 - E3 = 0. In that case, the
system is dependent.

The little stunt is that changing the value of [tex]k[/tex] rotates the third plane
around an axis that contains the solution point. So that point is always a
solution. At [tex]k=1[/tex] the rotating plane rotates to contain the line of
intersection of the first and second planes, so that the solution set is the
entire line, including the point that is always a solution.

Ordinarily, changing a coefficient and a constant in a system of equations
changes the solution too. So this is a misleading example.

Erin J. Schram

Just to let people know, I cut this thread some slack because it was plainly an informational question about curricular mathematics and not a challenge problem pertaining to problem-solving competitions. But to my mind, knowing the standard curriculum in the United States and east Asia, this kind mathematical topic, solving systems of equations in three variables, is more at the Intermediate Forum level than at the Getting Started level.

The kind of homework questions I would expect to see in Getting Started Forum would usually be prealgebra (the level of AMC 8), algebra 1, or "middle school geometry" questions, no higher level than what you would see in Mathcounts.

[quote="Erin J. Schram"]The writer of this homework problem pulled a little stunt that isn't fair to inflict on a beginner. Taking the system of equations and Guassian reducing it to upper triangular form (I left E1 alone, replaced E2 by E2 - 2E1, and replaced E3 by E3 - E1 - E2) gives

[tex]\begin{eqnarray*} x-y+3z &=&-8\\ 5y-7z &=& 21\\ (2k-2)z &=& 3-3k \end{eqnarray*}[/tex]

If [tex]k\ne 1[/tex], then we could divide the last equation by [tex]k-1[/tex] to get
[tex]2z = -3[/tex]. But if [tex]k=1[/tex], then the last equation is the all-zeros equation,
so the original E1, E2, and E3 have E1 + E2 - E3 = 0. In that case, the
system is dependent.

The little stunt is that changing the value of [tex]k[/tex] rotates the third plane
around an axis that contains the solution point. So that point is always a
solution. At [tex]k=1[/tex] the rotating plane rotates to contain the line of
intersection of the first and second planes, so that the solution set is the
entire line, including the point that is always a solution.

Ordinarily, changing a coefficient and a constant in a system of equations
changes the solution too. So this is a misleading example.

Erin J. Schram[/quote]

So if [tex]k=1[/tex] then there are infinitely many solutions, but if [tex]k\ne 1[/tex] the solution is [tex](-\frac{7}{5},\frac{21}{10},-\frac{3}{2})[/tex], and it is impossible to find a value of [tex]k[/tex] that leaves no solution to the system?

[quote="goldenfire6"]So if [tex]k=1[/tex] then there are infinitely many solutions, but if [tex]k\ne 1[/tex] the solution is [tex](-\frac{7}{5},\frac{21}{10},-\frac{3}{2})[/tex], and it is impossible to find a value of [tex]k[/tex] that leaves no solution to the system?[/quote]

Yes. And when [tex]k=1[/tex], the solution set is a line that contains the point [tex](-\frac{7}{5},\frac{21}{10},-\frac{3}{2})[/tex].


[quote="The Zuton Force"]If it's [hide]$f(n)=2^{n}$[/hide] I'll post my solution[/quote]
Congratulations!  :lol:

[quote="The Zuton Force"]If it's [hide]$f(n)=2^{n}$[/hide] I'll post my solution[/quote]
Solution?????  :lol:


analytic geometry
function
combinatorics unsolved
combinatorics

On the coordinate plane $ Oxy$, we consider the point $ M(a,b)$ where $ a,b$ are positive integers and a positive integer $ n > a \plus{} b$. How many paths from $ O$ to $ M$ with the length $ n$ (we can go back or go down, just need to reach $ M$).

Clearly $ n$ must have the same parity as $ a \plus{} b$.  I claim that the number of such paths is

$ {n \choose \frac {a \plus{} b \plus{} n}{2}} {n \choose \frac {a \minus{} b \plus{} n}{2}}$.

$ \frac {a \plus{} b \plus{} n}{2}$ is equal to the total number of moves up or to the right; in a path of length $ n$, color this subset red.  Now color a second subset blue of size $ \frac {a \minus{} b \plus{} n}{2}$ and assign as follows:

- A path segment that is red and blue is right.
- A path segment that is red but not blue is up.
- A path segment that is blue but not red is down.
- A path segment that is neither red nor blue is left.

It's straightforward to verify that the number of right segments minus the number of left segments is $ a$ and that the number of up segments minus the number of down segments is $ b$.

(You can get the same result by writing the generating function as $ (x \plus{} y \plus{} x^{ \minus{} 1} \plus{} y^{ \minus{} 1})^n \equal{} (x \plus{} y)^n(1 \plus{} x^{ \minus{} 1} y^{ \minus{} 1})^n$.)


Prove using induction, that sequence is correct for all positive integers:

$ a_{1}\equal{}1$
$ a_{2}\equal{}4$
$ a_{3}\equal{}12$

$ a_{n}\equal{}4a_{n\minus{}1}\minus{}4a_{n\minus{}2}$

I would like you to show me the whole proof, I am learning now the elementary maths and I do not know how to prove it exactly by induction...

Sorry for my english...

Help, me please

I don't quite understand what you mean by "proving it correct". You have some initial conditions, and a recursion. From this you get a sequence. What do you want to prove about it?

If you want a formula, I would go with generating functions.

I hope this is not too much of a bump.

[hide="Solution"] $ a_n \equal{} 4a_{n\minus{}1} \minus{} 4a_{n\minus{}2}$ has the characteristic equation

$ x^2 \minus{} 4x \plus{} 4 \equal{} 0 \implies (x \minus{} 2)^2 \equal{} 0$

The general solution then takes the form $ a_n \equal{} A \cdot 2^n \plus{} Bn \cdot 2^n$ (the reason why is somewhat complex).  Plugging in,

$ a_1 \equal{} 2A \plus{} 2B \equal{} 1 \implies A \plus{} B \equal{} \frac{1}{2}$
$ a_2 \equal{} 4A \plus{} 8B \equal{} 4 \implies A \plus{} 2B \equal{} 1 \implies B \equal{} \frac{1}{2}, A \equal{} 0$

Hence $ \boxed{ a_n \equal{} n \cdot 2^{n\minus{}1} }$. [/hide]


How do you calculate (121 ^ 0.41)(121 ^ 0.09)?

Without using a calculator?

Amazing... I never would have thought of that.  Mainly because of not having learned that method yet.
Maybe this should be moved to at least high school basics?

Thanks! We are just learning about sqrt and cbrt, so I did not know of this!!

You don't have to change it into integers.

$ 121^{.41}\cdot121^{.09}$.

A very simple rule of exponents is that $ a^b\cdot a^c \equal{} a^{b \plus{} c}$. We apply this to get:

$ 121^{.5} \equal{} \sqrt {121} \equal{} \boxed{11}$

nikeballa is right, and that's exactly how I did it.
A simple rule for powers to fractions - $ a^{\dfrac{b}{c}}$ is the cth root of $ a^b.$


This is a singular value decomposition SVD problem:

Let A be an m x n matrix.
What is the SVD of 4A , transpose of A , and inverse of A?

If $A$ has an [url=http://en.wikipedia.org/wiki/Singular_value_decomposition]SVD[/url] $A=U\Sigma V^{T}$, then $4A=U(4\Sigma)V^{T}$, $A^{T}=V\Sigma U^{T}$, and $A^{-1}=(V^{T})^{-1}\Sigma^{-1}U^{-1}$.


Every year Kazakhstan organizes IMPC (Intqenational Mathematical Project Competition),
and this year two of our students got Bronze medals, not good results for this year. Ravshan Mamatkarimov (me) :-) got his bronze in this competition (IMPC 2006) also he is going to IMO 2006, too.

[please continue the discussions, how many medals got your country?]

davron

I participated last year and I got bronze medal without any great effort! BTW Kazakhstan's great.

[quote="José"][quote="s1ck_psych0"][quote="José"]Are you taking part this year?[/quote]

No, this year I was dedicated to qualifying in IMO. So I'll be in Slovenia insead.[/quote]

Well, that's better ajajaj lol[/quote]

Of course this is better!  :D


function
limit
real analysis
real analysis unsolved

Let $ f,g: \mathbb{R}^n\to\mathbb{R}$ be functions, and for all $ x\in \mathbb{R}^n$ define
\[ k(x) \equal{} \max(f(x),g(x))
\]

\[ h(x) \equal{} \min(f(x),g(x))
\]
I already proved* that if $ \lim\limits_{x\to a}f(x)$ and $ \lim\limits_{x\to a}g(x)$ exist, then both $ \lim\limits_{x\to a}k(x)$ and $ \lim\limits_{x\to a}h(x)$ exist.
In fact, if $ \lim\limits_{x\to a}f(x) \equal{} p$ and $ \lim\limits_{x\to a}g(x) \equal{} q$, then $ \lim\limits_{x\to a}k(x) \equal{} \max(p,q)$ and $ \lim\limits_{x\to a}h(x) \equal{} \min(p,q)$, so in particular they exist.

Now, I have to prove or give a counterexample of the following statements:

[i]If $ \lim\limits_{x\to a}k(x)$ and $ \lim\limits_{x\to a}h(x)$ both exist and are [u]not equal[/u], then both $ \lim\limits_{x\to a}f(x)$ and $ \lim\limits_{x\to a}g(x)$ exist.

If $ \lim\limits_{x\to a}k(x)$ and $ \lim\limits_{x\to a}h(x)$ both exist and are [u]equal[/u], then both $ \lim\limits_{x\to a}f(x)$ and $ \lim\limits_{x\to a}g(x)$ exist.[/i]

My intuition said both are not true, but every counterexample I try to construct satisfies the statement.

* [hide][b]Lemma[/b]: Let $ p\in\mathbb{R}$. If $ \lim\limits_{x\to a}f(x) \equal{} p$, then $ \lim\limits_{x\to a}|f(x)| \equal{} |p|$
[u]Proof[/u]: For every $ \epsilon > 0$ there is $ \delta > 0$ such that $ \parallel{}x \minus{} a\parallel{} < \delta\Rightarrow |f(x) \minus{} p| < \epsilon$. But $ \left\parallel{}f(x)| \minus{} |p|\right|\leq |f(x) \minus{} p| < \epsilon$, therefore $ \lim\limits_{x\to a}|f(x)| \equal{} |p|\ \ \ \ \ \square$

Then use $ \max(x,y) \equal{} \frac {x \plus{} y \plus{} |x \minus{} y|}{2}$ and $ \min(x,y) \equal{} \frac {x \plus{} y \minus{} |x \minus{} y|}{2}$ for every $ x,y\in\mathbb{R}$ and the usual sum and product rules.[/hide]

An idea for the first: try to think if you can prove or provide a counter-example for the corresponding statement for sequences (instead of functions).  You might limit yourself to the case that $ (a_n)_{n \in \bf N}$ and $ (b_n)_{n \in \bf N}$ are sequences that only take on the values 0 and 1.

An idea for the second: how do $ h$ and $ k$ compare to $ f$?  If $ h$ and $ k$ approach the same limit, what does this tell you?

Thank you very much for your answer, it helped me.
[quote="JBL"]An idea for the first: try to think if you can prove or provide a counter-example for the corresponding statement for sequences (instead of functions).  You might limit yourself to the case that $ (a_n)_{n \in \bf N}$ and $ (b_n)_{n \in \bf N}$ are sequences that only take on the values 0 and 1.[/quote]
I think this will work: $ f(x) \equal{} \begin{cases} 1,\ \ x\mbox{ rational} \\
0,\ \ x\mbox{ irrational}. \end{cases}$ and $ g(x) \equal{} \begin{cases} 0,\ \ x\mbox{ rational} \\
1,\ \ x\mbox{ irrational}. \end{cases}$

Then for all $ x$ we have $ k(x) \equal{} 1$ and $ h(x) \equal{} 0$. Hence $ \lim k(x) \equal{} 1$, $ \lim h(x) \equal{} 0$, but $ \lim f(x)$ and $ \lim g(x)$ don't exist. (All independent of $ a$, but that's irrelevant.)

[quote]An idea for the second: how do $ h$ and $ k$ compare to $ f$? If $ h$ and $ k$ approach the same limit, what does this tell you?[/quote]
Hmm...if $ \max(f,g) \equal{} \min(f,g)$, then $ f \equal{} g$. But that's probably not what you mean.

You're very welcome  :)   For the second, we aren't so lucky to have $ h \equal{} k$, just that they are equal in the limit.  However, we do have $ h \leq f \leq k$, so ....

Ah, of course. The squeeze theorem. So the second statement is true (and if we replace $ f$ by $ g$ it is still true). Thanks again! :)


Alexandra played 15 games of Skee Ball for $ \$11.25.$ After each game, she earned either 2 tickets or 4 tickets. At the end of 15 games she had earned a total of 40 tickets. For how many games did she earn 4 tickets?

[quote="agentcx_deactivated"]Firstly, the 11.25 info is irrelevant. Secondly, we can turn this into a system of equations:
x+y=15
2x+4y=40
Dividing the second equation by 2 gives x+2y=20. Then we subtract the first from the second to get 5 as the number of games she earned 4 tickets.[/quote]

The answer's actually 5.


3D geometry
geometry
Locus
Locus problems
IMO
IMO 1961

Consider a plane $\epsilon$ and three non-collinear points $A,B,C$ on the same side of $\epsilon$; suppose the plane determined by these three points is not parallel to $\epsilon$. In plane $\epsilon$ take three arbitrary points $A',B',C'$. Let $L,M,N$ be the midpoints of segments $AA', BB', CC'$; Let $G$ be the centroid of the triangle $LMN$. (We will not consider positions of the points $A', B', C'$ such that the points $L,M,N$ do not form a triangle.) What is the locus of point $G$ as $A', B', C'$ range independently over the plane $\epsilon$?

Let $J \in\epsilon',\ J' \in \epsilon$ be the centroids of the triangles $\triangle ABC \in \epsilon'$, $\triangle A'B'C' \in \epsilon$. If we put equal masses m to the points $A, B, C, A', B', C'$, we can pile the 3 equal masses m at A, B, C into the centroid J of the triangle $\triangle ABC$ and and the 3 equal masses m at A', B', C' into the centroid J' of the triangle $\triangle A'B'C'$ without changing the center of mass of the mass points $A, B, C, A', B', C'$. Then we can pile the 2 equal masses 3m at the centroids $J, J'$ into the midpoint of the segment $JJ'$ without changing the system center of mass. Alternately, we can pile the 2 equal masses m at $A, A'$ into the midpoint K of AA', the 2 equal masses m at $B, B'$ into the midpoint L of BB' and the 2 equal masses m at $C, C'$ into the midpoint M of CC' without changing the center of mass of the system. Then we can pile the 3 equal masses 2m at the midpoints K, L, M into the centroid G of the triangle $\triangle KLM$ without changing the center of mass of the system. It follows that the centroid G of the triangle $\triangle KLM$ is the midpoint of the segment $JJ'$. The centroid $J \in \epsilon'$ is fixed, because the points $A, B, C \in \epsilon'$ are fixed. The centroid $J' \in \epsilon$ moves in this entire plane, because the points $A', B', C' \in \epsilon$ move in this entire plane. Thus the problem is reduced to finding the locus of the midpoint G of the segment $JJ'$, where the point $J \in \epsilon'$ is fixed and the point $J' \in \epsilon$ moves arbitrarily. Let F be the foot of a normal from the point J to the plane $\epsilon$. The locus of the midpoints G of the segments $JJ'$ obviously is a plane $\gamma$ through the midpoint N of the segment JF parallel to the plane $\epsilon$, because G is one endpoint of the midline $NG \parallel FJ' \in \epsilon$ of the right angle triangle $\triangle JFJ'$.


Consider the set $ M$ of all polynomials $ P(x)$ whose all roots are pairwise different integers and whose coefficients are integers less than $ 2007$ in absolute value. What is the highest power among all polynomials in $ M$?

The maximum degree of polynomials in $ M$ is $ 9$. For the proof, note that the polynomial $ \prod_{v=-4}^{4}(x+v)$ is contained in $ M$, i.e. the highest power is greater or equal to 9. Now let $ P$ be one polynomial in $ M$ which assumes the highest degree of polynomials in $ M$. Clearly, $ P$ must have $ 0$ as a zero, since otherwise $ Q(x)=x\cdot P(x)\in M$ would be of higher degree. We write $ P(x)=A\cdot x\cdot\prod_{v=1}^n(x+a_v)$ with $ n+1=\deg P$ and $ A,a_v\in\mathbb{Z}^*$. Moreover, we must have

$ 2007\geq |A\cdot\prod_{v=1}^na_v|\geq \prod_{v=1}^n|a_v|\geq \left\{\begin{array}{ll}\left[\left(\frac{n}{2}\right)!\right]^2 & n\mbox{ even}\\[0.2cm]
\frac{n+1}{2}\cdot\left[\left(\frac{n-1}{2}\right)!\right]^2 & n\mbox{ odd}\end{array}\right.$

This yields $ n\leq 8$, i.e. $ \deg P\leq 9$


Suppose $ s$ and $ t$ are directly proportional.  If $ s\equal{}14$ when $ t\equal{}12$, what is $ s$ when $ t\equal{}7$?

If $ s$ and $ t$ are directly proportional, it means their quotient is constant. We have $ \frac{s}{t} \equal{} \frac{14}{12} \equal{} \frac{7}{6}$. Now when $ t \equal{} 7$, we set up the equation $ \frac{7}{6} \equal{} \frac{s}{7}$. Multiplying both sides by $ 7$ gives $ s \equal{} \boxed{\frac{49}{6}}$.


hehe. Here's a video. I made this, so give it a shot.
[youtube]OiuQEsz3AXQ[/youtube]

Was that your voice? You seemed a bit hyperventilated.

Amazing! I love it! Please entertain us with more video. AUUUUUUUUGH!!!!!!

Yes, that's my voice. I'm recording into a fairly lame mic, but it's decent. I hope I don't sound terrible :P This is probably the first revision I'm going to do.

I have stumbled the site. Maybe it'll get views. :P

[quote="DiscreetFourierTransform"]And SLOPES![/quote]
[quote="DiscreetFourierTransform"]UHHHHHHHHH![/quote]

rofl

lol that was really funny
i sent it to all my friends
not that you actually get any math info from just watching it  :lol:


geometry
geometric transformation
reflection
geometry unsolved

If the primes A', B', C' are reflections of  A,B,C
   in the opposite sides,
   reconstruct ABC given the primes.

  T.Y.
  M.T.


Are given 20 points in the space, 7 of this points are in a line and 5 others are in another line. Find the largest number of planes that can pass through this points if :
a) these lines are parallell?
b)these lines are skew?

i am very bad at geometry :blush: , but how can 7 points are in a line (collinear right?) while no 4 are coplanar? i think if the points are collinear, then they must be coplanar.

I am sorry there was a mistake in my problem . I edited it now.


Find the efficiency of the proccess in the attachment.Be carefull. :)

Three problems from BAUPC in a row :) Unfortunately, I don't have the time to retype the (well-known :roll:) solutions... Maybe there's someone else who might want to do it...

The point in  the transformation we should be carefull about is (3/2V;3/2P).in this point the gas begins to release heat.
the final answer(hope it's right) is 1/(5Cv/2R+3/4) so it depends of what kind of gas it is

Hey...I said that be carefull. :) You forgot something.If you want to know your mistake ,tell me,I will post it.

When I was solving this problem ,I also made the same mistake :(

is it because i made the process 3-1 tangent with an isotherm and not with an adiabatic transformation?
is the point i am looking for (3V*gamma/gamma+1;3P/gamma+1)?

We have to know what kind of gas it is.
Here is what I got: if $\gamma = \frac{f+2}{f}$
then:
$\eta =\frac{16(f+1)^{2}}{5f^{3}+17f^{2}+28f+16}$
So if the gas is Oxygen, hydrogengas, nitrogen gas, then $f=5$
so:
$\eta = \frac{32}{67}$
if it is helium etc then $f=3$:
so :
$\eta = \frac{64}{97}$

I think you have one small mistake.Because the answer is $\frac{16}{97}$
The gas is monoatomic ideal gas.


calculus derivative geometry algebra function domain integration

Hello,

 Given x^4 = a^2(x^2 – y^2).

 I have been able to calculate the implicit form, the polar equation and the x and y intercepts.

 I'm unable to

 1) arrive at its parametric equations:
 x = a * sin(t)
 y = a * sin(t) * cos(t)?

Once I assume x = a * sin(t) getting to y = a * sin(t) * cos(t) is not difficult. I’m, however, specifically interested in knowing how do you come to a conclusion that for x a * sin(t) is a better choice than cos(t)? Are there any guiding principles or rules that make this procedure easy? Any good web resources on this?

 2) The horizontal and vertical tangents

What’s wrong with the following?

For horizontal tangents dy/dx=0

Using x^4 = a^2(x^2 – y^2)
x^4 = a^2 * x^2 – a^2 * y^2
a^2 * y^2 = a^2 * x^2 – x^4

Taking derivative:

a^2 * 2 * y * dy/dx = a^2 * 2 * x – 4 * x^3
a^2 * y * dy/dx = a^2 * x – 2 * x^3
dy/dx = (a^2 * x – 2 * x^3) / (a^2 * y)

Hence:
a^2 * x – 2 * x^3 = 0
x * (a^2 – 2 * x^2) = 0
a^2 – 2 * x^2 = 0
a^2 = 2 * x^2
x^2 = a^2 / 2
x = a / sqrt(1/2)
Hence horizontal tabgents at:
(a / sqrt(1/2), 0)

The correct answer is:
(±1/2 * sqrt(2 * a), ±1/2 * a)

 3) Area of the curve:
 4) arc length:
 5) Curvature
 6) tangential angle

 Thanks in advance for any response.

NB.: These are taken from http://mathworld.wolfram.com/EightCurve.html

Either I do not know the procedure or the values I calculate do not match.
=====

For your first question, it doesn't matter at all.  If your domain for t is any integral multiple of 2 pi, a simple shift of pi/2 will change sines into cosines.  There is no particular reason to choose one over the other except convention, habit, or the fact that you do have to make a decision.

For the question of the tangents, your method works except for two things:
First, you probably should have considered the case x = 0 before you divided by x.
Second, for a non-zero, the number $\frac{a^2}{2}$ has two square roots, not one.  That's the plus-or-minus.  The x coordinate is otherwise the same as yours, but written in a different form.  Then, you inexplicably chose y = 0 to go with this x value, but that gives you a point not even on the curve.  Instead, you need to go back to the equation of the curve to find the y values that match your x value.


[hide]Hint:
Those are 3 roots of a cubic polynomial. Using complex number to derive this polynomial+Vieta[/hide]

[hide="Hint"]$ m$ is integer numbers such that $ m\ge2$.
\[ \prod_{i\equal{}1}^{m\minus{}1}\sin\frac{i \pi}{m}\equal{}\frac{m}{2^{m\minus{}1}}\][/hide]

hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
Sonnhard.

Can you show solution using complex and polynomials?  :maybe:


vector
function
algebra
domain
calculus
calculus computations

Hello

It has been a long time since I have posted on this site but it has always been very helpful

I have two surfaces

$ z = \sqrt{9-x^{2}-y^{2}}$
and
$ x=y^{2}$

I am to represent the intersection of the surfaces with a vector function $ \vec{r}(t)$

What I have done is the following

I substituted the second equation into the first to get

$ z = \sqrt{9-x^{2}-x}$

Then I changed the variable to t...

$ \vec{r}(t) = \langle t, t^{2}, \sqrt{9-t^{2}-t}\rangle$

But I am not sure if that is correct

Any help is appreciated

Thanks

It's not complete; we should specify what values of $ t$ are allowed in this parametrization. The domain is just as important as the formula.

Thanks for the help

So to specified allowed $ t$ values, we know that $ 9-t^{2}-t \geq 0$

$ 9 \geq t^{2}+t$
$ 9 \geq t (t+1)$

And that is our domain

Thanks, jmerry and jorgeston


1.
[hide]
Supposed that $ x_1,x_2,x_3,\cdots,x_n$ is eigenvalues of $ A$, then :
$ S_i: \equal{}1\plus{}\frac {x_i^2}{2!}\plus{}\frac {x_i^4}{4!}\plus{}cdots$
Since every term of the summation is positive, we can conclude that, if $ S$ doesn't attain infinity, then $ S_1.S_2.\cdots.S_n \geq 0$.
[/hide]


In my thesis I have enumerated the sections as
1. Introduction
2. The basic of finite fields

and so on by \section -command.

But the references won't be enumerated similarly. My references chapter is the following
\begin{thebibliography}{9}
\bibitem{ax1} Ax, J: Solving Diophantine problems modulo every prime, The Annals of Mathematics, 2nd Ser., Vol. 85, No. 2 (Mar., 1967), 161-183.

\bibitem{ax2} Ax, J: The elementary theory of finite fields. The Annals of Mat-\newline hematics, 2nd Ser., Vol. 88, No. 2 (Sep., 1968), 239-271.
...
\end{thebibliography}
It just outputs Viitteet but it should output 18 Viitteet. I'm writing my thesis in Finnish. What am I doing wrong. If I put
\section{Viitteet}
\begin{thebibliography}{9}

it outputs the number 18 correctly but now there are two times the word Viitteet. And what does the part {9} means is the tag \thebibliography? I just saw it in one example and $ \text{\LaTeX}$ won't compile the file without it.

Add the section counter with
\renewcommand{\refname}{\thesection\ Viitteet} for the article class
or
\renewcommand{\bibname}{\thesection\ Viitteet} for chapter and book classes.
[quote="puuhikki"]And what does the part {9} means is the tag \thebibliography? I just saw it in one example and $ \text{\LaTeX}$ won't compile the file without it.[/quote]It gives the width of the widest label. LaTeX needs to know how wide the label will be at the start of each bibliography entry so it knows what indentation to use. If you are using a numbered list then any one digit number (such as 9) will do for less than 10 entries, a two digit number (such as 99) for at least 10 and less than 100 entries etc.

Hi puuhikki,

stevem has given a good direction. Let me just give some comments.
If the babel package is used then probably \renewcommand{\refname}{...} wouldn't work in the preamble. babel provides another way by \addto, if you need that then let us know.
If it's done like above then you may have to adjust the section counter, perhaps also the page headings if the section titles are printed there. And if you want to have Viitteet listed in the table of contents, \addcontentsline could be called.

Some time ago I've used a quick workaround, just changing the internally used \section* command to behave like the normal \section command, concerning numbering, headings, toc etc. Just two short lines of code would be necessary, see: [url=http://texblog.net/latex-archive/plaintex/unstarred-starred-section-schapter/]Change starred command into unstarred[/url]. I think today I would make it even shorter than shown there.

Stefan Kottwitz


quadratics
geometry
geometric transformation
rotation
complex numbers
continued fraction

see if you can generalize a way to get an integer A in Q[sqrt(-3)] (which means the integer will have the form a+b*sqrt(-3), where a and b are of the form n/2 where n is an integer) so that A and its conjugate are associates.

Associate means that two numbers have the property that A = eB and B = e'A, where e and e' are units.

And a side question: can anybody list ways to determine whether an element of Q[sqrt(d)] is prime?
the only one I have is that if the norm of an element is a rational prime (a regular 2, 3, 5, 7... type prime) then the element is a quadratic prime of the given field.

Remember we're dealing with a field, so there are no primes: everything is a unit, and all elements are associates of each other.  Do you want the set of elements $ \{\frac{a+b\sqrt{-3}}{2}\parallel a, b \in \mathbb{Z}\}$?

I don't understand what you mean by "a field has no primes." I think we define "field" in different ways, and my definition might be wrong. But for progress' sake let's just say a field is the set of numbers a+b*sqrt(d) where d is a square-free integer and a and b are rational.

It should have primes because if it contains integers, the integers can be factored down.

[quote="Riders on the Storm"]I don't understand what you mean by "a field has no primes." [/quote]
Because to be a prime element it needs to be a non-zero non-unit so that ..... But in any field all non-zero elements are units, so it doesnt have non-zero non-unit elements.

Primes are nonzero nonunit elements such that $ p|ab \implies p|a \vee p|b$.
Obviously there are no nonzero nonunit elements in a field.

And as sometimes unique factorisation is missing, factoring down integers is not always possible.

Even though a field has no irreducible or prime elements as explained by me and ZetaX it is still an Unique Factorization Domain.  :rotfl:
(Because if $ p\implies q$ and $ p$ is always false then the conditional statement is true. This is called being "vacouslly true".)

[quote="Riders on the Storm"]Isn't a unit any integer that divides 1?

Let Q[sqrt(-1)] be the field in question. Then wouldn't 2 be an integer of the field?
My definition of integer is any number A in a field whose defining equation (x-A)(x-(conjugate of A)) = 0 has 1 as the coefficient of the x^2 term.

So (x-2)(x-2) = x^2-4x+4 = 0 shows that 2 is an integer.

But 2 can be factored into (1+i)(1-i) right? And these clearly aren't units, and there's a certain theorem that says if the norm of a number is a rational prime, then the number itself is a prime element of the field.

Because the norm of 1+i and 1-i is 2, 1+i and 1-i must be prime.[/quote]

As I posted in the other thread, I think the problem here is that we are interpreting $ \mathbb{Q}[\sqrt{d}]$ as the set of $ a+b\sqrt{d}$ where $ a, b$ are rational: not integers.  I think the more common way to express it is $ \mathbb{Z}[\sqrt{d}]$, the set of $ a+b\sqrt{d}$ where $ a, b$ are integers.

If I am not mistaken, you are referring to a different ring: the ring of numbers which are roots of monic quadratic equations and have discriminant $ D$ (ie $ \frac{\mathbb{Z}}{2}[\sqrt{d}]$, if you will).  I apologize for misunderstanding you: I had a hunch that this was the case.

In this field with $ D=-3$, the only units are $ \frac{1+\sqrt{-3}}{2}$, $ \frac{1-\sqrt{-3}}{2}$, $ \frac{-1+\sqrt{-3}}{2}$, $ \frac{-1-\sqrt{-3}}{2}$, $ \pm1$ (the six roots of unity, which makes sense).  If you put it in the complex plane, then multiplication by any of these is just a natural number of rotations of $ \frac{\pi}{3}$ about the origin.  So if $ a+b\sqrt{-3}$ and $ a-b\sqrt{-3}$ are associates, then one is just a sufficient number of sixty-degree rotations of the other.  Thus, the angle $ a+b\sqrt{-3}$ makes with the $ x$ axis is a multiple of thirty degrees (I've moved away from the quadratic integers completely now and I'm just using complex numbers and De Moivre's Theorem).  Then this quadratic integer is a real multiple of one of the twelfth roots of unity.  Indeed, any such number works, so these are the numbers we want.

For the general $ D$, I know what to do iff $ D$ is negative: we know that the units of the ring are, in the complex plan, roots of unity, and from there we use the same argument.  For positive $ D$, the argument I used of size breaks down, because the equation $ a^{2}-Db^{2}=1$ can have infinitely many solutions (by a continued fraction argument), so we may have infinitely many units.

could you make that last part about "a real multiple of the twelfth roots of unity" clearer?

I think what confused everybody was me saying "the quadratic integers of the field Q[sqrt(d)]".
I was hoping this would appear to somebody else as Z[sqrt(d)] (or, as you said, Z/2[sqrt(d)] in the case that d = 1 mod 4).

[quote="Riders on the Storm"]could you make that last part about "a real multiple of the twelfth roots of unity" clearer?
[/quote]

So you know that the angle of $ a+b\sqrt{d}$ is half the angle between $ a+b\sqrt{d}$ and $ a-b\sqrt{d}$.  But we have already established that $ \frac{a+b\sqrt{d}}{a-b\sqrt{d}}$ is a sixth root of unity: that is, when raised to the sixth power, it yields 1.  (We established that the sixth roots of unity are exactly the units in this ring)

Therefore, $ \frac{a+b\sqrt{d}}{\sqrt{a-db^{2}}}$ is such that, when squared, we get a sixth root of unity.  Then this is a twelfth root of unity.  It then follows that $ a+b\sqrt{d}$ is a twelfth root of unity times its own absolute value.  Any number which satisfies this condition will work, as can be easily checked. (and of course, $ d=-3$)


limit
real analysis
real analysis unsolved

[quote="quantum leap"]I didn't know how to approach this one.  I think there's a neat trick to it, but i'm not sure what it is:
$\prod_{j=2} ^{\infty} (1-\frac{1} {j^2})$[/quote]
$1-\frac{1}{j^2} = \frac{j^2-1}{j^2} = \frac{(j-1)(j+1)}{j^2}$. Now write out, say,  $\prod_{j=2} ^{5} (1-\frac{1} {j^2})$ and cancel terms. Then guess and prove a formula for $\prod_{j=2} ^{N} (1-\frac{1} {j^2})$ and take the limit.

Duh! (i hate not seeing easy factors).  Good idea, here's what i got now:

$\prod_{j=2} ^{\infty} (1-\frac{1} {j^2}) \to \prod_{j=2} ^{\infty} (1+\frac{1} {j}) (1-\frac{1} {j}) \rightarrow \prod_{j=2} ^{\infty} (1+\frac{1} {j}) \prod_ {j=2} ^{\infty} (1- \frac{1} {j})=\prod_{j=2} ^{\infty} (\frac{j+1} {j}) \prod_{j=2} ^{\infty} (\frac{j-1} {j})$
Breaking this up,
$\prod_{j=2} ^{\infty} (\frac{j+1} {j})= \frac{3}{2}  \left \frac{4}{3}  \left \frac{5} {4}  \cdots =\frac{1} {2}$
and
$\prod_{j=2} ^{\infty}(\frac{j-1}{j})= \frac{1}{2}  \left \frac{2} {3}  \left \frac{3} {4}  \cdots = 1$
Therefore,
$\prod_{j=2} ^{\infty} (\frac{j+1} {j}) \prod_{j=2} ^{\infty} (\frac{j-1} {j})=  \frac{1}{2}  \left 1  =\frac{1}{2}$
                                                                               Q.E.D. :P
Thanks for your help hpe

im not sure you've done well quantum leap because for example
$P=\prod_{j=2} ^{n} (\frac{j+1} {j})=\frac{n+1}{2}$
and $\lim_{n \to \infty} P = +\infty$

uhm if what you say in your proof is true then
what i wrote here http://www.mathlinks.ro/Forum/topic-73970.html would
be true also
help someone im stuck on this products...

[quote="spx2"]$P=\prod_{j=2} ^{n} (\frac{j+1} {j})=\frac{n+1}{2}$
and $\lim_{n \to \infty} P = +\infty$[/quote]

You got the point  :D We would have the case $0 \cdot \infty$.

BTW, we have $\prod_{j=2}^n \left( 1 - \frac{1}{j^2} \right) = \frac{n+1}{2n}$ and that indeed tends to $\displaystyle \frac12$.

[quote="quantum leap"]Duh! (i hate not seeing easy factors).  Good idea, here's what i got now:

$\prod_{j=2} ^{\infty} (1-\frac{1} {j^2}) \to \prod_{j=2} ^{\infty} (1+\frac{1} {j}) (1-\frac{1} {j}) \rightarrow \prod_{j=2} ^{\infty} (1+\frac{1} {j}) \prod_ {j=2} ^{\infty} (1- \frac{1} {j})$
[/quote]
[b][color=red]Stop[/color][/b]. You need to cancel before you take the product apart, or do it for a finite product. The answer is correct, though.

wait, how do you find that the product's closed form is $\frac{n+1} {2n}$ ?  and why coudn't I break it up; multiplication is associative :huh:

We have

$\displaystyle \prod_{j=2}^n \left( 1 - \frac{1}{j^2} \right) = \prod_{j=2}^n \frac{(j-1)(j+1)}{j^2}$

$\displaystyle = \frac{\displaystyle \left( \prod_{j=2}^n (j-1) \right) \cdot \left( \prod_{j=2}^n (j+1) \right)}{\displaystyle \left( \prod_{j=2}^n j \right) \cdot \left( \prod_{j=2}^n j \right) }$

$\displaystyle = \frac{\displaystyle \left( \prod_{j=1}^{n-1} j \right) \cdot \left( \prod_{j=3}^{n+1} j \right)}{\displaystyle \left( \prod_{j=2}^n j \right) \cdot \left( \prod_{j=2}^n j \right)}$

$\displaystyle = \frac{1 \cdot (n+1)}{n \cdot 2} = \frac{n+1}{2n}$

quantum leap, watch out !!! When constructing the real numbers with their operations, we don't define infinite products, only finite ones.

Infinite products/sums appear in analysis, (they more of a convention, they actually represent the limit) and if they are convergent, addition/multiplication works. Your two products represented an undefined $0 \cdot \infty$ case. Also, note that you didn't compute correctly the first of the infinite products. It should have been $\infty$. (try some numerical examples if you are not convinced)


algebra
polynomial
induction
modular arithmetic
logarithms
function
pigeonhole principle

Find all polynomial $P\in\mathbb{Z}[X]$ such that $\{n\in\mathbb{N},2^n|P(3^n)\}$ is infinite.

If it is obvious, why don't you give your proof?

Collin! What is the source of the problem?

There is something peculiar with the first 3 posters in this thread :? ... I don't agree with multiple accounts of the same person! :mad: especially if they talk one with each other ... you have 1 day to explain :alien:

I don't remember what is the source of this problem...
I don't understand RobertuX solution. Can you explain, please :? ?

In my opinion, there is nothing to explain, since no solution.

RobertuY has been banned. If Collins and blang do not explain in the next 24 hours how come they share the exact same IP with RobertuY, I will ban that IP also.

[quote="Valentin Vornicu"]RobertuY has been banned. [/quote]

Cool. I was deleting some of his useless and quite insulting posts.

[quote] If Collins and blang do not explain in the next 24 hours how come they share the exact same IP with RobertuY [/quote]

???? How can this be?

Pierre.

It is true. All 3 of them have the same IP, and it does not seem to be a proxy  :?

[quote="harazi"]Anyway, it is a little strange to associate a person like Robertuy with someone like Blang.  :?[/quote]
I agree.

[quote="harazi"]Anyway, it is a little strange to associate a person like Robertuy with someone like Blang.  :?[/quote]Sorry, but I asked for an explanation. I have yet to do anything but I will in case anything like this happens again (i.e. the posts RobertuY has been making).

I have just now seen the posts RobertuY has been posting on the board. Next time this happens with someone please announce me so that I can stop it from happening in the beginning ;)

Enough about this let us turn our attention to the problem again :)

[quote="Myth"][quote]Can you see this?[/quote]
It is an interesting question ;) :)[/quote]

 :D  Indeed!

But, in fact, for any $c>0$ there is a treshold $n_0 = n_0(c)$ such that for $n>n_0$ the residue of $3^n \pmod{2^n}$ exceeds $c$.

Hmm... It is quite nonintuitive  :?

Actually, the idea I wanted to use for this problem is to show that for all large $n$ number $3^n$ has small residue modulo $2^k$ for some $k\leq n$ (small with respect to $2^n$).

[quote="Myth"]Hmm... It is quite nonintuitive  :?
[/quote]

The following result of Mahler is classical: if $\alpha>1$ is a rational number which is not an integer, and if $\ell \in (0,1)$ is arbitrary, then the fractional part $\{\alpha^n\} > \ell^n$ for all $n > n_0(\alpha,\ell)$.

Can you, please, give some references? I haven't so many number theory books or materials.
Or it would be great if you post the proof to Number Theory Theorems Section. :)

Wow! I imagine the proof is not for mere mortels... :(

[quote="harazi"]... mere mortels... :([/quote]
Is it an idiom? (just to improve my english)

In fact, I am not sure it exists... but I remember to have understood something like this in a film... and there it meant ordinary people.

And to complete the solution for the initial problem: so we have $ S(n) P(3^{an+b})=a_1 y_1^n+...+a_k y_k^n $ for all n. This can be written in the form $ a_1 y_1^n+...+a_k y_k^n+T_1(n) +T_2(n) 3^{an}+...+T_s(n) 3^{(s-1)n}=0$ for all n, where $T_i$ are polynomials. If we regard the topic "  vanishing linear sequences", I think we can conclude.

The idiom is 'mere mortals.'  :D  O.k., I'll say later in the day more about Mahler's result (which, by the way, has been recently generalized to the case when $\alpha>1$ is an arbitrary algebraic number which is not a rational power of neither an integer nor a Pisot number  :)  ... this last thing is indeed not for mere mortals!). But now I have to go.

Anyway, the problem that if $q \in \mathbb{Q}$ then $q3^n \equiv  1 \pmod{2^n}$ for at most finitely many $n$ has an easy solution, and we don't need Mahler's theorem to show this.

I'll be back later,
--Vesselin

Ok, [url=http://arxiv.org/pdf/math.NT/0403522.pdf?front]here[/url] is the paper (by Zannier and Corvaja again  :) ) in which the mentioned generalization of Mahler's result is proven. As for Mahler's result, it's proof uses a $p$-adic analog of the deep Roth theorem! But the following weaker result, which proves our problem for $\deg{P} \leq 1$  :) , is quite simple:

[b]Claim. [/b] Let $p$ be a prime and $a$ be an integer co-prime with $p$. If $r \in \mathbb{Q}$ is any rational number, then the congruence
\[
ra^n \equiv 1 \pmod{p^n}
\]
has only finitely many solutions $n$.

Sketch: Define $p' = p$, if $p \geq 3$, and $2' = 4$. For a $p$-adic integer $x$ with $p' \mid x$, the $p$-adic logarithm function $\ln_p(1 - x)$ is defined by the ($p$-adically convergent) power series $ - \sum x^k/k$. Useful properties of the $p$-adic logarithm are: (i) if $x,y \equiv 1 \pmod{p'}$ then $\ln_p{xy} = \ln_p x + \ln_p y$; (ii) if $x \equiv 1 \pmod{p'}$, then the exact power of $p$ dividing $x - 1$ coincides with the exact power of $p$ dividing $\ln_p x$.

Now, assume that $ra^n \equiv 1 \pmod{p^n}$, and write $n = (p-1)k + s$, $0 \leq s < p-1$. Then, $ra^s \equiv 1 \pmod{p}$, so
$\ln_p (ra^n) = \ln_p (a^{(p-1)k}) + \ln_p (ra^s) = k\alpha + \beta$, where $\alpha = \ln_p (a^{p-1})$ is a constant, and $\beta = \ln_p (ra^s)$ is a $p$-adic quantity that takes on only finitely many values. On the other hand, we have by assumption that $\ln_p (ra^n) \equiv 0 \pmod{p^n}$, and this certainly cannot hold if $n$ is large enough.

I can provide more details, if this is necessary  ;) .

--Vesselin

For you, such kinds of proofs are easy, for us I don't think so.  :(  :?

:?  :?  Why, this one is quite standard. Assume to the contrary that there are infinitely many $n$ satisfying $\ln_p (ra^n) \equiv 0 \pmod{p^n}$. Because the quantity $\beta = \ln_p (ra^s)$ takes on only $p-1$ (finitely many) values, we have, by the pigeonhole principle, infinitely many $n$ for which this $\beta$ is fixed. Thus, the problem is reduced to showing that if $\alpha, \beta \in \mathbb{Q}_p$ are any fixed non-zero $p$-adic numbers, then $\text{ord}_p (n\alpha + \beta) < n$ for $n$ sufficiently large. This is not that difficult, IMHO.

As I said, standard tools are for you, but for someone who does not use every day such tools is not at all standard. But this IMHO, of course.  :(

Well, I used the notion of $p$-adic logarithm just for convenience in notations, but what I wrote is more or less equivalent to the following triviality  :D . Write $n = k(p-1) + s$, $0 \leq s < p-1$. To show that there are only finitely many $n$ with $ra^n \equiv 1 \pmod{p^n}$, it suffices to show that there are finitely many such $n$ with $s \in \{0,1,\ldots,p-2\}$ fixed. Since $ra^n \equiv a^{k(p-1)} \equiv 1 \pmod{p}$, we also have that $ra^s \equiv 1 \pmod{p}$, and write $a^{p-1} = 1 + up, ra^s = 1 + vp$ with $u,v$ fixed $p$-integers (rationals with denominators co-prime with $p$). The condition $ra^n \equiv 1 \pmod{p^n}$ then implies
\[
(1 + up)^k(1  + vp) \equiv 1 \pmod{p^k},
\]
and by expanding the l.h.s. it is not difficult to see that this congruence, for any fixed $u,v \neq 0$, cannot hold for large $k$.


The [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]AoPS Problem Archive[/url] currently has all the AIME tests except for the 2000i test. I'd like to get that posted sometime soon.

If you have a computerized copy of this test, I'd appreciate it if you could send it to me via PM or e-mail. My e-mail is edwardjzhu@gmail.com.

Thanks.

Edit I: Thanks to Ravi B for sending me the alternate 2002 and 2003 tests!  :)
Edit II: Thanks to 1/(lnx) for sending me the alternate 2001 test!

Kalva has them. I'm adding them to my website.

Yes. That's the main problem with Kalva. I'm working on the 1999 one right now.  :wink:

EDIT: 1999 AIME is now available at http://theani.isgreat.org/tmp/contests.php.

It in a PDF file.

[quote="beta"][quote="13375P34K43V312"]It's in a bad format, though, it's pretty hard to read. Plus, the solutions are really, really, really badly written.[/quote]

I think kalva deserves more credit than that. Think about how much time and effort it would take to give solutions for about 2,300 Olympiad problems (and almost all the Putnam and AIME problems). Sure the solutions given by AoPSers are probably very well written, but there are 324908273 people on AoPS while kalva only has one person to do all that.[/quote]

that is very well true.

Kalva's solutions for 14/15 of the problems are fine if you catch the general gist of the problem.

Why? There may be copyright problems. Sure it works under fair use...but when they are selling it... :maybe:

Also, AoPS has 25487 people as of this post, which is not the inordinate amount you said. :roll:

I added [b]most[/b] of the AIME problems to my website, because only the 1983 one can be downloaded from aops.

yeah they sell the 21 century disk so i am not sure if well...

[quote="4everwise"]Joml and I decided to stick to the original wording of the problems.[/quote]
Why? I find some of Kalva's wordings to be much clearer. As long as it is the same mathematically, I don't see the difference. A completely authentic collection of problems would rock, but if you can't find some why not patch your collection with variations on kalva's versions, then correct it when you find the originals?

Maybe they don't have an abundance of free time?

Yeah, archiving problems is fairly time consuming. Also, if they're already on Kalva, we don't need another Kalva version on AoPS.  :)

I like original versions better so I can practice writing the answers in their correct format, but either way, I would like to see them in the contests section.

[quote="anirudh"]There may be copyright problems.[/quote]
AoPS is exempt from these problems somehow or other—after all, we have about 20 AIMEs, most of the USA TSTs, a handful of AMCs, and the USAMOs since 1989 on this site, all in the original wording :wink: .

And I agree with sticking to the originals.  You can find paraphrasings elsewhere pretty easily (Kalva), whereas original wordings are difficult, if not impossible, to find for free elsewhere on the Internet.  Plus, occasionally there are little things a paraphraser might overlook that could create some major issues.  And I know that occasionally the Kalva answers are not integers from 000 to 999.

Hi there. :bye:

So, does anyone have the AIME 2000I test? (Please? It's the last one...  :P)

[quote="anirudh"]I'm working on typing it up...[/quote]The original version?

[quote="anirudh"]No. A paraphrased one.[/quote]
If he wanted paraphrased ones he would've just gone to kalva...

[quote="4everwise"][quote="Phelpedo"]I have the test I think...I'll work on pdf-ing it for you asap. I must say, I'm pretty excited that AoPS will soon have a complete archive of all the AIMEs ever.[/quote]Awesome. Thanks!

I'm pretty excited too.  :lol:[/quote]
Wait, but aren't the problems listed at
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001
from the AIME I?

[quote="4everwise"][quote="bpms"]Wait, but aren't the problems listed at
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001
from the AIME I?[/quote]Stop scaring me like that! That's the AIME 2001I, not the AIME 2000I.  :P

Anyway, hopefully, we'll have this archive finished by um... Friday. :yup:[/quote]
Oh sorry, I can't read apparantly :D
Thanks for all the work you have put into this, I really appreciate the pdfs you make and the archiving.  :thumbup:

[quote="anirudh"]But they're not PDF's.[/quote]
Excuse me, but 4everwise worked really hard to make the pdfs and solutions documents that he has, it isn't easy to do this. You shouldn't pick away at the fact that a few of the aren't in pdfs while most of them are.

[quote="anirudh"]What I meant was that Kalva wasn't inpdfs.[/quote]
So what relevence did your post have, we weren't discussing kalva :huh: ? (unless I'm missing something)
I thought you were replying to my post since you hadn't quoted who you were actually replying to and your post was directly after mine.

I accidentally made a mistake somewhere. Sorry.

[quote="bpms"]You shouldn't pick away at the fact that a few of the aren't in pdfs while most of them are.[/quote]Actually, I have all the AIMEs as PDFs now (except the one test I'm missing). I'll post them all as soon as I get the last test.  :)

Umm, and does anyone have the AIME 2000I? Cmon guys, it's the [b]last one.[/b]  :wink: (disregard any connotations in this sentence)

Is it here? (I found this page last night)
[url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC12Problems]http://www.math.ksu.edu/main/handbook/ProblemSets/AMC12Problems[/url]
edit: nvm, apparently I can't read

[quote="Max300"][quote="4everwise"]
Umm, and does anyone have the AIME 2000I? Cmon guys, it's the [b]last one.[/b]  :wink: (disregard any connotations in this sentence)[/quote]

I got you covered bro   :wink:

so...do you want me to scan it and send it to you or what?[/quote]A scanned copy would be great. Thanks!  :lol:

Hmm, still not here.  :noo:

This AIME is [i]really[/i] elusive... argh... x_x :ninja:

hmm this is weird
i was pretty sure i scanned all those missing and got them to you; apparently not
i do have a copy of the 2000 AIME I; i'll get that in right now

Alrighty. I'll post those last 15 problems tomorrow. Thanks a million.  :)

Ok, this thread has served its purpose, so I'll lock it now.  :P

Thanks to everyone who contributed or thought about contributing! :yup:


quadratics
algebra
polynomial
Rational Root Theorem
calculus
calculus computations

How do I solve this equation:

$ y^{(4)} \plus{} 2y''' \plus{} 4y'' \plus{} 5y' \plus{} 4y \equal{} 1 \minus{} e^{ \minus{} x}sinx$

Possible with undetermined coefficients? How do I get the charachteristic equation from the left side?

Thanks.

Yes, this is definitely one for undetermined coefficients. You can just read the characteristic equation $ r^4\plus{}2r^3\plus{}4r^2\plus{}5r\plus{}4\equal{}0$ off the coefficients. Since you need the roots, look for rational roots first; the rational root theorem says that $ \pm 1,2,4$ are the only possibilities. Try these, including looking for multiple roots.
If that doesn't work, try simple quadratic factors. $ r\equal{}\minus{}1\pm i$ might appear, just because that complicates the procedure.


I'm having a hard time understanding it. For example how would you do this:

8a = 2 mod 9 ?

Thanks in advance

You keep adding $ 9$'s to the RHS until it's divisible by $ 8$, so $ 8a\equiv2\equiv56\pmod9\implies a\equiv7\pmod9$. However, you have to be careful in problems like $ 3a\equiv6\pmod9$. This doesn't mean that $ a\equiv2\pmod9$ is the only solution--$ a\equiv2\pmod3$ gives all solutions. In general, $ ax\equiv b\pmod m$, if $ a|b$, means that $ x\equiv\frac ba\pmod{\frac m{\gcd(a,m)}}$.

Or, for this particular problem, we note that the congruence is equivalent to \[ \minus{}a\equiv 2\text{ (mod 9)}\] Multiplying through by $ \minus{}1$, we get \[ a\equiv \minus{}2\equiv 7\text{ (mod 9)}\]
For the last statement math154 stated, this comes from the fact that $ a\equiv b\text{ (mod }m\text{)}$ if and only if $ \frac{a\minus{}b}{m}$ is an integer.  So if we take $ 3a\equiv 6\text{ (mod 9)}$, then \[ \frac{3a\minus{}6}{9}\equal{}\frac{a\minus{}2}{3}\] is an integer.  The RHS then tells us that $ a\equiv 2\text{ (mod 3)}$

I usually use the modular inverse, or the number you can multiply to another number to make it congruent to 1 in that modulo class. Then, you multiply this to each side. But it's not very convenient sometimes.

However, the inverse only exists iff the number is relatively prime to the modulus.

oohh...that might explain why i couldn't solve some problems.  :rotfl:


The trip from Carville to Nikpath requires $ 4\frac 12$ hours when
traveling at an average speed of 70 miles per hour. How many
hours does the trip require when traveling at an average speed
of 60 miles per hour? Express your answer as a
decimal to the nearest hundredth.

We must find how long the trip is by using D=rt. D=70*4.5 so D=315.

Using this same formula D=rt we have d=315 and r=60
so
315=60*t
solving we get t=5.25

Or you could speed up the process a bit by predividing. 70/60(7/6) and multiplying by 9/2. This yields 21/4=5.25