vector
linear algebra
matrix
linear algebra unsolved

Let $f: V \rightarrow V$ be an endomorphism of a real finite dimensional vector space $V$ with $tr(f) =0$ .

[b] 1.[/b] Show that if $f \neq 0$ , there is a vector $\bf{v}$ such that $\bf{v}$ and $f(\bf{v})$ are linearly independent.
Deduce that there is a basis of $V$ relative to which $f$ is represented by a matrix $A$ with all of its diagonal entries equal to $0$.

[b]2.[/b] Show that there are endomorphisms $g,h$ of $V$ with $f =gh-hg$ .

seach at "zero trace" in the forum and you will have a proof ..


[size=200]AHSME 1988 #30[/size]
(someone could link this to the database too - this is the exact text)

Let $f(x) = 4x-x^2$. Given $x_0$, consider the sequence defined by $x_n = f(x_{n-1})$ for all $n \ge 1$. For how many real numbers $x_0$ will the sequence $x_0,x_1,x_2,\ldots$ take on only a finite number of different values?
(A) $0$
(B) $1$ or $2$
(C) $3$, $4$, $5$, or $6$
(D) more than $6$ but finitely many
(E) infinitely many.

[hide]The sequence will take one finitely many values if there is a fixed point ($\exists\ n$ such that $x_{n+1}=x_n$) because then the sequence will be constant for sufficiently large $n$. $x_0$ is a fixed point when $x_0$ is a solution to the equation $x(4-x)=x\Rightarrow x(3-x)=0$. This occurs at $x=0,\ 3$.
$x_1$ will then be a fixed point when $x_0$ is a solution to $f(x)=0$ or $3$. $f(x)=3$ when $x=3,\ 1$. Similarly, if $f^n(x)=3$ for finite $n$, then there is a fixed point. An infinite decreasing sequence of fixed points can be generated through induction.
[b]Base Case:[/b] $x=1$, shown above to work.
[b]Inductive Step:[/b] Let $x=c$, where $0<c\leq 1$. Solutions to $f(x)=c\Rightarrow x^2-4x-c=0$ are given by $2\pm \sqrt{4-c}$. In order to complete the step, show that $0<2-\sqrt{4-c}<c$. The left inequality follows since $\sqrt{4-c}<2$. The right inequality works because $0<c\leq 1\Rightarrow c(c-3)<0$. Adding 4 and subtracting $c$, $(2-c)^2<4-c$. Both sides are positive, so take the square root, $2-c<\sqrt{4-c}\Rightarrow 2-\sqrt{4-c}<c$ as desired.

Since an infinite decreasing sequence of possible $x_0$ has been generated, the answer is [b](E)[/b].[/hide]


superior algebra
superior algebra unsolved

[quote="who"]every
group of odd order must be solvable. :|[/quote]

Feit, Walter; Thompson, John G., "Solvability of groups of odd order", Pacific Journal of Mathematics, 13:3 1963 775--1029.

Good luck.   :D


Έστω $ M$ μέσο τμήματος $ AB$. Έστω $ r$ ευθεία και $ R,S$ οι προβολές των $ A,B$ στην $ r$ αντίστοιχα. Υποθέτοντας ότι $ A,M,R$ δεν είναι συνευθειακά, να δείξετε ότι οι περιγεγραμμένοι κύκλοι των τριγώνων $ AMR$ και $ BSM$ είναι ίσοι.

Βάζω τη λύση σε hint

[hide]Εστω $ N$ η προβολή του $ M$ στην $ r$. Τότε $ MN\\BS\iff RN\equal{}NS$. Αυτό σημαίνει ότι ότι το τρίγωνο $ RSM$ είναι ισοσκελές με $ RM\equal{}MS$. Ετσι τώρα απ'το νόμο ημιτόνων εφαρμοσμένο στα ζητούμενα τρίγωνα κι απ'΄το γεγονός ότι $ sinA\equal{}sinB$ (αφού οι γωνίες είναι παραπληρωματικές) έχουμε το ζητούμενο[/hide]

Αλλη μια λυση:

[hide]Εστω $ MM'$ το υψος του τριγωνου $ \triangle{RMS}$. Τοτε ειναι $ BS \parallel AR \parallel MM'$ και απο το Θεωρημα του Θαλη εχεις: $ \frac {MA}{MB} \equal{} \frac {RM'}{M'S} \Leftrightarrow RM' \equal{} M'S$. Δηλαδη το υψος του τριγωνου $ \triangle{RMS}$ ειναι και διαμεσος του, αρα το τριγωνο $ \triangle{RMS}$ ειναι ισοσκελες, αρα $ \angle{MSR} \equal{} \angle{SRM} \Leftrightarrow \angle{BSR} \minus{} \angle{MSR} \equal{} \angle{SRA} \minus{} \angle{SRM} \Leftrightarrow \angle{MRA} \equal{} \angle{MSB}$ αφου $ \angle{BSR} \equal{} \angle{SRA} \equal{} 90$. Ετσι απο το θεωρημα των ημιτονων στα δυο τριγωνα, εφοσον $ MA \equal{} MB$ παιρνεις και $ 2R_1 \equal{} 2R_2$

Omoia ginetai kai i apodiksi stin periptosi pou ta A,B briskontai se diaforetiko imiepipedo os pros tin r, mono pou se ekeini tin periptosi oi gonies exoun isa imitona os parapliromatikes kai oxi os ises.[/hide]

[hide]Επίσης, αφού έχει αποδειχθεί ότι το τρίγωνο $ \bigtriangleup MRS$ είναι ισοσκελές και είναι $ AR\parallel MN\parallel BS,$ όπου $ N$ είναι η προβολή του $ M$ πάνω στην ευθεία $ r,$ τότε από $ \angle ARM \equal{} \angle RMN \equal{} \angle SMN \equal{} \angle BSM$ και $ AM \equal{} BM,$ συμπεραίνουμε ότι οι περίκυκλοι των τριγώνων $ \bigtriangleup AMR$ και $ \bigtriangleup BMS,$ είναι ίσοι ( σε ίσους κύκλους, ίσες γωνίες αντιστοιχούν σε ίσες χορδές και αντιστρόφως ).[/hide]
Κώστας Βήττας.


Rezultatele de la baraje sunt numai pentru seniori pe MathLinks.Pentru juniori, trage o privire pe olimpiade.ro :)
Succes in continuare!

rezultatele finale pentru seniori sunt afisate undeva?  :?

deci care sunt cei 6 magnifici care merg la inter la seniori ??? am gasit ceva pe olimpiade.ro , dar nu prea stiu cum sta treaba, ca aia au nr crt aiurea ... punctajele sunt puse la total in ordine, dar tot confused sunt :o

Cei calificati la internationala sunt chiar primii 6 de pe lista pe care ai vazut-o pe olimpiade.ro. Chiar e confuza lista aia...sa nu mai vorbesc de cea de la juniori... :?

zahariuc adrian
pachitariu marius
michnea dragos
ungureanu andrei
bazavan eduard
talau cristian

cred ca nr.crt. e ordinea de dupa barajul de geometrie

Lucian,

Personal nu te cunosc, dar ai fost printre cei carora le-am urmarit evolutia de-a lungul anilor, mai intai ca elev, si mai apoi ca student, care nu mai are alta legatura cu olimpiada de matematica in afara de rezultatele acesteia. Si acum tin minte entuziasmul cu care ai reactionat cand te-ai calificat pentru prima (oare o fi fost singura?) oara la JBMO, cred ca dintre toti copiii de acolo tu erai cel mai bucuros. Cred ca in spatele acestui entuziasm deosebit se ascunde o mare pasiune pentru matematica si stiintele exacte, care trebuie valorificata in sine, si nu doar in spatiul olimpiadelor.

Te felicit pentru fair play-ul de care dai dovada, atat pe acest forum, cat si pe forum-ul educational SEI, avand in vedere pozitia ta de "first runner-up" (pentru cititorii acestui forum: a ratat OIM la doar 1p din 140p!!), si iti doresc sa continui sa cresti de la an la an la fel de frumos cum ai crescut si pana acum, si fii sigur ca la anul vei obtine o medalie foarte frumoasa la OIM, daca nu renunti la acest vis.

In alta ordine de idei, felicitari unei generatii extraordinare de clasa a 12-a, care s-a dovedit a fi cea mai puternica din ultimii ani. Bineinteles, cele mai mari felicitari ii parvin lui Adrian Zahariuc, care a reusit si de aceasta data sa demonstreze ca este o forta in domeniul matematicii.

Succes mult tuturor, si sper sa aud de rezultate cel putin la fel de frumoase ca anul trecut, pe subiecte mult mai grele! Eu va urmaresc, si imi face placere sa aud de rezultatele voastre :)

Andrei Ismail

Scuze pentru intarziere, dar am fost, ca si Andrei mai sus, la un maraton de programare de 24 de ore si n-am avut cum sa postez nimic. Echipa de JBMO este

PADUREANU VICTOR
MOLDOVAN DORIN VASILE
CONSTANTINESCU ANDRA IOANA
RAITA BOGDAN
AMZA CATALIN ALEXANDRU
MANCE FELIX GABRIEL

Am postat clasamentul final in primul post (e un png - trebuie sa fiti logati pentru a-l vedea).

Felicitari celor calificati si bafta multa la Balk :D

In ce perioada se desfasoara Olimpiada Balcanica pt Juniori ?


LaTeX
function
analytic geometry
graphing lines
slope
induction
search

find all \[ f: \mathbb{R} \longrightarrow \mathbb{R} \\
f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)\]

I got proof for f(x)=2

Plug in x=y=0
f(0)+f(0)f(0)=f(0)+f(0)+f(0)
f(0)f(0)=2f(0) Therefore f(0)=0 or f(0)=2

Next plug in y=0
f(x)+f(x)f(0)=f(0)+f(x)+f(0)
f(x)=(f(0)+f(0))/f(0) ----> assumin that f(0)=2
f(x)=(2+2)/2=2

Check:
2+2*2=6
2+2+2=6

I'm sorry, but I don't have latex so it might be little harder to read.

[quote="kai_desmond11"]find all
\[ f: \mathbb{R} \longrightarrow \mathbb{R} \\
f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)
\]
[/quote]

I think : if find functions $ R \minus{}> R$ ,the problem will be easy so  we change  the problem  : find functions : $ R^{\plus{}} \minus{}> R^{\plus{}}$  .

Here is  solution for a f(x)=x

From the post above we took f(1)=1

Plug in x=x+n-1 y=1
f(x+n)+f(x+n-1)f(1)=f(x+n-1)+f(x+n-1)+f(1)
f(x+n)=f(x+n-1)+1
f(x+n+1)=f(x+n)+1

f(x+n+1)-f(x+n)=1
f(x+n)-f(x+n-1)=1
f(x+n+1)-f(x+n)=f(x+n)-f(x+n-1)=1 ------>Function is aritmetic, with constant slope of 1, and f(1)=1, f(2)=2
f(n+1)=n+1, and f(n+2)=n+2....
and f(n+x)=n+x.
Therefore f(x)=x

Can anybody gives a complete proof for this problem in R?

I know f(x)=0,2 and f(x)=x are solutions, it says we can first get f(x)=x for all Q using induction which is the procedure I failed to perform. :blush:

[color=green]Problem : Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that:
\[ f(x+y)+f(x).f(y)=f(x)+f(y)+f(xy)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\]

Solution:  First, we have the result : If a function such that :

i, $ f(x)+f(y)=f(x+y)$

ii, $ f(x).f(y)=f(xy)$

Then we have
\[ f(x)=x, \& \forall x\in \mathbb{R}
\]
Now i prove $ f(x)=x, \& \forall x\in \mathbb{R}$ in the Case : $ f(0)=0,\ \ f(1)=1$. It means we have $ f(x)=x,\ \forall x\in\mathbb{Q}$ and $ f(x+1)=f(x)+1$, too.

Let $ P(x,y)$ the property  $ f(x+y)+f(x).f(y)=f(x)+f(y)+f(xy)$.

$ \star\ \ P(-x,-1)$ implies $ f(-x-1)-f(-x)=f(-x)-1+f(x)$

Then $ f(-x)-1+-f(-x)=f(-x)-1+f(x)$ and so $ f(x)=-f(-x)$.

$ \star\ \ P(-x,-y)$ implies $ f(-x-y)+f(-x).f(-y)=f(-x)+f(-y)+f(xy)$

Then
\[ -f(x+y)+f(x)f(y)=-f(x)-f(y)+f(xy)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\]

Adding (1) and (2) we have $ f(x+y)=f(x)+f(y)$ and so $ f(xy)=f(x)f(y)$. We hence the result.

[/color]

[quote="quangpbc"][color=green]
If a function such that :

i, $ f(x) + f(y) = f(x + y)$

ii, $ f(x).f(y) = f(xy)$

Then we have
\[ f(x) = x, \& \forall x\in \mathbb{R}
\]

[/color][/quote]

It seems many condition can confine the solution of f(x)+f(y)=f(x+y) to f(x)=xf(1)
how do you actually prove the one above

[color=green]ah, if  a function such that:

1, $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

2, $ f(x)<f(y),\forall x<y$

Or

2', $ f(x)>f(y),\forall x<y$

Then $ f(x)\equal{}kx$. Do you know this result ? :) [/color]

Yes, but how about the one with additional condition f(xy)=f(x)+f(y) as in your solution


show that $ 30^{239}\plus{}239^{30}$ is not prime.
can you generalize?

FLT = $ a^{p\minus{}1} \equal{} 1 \pmod{p} \implies {239}^{30} \equal{} 1\pmod{31}$
$ 30^{239} \equal{} \minus{}1^{239} \pmod {31} \implies 239 ^{30} \plus{} 30^{239} \pmod{31} \equal{} \minus{}1 \plus{} 1 \equal{} 0 \pmod{31}$
Thus it is divisible by 31 and is not prime

What was the inspiration that led you to try 31?

edit: er.. nvrmd, I guess you could sort of see it seeing that 30 is -1.


Express as an improper fraction: \[ \frac{\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}}{\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{8}}\]

the whole point is to write a solution not just the answer.
we see that the LCM of 4,5,6 is 60.
Then we add them
15+12+10=47
or 47/60
LCM of 5,6,8 is 120.
24+20+15=59
59/120.
47/60 divided by 59/120=47/60*120/59=
47/1*2/59=94/59
And now you see why you should post a solution.([size=59]your answer is wrong.[/size]

Cgyao, your explanation is correct, but 15+12+10 does not equal 47. :P
15+12+10=37.
Following with cgyao's explanation, the answer is (37/60)/(59/120) or [b]74/59[/b]

dangit Jongy that was a typo. never mind joshisunny just keep in mind that u should post a solution.


Just sketch the graphs, see that they intersect when $ n \approx 173.6$, so $ n = 174$.


calculus
integration
floor function
logarithms
ceiling function
function
calculus computations

for $a>1$, find
${\int_{0}^{\infty}\lfloor{ \log_{a}\lfloor \frac{\lceil{x}\rceil}{x}}\rfloor}\rfloor dx$

For all $x>1$, $x\le\lceil x\rceil<x+1<2x$ and the integrand is $\lfloor\log_{a}1\rfloor=0$.

The integral reduces to $\int_{0}^{1}\lfloor \log_{a}\lfloor\frac1x\rfloor\rfloor\,dx$

This is still annoyingly complicated unless $a$ is actually meant to be an integer. In that case, the inner floor is redundant and the integral is $\int_{0}^{1}\lfloor \log_{a}\frac1x\rfloor\,dx=\sum_{n=0}^\infty \int_{a^{-n-1}}^{a^{-n}}n\,dx=\sum_{n=0}^\infty \frac{n}{a^{n}}-\frac{n}{a^{n+1}}$
$= \sum_{m=1}^\infty\frac{-1}{a^{m}}=-\frac1a\cdot\frac1{1-\frac1a}=\frac{-1}{a-1}$

There's not much here- just clear out the clutter, and it's easy.

yes, the $a$ is an integer,

a) i dont understand the part that you went from Integral to the series.

b) your answer doesnt seem right. i tried some numbers didnt quite work out.

plus this doesnt seem correct to me: (if it is could you elaborate?)

[quote="jmerry"]$\sum_{n=0}^\infty \frac{n}{a^{n}}-\frac{n}{a^{n+1}}$
$= \sum_{m=1}^\infty\frac{-1}{a^{m}}=-\frac1a\cdot\frac1{1-\frac1a}=\frac{-1}{a-1}$
 [/quote]

That last part is just off by a factor of $-1$; I was confused by signs.

The switch from an integral to a sum of integrals is essential- how else are you going to evaluate the integral of a piecewise-defined function?

could you explain how you went from

${\int_{0}^{1}\lfloor{\log_{a}\frac{1}{x}}\rfloor}$

to

$\sum_{n=0}^{\infty}\int_{a^{-n-1}}^{a^{-n}}ndx$

The function $\lfloor-\log_{a}x\rfloor$ is $n$ between $a^{-n-1}$ and $a^{-n}$.


The following (though not as strong as the above) is in my book
\[ \frac{a^2+bc}{b^2+c^2}+\frac{b^2+ca}{c^2+a^2}+\frac{c^2+ab}{a^2+b^2}\geq\frac52. \]
This was proposed by my co-author.

Is there a generalization?Something like $\sum \frac{a^2+kbc}{b^2+c^2} \geq \cdots$


I have problems writing math code in TeXnicCenter and then trying to move it to the forums.  It seems the dollar signs are needed here for all math code but only needed for in-line code in TexnicCenter.

Is there a means to write code so that I can seamlessly transfer it between forum and TeXnicCenter?

Thanks,
Shaw

You should be able to copy things seamlessly though there may be exceptions. For example: [code]This is in text mode but $x^2$ is in maths mode[/code] can be copied directly into the forum giving [quote]This is in text mode but $ x^2$ is in maths mode[/quote]Text should stay as text and only maths should be between dollar signs.

There are occasional more complicated cases like this which occurred on a recent thread
[code]\begin{tabular}{cccccccc} No. & q & w & $\Delta$U & $\Delta$H & $\Delta$ & $\Delta$S$_{\text{univ}}$ \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}[/code]which won't transfer to the forum as the dollar signs are misinterpreted (notice how using the code tag shows the dollar signs). You can solve this by replacing the dollar signs by their alternative \(...\) so
[code]\begin{tabular}{cccccccc} No. & q & w & \(\Delta\)U & \(\Delta\)H & \(\Delta\) & \(\Delta\)S\(_{\text{univ}}\) \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 & 0 \\
d) & + & - & 0 & 0 & + & - & 0
\end{tabular}[/code]works both in a document and on the forum.

If you have other examples you are having difficulty with then post them here using the code tags.

Thanks.  I just noticed that instead of dollar signs, some users here are using backslash bracket constructs which seem to transfer nicely between the two.  Here's a snippet of code (with the output below) I'm transferring from TexCenter. I seem to need the "noindent" in TexCenter which is not decoded here.  Also, too much line skipping.  Guess I'd like to have my stuff perfectly formatted.  After all, we actually like math in here :)

[code]
\noindent Let $h(z)$ be continuous on $\mathbb{C}$ with $U$ an open domain, and $\overline{D}(0,1)$ the closed unit disc such that $h\left[\overline{D}(0,1)\right]\subset U$.  Then there exists a $\delta>0$ such that: $h\left[\overline{D}(0,1+\delta)\right]\subset U$.  That is, compact subsets of an open set can always be enlarged.  This is nicely proven I think by noting that the distance between the boundary $\partial h\left[\overline{D}(0,1)\right]$ and the boundary of the set $\mathbb{C}\backslash U$ is non-zero so that $\partial h\left[\overline{D}(0,1)\right]$ can be covered with overlapping open discs of finite radii.  Consider then the following:
\begin{align*}h(z)&= 1/2(z+z^2) \\
U &= D(0,1)\cup D(1,\epsilon);\quad \epsilon>0
\end{align*}
\noindent Then if $|z|\le 1$, $h(z)\leq 1$ and is only $1$ when $z=1$.  Note then that:
\begin{align*}
h\left[\overline{D}(0,1)\right]\subset U
\end{align*}
\noindent and therefore there exists a $\delta>0$ such that:
\[
h\left[\overline{D}(0,1+\delta)\right]\subset U
\]
\noindent This is true for the general case:
\[H(z)=1/2(w^p+w^{p+1});\quad p\in\mathbb{N}^+.
\]
[/code]

\noindent Let $ h(z)$ be continuous on $ \mathbb{C}$ with $ U$ an open domain, and $ \overline{D}(0,1)$ the closed unit disc such that $ h\left[\overline{D}(0,1)\right]\subset U$.  Then there exists a $ \delta>0$ such that: $ h\left[\overline{D}(0,1+\delta)\right]\subset U$.  That is, compact subsets of an open set can always be enlarged.  This is nicely proven I think by noting that the distance between the boundary $ \partial h\left[\overline{D}(0,1)\right]$ and the boundary of the set $ \mathbb{C}\backslash U$ is non-zero so that $ \partial h\left[\overline{D}(0,1)\right]$ can be covered with overlapping open discs of finite radii.  Consider then the following:
\begin{align*}h(z)&= 1/2(z+z^2) \\
U &= D(0,1)\cup D(1,\epsilon);\quad \epsilon>0
\end{align*}
\noindent Then if $ |z|\le 1$, $ h(z)\leq 1$ and is only $ 1$ when $ z=1$.  Note then that:
\begin{align*}
h\left[\overline{D}(0,1)\right]\subset U
\end{align*}
\noindent and therefore there exists a $ \delta>0$ such that:
\[ h\left[\overline{D}(0,1+\delta)\right]\subset U
\]
\noindent This is true for the general case:
\[ H(z)=1/2(w^p+w^{p+1});\quad p\in\mathbb{N}^+.
\]

Unfortunately your code doesn't work in a document either with lots of errors.
1. You can't start an environment with align* and end with align. They must both be the same. You also cannot put the align environment inbetween dollar signs or \ [ ... \]. You must look at the errors and warnings that TexniCenter tells you.
2. The second align environment isn't doing anything as there is only one line and nothing to align it with
3.\ [...\ ] is used for display mode which is probably a good idea here
4. \noindent doesn't do anything in most of what you've written so should be abolished. The first line of any paragraph should be indented.

So use this code in your document:

[code]Let $h(z)$ be continuous on $\mathbb{C}$ with $U$ an open domain, and $\overline{D}(0,1)$ the closed unit disc such that $h\left[\overline{D}(0,1)\right]\subset U$.  Then there exists a $\delta>0$ such that: $h\left[\overline{D}(0,1+\delta)\right]\subset U$.  That is, compact subsets of an open set can always be enlarged.  This is nicely proven I think by noting that the distance between the boundary $\partial h\left[\overline{D}(0,1)\right]$ and the boundary of the set $\mathbb{C}\backslash U$ is non-zero so that $\partial h\left[\overline{D}(0,1)\right]$ can be covered with overlapping open discs of finite radii.  Consider then the following:
\begin{align*}h(z)&= 1/2(z+z^2) \\
U &= D(0,1)\cup D(1,\epsilon);\quad \epsilon>0
\end{align*}
Then if $|z|\le 1$, $h(z)\leq 1$ and is only $1$ when $z=1$.  Note then that:
\[
h\left[\overline{D}(0,1)\right]\subset U
\]
 and therefore there exists a $\delta>0$ such that:
\[
h\left[\overline{D}(0,1+\delta)\right]\subset U
\]
This is true for the general case:
\[
H(z)=1/2(w^p+w^{p+1});\quad p\in\mathbb{N}^+.
\][/code]and then here just add \ [ and \ ] before and after the align environment and, optionally some new lines, to get:

Let $ h(z)$ be continuous on $ \mathbb{C}$ with $ U$ an open domain, and $ \overline{D}(0,1)$ the closed unit disc such that $ h\left[\overline{D}(0,1)\right]\subset U$.  Then there exists a $ \delta>0$ such that: $ h\left[\overline{D}(0,1+\delta)\right]\subset U$.  That is, compact subsets of an open set can always be enlarged.  This is nicely proven I think by noting that the distance between the boundary $ \partial h\left[\overline{D}(0,1)\right]$ and the boundary of the set $ \mathbb{C}\backslash U$ is non-zero so that $ \partial h\left[\overline{D}(0,1)\right]$ can be covered with overlapping open discs of finite radii.  Consider then the following:
\begin{align*}h(z)&= 1/2(z+z^2) \\
U &= D(0,1)\cup D(1,\epsilon);\quad \epsilon>0
\end{align*}Then if $ |z|\le 1$, $ h(z)\leq 1$ and is only $ 1$ when $ z=1$.  Note then that:
\[ h\left[\overline{D}(0,1)\right]\subset U
\]  and therefore there exists a $ \delta>0$ such that:
\[ h\left[\overline{D}(0,1+\delta)\right]\subset U
\]This is true for the general case:
\[ H(z)=1/2(w^p+w^{p+1});\quad p\in\mathbb{N}^+.
\]

An alternative is to make more use of inline mode to get:

Let $ h(z)$ be continuous on $ \mathbb{C}$ with $ U$ an open domain, and $ \overline{D}(0,1)$ the closed unit disc such that $ h\left[\overline{D}(0,1)\right]\subset U$.  Then there exists a $ \delta>0$ such that: $ h\left[\overline{D}(0,1+\delta)\right]\subset U$.  That is, compact subsets of an open set can always be enlarged.  This is nicely proven I think by noting that the distance between the boundary $ \partial h\left[\overline{D}(0,1)\right]$ and the boundary of the set $ \mathbb{C}\backslash U$ is non-zero so that $ \partial h\left[\overline{D}(0,1)\right]$ can be covered with overlapping open discs of finite radii.  Consider then the following:
\begin{align*}h(z)&= 1/2(z+z^2) \\
U &= D(0,1)\cup D(1,\epsilon);\quad \epsilon>0
\end{align*} Then if $ |z|\le 1$, $ h(z)\leq 1$ and is only $ 1$ when $ z=1$.  Note then that $ h\left[\overline{D}(0,1)\right]\subset U$ and therefore there exists a $ \delta>0$ such that $ h\left[\overline{D}(0,1+\delta)\right]\subset U$. This is true for the general case $ H(z)=1/2(w^p+w^{p+1});\quad p\in\mathbb{N}^+$.

Thanks a bunch Steve.  Yes, I'm guilty as can be about ignoring the warning  messages.  Not a good thing I see.  Guess that's contradictory:  perfect format and ignoring the messages. :blush:  I need to work more with TeXCenter as I wish to begin saving my work and maybe even write a text book one day. :)

[edit]  oh yea, that last modification you made to my code is nicely put . . . perfect in fact in my opinion.  Thanks!

[edit] I see I made syntax error with the last line.  I corrected it in the thread this comes from (won't reformat it so nicely as you've done though as don't wish to use your work).


trigonometry
inequalities
geometry
trig identities
Law of Sines
geometry proposed

$A_1A_2...A_n$ is a convex n-agon. Let P be a point inside it.
Show that one of the angles $\measuredangle PA_1A_2$, $\measuredangle PA_2A_3$, ..., $\measuredangle PA_nA_1$ is less or equal than $\left(n-2\right)\frac{\pi}{2n}$.

(It's a generalization of an IMO problem)

Something tells me that you've gone to the IMO training camp last week....:D

Let $R_i = PA_i, x_i = \measuredangle PA_iA_{i+1}$ and $y_i = \measuredangle PA_iA_{i-1}$ where subscripts are considered modulo $n$.
Note that $\sum^n_1 x_i + y_i = (n-2)\pi$
From tha law of sines, we have $R_i/R_{i+1} = \sin (x_{i-1}) / \sin (y_i)$ for all $i$. Taking the product of all these relations, we deduce that $\prod^n_1 \sin (x_i) = \prod^n_1 \sin (y_i)$. Let p be this common value.
Thus, from AM/GM, and from the concavity of $\sin$ on $[0;\pi]$, it follows that:
$p^2 = \prod^n_1 \sin (x_i) \sin (y_i) \leq ( \frac{\sum^n_1 \sin (x_i) + \sin (y_i)}{2n} )^{2n} \leq (\sin ( \frac{(n-2)\pi}{2n}))^{2n}$
Thus $p\leq (\sin ( \frac{(n-2)\pi}{2n}))^n$, from which it follows that there is some $i$ such that $\sin (x_i) \leq \sin ( \frac{(n-2)\pi}{2n})$, and we are done.

Pierre.

Hey, Igor....look : No generalization of Erdös-Mordell inequality here!!

Pierre.


Harvard
college
calculus
real analysis
linear algebra

I have a question concerning college admissions: I am a highschool junior (as of septermber) and I'm going to be taking math classes at nearby universities such as linear algebra, number thoery, analysis, etc. My grades in my high school are pretty average, but I'm going to be doing some advanced math in high school. What are my chances of being accepted to somewhere like Caltech or Harvard?

Why don't you check out any of the 1237812 posts in this forum where the answer to that has already been posted.

Grades of C in your regular high school courses will be just about fatal to your chances at the most selective schools--because those schools have plenty of applicants with straight-A averages. Grades of B look a lot better if they are in genuine college courses that you take through open enrollment than if they are in regular high school courses, but you still have to compete against applicants who can get grades of A in both kinds of course.

On the other hand, compiling a strong record in college courses as a high school senior ought to help immensely for getting into the same level of college. So don't neglect any of your courses, but definitely try to do your best in the college-level courses.

Be sure to apply to some colleges that you are for sure going to get into. There is enough of a chance element, and a HUGE amount of competition, to applying to Harvard and to Caltech that you had better figure out your plan B before applying to plan A.

What university courses have you taken/are you planning to take?  Are you taking them at community colleges or real universities?  Moreover, how 'pretty average' is your GPA?  What are your SATs like?  Do you do extracurriculars?  Have you done the AMCs/AIMEs/USAMOs?  You really need to answer some of these questions before we can give you a reasonably informed opinion.

[quote="blahblahblah"][quote="1234567890"]Blah^3: Some intro courses (calc III, diff eq) are usually taught same at CC's and universities, so it probably won't matter much. But I think for aything else should be taken at universities.[/quote]

The student pool at most CCs is composed of people who cannot score higher than 1100 on their SATs (and I'm being generous, since there are *real* universities that have median SATs in the 900s).  Getting an A against such competition is hardly something to write home about.[/quote]

I must concur from experience.. The professors cannot make the courses too rigorous or no one would pay the few hundred bucks to end up failing it.. And there aren't too many math  majors at a CC (which is the only reason you would take Calc III at my CC.. I think)

But the upside to taking the course, is you get the book (it's all free where I am if you're a highschool student, so free books, even for a temporary time, are good) so you can just read up and do all of the exercises in the book if you want to make the course rigorous and make sure you know the material well, although I am not sure that will still bring you up to the level of a good university.. perhaps.

It's kind of immaterial, because at a good university, a good math major does not take Calculus III.  The progression of courses for a math major at the best university is something like Calculus/Analysis in a single variable -> Linear Algebra -> Calculus/Analysis in several variables, with an emphasis on rigor as opposed to rote plug and chug work.  In particular, you develop the techniques necessary to prove something like Stokes' theorems in its proper form (as in, say, Rudin or Spivak) as opposed to the heuristic arguments employed in a typical calculus text.

[quote="1234567890"]Blah^3: Some intro courses (calc III, diff eq) are usually taught same at CC's and universities, so it probably won't matter much.[/quote]

I actually rather doubt that there are many beginning undergraduate mathematics courses that are taught the "same" at community colleges and universities. That would be most likely in a system that requires a state flagship university to accept transfer credits from community colleges in the same state. (Indeed, my best friend from childhood, a very successful electrical engineer, did his first two years of college at a community college and then transfered all his credits to our state's flagship research university. He is of the opinion that the community college offered the advantage of smaller class sizes, teaching by Ph.D. professors rather than by graduate student T.A.s, and a generally more focused atmosphere. But I'm not at all sure that that is the typical community college experience.)

The way to demonstrate that two courses are the same is to examine the syllabus, homework assignments, and past examinations in each course. These days there is much material of that kind posted on the World Wide Web. At a minimum, I think there are SOME universities that are clearly more rigorous and more demanding of their best students than MOST community colleges. There are a lot of different calculus textbooks, and a lot of different linear algebra textbooks, and there is some clear market segmentation in both subjects. I suppose the great majority of community colleges might be as good as, or even better than, the majority of high schools in teaching the same subjects, but that is also something I would check on a case-by-case basis.


trigonometry
geometry
calculus
AMC
AIME
geometric transformation
Support

Hi,

I'm just posting this that is there any good competition where you can earn some scholarships to the college from math or science. I know Siemens-Westing house but that's only for SENIORS or at least people who are juniors and in RSI(I don't know what that is but heard it).

I'm planning to take biology as my major and despite what college I go, I hope I can get some scholarships. I'm currently freshman at high school and these are course I'm taking:

[b]Math[/b]
Trigonometry
Geometry
[b]Science[/b]
Biology
[b]English[/b]
CP English (College Prep. My school doesn't offer AP English)
[b]Social Studies or History[/b]
Civics (All 9th grade take this as freshman in my school
[b]Foreign Language[/b]
Spanish

Only thing is that by junior year, I will be done with my math in my school since my school doesn't offer calculus BC. Also, since I'm preparing to take SAT on junior year, I'll take two sciences on sophomore year, like Chemistry and Advanced Biology.

I know this year's gone pretty fast (since it's Dec.19) and at February, I'm in AMC-10. I know that AMC-10 certainly would help my application to school if I pass to AIME but that is only math. I'm looking for the competition that I can get scholarship to help myself on science and if can, have some math involve in it.

My plan, suggested by an undergraduate I know, is to have all credits on freshman and sophomore year and take SAT on junior year, and rest some and send application on my senior year so that I'll be ready on college.

I know this depends on what college I'm going, what scholarship I can find in the my school but is there anyway I can get some scholarship by my independent work? I know I should've tried Googling or something like that before posting here but I think since it's College Forum, I wonder if I can get some good advice.

Any competition during the summer would be nice. I'm also playing soccer on fall and tennis on the spring that I look not only good student but pretty good athlete (at least one that's trying).

Sorry for the long post. Any good advice please!  :)

First of all, Siemens-Westinghouse offers scholarships outside of their science competition for that hundred thousand dollar prize. You might want to try for their AP scholarships - might being the key word. Taking that many APs is a rather difficult thing to do, but it depends on your school/region/state.

RSI is a summer camp basically for incoming seniors, and although it is rather helpful for Westinghouse or Intel if you do go to RSI and work on your project, it is definitely possible to win without RSI.

As far as I know, besides the science scholarships, writing competitions also offer five digit scholarships; you might want to look into some of those if you write tremendously well.

There are also scholarships just for biology majors, but most important right now as a frosh is just to concentrate on school and extracurrics and not worry so much about what college money is out there. At least that's my opinion.

You might want to register for http://www.fastweb.org is you get a chance. That lists many of the best scholarships and colleges.

Here's a friendly suggestion: I think the single most important thing you need to do to become more competive for scholarships is work on your writing. I am aware that you are from a home in which some language other than English is spoken. (I'm not sure WHICH language that is, but I'll make you a deal: I'll work on the translation you asked about if you PM me telling me what other language is spoken in your family.) It's always tough to learn English if your home language background is different, but you DEFINITELY need to look competitive in English writing if you want to be competitive for those top-level programs and for the colleges they feed into. What I notice in your writing is enthusiasm for math and a lot of eagerness to help other people (which is how you became a moderator on this site) but also a LOT of strange grammar and usage. Sometimes it is very difficult to figure out what some of your posts mean--and I rarely encounter that level of grammatical difficulties in the online writings of other young people, even other young people I know are not native speakers of English. It's probably even more important for you to get extra grammatical help and correction of your writing than to double up on your science classes. Lots and lots and lots of reading in English-language books on various subjects (including grammar and usage) will help, of course, but you need to find someone with a strong native speaker of English's sense of grammar to help you learn idiomatic English writing. You have time to invest your time in doing that, and that will do the most to support your application for competitive programs.

Yeah, PM me to identify the other language(s) you encounter, and then I may have some even more specific suggestions. I used to be a teacher of English as a second language, which is why I am sensitive to learners' "interference" from their native language when writing English.

One other thought: the top colleges in the country have sufficient financial aid that IF you get into them at all, you will be able to afford to go to them. But many applicants with impressive backgrounds apply to those colleges. The reason to participate in certain activities as a teenager is NOT to get scholarships, or even to get into the top colleges, but simply to gain a good education through challenging learning activities. You should choose activities because they are important to you whether or not they have a scholarship prize. Being yourself is a good idea when applying to college.

Good luck in your studies.

[url=http://www.cee.org/apps/usabo/index.jsp]USA Biology Olympiad[/url]

As far as the grammar goes, tokenadult, would you care to tell him about the grammar course Stanford EPGY is offering? (I don't know much about it yet...but I'm planning on taking it sometime soon.)

[quote="MithsApprentice"]As far as the grammar goes, tokenadult, would you care to tell him about the grammar course Stanford EPGY is offering?[/quote]

Thanks for bringing that up. I'm happy to tell anyone about the Stanford University Education Program for Gifted Youth (EPGY) Grammar of the English Sentence course, course [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url], because we are satisfied customers. We have a second language going on in our family, namely Mandarin Chinese, and the very different sentence structure of that language, which is A+MATH's native language even more than English is, means that the kids here sometimes have a rather haphazard acquaintance with English grammar.

I had A+MATH take [url=http://epgy.stanford.edu/courses/english/EG20/]EG20[/url] last year soon after it was first offered. The thing I especially liked about the course is that most of the example sentences are part of an elaborate series of jokes that go through the whole course. The course is much more humorous than any English course I ever had in school. And I think the course does a good job of focusing on some of the issues of English grammar most important for students who have a bilingual background--which many EPGY students have. I highly recommend the EG20 course, and wil sign up A+MATH for the follow-on EG21 course whenever that is first offered. (The follow-on course is still in development.)

How difficult is the EG20? And also, is it free (I hardly doubt it) or how much it costs?

Plus, I'm getting A's on my engilsh class, usually 100 or 99 or 98, especially on parts of speech. Only part is on where I have to write essays, short story, biography (I have to do one for Jan. 7  :( ) and everytime for those writings, I get points off on grammar. Always.

The most thing is that I do have to concentrate on AMC and math. BUT, for SAT, beginning at year 2005, the writing takes 800 points so 2/3 of SAT points are determined by English.

Also, is there a summer program for this? Like summer class? That would be really nice because I don't have a lot of time during the school days.

 :)

[quote="Silverfalcon"]The most thing is that I do have to concentrate on AMC and math.[/quote]

You should definately take that course.  What is your native language, just out of curiosity?

this has probably been said by someone at sometime, but i'd like to echo it now- props to tokenadult for always stressing reading and writing on this math forum. a lot of times with a lot of people, that kind of stuff falls by the wayside, but you CAN'T GET BY ON MATH SKILLS ALONE. communication is the key to the business world! (yes, i realize the irony in that this entire thing is written with no caps, but its informal so there). anyways, three cheers to tokenadult! thanks, yo!  :)

to the original questioner,

I don't know if you were asking about the money side of
things specifically, but:

In some sense any scholarship that you aren't "obviously"
a strong contender for and that has a time-consuming application,
is, in financial terms, a waste of time, or a gamble at best.
Most of the Westinghouse finalists get paid at less than minimum
wage, if they win anything (say, 800 hrs work for $2000).
Unless the scholarship has broad intrinsic prestige such
as Westinghouse, the Putnam contest, or an olympiad, it's
a good idea to consider the opportunity costs (i.e. the value
of alternatives that you don't spend time on, in pursuing the
scholarship).   You could try and compute the value of the
scholarship divided by the effective number of applicants
you'd be competing against to assess the odds, for example.
Over-calculation is a bad idea too, but wasting a lot of time
hunting for minor scholarships is not that fruitful.

If the only thing you expect to gain from a competition is money, chances are it would be more fruitful to get a job, even a low-paying one. However, a contestant in a big competition such as Westinghouse will surely learn a lot about doing research, and that experience is likely a lot more valuable in the long run than the prize fund. (This is likely still true for the grand prize winner.) Therefore I would highly recommend competing in a competition such as Westinghouse over getting a job doing something you'd rather not be doing. (On the other hand, if you can get a job doing something you want, that may be a different story.)

Sure, there's nothing wrong with doing independent
work in high school.  Almost all the advanced math
people that I knew in high school had their own
projects (sometimes half-baked, sometimes more baked)
that they pursued, though these things usually were
never written up or submitted for publication/competition.
Usually these were rediscoveries of things that were
already known --- again, nothing wrong with that.

There's a separate issue, though, of whether it's worth
seeking out a project specifically for entering
a competition, or writing up or extending some already-done
for the Westinghouse, ISEF etc.  Unless it's
likely to win something substantial (in prestige, money,
or whatever), if the additional effort specifically
directed at the bureaucratic or competitive aspect of
things is significant,  it is debatable whether that's worth
it as compared to, say, learning and doing other things.

Professors at top universities are generally pretty
jaded about students' undergraduate and, especially,
high school research projects.  They know that a lot
of it is resume padding and they also know that it's
usually "research" in name only.   On the other hand, there
have been a lot of really good undergraduate research
projects where the faculty members put a strong student
onto a well-chosen subject.   See e.g. James Propp's
groups on algebraic combinatorics, or some of the
Harvard baccalaureate theses; some of these are
available on the WWW.

Well, I should ask this or look on the Siemens Westing House website itself but does your school have to support you when you're going to attend Siemens-Wesing House?

My school doesn't really support any of the competitions a lot. The reason might be that our school is that our school is new to math  (at least that's what I thought since I was the first one to beat the school record on MC) and even for AMC, I persuaded the school to do it.

I hope I can attend ARML because that sounds like a good experience, according to AoPS book, and perhaps, some other good competition into the science.

I don't really want to get a job now or some later unless I feel that I have to because unless it's something that I enjoy (like being a tutor or some substitute for math or other subject :P ), I don't really think of getting job.

Back to the original question, anyone can tell me any good science competition? I might need to post this on the science forum but still, I think the opinion from the AoPS is very good.

Thanks to all people who answered this. But more advices wouldn't hurt a bit! :D

fleeting_guest, on pure principle alone, of course you are right that one should not enter a competition if one must be bogged down with bureaucratic details. However, I wonder how many high school students would do proper research without the infrastructure.

Siemens-Westing house certainly looks like a good contest to participate, just for experience also.

But that's on senior :( I have then two years empty. Any suggestion?

I'll attend AMC from now to senior year for sure.. Is there any biology test that's available? It doesn't necessarily have to have a lot of scholarship. Just like to attend some contests. :D

Do you have a local science fair you could enter?  Prizes may not be scholarships directly, but something to look nice on your college applications, plus give you good experience.

here's a list of contests. Some involve scholarship prizes, some don't.
http://www.academiccompetition.org/contests.html

Thanks for the link.

I really appreciate it.

Maybe I should ask this topic on the science forum also. :D

This question should be in U.S. Competitions but someone in the science forum suggested me the contest "Intel."

I have no clue what that is. Can anyone tell me what this is OTHER than giving me link to it?

Thanks.

The Intel science fair program is the successor to the Westinghouse science fair program. The program has always been operated by Science Service. It involves putting together a teenager's research project in some subject (math is a possible subject) and entering a project presentation in a regional science fair. Regional winners can go on to national and international levels of competition. Top winners are very highly desired by top colleges; some AoPSers have won major awards in the Intel program.

You would be REALLY rushed to enter any phase of the Intel program this school year, but you have lots of time to get ready for next year.

http://www.sciserv.org/isef/

I think I'll enter it next year, and maybe prepare it during the summer. I'm RUSHED for AMC now anyway. ;)

Question: Do the school has to support it? My school isn't kind of school that has those activities going on a lot....

It's a shame that us Canadians are ineligible for such prestigious competitions such as Intel or Westinghouse .  :no:

here's a link to a list of competitions, some of which involve scholarship awards:
http://www.academiccompetition.org/contests.html

Then there's always the scholarship search engines like http://fastweb.monster.com/


Compute: $ 55\times1212\minus{}15\times1212$ .


Find the number of linear functions $ f(x)$ such that there exists another linear function $ g(x) \equal{} x \plus{} 1$ satisfying the equation:
\[ f(x)g(x) \equal{} \frac {[f(x)]^3}{g(x)}
\]

(ignore $ f(x) \equal{} 0$)

Yup, that's correct. It's an easy problem... I got bored in Christmas night.  :)

[quote="ManuelS"]Is it two?

because
[g(x)]^2 = [f(x)]^2
abs(g(x)) = abs(f(x))

so f(x) =  x+1
and     = -x-1[/quote]

This reasoning is incomplete...basically you have shown that for all $ x$, $ f(x)\equal{}g(x)$ or $ f(x)\equal{}\minus{}g(x)$

But there is nothing that says only one of them is true for all $ x$. Ex. $ f(x)\equal{}g(x)$ for $ x<0$ and $ f(x)\equal{}\minus{}g(x)$ for $ x\ge 0$ is certainly possible without further reasoning.

You need to use the fact that $ f$ is linear otherwise there are an infinite number of functions $ f$.


geometry
conics
parabola
analytic geometry
geometry solved

I had this problem in a test some years ago, at Laurentiu Duican inter-district contest:

prove that if two parabollas with perpendicular axes  intersect in 4 points, then the 4 points lie on the same circle.

This problem can be solved by following:

Let (P):   y = ax 2 + bx + c
and (Q):  x = dy 2 + ex + f

Then the coordinates of intersection points satisfy equation
              d(ax 2 + bx + c - y) + a(dy 2 + ex + f - x) = 0
which is equation of a circle.

This results can be generalize for all pairs of curves of second order. For example, in one exam some years ago there was following

Let (E): x 2/9 + y2 = 1
(P):  y = x 2 - 2x
(a) Show that (E) and (P) intersect in 4 dinstinct poitns A, B, C, D
(b) Show that A, B, C, D lie in a circle. Find center and radius of this circle.


For those of you who have not heard, the Winthrop University Math Meet will not be held this year. They will, however, have a Math Day in the Spring (I believe on March 3rd) where students can come see presentations and talk to professors about the Winthrop University Math program.

For information about other local Math contests see the resources section of the AOPS site: http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php#sc

Thanks for the info.  I hadn't heard yet.  Did that letter say why they weren't holding the contest this year?

The math day sounds interesting though :)

looks like there will be more competition will be at usc aiken this year (winthrop and aiken were on the same day last year).

How come last year we never went to The Citadel or the Coastal Carolina meets?

The Citadel is after school on Wednesday's at the Citadel.  Since you have to live fairly close to make it usually only the schools near Charleston come.

Also, Coastal is usually on a school day (though it was during spring break last year :( ).  Some schools might not allow for their students to miss a day for a math competitions.


My "Favorites" disappeared (when I try to open them I see the general heading but nothing else). I'm using Mozilla 1.7 run under Windows XP.


quadratics
trigonometry
geometry
similar triangles

In a certain circle, the chord of a $d$-degree arc is 22 centimeters long, and the chord of a $2d$-degree arc is 20 centimeters longer than the chord of a $3d$-degree arc, where $d<120.$  The length of the chord of a $3d$-degree arc is $-m+\sqrt{n}$ centimeters, where $m$ and $n$ are positive integers.  Find $m+n.$

i really like this problem,

[hide="answer"]one of the big things to notice is that the chord that intercepts the 3d arc is LESS than the chord that intercepts the 2d arc, so since that is true, we see that the 3d arc is actually talking about the major arc...(this confuses me a bit, but i got a satisfactory answer), we form a triangle, use the inscribed angles to find some of the angles, draw the angle biscector of the bigger angle, and use some similar triangles, and the isoceles triangle formed to get a quadratic, and you find the third side to be $-9+\sqrt{165} \implies m+n=\boxed{174}$[/hide]

maybe somebody else could attempt this because from my definition of intercepted arc, the intercepted arc is the minor arc, and is therefore less than or equal to 180, but in this, the 3d arc seems to be the major arc of that chord

I've forgooten Ptomely's Thm. What is it?

For any convex quadrilateral $ABCD$ \[ AB \cdot CD+BC \cdot AD \geq AC\cdot BC  \]
equality holds when $ABCD$ is cyclic.

[quote="linj"][quote="stupidkid"]This problem is quite simple using Ptolemy's theorem.

Call the length of the $ 3d$ chord $ x$. Then we know that $ 22x \plus{} 22^2 \equal{} (x \plus{} 20)^2$. Solve for $ x$ and we are done.[/quote]
where is (x+20) come from?[/quote]
x+20 is the length of the 2d arc.

how do you get that the length of the chord of the 2d arc is x+20?

Without Ptolemy

[geogebra]3593f29254c77336d076bce95215fb2300e0f9b0[/geogebra]

From the picture you can see, $ \bigtriangleup ABE \sim \bigtriangleup CDE \sim \bigtriangleup ACD$

So $ \frac{a}{AE} \equal{} \frac{a\plus{}20}{22} \equal{} \frac{22}{EC}$ with $ AE\plus{}EC \equal{} AC \equal{} a\plus{}20$

Rearranging yeilds $ AE \equal{} \frac{22a}{a\plus{}20}, EC \equal{} \frac{22^2}{a\plus{}20}$

Then $ a\plus{}20 \equal{} AE \plus{} EC \equal{} \frac{22a\plus{}22^2}{a\plus{}20} \Longrightarrow a^2\plus{}18a\minus{}84\equal{}0$

$ \boxed{a \equal{} \sqrt{165} \minus{} 9}$

[quote=crastybow]Could someone please tell me where I went wrong here?
I attempted to do this using LOC; setting the radius equal to $r$ and the length of the chord with measure $3d$ equal to $x$, we can generate the following equations:

\begin{align*}
2r^2-2r^2\cos d &= 22\\
2r^2 - 2r^2\cos 2d &= x + 20\\
2r^2 - 2r^2\cos 3d &= x\end{align*}. Taking the quotient of the former two equations, we have $\dfrac{2r^2(1 - \cos 2d)}{2r^2(1 - \cos d)} = \dfrac{x+20}{22}$. Using $\cos 2d = 2\cos^2d - 1$, we can simplify our expression to $\dfrac{2(1 - \cos^2d)}{1 - \cos d} = \dfrac{x+20}{22}$. Thus, we have $ 1 + \cos d = \dfrac{x+20}{44} \implies \cos d = \dfrac{x-24}{44}$.

Now we take the quotient of the third and first equations:
\[\dfrac{2r^2(1 - \cos 3d}{2r^2(1 - \cos d} = \dfrac{x}{22}\]
Whence using $\cos(a+b)$ for $\cos 3d$ we get \[\dfrac{2r^2(1 - 4\cos^3d + 3\cos d)}{2r^2(1-\cos d)} = \dfrac{x}{22}\], so $4\cos^2d + 4\cos d + 1 = \dfrac{x}{22}$, which we rewrite as $(2\cos d + 1)^2 = \dfrac{x}{22}$. Substituting our expression for $\cos d$ into this equation yields $\left(\dfrac{x-2}{22}\right)^2  = \dfrac{x}{22}$, or $x^2-26x+4 = 0$, hence we get $\boxed{x = 13 + \sqrt{165}}$, which is incorrect and does not even fit the form of the answer.[/quote]

Is this actually even right as LoC, requires the RHS to be squared as well in the start when crastybow set up the equations?

If this is right, can someone explain why and show me how to complete it and get the right answer? Thanks

[quote=Mathster360][quote=Math1331Math][quote=stupidkid]This problem is quite simple using Ptolemy's theorem.

Call the length of the $3d$ chord $x$. Then we know that $22x+22^2=(x+20)^2$. Solve for $x$ and we are done.[/quote]

I don't understand how you set up your equation it seems nice (much faster than trig lol) but how did u set up? Thanks[/quote][/quote]

sorry for second time


1) if cosx + (cosx)^2 + (cosx)^3 = 1 then find the minimum value of tanx.

2) if sinX = cotAtanC and tanX = cosAtanB, then show that, cosA = +-(cosB/cosC)


geometry
perimeter
function
calculus
calculus computations

For (a) and (b) use Lagrange multipliers to find the maximum and minimum values of f subject to the given constraints.

a.) f(x, y) = 1/x + 1/y. Constraint: 1/(x^2) + 1/(y^2) = 1

b.) f(x, y) = x^2 + 2y^2 + 3z^2. Constraint: x + y + z = 1 and x - y + 2z = 2.

c.) A package in the shape of a rectangular box can be mailed by the post service if the sum of its length and girth (the perimeter of a cross-section
perpendicular to the length) is at most 108cm. Find the dimensions of the package with largest volume that can be mailed.

Seems pretty straightforward.  Follow the method.  Find the gradient of the function, find the gradient of the constraint function, find $ \lambda$, and solve.  So for the first one, $ \nabla f\equal{} (\minus{}x^{\minus{}2},\minus{}y^{\minus{}2})$, and $ \nabla g\equal{}(\minus{}2x^{\minus{}3},\minus{}2y^{\minus{}3})$.  So write
$ \minus{}\frac{1}{x^{2}}\equal{}\minus{}\frac{2}{x^{3}}\lambda$ and
$ \minus{}\frac{1}{y^{2}}\equal{}\minus{}\frac{2}{y^{3}}\lambda$ and
$ \frac{1}{x^{2}}\plus{}\frac{1}{y^{2}}\equal{}1$.
You have three equations and three unknowns so just solve.


A triangle $ ABC$ is given with $ \left|AB\right| > \left|AC\right|$. Line $ l$ tangents in a point $ A$ the circumcirle of $ ABC$. A circle centered in $ A$ with radius $ \left|AC\right|$ cuts $ AB$ in the point $ D$ and the line $ l$ in points $ E, F$ (such that $ C$ and $ E$ are in the same halfplane with respect to $ AB$). Prove that the line $ DE$ passes through the incenter of $ ABC$.

$I$ is the incenter and $DI$ cuts $\ell$ at $E_0.$ Since $\angle BIC=\angle BDC=90^{\circ}+\frac{_1}{^2}\angle BAC,$ it follows that $B,I,C,D$ are concyclic $\Longrightarrow$ $\angle E_0IC=\angle ABC.$ Then $\angle ABC=\angle E_0AC$ $\Longrightarrow$ $A,I,C,E_0$ are concyclic. Since $\angle AID=\angle AIC,$ then chords $AC=AE_0$ are equal $\Longrightarrow$ $E \equiv E_0,$ as desired.

[b]Remark:[/b] Similarly, we show that $DF$ passes through the A-excenter of $\triangle ABC.$

Very easy. Call $ I$ is the incenter of $ \Delta ABC$. We have:
$ \angle ADE\equal{}\pi\minus{}frac{\angle A\plus{} \angle B}{2}\equal{}frac{\angle C}{2}\equal{}\\angle ACI\equal{}angle ADI\equal{}$
Hence we get the result.

[quote="djuro"] [color=darkred]Let $ \triangle ABC$ with the incenter $ I$ and  $ c>b$ . Denote $ D\in (AB)$ for which $ AD\equal{}b$ and $ E\in AA$ (the tangent in a point $ A$

to the circumcircle of $ \triangle ABC$) so that $ AE\equal{}b$ and the sideline $ AB$ doesn't separate $ C$ , $ E$ . Prove that the line $ I\in DE$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ X\in AI\cap DE$ and $ Y\in BC\cap AI\ \implies\ m(\angle BAE)\equal{}A\plus{}B$ , $ YD\equal{}YC\equal{}\frac {ab}{b\plus{}c}$ , $ m(\angle ADY)\equal{}C$ .

 Thus, $ DY\parallel AE\ \implies\ \frac {XA}{XY}\equal{}\frac {AE}{DY}\equal{}$ $ \frac {b}{\frac {ab}{b\plus{}c}}\ \implies\ \frac {XA}{XY}\equal{}\frac {b\plus{}c}{a}$ $ \stackrel{\mathrm{Van\ Aubel}}{\ \ \implies\ \ }X: \equal{}I$ . In conclusion, $ I\in DE$ .[/color]

Denote $\widehat{BAC}=2a,\widehat{ACB}=2c,\widehat{ABC}=2b$(after angle chasing) we have $\widehat{EAC}=2b$(using $\triangle{ACE},\triangle{ACD}$  are isosceles we have $\widehat{ADE}=90-a-b$,so$\widehat{EDC}=b,\widehat{ACI}=c,\widehat{ICD}=90+a-c$ and all we have to do is to aplt Ceva's reciprocal theorem in tre trigonometric form. The concluision follows only using thing like ${sin a sin b=-\frac{1}{2}(cos(a+b)-cos(a-b)}$

Let $\angle CAB=2\alpha$, $\angle ABC=2\beta$, $\angle ACB=2\gamma$. Suppose that $G\in BC$ such that $AG$ is bisector of $\angle A$. Let $I$ be the intersection point of $DE$ with $AG$. By hypothesis $\angle EAC=2\beta$, also the triangles $ADC$, $ADE$ and $ACE$ are isosceles, therefore making some calculations, $\angle AED=\gamma$, $\angle EDC=\alpha$ and $\angle IAC=\alpha$, hence AICE is cycle and $\angle ACI=\angle AEI=\angle AED=\gamma$, then $CI$ is bisector of $\angle C$ thus $I$ is the incenter.

[geogebra]5e812a8a2647e28f7ba0108b9e11c654a4a76154[/geogebra]

[quote="linboll"]In fact Where I first met this problem was in China's National High-school Math Competition(I don't konw the official English name of this contest, I just translated the words) in year 2005.
It was in the second test.The second test contains 3 olympiad-type problems, each problem worths 50 points.120 minutes are allowed in the second test.
I did this problem easily even though I was still a Junior student at that time.[/quote]
yes,it is.and it can be second-killed by just some angle substitutions.
by the way,the official name is China Second Round.

It can be also easily done with $S_{\triangle ADE}=S_{\triangle AID}+S_{\triangle AIE}$

Let $I$ be the intersection of the interior angle bisector with $\overline{DE}$. We have $\angle CEI = \angle CED = \frac12\angle CAD = \frac12\angle A = \angle CAI$, so $AICE$ is cyclic. Then, $\angle ICA = \angle IEA = \angle DEA = \frac12\angle DAF = \frac12\angle C$, so $\overline{IC}$ bisects $\angle C$. Hence, $I$ lies in two interior angle bisectors, so it is the incenter. $\blacksquare$


Το 10% του πληθυσμού μιας αφρικανικής χώρας πάσχει από σοβαρή ασθένεια. Ένα άτομο για το οποίο υπάρχει υποψία ότι πάσχει από την ασθένεια υποβάλλεται σε δύο ανεξάρτητα μεταξύ τους τέστ καθένα από τα οποία κάνει σωστή διάγνωση με πιθανότητα 90%. Ποια η πιθανότητα να πάσχει πράγματι από τη ασθένεια i) Δεδομένου ότι και τα δύο τέστ είναι θετικά, ii) Δεδομένου ότι μόνο το ένα τέστ είναι θετικό; Τι παρατηρείτε;

Υπόδειξη: Έστω Α το ενδεχόμενο «πάσχει από την ασθένεια»,   το ενδεχόμενο «το τέστ 1 είναι θετικό», και   το ενδεχόμενο «το τέστ 2 είναι θετικό» και αξιοποιήστε τις πιθανότητες:
P(A)=0,1
P(A')=0,9
P(Θ1/Α)=Ρ(Θ2/Α)=Ρ(Θ/Α)=0,9
P(Θ1'/Α')=Ρ(Θ2'/Α')=Ρ(Θ'/Α')=0,9
Ρ(Θ1Θ2/Α)=Ρ(Θ/Α)^2=0,81
Ρ(Θ1Θ2/Α')=Ρ(Θ/Α')^2=0,01

Τα Ρ(Ρ'1Ρ'2/Α) και  Ρ(Ρ'1Ρ'2/Α') πώς θα υπολογιστούν αριθμητικά;;;;


trigonometry
function
trig identities
Law of Cosines

A very massive cart accelerates down an inclined plane at angle $ \theta$. Inside, a very light pendulum of length $ l$ hangs in it. Find a formula for the period of this pendulum.

[hide="Answer?"]${ 2\pi\sqrt{\frac{l}{g\cos\theta}}}$ ?[/hide]

Draw the free-body diagram for the pendulum.  Once you get your forces, you can write a differential equation, from which you'll be able to get your formula.

Well, see, if you write the differential equation as JRav suggested, you'll actually find that the pendulum doesn't oscillate harmonically.

First of all, since the cart's acceleration down the incline is $ g\sin\theta$, the pendulum will experience a pseudoforce in the direction up the incline of magnitude $ mg\sin\theta$, where $ m$ is its mass (which doesn't matter).  The only other force on the pendulum is $ mg$, acting downward.  Then say the pendulum makes an angle $ \beta$ with the vertical; you can find the net torque on the pendulum about the pivot point from these two forces as a function of $ \beta$, and then equate that to $ I\frac{d^2\beta}{dt^2}$, where $ I \equal{} ml^2$.  This gives you a differential equation for $ \beta(t)$.  Can you take it from there?

Can one use the law of cosines on the triangle with sides $ g$ and $ g\sin\theta$ and included angle $ 90 \minus{} \theta$ to find the magnitude of the effective gravity is $ g\cos\theta$? The period of the pendulum is then $ 2\pi\sqrt {\frac {l}{a}}$ with $ a$ being the effective gravity.

I'm getting that the torque is $ \minus{} mgl\cos\theta\sin\beta$, so the differential equation (with small angle approximation) is $ \frac {d^2\beta}{dt^2} \plus{} \frac {g\cos\theta}{l}\beta \equal{} 0$. (By "vertical", I assume you mean a line perpendicular to the incline?)

It seems this is simple harmonic motion?

Oh, my mistake about saying that it's not simple harmonic, which it is (I didn't do the calculations carefully).

Yes, you are correct.

Edit: and also, LeFromage, I take back that statement about the effective gravity - that's actually the cleanest way to do this problem.  I'm just being stupid today.


[quote="bluejay206"]I am in Algebra 1 and in 9th grade (don't laugh :) ) I need help on this open ended problem, plz help.

[hide]Suppose you and your friends form a band, and you want to record a demo tape. Studio A rents for 100 dollars plus 50 dollars per hour. Studio B rents for 50 dollars plus 75 dollars per hour.
a. Solve the system
b. Explain what the solution of the system meants in terms of your band renting a studio (basically tell what the answer to the substitution means.)[/hide]

Thanks![/quote]
[hide]
Studio A: $y=50x+100$
Studio B: $y=75x+50$
solving, $-25x+50=0$
$x=2$
This basically means that after two hours, studio A becomes cheaper than studio B, and before two hours, studio B is cheaper than studio A. [/hide]


calculus
integration
trigonometry
calculus computations

Nice integral!

At first, I thought I should use $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ to evaluate the integral. Soon, I realised I was just making the problem even more complicated. The thing to note is that $99$ and $101$ are both close to $100.$ So, we should make use of that, also keeping in mind that integration by parts must appear somewhere in the course of the solution.

Note that $\sin (101x) = \sin (100x+x) = \sin 100x \cdot \cos x+\cos 100x \cdot \sin x.$

So, $\int \sin(101x)\sin^{99}x\,dx$

$= \int \sin 100x \cdot \cos x \cdot \sin^{99}x \,dx+\int \cos 100x \cdot \sin^{100}x \,dx$

$= \int \sin 100x \cdot \cos x \cdot \sin^{99}x \ dx+\frac1{100}\int \sin^{100}x \ d(\sin 100x)$

$= \int \sin 100x \cdot \cos x \cdot \sin^{99}x dx+\frac1{100}\left( \sin^{100}x \cdot \sin 100x-\int \sin 100x \cdot 100\sin^{99}x \cdot \cos x dx \right)$

$= \boxed{\frac1{100}\sin^{100}x \cdot \sin 100x+C}.$

Your second to last line; you subtract the first integral minus 1/100 of it...I dont see that it cancels out.

Look carefully -- there's a 100 in the middle of the integrand. They kill each other.

[quote="FieryHydra"]

$= \int \sin 100x \cdot \cos x \cdot \sin^{99}x \ dx+\frac1{100}\int \sin^{100}x \ d(\sin 100x)$

$= \int \sin 100x \cdot \cos x \cdot \sin^{99}x dx+\frac1{100}\left( \sin^{100}x \cdot \sin 100x-\int \sin 100x \cdot 100\sin^{99}x \cdot \cos x dx \right)$

[/quote]

Excuse me, How do you say this part, I can't understand why
$\frac1{100}\int \sin^{100}x \ d(\sin 100x)$=$\frac1{100}\left( \sin^{100}x \cdot \sin 100x-\int \sin 100x \cdot 100\sin^{99}x \cdot \cos x dx \right)$

It's integration by parts $\int f(x)g'(x)\, dx = g(x)f(x)-\int f'(x)g(x)\, dx$


See here for problem and solution
http://www.mathlinks.ro/Forum/viewtopic.php?p=630379#630379

This topic is in the wrong thread please move it.


Let $ \mathbb{F}$ be a field of $ q$ elements, and $ SL_2(\mathbb{F})$ the group of 2x2 matrices with entries in $ \mathbb{F}$ with determinant 1, the special linear group. What is the cardinality of  the special linear group?

The usual argument uses the fact that the set of bases of $ \mathbb F^n$ is a principal homogeneous $ \mathrm{GL}_n(\mathbb F)$-set, and the exact sequence $ 1\to\SL_n(\mathbb F)\to\GL_n(\mathbb F)\to\mathbb F^\times\to1$. owk

owk means $ 1\to SL_n(\mathbb F)\to GL_n(\mathbb F)\to\mathbb F^\times\to1$.


This may be asked before but I was wondering what subject material needs to be studied before you can attempt USAMTS problems?

More specifically I want to know if any of the problems requires trigonometry, calculus, or higher level math.

The problems are all "high school" level, meaning that some problems will involve trigonometry, but none will require calculus or anything more advanced.

The subject coverage on the USAMTS is very similar to that of the AMC 12 contest.


Let $y=(x-a)^2+(x-b)^2,a,b$ constants. For what value of $x$ is $y$ a minimum?
(A) $\frac{a+b}{2}$
(B) $a+b$
(C) $\sqrt{ab}$
(D) $\sqrt{\frac{a^2+b^2}{2}}$
(E) $\frac{a+b}{2ab}$


inequalities
JBMO
Junior Balkan
algebra
polynomial
function
algebra solved

Find all real numbers $ a,b,c,d$ such that \[ \left\{\begin{array}{cc}a \plus{} b \plus{} c \plus{} d \equal{} 20, \\
ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150. \end{array} \right.\]

Let $ S \equal{} ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150$.
We know that $ (a\plus{}b\plus{}c\plus{}d)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2S$, so $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 400 \minus{} 300 \equal{} 100$.

So we have to find all real numbers $ a, b, c, d$ such that
(1) $ a \plus{} b \plus{} c \plus{} d \equal{} 20$, and
(2) $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 100$.

By AM-QM inequality we have that

$ \frac {a\plus{}b\plus{}c\plus{}d}{4} \le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}$, with equality if and only if $ a\equal{}b\equal{}c\equal{}d$.

But these expressions are in fact equal, as they both equal 5.
So the only solution is $ a\equal{}b\equal{}c\equal{}d\equal{}5$  :lol:

[quote="uglysolutions"]Let $ S \equal{} ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150$.
We know that $ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2S$, so $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 400 \minus{} 300 \equal{} 100$.

So we have to find all real numbers $ a, b, c, d$ such that
(1) $ a \plus{} b \plus{} c \plus{} d \equal{} 20$, and
(2) $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 100$.

By AM-QM inequality we have that

$ \frac {a \plus{} b \plus{} c \plus{} d}{4} \le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}$, with equality if and only if $ a \equal{} b \equal{} c \equal{} d$.

But these expressions are in fact equal, as they both equal 5.
So the only solution is $ a \equal{} b \equal{} c \equal{} d \equal{} 5$  :lol:[/quote]Can we use AM-QM for negative numbers?

[quote="Johan Gunardi"][quote="uglysolutions"]By AM-QM inequality we have that

$ \frac {a \plus{} b \plus{} c \plus{} d}{4} \le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}$, with equality if and only if $ a \equal{} b \equal{} c \equal{} d$.

But these expressions are in fact equal, as they both equal 5.
So the only solution is $ a \equal{} b \equal{} c \equal{} d \equal{} 5$  :lol:[/quote]Can we use AM-QM for negative numbers?[/quote]

No, you can't use AM-QM for negative numbers.
But the gap in the solution is easily repaired:

If all four numbers are non-negative, then the above argument works.
If one of the four numbers is negative, then

$ \frac {a \plus{} b \plus{} c \plus{} d}{4}  < \frac {|a| \plus{} |b| \plus{} |c| \plus{} |d|}{4} \le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}$

and you get a contradiction.

That is really triviality
[hide]$ (a\minus{}b)^2 \plus{} (b\minus{}c)^2 \plus{} (c\minus{}d)^2 \plus{} (b\minus{}d)^2 \plus{} (a\minus{}c)^2  \ge 0$

$ \implies \frac{3}{8}(a\plus{}b\plus{}c\plus{}d)^2 \ge ab\plus{}bc\plus{}cd\plus{}ad\plus{}bd\plus{}ac$ with equality iff $ a\equal{}b\equal{}c\equal{}d$.

In this case we have equality, so $ a\equal{}b\equal{}c\equal{}d\equal{}5$.
[/hide]

With the given equations, we have $ \sum (a \minus{} 5)^2 \equal{} \sum a^2 \minus{} 10\sum ab \plus{} 100 \equal{} 0$ (the summations are cyclic), that is, $ a \equal{} b \equal{} c \equal{} d \equal{} 5$.

I suggest the following generalized problem:
Find all real numbers $ a_{1},a_{2},...,a_{n}$ which satisfy the conditions
$ \sum_{i\equal{}1}^{n}a_{i}\equal{}k>0$ and
$ \sum_{i,j\equal{}1, i\ne j}^{n}a_{i}a_{j}\equal{}\frac{k^{2}(n\minus{}1)}{n}$.

Yes! The main idea is that $ (\sum_{i\equal{}1}^{n}a_{i})^2\equal{}\sum_{i,j\equal{}1,i\ne j}^{n}a_{i}a_{j}\plus{}\sum_{i\equal{}1}^{n}a_{i}^{2}$. The final answer is $ a_{1},a_{2},\cdots,a_{n}\equal{}\frac{k}{n}$

with this generalization we can use a genaralization of the inequality FOURRIER proposed

[hide]
for reals $ a_1,a_2,...,a_n$

$ \sum_{i\ne j}{(a_i\minus{}a_j)^2}\ge 0$

$ (n\minus{}1)\sum{a_i^2}\ge 2 \sum_{i\ne j}{a_ia_j}$

$ (n\minus{}1)(\sum{a_i})^2\ge 2n\sum_{i\ne j}{a_ia_j}$
[/hide]

:)

No, I think AM-QM can works for negative numbers in the case $ p\equal{}2k$..

Because it's similar with Cauchy Schwarz Ineqauality :

$ 4(a^2\plus{}b^2\plus{}c^2\plus{}d^2) \ge (a\plus{}b\plus{}c\plus{}d)^2$

When (at least) one of the numbers is negative, the value will be smaller, so the ineq is still true..

yes, Ronald, you are right :)

$ \frac{a\plus{}b\plus{}c\plus{}d}{4} \le \sqrt{\frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2}{4}}$

if one (or more) of $ a,b,c,d$ is negative, then the value of $ LHS$ will decrease and won't ruin the system...

boxedexe's solution is good..

$ (a\minus{}5)^2\plus{}(b\minus{}5)^2\plus{}(c\minus{}5)^2\plus{}(d\minus{}5)^2\equal{}0$

[quote="Ahiles"]Find all real numbers $ a,b,c,d$ such that
\[ \left\{\begin{array}{cc}a \plus{} b \plus{} c \plus{} d \equal{} 20, \\
ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 150. \end{array} \right.
\]
[/quote]
$ 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\equal{}300$
$ 2(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)\equal{}300$ so
$ 3(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\minus{}2(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)\equal{}0$
$ (a\minus{}b)^2\plus{}(a\minus{}c)^2\plus{}(a\minus{}d)^2\plus{}(b\minus{}c)^2\plus{}(b\minus{}d)^2\plus{}(c\minus{}d)^2\equal{}0$
so $ a\equal{}b\equal{}c\equal{}d\equal{}5$

Suppose the problem was to solve for positive reals, it can be solved without any equation systems but by trivial AM- GM inequality.
a+b+c+d>= 4(abcd)^1/4---------->*
and ab+ac+ad+bc+bd+cd>=6 (abcd)^1/2 --------> **
which yields abcd=5 ^ 4 where AM=GM
implying the fact that a=b=c=d=5.

[hide="Trivial Solution using only AM-GM"]
$a+b+c+d=20$
$ab+ac+ad+bc+bd+cd=150$
Therefore $a^2+b^2+c^2+d^2=100$
By weighted AM-GM we have that:
$a^2/3+b^2/3>=2ab/3$
Therefore, we get that
$a^2+b^2+c^2+d^2>=(2/3)ab+ac+ad+bc+bd+cd$
$a^2+b^2+c^2+d^2>=100$
With equality when a=b=c=d.
Well, we already know that $a^2+b^2+c^2+d^2=100$ so $a=b=c=d=5$
[/hide]

You can also use the idea from the MMO '11, and look at it as a polynomial.
After we get $a^2+b^2+c^2=100-d^2$ and $a+b+c=20-d$, using Cauchy we get:
$(a^2+b^2+c^2)\cdot (1+1+1)\ge (a+b+c)^2$
$(100-d^2)\cdot 3\ge (20-d)^2$
$(d-5)^2\le 0$ for an arbitrary d, which implies $a=b=c=d=5$.

Similar Problem:
See here [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=424737&]1978 USAMO Problem 1[/url]

[quote="kunny"][quote="enescu"]The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$.
[/quote]

Let the quartic function $y=P(x)$ touch the line $y=l(x)$,

Rewrite $P(x)=(x^2-10x+p)^2+\cdots$, yielding $p=25$ by observing the coefficient of $x^2$, thus

We can write $P(x)=(x-5)^4+(m-500)x+n$, since $P(x)$ is a polynomial the fourth degree, therefore $P(x)=0\Longleftrightarrow x=5$,

that is to say, $(m-500)x+n\equiv0\Longleftrightarrow  m=500,\ n=0$.[/quote]
Slightly inintelligible (what is that "line" $y=l(x)$, and what role does it play?). Also slightly wrong; in fact $P(x) = (x-5)^4 + (m-500)x + (n-625)$. Now the follow-up (which I still fail to fully understand ...) at least reaches the correct values $m=500$, $n= 625$.

Thank you for the correction in replace of me, mavropnevma.

[quote="mavropnevma"][quote="enescu"]The numbers $a,b,c,d$ are the real roots of the polynomial $P(x)=x^4-20x^3+150x^2+mx+n$, for some reals $m,n$.
If they are not equal, then $P'(x)$ has three non-equal roots and hence $P''(x)$ has two non-equal roots. But $P''(x)=12(x^2-10x+25)$, whose roots are equal to $5$. We deduce that $a=b=c=d=5$.
I used a consequence of Rolle's Theorem[/quote]
A slight imprecision. If $a,b,c,d$ are pairwise distinct, or if two of them are of equal value $v$ and the other two are either of distinct values different from $v$, or of equal value different from $v$, then indeed $P'(x)$ has three non-equal real roots. But if three of them are equal, and the fourth is of a different value, then $P'(x)$ has two equal roots, and a third of a different value, still however enough for $P''(x)$ to have two non-equal real roots - leading to the above contradiction.[/quote]
I don't see the "imprecision". Three non-equal roots means that the three roots are not all equal, as I see it.

Then I must beg your pardon. I read "three non-equal roots" as meaning "three distinct roots", seeing that "three roots, not all equal" would have been an equivalent, clearer (to me) way of expressing it ...

[quote="mavropnevma"]Then I must beg your pardon. [/quote]

No, it's my fault. There was a better way to point out that idea.

It's very easy to prove that $(a+b+c+d)^2\ge \frac{8}{3}(ab+ac+ad+bc+bd+cd)$:
$(a+b+c+d)^2\ge \frac{8}{3}(ab+ac+ad+bc+bd+cd)\Leftrightarrow (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+(a-c)^2+(b-d)^2\ge 0$, which is obviously true.
Since, $a + b + c + d = 20$ and $ab + ac + ad + bc + bd + cd = 150 \Rightarrow$ In the inequality I've proved, the equality holds. $\Rightarrow a=b=c=d$
$\Rightarrow a=b=c=d=5$.

Note that \[ a^2+b^2+c^2+d^2 = 20^2 - 2 \cdot 150 = 100 \] but by Cauchy-Schwarz \[ 100 \cdot 4 = (a^2+b^2+c^2+d^2)(1+1+1+1) \ge (a+b+c+d)^2 = 400 \] so equality must hold meaning $a=b=c=d=5$, and this clearly works.

[hide=Same solution]Since $a^2+b^2+c^2+d^2=20^2-2\cdot150=100$, C-S gives $400=4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2=400$. Since equality holds, we get $(a,b,c,d)=\boxed{(5,5,5,5)}$, which works.[/hide]

This was A2 on the shortlist.

[b]Overcomplicated:[/b]

The only possible $4$-tuple is $(a, b, c, d) = (5, 5, 5, 5)$, which clearly works. Now, we will show this is the only solution.

It's clear that $$a^2 + b^2 + c^2 + d^2$$ $$= (a+b+c+d)^2 - 2(ab + ac + ad + bc + bd + cd) = 100.$$ Thus, we have $$\sum_{cyc} (a - 5)^2 = \left( \sum_{cyc} a^2 \right) - 10 \left( \sum_{cyc} a \right) + 4 \cdot 25$$ $$= 100 - 10 \cdot 20 + 100 = 0$$ so the trivial inequality implies $a = b = c = d = 5$ is the only possible solution. $\blacksquare$

[b]Remark:[/b] We actually have $$\sum_{cyc} (a - k)^2 = (10 - 2k)^2$$ for all real $k$. This result can be conjectured by plugging in small positive integers for $k$.

Let S=sum of permutations of ab from a to d = 150. Then $400=(a+b+c+d)^2=a^2+b^2+c^2+d^2+2S$, or $a^2+b^2+c^2+d^2=400-2*150=100$.
By AM-QM we have that
$ \frac {a\plus{}b\plus{}c\plus{}d}{4} \le \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}$, with equality if and only if $ a\equal{}b\equal{}c\equal{}d$.

But these expressions are equal, so equality holds at $a=b=c=d=5$.


Prove:
$\sum{\frac{a}{\sqrt{a^{2}+8bc}}}<2$ holds for all positive real numbers $a,b,c$

The inequality
\[\sum_\text{cyc}\frac{a}{\sqrt{4a^{2}+5bc}}\leq 1\]
is much stronger and equivalent to the one posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150022]here[/url].


[hide="big hint for (a)"]
\[\sin 3\theta=3\sin \theta-4\sin^{3}\theta\Rightarrow \sin^{3}\theta=\frac{3\sin \theta-\sin 3\theta}{4}\]
[/hide]
(b) is similar

[hide="Cool calculus to bash it!"]\[\begin{aligned}\sum_{i=0}^{\infty}\frac{1}{3^{i}}\sin^{3}3^{i}x & = \int\left(\sum_{i=0}^{\infty}3\sin^{2}3^{i}x\cos 3^{i}x\right)dx = 3\int\left(\sum_{i=0}^{\infty}\cos 3^{i}x-\cos^{3}3^{i}x\right)dx \\ & = \frac{3}{4}\int\left(\sum_{i=0}^{\infty}\cos 3^{i}x-\left(4\cos^{3}3^{i}x-3\cos 3^{i}x\right)\right)dx = \frac{3}{4}\int\left(\sum_{i=0}^{\infty}\cos 3^{i}x-\cos 3^{i+1}x\right)dx \\ & = \frac{3}{4}\int \cos xdx = \frac{3}{4}\sin x+C\end{aligned}\]For $x= 0$ we get $C = 0$, so our expression is equal to $\frac{3}{4}\sin x$.[/hide]

What is this problem doing in "high school intermediate topics"? This looks to me like it belongs in the "calculus computations and tutorials" section.

Well, I think the problem is solvable with intermediate topics using scorpius's hint.

[hide="like this"]
$\sin^{3}\theta=\frac{3\sin \theta-\sin 3\theta}{4}$
Plugging this into the problem,
$\frac{3\sin \theta-\sin 3\theta}{4}+\frac{3\sin 3\theta-\sin 9\theta}{12}+\cdots$
$=\frac{3\sin \theta-\sin 3\theta}{4}+\frac{\sin 3\theta-(\sin 9\theta)/3}{4}+\cdots$
The second part of the first term and first part of the second term cancels out, so we see that the whole thing telescopes to
$=\frac{3\sin \theta}{4}$
 (b) is solved similarly(I think), with the identity for $\cos3\theta$
[/hide]


[quote="bpms"]At the grading of the Advanced Placement Calculus exams,  470  math teachers graded  103000  tests in  50  hours. Each test had  6  problems. What is the average number of minutes it took one teahcer to grade  15  problems? Express your answer to the nearest minute.[/quote]
[hide]
103000 / 470 * 6 / 50 * 15 DIVIDED BY 60 ~ 7 minutes. OK, corrected.[/hide]


trigonometry
inequalities open
inequalities

Let [tex]a,b,c,d \in[\frac{-\pi}{2},\frac{\pi}{2}][/tex] be real numbers such that:
[tex]\sin a+\sin b+\sin c+\sin d=1[/tex]

and:
[tex]\cos 2a+\cos 2b+\cos 2c+\cos 2d\geq \frac{\ 10}{3}[/tex]
Prove that [tex] 0\leq a,b,c,d \leq \frac{\pi}{6}[/tex]

It  is not hard.
 the solution it here.
[hide]
let $p=\sin a,q=\sin b,r=\sin c,s=\sin d$
the problem is the same as $p+q+r+s=1,p^{2}+q^{2}+r^{2}+s^{2}\leq\frac{1}{3}$
we should prove $0\leq p,q,r,s\leq\frac{1}{2}$
because  $(1-p)^{2}=(q+r+s)^{2}\leq 3(q^{2}+r^{2}+s^{2})\leq 1-3p^{2}$
it is clear.[/hide]


quadratics
function
linear algebra
matrix
algebra
polynomial

I'm struggling with starting this project i have drawn up the graphs but i dont understand where or how i get the functions of each graph!

[u]Question[/u]:
Using the data, using a matrix or otherwise, find a quadratic function and a cubic function which model this situation. Explain the process used.

data = Gude numbers (from tip)  1     2     3     4     5    6     7      8
          Distance from tip (cm)     10 -  23 -  38 -  55 -  74 -  96 -  120 - 149

If anyone could help out or even maybe give me a hint of how to start the problem i would be very grateful
thank you for your time and help
Magic Man

Are you allowed to use a graphing calculator?

yes your allowed to use a graphing calculator, excel or any program which can generate graphs.

You'll find that Microsoft Excel is very easy to use, especially if you're going to be taking a lot of future math/science classes that will do a lot of modeling.

First, open Excel and type your data as in the first picture below. [b]Highlight[/b] all of your data and have Excel give you a [b]scatterplot graph[/b]. Depending on what version of Excel you have, this will be done in various ways. If you don't know how to do this, press F1 for help, or ask that paper clip guy to tell you how to do it.

Then, right click one of the data points in your graph, and click Add Trendline..., as seen in the second picture below.

Next, in the window that pops up, make sure that a polynomial-type trendline is selected. Depending on what degree polynomial you want (2 for quadratic, 3 for cubic), make sure the proper selection is made. You also want to make sure that the "Display Equation on Chart" box is selected. See the third picture below.

Your equation will then pop up on the graph.For example, I found the quadratic to be $ y \equal{} 1.244x^2 \plus{} 8.458x \plus{} 0.839$. Make sure you can do this on your own by finding the cubic.


Prove that $ 2^p\plus{}3^p$ is not a perfect power if $ p$ is a prime number

Check $ p\equal{}2,5$ manually. Then note that for all other primes, $ 2^p\plus{}3^p$ is divisible by $ 5$ (as $ p$ is odd) but not by $ 25$ (as $ p$ is not divisble by $ 5$).

or modulo $ 4$ if $ p$ is odd than $ 2^p\plus{}3^p\equal{}3 (mod4)$

Why does this suffice¿ I think you misunderstood "perfect power" for "perfect square".

lol true:D didnt read that well. my bad. than what u said. a bit longer but works


search
number theory unsolved
number theory

Prove that this equation doesnt have any result in set of Natural numbers:
x^2 y^2 =z^2(z^2 - x^2 - y^2)


integration
logarithms
trigonometry
function
algebra
polynomial
rational function

$\int \frac{e^{x}dx}{\sqrt{e^{2x}+1}}\,dx$

How do you know what to let?

MY answer [hide]$2\sqrt{e^{2x}+1}$[/hide]

Actual answer [hide]$log_{e}(e^{x}+\sqrt{e^{2x}+1}$[/hide]

Start by changing variables according to $e^{x}= t \Rightarrow x = \ln t \Rightarrow x' = 1/t$:

$P = \int \frac{e^{x}}{\sqrt{1+e^{2x}}}\,dx = \int \frac{1}{\sqrt{1+t^{2}}}\,dt$.

Now change variables again as $t = \tan u \Rightarrow t' = \sec^{2}u$:

$P = \int \frac{\sec^{2}u}{\sqrt{\sec^{2}u}}\,du = \int \sec u\,du = \ln|\sec u+\tan u|+C$.

Remembering that $t = \tan u$ and $\sec u = \sqrt{1+\tan^{2}u}$ then

$P = \ln|t+\sqrt{1+t^{2}}|+C$.

But $t = e^{x}$, and so we have

$P = \ln(e^{x}+\sqrt{1+e^{2x}})+C$

where I have dropped the modulus because the exponential is always positive.

Alternatively, one could recognize that we have a primite of the type

$\int\frac{f'}{\sqrt{1+f^{2}}}$

which equals $argsinh(f)+C = \ln|f+\sqrt{1+f^{2}}|+C$

Same idea: make the substitution you can make. Let $u=e^{x},$ since you have available for use $du=e^{x}\,dx.$ That would convert this to a rational function of $u.$ You could do several things with it, including factoring the denominator and using partial fractions.

$\frac1{u^{2}-1}=\frac{1}{(u-1)(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}$

There are several ways to proceed from here to finding the unkown constants $A$ and $B.$ One way, and it should work in all instances, is to multiply both sides of the equation by the common denominator - in this case, $(u-1)(u+1).$ That makes both sides polynomial, and the polynomials must be identical. Equate the coefficients of the constant terms on both sides, the $u$ terms on both sides, the $u^{2}$ terms on both sides (none of those in this case), and so on. You should have as many linear equations as you have unknowns, and you can solve them.

There's a special trick for simple linear factors: multiply by just one factor, then take the limit as the variable tends to the point at which that factor would be zero. Repeat for the other factors.

$\frac{1}{(u-1)(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}$

Multiply by $u-1.$

$\frac{1}{u+1}=A+\left(\frac{B}{u+1}\right)(u-1)$

Take the limit as $u\to 1.$

$\frac12=A+0,$ so $A=\frac12.$

Go back to the top, then multiply by $u+1.$

$\frac{1}{u-1}=\left(\frac{A}{u-1}\right)(u+1)+B.$

Take the limit as $u\to-1.$

$\frac{1}{-2}=0+B,$ so $B=-\frac12.$

We have found the following algebraic identity. Now that we have it, it would be easy to check.

$\frac{1}{u^{2}-1}=\frac12\left(\frac1{u-1}-\frac1{u+1}\right).$

[In my own personal case, I've done this one and its close relatives often enough that I would just write down the answer without writing out any intermediate steps. You learn to recognize some things.]

thanks doc for your detailed proof and the tricks

appreciate it!


S25

Let $ (\sqrt{2}+1)^{2007}= a+b\sqrt{2}$, where $ a$ and $ b$ are integers. What is the highest common factor of $ b$ and 81?

[hide="Hint 1"] $ (\sqrt{2}-1)^{2007}=-a+b \sqrt{2}$ [/hide]
[hide="Hint 2"] There is a second-order linear recurrence for $ b$.  Thus it's only necessary to find the period of the recurrence $ \bmod 81$. [/hide]

What's a second-order linear recurrence?

Let $ (\sqrt{2}+1)^{n}= a_{n}+b_{n}\sqrt{2}$.  Then there are constants $ A, B$ such that

$ b_{n+2}= A b_{n+1}+B b_{n}$

My answer is 1.
 Note that the highest common factor repeat after a cycle of 4.

 (1+root2)^2=3+2root2
 (1+root2)^3=7+5root2
 (1+root2)^4=17+12root2

Continue to do this and you will discover that the HCF repeat in this cycle 1,1,1,3,1,1,1,3,1,1,1,3,...................

t0rajir0u, where do you read about these second-order linear recurrences? I read somewhere else they can be used to generate solutions to equations like $ x^{2}-2y^{2}= 0$

I also put 1, how did you get it?


geometry
geometric transformation
reflection
angle bisector
geometry proposed

Incircle $ (I)$ of $ \triangle ABC$ is tangent to $ BC,CA,AB$ at $ D,E,F.$ The A-excircle is tangent to $ BC$ at $ P.$ Let $ Q$ be the antipode of $ D$ WRT $ (I)$ and define the points $ M \equiv QE \cap BC,$ $N \equiv QF \cap BC$ and $R \equiv IA \cap EF.$ Prove that the lines $ QR$ and $ PI$ meet at the symmedian point of $ \triangle QMN.$

[b]Solution:[/b]

$ \boxed{\widehat{QMN}} \equal{} 90^o \minus{} \widehat{QDE} \equal{} \boxed{\widehat{QFE}} \equal{} \boxed{\widehat{QEA}} \equal{} \widehat{QDE} \equal{} \boxed{\widehat{ICD}}$

$ \Longrightarrow EFMN$ is concyclic ;$ QM \parallel IP$; $ \ QN \parallel IB$ ;  $ \bigtriangleup NBF$ is isosceles ; $ \bigtriangleup MCE$ is isosceles

Because $ I_aB \perp IB$ ; $ IB \parallel QN$ and  $ \bigtriangleup NBF$ is isosceles $ \Longrightarrow I_aB$ is the midpenperdicular line of $ NF$ . Similarly, $ I_aC$  is the midpenperdicular line of $ ME$. It's mean  $ I_a$ is the center of circle $ (MNFE)$.

$ \Longrightarrow P$ is midpoint of $ MN$

$ \bigtriangleup QEF \sim \bigtriangleup QNM$ and $ R$ is midpoint of $ EF$ . So the symmedian point of $ \triangle QMN$ lies on $ QR$ (the reflection of $ QP$ throught the angle bisector of angle $ Q$ of $ \bigtriangleup QMN$)

We also knew that the the symmedian point of $ \triangle QMN$ is collinear with $ I$ (midpoint of $ Q$-altitude of $ \bigtriangleup QMN$) and $ P$ ( the midpoint of side $ MN$)

As the consequence, $ G' \equiv IP \cap QR$ is the symmedian point of $ \triangle QMN$

We end the proof .

[hide]@Thanx Luis, your problem and solution with inversion are very nice  :) [/hide]

Thanks for your solution hophinhan, mine is similar but I used inversion to prove that $ P$ is the midpoint of $ MN.$

Inversion with respect to the circumference centered at $ Q$ with radius $ QD$ takes the incircle $ (I)$ into the sideline $ BC.$ Thus $ M,N$ are inverse images of $E,F$ and the circle $ (A,AF)$ orthogonal to $ (I)$ is transformed into the circle with diameter $MN$ $\Longrightarrow$ its center must lie on $ QA$ $ \Longrightarrow$ $ P \equiv QA \cap BC$ is midpoint of $ MN.$ It's easy to see that $ QR$ is the Q-symmedian of $ \triangle QMN,$ since $ R$ is the midpoint of the antiparallel section $ EF$ to $ MN.$ Then the line connecting the midpoint $ I$ of its Q-altitude $ QD$ and the midpoint $ P$ of $ MN$ is a [url=http://xtec.net/~qcastell/ttw/ttweng/definicions/d_Schwatt_r.html]Schwatt line[/url] of $ \triangle QMN$ containing its symmedian point, i.e. $ J \equiv QR \cap PI$ is the symmedian point of $ \triangle QMN$ and the proof is completed.

“Schwatt line”，I first heard that，it's good，thanks，you guys .


limit
integration
calculus
function
calculus computations

My Analysis teacher wrote this problem on his website and it looks like it is not true. I cannot reach him for clearification. Can anyone correct this and also I would appreciate any hint on how to proceed.

Show that if $-1 < x < 1$ then:
$\lim_{n\to\infty}\sum_{k = 0}^{n}\frac{x^{k}}{k}=-ln(1-x)$

Shouldn't $k$ in the summation be $k!$ so as to avoid division by zero? Or one could start index k at 1 instead of 0, but I am not sure which one is the correct one.

It should be $k,$ not $k!,$ but the summation should run from $k=1$ to infinity.

That looks like a simple typo to me.

$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{x^{k}}{k!}=e^{x},$ which is a very different function.

[quote="Extremal"][quote="akech"]thanks!!!![/quote]it's look like something this $\int\frac{A^{u}}{u^{a}}du$ in wich $a=1$ and integral interval is $(0,+oo)$ and $A=e^{lnx}$. let's   research it. Problem number one: find equivalent function of this integral $\int\frac{A^{u}}{u^{a}}du$ , find in wich interval it will be equivalent: in few words: research this problem :lol:[/quote]
It's wellknow , and I know what is this equivalent function  :wink:

Extremal,
That was a good suggestion, but I just used epsilon definition to show that if $|x|< 1$ then:
$\lim_{n\to\infty}\sum_{k = 0}^{n}x^{k}= \frac{1}{1-x}$.
Then I just integrate both sides of the above equation to get the following:
$\lim_{n\to\infty}\sum_{k = 1}^{n}\frac{x^{k}}{k}=-ln(1-x)$.


Find all positive integers $n$ such that $2^{n-1} \mid n!$.

I think it's quite easy. The number of two's in the prime extension of $n!$ is:

$\sum_{i=1}^{infinity}\frac{n}{2^i}$.

The last non-zero term of the expansion is $\frac{n}{2^{\left[log_2n\right]}}$.

Then from $2^{n-1}|n!$ it follows:

$n(\frac{1}{2}+...+\frac{1}{2^{\left[log_2n\right]}})\geq n-1$

From this we have:

$2^{\left[log_2n\right]}\geq n$.

To assure this $log_2n$ must be an integer and $n$ a power of $2$. It's obvious that all such $n$ really solve the problem.

Misha


inequalities
modular arithmetic
algebra unsolved
algebra

Let $ x_1, x_2,..., x_n$ are reals satisfy:
1. $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 0$.
2. Exist unique i such that  $ x_i \equal{} 1$.
Prove that Max$ (|x_i \minus{} (x_{i \minus{} 1} \plus{} x_{i \plus{} 1})/2)$ >= 8/$ n^2$   with i=1, 2,..., n.

Plugging $ i \equal{} 1$ in the inequality gives us an unknown value $ x_0$. Recheck your question.

i think the indices are defined $ \pmod{n}$.

Yes $ x_{n\plus{}1}\equal{}x_1$.
Let $ y_k\equal{}x_{k\plus{}1}\minus{}x_k,z_k\equal{}y_{k\plus{}1}\minus{}y_k\equal{}$. Then $ \sum_i y_i\equal{}0, \ \sum_i z_i\equal{}0$. If $ |z_k|<C \ \forall k$, then $ x_j\equal{}x_i\plus{}y_i\plus{}....\plus{}y_{j\minus{}1}\equal{}x_i\plus{}(j\minus{}i)y_i\plus{}(j\minus{}i\minus{}1)z_i\plus{}...\plus{}z_{j\minus{}2}$. If $ C<\frac {8}{n^2}$, then $ x_{i\pm n/2}>\minus{}1$ and $ \sum_i x_i >0$ give contradition.

[quote="Rust"] If $ |z_k| < C \ \forall k$, then $ x_j \equal{} x_i \plus{} y_i \plus{} .... \plus{} y_{j \minus{} 1} \equal{} x_i \plus{} (j \minus{} i)y_i \plus{} (j \minus{} i \minus{} 1)z_i \plus{} ... \plus{} z_{j \minus{} 2}$. If $ C < \frac {8}{n^2}$, then $ x_{i\pm n/2} > \minus{} 1$ and $ \sum_i x_i > 0$ give contradition.[/quote]

What is C?

[quote="Rust"]. If $ C < \frac {8}{n^2}$, then $ x_{i\pm n/2} > \minus{} 1$ and $ \sum_i x_i > 0$ give contradition.[/quote]
I see. But I still don't understand this  :(

duytungct! have you had the answer of the problem? please post it here!


linear algebra
matrix
superior algebra
superior algebra unsolved

calculate rank of matrix
$\left( \begin{array}{ccccc} e_1 \\ e_2 \\ e_3 \\ e_4 \\ e_5 \end{array} \right)$
for b)
for a) i dont know how could you do it


Hello everyone. I found this site a few days ago through googling "AMC 10/12" in search of methods to score higher. Every year, about 8 kids from my high school progress onto the AIME. I consider myself competently intelligent to achieve most intellectual feats, if i put time into the learning, but my AMC 10 scores were... unsatisfactory (I didn't qualify). Now that i know a bit more about the AMC type tests, I realize how little i know. I didn't even know exactly what "Number Theory" was, until recently! Now my head is spinning with how much I didn't know about this broad world of "pure mathematics". I ordered a textbook off amazon, "Mathematical Olympiad Challenges (Paperback)
by Mark Saul (Foreword), Titu Andreescu, Razvan Gelca", perhaps rather hastily, but I am very eager to delve into this intriguing field.

I would appreciate some help with the best ways of approaching this new field to me.

-Where do I start? I've been reading some stuff on Wikipedia; but so much of it seems esoteric to me right now. Which theorem, conjecture, or hypothesis should I start with to build a foundation?
-What are the best resources to take advantage of for a thorough introduction to number theory? (Books only if they aren't too spendy please, and i would much prefer websites)
-I'm a junior next year (I have to take the AMC 12). I would love to qualify for AIME. Can someone guesstimate approximately how much work i will need to put in to achieve this, in the next ~9 months?

Any additional help anyone could provide me with in regard to advice for beginners to these types of problems (number theory) would be VERY appreciated.

Thank you very much,
Miller Williams

Well, if you're not qualifying for the AIME, my guess is that you don't have a basic foundation of knowledge - honestly, I'd recommend the Art of Problem Solving books (Both volumes help, but if money is an issue, just get volume 1, as it is about five times as useful as volume 2 for the AMCs.)

The thing is ... you already started it and you didn't know!!!

Begin to train your problem solving skills right here in this forum!!!

There are tons of old posts that you can browse and learn. Begin from the easier steps (you did the right thing, begin right here in the subforum Getting Started). Check out the Resources page under USA and check out the past AMC tests so that you can get used to the type of problems that might ask. Also, try to solve as many problems as you can, the more you try to tackle (even if some are too hard, at least try and think you've seen it and done it) the more experienced you become and the better problem solver you will be.

Lastly,

 :welcomeani:


So as you might remember, last year I posted a list of all of the major contests in the Bay Area region.
I'm going to be doing the same thing shortly, and I have to update my personal list.

I'll be posting this list up sometime next week, so if you have any information that might be useful to me (contest directors for the Julia Robinson thing, I'm looking at YOU) then if you could let me know (either on these boards, via PM, or email me at matt.redmond@menloschool.org) that would be great.

Having a centralized schedule like this helps to plan things like trips, tests, even vacations and finals around the contest dates. (Because I KNOW that nobody likes to miss a contest (ARML) due to finals, or SATs (I know it's not an issue this year, yes). That really sucks.)

So any help I can get will be awesome.

Local bay area contests: Tokutomi, SCU HS math competition, SCVMA math field day

Things to do at school: the 5 NA&T contests, Mu Alpha Theta, Mandelbrot, CA Math League, AMC/AIME/USAMO

Things to do at home: USAMTS

And don't forget ARML.

That's all I know (I think), but you probably know most or all of that though.

I knew all of them except Tokutomi and the SCVMA math field day. I'll look those up.

So yeah, all of those are good, and I'll be posting dates for most of them.

Thanks for the initiative, I'm looking forward for that list!

Sorry that it's not up yet.

An update on progress:
I'm still compiling the list, but progress is going slowly.
I'll see if I can enlist my math teacher's help in getting some of the more obscure dates listed.


[quote="hungkhtn"]Trurthly, It was appeared as example $3.3.4$ in my book, and there are two solutions for it, one is S.O.S method an one is mixing variables method.[/quote]
what is S.O.S method?
and what is  mixing variables method?
can you explain a little more?

hi,

i don't know either  :blush: what S.O.S method is ,could anyone

explain it generally ?

thanks in advance.

I am very interested in using SOS method,but with this ineq,using it is longer than using mixing variable method. ;)
[b][u]generalization[/u][/b]:With $k\ge 0,x+y+z=6,x,y,z\ge 0$.Prove that
$\sum x^2y^2+k\sum xy\le max(81+9k;48+12k)$

$a^2+b^2+c^2-(ab+bc+ca)=\frac{1}{2}.((a-b)^2+(b-c)^2+(c-a)^2) \newline =(a-\frac{b+c}{2})^2+\frac{3}{4}.(b-c)^2$
The end of this symtem is mixing variable method and the first is SOS method  :lol:


number theory
greatest common divisor
number theory solved

Solve the equation in Z:
x^4+6x 2 y 2 +y^4=z 2

(x 2 +y 2 ) 2  + (2xy) 2  = z 2  (given)
(x 2 +y 2 ) 2 -(2xy) 2  = (x 2 -y 2 ) 2
denote a=x 2 +y 2  and b=2xy
the above two equations imply
a^4 - b^4 is a perfect square.
we will prove that this is not possible
assume there xists a nonempty set S s.t when a \varepsilon  S ,then there xist natural numbers b and c  satisfying a^4 - b^4 = c^2.by well ordering there is a least element ,suppose that element is a .which implies
gcd (a,b,c) = gcd(a,b)=gcd(b,c)=gcd(c,a) = 1
consider cases
CASE 1 (c is even)
=>a and b are odd. and 4lc 2
gcd(a,b)=1 and gcd(a 2 +b 2 ,a 2 -b 2)=2
let x= \sqrt  [(a 2 +b 2)/2] and y = \sqrt [(a 2 -b 2)/2] then x and y are both integers and x^4-y^4=(ab) 2 .but a>b implies x wlog ,a is odd and b is even .now use pythagorean substitution and there xist integers r and s such that gcd(r,s)=1 and  a 2 =r 2+s 2  ,c 2 =r 2-s 2  and b 2 =2rs
xactly one of r and s is even.assume r' to be even  and s' to be odd.
r'/2=x 2  and  s'=y 2  then
a 2 =4x^4+y^4
again by pythagoras,there xist integer m and s.t gcd (m,n)=1 and
a=m 2 +n 2 , 2x 2 =2mn and
y 2 =m 2 -n 2 .since x 2 =mn and gcd(m,n)=1 m=p 2 and n=q 2
.so y 2 = p^4-q^4.thus (p,q,y) is a new solution with p= \sqrt m \leq x= \sqrt (r'/2)<r' \leq  a,which is a contradiction.
thus there do not xist solutions to this equation
QED
(Hope tis correct)
cheers

[quote="grobber"]Nice soln Galois! This a well known problem: there are no numbers x and y s.t. x^2+y^2=z^2 and x^2-y^2=t^2. This was used before on this forum I think. I don't remember for which problem though.. Bye! :D[/quote]

my friend solved the xx+yy=zz and xx-yy=tt problem be4, but i forgot his solution, can you tell me how to solve it?

Sure, but it's kind of long. A bit later, OK?

[quote="Arne"]In fact you can find the solution above, siuhochung.
Galois proved that the equation a^4 - b^4 = c^2 doesn't have solutions.
So suppose a + b = z, a - b = t. Then a^4 - b^4 = (zt).
Now put zt = c and you will get the result :D[/quote]

oh sorry, i didn't notice that... thx~ :D

There are many simmilar problems regarding this one which are done using [b]galois [/b] method :D

1)Prove that x^2+xy+y^2=36^2 has no solutions in Natural numbers \{0}
2)Solve in N* the equation 3x^4+10x^2y^2+3y^2=z^2.

cheers! :D


geometric series
algebra unsolved
algebra

1) #1 is quite difficult without a calculator.  You could rearrange the products into $(8)24^{10}$ and $(5)25^{10}$.  Now it is a question of which is larger: $\frac{8}{5}$ or $\left(\frac{25}{24}\right)^{10}$.

It is close, but $\frac{8}{5}$ turns out to be the larger of the two.  You could use the approximation $a^n\approx1+n(a-1)$ (for $a$ close to $1$), but the error associated with this approximation is too large, considering the magnitude of the difference you're dealing with.

Another (although admittedly more difficult) way is to reorganize the quantities into $\frac85$ and $\left(\frac{25}{24}\right)$ and calculuate the latter by using the Taylor series centered around 1:
$a^k=1+k(a-1)+\frac{k(k-1)}{2}(a-1)^2+\frac{k(k-1)(k-2)}{6}(a-1)^3+...$

The first degree approximation is:
$1+10\left(\frac{25}{24}-1\right)=1+\frac{10}{24}=\frac{17}{12}$, which is smaller than 8/5 (since $(17)(5)=85<96=(8)(12)$).

The error in the appoximation is:
$10(10-1)(25/24-1)^2/2+10(10-1)(10-2)(25/24-1)^3/6+...$

The nth degree term is the (n-1)th term muliplied by $\frac{10-n+1}{n}(25/24-1)$. As $n$ increases, the factor decreases.  Therefore, the smallest such factor in the error series occurs when $n=3$ ($n=2$ does not have a term before it).
$\frac{10-3+1}{3}(25/24-1)=\frac83(1/24)>3(1/24)=\frac18$
The first term in the error series is $10(10-1)(25/24-1)^2/2=45(1/24)^2>2(1/24)=1/12$

By using the formula for a geometric series, the error is less than:
$\frac{1/12}{1-1/8}=\frac{8}{(7)(12)}=\frac{2}{21}<2/12$

When we add 2/12 (more than our error) to our approxiamation, we obtain 19/12 which is still less than 8/5 (since $(19)(5)=95<96=(8)(12)$).
Since $\frac{8}{5}>\left(\frac{25}{24}\right)^{10}$, $2^{33}\times3^{10}>5^{21}$.