geometry
circumcircle
parallelogram
similar triangles

I used the following theorem(I think it belongs to Brocard):

Given 2 triangles ABC and A'B'C', if you inscribe a triangle similar to A'B'C' in triangle ABC (call this one A1B1C1, with A1 on BC etc.), the circles AB1 C1, BA1C1, CB1A1 intersect an the same point, no matter which triangle A1B1C1 you take (it's called the Brocard point of triangle ABC with respect to triangle A'B'C' I think).

I'm not going to prove it, but if you check it out you'll see that it's not hard at all, and I think it's accepted in contests anyway(I used it once and I got all the points).

Here goes:
Let's assume that BC<=AB<=AC (so that we have the same drawing), and I also assumed that K is on the segment BC and M is also on segment AB, because the other cases are treated similarly.

Take the circumscribed circle of triangle KMN. You can easily prove that it intersects again the side AC at T (you use the fact that AC is the largest side). The quadrilateral MKNT is inscribable (is that the word? :D ) so angle TMK =angle KNC=angle C & angle TKM=angle TNM=angle A, so triangle KTM and triangle ABC are similar.
Now, if you take A'B'C'=ABC in the theorem above, that point at which those circles intersect is O(the circumcircle):just take the median triangle. But KTM is one of those triangles inscribed in ABC and similar to A'B'C' (which is also ABC), so according to the theorem, circles BKM,CKT and ATM intersect at O, so O is on circle BKM.

here is my proof of what you called Brocart theorem :
I take the same notation as above and I called O the point
of intersection of circles AB1C1,BA1C1,CB1A1.
let
Z=angle(O C1 B1)
E=angle(O C1 A1)
R=angle(O A1 C1)
T=angle(O A1 B1)
Y=angle(O B1 A1)
U=angle(O B1 C1)
then we have the following system of equation:
Z+E                =C1
        R+T        =A1
               Y+U =B1
Z               +U =A
    E+R            =B
            T+Y    =C

so there are 6 equation, 6 unknown value and
we easily see there is a unique solution.

What is the usual(synthetic) solution?
(i'm not sure this theorem is called brocart theorem)

Well, when I thought of proving this theorem the same proof came to me, the one you posted. I'll try to solve it in some other way.

Circumcircle KMN  intersects AC  at T - orthocenter of KMO.

   Maj.  Pestich

Can we prove this using Simpson's theorem? The perpendiculars from O to BK and BM are the midpoints X and Y of AB and BC.

Which AC is the longest side is not neccessary :)

I found another way to solve it, using transformations.
[hide="Solution"]We look at three rotations (all angles are clockwise):
Rotation R: centered at M taking C to N with angle 180-2C
Rotation S: centered at K taking N to A with angle 180-2A
Rotation T: centered at O taking C to A with angle 2B

Since the sum of two rotations is another rotation, we can say that R+S=T if the angles add up and if at least one point is taken to the correct image. Both conditions happen:
[hide="checking the two conditions"]angles add up: (180-2C)+(180-2A)=2B
image of C for the rotation R+S is: A, since R takes C to N and K takes N to A
image of C for the rotation T is: A, since <AOC = 2<ABC=2B and OA=OC[/hide]
So we know that R+S=T. So now we take the image of O under these two rotations. Clearly, O will not move since T is centered at O. So the sum of R and S must also fix O. That means that if rotation R takes O to O', then: O'M=OM and O'K=OK. There's only one such point O', that point is the reflection of O over KM. So because of the rotation, <OMO'=180-2C, and since we know that's just a reflection,

<KMO = (1/2)<OMO' = 90-C.

We also know that

<KBO = <ABO = (180-<AOB)/2 = (180-2C)/2 = 90-C

So <KMO = <KBO, implying that KBMO is cyclic.[/hide]

I have a solution for this nice problem.Let $(K,KA)\cap (M,MC)=P$
We have
$\angle APC=\angle APN+\angle NPC=\frac {1}{2}\angle AKN+\frac {1}{2}\angle NMC=$ $(90^0-\angle BCA)+(90^0-\angle BAC)=\angle ABC$
Hence $ABPC$ is cyclic so $P\in (O)$.
Since $P\in (O),P\in (K)$ and $A\in (O),A\in (K)$ so $OK\perp AP$. Similarity we have $OM\perp CP$
Thus $\angle KOM+\angle APC=180^0$ or $\angle KOM=\angle APC$
$\bullet $If $\angle KOM+\angle APC=180^0$ then $\angle KOM+\angle ABC=180^0$ so $OKBM$ is cyclic.
$\bullet $If $\angle KOM=\angle APC$ then $\angle KOM=\angle ABC=180^0-\angle BCA-\angle BAC=$ $180^0-\angle MNC-\angle KNA=\angle KNM$
$\Rightarrow O\in [AC](?)\Rightarrow \angle KOM=\angle KBM=90^0$ so $OKBM$ is cyclic.
$Q.E.D$
------------
Actually if see picture we can see it must be $\angle KOM+\angle APC=180^0$ :)
Sorry if i wrong

Here is another solution, using just similar triangles.
[hide]
Let $P$ and $Q$ be the midpoints of $AN$ and $NC$ respectively. Also, let $PK$ and $AO$ intersect at $U$, while $QM$ and $CO$ intersect at $V$. We claim that $\triangle KNM \sim \triangle OVM$.

First, for simplicity, let $\angle ABO=\alpha$ and $\angle CBO=\beta$. A bit of angle-chasing (using the fact that $O$ is the circumcenter and that $\triangle AKP \cong \triangle NKP$, $\triangle NMQ \cong \triangle CMQ$) gives the angles:

$\angle NKU= \angle MNV = \beta$
$\angle UNK = \angle VMN = \alpha$
$\angle KUO = \angle KNM= \alpha + \beta$
$\angle UOV= \angle VNU = 2 \alpha + 2\beta  $
$\angle OVN= \angle NUO =180-  2 \alpha - 2\beta$.

Immediately, we get $\triangle KUN \sim \triangle NVM$ and that $OVNU$ is a parallelogram, so $UN=OV$. But by similar triangles, we have
$\frac{NK}{NM}=\frac{UN}{VM}$, so $\frac{NK}{NM}=\frac{OV}{VM}$, implying by SAS that $\triangle KNM \sim \triangle OVM$. Now, note that $\angle KMN = \angle OMV \Longrightarrow \angle KMO = \angle NMV = \alpha = \angle KBO$, so $KBMO$ is cyclic, as desired.

[/hide]

Here's a slightly different solution than the ones above. Honestly, I feel it is hilarious this idea can even be used, though this problem is nice to do synthetically.

[hide=Solution]
Let $T$ and $S$ be the midpoints of $AB$ and $BC$. $BTOS$ is cyclic, so it is enough to show that $\angle KOT = \angle MOS$, or $\triangle KOT \sim \triangle MOS$.

First, observe $\frac{OT}{OS} = \frac{\cos C}{\cos A}$. Now, observe that if $A, K, T$ are in this order on side $AB$, then $B, M, S$ are in this order on side $BC$, and if $A, T, K$ are in this order on side $AB$, then $B, S, M$ are in this order on side $BC$. Therefore, where $AN=x, CN=y$,
\[ \frac{KT}{MS} = \frac{c-\frac{x}{\cos A}}{\frac{y}{\cos C} - a} = \frac{\cos C}{\cos A} = \frac{OT}{OS} \]
because multiplying yields that this is equivalent to $x+y=a \cos C + c \cos A = 2R \sin(A+C) = b$, which is true.
Thus, $\triangle KOT \sim \triangle MOS$, so done.
[/hide]

Note that as $N$ varies [i]uniformly[/i] on $\overline{AC}$, points $K, M$ vary [i]uniformly[/i] on $\overline{AB}, \overline{BC}$ respectively, so the circumcircle of $\triangle KBM$ passes through a fixed point other than $B$. For $N \in \{A, C\}$ it is clear that $O$ lies on $\odot (KBM)$ so we are done!

[quote=anantmudgal09]points $K, M$ vary [i]uniformly[/i] on $\overline{AB}, \overline{BC}$ respectively, so the circumcircle of $\triangle KBM$ passes through a fixed point other than $B$[/quote]

What is the definition of uniform movement and why is this true? Is this the result of some theorem?

Uniform movement means that at time $t$ point $P$ lies on a segment $P_0P_1$ with $\overrightarrow{P_0P}=t \overrightarrow{P_0P_1}$ or anything equivalent. Now perpendicular bisectors of $AN, CN$ are essentially a pencil of lines that move uniformly too (it's easy to work out what that means), so their intersections with fixed lines also move in the same fashion.

It can also be shown using spiral similarity, that $(BKM)$ passes through a fixed point in this case. So, we're good by checking few cases.

[quote]It can also be shown using spiral similarity, that $(BKM)$ passes through a fixed point in this case.[/quote]

How do you incorporate the fact that K and M move uniformly into proving this?

If $K$ divides $P_1Q_1$ in the same ratio as $M$ does $P_2Q_2$; then let $S$ be the spiral center $P_1Q_1 \mapsto P_2Q_2$. Then $S$ also sends $K$ to $M$, so $(BKM)$ passes through $S$.

[quote]Let $N$ be a point on the longest side $AC$ of a triangle $ABC$.  The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$.  Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$.[/quote]
[b]Solution:[/b] Assume $NA<NC$. Let $CO \cap NK=G$. Let the circle at $K$ with radius $KA=KN$ intersect $OA$ $=$ $D$ $\implies$ $\angle KAD$ $=$ $\angle KDA$
$$\angle KGO=180^{\circ}-(180^{\circ}-A+90^{\circ}-B)=\angle KDA \implies GKDO \text{ is cyclic}$$
And, $\angle KBO=90^{\circ}-C=\angle KDA \implies B \in \odot (GKDO)$. Also,
$$\angle KBO=90^{\circ}-C=\angle NGC=\frac{1}{2}\angle NMC \implies M \text{ is the center of } \odot (GNC)$$
Hence, $\angle OBM=\angle GCM=\angle OGM$ $\implies$ $M \in \odot (BGKDO)$ $\implies$ $O \in \odot (KBM)$

Close to usatst 2012
Switch for $A$ labbeling.
Let $BN=x$ ($CN=a-x$)
From triangles $NKC$ and $NMB$ we have $AK=b-\frac{a-x}{2cos \gamma}$ and $AM=c-\frac{x}{2cos \beta}$ so $AK \cdot  cos \gamma + AM \cdot  cos \beta =R \cdot sin \alpha \leftrightarrow b \cdot cos \gamma + c \cdot cos \beta = a$ is true.


is it possible for me to use color latex tags here? Like [img]6924[/img]?

With the code:

[code]\blue\Large e^x=\sum_{n=0}^\infty\frac{x^n}{n!}[/code]

It works on my site but not on aops

Can it work in texniccenter?

It requires the color package which is not installed on AoPS so won't work here. But it will work fine in your document. Try this:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{color}
\begin{document}
\noindent
\color{red}$1+1=2$ \\
\color{green}$1+1=2$ \\
\normalcolor $1+1=2$
\end{document}[/code]

Thanks. Works fine now.

If I need color here, I guess I'll just have to do what i did with that


$ a_1,a_2,...,a_n\ge0$    and  $ \sum_{i\equal{}1}^{n}a_i\leq \frac{n}{k}$           with         $ k\ge1$

Find the minimum of :      $ \prod_{i\equal{}1}^{n}a_i\plus{}(\prod_{i\equal{}1}^{n}a_i)^{\minus{}1}$

:)


Let $ S$ be a set of all positive integers which can be represented as $ a^2 \plus{} 5b^2$ for some integers $ a,b$ such that $ a\bot b$. Let $ p$ be a prime number such that $ p \equal{} 4n \plus{} 3$ for some integer $ n$. Show that if for some positive integer $ k$ the number $ kp$ is in $ S$, then $ 2p$ is in $ S$ as well.

Here, the notation $ a\bot b$ means that the integers $ a$ and $ b$ are coprime.

What is the meaning of $ a\bot b$ :blush:

It means: the numbers are coprime.

[Moderator edit: Added to the first post. I love this notation, but I wouldn't use it without explanation. ;) - darij]

Here is my solution :
From the condition we have property ,there exist $ x,y\in N$ such that
$ p|x^2+5y^2$
Exist $ (\frac{-5}{p})=1$
It mean that there exist a number $ a\in N$ such that $ p|a^2+5$
Let $ m=[\sqrt {p}]$
Consider ${ A=\{x+ay}$ where $ x,y=0,..,m$
Because $ |A|=(m+1)^2>p$ so exist $ (x_1,y_1),(x_2,y_2)$ such that
$ x_1+ay_2\equiv x_2+ay_2 (\mod p )$
$ \Leftrightarrow x_1-x_2\equiv a(y_2-y_1) (\mod p )$
Take $ x=|x_1-x_2| ,y=|y_1-y_2|$ then
$ p|x^2+5y^2$
But $ 0<x^2+5y^2<6p$ so
$ x^2+5y^2\in \{p,2p,3p,4p,5p\}$
But $ 4\not |x^2+5y^2$ and $ 5\not |x^2+5y^2$
It follow that $ x^2+5y^2 \in \{p,2p,3p\}$
It $ p=x^2+5y^2$ then $ p\equiv 1 (\mod 4)$ (traddition from the condition )
[u]Case 1[/u] $ 2p=x^2+5y^2$ problem claim .
[u]Case 2[/u] $ x^2+5y^2=3p$
Have four case
1)$ x \equiv 1 (\mod 3 ) ,y\equiv 1 (\mod 3 )$
Then exist $ k,l \in N$ such that $ x=3k+1 ,y=3l+1$ :
$ p=3k^2+2k +5(3l^2+2l)+2$
$ \Rightarrow 2p=6k^2+4k+10 (3l^2+2l)+4=(k+2+5l )+5(k-l)^2$
2)Case $ x\equiv (\mod 3 ) ,y\equiv -1 (\mod 3)$ then
$ \Rightarrow x=3k+1,y=3l-1$
$ \Rightarrow  2p=(k+2-5l )^2+5 (k+l)^2$
3) $ x\equiv -1 (\mod 3 ),y\equiv 1 (\mod 3 )$
$ \Rightarrow x=3k-1,y=3l+1$
$ \Rightarrow 2p=(k-2-5l)^2+5(k+5l)^2$
4) $ x=3k-1,y=3l-1$
$ \Rightarrow 2p=(k-2+5l )+5 (k-5l)^2$
So exist $ x,y\in N$ such that $ 2p=x^2+5y^2$
Problem claim.

[quote][Moderator edit: Added to the first post. I love this notation, but I wouldn't use it without explanation. ;) - darij][/quote]

BTW, this notation is discussed quite prominently in the book Concrete Mathematics by Graham, Knuth, Patashnik. If I remember right, the authors seem quite proud of this notation and wish that more people use it in the future :).

first,we can trivially choose $a,b$ such that $p|a^2+5b^2,a^2+5b^2<6p$
mod 4,$a^2+5b^2\ne 4p,5p,p$
hence $a^2+5b^2=2p or 3p$
if $a^2+5b^2=3p$,then
$(\frac{a+5b}{3})^2+5(\frac{a-b}{3})^2=2p$ and$(\frac{a-5b}{3})^2+5(\frac{a+b}{3})^2=2p$,with at least one pair of integer solutions between them.


inequalities
function
trigonometry
calculus
derivative
real analysis
144

If $A\leq B\leq C$ are the angles of a  triangle, then

$\cot C\leq\frac{\sqrt{3}}{\pi}A$.

$cot C\leq cot A$ and note that $f(x)=\cot x-\frac{\sqrt{3}}{\pi}x$ is increasing on $[0;\pi/3]$, so $f(A)\leq f(\pi/3)=0$.

[quote="goldhp"]$f(x)=\cot x-\frac{\sqrt{3}}{\pi}x$ is increasing on $[0;\pi/3]$[/quote]
decreasing

Oh sorry Chen, i'm really tired when i wrote the solution :D, here's true,
$\cot C=-\cot(A+B)$ with $x=A+B\in(0;2\pi/3]$. The function $f(x)=\cot x-\frac{\sqrt{3}}{\pi}x$ is decreasing on the inteval. So $f(A+B)\geq f(2\pi/3)=-\sqrt{3}$

[quote="goldhp"]here's true,$f(A+B)\geq f(2\pi/3)=-\sqrt{3}$[/quote]
the inequality is equvalent to
$\cot C\leq\frac{\sqrt{3}}{\pi}C$
So am I still waiting

oh I was very fool lastnight  :blush:. Here's my last try or I'll give it up:
Let $B=A+x$ and $C=A+y$ then $0\leq x\leq y$. So we have
$\cot C=\cot (A+y)\leq \cot \Big(A+\frac{x+y}{2}\Big)=\tan\frac{A}{2}$. So let me show that $\tan \frac{A}{2}\leq \frac{\sqrt{3}}{\pi}A$. In fact the function $f(x)=\tan\frac{x}{2}-\frac{\sqrt{3}}{\pi}x$ on $[0;\pi/3]$ has derivative $f'(x)=\frac{1}{2\cos^{2}x/2}-\frac{\sqrt{3}}{\pi}$ which has unique solution on the inteval $[0;\pi/3]$, and $\pi-2\sqrt{3}\cos^{2}x/2$ is increasing. So $f(x)$ attaines maximal at $0$ and $\pi/3$. But note that $f(0)=f(\pi/3)=0$ so we obtain the result.

Good luck in finding my mistake Chen :D

[quote="goldhp"]$\cot C\leq \cot \Big(\frac{B+C}{2}\Big)=\tan \frac{A}{2}\leq \frac{\sqrt{3}}{\pi}A$. [/quote]
Very nice proof!


calculus
integration
induction
real analysis
real analysis unsolved

[quote="FieryHydra"]Let $k$ be an integer. Prove that the formal power series \[\sqrt{1+kx}= 1+a_{1}x+a_{2}x^{2}+\ldots\] has integral coefficients if and only if $k$ is divisible by 4.[/quote]
Set $a_{0}= 1$ for ease of notation. Squaring both sides, we see that $a_{1}= \frac{k}{2}$ and $a_{2}=-\frac{a_{1}^{2}}{2}=-\frac{k^{2}}{8}$. Therefore, if $a_{1}$ and $a_{2}$ are integers, then $8 | k^{2}$, and $4 | k$.

Suppose now that $4|k$. Note that  $\sum_{j=0}^{m}a_{j}a_{m-j}= 0$ for $m>1$. Therefore, if $m = 2n$ is even, then \[a_{m}=-\frac{a_{n}^{2}}{2}-\sum_{j=1}^{n-1}a_{j}a_{m-j}\quad (*)\] and if $m=2n+1$ is odd, then \[a_{m}=-\sum_{j=1}^{n}a_{j}a_{m-j}.\quad (**)\] Then if $4|k$, one shows by induction that all $a_{i}$ are [b]even[/b], as follows:

Obviously, $a_{1}$ and $a_{2}$ are even. Suppose $a_{1}, \dots, a_{m}$ are all even, and wlog $m$ is even. Then by (**), $a_{m+1}$ is even, and by (*),also $a_{m+2}$ is even.   :lol:


modular arithmetic
induction
combinatorics proposed
combinatorics

Suppose $ n$ is a positive integer such that $ f(n)$ represent the number of a kind of n-digit number $ \overline{a_1a_2...a_n}$ which satisfies
[list]
(i) $ a_i\in \{1,2,3,4\}$ and $ a_i\neq a_{i\plus{}1}$, $ i\equal{}1,2,...$.
(ii) if $ n\leq 3$, then $ (a_i\minus{}a_{i\plus{}1})\times (a_{i\plus{}1} \minus{} a_{i\plus{}2})<0$, $ i\equal{}1,2,...$.
[/list]
Find:
[list]
(a) $ f(10)$
(b) The remainder when $ f(2008)$ is divided by $ 13$
[/list]

I assume $ n \leq 3$ should really be $ n \geq 3$.  So, we're trying to count some word analogue of alternating permutations.  By replacing $ a_1a_2\cdots a_n$ with $ (5 - a_1)(5 - a_2)\cdots(5 - a_n)$ if necessary, we may as well assume that $ a_1 < a_2$, so $ a_1 < a_2 > a_3 < a_4 > a_5 < \ldots$.

[hide="One approach:"] Let $ f_i(n)$ be the number of such words of length $ n$ that end in $ i$, $ 1 \leq i \leq 4$.  Then we have the following recursive system:
\begin{align*} f_{1}(2n) & = 0 \\
f_{2}(2n) & = f_1(2n - 1) \\
f_3(2n) & = f_1(2n - 1) + f_2(2n - 1) \\
f_4(2n) & = f_1(2n - 1) + f_2(2n - 1) + f_3(2n - 1) \\
f_4(2n + 1) & = 0 \\
f_3(2n + 1) & = f_4(2n) \\
f_2(2n + 1) & = f_4(2n) + f_3(2n) \\
f_1(2n + 1) & = f_4(2n) + f_3(2n) + f_2(2n) \end{align*}
with initial values $ f_1(1) = f_2(1) = f_3(1) = 1$, $ f_4(1) = 0$.  Then it's "just computation" to solve part (a) to get $ (f_1(10),f_2(10),f_3(10),f_4(10)) = (0, 793, 1429, 1782)$ and so $ f(10) = 8008$, and similarly to demonstrate that $ f_i(n) \equiv f_i(n + 12) \pmod{13}$ by induction after computing the first 13 terms of the sequence.[/hide]

[hide="Another incomplete approach"]For $ k \geq 2$, let $ f_k(n)$ be the number of such numbers of length $ n$ with digits between 1 and $ k$.  It shouldn't be too hard to figure out a recursion for $ f_k(n)$ that will be nicer to work with than the preceding ones by finding the first position in the word where $ k$ appears.  The details have been giving me a headache, though ... I'll try again after lunch.
[/hide]

The associated sequence is [url=http://www.research.att.com/~njas/sequences/A006356]A006356[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


AP Stats
AP Gov
AP Econ
AP Physics C Mech
AP Physics C E&M
AP Comp Sci

Ok, that's not me, that's my friend. I'm only taking Stats, Gov, and Econ; my college will take the physics credit, but the professsional school AFTER that doesn't want a lot of AP credits counting towards my major, so I'm not taking it. But yeah, so my friend can top you. So there.
What does WH stand for?
I wouldn't worry too much. I took three last year and I did pretty well (two 5s and a 4 but man I studied my BUTT off for that 4). If your teacher covers the material well you've got nothing to worry about. I know a lot of AP teachers will "teach to the test" by giving normal exams and essays in the form of the AP test-- this helps majorly, because you know what to expect. Notecards are a big help, especially for calculus, believe it or not. And don't forget +c.

Your friend's classes honestly looks like the courses i'm planning on taking next year :|

WH= world history, i believe.  dunno what else it could stand for

[quote="quantum leap"]ooh, i think i win: i got ap calc bc, ap physics b, ap us history, [i]and[/i] ap chemistry

can anyone top that :roll:[/quote]

That's basically what I took last year with stats in place of chemistry.

Oh, and btw...AoPS is quite humbling.  4 AP's is hardly anything for some people on here...I remember that towersfreak2006 took around 7 AP exams last year ;)

Well, I'm a sophomore taking:

AP World History
AP Computer Science A

Studying for them by reading the review books.

[quote="kimby_102"]How do u guys study for AP tests?
This year, I'm taking W.H, European History, and Physics B.
I'm kind of worried...[/quote]

I cram like there's no tomorrow. Basically read through each review book thrice in the last three weeks before the AP test. Last year I did this and got 5's on all four of my AP tests (in addition to studying the entirety of AP Stats in less than a week... not fun). I hope this strategy works even though it means I know nothing... I'm taking:

AP US History
AP American Language & Composition
AP Physics C - Mechanics and E&M
AP Chemistry

Aw man, I only have three AP's this year.

W.H., stats, bio.

No APs this year (I'm a freshman), but next year I'll be taking:
AP Chemistry
AP Computer Science
AP BC Calculus (hopefully)

Wow, this is humbling.  My school has loads of prerequisites and only offers a few AP's (Chemistry, Calculus AB, English, French, Spanish, German; Calculus BC has only been offered twice because that's how many times people have qualified to take it before graduating).  So I'm only taking 1 AP this year, Calc BC.  And probably two next year (French, English) since Physics isn't offered and the chemistry teacher is absolutely terrible.

[quote="paladin8"][quote="kimby_102"]How do u guys study for AP tests?
This year, I'm taking W.H, European History, and Physics B.
I'm kind of worried...[/quote]

I cram like there's no tomorrow. Basically read through each review book thrice in the last three weeks before the AP test. Last year I did this and got 5's on all four of my AP tests (in addition to studying the entirety of AP Stats in less than a week... not fun). I hope this strategy works even though it means I know nothing... [/quote]
That's exactly what i'm doing right now... :D

for those who already took Ap comp. sic., what do u think about the couse?
Is it fun?
I'm planning to take it next year (junior)... so i can participate in robotics in my senior year....

[quote]Oh, and btw...AoPS is quite humbling. 4 AP's is hardly anything for some people on here...I remember that towersfreak2006 took around 7 AP exams last year ;) [/quote]

I don't really think much of my four; this being an international forum, there are some people who go to schools with fewer "prerequisites" (I so hate those) and more opportunities to take harder (yes, and better :roll:) classes.  I  [i]wish[/i] my school had just Calculus BC for seniors (it's not a class at all now) and a good computer science teacher (it'll be offered for the first time next year).

[quote]I just signed up for my sophomore year classes-

AP Comp. Sci AB
AP Physics B
AP Calc BC
AP Chem.
fun? [/quote]

Yes, very. :roll:  Have fun!

[quote="surge"]I just signed up for my sophomore year classes-

AP Comp. Sci AB
AP Physics B
AP Calc BC
AP Chem.
fun?

for this year, should I take notes when reading the AP Review books thsi year, or just read straight thruogh? I plan on reading the review book at least twice.[/quote]

You have physics somphomore year? At TJ, only about one or two sophomores take AP physics each year. I was gonna try, but then realized that that would require me to skip precalc (which I'm trying to do anyway) But as there's no guarentee of that, there's no way the AP physics teacher would let me.

[quote="randomdragoon"]We took a practice AP exam in Statistics last week over three days. I got one wrong on the multiple choice and one point off the free response. [i]Possibly[/i] a five. At a $\alpha=0.001$ significance level.[/quote]

5??
No, you are definitely getting a 1.

(That is, if you fall asleep during the free response, a horde of elephants comes marching through the testing room and eats your #2 pencils, your sadistic teacher changes all your answers after the test, they accidentally administer last year's test, AND the statisticians suddenly realize that all the formulas they came up with were wrong so they change all the AP tests a week before and they tell everyone but you.)

[EDIT]: If only I had exactly 1000 more posts...

[quote="eryaman"][quote="paladin8"][quote="kimby_102"]Wait, u can retake the AP's??

Will it affect u when ur applying for colleges?

One mor question..

How many do u have to get in MC in World History to get 5(generally)??[/quote]

I think you can retake them... they are helpful for college apps. I think I got around 50-55/70 on the AP World MC and got a 5.[/quote]Wait, 50-55 points or questions? (I'm taking World History Wednesday...nervous. )[/quote]

Uhh, I usually answer most of them... so probably questions? I can't remember. I do recall planning out exactly how many points I wanted to get on the MC and each essay question to just push me over the 5 mark, though. I'm almost sure I barely made a 5, so you might want to shoot a little higher.

Bah, AP English today =  :(.

[quote="Adunakhor"][quote="eryaman"]Haha, I just took the AP Comp Sci A today...there was a bomb threat at our school (and an evacuation) DURING the exam.

What a whacky day.[/quote]

Whoa. Surely you'll be able to make-up the test if the evacuation took away a considerable amount of the test time, right?

As for a bomb threat, there was one in my school last year... there are some weird people out there  :o[/quote]Well what happened was, the proctor wrote down the time in which the alarms went off, and we were able to make up that amount of time after coming back in.  So hopefully it all worked out alright.

[quote="paladin8"][quote="quantum leap"][quote="kimby_102"]OMG, WH is on Wed... :([/quote]
If it's this Wednesday, that's the same day as the cal exam :D Was I the only one who thought it's ironic how Viscardi, National Winner of the Siemens Research competition on an awesome math project, is the guest of a mathjam [i]the night of[/i]the ap cal exam :roll:[/quote]

Oh yeah, I totally remember Calc BC + World History consecutively without lunch to make my day last year...[/quote]

Calc BC + World History + Baseball game consecutively with no lunch and a scratchy turkey sandwich for dinner. Tomorrow is gonna be so cool!

[quote="yif man12"][/quote][quote="paladin8"][quote="quantum leap"][quote="kimby_102"]OMG, WH is on Wed... :([/quote]
If it's this Wednesday, that's the same day as the cal exam :D Was I the only one who thought it's ironic how Viscardi, National Winner of the Siemens Research competition on an awesome math project, is the guest of a mathjam [i]the night of[/i]the ap cal exam :roll:[/quote]

Oh yeah, I totally remember Calc BC + World History consecutively without lunch to make my day last year...[/quote][quote="yif man12"]

Calc BC + World History + Baseball game consecutively with no lunch and a scratchy turkey sandwich for dinner. Tomorrow is gonna be so cool![/quote]

Lucky!

[quote="drunner2007"]I took 5 last year, 4 this year (I'm a Junior):
Soph.
Cal AB: 5
Stat: 5
US H: 5
Physics B: 5
Lang: 4

Junior:
Cal BC
Chem
Physics C Mech
Physics C E&M

I'm worried about chemistry and physics C. I screwed myself over in AP Chem by practicing for math competitions. In Physics C I went at my own pace, which happened to be a little slow since I tried studying theory. Now I'm cramming.[/quote]
That's how I got screwed up this year! I spent my whole spring break studying math comp. :(

[quote="kimby_102"]Hey guys,

Today, i took AP World, and I think I messed up on one of
the essay questions.. I thinks I did OK on the rest of them....
How would it affect my grade?? (yes, I'm worried). Is it still possible to get 5?[/quote]

Yeah, it's possible. Basically everyone at my school who took the AP World test last year (150-ish maybe?) bombed the Compare and Contrast essay last year on the Mongols, but a bunch of us still got 5's. I had no idea what I was talking about; I probably got at most a 4 on that essay.


calculus
integration
number theory proposed
number theory

Prove that Diophantine equation $y^2=4x^3+4x^2+4x+1$ has no integral points if $x>0$.

Firstly, $y$ must be odd so let's put $y=2k+1$ what gives us $k^2+k=x^3+x^2+x$, what is equivalent
to $(k-x)(k+x+1)=x^3$. There are two possibilities:
- There exists a prime divisor $p$ of $x$ for which $p|k-x$ and $p|k+x+1$ - but that is nonsens cause
that implies $p|k$ and $p|2k+1$.
- There is no such prime $p$ so $gcd(k-x,k+x+1)=1$ so there exist such positive integers $u$,$v$ such that
$x=uv$ where $gcd(u,v)=1$ as also $k-x=u^3$ and $k+x+1=v^3$.That implies $k=u^3+uv$ so
$u^3+2uv+1=v^3$. Let's notice that for every positive integers $u,v$ where $v>u$ there is
$v^3-u^3>2uv+1$ cause for maximum $u=v-1$ we have $v^2>v$ what is true cause $v>1$,so we get a contradiction, q.e.d.

It looks so easy. My solution based on computing rank and torsion group of given elliptic curve and I got the result that way. But nice to see that there is an elementary solution too!


Pete chose a positive integer $ n$. For each (unordered) pair of its decimal digits, he wrote their difference on the blackboard. After that, he erased some of these differences, and the remaining ones are $ 2,0,0,7$. Find the smallest number $ n$ for which this situation is possible.

From the two $ 0$ there are at least $ 3$ equal digits.
From the $ 2$ and the $ 7$ there are at least $ 2$ more digits
so the minimum number of digits is $ 5$ with $ {a,a,a,a\plus{}\minus{}2,a\plus{}\minus{}7}$, $ {a,a,a,a\plus{}\minus{}2,a\plus{}\minus{}9}$ in some order.


function
calculus
derivative
inequalities unsolved
inequalities

In isosceles right-angled triangle $ABC$, $CA = CB = 1$. $P$ is an arbitrary point on the sides of $ABC$. Find the maximum of $PA \cdot PB \cdot PC$.

I like this kind of problems!  :lol:

1) $P \in BC$

Let $PC=x \Rightarrow PB=1-x$

$\triangle CPA: \ \ PA^2=PC^2+CA^2=x^2+1\Rightarrow \ \ PA=\sqrt{x^2+1}$

$PA \cdot PB \cdot PC = x(1-x)\sqrt{x^2+1}$

So we want to maximize the function $h(x) = x(1-x)\sqrt{x^2+1}$, when $x \in [0,1]$

2) $P \in AB$

$PH \perp AC, H \in AC$

$CH=x \Rightarrow HA=PH=1-x$

$\triangle CPH: \ \ PC^2 = CH^2+PH^2 = x^2+(1-x)^2$

from the 45-45-90 triangle $PHA$ we have $PA = \sqrt{2}\cdot HA \Rightarrow PA = \sqrt{2}(1-x)$

similarly $PB = \sqrt{2}\cdot x$

$PA \cdot PB = 2x(1-x)$

$PA \cdot PB \cdot PC = 2x(1-x) \sqrt{x^2+(1-x)^2}$

In order to find the derivative in a shorter way, we define the functions:

$f(x) = 2x(1-x)$
$g(x) = x^2+(1-x)^2$

product: $F(x) = f(x) \sqrt{g(x)}$

(of course $g(x)>0$)

now, notice that $g(x)+f(x) = x^2+(1-x)^2+2x(1-x) =[x+(1-x)]^2 = 1\Rightarrow$

$g(x) = 1-f(x)$
$g'(x) = [1-f(x)]' = -f'(x)$

$F(x) = f(x) \sqrt{g(x)}$

$F'(x) = f'(x) \sqrt{g(x)} + f(x) \frac{g'(x)}{2\sqrt{g(x)}}$

$F'(x) = f'(x) \sqrt{g(x)} - f(x) \frac{f'(x)}{2\sqrt{g(x)}}$

$F'(x) = f'(x) [\sqrt{g(x)} - \frac{f(x)}{2\sqrt{g(x)}}]$

$F'(x) = f'(x) [\frac{2g(x)-f(x)}{2\sqrt{g(x)}}]$

$F'(x) = f'(x) [\frac{2[1-f(x)]-f(x)}{2\sqrt{g(x)}}]$

$F'(x) = f'(x) [\frac{2-3f(x)}{2\sqrt{g(x)}}]$

The fraction $\frac{2-3f(x)}{2\sqrt{g(x)}}$ is always positive. Indeed:

the denominator is always positive

also:
$f(x) \leq \frac{1}{2} \Rightarrow$

$-3f(x) \geq -\frac{3}{2} \Rightarrow$

$2-3f(x) \geq 2-\frac{3}{2} \Rightarrow$

$2-3f(x) \geq \frac{1}{2}>0$

$F'(x)=0 \Leftrightarrow f'(x)=0$

$f'(x) = 2-4x$

It has a root at $x=\frac{1}{2}$ and

$x<\frac{1}{2}\Rightarrow f'(x)>0$

$x>\frac{1}{2}\Rightarrow f'(x)<0$

The same happens from $F'(x)$

Hence $x=\frac{1}{2}$ is a local maximum of $F(x)$

$f(\frac{1}{2}) = \frac{1}{2} = g(\frac{1}{2})$

applying these values to $F(x) = f(x) \sqrt{g(x)}$ we get

$F(\frac{1}{2}) =\frac{1}{2} \cdot \frac{\sqrt 2}{2} = \frac{\sqrt 2}{4}$

Now let's see why this maximum is greater than the maximum of $h(x) = x(1-x)\sqrt{x^2+1}$

$h(x) = x(1-x)\sqrt{x^2+1}$
$F(x) = 2x(1-x) \sqrt{x^2+(1-x)^2}$

We will show that $h(x)<F(x)$ in $(0,1)$

$h(x)<F(x) \Leftrightarrow$

$x(1-x)\sqrt{x^2+1} < 2x(1-x) \sqrt{x^2+(1-x)^2} \Leftrightarrow$ (divide with the positive $x(1-x)$)

$\sqrt{x^2+1} < 2 \sqrt{x^2+(1-x)^2} \Leftrightarrow$

$x^2+1 < 4 [x^2+(1-x)^2] \Leftrightarrow$

$x^2+1 < 4x^2+4(1-x)^2 \Leftrightarrow$

$x^2+1 < 4x^2+4(1+x^2-2x) \Leftrightarrow$

$x^2+1 < 4x^2+4+4x^2-8x \Leftrightarrow$

$0 < 7x^2-8x+3,$ true because

discriminant: $D = 8^2-4\cdot 7\cdot 3<0,$ so $7x^2-8x+3$ is always positive

$\Rightarrow h(x)<F(x), \ \forall x\in (0,1)$

hence, if $h(x_o)$ is maximum of $h(x)$ then $h(x_o)<F(x_o)$

Finally, the maximum value of $PA \cdot PB \cdot PC$ for all possible positions of $P$ is $F(\frac{1}{2}) = \frac{\sqrt 2}{4}$

$x=\frac{1}{2}$ means that the point $P$ is the midpoint of $AB$

My solution was essentially the same as above.
Would like to other solutions as well.

[quote=mihaig][b][size=150]Given $\Delta ABC,$ with $AB=AC=1~\text{and}~BC=\sqrt2.$ Consider the set $\mathbb{S}=\Delta ABC\cup\text{Int}(\Delta ABC).$[/size][/b]
[b][size=150]Determine[/size][/b]
$$\max_{M\in\mathbb{S}}{\left(MA\cdot MB\cdot MC\right)}.$$
[b][size=150][color=#600]Preluata[/color][/size][/b][/quote]

Anyone?


Let $ a,b,c \geq 0$ such that $ a\plus{}b\plus{}c\equal{}3$. Prove that:
 $ \sqrt{a(b\plus{}c)}\plus{} \sqrt{b(c\plus{}a)}\plus{} \sqrt{c(a\plus{}b)} \geq 3 \sqrt{2abc}$

$ \sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)} \geq 3 \sqrt {2abc} \Longleftrightarrow (\sqrt {a(b \plus{} c)} \plus{} \sqrt {b(c \plus{} a)} \plus{} \sqrt {c(a \plus{} b)})^2 \geq 18abc$
$ \Longleftrightarrow \sum (ab \plus{} ac) \plus{} \sum 2\sqrt {ab(b \plus{} c)(c \plus{} a)}\geq 18abc.$
but $ \sum (ab \plus{} ac) \equal{} 2(ab \plus{} bc \plus{} ca) \geq 6(abc)^{2/3} \geq 6abc$(because $ abc\le1$)
now we prove that $ \sum \sqrt {ab(b \plus{} c)(c \plus{} a)}\geq 6abc$ wich equivalent to  $ \sum \sqrt {3abc \plus{} a^2b^2}\geq 6abc$ $ \Longleftrightarrow \sum \sqrt {3 \plus{} \frac {ab}{c}}\geq 6\sqrt {abc}$.
and by AM-GM we have $ 3 \plus{} \frac {ab}{c} \equal{} a \plus{} b \plus{} c \plus{} \frac {ab}{c}\geq 4\sqrt {ab}$
so $ \sum \sqrt {3 \plus{} \frac {ab}{c}} \geq \sum 2(ab)^{1/4}$ now let $ a \equal{} x^2,b \equal{} y^2$ and $ c \equal{} z^2,$
we have $ \sum 2(ab)^{1/4} \equal{} \sum 2\sqrt {xy} \geq 6(xyz)^{1/3}\geq 6xyz$ (because $ xyz \le 1$) =$ 6\sqrt {abc}$

My solution:

We have:
$ \sqrt{ab \plus{} ac} \geq \frac{\sqrt{2}}{2}\cdot (\sqrt{ab} \plus{} \sqrt{ac})$
And so on. Now it sufficies to prove that:
$ \sqrt{ab} \plus{} \sqrt{bc} \plus{} \sqrt{ca} \geq 3\sqrt{abc}$
But from AM-GM we have:
$ \sqrt{ab} \plus{} \sqrt{bc} \plus{} \sqrt{ca} \geq 3\sqrt[3]{abc}$
But $ \sqrt[3]{abc} \geq \sqrt{abc}$ because $ abc \leq 1$
So the result follows.

Ok!
 It has other solution,using counter-evidence method...


Find the value of $ (x \plus{} y)$ if $ 8(x \plus{} y) \equal{} 48$.

We divide by 8 on both sides:

8(x+y)=48
x+y=6

We get exactly what we want, and the answer is $ \boxed{6}$


Those have been around a long time and have been posted several times...

Search "Harry Potter Parody" on Youtube and you will piss your pants laughing.

Long live Saturday Night live.


advanced fields
advanced fields unsolved

I'm trying to prove what the title says.
I've found that can be easily shown using Nash-Williams theorem:

[i]A multigraph $ G\equal{}(V,E)$ can be partitioned into at most k forests if and only if $ \parallel{}G[U]\parallel{} \leq k(|U|\minus{}1)$ for every non-empty set $ U \subseteq V$.[/i]

Well... I'm having difficulties in applying the theorem. :oops:

By the way I do not understand the meaning of : $ \parallel{}G[U]\parallel{} \leq k(|U|\minus{}1)$
What exactly is $ \parallel{}G[U]\parallel{}$ ? :|

Thanks in advance. :)

i've heard that:
A graph $ G$ can be an union of at most $ k$ forests if and only if for any non-empty set $ U$ ,which has $ m$ elements, of the vertices of the graph $ G$ the number of edges between the vertices in this set is at maximum $ k(m\minus{}1)$.
to prove what u ask just see that if the graph is a planar one then we must have $ |E| \le 3|V|\minus{}6$.also u can easily see that the condition that a graph must have to be an union of at most $ 3$ forests is hold for a planar graph since for any non-empty set of its vertices with $ m$ elements the number of edges between these vertices is at maximum $ 3m\minus{}6<3(m\minus{}1)$ (if not then the graph won't be a planar one).
so we r done!


Well, now all contests are officially over. Does it mean we can start to predict this year's IMO team members?
I guess, Robin..Jarno..Jon..Danny..Chen..
and a sixth.? who do you think will be the sixth? maybe Frank Ban?

[b]
[color=red][Moderator edit (see #5 below): Please note that IMO team members, when informed of their selection, are told not to release the information until an official media release has been made by the CMS. Until that time, please keep this discussion to speculations only. Thank you.][/color][/b]

when do they usually select the members and sent out the invitations?

Isn't Robin going to IOI? Can one go to both IOI and IMO?


complex analysis
complex analysis unsolved

$U$ an open connected subset of $\mathbb{C}$, $D$ the unit disk, $f: U\to D$ a non constant analytic function.
Suppose that $|f(z)|=1$ for every $z$ on the boundary of $U$. Shows that $f$ is surjective.

By contradiction. If $f$ is not surjective, say, $a\notin f(U)$ with $|a|<1$, then by composing with an automorphism of $D$ we may assume $a=0$. Note: this composition also has moduls one at the boundary of $U$. Hence $\frac 1f$ is also analytic on $U$. Use Maximal principle to conclude that $f$ must be constant, which is absurd.


The digits of a number are 1,2,3,4,5,6,7,8,9 written in any order.

Of all the numbers that can be so formed  ( that is all the nine digit numbers where each of the integers 1 through to 9 must appear without repitition)  how many are divisible by 11 ?

[hide="Hint"] $ abcdefghi \equiv (a \plus{} c \plus{} e \plus{} g \plus{} i) \minus{} (b \plus{} d \plus{} f \plus{} h) \bmod 11$ [/hide]

[hide="Solution"]Going from t0rajir0u's hint, we require $ (a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g\plus{}h\plus{}i) \equiv 2(b\plus{}d\plus{}f\plus{}h) \pmod{11}$. That is, $ b\plus{}d\plus{}f\plus{}h \equiv 6 \pmod{11}$. Of $ 6,17,28,39$, $ 6$ is too low and $ 39$ is too high, so we check $ 17$ and $ 28$. $ 28$ has only two (unordered) ways, $ 9,8,7,4$ and $ 9,8,6,5$. $ 17$ has a lot more though. $ 9,5,2,1$, $ 9,4,3,1$, $ 8,6,2,1$, $ 8,5,3,1$, $ 8,4,3,2$, $ 7,6,3,1$, $ 7,5,4,1$, $ 7,5,3,2$, $ 6,5,4,2$, $ 10$ if I didn't miss any. So $ 10 \plus{} 2 \equal{} 12$ unordered sets. Then since we can arrange those $ 4$ digits and the other $ 5$ digits in any fashion, we need to $ 12 \cdot 4! \cdot 5!$, which we can compute to be $ \boxed{57600}$.

Unless I missed an easier way of doing this, I have to say experience doing Kakuro puzzles really helps for this problem.[/hide]


at a sporting event tickets are 4.50 for the first person and 2.50 for all remaining family members. in a section there is 480 people who have in total paid 1550. how many 4.50 tickets were sold?

Equation 1: 4.5x + 2.5y = 1550

Equation 2: ??????????????

 :oops:


geometry
trapezoid
cyclic quadrilateral
geometry unsolved

[quote="dima ukraine"]Let ABCD-cyclic quadrilateral. F,E points on arc AB. Let P1 and Q1 points of intersection AB with FC and FD, P2,Q2 points of intersection AB with EC and ED. Point S1 lie on segments P1Q1, S2 lie on segments P2Q2.And P1S1/S1Q1=P2S2/S2Q2. Prove that
FS1,ES2,CD are concurrent.[/quote]
Did I make a mistake?  I seem to get that the problem is false:

We may rephrase the problem:

[i]Consider cycic quadrilateral $ ABCD$ with points $ F$ and $ E$ on the arc $ AB$ and point $ X$ on $ CD$.  Let $ P_1$, $ Q_1$, and $ S_1$ be the respective intersections between $ AB$ and $ FC$, $ FD$, and $ FX$.  Define $ P_2$, $ Q_2$, and $ S_2$ similarly, but with respect to $ E$ rather than $ F$.  Prove that $ \frac {Q_1S_1}{P_1S_1} \equal{} \frac {Q_2S_2}{P_2S_2}$.  [/i]

We assume the truth of the problem.  Notice that\[ \frac {Q_1S_1}{P_1S_1} \equal{} \frac {[FQ_1S_1]}{[FS_1P_1]} \equal{} \frac {\frac {FQ_1}{FD}\cdot \frac {FS_1}{FX}\cdot [FDX]}{\frac {FS_1}{FX}\cdot \frac {FP_1}{FC}\cdot [FXC]} \equal{} \frac {DX}{XC}\cdot \frac {FQ_1}{FP_1}\cdot \frac {FC}{FD}\]Applying the same reasoning, we have that\[ \frac {Q_2S_2}{P_2S_2} \equal{} \frac {DX}{XC}\cdot \frac {EQ_2}{EP_2}\cdot \frac {EC}{ED}\implies \frac {Q_2S_2}{P_2S_2} \equal{} \frac {Q_1S_1}{P_1S_1}\iff \frac {FQ_1}{FP_1}\cdot \frac {FC}{FD} \equal{} \frac {EQ_2}{EP_2}\cdot \frac {EC}{ED}\]This is equivalent to having that $ \frac {FQ_1}{FP_1}\cdot \frac {FC}{FD}$ is constant as $ F$ moves around the arc $ AB$.  We may thus consider the equivalent problem:

[i]As a point $ F$ moves along arc $ AB$ of cyclic quadrilateral $ ABCD$, $ \frac {FQ}{FP}\cdot \frac {FC}{FD}$ is constant where $ Q$ and $ P$ are the respective intersections between $ AB$ and $ FD$ and $ AB$ and $ FC$.  [/i]

Consider $ C'$ and $ D'$ on the circle so that $ CC'\parallel DD'\parallel AB$.  We have that  $ \angle FC'D'\equal{}\angle FDD'\equal{}\angle FQP$.  Similarly, $ \angle FD'C'\equal{}\angle FPQ$, meaning that $ \triangle FQP\sim \triangle FC'D'$, from which we may note that $ \frac{FC'}{FD'}\equal{}\frac{FQ}{FP}$, so we wish to have to have $ \frac{FC'}{FD'}\cdot \frac{FC}{FD}$ be a constant.  Thus, we may again rephrase the problem:

[i]Consider cyclic trapezoid $ DD'CC'$ with $ DD'\parallel CC'$.  Let $ F$ be a point on its circumcircle.  Prove that $ \frac{FC'}{FD'}\cdot \frac{FC}{FD}$ is invariant. [/i]

Yet, this is clearly not true when $ F$ goes to $ D$.


Also try referring to the LaTeX manual in this site.  It's a very useful resource!


Let $ B$ be the set of the roots of $ z^{6}+z^{4}+z^{3}+z^{2}+1=0$ with positive imaginary parts. Let the product of the elements in $ B$ be $ r(\cos\theta+i\sin\theta)$. Find $ \theta$.

[hide="Factored"]
I realized that $ \zeta$ being a 5-th root of unity solves the equation.
So,
$ (z^{6}+z^{4}+z^{3}+z^{2}+1)=(z^{4}+z^{3}+z^{2}+z+1)(z^{2}-z+1)$
[/hide]

You can just solve the problem directly by making a substitution: $ y=z+\frac{1}{z}$.


function
calculus
integration
real analysis
inequalities
real analysis solved

Note L^2 norm is defined  by Lebegue Integeral ( (f)^2 )0.5
if lim n---> infinity L^2 norm of (fn -f) =0
and
lim n---> infinity L^2 norm of (gn -g) =0
Prove that
Lim n-->infinity Lebesgue Integral  (fn*gn )=  Lebesgue Integral  (f*g )

$f_n=f+F_n$, $g_n=g+G_n$. $\int f_ng_n=\int (fg+F_ng+fG_n+F_nG_n)$.
By Cauchy's inequality, $\left|\int f_ng_n-\int fg\right|\le ||F_n||_2||g||_2+ ||f||_2||G_n||_2+ ||F_n||_2||G_n||_2$. The standard argument for the limit of a product works here. I could even pull the integral outside the absolute values on the left side- $f_ng_n$ converges to $fg$ in $L_1$ if $f_n\to f$ and $g_n\to g$ in $L_2.$

There are some typos in your proof, jmerry.

Jmerry thank you again,  I miss understand ur prove, do u mind if u clearified better. I believe there are some gaps , pls do not forget the fact that that integral is Lebesgue Integral

The typos are fixed now.
It doesn't matter what kind of integral this is; all I need is Cauchy's inequality so that the integral of the product is an inner product. The rest is just chasing the details, and it's a very easy argument.


AMC
USA(J)MO
USAMO
AMC 8
AIME
ARML
geometry

Hi, what school are you from? I'm in 8th grade and hope to make USAMO this year. I went to MC states  last year but due to sickness I wasn't even able to make the countdown round.

Do you know if the AMC8 perfect scorer from our state has an AoPS account?

Dude! Two other AOPSers in Iowa! Where are you guys from and what are your names?

I'm from Iowa City West and I'm a senior. My name is Paul Hlebowitsh, maybe you know me?

I'm Patrick, and I'm from central Iowa, about an hour away from Des Moines. I'm in 8th grade and currently get practice AIME scores of >5 and <10.

I'd like to join the ARML team, but I'm not sure exactly what this entails...

(ARML 2007 is in June, right?)

Maybe we could hold an Iowa board marathon sometime.

Paul, I've been reading some of your F^3 stuff, and it's really good...

Also, does anyone here play chess? I'm pretty horrible (1200-1400 USCF) but I'd be up for a few games on FICS or USCL.

Yes, ARML is in June. I know the coach who runs the Iowa team, so I can talk to him about getting you on the team. I'll need to know a bit more information, like your last name and school, but after that I can definately try to get you on the team. We had an 8th grader last year, if I recall, so I see no reason why you couldn't be on the team. Especially if you're getting 5+ on AIME as an 8th grader. (which is pretty good)

As for f^3, thanks.  :) I've worked hard on that with shogia. You're welcome to come to the meetings if you can make it. Also, where did you find them? I'm pretty sure our math club site was deleted because of security reasons...

An Iowa board marathon would be fun, and I would definately do it.

Finally, I don't really play chess, sorry. I do play bughouse, though.

Is there some formal sign up process I could go through? I don't think my parents would take kindly to some high schooler they don't know inviting me to Iowa City(or knowing my last name, or where I live, etc)...

True! Er... so I talked to Mr. Kirpes today, who is the person who puts the team together. He said that you should take the AMC 10/12 (preferrably the 12) and then based on the scores we would talk about putting you on the team. Basically, he just wants to make sure you're good.

The process for choosing the team is just whoever he thinks are the best mathematicians in Iowa. So do well on the AMC!  :P

Pfft, I'll do horrible as always on speed math competitions. Probably waste an hour rederriving things, [i]just to make sure[/i], like I always do. Having no self-confidence sucks.

EDIT: I'm aiming for >140 on the 10 and >125 on the 12, is this a reasonable goal?

Well, that depends on what kind of a solver you are. I was not as good as you in 8th grade, but here are my scores from the last two years.

10th grade: AMC 12- 105 AIME - 2
11th grade: AMC 12 - 115.5 AIME - 10

So you see, although I improved my AIME score a lot, and improved a lot in mathematics, I still didn't do all that well on the AMC 12. I would advise you just to not worry. I don't know how strong you are at solving quickly, so just chillax. Even if it's not a reasonable goal, still try for it. My goal is 150 AMC 15 AIME 42 USAMO this year.  :P

But I think for the average person who can get 5+ on AIME, 125+ is a good goal for the AMC 12.

A ten!? What kind of practice schedule did you need for that?

And what did you get on the USAMO?

Well, 115.5 + 10*10 = 215.5 < 217, so I didn't qualify for the USAMO.  :blush: This is why the AMC is important.

And for practice, I basically went back to what I had learned for GPML and proved all of it. I didn't have a "schedule", I just did math when I felt like it. I did a lot of problems, true, but it was basically just stuff from Kalva. Basically, the major thing I learned was how to think. And it certainly helped a lot.  :)

It's unlucky that you got problems with the AMC 12; most people struggle with the AIME.

For PenguinIntegral's case, since you are 10th grade or below, if you don't qualify via the USAMO index, I believe they take any 10th graders or below that get the lowest AIME score or above of the person that qualified via the index (last year this was 8).
Since more people qualify for the USAMO this year, it'll probably be a 7.

What did you learn for the GPML and where did you learn it? How can I get my high school involved next year? Where can I find past problems?

Have you looked at a copy of ACoPS? All the problems in that are in the 7-15 AIME range, and have some very instructive solutions. That might boost your score on the AIME a bit. I would grab the new second edition because of the extra problems and geometry chapter, but it might be hard because libraries wouldn't carry it yet.

If all else fails, you can always sacrifice a goat to the math gods...

Registration information can be found at the membership link at http://www.mathleague.org, and it costs $20 for a large school (>250 in each grade). I think it's a dumb rule, but the GPML people don't let middle schoolers qualify for the state competition (really easy). However, you can still go to a local meet for fun.

The mathleague store sells past tests, but I believe a coach collecting all the past tests would have more than they sell. I'll see if I can get any to send to you.

And no, I have not looked at ACoPS (although I might buy one), although Paul might have.

Basically I'm going to back up hillarryous here and say that if you want to join GPML, you and your school should look at mathleague.org. However, if you're taking classes at a high school currently, you can compete on the high school's GPML team.

You can get past test from them (I remember they had a bunch of old tests put in these huge binders at regionals last year), but I own a lot of the tests just from the competitions, so we could start a marathon on here with questions from the tests or something.

As for what to learn, I basically went back to the basics. I had memorized all these weird formulas for math club, like the number of zeroes at the end of n! is floor(n/5) + floor(n/25) + ..., but I had no idea how any of them worked. So I went back and started proving things. And that's when I like to say I learned to think.  :)

And I own ACoPS! It's a wonderful book, chock full of interesting stuff. However, I'm not exactly worried about making USAMO this year. I'm a lot better than I was last year.  :P  The trick this year will be doing well at USAMO.

[quote="The Original Pi Guy"]
And I own ACoPS! It's a wonderful book, chock full of interesting stuff. However, I'm not exactly worried about making USAMO this year. I'm a lot better than I was last year.  :P  The trick this year will be doing well at USAMO.[/quote]

Is doing USAMO problems just a regular extension of doing AIME problems after one is good enough? Because I don't think I'm that good yet, and I think I just need more AIME practice...

For ACoPS, do you have the first or second edition?

EDIT: Do you mind starting a GPML marathon with problems of AIME difficulty? I can't post problems in it because I lack GPML problems, but I would love to solve them :P

Hello.  I've been on AoPS for a while now and just found this forum... fail.  Never knew there were any other Iowans here.
I'm going into 8th grade at Northwest, and hoping to do better at AMC and Mathcounts ([i]totally[/i] failed this year at nats).
I also did my first GPML last year.  I was an epic fail.

I found the forum about a month ago...
6th grade at weber, and first year of MC and second year of AMC 8 (24)

Tried GPML this year and failed

I am a 6th grader in Weber and I am trying out for AMC 8 this year and MC next year.

Wow, it's been a long time since I've wandered around to the Iowa forum. I feel so old now, having finished my first year at MIT. I'm taking several math classes next year that I'm really looking forward to. (one of them involves secret codes!) I'm glad I'm back in Des Moines over the summer though! I visited Central Academy where I went to high school, and the senior prank was up! They turned the school into Hogwarts! It was pretty awesome. :)

Hi!!! I'm Paolo, from the Burlington. I'm really new to the site, can someone help me? I'd like to become a part of this community. I don't really know what I'm doing and I'm looking for guidance. Could I have some help? Thanks to all Iowans!!!

Hi, Paolo.  Welcome to the Iowa forum.  As you can see, there's not much of a community around here, because unfortunately almost nobody posts here anymore.  I tried to revive the forum a bit with a problem marathon a while ago (which was not the biggest success in the world).

What do you need help with?  I'm not really an expert myself, but I've been here a bit longer, so... maybe I can help a bit?

Well, I'm having trouble navigating the site. Also, what features are on here? I'd appreciate your help!
Thanks!

If you're not taking any classes through AoPS at the moment, the biggest two resources on this website are [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]Alcumus[/url] and the [url=http://www.artofproblemsolving.com/Forum/index.php]Forums[/url]. Everything can be accessed in the tabs at the top of the screen. Hover over one of them, and a drop down menu will show you your options. The Bookstore is where you can buy books; the School tab is for online classes; the Community tab is for forums, blogs, and websites; the Alcumus tab is for Alcumus-related stuff (Alcumus is an online database of problems that will work you through different topics and move you on when it thinks you've mastered it); FTW is a multiplayer speed-based math contest; Videos gives you access to many of the math videos AoPS has made; Resources gives you access to AoPS Wiki, an online collection of past contest problems and other things, and the LaTeX guides.

Does that make sense? Let us know if you have any questions! I am a teaching assistant for a lot of different classes here at AoPS, so if you have any questions about classes, I would love to help. :laugh:

Is there any place on this website where I can learn calculus?

Woohoo :O It's pretty crazy looking back at these posts I was making in this thread 8 years ago o.O

How are you today?

Lol i'm good just started high school it sucks


trigonometry
vector
projective geometry
geometry unsolved
geometry

As i know, apply Pascal's theorem, we just can get $X_{1}, Y_{2}, Y_{3}$ colinear, do you mean that? Have you proved all 4 points are colinear?

Thank you for pointing this out. Like you said, Pascal's theorem just yields the collinearity of $X_{1}, Y_{2}, Y_{3}$.

To prove $A, Y_{2}, Y_{3}$ are collinear, you need to prove the following equivalence form $\frac{O_{1}Y_{3}}{O_{1}O_{2}}+\frac{O_{1}Y_{2}}{O_{1}O_{3}}= 4$ (notice that by $AB$, I mean the algebraic length of the segment $AB$, i.e $AB+BA = 0$). But this is easily proved by Ceva's theorem.

Thank you all! I will read your solutions carefully. Thank you.

Let us to present a solution, based on the (lovely to me) [b][size=100]Desarques theorem[/size][/b].

[b][size=100][color=DarkBlue]LEMMA. Let $ A,\ P,\ B,$ three collinear points ( $ P,$ between $ A,\ B$ and $ AP < BP$ ) and we denote the point $ Q,$ as the harmonic conjugate of $ P,$ with respect to $ A,\ B.$ Prove that $ \frac {PA}{AQ} \equal{} \frac {PM}{MB}$ $ ,(1)$ where $ M$ is the midpoint of the segment $ AB.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Because of the points $ Q,\ A,\ P,\ B,$ are in harmonic conjugation, we have $ \frac {QA}{QB} \equal{} \frac {PA}{PB}$ $ \Longrightarrow$ $ \frac {PA}{AQ} \equal{} \frac {PB}{QB}$ $ ,(2)$

From $ (1),$ $ (2),$ it is enough to prove that $ \frac {PB}{QB} \equal{} \frac {PM}{MB}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ \frac {PM}{PB} \equal{} \frac {MB}{QB}$ $ \Longrightarrow$ $ \frac {PM}{PB \minus{} PM} \equal{} \frac {MB}{QB \minus{} MB}$ $ \Longrightarrow$ $ \frac {PM}{MB} \equal{} \frac {MB}{QM}$ $ ,(4)$

From $ (4)$ $ \Longrightarrow$ $ (MB)^{2} \equal{} (MP)\cdot (MQ)$ $ ,(5)$ which is true.

Hence, we conclude that $ (1)$ is also true and the proof is completed.

$ \bullet$ In our problem now, let us to change its notations as follows:

[b][size=100][color=DarkBlue]PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ AD,\ BE,\ CF$ be, three  concurrent cevians at point $ H$ and we denote as $ M,\ N,\ P,$ the midpoints of the side-segments $ BC,\ AC,\ AB,$ respectively. Prove that the points $ A,\ X\equiv EP\cap FN,\ Y\equiv DF\cap MP$ and $ Z\equiv DE\cap MN,$ are collinear.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Let be the points $ Q\equiv AB\cap DE,\ R\equiv BC\cap FN$ and we will prove that $ QR\parallel MP\parallel AC.$

Applying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \bigtriangleup ABC$ with transversal $ EDQ,$

we have that $ \frac {QB}{QA}\cdot \frac {EA}{EC}\cdot \frac {DC}{DB} \equal{} 1$ $ \Longrightarrow$ $ \frac {QB}{QA} \equal{} \frac {EC}{EA}\cdot \frac {DB}{DC}$ $ ,(1)$

Applyin the [b][size=100]Ceva theorem[/size][/b], with respect to the point $ H,$ we have that  $ \frac {EC}{EA}\cdot \frac {FA}{FB}\cdot \frac {DB}{DC} \equal{} 1$ $ \Longrightarrow$ $ \frac {EC}{EA}\cdot \frac {DB}{DC} \equal{} \frac {FB}{FA}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {QB}{QA} \equal{} \frac {FB}{FA}$ $ ,(3)$

Taken the line segment $ NFR,$ as transversal of $ \bigtriangleup ABC,$ applying the [b][size=100]Menelaus theorem[/size][/b],

we have also that $ \frac {RB}{RC}\cdot \frac {NC}{NA}\cdot \frac {FA}{FB} \equal{} 1$ $ \Longrightarrow$ $ \frac {RB}{RC} \equal{} \frac {FB}{FA}$ $ ,(4)$ (because of $ NA \equal{} NC$ ).

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {QB}{QA} \equal{} \frac {RB}{RC}$ $ ,(5)$

From $ (5)$ $ \Longrightarrow$ $ QR\parallel MP\parallel AC.$

$ \bullet$ Because of now, the collinearity of the points $ Q\equiv DE\cap FP$ and $ R\equiv CD\cap LF$ and $ \infty \equiv CE\cap LP$
(we say that two parallel lines, are concurrent at infinity ), where $ L\equiv MP\cap FN,$ based on the [b][size=100]Desarques theorem[/size][/b],
we conclude that the triangles $ \bigtriangleup CDE$ and $ \bigtriangleup LFP,$ are perspective and so, we have that the line segments $ DF,\ CL,\ EP,$
are concurrent at one point, so be it $ S.$

Then, because of the collinearity of the points $ C,\ L,\ S,$ we conclude that the triangles $ \bigtriangleup MDY$ and $ \bigtriangleup NEX,$ are also perspective $ ($ $ C\equiv MD\cap NE$ and $ L\equiv MY\cap NX$ and $ S\equiv DY\cap EX$ $ ).$

So, we conclude that the line segments $ DE,\ MN,\ YX,$ are concurrent at one point.

That is, the line segment $ XY,$ passes through the point $ Z,$ as the intersection point of $ DE,\ MN.$

Hence, it has already been proved the collinearity of the points $ X,\ Y,\ Z$ and we will prove the collinearuty of $ A,\ X,\ Y.$

$ \bullet$ Let be the point $ T\equiv AC\cap DF$ and because of $ MP\parallel AC,$ based on the [b][size=100]Thales theorem[/size][/b], we have that $ \frac {PL}{LY} \equal{} \frac {EC}{CT}$ $ ,(6)$

The points $ T,\ C,\ E,\ A,$ are in harmonic conjugation as well and based on the above [b][size=100]Lemma[/size][/b], we have $ \frac {EC}{CT} \equal{} \frac {EN}{NA}$ $ ,(7)$

From $ (6),$ $ (7)$ $ \Longrightarrow$ $ \frac {PL}{LY} \equal{} \frac {EN}{NA}$ $ ,(8)$

From $ (8),$ we conclude that the points $ A,\ X,\ Y,$ are collinear and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Cosmin Pohoata.[/size][/b]

Kostas Vittas.

[size=150][color=darkblue][b]Very nice problem ![/b][/color][/size]

[quote="libra_gold"][size=100][color=darkred] [b]A nonisosceles triangle $ ABC$ is given and let $ AD,\ BE,\ CF$ be its altitudes with orthocenter $ H$ and we denote the midpoints

$ M,\ N,\ P$ of  $ [BC],\ [CA],\ [AB]$ respectively. Denote $ \|\begin{array}{c} X\in EP\cap NF \\
\ Y\in FD\cap PM \\
\ Z\in DE\cap MN\end{array}$ . Prove that the points $ A$ , $ X$ , $ Y$ , $ Z$ are collinearly.[/b][/color][/size][/quote]

[quote="Virgil Nicula"][size=100][color=darkred][b][u]Lemma (memorize !).[/u][/b] Let $ ABC$ be a triangle. For the points $ \|\begin{array}{c} D\in (BC) \\
 \\
E\in (CA) \\
 \\
F\in (AB)\end{array}$ denote $ P\in EF\cap AD$ . Then $ \boxed {\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}\ }$ .[/color][/size]
[hide="Proof."]
$ \|\begin{array}{c} \frac {PF}{PE} = \frac {AF}{AE}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAE}} \\
 \\
\frac {DB}{DC} = \frac {AB}{AC}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAE}}\end{array}\|\ \implies\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ .[/hide][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Apply [b]Pascal's theorem[/b] to $ FDEPMN$ (in the [b]Euler's circle[/b] of $ \triangle ABC$ ) : $ \|\begin{array}{c} Y\in FD\cap PM \\
\ Z\in DE\cap MN \\
\ X\in EP\cap NF\end{array}\|\ \implies\ \boxed {\ Z\in XY\ }$ .

$ \blacktriangleright$ Denote $ \|\begin{array}{c} S\in AX\cap EF \\
\ U\in AX\cap BC \\
\ W\in BY\cap AC\end{array}$ . Apply [b]Ceva's theorem[/b] to $ X$ in $ \triangle AEF\ \ : \ \ \frac {SE}{SF}\cdot\frac {PF}{PA}\cdot\frac {NA}{NE} = 1$ $ \implies$ $ \frac {SF}{SE} = \frac bc\cdot\frac {PF}{NE}\ (1)$ . Suppose $ a < b < c$ .

Apply upper [b]lemma[/b] : $ \frac {SF}{SE} = \frac {UB}{UC}\cdot \frac {AF}{AE}\cdot\frac {AC}{AB}$ . Using $ (1)$ obtain $ \frac {UB}{UC} = \frac bc\cdot\frac {PF}{NE}\cdot\frac {AE}{AF}\cdot\frac cb =$ $ \frac {b^2 - a^2}{2c}\cdot\frac {2b}{c^2 - a^2}\cdot\frac cb$ $ \implies$ $ \boxed {\frac {UB}{UC} = \frac {b^2 - a^2}{c^2 - a^2}\ }\ (2)$ .

$ \blacktriangleright$ Apply [b]Menelaus' theorem[/b] to $ \overline {DYF}$ in $ \triangle BPM\ \ : \ \ \frac {DM}{DB}\cdot\frac {FB}{FP}\cdot\frac {YP}{YM} = 1$ . Since $ Y\in PM\parallel AC$  obtain  $ \frac {WA}{WC} = \frac {YP}{YM} = \frac {BD}{BF}\cdot\frac {FP}{DM}$ $ \implies$

$ \frac {WA}{WC} = \frac ca\cdot\frac {b^2 - a^2}{2c}\cdot \frac {2a}{c^2 - b^2}$ $ \implies$ $ \frac {WA}{b^2 - a^2} = \frac {WC}{c^2 - b^2} = \frac {b}{c^2 - a^2}$ . Apply [b]Menelaus' theorem[/b] to $ \overline {AYV}$ in $ \triangle BCW\ :$

$ \frac {AW}{AC}\cdot\frac {VC}{VB}\cdot\frac {YB}{YW} = 1$ . Since $ Y\in BW\cap PM\implies YB = YW$ obtain $ \frac {VB}{VC} = \frac {AW}{AC}$ , .e. $ \boxed {\ \frac {VB}{VC} = \frac {b^2 - a^2}{c^2 - a^2}\ }\ (3)$ .

From the relations $ (2)$ , $ (3)$ results $ U\equiv V$ , i.e. $ \boxed {\ A\in XY\ }$ . In conclusion, the points $ A$ , $ X$ , $ Y$ , $ Z$ are collinearly.[/color]

Dear Mathlinkers,
this is the third Schroeter's result that you can see on
http://perso.orange.fr/jl.ayme  vol. 3  Les deux points de Schroeter p. 8-11.
Sincerely
Jean-Louis


i've been lookin for websites talkin bout the notion of hashtabes but i've found links with codes without comments so i wonder if some1 might do me a favour n explain me how and why we can use the hashtables till the moment i only know that it might help in order to find informations from a table that its elements are named by characters and that we need a function h but franckly i didn't figure out the use of this function.
thanx in advance

Suppose you have some data associated with strings of characters - for example, you have an address book with names and addresses. You would like to store this data in such way that you can find any person's address in constant time. One way to do this is to have one huge array with one entry for every possible name:

string address[a..zzzzzzzzzz];

address[sherlock] = "baker street";
writeln (address[sherlock]);

etc.

However, this array would have to store $27^k$ entries (if we also consider spaces; $k$ is the maximum length of name this address book would support.) If you know that you won't need to store more than 1000 people, you could instead keep the addresses in an array of length 1000 but you'd have to search through the array every time you want to find somebody's address. A better solution is to use hashing. A hash function is a function that takes as input in this case the names of people and returns integers from 1 to 1000. When you want to store information about James Maxwell you store it on location h ("James Maxwell"):

address[h ("James Maxwell")] = "Cambridge"
writeln (address[h ("James Maxwell")]);

where h is your hash function. Of course, there are many more than 1000 different names so it may sometimes happen that the hash function returns the same value for two different names. This is called a collision and because of this "address" should be actually an array of arrays - if you find out that there is already some other person stored on the index your hash function has returned, you'll add him to that array:

h ("Adam") = 123, h ("Eve") = 456, h ("Tony") = 123

After adding all three people, address contains:

address[123] = {("Adam", "1 Paradise Street"), ("Tony", "10 Downing Street")}
address[124] = {}
...
address[456] = {("Eve", "2 Paradise Street")}

This way you still have to search an array but with much less entries. The hash function is usually designed in such way that the probability of collision is as small as possible (1:1000 is the best you can get with a 1000 array) and the size of the array should be about the same as the expected size of input so that on average you have to search only through 1 or 2 entries when you extract data.

Of course, this doesn't only work with strings - you could hash for example numbers from <1, 1000000000> to <1, 1000> or just any kind of data.


2.75 g of magnesium oxide are added to 70mL of a 2.40 M solution of nitric acid. Is the resulting solution acidic or basic? What is the concentration of the ion that is responsible for the acid-base character of the solution?


i'm new and not that smart. but i got AoPS1 and am reading it. just got to the "clock problem". i understand it except for one thing. why does the word "of" mean multiply? the solution was like $\frac{x}{60}$ of the way from $4$ to $5$ o'clock is the same place as $20+5\left( \frac{x}{60}\right)$. i just don't see why you multiply by $5$.

'of' does mean multiply.
think about when you were learning to multiply... 3 times 5 was 3 groups [b]of[/b] 5, or 3 plates [b]of[/b] 5 oranges, or 3 boxes [b]of[/b] 5 balls, etc.

in you problem, I never got the solution manual but I think the '20' is the 20 minutes before the '4' on the clock, and the '5' is the 5 minutes tha are between the '4' and the '5' on the clock, and you are multiplying that by $\frac{x}{60}$ because you want to know what part of 5 minutes have already past...

oh i think i get it.

there are $5$ minutes between $4$ o'clock and $5$ o'clock. if you divide these $5$ minutes into $60$ equal units, there are $\frac{5}{60}$ of them. then, the hour hand has traveled $x$ of these units, which is $\frac{5x}{60}$.


Find the first $5$ consecutive integers such that each is expressible as a sum of two squares.

(Can you generalize for $n$ consecutive integers?)

Oh, duh. I wasn't thinking when I posted the question, sorry. :blush:

Ok, can you find the first $3$ consecutive integers that are expressible as the sum of two squares? (the squares must be $\in \mathbb{N}$)

All odd squares are congruent to $1 \mod 8$, and all even squares are congruent to $0$ or $4$ $\mod 8$, so the least one of the triplet must be divisible by $8$.

Then there are only $9$ cases to check before I found those three- a bit ugly, but it didn't take long.

[quote="drunner2007"][hide]Not possible. If this is possible, then at least one of the integers is congruent to $3$ modulo $4$. But the sum of two squares is congruent to $0,1,$ or $2$ modulo $4$. Thus this is impossible.

In fact, for $n\geq 4$ this is not possible.[/hide][/quote]

Oh, right because each square is 1(mod 4) or 0(mod 4).

Interesting question:  do there exist infinite such triples?

Edit:  [hide="Answer"] Yes.  If $n-1, n, n+1$ are all expressible as a sum of two squares then so are

$n^{2}-1, n^{2}, n^{2}+1$ [/hide]


integration
calculus
calculus computations

[quote="orangedude654"]Find the volume of the solid created when rotating the region under the graph $y=x^{2}$ from x=0 to x=1 about the x-axis.

So  is it $\pi \int^{0}_{1}x^{4}$

then $x^{5}/5$ from 0 to 1
then plugging in for 0 and 1 you get 1/5 $\pi$?[/quote]

$y=x^{2}$ from $0$ to $1$:

$V= \pi \int_{0}^{1}(x^{2})^{2}dx$
$V= \pi \cdot \frac{x^{5}}{5}|_{0}^{1}$
$\boxed{V= \frac{\pi}{5}}$

So yes.


A plane flies at a height of 10 km and the state of weightlessness is created for 20 seconds. The plane ascends for 10 seconds and then descends for 10 seconds and returns to the initial height of 10 km.

a) Calculate the angle of elevation at the beginning of the weightlessness, if the speed at this time is 900 km/h?

b) To what height does the plane ascend?

c) What is the speed of the plane at the highest point?

To achieve weightlessness, the plane must decelerate at the rate of $9.8 \frac{m}{s}$ To simplify calculations, I'll use 10.
[hide]
First, I solved for the vertical component of velocity via Galileo's.
$0 = v_{y}\cdot 20+\frac{1}{2}\cdot-10 \cdot 20^{2}$
$v_{y}= 100 \frac{m}{s}$

Then simply do $\sin^{-1}(\frac{100}{900})$
Therefore the angle is: $\approx 6.3794^\circ$

For part (b), we can assume that the plane will reach max height midway through its descent.
$d = 10 \cdot 10^{3}+100 \cdot 10+\frac{1}{2}\cdot-10 \cdot 10^{2}$
$d = 10500 m.$

For part (c), the vertical velocity is 0, so the horizontal velocity is $900^{2}-100^{2}$, by Pythagorous. We can use this value from the beginning of the fall because the horizontal velocity remains constant, barring air resistance.
$v \approx 894.42719$
[/hide]

Well stuck, you made a mistake by mixing up the dimensions. you wrote: $\frac{100\frac{m}{s}}{900\frac{km}{h}}$
But the write answer is:
$\sin^{-1}\frac{360}{900}\approx 23,6^{\circ}$

Thank you, I'll fix the solutions up.

[hide="Rectification of Solution"]
Converting $100 frac{m}{s}$ to $360 \frac{km}{h}$
(a) Angle is as Thaakisfox said, $\approx 23.6^\circ$
(b) This one seems correct.
(c) It should be $900^{2}-360^{2}= v^{2}$ , $v \approx 824.8636$
[/hide]


[quote="niyo"]x,y,z are a pozitive number,
Prove that the inequalities,
 (xy+yz+xz)(1/(x+y)^2 + 1/(y+z)^2 + 1/(x+z)^2 ) >=9/4[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3547


[quote="Taoli"]Follow your teacher to get a text for you to master elementary Number Theory.[/quote]
you didn't understand me!

Assume that exist $ n$ such as
$ k\equal{}\frac{80^n\minus{}1}{8^n\minus{}1}$, with $ k \in \mathbb{Z}$
$ j\equal{}k\minus{}10^n\equal{}\frac{10^n\minus{}1}{8^n\minus{}1}$ , with $ j \in \mathbb{Z}$
$ j\minus{}1\equal{}\frac{2^n (5^n\minus{}4^n)}{8^n\minus{}1}$ notice that $ (2^n,8^n\minus{}1)\equal{}1$ for $ n \ge 1$
$ l\equal{}\frac{j\minus{}1}{2^n}\equal{}\frac{5^n\minus{}4^n}{8^n\minus{}1}$ with, $ l \in \mathbb{Z}$
Notice that $ 0< 5^n\minus{}4^n < 8^n\minus{}1$ for all natural $ n$ Contradiction

FOR mszew:
ok it's true that j is odd but you can't say that j-1 can be divided by 2^n

Of course he can.

I believe you agree that $ j\minus{}1\equal{}\frac{2^n(5^n\minus{}4^n)}{8^n\minus{}1}$. Since LHS is an integer, then RHS is also an integer. $ 2^n$ and $ 8^n\minus{}1$ don't have common factors, so $ (8^n\minus{}1)|(5^n\minus{}4^n)$. Hence, $ j\minus{}1\equal{}2^n\cdot\text{integer}$, i.e. $ 2^n|(j\minus{}1)$.

[quote="Ulanbek_Kyzylorda KTL"]Find all natural n with 8^n-1|80^n-1.[/quote]
$ 8^n \minus{} 1 \mid 80^n \minus{} 1 \mid 320^n \minus{} 4^n \equal{} 8^n \cdot 8^n \cdot 5^n \minus{} 4^n \iff 8^n \cdot 8^n \cdot 5^n \equiv 4^n \bmod 8^n \minus{} 1$. Since $ 8^n \equiv 1 \bmod 8^n \minus{} 1$ we get $ 5^n \equiv 4^n \bmod 8^n \minus{} 1 \iff 8^n \minus{} 1 \mid 5^n \minus{} 4^n$.
Then $ |8^n \minus{} 1| \le |5^n \minus{} 4^n| \iff 8^n \minus{} 1 \le 5^n \minus{} 4^n \iff 8^n \plus{} 4^n \le 5^n \plus{} 1$, which is obviously impossible when $ n \in \mathbb{N}$, since $ 8^n > 5^n$ and $ 4^n > 1$.

Edit: Oh.. It was the same solution as all the others..  :maybe:


On Firefox there is a search engine list in your upper right hand corner...where you can get yahoo, google, wiki, etc....can you make an AoPSWiki one??

Not official AoPS content, but you could create your own:
http://developer.mozilla.org/en/docs/Creating_OpenSearch_plugins_for_Firefox

If you're not that into coding, there are links to easy tools at the bottom.

I can code one and give you the directions for installation...

You can also get ones for dictionary, encyclopedia, and thesaurus here:
[url]http://dictionary.reference.com/tools/firefox.html[/url]


Can anyone help me solve the following equation:

x = y * ((a - b)^2)

I need help solving for "a"

Thanks

[hide]$x=y(a-b)^2$
$\frac{x}{y}=(a-b)^2$
$\pm\sqrt{\frac{x}{y}}=a-b$
$b\pm\sqrt{\frac{x}{y}}=a$
$b\pm\frac{\sqrt{xy}}{y}=a$[/hide]

Wow, you made that look so easy.  I understand now how it was done.  Thank you so much for helping me.

Thanks


limit integration factorial trigonometry logarithms calculus derivative

Find the radius and the interval of convergence and calculate the sum of the following series.

$ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^{2n \plus{} 1}}{2n \plus{} 1} \equal{} x \plus{} \frac {x^3}{3} \plus{} \frac {x^5}{5} \plus{} \frac {x^7}{7} \plus{} \cdots$

Attempt:

$ \sum_{n \equal{} 0}^{\infty}\frac {x^{2n \plus{} 1}}{2n \plus{} 1} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {x^k}{k}$

$ R \equal{} \lim_{k\rightarrow\infty}\frac {k \plus{} 1}{k } \equal{} 1$

I don't know exactly how to calculate sum

I tried in this way

$ f'(x)\equal{}\sum_{n\equal{}0}^{\infty}x^{2n}\equal{}\frac{1}{1\minus{}x^2}$

$ f(x)\equal{}\int_{0}^{x}\frac{1}{1\minus{}x^2}dx\equal{}\int_{0}^{x}\frac{dx}{2(1\plus{}x)}\plus{}\int_{0}^{x}\frac{dx}{2(1\minus{}x)}\equal{}\frac{1}{2}ln\frac{1\plus{}x}{1\minus{}x}$

[quote="azatkgz"]Find the radius and the interval of convergence and calculate the sum of the following series.

$ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^{2n \plus{} 1}}{2n \plus{} 1} \equal{} x \plus{} \frac {x^3}{3!} \plus{} \frac {x^5}{5!} \plus{} \frac {x^7}{7!} \plus{} \cdots$

Attempt:

$ \sum_{n \equal{} 0}^{\infty}\frac {x^{2n \plus{} 1}}{2n \plus{} 1} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {x^k}{k}$

$ R \equal{} \lim_{k\rightarrow\infty}\frac {k \plus{} 1}{k } \equal{} 1$

I don't know exactly how to calculate sum[/quote]

your series in question is inconsistent, on the LHS, you don't have factorial, and on the expanded RHS, you have factorials, so, which one is it ?  :(

apart from the fact that the given series is simply $ \sinh\,(x)$ [if it has factorials] and $ \tanh^{\minus{}1}\,(x)$ [if it doesn't], let us actually try and find the radius of convergence using limits. you will find it is the same as that of $ \sin\,(x)$, which is the entire real line.

$ R\;\equal{}\;\dfrac{1}{ \lim_{n\rightarrow\infty}\; \bigg|\; \frac {a_{n \plus{} 1}}{a_n} \;\bigg|\;} \;\equal{}\; \lim_{n\rightarrow\infty}\; \bigg|\; \frac { (2n\plus{}3)! } { (2n\plus{}1)! } \;\bigg| \;\equal{}\; \infty$

( of-course, another way to look at it is $ \sinh\,(x)\;\equal{}\;\frac{1}{2}\left(e^x\minus{}e^{\minus{}x}\right)$ )

i will leave it to you to show that $ R$ for $ \tanh^{\minus{}1}\,(x)$ is equal to $ 1$

and yes, $ \tanh^{\minus{}1}\,(x) \;\equal{}\;\dfrac{1}{2}\;\ln\,\dfrac{1\plus{}x}{1\minus{}x}$

i suppose one can also take the [b]derivative[/b] and measure the distance from $ 0$ to the nearest [i]non-removable[/i] singularity to get $ R$.  :|


inequalities
integration
vector
calculus
real analysis
real analysis unsolved

Let $\Omega$ be a bounded open subset of $\mathbf R^{n}$ with smooth boundary and $u \in C^{2}(\overline\Omega)$ be such that $u=0$ on $\partial\Omega$. Prove that for each $\epsilon>0$ there holds
\[\int_\Omega |\nabla u|^{2}\le \epsilon \int_\Omega (\Delta u)^{2}+\frac 1{4\epsilon}\int u^{2}.\]

A vector calculus computation:

$\Delta(u^{2})=\nabla\cdot\nabla(u^{2})=2\nabla\cdot(u\nabla u) =2|\nabla u|^{2}+2u\Delta u$

Hence by the divergence theorem,

$\int_{\Omega}(|\nabla u|^{2}+u\Delta u)=\int_{\partial\Omega}u\nabla u\cdot\vec{n}$

With $u$ being zero on the boundary, the integral on the right is zero. That leaves us with

$\int_{\Omega}|\nabla u|^{2}=-\int_{\Omega}u\Delta u$

The result the follows from the simple inequality $ab\le\frac12(a^{2}+b^{2})$ in the form

$-u\Delta u\le \epsilon(\Delta u)^{2}+\frac1{4\epsilon}u^{2}.$


pigeonhole principle
geometry
combinatorics unsolved
combinatorics

Prove that among any ten points inside a cirlce of diameter 5 there exist two whose distance is less than 2.

:)  No one can solve and explain this problem..  :(  ok

You can break it up into 9 pigeonholes so that the maximum distance between two points in any one pigeonhole is 2.  You can first take a circular area around the center be be the first pigeonhole, and then break up the rest of the circle into 8 equal sectors.  Then, two points must lie in the same pigeonhole, and we are done.


function
probability
algebra unsolved
algebra

Let $\mathbb{R}$ denote the set of real numbers. Find all function $f : \mathbb{R}\rightarrow \mathbb{R}$ such that
$f(x) \cdot f(yf(x)-1) = x^{2}\cdot f(y)-f(x)$
for all real numbers $x$ and $y$.

Letting $x=y$ we get $f(x)f(xf(x)-1)=x^{2}f(x)-f(x)=f(x)(x^{2}-1)$
from which its clear that either $f(x)\equiv 0$ or $f(x)=x$ :D

Some progress...

If f(x) = 0 for some $x \ne 0$, then we have $x^{2}f(y) = 0$ for all y, so f(x) is identically zero. Now set x = 0 and observe that $f(0)f(f(0)y-1) =-f(0)$. If f(0) is not zero, then we divide both sides by f(0) and get that f(x) is identically -1, which clearly fails. So f(0) = 0. Plugging in y = 0 then yields f(-1) = -1.

Then plugging in x = -1, we have $-f(-1-y) = f(y)+1$ (*), so that $f(x)f(yf(x)-1) =-f(x)f(-yf(x))-f(x)$. Using that in the original equation, $-f(x)f(-yf(x)) = x^{2}f(y)$ (**). Taking y = x, we have $f(-xf(x)) =-x^{2}$, proving that the image of f contains the negative reals. Note also that if f(x) = f(y), then

$-f(x)f(-yf(x)) = x^{2}f(y)$
$y^{2}=-f(-yf(y)) = x^{2}$

so basically if we consider only negative reals to negative reals, f is bijective.

Edit--In fact, looking again at (*), we can conclude that since f is surjective onto negative reals, f is surjective onto all reals. Let t be such that f(t) = 1. Then putting x = t in (**) yields $-f(-y) = t^{2}f(y) \implies t^{2}= 1$ from which $t = \pm 1$, and obviously t is not -1, so t = 1. Now we can get rid of the negative signs in (*) and (**):

$f(y+1) = f(y)+1$ (*')
$f(x)f(yf(x)) = x^{2}f(y)$ (**').

Note also that since $f(y) =-f(-y)$, we now know that f is fully bijective from $\mathbb{R}$ to $\mathbb{R}$.

I'll try to do more later...

I am also able to prove f(x) = x for many rational numbers x (the set for which this holds is probably dense), so if someone can prove monotonicity or continuity the problem can be solved.

Yes, I can also prove that $f(x) = x$ for all rational numbers.. (other than $f(x) = 0$)
But I hate such "density thing" my teacher taught me since I can't find out clearly what's the idea..
Can someone explain with a little example ?
Also how can we prove the continuity ? By limit only ? Thanks..

Argh, so how hard can this be? I reduced it to the following and I can't believe it's not solved yet:

f is bijective.
$f(y) =-f(-y)$ (so we can just think about positive numbers)
$f(1) = 1$
$f(y+1) = f(y)+1$
$f(x)f(yf(x)) = x^{2}f(y)$

Is my following proof true ($f$ is bijective) ?

First, we will prove that $f$ is injective.
Since $f$ is an odd function, then we can consider only positive values.
By taking $y = 1$, we have $f(x) \cdot (f(f(x)-1)+1) = x^{2}$.
So if $f(x) = f(y)$, then $x^{2}= y^{2}\Rightarrow x = y$.

Now, we will proceed for the surjective one.
Set $y = x \neq 0$, then $f(xf(x)-1) = x^{2}-1$.
Let $a$ be a arbitrary real positive numbers.
By setting $b = \sqrt{a+1}\cdot f(\sqrt{a+1})$, we can prove that there exists $b > 0$ so that $f(b) = a$.

[quote="e.lopes"]this problem is in reid barton's site! i solved this recently!

to complete your solution, probability 1.01, let $f(y) = \frac{f(x)}{x^{2}}$ (we can do this, because f is bijective)

now, we have $f(\frac{f(x)^{2}}{x^{2}}-1) = 0$!

(easy to complete, because f is injective!)[/quote]
I guess I've misunderstood something, but this gives $f(\frac{1}{f(x)})=\frac{f(x)}{x^{2}}$ and not $f(\frac{f(x)^{2}}{x^{2}}-1) = 0$

Anyway, I guess this completes the solution:
I'll use the bijectivity, $f(1)=1$, $f(x)f(yf(x))=x^{2}f(y)$ (1) and $f(y+1)=f(y)+1$ (2), which were nicely proven by probability1.01.
For any $x \neq 0$ let's take $y=\frac{1}{f(x)}$ in (1): $f(x)=x^{2}f(\frac{1}{f(x)})$.
Thus we can rewrite (1) as: $f(\frac{1}{f(x)})f(yf(x))=f(y)$. (for any $x \neq 0$)
$f$ is surjective, so we can deduce that $f(a)f(b)=f(ab)$ (3) for any $a \neq 0$ (but, of course it's true for $a=0$, because $f(0)=0$)
So f is multiplicative
Substitute $a$ -> $a+1$ in (3) and get $f(ab+b)=f(a+1)f(b)=(f(a)+1)f(b)=f(a)f(b)+f(b)=f(ab)+f(b)$ (3')
Obviously, (3') leads to additivity.
So $f$ is both multiplicative and additive, the rest is easily done.
I hope there are no mistakes in above.

[quote="e.lopes"]this problem is in reid barton's site! i solved this recently!

to complete your solution, probability 1.01, let $f(y) = \frac{f(x)}{x^{2}}$ (we can do this, because f is bijective)

now, we have $f(\frac{f(x)^{2}}{x^{2}}-1) = 0$!

(easy to complete, because f is injective!)[/quote]

Hm.. Why is this wrong ?
The problem is $f(x) \cdot f(yf(x)-1) = x^{2}\cdot f(y)-f(x)$
If $f(y) = \frac{f(x)}{x^{2}}$, then
$LHS = f(x) \cdot f( ( \frac{f(x)}{x^{2}})^{2}-1)$
$RHS = 0$
So, $f(x) = 0$ or $f( ( \frac{f(x)}{x^{2}})^{2}-1) = 0$.

EDIT : I see, it must be $f(yf(x)-1) = 0 \Rightarrow f(x) = \frac{1}{y}$.. Sorry.

Well, if we have $f(y_{0})=\frac{f(x)}{x^{2}}$ in original problem, we get
$LHS = f(x)f(y_{0}f(x)-1)$
$RHS = 0$
so, $f(x)=0$ or $f(y_{0}f(x)-1)=0$, which only leads to $y_{0}f(x)=1$.
Still, maybe I'm just missing something.


[hide]We know that the first digit will be 2. Now for the next digit, we want the next lowest prime number, which is 3. So the next digit will be the next highest prime number, which is 5. By the same logic, 7 is the next digit. So the answer is 2357.

By the way, 1 isn't a prime, is it? If it is, my solution is wrong.[/hide]

2357 just happens to be prime itself. i don't think there's really fast way to do this.....and to anyone, 1 is not prime

[quote="i_like_pie"]If the digits did not have to be distinct, then the next one would be 2333. As it is, 2357 is the answer.[/quote]If the digits weren't distinct it would be 2237. But yes you are right with 2357.

[quote="anyone"]We know that the first digit will be 2. Now for the next digit, we want the next lowest prime number, which is 3. So the next digit will be the next highest prime number, which is 5. By the same logic, 7 is the next digit. So the answer is 2357.[/quote]
That logic will not always work, but it did for this one.
For instance, with that logic you would think that the smallest 3 digit prime that each digit is prime would be 235 but it's not. (It's 257 if you want to know...).

[quote="Quevvy"][quote="ProtestanT"][quote="Quevvy"]What is the first year after 2007 where each digit is prime and the year is also prime (i.e. 1753 would work (but that's not the answer...))? and the next one? and the next one?

What is the first year after 2007 where each didit is prime and distinct and the year is also prime (i.e. 257 would work (but that's not the answer...))?

Is there a quick way to figure this out?[/quote]
um why would 1753 work? 1 is not prime

the first year after 2007 would be 2222 with all the digits prime, and it would be 2357 with all the digits prime and distinct. 2237 is the first allprime digits prime, and 2347 is the first allprime and distinct digits prime (after 2007)[/quote]
4 is not prime and 2222 is not prime

And I know, 1 is not a prime...[/quote]
i didn't say 2222 is prime, i said all the digits are prime. its supposed to be 2237 for the first 1. and yes i see my mistake for the second one


calculus
derivative
function
ratio
calculus computations

Okay, my knowledge of calculus is very limited, but i understand basic facts about derivatives.
Recently, my friend, who is taking calculus, proposed the following paradox.  neither one of us understands why it is not allowed.  here it is:

it is relatively simple to prove the power rule for derivatives, stating that the derivative of $ax^{n}$ is $nax^{n-1}$.
We all know that the derivative of $x^{2}$ is simply $2x$, this follows from the power rule when we let $a=1$ and $n=2$.
However, if we write $x^{2}$ as $x\cdot x$ we have $a=x$ and $n=1$.  This gives us a derivative of $nax^{n-1}=1\cdot x\cdot x^{0}=x$.
Why are the two answers different?  are we not allowed to let $a$ be a variable?

thanks
Eric

Let's go even further and take $a=x^{2}$, $n=0$. Then the derivative will be $0$. Also, there is no need to restrict $a$ to power functions. Let's write [b]any[/b] function $f(x)$ as $ax^{0}$ with $a=f(x)$. Then, by the power rule, we'll get $0$ again as the derivative of anything we wish. Of course, you see yourself that it is an ultimate nonsence and the reason is that you cannot take $a$ to be a function in the "constant multiple rule" $(af)'=af'$. Indeed, the reason the rule works is that you can pull $a$ out as a factor in the difference ratio $\frac{(af)(x+h)-(af)(x)}{k}=a\frac{f(x+h)-f(x)}{h}$. If $a$ is a function of $x$, this algebraic trick fails (you now have $a(x+h)f(x+h)-a(x)f(x)$ in the numerator with
no common factor) and you have to use the product rule, which eliminates the "paradox" you asked about. So, shortly put, yes, $a$ cannot be taken a function.

Two things you should look at are the Product Rule and the Chain Rule; understanding (cf. knowing) these will prevent you from running into these kinds of "paradoxes". The Chain Rule is what your question really addresses, but the product rule also shows up often in these kinds of situations.

[b]
Chain Rule:[/b]

The chain rule basically says that if Bob is running $ a$ times as fast as Jen and Jen is running $ b$ times as fast as Eric, then Bob is running $ ab$ times as fast as Eric.

In terms of derivatives, if $ z$ is increasing $ \frac{dz}{dy}$ times as fast as $ y$, and $ y$ is increasing $ \frac{dy}{dx}$ times as fast as $ x$, then $ z$ is increasing $ \frac{dz}{dy}\frac{dy}{dx}$ times as fast as $ x$. In other words:

$ \frac{dz}{dy}= \frac{dz}{dy}\frac{dy}{dx}$

Or if you prefer, $ [f \circ g]'(x) = f'(g(x))g'(x)$

In your problem, you cannot say that $ a$ is independent of $ x$. You must consider $ a$ as a function of $ x$, because $ a$ changes with $ x$.

[b]Product Rule:[/b]

Here's a heuristic argument for the product rule. Suppose you want to find $ [y \times z] '(x)$

Then, you want the limit value of $ \frac{(y+\Delta y)(z+\Delta z)-yz}{(x+\Delta x)-x}$ $ =\frac{yz+y \Delta+z \Delta y+\Delta y \Delta z-yz}{\Delta x}$ $ =y \frac{\Delta z}{\Delta x}+z\frac{\Delta y}{\Delta x}+\Delta y\frac{\Delta z}{\Delta x}$

As your sampling gets closer, this approaches $ y(x)z'(x)+z(x)y'(x)$, where the last term drops out because it approaches 0.


Diophantine equation
algebra proposed
algebra

given c is natural number.
Prove that exist infinite (x,y,z) \in  N*  s.t
     1/x + 1/y +1/z+c/(xyz)=3(c+3)/(x+y+z)

x=a-ck
y=a
z=a+ck
given equation reduces to
a 2 -c(c+2)k 2 = 1
the pell equation with infintely many positive solutions as c(c+2) is not a perfect square :D and by the equation we can easily see x,y,z are positive too (as a-ck  \geq 0)


function
Asymptote
limit
complex analysis
calculus
calculus computations

Hello,

I am having some trouble with the following problems and would appreciate help.

Thank you.

--

1. Determine the discontinuities of this function and explain the discontinuities:

$ f(x) \equal{} \left\{
\begin{array}{ccc}
4 \minus{} x ,&\mbox{ if }
x < 3  \\
x \minus{} 2,&\mbox{ if }
0<x<3\\
x \minus{} 1,&\mbox{ if }
x\leq 0 \\
\end{array}
\right.$

2. Sketch the graphs of the following functions and determine whether they are continuous on the closed interval $ [0, 1]$:

a) $ f(x) \equal{} \left\{
\begin{array}{cc}
\cfrac{1}{x},&\mbox{ if }
x > 0  \vspace{5pt}\\
1,&\mbox{ if }
x\leq 0\\

\end{array}
\right.$

b)
$ f(x) \equal{} \left\{
\begin{array}{ccc}
x,&\mbox{ if }
x\leq 0\\
0&\mbox{ if }
0<x<1\\
x,&\mbox{ if }
x\geq 1 \\
\end{array}
\right.$

-----

1. I am not entirely sure how to find the discontinuity of this entire function, with three piecewise-defined functions. I graphed it out but I do not see any discontinuities:

[img]http://img184.imageshack.us/img184/6761/problem1un6.jpg[/img]

2. a) Graph: [img]http://img55.imageshack.us/img55/3700/problem2my8.jpg[/img]

I just want to check my reasoning. Would this function be discontinuous because the function $ f(x) \equal{} 0$ and $ f(x) \equal{} \cfrac{1}{x}$ do not intersect each other as the latter is a horizontal asymptote?

b) Like question #1, I am not sure how to determine the discontinuity of this function. To me, nothing seems to be disconnected on the graph.

There are three criterion for continuity at a point $ a$:
1) $ f(a)$ exists
2) $ \lim_{x\rightarrow a}f(x)$ exists
3) $ f(a)\equal{}\lim_{x\rightarrow a}f(x)$

Test each of your functions for these three criterion and you should have no trouble determining continuity.

Thanks for your reply, JRav.

Unfortunately, this book's attempt to explain these concepts are quite feeble; so I am using another book to understand these concepts, which does not talk about the aforementioned type of functions.

Just on #1:

I can see that there is a jump discontinuity at x = 0 because:

$ \lim_{x\rightarrow 0} x \minus{} 1 \equal{} \minus{} 1$.

Therefore:

$ \lim_{x\rightarrow c} f(x) \neq f(c)$.

However, the answer states that there are [hide]removable discontinuities at $ x \equal{} \pm 1$[/hide]. I do not understand this part.


[b]Pro 1.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=144074
[b]Pro 2.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=144075
[b]Pro 3.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=144076
[b]Pro 4.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=144077

April, do you have the Seniors also? I've put up the Juniors at http://www.mathlinks.ro/Forum/resources.php?c=66&cid=62&year=2007 Thanks.

See here
(But the first problem is wrong . The problem in the contest stated 4^n and not n^4)
http://www.mathlinks.ro/Forum/viewtopic.php?t=137037


articles function calculus integration vector

As you might have seen in some mathematics articles, some conclusions are not directly based on mathematical facts. For instance, suppose one gives a physical equivalent of a math problem and then uses physics laws, say, conservation of energy to reach a conclusion.

The question is; are these arguments considered as proofs to a given problem or not?
As far as I'm concerned, the law of conservation of energy is not something provable. So what if it collapses some time? Can we use a "[u]fact[/u]" which is not in the scope of math to prove a math statement?

P.S By "fact" I mean, it is considered true in its own area.
The original page is ([url]http://www.mathlinks.ro/weblog_entry.php?p=924105#924105[/url])

If we can precisely define physics in terms of more basic mathematical ideas, i.e. in terms of functions and $ \Re^{n}$ space, then we could define the energy of an object based on those ideas and proceed from there to prove the law of conservation of energy. I'm thinking somewhere along the lines of defining an object as a function from $ \Re$ (representing "time") to $ \Re^{3}$ (representing "position") with possibly more variables to determine the mass and shape of the object. We might then need some basic axiom as to how objects interact, which we would then hopefully prove is consistent with everything else. Then we could proceed to define energy as some sort of integral, and the law of conservation of energy would hopefully follow. This would essentially turn energy into a mathematical concept, which we could then use to prove other mathematical ideas.

However, I don't think that constitutes a mathematical proof until you've rigorously defined what you mean by energy or whatever physical law you're talking about.

Of course it's not a proof.

Physics is a mathematical model of the world. This means the physics assumes the math. If you go back and try to prove the math with the physics, it's circular reasoning. And it's pretty lame, too... most likely, it could be proven much more easily without physics.

But yes, physical concepts can be used to nicely [b]demonstrate [/b]a mathematical concept in a manner that anyone can relate to.

[quote="JBL"]It would be helpful if you could provide an actual example of what you are talking about.  (The link in your original post is dead.)[/quote]

For example, suppose you're given a triangle $ ABC$ and you want to find $ P$ such taht $ AP\plus{}BP\plus{}CP$ is minimized.  Take three lengths of string.  Tie a 1 kg mass to one end of each.  Poke a small hole in each vertex, and thread each string through one of the holes.  Tie the three ends toghether.  Now hold the triangle in place.  The system will tend towards a position with the lowest potential energy, which will happen iff the point where the strings meet is the point we want.  This happens when force is zero.  For this to happen, we must have unit vectors pointing in the directions of $ PA$, $ PB$ and $ PC$ sum to zero.  This happens iff $ m\angle{APB}\equal{}m\angle{BPC}\equal{}m\angle{CPA}$.  So we have the desired $ P$ is that point. (the actual physical model undergoes oscillation, but that doesn't matter)

[quote="JBL"](The link in your original post is dead.)[/quote]

http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=924105#924105

(S)he forgot the "Forum" dir.

[quote="The Captain"]As you might have seen in some mathematics articles, some conclusions are not directly based on mathematical facts. For instance, suppose one gives a physical equivalent of a math problem and then uses physics laws, say, conservation of energy to reach a conclusion.[/quote]

This is not a proof unless the physical principles involved can be quantified mathematically.  In K81o7's example, potential energy and forces are quantities that can be defined mathematically without any reference to the real world, and can be used to solve the problem in an entirely mathematical way.

I think it is important here to distinguish between [b]physics[/b] and [b]the mathematical tools that physics employs.[/b]  The former is an attempt to model aspects of the real world; this has nothing, intrinsically, to do with math until we use math in those models.  In that sense, [b]mathematics is a tool of physics[/b], which means that the mathematical tools we use to do physics are equally valid to be used in a purely mathematical sense because those tools are put on a rigorous mathematical foundation.

By the way, I have not seen an example of a published mathematical paper which uses a physical concept to prove a mathematical principle.

Edit:  For an example of what I mean, the principle of conservation of energy is not, in and of itself, a true statement about the world.  It is a true statement in some mathematical framework which approximates the world (for example, Newtonian mechanics, or depending on what other kinds of energy you're using, thermodynamics).  It is [b]empirical observation and the scientific method[/b] which provides any kind of connection between that mathematical framework and the real world.


1st place had 42
2nd place had 41
3rd place had like 38 or 39 ( i think)
4th place had 37 or 38 (1 or the other i'm sure)

i guessed 42, 39, 37, 37 or something and i was told one was more than two off

jorian

(i had a feeling that post would get a bad rating :rotfl: )

[quote="jhredsox"]Now just wondering, but can someone from Indiana (or someone who knows) tell me/us what the state scores for Indiana were? I mean, you had two VERY HIGH placers (and I'm sure the others weren't far from them), and it'd be nice to know how their state scores had correlation with their national scores. Thank you,
jorian[/quote]

hmm i dont' know yet, but i'm pretty sure one of the members did not perform well. let's just say that he lucked out at states.

Ok, sorry for the spamming guys.

(you can't really say people lucked out, cause if they weren't good they wouldn't be in mathcounts in the first place. you can say they had a good day.)

ok, really sorry for this topic,
jorian


Plot it out. Its logical, and it certainly looks true to me.

The basic transformations (what happens when you add or subtract to x or y, or when you multiply x or y by a constant) are always the same, no matter what the graph is.

By the way, I suggest you stop using step and start using the proper floor or just square brackets.. I think you said ages ago thats how you learnt it but if you used it in a competition or something I don't think its official enough to be understood properly..

But yes, those transformations are exactly the same, no matter what graph you have. Even if it some crazy graph like f(x) is the [x]th digit of pi. Or anything.


Can anyone tell me how to get the little animation below the logout??? They r just so cool. Thx in advance.

what do you mean?  Don't you just see them on the screen>

Maybe his forum is in the MathLinks layout like mine is...

Do you mean the "proof without words" animations?  You can view all of them at [url=http://www.usamts.org/Gallery/G_Gallery.php]http://www.usamts.org/Gallery/G_Gallery.php[/url]

i meant like where can u download those files.


Given an angle of 13 degrees, construct with ruler and compass an angle of 1 degree.

Generalization:  For any angle $\theta$ (in degrees) such that $(\theta, 180) = 1$ we must have that $\exists a, b \in \mathbb{Z} : a (\theta) + b( 180 ) = 1$.  :)

Isn't that essentially a result/restatment of the Euclidean Algorithm?

[quote="Farenhajt"]$7\cdot 13^\circ-90^\circ=1^\circ$[/quote]

Sorry, can you explain with detail? Thanks.

[quote="Farenhajt"]Use ruler and compass to add the angle of $13^\circ$ six more times to itself. Then construct the right angle and subtract it from the obtained $7\cdot 13^\circ=91^\circ$ angle.[/quote]

And you can construct and angle of $13^\circ$ with ruler and compass?? how?

[quote="José"][b]Given an angle of 13 degrees[/b], construct with ruler and compass an angle of 1 degree.[/quote]

(Emphasis added.)  :)

The angle is given, correct?  Then it must be given as an actual physical angle.  So you trace it, and then copy it six more times.

[quote="Farenhajt"][quote="José"][quote="Farenhajt"]Use ruler and compass to add the angle of $13^\circ$ six more times to itself.[/quote]

That's the step I don't understand......... sorry if I'm unclear....[/quote]

To copy an angle:

Take a compass and draw an arc of an arbitrary radius inside the angle. Let it intersect the sides of the angle at points $A$ and $B$. Now draw a ray $Or$, put the compass at $O$ and use [u]the same radius[/u] to draw a new arc. Let it intersect the ray at $C$. Then go back to the given angle and take the chord $AB$ into the compass. Put the compass into $C$ and draw the small arc which will intersect the big arc at $D$. Draw the ray $OD$. Then $\angle DOC$ is equal to the given angle.[/quote]

I feel stupid....... what's ray?????

ray is a type of line, that starts on one fixed point and goes infinitely out to some other direction. (you can think out it as a trucated line or a vector with magnitude of infinity)

[quote="pkerichang"]ray is a type of line that starts on one fixed point and goes infinitely out to some other direction. (you can think out it as a trucated line or a vector with magnitude of infinity)[/quote]

Ohhh I seee

[quote="Farenhajt"][quote="José"][quote="Farenhajt"]Use ruler and compass to add the angle of $13^\circ$ six more times to itself.[/quote]

That's the step I don't understand......... sorry if I'm unclear....[/quote]

To copy an angle:

Take a compass and draw an arc of an arbitrary radius inside the angle. Let it intersect the sides of the angle at points $A$ and $B$. Now draw a ray $Or$, put the compass at $O$ and use [u]the same radius[/u] to draw a new arc. Let it intersect the ray at $C$. Then go back to the given angle and take the chord $AB$ into the compass. Put the compass into $C$ and draw the small arc which will intersect the big arc at $D$. Draw the ray $OD$. Then $\angle DOC$ is equal to the given angle.[/quote]

Have you got a drawing or picture?????