modular arithmetic
number theory proposed
number theory

Prove that for each prime number $ p>3$ and for each positive integer $ k$ there exists a positive integer $ n$ such that $ p^{k}\parallel{} n^{p\minus{}2}\plus{}2,$ i.e. $ p^{k}| n^{p\minus{}2}\plus{}2$ but $ p^{k\plus{}1}\not | n^{p\minus{}2}\plus{}2.$

Choose a number $ s$ such that $ 2s\equiv 1(\mod p^k)$ so $ s^{p\minus{}2}\equiv 2(\mod p^k)$

Now choose the numer $ n$ such that $ p^k\parallel{}n\plus{}s$
Now we know that if $ p^k\parallel{}a\minus{}b$ and $ p^l\parallel{}n$ then $ p^{k\plus{}l}\parallel{}a^n\minus{}b^n$

Using this fact we have that $ p^k\parallel{}n^{p\minus{}2}\plus{}s^{p\minus{}2}\equiv n^{p\minus{}2}\plus{}2$ and we are done

[quote="silouan"]Choose a number $ s$ such that $ 2s\equiv 1(\mod p^k)$ so $ s^{p \minus{} 2}\equiv 2(\mod p^k)$[/quote]
Why is that? But it's clear that such $ s$ exists anyway, since the map from $ \mathbb{Z}^*_{p^k}$ to $ \mathbb{Z}^*_{p^k}$: $ x \minus{}> x^{p\minus{}2}$ is bijective.

I believe it's because we know that $ s^{p - 1} \equiv 1 \pmod {p^k}$, so:

\begin{eqnarray*} 2*s^{p - 1} & \equiv & 2 \pmod {p^k} \\
2*s*s^{p - 2} & \equiv & 2 \pmod {p^k} \\
s^{p - 2} & \equiv & 2 \pmod {p^k}\end{eqnarray*}

On the other hand...
[quote]Now we know that if $ p^k\parallel{}a - b$ and $ p^l\parallel{}n$ then $ p^{k + l}\parallel{}a^n - b^n$ [/quote]
What result tells us this? Can someone name it please?

EDIT: nevermind, that top part is wrong... we can't actually say that $ 2s \equiv 1 \pmod {p^k} \Rightarrow s^{p - 2} \equiv 2 \pmod {p^k}$, but your point about bijective mapping still holds so that's ok.

[quote="kops723"]
EDIT: nevermind, that top part is wrong...[/quote]

Why ? :huh:

Because $ s^{p\minus{}1} \equiv 1 \mod{p^k}$ does not necessarily hold.

In fact I thought that the system of congruences $ 2s\equiv 1(\mod p^k)$ and $ s^{p\minus{}1}\equiv 1(\mod p^k)$

has a solution ... But in fact this is not trivial (and maybe wrong .Is it? )

Hopefully the bijection saves my proof  :)

[quote="silouan"]But in fact this is not trivial (and maybe wrong .Is it? ) [/quote]

Yes. Note that existence of such solution implies that $ 2^{p-1} \equiv 1 \mod{p^k}$ which usually doesn't hold even for $ k=2$ (primes for which congruence hold for $ k=2$ are called Wieferich primes and they occur very rarely and we don't know if there are infinitely many of them). But since $ (p-2, \varphi(p^k))=1$ map $ x^{p-2}$ is an automorphism of multiplicative group of ${ \mathbb{Z}_p^k}$ so it's not a big deal.

[quote]On the other hand...
[quote]Now we know that if $ p^k\parallel{}a \minus{} b$ and $ p^l\parallel{}n$ then $ p^{k \plus{} l}\parallel{}a^n \minus{} b^n$ [/quote]
What result tells us this? Can someone name it please?
[/quote]
See for example here : http://reflections.awesomemath.org/2007_3/Lifting_the_exponent.pdf


Has this problem been discussed/solved on the forum?

Find the sum:

$ \sum\limits_{k\equal{}0}^{n}\biggm|\binom{n}{k}\minus{}2^{k}\biggm|\equal{}?$

In other words when is $ 2^{k}>\binom{n}{k}$?

when 2^k>nCk
in such cases the sum is
| 2^n - (2^n+1 - 1)|
[/code][/hide][/list]


Just curious, but how much do you think one could raise their score by working through all the past Aime tests?

I'd say about 2-6 pts. For me, I raised my AIME score from a 6 in 2004 to a 12 in 2005, mostly by just doing old AIME tests. In my opinion, doing old AIMEs is BY FAR the most effective way to practice for the AIME.

Hmm..

I only took one AIME in my life ( :D ) so far and that was 2002 AIMEA. I made two to three careless mistakes on that exam and got 5. If I practice AIME for next two months, will there be any chance for me to reach the floor value?

I mean, you don't know what that will be but let's just say around 7-8.  :) In another word, do you think I can raise my score by 3-4 points at least in next two months by practicing old AIMEs?

Thanks a lot!

Well...

The first time I ever saw an AIME was when I took the 2005 AIME 1. I got a 3. After joining AoPS at the end of April, after about six months I could solve over 10 problems on the same AIME. I think it varies between people, but you should expect probably at least 4 points improvement over the course of a few months.

[quote="1234567890"]Well, in around 4 months (May-September), I slowly but steadily raised my scores around 5 points by taking a [b]lot [/b]of AIMEs. See:
[url=http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=352]My AIME Scores[/url][/quote]

Lol, I'm at where you were in August right now :blush: I need to get my score up at least 5 points if I want to get a 10. I'm modifying my schedule to include an AIME every week :!: :D

Hmm... I'm doing an AIME a day during breaks. :)  :lol:

I'm just doing any AIME problems that are posted on AoPS... :roll:

I do an AIME a week... I get about a 10 on them. But that's with the pressure off, with the pressure on I'll probably get like a 6 or something.


probability
complex analysis
real analysis

I will be graduating next quarter with a degree in mathematics (God willing).  Problem is: during my first and second years in UCSD(after transferring from WLAC), I really messed up my gpa and I was not really focused.  I've been trying to clean up my act ever since.
I started researching for jobs and stuff and I stumbled across this place and I learnt more about careers in math than I ever knew.

I'm interested in Actuaries since that's a path that I can still take (due to the fact that I'm graduating soon)...

Can one of you please point me in the right direction?

[quote="MCrawford"]If you look through this forum you'll find a fair amount of discussion about actuarial work.

Otherwise, you might do better to ask specific questions.[/quote]

Thanks for your reply.  I'll be more specific.
After I get my B.A. in Applied Mathematics, will I be able to become an actuary?
I haven't taken any classes in finance, but I have probability and statistics in my bag along with some programming and a bunch of mathematics (complex analysis, real analysis, abstract algebra, numerical analysis, numerical pde's).

I don't know enough about the actuarial world to have an opinion on if you're screwed or not, but I strongly suggest you just proceed at this point as if you are not screwed.  Start researching jobs.  Check out the major actuarial societies for guidance.

[quote="MCrawford"]I also recommend taking the first exam.  If you're solid with your prob/stats, then you'll pass it easily with a high score.  It should be easier than a 200 level math course exam at UCSD, but you'll want to make sure you get a high score, so looking at a couple of practice tests is probably a good idea.[/quote]

Thank you.  I've already looked at some of the practice tests and they don't seem that bad.  I'll try to take the exam.

A good place that answers a lot of questions about actuarial careers is the Actuarial Advice page:

http://users.aol.com/fcas/advice.html


advanced fields
advanced fields theorems

Hi, I am just interested in a simple formula. I just need it to do other stuff.

Suppose you are in 3-dimension Euclidean with f(x,y,z)=k where k>0.
Suppose that I have two points (a1,b1,c1) and (a2,b2,c2) such that ai,bi,ci>0 and
f(ai,bi,ci)=o for i=1,2. Then what is the exact equation of the geodesics (smallest curve on the plane given by f) that go from one point to the other on f? Thanks.

It's hard in general to find the exact solution of the geodesic-problem, even in this simple case. The geodetic that connect two points on a surface is the solution of a second order sistem of linear differential equations that involves coefficients of the metric tensor of the surface. You can find it in any text of Riemannian geometry...


geometry
trapezoid
geometric transformation
reflection
Gauss
cyclic quadrilateral
perpendicular bisector

Given a cyclic quadrilateral ABCD, and the points of intersection $F=AB\cap CD$ and $G=BC\cap DA$. Let M be the midpoint of the segment FG. Denote by A', B', C' and D' the nine-point centers of triangles AFG, BFG, CFG and DFG.

Prove that these points A', B', C', D' and M lie on one circle. More precisely, the quadrilateral A'D'B'C' is an isosceles trapezium, and the perpendicular bisectors of the segments A'C' and B'D' coincide and pass through the point M.

This was indicated as a conjecture in

Christopher J. Bradley, [i]The Diagonal Points of a Cyclic Quadrangle[/i], Crux Mathematicorum 3 / 31 (2005), pages 168-172.

However, I must say the result is true and rather easy to prove. Enjoy!

  Darij

If you reflect $B$ in $FG$ to get $B''$, the quadrilateral $B''GDF$ will be cyclic, so the circumradii of $BFG,DFG$ are equal, which means that so are the radii of their nine-point circles. We thus get $B'M=D'M$. In the exact same way we find $C'M=A'M$.

Let $X,Y,Z$ be the midpoints of $BG,BF,BD$ respectively. We have $\angle XMY=\angle GBF$ and $\angle XZY=\angle GDF=\pi-\angle GBF=\pi-\angle XMY$, which means that $X,Y,Z,M$ are concyclic, i.e. the midpoint of $BD$ lies on the nine-point circle of $BGF$. It must also lie on the nine-point circle of $DGF$ then, which means that $B'D'$ is perpendicular to the line connecting the midpoints of $FG,BD$, i.e. the Gauss line of $ABCD$. Of course, $A'C'$ must also be perpendicular to this line, and we have $A'C'\|B'D'$. Together with what we have from the paragraph above, we get the stronger conclusion written by Darij.

You forgot to show that the points A', B', C', D' and M lie on one circle. Funnily enough, so did I when I first saw the problem.

Anyway, now I have a [i]complete[/i] solution of the problem:

We will work with directed angles modulo 180°.

Let U and V be the midpoints of the segments BD and AC. Then, the Gauss line g of the complete quadrilateral formed by the lines AB, BC, CD, DA passes through the midpoints U, V and M of its diagonals BD, AC and FG.

Let X, Y, R, W be the midpoints of the segments BG, BF, DG, DF, respectively.

Since the points M and Y are the midpoints of the sides FG and BF of triangle BFG, we have MY || GB. Similarly, UY || DF, MX || FB and UX || DG. Thus, < (MY; UY) = < (GB; DF) and < (MX; UX) = < (FB; DG). Since the quadrilateral ABCD is cyclic, < BCD = < BAD. Hence,

  < MYU = < (MY; UY) = < (GB; DF) = < BCD = < BAD = < (FB; DG) = < (MX; UX) = < MXU.

Thus, the points M, X, Y and U lie on one circle; call this circle $k_B$. This circle $k_B$, passing through the midpoints M, X and Y of the sides FG, BG and BF of the triangle BFG, must be the nine-point circle of triangle BFG; hence, its center is the nine-point center B' of triangle BFG. Similarly, the points M, R, W and U lie on one circle $k_D$, which is the nine-point circle of triangle DFG and has center D'. Now, as we saw above, the chordal angle of the chord MU in the circle $k_B$ equals < MYU = < BAD; similarly, we can see that the chordal angle of the chord MU in the circle $k_D$ equals < DCB. Since < DCB = - < BCD = - < BAD, it follows that the chordal angles of the chord MU in the circles $k_B$ and $k_D$ are opposite. Thus, these circles are symmetric to each other with respect to the line MU. Hence, their centers B' and D' are also symmetric to each other with respect to the line MU. Since the line MU is the Gauss line g of the complete quadrilateral ABCD, it follows that the points B' and D' are symmetric to each other with respect to the Gauss line g. Thus, the Gauss line g is the perpendicular bisector of the segment B'D'. Similarly, the Gauss line g is the perpendicular bisector of the segment A'C'. Hence, we have shown that the perpendicular bisectors of the segments A'C' and B'D' coincide - namely, they both coincide with the Gauss line g -, and pass through the point M (since g passes through M).

Since the perpendicular bisectors of the segments A'C' and B'D' coincide, the quadrilateral A'D'B'C' is an isosceles trapezium.

Now, it remains to show that the points A', B', C', D' and M lie on one circle. Prepare for another angle chase:

First, since the points M, U and Y lie on the circle $k_B$, while the center of this circle is B', the central angle theorem yields < UMB' = 90° - < MYU. Since < MYU = < BCD, this becomes < UMB' = 90° - < BCD. Similarly, < UMD' = 90° - < DAB. Thus,

  < B'MD' = < UMD' - < UMB' = (90° - < DAB) - (90° - < BCD) = < BCD - < DAB = < BCD + < BAD.

Since < BCD = < BAD, this becomes < B'MD' = 2 < BCD.

Let S and T be the feet of the perpendiculars from the points F and G to the lines BC and CD. Then, < FSG = 90° and < FTG = 90°; thus, the points S and T lie on the circle with diameter FG. The center of this circle is, of course, the midpoint M of the segment FG. Hence, by the central angle theorem, < SMT = 2 < GST. But < GST = < (BC; GT) = < (BC; CD) + < (CD; GT), and $GT\perp CD$ yields < (CD; GT) = 90°, so that we get < GST = < (BC; CD) + 90° = < BCD + 90°, and thus

  < SMT = 2 < GST = 2 (< BCD + 90°) = 2 < BCD + 180° = 2 < BCD        (since 180° = 0°).

Since < B'MD' = 2 < BCD, this becomes < SMT = < B'MD'.

The nine-point circles of triangles BFG and CFG have two points in common: the point M, being the midpoint of the common side FG of the two triangles, and the point S, being the foot of the altitude from the vertex F to the side BG aka CG. Hence, the line MS is the common chord of these two nine-point circles. But the centers of these two nine-point circles are B' and C', respectively. Since the common chord of two circles is perpendicular to the line joining their centers, we thus have $MS\perp B^{\prime}C^{\prime}$, so that < (MS; B'C') = 90°. Similarly, < (MT; C'D') = 90°. Hence,

  < B'C'D' = < (B'C'; C'D') = - < (MS; B'C') + < (MS; MT) + < (MT; C'D')
                 = - 90° + < SMT + 90° = < SMT = < B'MD'.

Hence, the point C' lies on the circle through the points B', D' and M. Similarly, the point A' lies on the circle through the points B', D' and M. Hence, the points A', B', C', D' and M lie on one circle. And the problem is solved.

  darij


linear algebra
matrix
induction
linear algebra unsolved

Consider $ A, B \in M_n (\mathbb{R})$ such that $ B\minus{}A\equal{}I_n$. What conditions must the matrix $ A$ have such that for any $ k \in \mathbb{N}^*$, the equality $ B\plus{}B^2\plus{}...\plus{}B^k\equal{}kI_n\plus{}\frac{k(k\plus{}1)}{2}\cdot A$ holds?

for $ k\equal{}2$ we find $ A^2\equal{}0$.
now suppose that $ A^2\equal{}0$ then $ B^2\equal{}2B\minus{}I$ ,and by induction $ B^k\equal{}kB\plus{}(1\minus{}k)I$ for every natural number $ k$.
and in this case we have : $ B\plus{}B^2\plus{}...\plus{}B^k\equal{}kI_n\plus{}\frac{k(k\plus{}1)}{2}\cdot A$.


Provide reasonable synthesis of each of the following amine drugs (click to see structure and relevant information):

1) [url=http://en.wikipedia.org/wiki/Pargyline]Pargyline[/url]

2) [url=http://en.wikipedia.org/wiki/Tranylcypromine]Tranylcypromine[/url]

3) [url=http://en.wikipedia.org/wiki/Rasagiline]Rasagiline[/url]

From readily available reagents, like benzene or toluene, alcohols, etc.

for the second one, is it possible to use a simmons-smith reaction without affecting the benzne ring?

i just realized that you wud probably get tropilidene as well.
neways how abt this for the frist one?
n,nmethyl,benzylamine + formaldehyde + ethyne(one end protected by TMS) in presence of CuBr
 and foll by deprtoection of the TMS group?
something like this?
http://www.orgsyn.org/orgsyn/prep.asp?prep=v84p0001


limit
linear algebra
matrix
real analysis
real analysis unsolved

With a) we can use Binet's theorem and $ 1+x+...+x^{n}=\frac{x^{n+1}-1}{x-1}$.  :D

Hmm, Kuba told me a solution for a). I would be interrested if you can post a full solution. :)

Let $ A = \begin{pmatrix}1 & 1 \\ 1 & 0 \end{pmatrix}$. For appropriate* $ x$ we have that
\[ \sum_{n=0}^\infty \left( xA \right)^{n}= \left( I-xA \right)^{-1}. \]
Now let's look at the upper-right element of both sides. In the left, we have $ \sum_{n=1}^\infty x^{n}F_{n}$. In the right, it's $ \frac{x}{1-x-x^{2}}$.
[I assumed that we all know that $ A^{n}= \begin{pmatrix}F_{n+1}& F_{n}\\ F_{n}& F_{n-1}\end{pmatrix}$.]

Plug in $ x = \frac12$ and there you go :)

*We do have to find out what the range of $ x$ can be. The condition is $ (xA)^{n}\to 0$, i.e. $ |x| < \frac{\sqrt 5-1}2$.

Here is the solution for number 1)

$ \sum_{n=1}^{\infty}\frac{F_{n}}{2^{n}}=\frac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left(\left(\frac{\alpha}{2}\right)^{n}-\left(\frac{\beta}{2}\right)^{n}\right)$

$ =\frac{1}{\sqrt{5}}\left(\frac{\alpha}{2-\alpha}-\frac{\beta}{2-\beta}\right)=2$,

where $ \alpha=\frac{1+\sqrt{5}}{2}$ and $ \beta=\frac{1-\sqrt{5}}{2}$.

The simple trick without Binet's formula for number 1 can be found, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160771]here[/url].

I think you meant

$ \sum_{n=0}^{\infty}\arctan\frac{1}{F_{2n+1}}=\frac{\pi}{2}$.

We have that

$ \arctan F_{2k+2}-\arctan F_{2k}=\arctan\frac{F_{2k+1}}{1+F_{2k+2}F_{2k}}=\arctan\frac{1}{F_{2k+1}}$, since

$ 1+F_{2k+1}F_{2k}=F_{2k+1}^{2}$.

Iterate and the result would follow.


Provide the shortest way of making the following functional group interconversion:

$ \ce{R\minus{}NH2} \longrightarrow \ce{R\minus{}CHO}$

how about Hofmann elimination (excess methyl iodide; silver oxide, water, heat), addition of HBr, conversion to a Grignard (oxidative insertion of magnesium metal), attack on formaldehyde and quench, and oxidation via PCC ?

how about converting $ \ce{RNH2}$ to $ \ce{RCN}$ and then convert into $ \ce{RCN}$ and then hydrolyse to $ \ce{RCOOH}$ and then reduce to $ \ce{RCHO}$ :maybe:

[b]MysticTerminator[/b], the conversion should be independent of the existence or not of a beta hydrogen. And how about if R is aryl?

[b]Madness[/b], your synthesis envolves at least 6 steps. Anything shorter?

1. $ \ce{R \minus{} NH_2}$ + $ \ce{CHCl_3}$ + $ \ce{KOH \longrightarrow R \minus{} NC}$

2. $ \ce{R \minus{} NC}$ +$ \ce{\Delta}$ $ \longrightarrow$ $ \ce{R \minus{} CN}$

3. $ \ce{R \minus{} CN}$ + $ \ce{SnCl_2}$ /$ \ce{HCl \longrightarrow R \minus{} CHO}$

That method of reduction is almost never used nowadays. A more modern one is to simply use LiAlH(O-tBu)3, for example. Also, isonitriles are seldom made that way. But the idea is ok.


http://www.artofproblemsolving.com/NewClassroom/ClientDoc.php#latex

Note that we do in fact need dollar signs around math and it is *not* different from the forum procedure (except the semicolon).

I have just tried it without the dollar signs and it works properly. Can you provide the formula you are using which is not working?

Ok, nvm. I was confused by
[quote]This differs from the forum in that you should not use dollar signs around the LaTeX code.[/quote]

but then having to use \text{}. Apologies :).


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let n>4, and N a normal subgroup of An (a group of even permutations of Sn), and N $ \neq$ {(1)}.
Prove that there is at least one three-cycle in N.

$ A_n$ is simple for $ n > 4$; so $ N$ is necessarily either $ 0$ or $ A_n$.

Though I suppose you're not supposed to be using that fact...?


help me :

http://www.mathlinks.ro/Forum/topic-28678.html

i can't understand why we can assume a+b+c=1 in this problem,

sory if i post in the wrong sub forum :D  ;)

Hi ,
    This is a homogeneous ineq !
  Bat dang thuc nay thuan nhat ( co nghia la f(a,b,c) = f(ka,kb,kc) )nen ta co the gan cho a+b+c 1 gia tri nao do (khac 0 ) (them gia thiet de bai)


Ooops. We will copy the images over. Sorry about that :D

[quote="azjps"]It's probably related to the server move, but all images on the wiki (with the exception of those uploaded extremely recently) seem to have disappeared.  :(  Thumbnails have the error message "Error creating thumbnail: sh: /usr/bin/convert: No such file or directory".

[I wasn't sure which forum it would be better to post this][/quote]

I copied over the image, but there were two that were posted after the server switch but before I moved the images directory that were lost.  Azjps, I think you posted both; can you repost (glance through the recent entries, you'll see 'em).


abstract algebra
superior algebra
superior algebra solved

Let $G$ be a non-abelian group of order $p^3$ for some prime $p$. Prove that $Z(G)=G'$.

I can prove that $Z(G)$ has order $p$, but I guess that is not sufficient?

Since $G/Z(G)$ has order $p^2$, it's abelian, which means that $G'\le Z(G)\ (*)$. Now, $Z(G)$ has order $p$ and $G'$ cannot be trivial, so $(*)$ implies the desired result.


[quote="tranthanhnam"]cho minh hoi muon xem de thi cac nuoc cac nam cach day nhieu nam thi vao cho nao
minh nghe noi trong mathlinks.ro co de chon doi tuyen vong 2 tu nam 64 minh khong biet cho nao? giup minh nhe
  ai co trang web toan gi hay noi cho minh , minh cam on nhieu .
                                                          de thi cac nuoc ban nao co the tot bung cung cap cho minh khong

      WTO_tientethegioi@yahoo.com
               lien he voi minh nhe[/quote]
Vao cho High shool contest


[quote="rebirth"]what a mean thing to say because I love jesus and god and dont like saying people die in real life when they ddint.[/quote]

Like Jesus?

oh wait, i interpreted your statement backwards

[quote="rebirth"]what a mean thing to say because I love jesus and god and dont like saying people die in real life when they ddint.[/quote]

He never mentioned death. He only mentioned the accident.

[quote="Jesus sucks"]Did you hear the one about how Arnav's dad got ran over by a truck?  Now, Delong has to take his place.[/quote]

 Mystic Terminator hater, you should be ashamed of your self.


calculus
integration
trigonometry
calculus computations

Using the substitution $ x =-t$,

$ \int_{-\pi}^{\pi}\frac{\cos nx}{1+2^{x}}\; dx = \int_{-\pi}^{\pi}\frac{2^{t}\cos nt}{1+2^{t}}\; dt$

So $ \int_{-\pi}^{\pi}\frac{\cos nx}{1+2^{x}}\; dx = 0$ for all nonzero integer $ n$.

Now let ${ I_{n}= \int_{-\pi}^{\pi}\frac{\sin nx}{(1+2^{x})\sin x}}\; dx$. Then $ I_{n+2}-I_{n}= \int_{-\pi}^{\pi}\frac{2\cos(n+1)x}{1+2^{x}}\; dx = 0$ for all $ n \geq 0$, with initial values $ I_{0}= 0$ and $ I_{1}= \pi$.


Resistor network:
If the resistors are in series, current is the same and voltages add. Resistance values add.
If the resistors are in parallel, currents add and voltage is the same. Reciprocal values of resistances add.

Capacitor network:
If the capacitors are in series, charge is the same and voltages add. Reciprocal values of capacitances add.
If the capacitors are in parallel, charges add and voltage is the same.  Capacitance values add.

(1) C = 4.4 uF, V = 9.8 V.

Initial charge $Q_i = VC_i = V \frac{C \cdot C}{C + C} = \frac 1 2 VC$

Final charge $Q_f = VC_f = V \frac{2C \cdot C}{2C + C} = \frac 2 3 VC$

Added charge $\triangle Q = Q_f - Q_i = \frac 1 6 VC$

(2) V = 88 V. $C_1, C_2$ are in parallel, $C_{12} = C_1 + C_2$

$C_3, C_{12}$ are in series, the total capacitance is $C = \frac{C_{12}C_3}{C_{12} + C_3} = \frac{(C_1 + C_2)C_3}{C_1 + C_2 + C_3}$

Total charge is $Q = CV$. Charges on $C_{12}, C_3$ are the same, $Q_{12} = Q_3 = Q$. Voltages add, $V_{12} + V_3 = V$.

$V_3 = \frac{Q_3}{C_3} = \frac{Q}{C_3} = V \frac{C}{C_3} = V \frac{C_1 + C_2}{C_1 + C_2 + C_3}$

$V_{12} = \frac{Q_{12}}{C_{12}} = \frac{Q}{C_{12}} = V \frac{C_3}{C_1 + C_2 + C_3}$

Charges on $C_1, C_2$ add, $Q_1 + Q_2 = Q_{12} = Q$. Voltages are the same, $V_1 = V_2 = V_{12}$.

$Q_1 = C_1 V_1 = C_1 V_{12} = V \frac{C_1 C_3}{C_1 + C_2 + C_3}$

$Q_2 = C_2 V_2 = C_2 V_{12} = V \frac{C_2 C_3}{C_1 + C_2 + C_3}$

(3) C = 9.88 uF, V = 10 V. Label $C_3, C_4, C_5$ the remaining capacitors. $C_2, C_3$ are in series, $C_{23} = \frac{C^2}{2C} = \frac 1 2 C$. $C_{23}, C_4$ are parallel, $C_{234} = C_{23} + C_4 = \frac 3 2 C$. $C_{234}, C_5$ are in series, $C_{2345} = \frac{\frac 3 2 C^2}{\frac 3 2 C + C} = \frac 3 5 C$. $C_1, C_{2345}$ are parallel, $C_{12345} = C_2 + C_{2345} = \frac 8 5 C$.

Total charge is $Q = C_{12345}V = \frac 8 5 CV$. $C_1, C_{2345}$ are parallel, charges add and voltages are the same, $Q_1 + Q_{2345} = Q, V_1 = V_{2345} = V$.

$Q_1 = C_1 V = CV$

$Q_{2345} = C_{2345} V = \frac 3 5 CV$

$C_{234}, C_5$ are in series, charges are the same and voltages add, $Q_{234} = Q_5 = Q_{2345}, V_{234} + V_5 = V$.

$V_{234} = \frac{Q_{234}}{C_{234}} = \frac{Q_{2345}}{C_{234}} = \frac{\frac 3 5 CV}{\frac 3 2 C} = \frac 2 5 V$

$V_5 = \frac{Q_5}{C_5} = \frac{Q_{2345}}{C_5} = \frac{\frac 3 5 CV}{C} = \frac 3 5 V$

$C_{23}, C_4$ are parallel, charges add and voltages are the same, $Q_{23} + Q_4 = Q_{234}, V_{23} = V_4 = V_{234}$.

$Q_{23} = C_{23}V_{23} = C_{23}V_{234} = \frac 1 2 C \cdot \frac 2 5 V = \frac 1 5 CV$

$Q_4 = C_4V_4 = C_4V_{234} = C \cdot \frac 2 5 V = \frac 2 5 CV$

$C_2, C_3$ are in series, charges are the same and voltages add, $Q_2 = Q_3 = Q_{23}, V_2 + V_3 = V_{23}$.

$V_2 = \frac{Q_2}{C_2} = \frac{Q_{23}}{C_2} = \frac{\frac 1 5 CV}{C} = \frac 1 5 V$

$V_3 = \frac{Q_3}{C_3} = \frac{Q_{23}}{C_3} = \frac{\frac 1 5 CV}{C} = \frac 1 5 V$

The answer is $Q_1 = CV$ and $Q_2 = \frac 1 5 CV$.

Thank you so much for your help. I think I understand it now. However, I was still unable to answer the second part of the first question which read:

(b) what is the increase in the sum of the charges: the charge on the positive plate of one capacitor plus the charge on the positive plate of the other capacitor?

What am i doing wrong? Thanks

Initially, the positive charge $+Q_i = \frac 1 2 CV$ is on the positive plate of the capacitor $C_1$ and the negative charge $-Q_i = -\frac 1 2 CV$ of equal magnitude is on its negative plate. The same is true for the capacitor $C_2$, because the 2 capacitors are in series and their charges are equal. The initial sum of charges on the positive plates of the 2 capacitors is $+2Q_i = CV$. Afterwards, the positive charge $+Q_f = \frac 2 3 CV$ is on the positive plate of the capacitor $C_1$, the negative charge $-Q_f = \frac 2 3 CV$ on its negative plate and the same is true for the other capacitor. The final sum of charges on the positive plates of the 2 capacitors is $+2Q_f = \frac 4 3 CV$. The increase in the sum of the positive charges is $2Q_f - 2Q_i = \frac 4 3 CV - CV = \frac 1 3 CV$.

Since [b]yetti[/b] has provided sufficiently detailed solutions of the past problems, I will present only the main steps of the work here:

1) In short: \[ V_3 = V, \qquad q_3 = V_3C_3 = VC_3, \qquad U_3 = \frac{V^2C_3}{2},  \]
\[ q_1 = q_2 = V\frac{C_1C_2}{C_1+C_2}, \qquad V_1 = \frac{q_1}{C_1} = V\frac{C_2}{C_1+C_2}, \qquad V_2 = \frac{q_2}{C_2} = V\frac{C_1}{C_1+C_2},  \]
\[ U_1 = \frac{V_1^2 C_1}{2} = \frac{V^2C_2^2C_1}{2(C_1+C_2)^2}, \qquad U_2 = \frac{V_2^2 C_2}{2} = \frac{V^2C_1^2C_2}{2(C_1+C_2)^2}.  \]

2) When $C_3$ breaks down, the voltage on both capacitors left will be $V$, so that is the answer for b). As for a), the total charge on the system $C_1||C_2$ ($C_1$ parallely connected to $C_2$) will be $q_{12} = V\frac{C_3(C_1+C_2)}{C_1+C_2+C_3}$ which is further equal to $q_3$. The voltage on this system is $V_{12} = V\frac{C_3}{C_1+C_2+C_3} \Rightarrow q_1 = V_1C_1 = V\frac{C_1C_3}{C_1+C_2+C_3}$. After the breakdown the charge on $C_1$ will be $q'_1 = VC_1$, which means that the change of the charge is $\Delta q = q'_1 - q_1 = V\frac{C_1(C_1+C_2)}{C_1+C_2+C_3}$.

Thank you so much. I believe you were slightly incorrect with the algebra in determining U1 and U2 (when you squared V1 and V2, respectively) but I figured it out nonetheless. Thank you very much.

Oh, yes, I forgot to write down the two-s in the final expressions... :oops:

its ok i figured it out anyway..

thank you very much. i really appreciate your help

[quote="jeter123"]Thanks a lot[/quote]

It is not a good idea to erase the initial post of  a thread and especially not the central question.!  :mad:  Try to restore these.
 Others have to search the whole topic to find out what it is all about, so please
dont edit without consideration.


function
number theory
greatest common divisor
combinatorics proposed
combinatorics

There are $ n$ runners waiting at distinct points of a circular race track. At the starting signal each runner takes off. Each person arbitrarily chooses which of the two possible directions to go. Each has the same constant speed and whenever two of them meet they both change direction (instantaneously). This continues indefinitely. Show that there is a moment (after the start) at which each runner is back at his/her starting point.

Given the values of $ n$ and the time $ t$ it takes a runner to run a full lap (unobstructed), is there an upper bound on the amount of time it can take for the above to happen, and if there is, what is it (as a function of $ n$ and $ t$)?

We can think of the situation in a different way: when two runners encounter each other, rather than each changing directions, the two continue as they were, but trade identities. Considering the situation like this, it is clear that after every period of time $ t$, there is exactly one runner in each of the starting locations. However, after time $ t$, the runners in those initial positions have possibly changed identities. Yet, it is clear that the identities of the runners in each starting position cycle the same way every time $ t$. Since this is represented by some finite set of finite cycles, after the GCD of the sizes of these cycles times $ t$, all the runners will be in their initial positions, as desired.

To find an upper bound on the amount of time it can take for all the runners to return to their initial positions, we make one more observation: the order of the runners never changes. If we look around the track, the runners will always appear in the same order. Thus, our set of cycles from before is actually a single cycle containing all $ n$ runners, and it follows that the maximum amount of time it can take for all the runners to return to their initial positions is $ nt$, which can be obtained, for instance, when all the runners are running the same way except for one.


algebra
polynomial
modular arithmetic
search
superior algebra
superior algebra unsolved

Let $ f$ be an irreducible polynomial in $ \mathbb{Z}[x]$. Let $ \mathcal{P}$ be the family of prime ideals in $ \mathbb{Z}[x]$ that contain $ f$. Is $ \#\mathcal{P}$ finite?

Not in general.
Put $ f: \equal{} x^{2} \plus{} 1.$
Then for every $ p\in \mathbb{Z}$ prime st $ p \equiv \minus{} 1\pmod{4} \ f$ is irreducible in $ \mathbb{Z}_{p}$, thus ideal $ I_{p}: \equal{} (f, p) \subset \mathbb{Z}[x]$ is prime
(it's even maximal, see: http://www.mathlinks.ro/viewtopic.php?search_id=1643396135&t=72993).
Since we have infinitely many primes $ p\in \mathbb{Z}$ satisfying $ p\equiv \minus{} 1\pmod{4},$ and $ I_{p_{1}}\neq I_{p_{2}}$ for such primes $ p_{1} \neq p_{2},$ then the set $ \mathcal{P}$ is infinite.


calculus
algebra
partial fractions
real analysis
real analysis solved


What is $ |x^{3}|\div{x^{2}}$? Is the answer $ x$ or $ |x|$?

Then what is $ |x^{3}|^{3}$?

How about $ 5|x|y^{2}\minus{}4xy^{2}$? Is it $ |x|y^{2}$, $ xy^{2}$, or something completely different?

For the first one, we have \begin{align*}
|x^3|\div x^2 &= |x^3|\div |x^2| \\
&= |x^3\div x^2| \\
&= |x| \\
\end{align*}
The second one is just \begin{align*}
|x^3|^3 &= |x^3|\cdot |x^3|\cdot |x^3| \\
&= |x^3\cdot x^3\cdot x^3| \\
&= |x^9| \\
\end{align*}
For that last one, if $ x\geq 0$, then we just get $ 5xy^2-4xy^2=xy^2=|x|y^2$.  However, if $ x\leq 0$, we end up with $ -9xy^2$

Why $ \minus{} 9xy^{2}$?

And multiplying absolute value quantities is like multiplying radicals with similar indexes?

$ |x|*|y| \equal{} |xy|$?

But:

$ |x^{3}|*|x^{2}| \equal{} x^{6}$?

Since the power in the product is even?

okay, think about it. how can $ |a^2|\cdot|a^3|$ equal $ |a^6|$? suppose they are all positive:

$ a\cdot{a}$ is $ a^2$ and $ a\cdot{a\cdot{a}}$ is $ a^3$. therefore, $ |a^2|\cdot|a^3|\equal{}|a^5|$

i hope that made sense?

I made a typo. I meant:

$ |x^{3}|*|x^{3}|\equal{}x^{6}$

Since even powers are always positive?


I need the old mathcounts tests from about the first one to the the year 2000. does anybody know where I can get them?

look on the stickie " MATHCOUNTS Resources Here"


algebra
polynomial
Gauss
algebra proposed

Greetings. It's been quite a long time since I last entered the forum,
so I decided to share the following original irreducibility criterion, which
has a nice (btw classical) corollary.

Theorem. (Original)  Let $p$ be a prime and $l$ be an integer not divisible by
$p$. Assume that a polynomial
\[
f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0 \, \in \mathbb{Z} [x]
\]
has the following properties.

(i) $p \, \nmid \, a_0$ and $p \, \nmid \, a_{n-1}$;
(ii) The leading coefficient $a_n$ is of the form $lp^k$.
(iii) All complex roots of $f(x)$ lie in the annulus $|z| <
  l^{-1}$.

Then, $f$ is irreducible over $\mathbb{Z}$.

With $l=1$, $n=p^k$ and $f(x) = (1 + x + \cdots + x^n)' = 1 + 2x + 3x^2 + \cdots + nx^{n-1}$, we obtain the following cute corollary.

Corollary. If $n$ is a prime power, then the polynomial
\[
1 + 2x + 3x^2 + \cdots + nx^{n-1}
\]
is irreducible.

It has been conjectured that this polynomial is irreducible for all values of $n$, and, to the best of my knowledge, this conjecture is still open. (Although it has been shown that the above polynomial is irreducuble for "almost all" $n$.)

I would welcome other nice corollaries of my criterion.

Have a great day!

--Vesselin

Welcome back, Vess, we missed you. Please, do not post the proof and let us the pleasure to try. Of course, I will fail, but let us think a little bit. I also knew that this is an open question and that is solved for almost all n, but I forgot where I wrote this.

Here is a little ammendment. In condition (iii), it actually suffices to have $l^{-1}$ instead of $l^{1-n}$, sorry for the misjudgment. This is actually much better, and now condition (iii) does not depend on $n$.

To answer iura's question, here is an outline. Assume that $f=gh$ with $g,h \in \mathbb{Z}[x]$ --- non-constant.

1. Using (i) (only), show that either $g$ or $h$ has leading coefficient which is not divisible by $p$.
2. Assume that the leading coefficient of $g$ is not divisible by $p$. By (ii), this coefficient is at most $l$.
3. Since, by (iii), all complex roots of $f$ are in the annulus $|z|<l^{-1}$, then so are all roots of $g$. Now, if $z_1,\ldots,z_k$ are those roots ($1 \leq k \leq n-1$), we get to the contradiction
\[
1 \leq |g(0)| \leq l |z_1 \cdots z_k| < l \cdot l^{-k} \leq l \cdot l^{-1} = 1,
\]
which is clearly a contradiction.

So, this is a pretty simple criterion. However, look at the beautiful corollary it implies. Also, this may easily be generalized in many directions via the theory of Newton polygons.

[Note. The condition $p \nmid a_0$ is actually redundant; we just need that $a_0 \neq 0$ and $p \nmid a_{n-1}$.]

Hi harazi!

Sorry, I posted the outline of this proof before seeing your message. However, this is a very easy and simple criterion, as the proof shows. [I initially proved it with Newton polygons, but then laughed at myself when I saw how trivial Part 1 is :) .]

It's more interesting to think of other good corollaries, as well as appropriate generalizations. Here is, for instance, one other corollary:

Corollary. If $n$ is a prime power and $k$ is arbitrary, then
\[
\sum_{1 \leq j \leq n} j^k x^{j-1}
\]
is irreducible.

[One just has to check that these polynomials have all their zeros inside the unit circle, and this is easily done, for instance, by the Gauss-Lucas theorem.]

So, I'm waiting for other corollaries, and for generalizations of the criterion.  :)

As for the conjecture that $1 + 2x + \cdots + nx^{n-1}$ is always irreducible, this has been also proven when $n=2p-1$ with $p$-prime. If $n=p-1$ the irreducibility is trivial, since the polynomial is then Eisensteinian. Here is a reference:

http://www.math.psu.edu/borisov/Documents/papers/noncycl.ps

This also has a reference to another most interesting paper in Acta Arithmatica (by Borisov) which is examines irreducibility of a more general class of polynomials.

--Vesselin

Dear Vess, please explain to me where Im idiot. Let us suppose that $ f(x)=( b_m\cdot x^m++b_1\cdot x+b_0 )\cdot (c_{n-m}\cdot x^{n-m}++c_1\cdot x+c_0) $ . We identify the leading coefficient and find that $ b_m\cdot c_{n-m}=l\cdot p^k $. Now, the numbers $ b_m, c_{n-m} $ cannot be both multiples of $ p$ since it would follows that $ p|a_{n-1} $. Thus, we may assume that $ p^k| c_{n-m} $. Let $ z_1,,z_m $ be the roots of the first polynomial in the decomposition. Then $|\frac{b_0}{b_m}|=|z_1\cdot\cdot z_m |<l^{m(1-n)} $. But since $ p^k|c_{n-m} $ it follows that $ b_m|l $ and so $|b_m|\leq l $. Thus, $|b_0|<l^{1+m(1-n)}<1 $ and so $ b_0=0 $ false. Where am I wrong?

Well, this is the first time a post a question after the answer was already given. And how proud I was after solving the problem...  :( . Anyway, a cool problem and especially some cool consequences. Congratulations!

A trivial problem with cool consequences, actually.  :D

Notice the ammendment: $l^{-1}$ in place of $l^{1-n}$ (how stupid of me to make such a mistake.  :cool: ).

Initially, I used the Newton polygon of $f$ with respect to $p$ to show such a thing as obvious as that one of $g$ and $h$ has a leading coefficient which is prime to $p$.  :maybe: !!!!

However, Newton polygons will surely by a useful tool in generalizing this "criterion."

One more thing. The paper in Acta Arithmetica I mentioned can be found at

http://www.math.psu.edu/borisov/Documents/papers/abc.ps


A vibrating string vibrates at 100Hz in oxygen.What is the frequency of its vibration in Hydrogen?[assume same temperature and pressure]

Equating the velocities of sound gives 400Hz...But what pokes me is that the frequency is the property of the source.Will it change with ambient condition?
In other words should not the answer be still 100 Hz?

Yes. The string's vibration has nothing to do with the medium; it would be the same in a vacuum. The hydrogen only affects the sound transmitted away from the string, which would have a different speed and wavelength than in oxygen.

[Fixed a mistake]

the speed of the sound would be different,I suppose.

Take an orchestra and move them up to a high-altitude area where the air pressure is less. The string players may worry about the ill effects of low humidity on their instruments, but they don't have to re-tune. The wind players do have to re-tune, since the pitch produced by a wind instrument of a certain size and shape does depend on the speed of sound.

(Note: using 440 Hz for an A, 100Hz would be somewhere between G and G#, as played on the second-lowest string of a cello. A human could sing that pitch, but anyone who could do so comfortably or with any power would be called a bass.)


what does this have to do with AIME????

F*ck you for spamming. You're a mod. And I won't expect this from a dirty person like you.

Otherwise if you really wanna talk about AIME, I got an official 6 on the AIME I and an unofficial 7 on the AIME II.

There are many poll choices I want to check...

[quote="gauss1181"]a dirty person like you.[/quote]

indeed, piano, you need to practice more cleanliness.

LOLOLOLOLOLOLOLOLOLOLOLOL at the poll's choices

I answered Shyam who probably got  a 15. (in 6th grade)

Greetings gauss1181:

While I value your opinion, if you may have noticed, this is the Missouri forum. Subject to different standards than other forums, as a short perusal of the other threads would reveal. The thread title was a hook to a much much more important topic than the AIME, which of course, all the MO and KS people probably know all of each others scores by now, and was never an invitation for those of states not greatly endowed in cornfields or bad roads.

The word "luv" does connote anything wrong, it is cute and adorable. Clearly. Nor does the word "BEASTLY," which clearly Mewto55555 is. The suggestion that math154 luvs Mewto55555 is clearly platonic.

I respectfully decline your offer that suggests some expletive action against my person.

With regards,
pianoforte

P.S. no hard feelings?

EDIT: I MISSED MY 2000th POST. GRAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAR

Correction: Mewto55555 is [/b]NOT BEASTLY[b].[/b]

I am not understanding how few people chose option 1.

PS gauss you cussed out a mod in the MO forum. That is against the rules. You can only do it to admins here.

[quote="justinahmann"]We need to have a nice little talk, Mewto... :mad:  :mad:[/quote]

I rickrolled you. Get over it.

Fine, but was there anything bad in the rickroll?

Me is getting too worried.

Anyone else here wanted to check 2,3,4,5?

ah, this poll and its replies reaally made my day

priceless poll choices and its hilarious replies

I picked "Mewto55555 luvs dragonlaird" cuzz it just sounded wrong. A brother likes his own sister..........


Source: de.rec.denksport, [url=http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&frame=right&th=1ddcd88ff051465f&seekm=a3a3974c.0403212217.4bedc2de%40posting.google.com]at Google[/url]

Find a quadrilateral inscribed into a circle of radius 9, such that all of
its side and diagonal lengths are different positive integers.

I start guessing it is impossible...

hallo lukas, schn da du auch den weg in dieses nette forum gefunden hast...

[quote="luqash"]I start guessing it is impossible...[/quote]

At [url=http://groups.google.com/groups?dq=&hl=de&lr=&ie=UTF-8&oe=UTF-8&threadm=405EBFCE.3090406%40t-online.de&prev=/groups%3Fhl%3Dde%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26q%3Dde.sci.mathematik%26meta%3D]de.sci.mathematik[/url], people were of the same opinion, and Ingmar meant it shouldn't be necessarily *cyclic*, it should just fit into a circle with radius 9. Strange terminology, of course.

Actually, 3 solutions were given, none of them being a cyclic quadrilateral.

  Darij

all those sites are in German :(

anyway, it's clearly impossible, and even if you got no clue about this you could just write out the options :? it's not so long

Not such a good question thus... :(


geometry
similar triangles
Pythagorean Theorem

In quadrilateral $ ABCD$ the point of intersection of the perpendicular bisectors of $ AD$ and $ BC$ lies on $ AB$. Prove that $ AC\equal{}BD$ if and only if $ \angle A\equal{}\angle B$.

Sorry...but don't have a diagram

[hide]
Draw the quadrilateral ABCD. Let the perpendicular bisectors AD and BC meet at E. Drop perpendiculars from D and C on to AB. Let the perpendicular from D be F and the perpendicular from C be G.

Since $ \angle A \equal{} \angle B$, $ \triangle AED\sim \triangle BEC$. Let $ AE \equal{} DE \equal{} a$ and $ BE \equal{} CE \equal{} b$, and $ EG \equal{} x$. Use similar triangles and pythagorean theorem to find out that $ AC \equal{} BD \equal{} \sqrt {a^2 \plus{} 2ax \plus{} b^2}$.

To prove the "only if" part, assume the same values, except let $ EF \equal{} y$.

Since A doesn't equal B now, $ \frac {a}{b}$ doesn't equal $ \frac {y}{x}$, so $ ax$ doesn't equal $ by$.

Using pythagorean theorem again, we find out that:

$ AC \equal{} \sqrt {a^2 \plus{} 2ax \plus{} b^2}$

$ BD \equal{} \sqrt {a^2 \plus{} 2by \plus{} b^2}$, which aren't equal, which is what we wanted.[/hide]

[quote="Carbon57"]Since A doesn't equal B now, $ \frac {a}{b}$ doesn't equal $ \frac {y}{x}$, [b]so $ ax$ doesn't equal $ by$.[/b]
[/quote]
Is there justification?

[quote="Carbon57"]Using pythagorean theorem again, we find out that:

$ AC \equal{} \sqrt {a^2 \plus{} 2ax \plus{} b^2}$

$ BD \equal{} \sqrt {a^2 \plus{} 2by \plus{} b^2}$, which aren't equal, [b]which is what we wanted.[/b][/quote]

Are you so sure you wanted that? This is not the usual manner of proving, and I'm not sure I follow you. It's a lot easier proving the converse (hint, hint).

Edit: It's AA similarity, not A similarity. You might have wanted to mention the right angles somewhere too.

They're not equal because the triangles aren't similar anymore.

I know it's kind of weird...but it uses basically the same method as the "if" part.

I suppose that works but it is still not the best way of solving the problem, but I guess it is correct...

That is

If $ a\ne b$, then $ ac\ne bc$ if $ c\ne 0$
If $ a\ne b$, then $ a\plus{}c\ne b\plus{}c$.
If $ a\ne b$, then $ \sqrt{a}\ne \sqrt{b}$.


algebra
polynomial
limit
algebra unsolved

[quote="manjil"]If a third degree polynomial in $ x$ is always greater than zero for all real values of $ x$ then prove that the co efiicient of the third degree term is zero.[/quote]

No polynomial of odd degree can have a constant sign over $ \mathbb R$ since $ \lim_{x\to\plus{}\infty}P(x)P(\minus{}x)\equal{}\minus{}\infty$

And ... you are not in the good forum.

[quote="pco"][quote="manjil"]If a third degree polynomial in $ x$ is always greater than zero for all real values of $ x$ then prove that the co efiicient of the third degree term is zero.[/quote]

No polynomial of odd degree can have a constant sign over $ \mathbb R$ since $ \lim_{x\to \plus{} \infty}P(x)P( \minus{} x) \equal{} \minus{} \infty$

And ... you are not in the good forum.[/quote]

Thats what I thought, but this was asked in a  test, can you maybe revise the question to something correct?


advanced fields
advanced fields unsolved

Prove that every closed (open) set in $ \mathbb{R}^{n}$ is type $ G_{\delta}$ ($ F_{\sigma}$).

If $ F$ is closed, then the sets $ G_{n}\equal{}\left\{ x\in\mathbb R^{n}:\exists y\in F\text{ such that }|x\minus{}y| <\frac{1}{n}\right\}$ should work.
For the other part, just take complements.

Thank you,I had the same idea, but how do I actually prove that that intersection is a closed. I am probably missing something obvious.

You were given that $ F$ is closed. What you have to prove is that the intersection is $ F.$


complex analysis
function
complex analysis unsolved

Let U1 and U2 be plane sets which are open and connected. ("regions") If f(U1) is in U2 (is in U2)
and if we set h(z) = f(g(z)) then we know h is holomorphic if f and g are. What if I tell you
h is holomorphic and g is also. Does f have to be? What if h is holomorphic and f is also. Does g have to be?

So, the question is like this: if $ h\equal{}f\circ g$ is holomorphic and $ h$ and $ f$ are holomorphic, has $ g$ to be holomorphic? The formal answer is "No" because you can take $ f$ a constant function and then $ g$ can be just anything. If we exclude this trivial case, we may consider something like $ f(z)\equal{}z^n$. If $ g^n$ is holomorphic, must $ g$ be holomorphic? Apparently again "No" without further assumptions (we can use some ugly pointwise $ \sqrt[n] z$ for $ g$). Suppose now that we require $ g$ to be at least continuous. Then we can say that $ g$ is holomorphic except when $ g(z)$ belongs to some discrete set $ D$ (the set of zeroes of $ f'$). This, indeed, does imply that $ g$ is holomorphic everywhere. Can anybody see why? ;).


[quote="Mathias_DK"][quote="Stephen"]Find all polynomial $ f$ that all coefficients are all intergers which satisfies the term below.

There are infinitly many natural numbers $ a, b$ that satisfies $ gcd(a,b) \equal{} 1$ and $ a \plus{} b|f(a) \plus{} f(b)$[/quote]
It is obvious that $ a \plus{} b \mid a^n \plus{} b^n$ when $ n$ is odd. So all odd polynomials satisifies the condition.
Let $ f(x) \equal{} x r(x^2) \plus{} q(x^2)$ (where $ xr(x^2)$ is the odd part of $ f$ and $ q(x^2)$ is even part of $ f$)
$ a \plus{} b \mid f(a) \plus{} f(b) \equal{} (ar(a^2) \plus{} br(b^2)) \plus{} q(a^2) \plus{} q(b^2) \iff$
$ a \plus{} b \mid q(a^2) \plus{} q(b^2)$
$ q(b^2) \equiv q(( \minus{} a)^2) \equal{} q(a^2) \bmod a \plus{} b$
So $ a \plus{} b \mid 2q(a^2),2q(b^2)$
Assume that $ q(x) \equal{} c \cdot x^m, c \neq 0$ for some $ m$. Then
$ a \plus{} b \mid 2c \cdot a^{2m} \iff a \plus{} b \mid 2c$. But there are not infinitly many pairs $ (a,b)$ such that $ a \plus{} b \le 2c$ so this is impossible.
Hence we rule out $ q(x) \equal{} c \cdot x^m, c \neq 0$.
Now let $ q(x) \equal{} x^t \cdot r(x)$ s.t. $ r(0) \neq 0$.
Now $ a \plus{} b \mid 2q(a^2) \iff a \plus{} b \mid 2a^2t r(a^2) \iff a \plus{} b \mid 2r(a^2)$. Now we can find infinitly many pairs by choosing $ b \equal{} |2r(a^2)| \minus{} a$. Fore then $ |2r(a^2)| \minus{} a \equiv |2r(0)| \bmod a$. And hence we choose $ a \equal{} k|2r(0)| \plus{} 1$ for some $ k$. Then $ \gcd(a,b) \equal{} 1$. Since $ r(x)$ isn't constant, because then $ q(x)$ would have been on the form $ c \cdot x^m$, we see that $ |2r(a^2)|$ grows faster than $ a$, and hence $ |2r(a^2)| \minus{} a$ must be positive for all $ a > n$ for some $ n$.
So we ruled out $ f(x) \equal{} c \cdot x^{2m} \plus{} r(x), c \neq 0$ where $ r(x)$ is an odd polynomial, but every polynomial which can not be written on that form is a solution.[/quote]

Thank you so much! :lol:


I've gone through all of AOPS vol. 1 twice now. Looking back, I can solve all questions requiring a simple answer consisting of a number or a few variables.

However, I have skipped quite a few proofs, even the fundamental ones in the exercises throughout the chapter (I've gone through some of the thought processes needed to prove a statement, but did not formulate anything onto paper).

Now I am going back to the proofs found in the problem section
at the end of each chapter, beginning with 11 (triangles). I think I can remember the proofs that were part of the exercises, and I was wondering if most of their results are considered well-known?

For instance, "median to opposite side of a triangle is half that side, then the triangle is right" or "sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagnols" are examples. Are some of these considered well-known theorems? In other words, when working on the proofs from selected contests such as M&IQ or AHSME (these were MC questions turned into proofs by authors?), can I state most of the results from the exercises found throughout the first part of the chapter?

[quote="PenguinIntegral"]Yes, though you should know how to derive them in case you run into something you can't cite. (Like in Aops 2).

But you should REALLY REALLY REALLY know the proofs by heart. People won't always be there to tell you what is trivial and what isn't, and you should err on the side of caution.[/quote]
I agree, and once you do the proofs, you don't forget the stuff you learned. I had to learn the hard way. :blush: (not the whole book, only one chapter)
btw, nice quote snipez, Harlan Ellison should have meet Einstein. :D


What is "inner" point ? Can someone explain it in Russian, please?
http://www.mathlinks.ro/Forum/viewtopic.php?t=60810

Inner point of a figure = внутренняя точка фигуры.

Если вопрос о строгом определении этого понятия - вот определение для многоугольников:

Несамопересекающийся многоугольник разделяет плоскость на две части (такие, что любой отрезок, соединяющий точку в одной части с точкой в другой части обязательно пересекает некоторую сторону многоугольника). Среди этих двух частей одна - ограниченная, другая - неограниченная. Ограниченная часть называется [i]внутренностью[/i] многоугольника. Внутренняя точка многоугольника - это точка, лежащая во внутренности многоугольника.

[quote="Mamat"] http://www.mathlinks.ro/Forum/viewtopic.php?t=60810 [/quote]

А тут речь идёт просто о внутренней точке отрезка...

  Darij


induction
superior algebra
superior algebra unsolved

I prove the theorem with induction.
It is obvious to see when $n=1$ is ture.
And if we have $a^{-(n-1)}ba^{n-1}=b^{i^{n-1}}$
then we have:$a^{-n}ba^{n}=a^{-1}b^{i^{n-1}}a$
but it is easily to know:
$a^{-1}b^{i^{n-1}}a=\underbrace{b^i b^i \cdots b^i}_{i^{n-1} times}=b^{i^n}$

For your problem:
I don't what do you mean by "stated",it is mean $i \ge 0$
if it is I also think it is unnecessary.
PS:It is my first post in this section. :)

[quote="The Captain"]The following problem is very famous and easy to solve. But sth is not clear for me.So I decided to discuss about it.

:Let $a, b$ be elements of a group such that $a^{-1}ba=b^i$. Prove that $a^{-n}ba^n=b^{i^n}$for all $n \in N \ and\ i\ge 0$.
In all my sources the condition on $i$ is stated, whereas in my opinion this is unnecessary. What do you think?[/quote]

I can prove it for $i=-1$.  For $i=-1$, the condition of the problem becomes $a^{-1}ba=b^{-1}$.

We can invert both sides of the original premise to give $a^{-1}b^{-1}a=b$.  We'll refer to this as the derived premise.

We proceed by induction.  The result is trivially true for $n=1$.  We assume the result is true for $n=k$ and consider the $n=k+1$ case.

We consider two subcases.  If $k$ is even then:

$a^{-k-1}ba^{k+1}=a^{-1}b^{i^{k}}a=a^{-1}ba=b^{-1}$

using the original premise.

And if $k$ is odd then

$a^{-k-1}ba^{k+1}=a^{-1}b^{i^{k}}a=a^{-1}b^{-1}a=b$

using the derived premise.

So in either case the result is true for $n=k+1$ so by mathematical induction it is true for all positive integers $n$.

I suspect a similar argument works for $i<-1$ but I haven't worked it out yet.


How is it possible that the infinite series:

1/(x+3) + 1/[(x+3)(x+4)] + 2/[(x+3)(x+4)(x+5)] + 6/[(x+3)(x+4)(x+5)(x+6) + ....

is simply 1/(x+2)?

However, I am applying the methods in numerical analysis.

It should be  the summation (from n = 0 to +oo)  n! / [(x+3)(x+4)...(x+3 + n)]

From here, I am clueless.
[/list]

We are looking at $f(y)=\sum\limits_{n=0}^{\infty}\dfrac{n!}{y\cdot(y+1)\cdots(y+n)}$
(I substituted $y=x+3$). By comparison to $\sum\dfrac1{n^y}$, this series diveges for $x\le1$ and converges for $y>1$.

We have ${\dfrac{n!}{y\cdot(y+1)\cdots(y+n)}=\dfrac{(n-1)!}{y\cdot(y+1)\cdots(y+n-1)}- \dfrac{(n-1)!}{(y+1)\cdot(y+2)\cdots(y+n)}}$ for $n\ge1$, so $f(y)=\dfrac1y+f(y)-f(y+1)$, and $f(y+1)=\dfrac1y$, at least for $y>1$ where everything converges absolutely and this rearrangement works. This result should hold in general, but we need a slightly different argument to get at $f(y)$ for $1<y\le2$.

Numerical analysis isn't about methods for finding exact values; this is an entirely different bag of tricks.


algebra
polynomial
linear algebra
matrix
linear algebra theorems

I was reading the IMC solutions and there was something new for me. Namely we have $A$ be an $n\times n$-matrix such that $3A^{3}=A^{2}+A+I$. The minimal polynomial of $A$ is a divisor of $3x^{3}-x^{2}-x-1$. Now it was said that the polynomial has three different roots so $A$ is diagonizable. Is it in general true that if the minimal polynomial of $A$ is of degree $n$ and $A$ has $n$ different roots, then $A$ is diagonizable? And if $A$ is diagonizable, does it tell something about its minimal polynomial?

The theorem: A matrix $A\in M_{n}(F)$ is diagonalizable over a $F$ if and only if its minimal polynomial factors into a product of distinct (monic) linear factors with coefficients in $F$.


I will assume that the fly was initially at rest
thus Energy $ E$ released after time $ \tau$ is $ P \tau$
but this corresponds to certain number of photon of wavelength $ \lambda$ say $ x$ number
thus $ P \tau \equal{} \frac {xhc}{\lambda}$
thus $ x \equal{} \frac {P \tau \lambda}{hc}$
conserving momentum we have the final momentum of fly be $ mv$
so $ mv \equal{} \frac {xh}{\lambda} \equal{} \frac {P \tau}{c}$
so $ \boxed{mv\equal{}\frac{P \tau}{c}}$

[quote="Cyber-Shreyas"]Perfect. Full score. Yours is the simplest soln.

BTW, did you guys notice that it is totally independent of lambda![/quote]
is there any other solution except this  :maybe:


function
floor function
logarithms
algebra proposed
algebra

Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x \plus{} y \plus{} xy) \equal{} f(x) \plus{} f(y) \plus{} f(xy)\]
happens for all $x,y \in \mathbb R$.

[quote="imedeen"]Determine all function $ f$ defined and continuous on $ R$ such that
$ f(x \plus{} y \plus{} xy) \equal{} f(x) \plus{} f(y) \plus{} f(xy)$ , $ \forall x,y \in R$[/quote]

I think some shorter method might exist :

Let $ P(x,y)$ be the assertion $ f(x\plus{}y\plus{}xy)\equal{}f(x)\plus{}f(y)\plus{}f(xy)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,\minus{}1)$ $ \implies$ $ f(\minus{}x)\equal{}\minus{}f(x)$
$ P(1,1)$ $ \implies$ $ f(3)\equal{}3f(1)$
$ P(x,1)$ $ \implies$ $ f(2x\plus{}1)\equal{}2f(x)\plus{}f(1)$

$ P(x,3)$ $ \implies$ $ f(4x\plus{}3)\equal{}f(x)\plus{}f(3)\plus{}f(3x)\equal{}f(3x)\plus{}3f(1)\plus{}f(x)$
$ P(2x\plus{}1,1)$ $ \implies$ $ f(4x\plus{}3)\equal{}2f(2x\plus{}1)\plus{}f(1)$ $ \equal{}4f(x)\plus{}3f(1)$
Comparing these two lines, we get $ f(3x)\equal{}3f(x)$

$ P(x,\minus{}3)$ $ \implies$ $ f(\minus{}3\minus{}2x)\equal{}f(x)\plus{}f(\minus{}3)\plus{}f(\minus{}3x)$ and so $ f(2x\plus{}3)\equal{}2f(x)\plus{}3f(1)$ $ \equal{}(2f(x)\plus{}f(1))\plus{}2f(1)$ $ \equal{}f(2x\plus{}1)\plus{}2f(1)$

So $ f(x\plus{}2)\equal{}f(x)\plus{}2f(1)$

$ P(1,\minus{}2)$ $ \implies$ $ f(\minus{}3)\equal{}f(1)\plus{}f(\minus{}2)\plus{}f(\minus{}2)$ and so $ f(2)\equal{}2f(1)$
$ P(x,\minus{}2)$ $ \implies$ $ f(\minus{}2\minus{}x)\equal{}f(x)\plus{}f(\minus{}2)\plus{}f(\minus{}2x)$ and so $ f(x\plus{}2)\equal{}f(2x)\minus{}f(x)\plus{}2f(1)$ and, since $ f(x\plus{}2)\equal{}f(x)\plus{}2f(1)$, we get : $ f(2x)\equal{}2f(x)$

So $ f(x\plus{}1)\equal{}f(x)\plus{}f(1)$ and $ f(x\plus{}n)\equal{}f(x)\plus{}nf(1)$ and $ f(n)\equal{}nf(1)$

Then $ P(x,n)$ $ \implies$ $ f((n\plus{}1)x)\equal{}f(x)\plus{}f(nx)$ and so $ f(nx)\equal{}nf(x)$

This immediately gives $ f(x)\equal{}f(1)x$ $ \forall x\in\mathbb Q$ and continuity implies then $ \boxed{f(x)\equal{}ax}$ $ \forall x$
And it is easy to check back that this indeed is a solution.

[quote="imedeen"]Determine all function $ f$ defined and continuous on $ R$ such that
$ f(x \plus{} y \plus{} xy) \equal{} f(x) \plus{} f(y) \plus{} f(xy)$ , $ \forall x,y \in R$[/quote]
As in pco's solution let $ P(x,y)$ be the assertion that $ f(x\plus{}y\plus{}xy)\equal{}f(x)\plus{}f(y)\plus{}f(xy)$.
$ P(0,0): f(0)\equal{}0$
$ P(x,\minus{}x): f(x) \equal{} \minus{}f(\minus{}x)$
$ P(1,1): f(3)\equal{}3f(1)$

$ P(1,y): f(1\plus{}2y) \equal{} f(1)\plus{}2f(y)$
$ P(3,y): f(3\plus{}4y) \equal{} f(3)\plus{}f(y)\plus{}f(3y) \equal{} 3f(1)\plus{}f(y)\plus{}f(3y)$
$ P(1,2y\plus{}1): f(3\plus{}4y) \equal{} f(1) \plus{} 2f(2y\plus{}1)$ $ \equal{} f(1) \plus{} 2(f(1)\plus{}2f(y)) \equal{} 3f(1)\plus{}4f(y)$ and hence $ f(3y)\equal{}3f(y)$

$ P(3,1): f(7) \equal{} f(3)\plus{}f(1)\plus{}f(3) \equal{} 7f(1)$
$ P(7,y): f(7\plus{}8y) \equal{} f(7) \plus{} f(y) \plus{} f(7y)$

$ P(1,4y\plus{}3): f(7\plus{}8y) \equal{} f(1) \plus{} 2f(4y\plus{}3)$ $ \equal{} f(1) \plus{} 2(3f(1)\plus{}4f(y)) \equal{} 7f(1)\plus{}8f(y)$ and hence $ f(7y) \equal{} 7f(y)$.

Now we have $ f(0)\equal{}0, f(x)\equal{}\minus{}f(\minus{}x), f(3x)\equal{}3f(x), f(7x) \equal{} 7f(x)$, and this is enough to prove that $ f(x) \equal{} ax$ for some $ a \in \mathbb{R}$. Let $ f(1)\equal{}a$. Since $ f(0)\equal{}0$ and the rest of the function is uniquely defined by $ f(x), x \in [1;3)$ it is enough to prove that $ f(x) \equal{} ax \forall x \in [1;3)$.

From $ f(3x)\equal{}3f(x), f(7x) \equal{} 7f(x)$ we easily get $ f(3^nx)\equal{}3^nf(x), n \in \mathbb{Z}$ and $ f(7^nx) \equal{} 7^nf(x), n \in \mathbb{Z}$.

From $ f(3^nx)\equal{}3^nf(x)$ we get: $ f(x) \equal{} 3^{\left \lfloor \log_3(x) \right \rfloor} f( x 3^{\minus{}\left \lfloor \log_3(x) \right \rfloor} )$.

Since $ f(7^n) \equal{} 7^n f(1) \equal{} 7^n a$ we have $ 7^n a \equal{} f(7^n) \equal{} 3^{\left \lfloor \log_3(7^n) \right \rfloor} f( 7^n \cdot 3^{\minus{}\left \lfloor \log_3(7^n) \right \rfloor} )$.

Let $ g(n) \equal{} \frac{7^n}{3^{\left \lfloor \log_3(7^n) \right \rfloor}}$. Then $ f(g(n)) \equal{} ag(n)$. Let $ h(n) \equal{} \log_3(g(n)) \iff$

$ h(n) \equal{} \log_3(7) \cdot n \minus{}\left \lfloor \log_3(7) \cdot n \right \rfloor$. Let $ \alpha \equal{} \log_3(7)$ which is clearly irrational, then $ h(n) \equal{} \alpha n \minus{} \left \lfloor \alpha n \right \rfloor \equal{} \{ \alpha n \}$, where $ \{x\} \equal{} x \minus{} \lfloor x \rfloor$ denotes the fractional part of $ x$.

Since $ \alpha$ is irrational we can easily prove that in any interval $ (a;b) \subset (0;1)$ there exists a $ n$ such that $ h(n) \in (a;b)$. So for any $ (a;b) \in (1;3)$ there exists $ n$ such that $ g(n) \in (a;b)$. Because of continouity we find that $ f(x) \equal{} ax, 1 < x < 3$, and hence $ f(x) \equal{} ax \forall x$ using $ f(3x)\equal{}3f(x)$ and $ f(x)\equal{}\minus{}f(\minus{}x)$. So $ f(x) \equal{} ax \forall x \in \mathbb{R}$ is the only solution - and it is easily seen to be a solution.


calculus
integration
trigonometry
algebra
polynomial
function
rational function

pronaam ! :D

can you sum the series:

$ \sum_{j \equal{} 1}^\infty\dfrac{( \minus{} 1)^j}{(j^4 \plus{} 2j^2\plus{}3)}$

Consider integral  $ \int_{C}\left( \frac {( \minus{} 1)^{z}}{z^{4} \plus{} 2z^{2} \plus{} 3}\right) \frac {1}{\tan(\pi z)}dz$ where $ C$ is big circle  :lol: .
and prove that:
$ \frac {1}{2\pi i} \int_{C}\left( \frac {( \minus{} 1)^{z}}{z^{4} \plus{} 2z^{2} \plus{} 3}\right) \frac {1}{\tan(\pi z)}dz \equal{} \frac {1}{\pi}\sum_{k \equal{} \minus{} \infty}^{ \plus{} \infty}\frac {( \minus{} 1)^{k}}{k^{4} \plus{} 2k^{2} \plus{} 3} \plus{} \sum_{j \equal{} 0}^{3}res_{z \equal{} a_{j}}\left(\left( \frac {( \minus{} 1)^{z}}{z^{4} \plus{} 2z^{2} \plus{} 3}\right) \frac {1}{\tan(\pi z)} \right)$
where $ a_{j}$ is the root of equation :
$ z^{4} \plus{} 2z^{2} \plus{} 3 \equal{} 0$

The polynomial j^4+2j^2+3 has roots
at the square roots of -1 + sqrt(2)i and -1 - sqrt(2)i.

Then the rational function 1/(j^4+2j^2+3) can be decomposed into a/(j-root1) + b/(j-root2) + c/(j-root3) + d/(j-root4).

The sum 1/(1-root1) + 1/(3-root1) + ... =0.5 * (1/(0.5-0.5*root1) + 1/(1.5-0.5*root1) + ...) = 0.5*(Digamma(infinity) - Digamma(0.5-0.5*root1))

So the solution can be expressed as -0.5a*(Digamma(1.0-0.5*root1) - Digamma(0.5-0.5*root1))-0.5b*(Digamma(1.0-0.5*root2) - Digamma(0.5-0.5*root2))
-0.5c*(Digamma(1.0-0.5*root3) - Digamma(0.5-0.5*root3)) - 0.5d*(Digamma(1.0-0.5*root4) - Digamma(0.5-0.5*root4))

yes, but How to calculate digamma at point (1.0-0,5root)..?It's same as to calculate res ..   :wink:

$ \frac{1}{j^4\plus{}2j^2\plus{}3}\equal{}\frac{i}{2\sqrt{2}}\, \left(\frac{1}{j^2\plus{}\sqrt{1\plus{}i\sqrt{2}}^{\,\, 2}}\minus{}\frac{1}{j^2\plus{}\sqrt{1\minus{}i\sqrt{2}}^{\,\, 2}}\right)$

whereat $ \sqrt{1\pm i\sqrt{2}}\equal{}\sqrt{\frac{\sqrt{3}\plus{}1}{2}}\pm i \sqrt{\frac{\sqrt{3}\minus{}1}{2}}\equal{}: x\plus{}iy$

Use the formula $ \sum_{k\in\Bbb{Z}} \frac{(\minus{}1)^j}{j^2\plus{}\alpha^2}\equal{}\frac{\pi}{\alpha\, \sinh\alpha\pi}$ to obtain

$ S\equal{}\sum_{j\in\Bbb{Z}} \frac{(\minus{}1)^j}{j^4\plus{}2j^2\plus{}3}\equal{}\frac{i\pi}{2\sqrt{2}} \left(\frac{1}{(x\plus{}iy)\, \sinh(x\plus{}iy)}\minus{}\frac{1}{(x\minus{}iy)\, \sinh(x\minus{}iy)}\right)$

which can be written as $ \pi\sqrt{\frac23}\,\frac{x\, \sin(\pi y) \, \cosh(\pi x)\plus{}y\, \cos(\pi y)\, \sinh(\pi x)}{\cosh(2\pi x)\minus{}\cos(2\pi y)}$

And $ \frac12 \left(S\minus{}\frac13\right)$ is the sum we are looking for.


calculus
integration
logarithms
calculus computations

Substitute $ u \equal{} xy$ then $ I \equal{} \int_0^1 \int_0^y \underbrace{\frac {2 \minus{} 4u}{(9 \minus{} u)(8 \plus{}u)}}_{ \equal{} f(u)} \, \frac {du}{y}\, dy$.

Reverse integration order $ I \equal{} \int_0^1 \underbrace{\int_u^1 \frac {dy}{y}}_{ \minus{} \ln u} \, f(u)\, du$ , which is

$ 2\int_0^1 \left(\frac {\ln u}{9 \minus{} u} \minus{} \frac {\ln u}{8 \plus{} u}\right)\, du$ because of $ f(u) \equal{} \minus{} 2\,\left(\frac {1}{9 \minus{} u} \minus{} \frac {1}{8 \plus{} u}\right)$ .

For the last integral see this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185149]link[/url].


If $ x^{2} \plus{}y^{2} \plus{}z\equal{}15$, $ x\plus{}y\plus{}z^{2} \equal{}27$ and $ xy\plus{}yz\plus{}zx\equal{}7$, then

$\textbf{(A)}\  3\le \left|x\plus{}y\plus{}z\right|\le 4 \\ \textbf{(B)}\  5\le \left|x\plus{}y\plus{}z\right|\le 6 \\ \textbf{(C)}\  7\le \left|x\plus{}y\plus{}z\right|\le 8 \\ \textbf{(D)}\  9\le \left|x\plus{}y\plus{}z\right|\le 10 \\ \textbf{(E)}\ \text{None}$

[hide]Denote $ S\equal{}x\plus{}y\plus{}z$. Multiply the second equation by $ 2$ and add them all up to get $ S^2\plus{}S\equal{}56\iff S\in\{7,\minus{}8\}$, hence the answer is $ C$.[/hide]


linear algebra
matrix
linear algebra unsolved

Suppose $ A\in C^{n\times n}$. Show that $ A$ is nilpotent iff $ tr(A^k)\equal{}0$ for $ k\equal{}1,2,\cdots, n$. :blush:

I claim that the eigenvalues of such a matrix are uniquely determined by the values of $ \text{tr}(A^k), 1 \le k \le n$, and of course the matrices with zero eigenvalues satisfy the condition.  This is a standard exercise in Newton's sums.

$ \mathbb{C}$ can be replaced by any field of characteristic zero; the identities we get from the Newton sum formulas require division by everything up to $ n$, but they don't actually require the eigenvalues to be in the base field.


conics
parabola
geometry
circumcircle
geometry unsolved

i am having diffculty to prove that
the are infinite number of triangles whose vertices lie on $ y^2 \equal{} 4ax$ in which the sides are tangent to $ x^2 \equal{} 4by$
will anyone give me a clue please??

$ \mathcal P_a$ is a parabola with the axis $ y$ and vertex $ O.$ $ \mathcal P_b$ is a parabola with the axis $ x \perp y$ and the same vertex $ O \equiv x \cap y.$ Forget about the parabola $ \mathcal P_b$ for a moment. $ x$ is tangent of $ \mathcal P_a$ at $ O.$ Let $ A, B$ be 2 arbitrary points on $ \mathcal P_a,$ let the line $ AB$ cut $ x$ at $ P$ and $ y$ at $ Q.$ Let circumcircle $ (Z)$ of the $ \triangle OAB$ cut $ x$ at $ F.$ Then $ PO \cdot PF \equal{} PA \cdot PB \equal{} PQ^2$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=209493[/url] for detailed proof) and $ QO$ is altitude of the $ \triangle QPF$ $ \Longrightarrow$ the angle $ \angle PQF$ is right. Circle with diameter $ AF$ cuts $ y$ at $ Q$ and again at $ T.$ Let the line $ AT$ cut $ x$ at $ S$ and the circle $ (Z)$ again at $ C.$ Then $ SA \cdot SC \equal{} SO \cdot SF \equal{} ST^2,$ because the angle $ \angle STF$ is also right and $ TO$ is altitude of the right $ \triangle STF.$ If follows that $ C$ lies on the parabola $ \mathcal P_a.$ The $ \triangle ABC$ with circumcircle $ (Z)$ is also inscribed in the parabola $ \mathcal P_a.$ Let $ BC$ cut $ y$ at $ V.$ Since $ FQ \perp AB, FT \perp AC$ and $ F \in (Z),$ $ y \equiv QTV$ is Simson line of the $ \triangle ABC$ with the pole $ F$ and $ FV \perp BC.$ Let $ \mathcal P_b$ be parabola with focus $ F$ and vertex tangent $ y.$ The vertex tangent $ y$ is locus of the feet of perpendiculars from the focus $ F$ to the tangents of $ \mathcal P_b,$ hence $ AB, AC, BC$ are all tangents of $ \mathcal P_b.$ Conversely, if $ \mathcal P_b$ with focus $ F$ is given, any circle $ (Z)$ through $ O, F$ cutting $ \mathcal P_a$ at $ O$ and 3 more points $ A, B, C$ yields a $ \triangle ABC$ with the desired properties.


지금 한국? 살고 계시는 분 있나요?

저는 한국? 살고 있지는 않습니다.....

저는 굉장히 비밀스러워요 :D

저는 한국? 살고 있습니다. ^^

[code]My mom asks the same thing.  What is that? Han kuk yut????? and how do I get korean font?[/code]

앞분이나 저나 "한국에" 라고 쓴 것입니다.
이곳은 한글처리에 조금 문제가 있는 것 같아요.
백업 후 복구되거나 하면 글자가 바뀌는 것들이 몇몇 생기더군요.

[quote="lightrhee"]"한국엿 사시는 분"이 무슨 뜻이죠? "한국에서 사시는 분" 이라는 뜻인가요?
저는 캐나다에서 살고 있습니다(이민온지 $2 \frac{11}{12}$년 조금 넘었음).[/quote]

semy님은.. 과수원 BBS에서...^^

저도 한국엿 살고있습니다.ㅋㅋ

반갑습니다.
Unique님은 아마도... 나주한씨?
우리 만난 적은 없지요? ㅎㅎ

꾸준히 발전하여 훌륭한 성과 얻으시길 바랍니다.

전 캐나다 핼리팩스 살아요 ^^

KAIST93이구요.. 지금은 수학경시 학원에 있습니다. ㅎ

반갑습니다....

안녕하세요, 저는 한국에 살고 있지는 않습니다.

Same here...
Don't have korean keyboard installed yet. :(
Lived here for 4 years...

저는 한국에 살고있지 않습니다.

(토론토 거주 )
( 제발 저랑 수학공부좀 같이 해주실분없나요?)

저도 과수원  다녓던 기억이 스멀스멀납니다만..

저는 한국에 안 살지만 오클라호마에 살아요.. :lol:

저는 한국�?

i'm not living in korea

by the way, i like rice krispies

[hide][list]Duh? WHat? Huh? sup?[/list][/hide]

저는 한국에 살진않지만 1년전에는 살고 있엇어요. 그리고 이상하게 한국어가 잘 안써지고 네모난게 나오네요.

제가 서울살고 있죠 ㅎㅎ :wink:

에휴....... 신촌에서 놀고싶당 ㅋㅋ --"

뭐 옛날엔 일산 거주했지만 지금은 뉴저지쪽에 살고 있음당~~

I'm chinese and I'm getting killed here DonkeyKong!! Argh!!!!

저는 한국 사람이지만 미국에서 살아요.

I lived in Seoul,Korea but I immigrated to US :(. kemy님 부럽네요. 제 꿈은 KAIST 가는거였는데... 이제는 Phillps Exeter Academy 를 가도록 노력하고 있슴니다

저는 미국에서 태어났지만 한국에서 2년 쯤 살아서 한국어도 합니다. 저도 지금 gamjawon님 처럼 exeter를 갈려고 노력하고 있습니다.

ㅋ 나 진짜 머가 웃긴지 모르겠는데 ㅋㅋ

[quote="lykarg44"]ㅋ 나 진짜 머가 웃긴지 모르겠는데 ㅋㅋ[/quote]

ㅋㅋㅋㅋㅋ

Philips Exeter Academy 좋죠. 우리나라에서도 아는 고등학교가 그 곳 밖에 없어요. ㅋㅋㅋ

Um i cant type korean rip me
i live in california


This one was NOT made by me... I give credit to meewhee009. Thanks!

This is the Avatar I made myself! It's really fun!

you might have to click the picture to see the animated part of it!


I need help with this problem:

[b]Graph x > sqrt(2) on a number line.[/b]

My teacher said that the answer is:

[b]<======!===O[/b]--I-------|-------I--[b]O===!=======>[/b]
            -2        -1         0         1         2

but I think that the answer is:

<-------!---[b]O==I======|======I======!=======>[/b]
           -2       -1          0          1           2

because x can be any number greater than -1.4142 ...

(=====) means shaded in,
(------) means unshaded,
( O ) means greater than, but not including
( ! ) means 2 and -2
( I ) means 1 and -1
( | ) means 0.

Thanks,
Kevin K.

Wait, is the problem asking you to graph $x > \sqrt2$ or $x >-\sqrt2$? Either way, your teacher is wrong. If the first is the problem, you're both wrong. It should be all points to the right of $\sqrt2$. If it's the second, then you're right.

Your teacher graphed $|x| > \sqrt2$.

I meant to graph x > sqrt(2) for both the negative and positive values, -1.414 and 1.414.

Also, I thought a radical sign without anything meant both roots, positive and negative. One with a negative in front would denote the negative square root.

the radical sign usually denotes the positive.
i see why you think that your way does both the positive and negative values, but think about it this way:  0 is on your graph, but $0<\sqrt{2}$.  Anything bigger than $\sqrt{2}$ is also bigger than the negative value.  also, your teacher is wrong unless he/she is talking about $|x|$

the problem was probably supposed to be: graph x^2>2


I need help on several 2000 National Sprint questions...

1. In triangle ABC, angle C is a right angle. Point M is the midpoint of segment AB, point N is the midpoint of segment AC, and point O is the midpoint of segment AM. The perimeter of triangle ABC is 112 cm, and ON = 12.5 cm. What is the number of square centimeters in the area of quadrilateral MNCB?
[hide]I've gotten to the point where I've figured out it's a 7-24-25 triangle, so the side lengths are 14, 48, and 50. But I still don't know how to find the quadrilateral area.  :( [/hide]

2. Alyssa rode her bike for 7 hours. She started on level ground and rode at a rate of 8 mph. She came to a hill that slowed her down to a rate of 5 mph. Upon reaching the top of the hill, she turned around and descended at a rate of 20 mph. Finally, she returned home on the flat portion at 8 mph. What is the total number of miles she rode?

3. For positive integers a, b and c, what is the value of the product abc?
$ \frac{1}{a\plus{}\frac{1}{b\plus{}\frac{1}{c}}} \equal{} \frac{3}{8}$

2: Hint. On the hill find the average speed.
[hide]It's 8mph as well. Therefore her average speed is 8mph, *7hr=56 miles[/hide]

3: continued fractions.

a+(1/(b+1/c))=8/3, so a=2
1/(b+1/c)=2/3
b+1/c=3/2 so b=1, c=1

[hide="Hint for 1"]What kind of a quadrilateral is MNCB?[/hide]
[hide="Bigger hint for 1"]Show via similar triangles that MNCB is a trapezoid.[/hide]
[quote]a+(1/(b+1/c))=8/3, so a=2
1/(b+1/c)=2/3
b+1/c=3/2 so b=1, c=1[/quote]
You mean c=2, and the product is $ 2\cdot 1\cdot 2\cdot\equal{}\boxed{4}$, right? :wink:


Hi all of my dears in spain community, :)

sorry for interferring , but i had to ask this question. :wink:

could any body plz show me , how to buy all of the salvatore dali's

pictures?

or even show me any links about it?

thanks alot my dears.

This will maybe help you: http://ar.search.yahoo.com/search/images?p=salvador+dal%ED&sm=Buscar&fr=FP-tab-img-t&toggle=1&ei=ISO-8859-1&meta=all%3D1

oh tanks dear jose,

i searched alot ,but

infact i want to buy a book from salvatore dali, if it exists.

deos in spain exist any book from him?

or if it is , how can i buy? :wink:


I found something about a calendar and setting options for it in my profile. I haven't seen anything like a calendar on AoPS, so could anyone tell me what it is?

The settings reqard the dating system on the board. Maybe Calendar wasn't the best naming but that's about all there is to it.


Prove that
$\sum_{A,B,C}\sqrt{2Sin^{2}A+2Sin^{2}B-Sin^{2}C}\geq 2 \sum_{A,B,C}Sin^{2}A$
Where $A,B,C$ are angles of triangle $ABC$

Any trouble? Your ineq is not homogenized.

The inequality is equivalent to
$\sum_{a,b,c}m_{a}\geq \frac{\sum_{a,b,c}a^{2}}{2R}$
where $m_{a}, m_{b}, m_{c}$ are the lengths of the medians of triangle and $R$ is the circumradius.($a,b,c$ are the sides) :idea:

or

$\sum_{a,b,c}\sqrt{2a^{2}+2b^{2}-c^{2}}\geq 2 \frac{\sum_{a,b,c}a^{2}}{abc}\sqrt{s(s-a)(s-b)(s-c)}$ :idea:

Now substituting,

$a= x+y$,$b = y+z$ ,$c = z+x$,
we get a homogenous inequality in $x,y and z$

$Sin A$ etc are dimensionless and do not have any relationship with homogeneity of any inequality :wink: .....


I have two problems I can't solve!!! D:

The 1st term of a sequence is 2. Each term after the 1st is obtained by multiplying the preceding term by 5 and then subtracting 6. What is the sum of the first five terms of the sequence?
Here I got 398... It should be right... but the book says 458...

Ronnie, Ben, and Jevon are in the same math class. There are 24 students in the class, and the teacher has announced that she will be dividing the class into groups of 3. What is the probability that Ronnie, Ben, and Jevon will be in the same group? Express as a common fraction.

I'm still a little shaky on THESE types of probablilty questions, so I don't know where to start.

[hide="Number One"]
Let's list out the terms.

1st Term: 1
2nd Term: $ (1 \cdot 5) \minus{} 6 \equal{} \minus{}1$
3rd Term: $ (\minus{}1 \cdot 5) \minus{} 6 \equal{} \minus{}11$
4th Term: $ (\minus{}11 \cdot 5) \minus{} 6 \equal{} \minus{}61$
5th Term: $ (\minus{}61 \cdot 5) \minus{} 6 \equal{} \minus{}311$

$ 1 \minus{} 1 \minus{} 11 \minus{} 61 \minus{} 311 \equal{} \boxed{\minus{}383}$

Hmm, I'm getting an answer completely different from both your and the book's answer.
[/hide]

[hide="Number Two"]
There are $ 24 \div 3 \equal{} 8$ total groups, so there are $ 8 \times 3! \equal{} 48$ total ways to divide the class into groups of 3. There's only 1 way for Ronnie, Ben, and Jevon to be in the same group, so our answer is $ \boxed{\frac{1}{48}}$.
[/hide]

Nooo... wait a minute...

1) The first term of the sequence is two.

2) Aren't there three ways for Ronnie, Ben, and Jevon to be in the same group? They could be in the first group together, the second group together, and the third group together... right?  :huh:

1.
1st-2
2nd-2*5-6=4
3rd-4*5-6=14
4th-14*5-6=64
5th-64*5-6=314
314+64+14+4+2=398
i believe you are right.
#2 Suppose your were those three were picked right away then that would be 3/24x2/23x1/222=1/2024
Suppose that they weren't picked first and picked second
21/24x20/23x19/22x3/21x2/20x1/19=1/2024
so order they are picked in they have a 1/2024 chance. There are eight different places the group can be picked 1st, 2nd, 3rd, 4th, 5ht, 6th, 7th or 8th. So 1/2024x8=1/253.

But for problem number two, there are only three groups. How did you get eight?  :huh:

Oh yeah. There is another problem I don't get...  :blush:

Keith and Chi-An are playing a game. Keith chooses a poistive integer less than 100. Chi-An then tries to guess Keith's number by asking questions to which Keith answers yes or no. What is the number of questions Chi-An must ask so that, even if she is unlucky, she will always be able to identify the number?

eh this is like common sense

so keith chooses a number among 1~99.

if the number was 99

chi-an can ask 1, 2, 3, ..., 98, which is asking 98 times.

answer : 98

Nah... She could ask, "is the number greater than x", etc. So, rounding up, you start with 50 possible, than 25, 13, 7, 4, 2, 1. Thus, you have to ask at the very most $ 7$ questions.

For number 2, one of the three has to first be in that group to "pick out" that group. Then there are 23 people left, so there's a $ \frac{2}{23}$ chance of one of them being in that group. Finally, there are 22 people left, so there's a $ \frac{1}{22}$ chance of the group containing Robbie, Ben, and Jevon. Multiplying gives the answer $ \frac{1}{253}$.

Wait, I think I messed up somewhere...

Well, on number three it can't be seven, because If it is greater than fifty, she'll ask six questions...

Yes, we are [b]always[/b] looking at the worse-case scenario to find out the number of times that we must do something to assure that it will work every time, not just when you're lucky. All of the numbers thus must be taken into account.

Lol nick :D.

Can anyone solve #2 with a solid solution?

blah i suck at counting but

there are total $ \binom{24}{3}\cdot\binom{21}{3}\cdots\binom{6}{3}\cdot\binom{3}{3}$ ways to divide 24 students to 8 groups of 3.

and if ronnie, ben and jevon are in the same group, there would be total $ \binom{21}{3}\cdot\binom{18}{3}\cdots\binom{6}{3}\cdot\binom{3}{3}$

so the probability would be $ \frac{\binom{21}{3}\cdot\binom{18}{3}\cdots\binom{6}{3}\cdot\binom{3}{3}}{\binom{24}{3}\cdot\binom{21}{3}\cdots\binom{6}{3}\cdot\binom{3}{3}}\equal{}\frac{1}{\binom{24}{3}}$

hmm i got $ \frac{1}{2024}$.

Wow. That makes sense! :)

The [i]transient Triangle[/i] is a region with three cities: A-ville, B-burg andC-town. The ppopulation of each of these cities was the same at the end of 1996 as it was at the end of 1995, but the residents had moved between cities. What percent of the residents of the Transient Triangle were residents of A-ville at the end of 1996 if:

- 10% of the 1995 residents of A-ville moved to B-burg in 1996;
- 10% of the 1995 residents of A-ville moved to C-town in 1996;
- 20% of the 1995 residents of B-burg moved to A-ville in 1996;
- 10% of the 1995 residents of B-burg moved to C-town in 1996;
- 10% of the 1995 residents of C-town moved to A-ville in 1996; and
- 10% of the 1995 residents of C-town moved to B-burg in 1996?

[quote="FantasyLover"][quote="shentang"]#2 Suppose your were those three were picked right away then that would be 3/24x2/23x1/222=1/2024
Suppose that they weren't picked first and picked second
21/24x20/23x19/22x3/21x2/20x1/19=1/2024
so order they are picked in they have a 1/2024 chance. There are eight different places the group can be picked 1st, 2nd, 3rd, 4th, 5ht, 6th, 7th or 8th. So 1/2024x8=1/253.[/quote]

blah, didnt see that
...you said the answer was 1/253.
besides, i solved it differently anyways[/quote]

Oh, you just missed that the trio could be picked 1st 2nd 3rd....8th you assumed they were picked 1st. so 1/2024*8=1/253

actually, it doesn't matter.

you have to pick 3 out of 21, out of 18, etc. because you must not pick one of 3 guys.
so eventually it becomes JUST 21C3*18C3*...*3C3.

Successful: pick a group, order the three people. 8*3! = 48
Total: place the three people in 24 slots. 24P3 = 24*23*22 = 48*23*11
48/24P3 = 1/253

I thought of previous solutions, but here's a different one for variety.

Sooo... is

[u]2+SQRT(2)[/u]
2+2SQRT(2)

in simplest radical form as a common fraction?

(Irrelevant question)

No, because you can't have radicals in the denominator. Multiply both the numerator and the denominator by the conjugate(in this case, it's $ 2 \minus{} 2\sqrt {2}$), and the final answer would be $ \frac {\sqrt {2}}{2}$.

This is called rationalizing the denominator.

Alternatively, in this case, you can do something sneaky with this special form of fraction. If you have $ \frac {n \plus{} \sqrt n}{n \plus{} n\sqrt n}$, factor our $ \sqrt n$ from the numerator and $ n$ from the denominator, we get \[ \frac {\sqrt n(\sqrt n \plus{} 1)}{n(1 \plus{} \sqrt n)} \equal{} \frac {\sqrt n}n \left(\frac {1 \plus{} \sqrt n}{1 \plus{} \sqrt n}\right) \equal{} \frac {\sqrt n}n\]


modular arithmetic
combinatorics proposed
combinatorics

$ (a)$ Prove that $ \mathbb{N}$ can be partitioned into three (mutually disjoint) sets such that, if $ m,n \in \mathbb{N}$ and $ |m\minus{}n|$ is $ 2$ or $ 5$, then $ m$ and $ n$ are in different sets.
$ (b)$ Prove that $ \mathbb{N}$ can be partitioned into four sets such that, if $ m,n \in \mathbb{N}$ and $ |m\minus{}n|$ is $ 2,3,$ or $ 5$, then $ m$ and $ n$ are in different sets. Show, however, that $ \mathbb{N}$ cannot be partitioned into three sets with this property.

[quote="moldovan"]$ (a)$ Prove that $ \mathbb{N}$ can be partitioned into three (mutually disjoint) sets such that, if $ m,n \in \mathbb{N}$ and $ |m - n|$ is $ 2$ or $ 5$, then $ m$ and $ n$ are in different sets.
$ (b)$ Prove that $ \mathbb{N}$ can be partitioned into four sets such that, if $ m,n \in \mathbb{N}$ and $ |m - n|$ is $ 2,3,$ or $ 5$, then $ m$ and $ n$ are in different sets. Show, however, that $ \mathbb{N}$ cannot be partitioned into three sets with this property.[/quote]

[b](a)[/b] $ n \equiv 1 \pmod{3} \Rightarrow n \in A_1$
$ n \equiv 2 \pmod{3} \Rightarrow n \in A_2$
$ n \equiv 0 \pmod{3} \Rightarrow n \in A_3$

Now clearly $ A_1,A_2,A_3$ are partions of $ \mathbb{N}$ satisfying the condition as difference of any two numbers in the same set must be divisible by $ 3$.

[b](b)[/b] Similarly to (a) we take $ n \equiv 1 \pmod{4} \Rightarrow n \in B_1$
$ n \equiv 2 \pmod{4} \Rightarrow n \in B_2$
$ n \equiv 3 \pmod{4} \Rightarrow n \in B_3$
$ n \equiv 0 \pmod{4} \Rightarrow n \in B_4$

and $ B_1,B_2,B_3,B_4$ are partitions of $ \mathbb{N}$ satisfies the condition.

For second part of (b), assume that the $ 3$ partions $ C_1,C_2,C_3$ exist and let $ 1 \in C_1$, then $ 3 \nin$ subset containing $ 1 \Rightarrow 3 \in C_2$ WLOG.

Similarly $ 6 \notin$ subsets containing $ 1,3 \Rightarrow 6 \in C_3$

$ 4 \notin$ subsets containing $ 1,6 \Rightarrow 4 \in C_2$

$ 8 \notin$ subsets containing $ 3,6 \Rightarrow 8 \in C_1$

$ 5 \notin$ subsets containing $ 3,8 \Rightarrow 5 \in C_3$

$ 7 \notin$ subsets containing $ 4,5 \Rightarrow 7 \in C_1$

$ 2 \notin$ subsets containing $ 4,5,7 \Rightarrow 2 \notin C_1,C_2,C_3$ , contradiction, done.


calculus
derivative
real analysis
real analysis unsolved

[quote="numero1"]Consider f(x) = 4/(1+x^2)

By finding a bound on f(4) show that

| pi - SIMPn | <= 8/15n^4.[/quote]
Do you mean a bound on $f(4)$ or on $f^{(4)}$, the fourth derivative? That would be the right kind of bound for this problem.

Btw, this is a standard homework problem.

Yes the fourth derivative.
I am having difficulty taking the fourth derivative..LOL.
when I use the quotoent rule the denominator just blowa up too big.


The AMC 10 has 25 questions. If there are an equal number of  correct answers being (A), (B), (C), (D), and (E), what is the number of questions will I get correct that if I select (C) for each question?

$ \textbf{(A)} 4\qquad\qquad\qquad\textbf{(B)} 5\qquad\qquad\qquad\textbf{(C)} 6\qquad\qquad\qquad\textbf{(D)} 7\qquad\qquad\qquad\textbf{(E)} 8$

A rhombus is given with diagonals of which one is double the length of the other. If the length of the longer diagonal is x, what is the area of the rhombus?

$ \textbf{(A)} {x^2}\over {2} \qquad\qquad\qquad\textbf{(B)} x^2\qquad\qquad\qquad\textbf{(C)} x^2-4\qquad\qquad\qquad\textbf{(D)} {x^2}\over 4\qquad\qquad\qquad\textbf{(E)} x$

Joe uses a ridiculous phone provider which charges an amount x cents per minute for which he talks in a month. The phone company also has a flat rate for which it charges all customers regardless of how much they talk. In January, the phone company charges Joe $ \$$14.20. The following month, Joe does a lot of business conferences and talks on the phone for four times as long as he did in January. His monthly bill is $ \$$37.30. What is the fixed rate that the phone company charges to the nearest cent?

$ \textbf{(A)} \$3.55\qquad\qquad\qquad\textbf{(B)} \$6.50\qquad\qquad\textbf{(C)} \$9.33\qquad\qquad\qquad\textbf{(D)} \$5.78\qquad\qquad\textbf{(E)} \$23.10$

What is the maximum possible number of intersection points between two circles and two lines?

$ \textbf{(A)} 8\qquad\qquad\qquad\textbf{(B)} 9\qquad\qquad\qquad\textbf{(C)} 11\qquad\qquad\qquad\textbf{(D)} 12\qquad\qquad\qquad\textbf{(E)} 15$

\item In a standard cube, an ant is set at a corner. What is the maximum possible lengths an ant can walk if it cannot visit any point twice or overlap any of the lengths, if the ant must end up in the point it started at?

$ \textbf{(A)} 5\qquad\qquad\qquad\textbf{(B)}
6\qquad\qquad\qquad\textbf{(C)}
7\qquad\qquad\qquad\textbf{(D)}
8\qquad\qquad\qquad\textbf{(E)}
9$

[hide="Problem 1"]We know that there are 25 questions. When we select (C) for each question, and there are an equal number of (A)'s,(B)'s,(C)'s,(D)'s, and (E)'s, there are 5 (C)'s, which implies that the answer is $ \boxed{\textbf{(C)}}$.[/hide]

[hide="Problem 2"]If the length of the longer diagonal is $ x$, then the length of the shorter diagonal is $ x/2$. The area of a rhombus is $ \dfrac{d_1d_2}{2}$. Applying this gets us $ x^2/4\implies \boxed{\textbf{D}}$.[/hide]

[hide="Problem 3"]We translate these statements into equations. It now becomes quite clear that the correct answer is $ \boxed{\textbf{(B)}}$.[/hide]

[hide="Problem 4"]We can simply draw this out and verify that the correct answer is $ \boxed{\textbf{(C)}}$. For example, we can see such a situation in the below image:

[img]http://myfiles.zapto.org/AMC10Images/CirclesandLine.jpg[/img]
[/hide]

[hide="Problem 5"]One method of solving this problem is drawing a cube, starting the ant at the foremost top left hand corner, travel around the top of the cube to the point at the foremost top right hand corner, then travelling downwards, making a similar travelling, and then heading back to the point it started at. The answer is $ \boxed{\textbf{(D)}}$.[/hide]

Wasn't there also a typo for problem 1 or its answer?
[quote]The AMC 10 has 25 questions. If there are an equal number of correct answers being (A), (B), (C), (D), and (E), what is the number of questions will I get correct that if I select (C) for each question?
(A)4   (B)5   (C)6   (D)7   (E)8[/quote]
[quote]...there are 5 (C)'s, which implies that the answer is (C).
[/quote]
Anyway, those were enjoyable problems with or without the typos.

yea i liked the telephone problem..  Solution:
let m be the minutes used in January, and f be the fixed rate:
$ xm + f = 14.20$
 $ 4xm + f = 37.30$
multiplying the first equation by 4 and subtracting we get $ 3f = 19.50 \implies f = 6.50   \implies \boxed {B}$

however, these were more difficult than the average 1-5 on the AMC 10


you have a stick of length $x$, and a paper with parallel lines $y$ distance apart. suppose $x>y$.
when you throw the stick on the paper $n$ times, you find, out of those $n$ times, that the stick crosses
[u]at least[/u] one of those parallel lines, $m$ times.

show:

 $\frac{n}{m}\left[ \Omega\;+\;\sec^{-1}{\Omega}\;-\sqrt{\Omega^{2}-1}\;\;\right] \;=\;\boxed{\frac{\pi}{2}}\;\;\;\quad\;\;\;\quad where\;\;\;\;\;\Omega\;=\;\frac{x}{y}$

This is called Buffons's Needle Problem.

More is to read on this on say [url=http://mathworld.wolfram.com/BuffonsNeedleProblem.html]mathworld[/url] and   http://en.wikipedia.org/wiki/Buffon's_needle

(sorry for the wiki link, couldn't fix it )


i am looking for a way to do te volume of a rectangular prisim knowin only side areas.  i know how to work the problems, but am looking for a less trial and error method that would work with large numbers and no calculator

heres what i have so far

factor all three numbers
EX. side areas = 70, 30, 21.
      70=2*5*7
      30=2*3*5
      21=3*7
notice 70 and 21 both have a seven in common, so the edge they meet at=7
continue this with the other numbers and you can ultimantly find the volume
BUT, ive noticed that doesnt always work

i was wondering if any aopsers could think up some cool methods to solve these kinds of problems without guess and check.

thanks

here you go.

If the dimensions are x, y, and z then your face areas would be xy, yz, and xz.  Those would be equal to the 70, 30, 21.

If you take the product of the face areas both with the variables and the numbers you get $x^{2}y^{2}z^{2}= 70\cdot30\cdot21$

If you take the square root of both sides you get

$\sqrt{x^{2}y^{2}z^{2}}= \sqrt{70\cdot30\cdot21}$ =
$xyz = \sqrt{70\cdot30\cdot21}$
$xyz = \sqrt{7\cdot10\cdot10\cdot3\cdot3\cdot7}$
$xyz = \sqrt{7^{2}\cdot10^{2}\cdot3^{2}}$
$xyz = 7\cdot10\cdot3$
$xyz = 210$

and since $xyz$ is the volume of the prism, there you go.  You could probably simplify the radical faster than I did, I just wanted to show all the steps out for you.

The side lengths are $a$, $b$, and $c$.

You are given $ab=70$, $bc=30$, and $ac=21$, and you know $V=abc$.

$V=abc=\sqrt{(abc)^{2}}=\sqrt{ab\cdot bc\cdot ac}$

Using this formula, you can find the volume.

Wow, do those two solutions look identical... I wonder why...
(*cough*i_like_pie*cough*)

Yeah, but it's still worth noting that this solution works in any case, not just the surface area/volume ones...

But the s. area and volume is by far the most common in mathcounts


Noah has 8 species of animals to fit into 4 cages of the ark. He plans to put the species in each cage.

It turns out that, for each spieces, there are at most 3 other species with which it cannot share the accomodation.

Prove that there is a way to assign the animals to their cages so that each species shares with compatible species.

total possible pairs = 8C2=28
maximum incompatible pairs has to be less than 24 (8*3,assuming incompatibility relation is symmetric)
Thus compatible pairs has to be greater than 4

Consider the graph whose vertices are the species and the edges join any two species which cannot share accomodation.
Thus, each vertex has degree at most $ 3$.
It is a basic result to prove that any simple graph with maximal degree $ \Delta$ can be properly colored with at most $ \Delta \plus{}1$ colors.
Here, one color is one cage.

Pierre.

In the "sledgehammer" category:

Consider the graph whose vertices are the species, with edges joining species that can share accommodation (the complement of pbornsztein's graph).  It has 8 vertices and minimum degree 4, so by Dirac's theorem it has a Hamiltonian circuit.  But the 8-cycle contains two natural perfect matchings, so we're done.


Does there exist 10 integers $ a_1,a_2,\cdots,a_{10}$ satisfying $ a_i\ne a_j$ for $ i\ne j$ and $ S\minus{}a_i$ is perfect square for every $ i\equal{}1,\cdots,10$ with $ S\equal{}\sum_{i\equal{}1}^{10} a_i$

Let $ b_1,b_2\dots,b_{10}$ be integers such that $ \sum_{i\equal{}1}^{10}b_i^2\equal{}9S$, for some $ S$. Then $ a_i\equal{}S\minus{}b_i^2$ for $ i\equal{}1,2,\dots,10$ work.

For example: 376, 349, 304, 241, 160, 61, -56, -191, -344, -515 is such a sequence. Here $ a_i\equal{}385\minus{}9i^2$ and $ S\minus{}a_i\equal{}(3i)^2\,\forall i\in\{1,2,\dots,10\}$.


probability factorial AoPSwiki MATHCOUNTS geometry trigonometry calculus

I have a problem . How can I improve on counting and probability problems besides Alcumus. I can never et those.

I suggest reading Introduction to Counting and Probability.

Tip: Just get used to as many different types of those problems. Counting and probability problems are generally very limited in their types so just practice getting used to them. (Similar to stoichiometry problems). For example, identify which are independent events, which are dependent, which need combinations, which are permutations, when to use factorials, when to use combinatorics, etc.

izzy how can i geet this book. How much does it cost?

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]This[/url] is Introduction to Counting and Probability. You can look at the excerpts/tests to see if it suits you. If it's too expensive, there are many places to find problems, including:
[url=http://www.artofproblemsolving.com/Wiki/index.php/Category:Introductory_Combinatorics_Problems]AoPSWiki[/url]
[url=http://mathcounts.saab.org/mc.cgi]Elias Saab's MC website[/url]

If you look hard enough, you can find many more. :wink: The advantage of the text is that it gives thorough examples and problems. However, it is possible to just look at the table of contents and learn the topics by yourself.

If you can find a cheaper, used copy at a local bookstore, that's even better. I know someone who has all the books, so I can borrow them. There are lots of ways to get a copy.

If you got a gift card or something, you could specially order an AoPS book and pay less than the normal price. That's what I'm doing for All-Time Greatest MathCounts Problems.

Lots of people will give you gift cards for AoPS. :wink:

[quote="math154"]Not that many bookstores have AoPS books. :wink: But if you're lucky, you can borrow them from somewhere.[/quote]
I believe some Barnes and Nobles have them, and Bookmans (depending on where you live)...but don't try eBay =]

Since when could you find AoPS books in typical libraries...

jjx never specifically asked for aops books.

edit: oh wait he did haha

I suggest looking at different counting problems on the web, and analyzing there solutions. That is how I learned, and I got pretty good at it.

[quote="math154"]Eh...My Barnes & Nobles has a single math section...like one shelf...and it's filled with...Pre-Algebra for Dummies, Algebra for Dummies, Geometry for Dummies, Algebra II for dummies, Trigonometry for dummies, Precalculus for dummies, Calculus for dummies, ...[/quote]
As I said, it depends on where you live. Some have bigger sections. (and both of the B&Ns in my city have ______ for Dummies too =P plus more!!!)

[quote="math154"]Eh...My Barnes & Nobles has a single math section...like one shelf...and it's filled with...Pre-Algebra for Dummies, Algebra for Dummies, Geometry for Dummies, Algebra II for dummies, Trigonometry for dummies, Precalculus for dummies, Calculus for dummies, ...[/quote]

Lol. The exact same thing happened to me, I wasn't able to find a single helpful book at my nearest Barnes and Noble book store.

[quote="isabella2296"]While AoPS books are very good, keep in mind that they're not the only good books in the world. I've found some pretty good math books at B&N that show proofs and such.[/quote]
Even the Dummies series is okay. (but it's great for beginners)

i like the ____for dummies books. they're easy to understand but they cover the information. the aops books are very brief, though they give the info too.

[quote="isabella2296"]Actually, I find that Dummies books and The Complete Idiot's Guide to ______ books are pretty good if you're doing something totally new/difficult for you. Especially if there's no AoPS book on it yet, like Calculus.[/quote]
But sometimes it just breezes over the concept.
Which isn't always a good thing

Okay, Thank you for your help everyone! :D I went to Barnes and Nobles and got some really good AOPS books. I hope they help me because currently I am horrible at counting and probabilty. I can never get them(seriously) unless I do guess and check which doesn;t work

try Counting and Probability for Smarties

[quote="isabella2296"]Really? That's never happened in my experience.[/quote]
It doesn't happen that often though.

i usually just go to AoPSWiki and check out the AIME Problems and Solutions. Those are pretty helpful

[quote="AIME15"][quote="Wickedestjr"]How do you guys practice geometry?[/quote]

AoPS geometry class.

jjx - how'd you find aops books at barnes and noble!?!?!?!?  i could never find them at my local bookstore...[/quote]

yeah, read the rest of the thread some people just have them and some people don't. i personally have never seen them at my bookstore.

[quote="AIME15"]I know you can purchase them from "Bookstore" on the top of the sidebar.[/quote]

of course, they are AoPS books, if they didn't sell them here, where else would they sell them?

I think that book stores would get a lot of money by selling AoPS books.

[quote="Poincare"][quote="Wickedestjr"][quote="isabella2296"]There are tons of resources on the web - a quick Google search will yield a bunch. Elias Saab's website has some pretty good geometry problems, and there's a lot of books out there too.  :)[/quote]

Thanks. I'll check it out.  :lol:  :)[/quote]
if you have intro to geo DO THE CHALLENGE PROBLEMS!!! they are gonna make you beast in geometry.
Also, search Math books in AoPs wiki and you can get some pretty satisfying results. I have found that for counting and probability (especially the latter) it helps to do A LOT of problems even the easy ones.[/quote]

I really wanted to get the geometry intro book, and I was planning on buying it soon. My MC coach let me borrow a copy of Intro to C & P, and it was really helpful. The challenge problems in that book were good too.

If you've already taken the geometry class, or any aops class for that matter, over a year ago, best to review everything by doing the challenge probs in the book.

Wicked, buy the books.  It'll be worth the mony.  Why do you use the past tense "I really WANNTED", "I was PLANNING"?

I don't know, it was a long time ago, and it sort of slipped past my mind. I was waiting til Christmas.


Suppose that 7 boys and 13 girls line up in a row. Let S be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row GBBGGGBGBGGGBGBGGBGG we have S = 12. If all possible orders of these 20 people are considered, what is the average value of S?

I forget if we ever solved this, but it was one of the harder ones there.

Also, if you have a problem to post, post it here, so we can revive the SC Forum!! :P

We have $7$ boys and $13$ girls, however, if we find out the ways the boys can be positioned, this will determine the position of girls as well (since there are no extra spaces in the row).

Because BG is the same as GB, we can select two side by side spaces to put these. For instance, if we select the first two spaces, we have:

BG_ _ _ _ .... (18 blank spaces)

We now have $18$ spaces and $6$ remaining boys, so we can determine the number of ways of determining the sequence of the remaining spaces as $\binom{18}{6}$.

For the second case, we have

_BG_ _ _ _ (18 blank spaces)

Similarily, we get $\binom{18}{6}$

Continuing this way, we find the total number of arrangements with BG are $19 \cdot \binom{18}{6}$ and the total number of ways we can arrange the sequence (regardless of having BG) is $\binom{20}{7}$, thus our average would be $\boxed{\frac{91}{20}}$