AMC
AMC 12
AMC 12 A
algebra
polynomial
quadratics
geometry

I have the solution key from the official place, but I dont understand the solutions they give, anyway here are the questions:

24. Let P(x) = (x - 1) (x - 2) (x - 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 s.t. P(Q(x)) = P(x)R(x)?
(Its P(x) times R(x))
25. Let S be the set of all points with corrdinates (x, y, z), where x, y, and z are each chosen from set {0, 1 ,2}. How many equailateral traingeles have all their vertices in S?

I donno if this is right, but:

[hide="24"]
If $d$ is the degree of $Q(x)$, and $P(x)$ and $R(x)$ are both cubics $\implies$ have degree $3$, we get $3d=3+3 \implies d=2$, so we're looking at a quadratic.

For the three zeros of $P(x)$, $P(x)R(x)=0 \implies P(Q(x))=0$.

$P(x)$ has zeros at $1,2,3$. Thus, $Q(x)$ must map $\{1,2,3\}$ onto itself. 3 non-collinear points define a unique quadratic, and a quadratic cannot have the same value at $3$ different points. So, there are $3^{3}-5=\boxed{22}$ such functions.
[/hide]

[hide="25"]
This is a 2 by 2 by 2 cube. For any cube, connecting the three vertices adjacent to a given vertex forms an equilateral triangle. Each cube has 8 vertices, and thus 8 equilateral triangles.

There are $(2)(2)(2)=8$ 1 by 1 by 1 cubes, and one large 2 by 2 by 2 cube to consider, for a total of $9$ cubes and a total of $(9)(8)=72$ equilateral triangles.

[b]Edit:[/b] And then there are the equilaterals that are formed by joining the midpoints of side of the large cube. There are 8 of these for a grand total of $\boxed{80}$
[/hide]

25 is wrong, you missed 8 triangles that can be gotten by connecting the midpoints of the sides of the larger cube.

Nope, the answers are :
[hide] 22 [/hide]
and
[hide] 80 [/hide][/code]

Forgot about those  :(

I'm curious to see the solution to #24... my approach was probably way off, I donno that much about function theory

Your approach for #24 is perfect, there's just five linear functions that are map of 1,2,3 onto itself (diagonals and horizontals).

[quote="danee 1nonly"]for 24, im confused where u got the 3^3 and 5, whats this 3 collinear points thing?[/quote]

[hide="If you followed the rest of the argument..."] We are looking for a quadratic polynomial $Q(x)$ such that $Q(1), Q(2), Q(3)$ are all either $1, 2$, or $3$ (not necessarily all of them!)

So there are three possibilities for each value.  That is to say, $Q(1) = 1, 2, 3$, and so forth.  This gives us $3^{3}$ possibilities.

Three points uniquely determine a quadratic.  However, if the three points are on the same line, then the quadratic that passes through them is in fact [b]linear.[/b]  (It's the line that passes through them.)  How many lines are involved?

The set of points that we're dealing with are the points that have both coefficients either $1, 2, 3$.  So we're looking at a $3 \times 3$ lattice square of $9$ points.

Well, we have the three lines $y = 1, 2, 3$.  Those pass through three points.  We also have the two diagonals of the square.

The three lines $x = 1, 2, 3$ don't count because they only involve one $x$-value, and we need to use all of them.  Hence we have counted $5$ linear functions with our quadratic ones, so the number of quadratic answers is

$3^{3}-5 = 22$ [/hide]

EDIT: Never mind, t0ra beat me to it  :)

Essentially the same explanation:

[hide]
Well, $Q(x)$ is a function that maps from $\{1,2,3\}$ to itself.

So for each of these 3 inputs, we must get 1 distinct output for each.

This makes $(3)(3)(3)=27$ possibilities.

Amy has proven that $Q(x)$ is quadratic. 3 non-collinear points define a unique parabola

(a parabola, $ax^{2}+bx+c$, is defined by 3 parameters, and we have 3 points, so we can solve uniquely for $a$, $b$, and $c$)

Now how many of these 27 possibilities are non-collinear?
We know from the way we constructed them that none of them form vertical lines. However, if the output is the same for all 3 inputs {1,1,1}, {2,2,2}, or {3,3,3}, then you have a horizontal  line, and this cannot be described by a quadratic. Similarly, output orders {1,2,3} and {3,2,1} form diagonal lines, which also cannot be described by a quadratic. So we must eliminate 5 from the 27 possibilities: $\text{Total}=27-5=\boxed{22}$[/hide]


inequalities
absolute value
probability
expected value
real analysis
real analysis solved

That's not particularly easy to read. Let me try re-doing it. Michael - tell me if it's not the problem you intended:

Define $F(x)=e^{-x^2}$ for $|x|\le 1$ and zero otherwise.

Suppose $a_1,\dots,a_n,x_1,\dots,x_n$ are real numbers such that

$a_1F(x-x_1)+a_2F(x-x_2)+\cdots+a_nF(x-x_n)\ge 0$ for all real $x.$

Prove that $a_1+a_2+\cdots+a_n\ge 0.$

I think a probabilistic argument will work.  Let x be a random number chosen uniformly from $\min(x_i) - 1$ to $\max(x_i) + 1$.  Taking expected value of the given inequality, we have
\[\sum_{i=1}^n a_i E(F(x - x_i)) \geq 0.\]
But each $E(F(x - x_i))$ is the same positive constant.  Thus $\sum_{i=1}^n a_i \geq 0$.


Ce inseamna n!! exact? Am gasit o problema cu o serie cu termenul general (2n-1)!!/(2n+2)!! . Sa se afle suma. (n de la 1 incepe). Cum se face?

P.S. Daca am postat in loc gresit scuze, nu stiu unde sa postez matematica de anul I.


Congratulations to all of the board members that took part in the ACM international collegiate programming contest--I recognize a few names!

[url=http://icpc.baylor.edu/icpc/Finals/default.htm]icpc.baylor.edu/icpc/Finals/default.htm[/url]

My congratulations to all Russian teams!

[quote="liyi"]My univ got the champion.  :D Congratulations to them
Also congratulations to Moscow State Univ and St Petersburg Institute of Fine Mechanics and Optics![/quote]
Congratulations!!!!!!! :first:  :D

Wow Politehnica Bucharest on 10th place! :D


calculus
derivative
blogs
LaTeX
limit
calculus computations

Hello. I have 2 problems here:
1) If f:(a,b)->R is derivative in xo belogs to (a,b) then I have to calculate
 lim (n->infinity) [f(xo+1/n)/f(xo)]^n ,  f(xo)>0.
I realy don't know how to begin this.

2) If f:(a,b)->R is derivative in xo belogs to (a,b) then I have to calculate
 lim (x->xo) [xo*f(x)-x*f(xo)]/(x-xo)

All help appreciated.

$ \text{\LaTeX}$ please!

Assume $ f \, : \, ]a,b[ \to \mathbb{R}$ is differentiable in $ x_0 \in ]a,b[$.

[b]1):[/b]
Calculate $ \lim_{n \to \infty} \left(\frac{f(x_0 + \frac1n)}{f(x_0)}\right)^n$, assuming $ f(x_0) > 0$.

[b]2):[/b]
Calculate $ \lim_{x \to x_0} \frac{x_0 f(x) - x f(x_0)}{x- x_0}$.

thank you Tyl. :D.. does anyone know how to solve this?


Find all functions $f : R =\Rightarrow  R$ satisfying the condition:
$f(f(x-y))=f(x)f(y)-f(x)+f(y)-xy$ for all $x, y$.

There are some problems at the past solution.

The solution hasn't proved that $f(x)=+\sqrt{x^{2}+k}$ for all $x$ or $f(x)=-\sqrt{x^{2}+k}$ for all $x$.
So far, it only has proved that, for each separate value $x$, $f(x)=+\sqrt{x^{2}+k}$ or $f(x)=-\sqrt{x^{2}+k}$.
I mean, so far there could be a set $P\subset\mathbb{R}$, where
$f(x)=+\sqrt{x^{2}+k}$ if $x\in P$,
$f(x)=-\sqrt{x^{2}+k}$ if $x\notin P$.

Also, not every $k\ge 0$ is possible, since $k=f(f(0))=f(\pm\sqrt{k})=\pm\sqrt{2k}$, so
$k=2$ (only if $f(0)\in P$),
or $k=0$.

Every possibility should be checked.

[Edited: I added the following]
This is my conjecture:
- No solution with $k=2$ works.
- The only solution with $k=0$ is $f(x)=-x$.
I'll try to prove it strictly later. It doesn't seem so difficult.

Put $f(0)=k$ :arrow: $f(k)=k^{2}$
1)Take$x=y\neq 0$ :arrow: $f(x)^{2}=x^{2}+k^{2}=f(-x)^{2}$
2)Take $y=0$ gives $f(f(x))=(k-1)f(x)+k$ then $f(f(-x))=(k-1)f(-x)+k$
3)Take $x=0,y=x$ gives $f(f(-x))=(k+1)f(x)-k$ with 2) then $(k-1)f(-x)+k=(k+1)f(x)-k$ (E)
if $k^{2}=1$ then $f(x)=cste$ not indeed solution so $k^{2}\neq 1$
4) Square (E) gives:$(k-1)^{2}f(-x)^{2}+2k(k-1)f(-x)=(k+1)^{2}f(x)^{2}-2k(k+1)f(x)$ then $2k((k-1)f(-x)+(k+1)f(x))=4k(x^{2}+k^{2})$
if $k\neq 0$  then $f(x)=\frac{x^{2}}{k+1}+k$ thus we must have:
$\forall x$  : $(\frac{x^{2}}{k+1}+k)^{2}=x^{2}+k^{2}$ not possible thus $k=0$

[quote="lordWings"]- The only solution with $k=0$ is $f(x)=-x$.
I'll try to prove it strictly later. It doesn't seem so difficult.[/quote]
I let this part to lordwings  :)


[quote="aufha"]Where would a good place strings or cello competitions in Virginia?
Do music competitions exist where you send a sound file or recording?[/quote]

Yes there are competitions where you send a sound file or recording.  There's a Menuhin Competition hosted in Europe and I think the preliminary round is all based on a recording you send in.  After that, however, you're going to have to audition.  AND IT'S REALLY REALLY REALLY REALLY HARD.  :wallbash:
And I don't know any competitions in VA cuz I'm in NC.  :rotfl:  :roll:


I'm on Aops mostly, and I chat some.

I don't really do much else online.

I had a xanga, but I don't update anymore.
I used to play computer games A LOT but I've uninstalled basically all games until national mathcounts is over.
I used to be a mega chatter but now I only IM people when I'm bored or when I need to talk about/ask something.

So AOPS, then schoolwork.

I'm usually on Myspace, but that is beginning to be overtaken by the time I spend on AoPS.

[quote="yif man12"][quote="sugarwergoingdown"]as for me, MYSPACE!/chatting

hey do u guys hav myspaces?....mebbe not but theres this really gay virus going round and my profile like decapitated...oh well if you guys know anything about that then can you plz tell me?

Oh and if any of you guys have one, then put it down! (url)

reason for this poll is i just wanted to see what AoPsers do in spare time on net[/quote]

Computer viruses can't be gay, they reproduce asexually by splitting. No sex or infatuations. Sorry to disappoint you.[/quote]

 :rotfl:  :rotfl:  :rotfl:
Im probably on AoPS mostly.

I'm on ML/AOPS most of the time. I keep YM open all the time to know when I get replies here, but I rarely chat with anyone.

I normally chat mostly, and I use xanga sometimes and used to be on AoPS a lot.

[quote="yif man12"][quote="sugarwergoingdown"]as for me, MYSPACE!/chatting

hey do u guys hav myspaces?....mebbe not but theres this really gay virus going round and my profile like decapitated...oh well if you guys know anything about that then can you plz tell me?

Oh and if any of you guys have one, then put it down! (url)

reason for this poll is i just wanted to see what AoPsers do in spare time on net[/quote]

Computer viruses can't be gay, they reproduce asexually by splitting. No sex or infatuations. Sorry to disappoint you.[/quote]

no reason 2 be sarcastic lol......
hmmmm my school like banned myspace....but scru that.
is the situation same 4 ur scools too?
cuz now myspace is on news alot about being "dangerous"
i would like some opinions on this.

[quote="sugarwergoingdown"][quote="yif man12"][quote="sugarwergoingdown"]as for me, MYSPACE!/chatting

hey do u guys hav myspaces?....mebbe not but theres this really gay virus going round and my profile like decapitated...oh well if you guys know anything about that then can you plz tell me?

Oh and if any of you guys have one, then put it down! (url)

reason for this poll is i just wanted to see what AoPsers do in spare time on net[/quote]

Computer viruses can't be gay, they reproduce asexually by splitting. No sex or infatuations. Sorry to disappoint you.[/quote]

no reason 2 be sarcastic lol......[/quote]

Of course, of course. But then again I could say that there's no reason to use gay as a derrogatory term. Or was that the point of my response in the beginning? The questions never end...

after i do my homework, i like to play a game called Battlefield 2.

I like playing on kurnik.org. Sometimes I'm reading news. In the meanwhile chatting :)
I had some time ago a weblog but I had troubles with this so deleted this.

I am in an Virtual Online Acadamy (CAVA and Apex Learning)  So I spend 6-7 hours a day on schoolwork online :rotfl:  :rotfl:

I don't see how a virus would ruin someone's life...most don't even leave a hard hit on the infected computer.

I used to do random online account games before the era of math in my life, then I would chat a lot and now I mainly just do AoPS


algebra
polynomial
induction
algebra unsolved

let f(x) and g(x) be polynomials with non-negative integer coefficients, and let m be the largest coefficient of f(x), suppose that there exist natural numbers a<b such that f(a)=g(a) and f(b)=g(b). show that if b>m, then f(x)=g(x).

We have $ g(x)\equal{}f(x)\plus{}(x\minus{}a)(x\minus{}b)h(x)$, $ h(x)\in\mathbb Z[x]$. Write $ f(x)\equal{}\sum_{n\in\mathbb Z}f_nx^n$, $ g(x)\equal{}\sum_{n\in\mathbb Z}g_nx^n$, $ h(x)\equal{}\sum_{n\in\mathbb Z}h_nx^n$. Then $ g_n\equal{}f_n\plus{}abh_n\minus{}(a\plus{}b)h_{n\minus{}1}\plus{}h_{n\minus{}2}$. Since $ g_{n\plus{}2}\geqslant0$ and $ f_{n\plus{}2}\leqslant m\leqslant b\minus{}1$, then for $ n\in\mathbb Z$, $ h_n\geqslant (a\plus{}b)h_{n\plus{}1}\minus{}abh_{n\plus{}2}\minus{}(b\minus{}1)$. It easily follows from this that, for $ n\in\mathbb Z$, $ k\in\mathbb N$, $ h_n\geqslant \frac{b^{k\plus{}1}\minus{}a^{k\plus{}1}}{b\minus{}a}h_{n\plus{}k}\minus{}ab\frac{b^k\minus{}a^k}{b\minus{}a}h_{n\plus{}k\plus{}1}\minus{}(b\minus{}1)\sum_{l\equal{}1}^k\frac{b^l\minus{}a^l}{b\minus{}a}$.

Suppose that there exists $ h_n>0$ and $ N\equal{}\max\{n\in\mathbb N_0\mid h_n>0\}$. Then $ h_{\minus{}1}\geqslant\frac{b^{N\plus{}2}\minus{}a^{N\plus{}2}}{b\minus{}a}\minus{}(b\minus{}1)\sum_{l\equal{}1}^{N\plus{}1}\frac{b^l\minus{}a^l}{b\minus{}a}>0$ which is impossible. Hence all $ h_n\leqslant0$.

Prove by induction that $ h_n\equal{}0$, $ n\in\mathbb N_0$.
1) $ n\equal{}0$. $ 0\leqslant g_0\equal{}f_0\plus{}abh_0<ab(1\plus{}h_0)$, hence $ \minus{}1<h_0\leqslant0$, $ h_0\equal{}0$.
2) Suppose that $ h_0\equal{}h_1\equal{}\ldots\equal{}h_{n\minus{}1}\equal{}0$. Then $ 0\leqslant g_n\equal{}f_n\plus{}abh_n<ab(1\plus{}h_n)$,...

Consequently, $ g(x)\equal{}f(x)$.

[quote="RIP"]It easily follows from this that, for $ n\in\mathbb Z$, $ k\in\mathbb N$, $ h_n\geqslant \frac {b^{k \plus{} 1} \minus{} a^{k \plus{} 1}}{b \minus{} a}h_{n \plus{} k} \minus{} ab\frac {b^k \minus{} a^k}{b \minus{} a}h_{n \plus{} k \plus{} 1} \minus{} (b \minus{} 1)\sum_{l \equal{} 1}^k\frac {b^l \minus{} a^l}{b \minus{} a}$.
[/quote]

Hello, can you show how you get this in detail?

Induction on $ k$. Step of induction:
$ h_n\geqslant\frac{b^{k\plus{}1}\minus{}a^{k\plus{}1}}{b\minus{}a}\bigl((b\plus{}a)h_{n\plus{}k\plus{}1}\minus{}abh_{n\plus{}k\plus{}2}\minus{}(b\minus{}1)\bigr)\minus{}$
$ \minus{}ab\frac{b^k\minus{}a^k}{b\minus{}a}h_{n\plus{}k\plus{}1}\minus{}(b\minus{}1)\sum_{l\equal{}1}^k\frac{b^l\minus{}a^l}{b\minus{}a}.$
That's it.

Thanks,  :D

also, if I have a sequence defined by $ a_1\equal{}x, a_2\equal{}y$ and $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}a_{n}$ for positive integer n, then what is the formula for this sequence. I think it related to your solution.


AMC
AIME
AMC 12
AMC 12 A
USA(J)MO
USAMO
MATHCOUNTS

What's your index? I got 141 on AMC 12A and a 10 on the AIME I, so 241

I'm assuming AMC 12 index.
$ 112.5 \plus{} 90 \equal{} 202.5$

But I'm a sophomore, and 135 AMC 10, so I should get in anyway.

141 AMC 12B, 13 AIME I. So 271 index.

Note: there's nowhere for someone who had a 280-288 to post.

[quote="CatalystOfNostalgia"]Note: there's nowhere for someone who had a 280-288 to post.[/quote]Fixed.

150, 13.

Is this to get some idea of where the index cutoff will be?  I'm guessing that will be around 200-205, maybe?

There's also no choice for less than 190, 199.5, 209.5, 219.5, 229.5, 239.5, 259.5 or 279.5. (It is, as far as I know, impossible to get 299.5. Then again, someone apparently got 101 last year, according to [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/2007--AIMEindex.pdf]this[/url].)

141+10*10 =241...

so i'm happy with that I guess...I made only three careless errors on the aime (YAY)...i spent a while checking so i didn't even attempt those combo problems...

[quote="t0rajir0u"]$ 150 \plus{} 10 \cdot 12 \equal{} 270$.  I'd like to point out that AoPS index polls are notoriously bad at predicting the actual cutoff  :P[/quote]

I think a better method would be the question "What do you think the index cutoff will be?" Many AoPSers have above average indexes but also have an idea of how other people in their schools did, etc.

Also, is there data on previous year's polls and what the indexes turned out? that would be an interesting correlation to take a look at...

150+10*9=240 (cough only 2 stupid mistakes).

Only AMC12 scores count for indices, by the way. 135+10(10)=235.

97.5 + 70 = 167.5  :wallbash:  (down from my 182 last year 132/50)

at least if i had made one more mistake on the AMC12, my index would be 76 points less than that, so I'm somewhat happy...

Fourierseries also got a 300 index this year.

[quote="cognos599"][quote="the future"]yea nice job.
Btw to answer your question, i believe David Benjamin got a perfect index a couple years back.

oh yea and the reason why this poll isnt an accurate representation is because 1. its AoPS but 2. there are alot of AMC 10 scores in here being used as the index.[/quote]

in 2006 i think, david benjamin got a perfect 300, in his 9th grade. that's pretty good, i am surprised that he didn't make IMO.[/quote]

Just because someone does well on the AMC and the AIME doesn't necessarily mean they are superior to others at olympiad problems as well...being able to make IMO, as far as I know, is about mathematical experience (as in, problem experience, not competition experience).

That's like saying all the people good at MathCounts do exceptionally well on the AIME, and vice versa.

[quote="unimpossible"][quote="cognos599"]um i am like the dumbest one here but

2007(7th grade): 137
2008(8th grade): 186

improved 49.  :blush:[/quote]

index is for AMC 12.[/quote]

that's a jump to a conclusion.

For all you know, cognos could have taken the 12 in both years if her/his school didn't offer it, and also if he/she took it at a college that for some reason only offered the amc 12.

[quote="mchoi815"]
that's a jump to a conclusion.

For all you know, cognos could have taken the 12 in both years if her/his school didn't offer it, and also if he/she took it at a college that for some reason only offered the amc 12.[/quote]
that's also a jump to a conclusion too (that that was a jump to a conclusion) because it's possible that unimpossible personally knows cognos599

you are correct randomdragon, unimpossible does know who i am and he knows i took 10 both years. if he didn't know who i was, then mchoi815 is correct he was jumping to conclusions. btw cognos is a he.

Cognos's signature also indicates that he took the AMC10 this year, and it's unlikely that someone would take the AMC12 one year but only the AMC10 the next.

Another thing to note is that it is not always best to shoot for a perfect index. It is better to focus solely on olympiad problems if you can get 220+. I know that I didn't shoot for a perfect score on the aime, and wasted a fair amount of time checking problems instead of attempting another two. I made an additional three stupid errors but oh well. There is no reason for me to try and improve my AIME test taking ability to get just a bit higher index.

that's because Altheman knows he's gonna pwn on USAMO :D

Yeah, now that I've proved this year that I am capable of making USAMO, pretty much all the prep I'm planning for next year is gonna be olympiad stuff, as I'm atrocious at it right now.
I'll just be shooting for a 125+ on the AMC 12 and 10+ on the AIME next year, enough to get me safely into USAMO.
In case you're wondering, yes, I suck at fast contests like the 12.

Yeah, same here...since I'm pretty sure I can make it into USAMO. I'll probably start doing a lot more olympiad prep for next year, since this past year, I haven't really done much except AIME prep. I'll probably be shooting for 144/120 for AMC 10/12 and 9+ for AIME. I also suck at faster contests, like the AMCs.

$ 123 \plus{} 50 \equal{} 173$

$ \text{phailure}$

5 b/c of stupid mistakes. could have been 8.

[quote="physisnu"]Well....what about this?  :)

150 on the AMC 10B, and 8 on the AIME  II,

[b]Index: 230[/b]

OMG I made 2 STUPID mistakes on the AIME and could've gotten 1 more right by guessing..

I put 502 for # 8 you know hehehehehe....b.b

[b]Anyway, I am an 8th grader...does grades seriously influence the possibility for USAMO?[/b][/quote]

No, and most people could have increased their scores by guessing correctly.

looks like the poll should've focused much more on the 170 to 230 range than 200-300 :)


If S is a set of positive integers less than or equal to 100 with no two distinct elements of S summing to an element of {7, 12, 33, 45, 69, 81}, then the maximum number of elements S can have is

a) 56
b) 60
c) 64
d) 65
e) 68

This is a guess...

[hide]For my purposes, there can be three numbers (1,3,7) less than twelve and over twelve, we can include all positive odd integers. That would include 13 through 81 (for my purposes) and that includes 35 integers.  We can also include anything between 82 and 100 because they are already bigger than any number in set S. So, $19+35+3\approx56$ which is $a)$.  (I approximated because there might have been one that wouldn't work that I included?)[/hide]

[hide]I can easily create a set of 60 elements that works, namely the positive integers between 41 and 100, inclusive (the sum of any two is at least 83).

Trying to add another element to the set would require taking away at least one element first, so it doesn't seem possible to increase the size of the set. So it is [b](B)[/b] then?[/hide]

Oh yeah.  I didn't think of that.  Scorpius, I think you're right.

I think that scorpius is right too. That was a lot easier than I thought it was...

The answer is (b). Nice solution, scorpius  :)


If $ (1\plus{}\tan(1^{\circ}))(1\plus{}\tan(2^{\circ}))(1\plus{}\tan(3^{\circ}))\ldots(1\plus{}\tan(45^{\circ}))\equal{}2^N$, find the sum of the digits of $ N$ (obviously, no calculator).

[quote="gfour84"]If $ (1 \plus{} \tan(1^{\circ}))(1 \plus{} \tan(2^{\circ}))(1 \plus{} \tan(3^{\circ}))\ldots(1 \plus{} \tan(45^{\circ})) \equal{} 2^N$, find the sum of the digits of $ N$ (obviously, no calculator).[/quote]

We have $ (1\plus{}tan x^{\circ})(1\plus{}tan (45^{\circ}\minus{}x^{\circ}))$
$ \equal{}(1\plus{}tan x^{\circ})(1\plus{}\frac{tan 45^{\circ}\minus{}tan x^{\circ}}{1\plus{}tan 45^{\circ}tan x^{\circ}})$
$ \equal{}(1\plus{}tan x^{\circ})(1\plus{}\frac{1\minus{}tan x^{\circ}}{1\plus{}tan x^{\circ}})$
$ \equal{}2$

So $ (1 \plus{} \tan 1^{\circ})(1 \plus{} \tan 2^{\circ})(1 \plus{} \tan 3^{\circ})\ldots(1 \plus{} \tan 45^{\circ})$
$ \equal{}[(1 \plus{} \tan 1^{\circ})(1 \plus{} \tan 44^{\circ})] \cdot [(1 \plus{} \tan 2^{\circ})(1 \plus{} \tan 43^{\circ})] \cdot  \ldots \cdot [(1\plus{}\tan 22^{\circ})(1\plus{}\tan 23^{\circ})] \cdot (1 \plus{} \tan 45^{\circ})$
$ \equal{}2^{22}(1\plus{}1)\equal{}2^{23}$

So sum of digits of $ N$ is $ 5$.

[hide]

We have
 $ (1 \plus{} \tan 1^{\circ})(1 \plus{} \tan 2^{\circ})...(1 \plus{} \tan 43^{\circ})(1 \plus{} \tan 44^{\circ})(1 \plus{} \tan 45^{\circ})$.

Then, the last term is $ 1 \plus{} 1 \equal{} 2$.

Consider:

$ (1 \plus{} \tan 1^{\circ}) (1 \plus{} \tan 44^{\circ}) (1 \plus{} \tan 2^{\circ}) (1 \plus{} \tan 43^{\circ})...$.

Rewriting  $ (1 \plus{} \tan 44^{\circ})$ as $ (1 \plus{} \tan (45^{\circ} \minus{} 1^{\circ}))$, we find it equals

$ 1 \plus{} \frac{1\minus{} \tan 1^{\circ}}{1 \plus{} \tan 1^{\circ}} \to \frac{2}{1 \plus{} \tan 1^{\circ}}$.

We notice that the terms cancel in pairs:

$ (1 \plus{} \tan 1^{\circ} ) * \frac{2}{1 \plus{} \tan 1^{\circ}} \equal{} 2$.

There are $ 22$ such cancellations, meaning the expression so far equals $ 2^{22}$.

Multiplying on the final two makes the final expression to be $ 2^{23}$.

Thus, $ N \equal{} 23$, and the sum of the digits of $ N$ is $ 2\plus{}3 \equal{} 5$.
[/hide]

By the way, where did you get this problem?

THis question is from Purple Comet 2003.


function
integration
trigonometry
calculus
limit
complex analysis
calculus computations

why does
$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$
because the infinite sum
$\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}$
yields the differential equation for the power series
$x^{2}\ddot{y}+x\dot{y}= \frac{1}{1-x}$
the homogeneous solution would be
$c_{1}ln{x}+c_{2}$
I won't do out the nonhomogeneou solution but it definately does not involve any $\pi$'s especially $\pi^{2}$'s

can anybody explain?

You do not have to use differential equations for calculating this sum.
The function you are playing with is the famous polylogarithmic function of order two or the so-called dilogarithm function $Li_{2}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}=-\int_{0}^{z}\frac{\ln(1-t)}{t}dt$.
Now it is known that
$Li_{2}(1)=\frac{\pi^{2}}{6}$.
Or

$\sum_{n=1}^{\infty}\frac{\cos n\theta}{n^{2}}=\frac{\pi^{2}}{6}-\frac{\theta(2\pi-\theta)}{4}$, for $0\leq \theta\leq 2\pi$.
Set $\theta =0$ and you should get the desired result.
Didi
PS Even if you play with differential equations when you calculate the nonhomogenous part still you will not get anything involving $\pi^{2}$ , however when you  set $x=1$ you get some integrals from 0 to 1 and after some manipulations the result follows , but this approach is long and not efficient.

alright, so what general methods does one have of calculating sums, i know differential equations, and some sums can be expressed as reimann sums - what about with definite integrals or are there any other tools(maybe with complex analysis even though i don't know it)?
oh, and p.s. when doing out the differential equation on my own, part of my answer did include that integral, however I quit before I finished it out.

well, in general there are some classical methods as for instance

1) complex integration involving special choosen functions

2) Mellin transform

3) Riemann integrals via Riemann sums

4) Fourier analysis

However, in my opinion there is no general technique that can be applied for calculating sums, each problem should be considered as a new problem and, in order to solve it, of course if the problem is hard enough, try to look at the problem from a point of view that no one before has viewed it, namely come up with nonstandard techniques---but this means that one should have trained before with the so called standard results.

Alright, thanks, it's obvious I have a lot to learn considering I only know reimann sums.  Could you use those listed techniques to evaluate nonpower series like
$\sum_{n=0}^{\infty}x^{\frac{1}{n}}$
I don't know if this is even possible, sorry if I am supplying absurd problems

Your last series is divergent. Think how its terms look like when $n$ is large.

Doesn't matter. $\lim_{n\to\infty}x^{1/n}=1$ for any $x>0.$ (And you have definition problems for $x\<0,$ but that's another story.)


In $\triangle{ABC}$, we have $\displaystyle\frac{AG}{DE}=\frac{AH}{BC}$. Can we imply that $DE || BC$?

[hide]no.  e.g., take $AH$ to be the angle bisector, and take $\triangle ABC$ non-isosceles...[/hide][/hide]

[hide="no"]no, for the same reason as pleurestique[/hide]


Alice, Bennita, and Chamique are in jail, but are nearing the end of their term.  The warden wants to release one of them, but wants them to prove that they are logical human beings, so she devises a little test.

There are 3 black hats and 2 white hats in a box.

Alice, Bennita, & Chamique each reach into the box and place one of the hats on her own head. They cannot see what color hat they have chosen. The first woman to correctly identify the color of her hat goes free.

The women are situated in a way that Alice can see the hats on Bennita & Chamique's heads, Bennita can only see the hat on Chamique's head, and Chamique cannot see any hats. When Alice is asked if she knows the color of the hat she is wearing, she says no. When Bennita is asked if she knows the color of the hat she is wearing, she says no. When Chamique is asked if she knows the color of the hat she is wearing, she says yes and she is correct.

What color hat and how can this be?

[hide]Since Alice can't tell what her hat is, Bennita & Chamique must either have both black hats or one white and one black. If they were both white, then Alice would know she has a black hat, since only black hats are left.
Since Bennita can't tell what her hat is, Chamique's hat couldn't of been white, (in the one white, one black case) because that would mean Bennita would know she has black. Therefore Chamique's hat is black.[/hide]


I get the error
[Doc open("%bm.pdf)][File open("%bm.pdf]
Cannot execute the command

when I try "build and view current file"
I recently downloaded adobe reader 9.1, and I've had this error. before downloading, it was perfectly fine
Any ideas why this isn't working?

I believe that this is due to the fact that the execute command (for Adobe Reader) is not linked to the correct location.

How do I fix this?

Nevermind, I figured it out, by going to format, output details or something.

Thanks though:) It works now


(Mathscope 262.5)
Without using a calculator, determine:

$\sin(\frac{\pi}{14})+6\sin^2(\frac{\pi}{14})-8\sin^4(\frac{\pi}{14})$.

--
(me)

Is there a generalization for simplifying $\sin(\frac{\pi}{n})$ for real $n$?

[hide]I don't know if there is...  :D good luck![/hide]
(edit: Changed sign... thanks alex  :) )

[quote="cincodemayo5590"](Mathscope 262.5)
Without using a calculator, determine:

$\sin(\frac{\pi}{14})+6\sin^2(\frac{\pi}{14})+8\sin^4(\frac{\pi}{14})$.

--
(me)

Is there a generalization for simplifying $\sin(\frac{\pi}{n})$ for real $n$?

[hide]I don't know if there is...  :D good luck![/hide][/quote]

the problem is

$\sin(\frac{\pi}{14})+6\sin^2(\frac{\pi}{14})-8\sin^4(\frac{\pi}{14})$

and to answer your other question, there is no simple solution...these problems work specifically on the numerator being a certain value, graph $f(x)=\sin\frac{\pi}{x}$, it only attains simple values rarely

[hide]I got $\frac{1}{2}$ using triple-angle formula for sine and sum-to-product formula for sinx+siny  :? [/hide]


linear algebra
matrix
algebra
system of equations

[quote="jli"][quote="ProblemSets"]How would you solve this system of equations?

(1) a+b+c+d=119
(2) a+b+c+e=130
(3) a+b+d+e=134
(4) a+c+d+e=149
(5) b+c+d+e=152

Thanks

_______________________________

ok, i figured it out:

a,b,c,d,e = 19, 22, 37, 41, 52

sorry for bothering[/quote]

Add them all together and you get 4(a+b+c+d+e)=?

Then divide by 4 to get a+b+c+d+e=?/4

Then subtract equation 1 from ?/4 to get e.[/quote]

yes, that's the way i would do it. to make it easier to comprehend, i usually line up the variables in each equation (it sounds petty, but i found it helps in more complicated things).

EDIT:
[quote="7h3.D3m0n.117"]or you can do cramer's rule  :D[/quote]

i tried looking up cramer's rule in Wikipedia, and there was nothing on the subject. what is cramer's rule?

no, if you arrange it, you get four a's, b's, c's, d's and e's. therefore, you can just factor out the four and carry on from there....

[quote="vishalarul"][quote="7h3.D3m0n.117"][quote="Art of Owna"]cramers rule?  But for that problme..wouldnt you have to calculate a determinant for a 5X5 matrix?[/quote]

5x6 if ur using a cheap shortcut: reverse matrices on the calculator =)[/quote]

How do you calculate the determinant of a $5 \times 6$ matrix?

You can only calculate the determinants of square matrices.

I think you are talking about the augmented matrix method.[/quote]

yeah...that's it...im not thinking =)

[quote="mwpl11"]don't laugh, but what are determinants? i remember knowing once upon a time...[/quote]

 :rotfl: [url=http://en.wikipedia.org/wiki/Determinants]Click here[/url]

I think [url]http://mathworld.wolfram.com/Determinant.html[/url] gives a better description  :wink: .

A classic.
[hide="Solution"] Add all five equations together, giving

$4a + 4b + 4c + 4d + 4e = 684$

Divide both sides of the equation by $4$:

$a + b + c + d + e = 171$

Now subtract each of the original five equations from this one to get the variables' values.

Subtracting 1st equation from: $e = 52$
Subtracting 2nd equation from: $d = 41$
Subtracting 3rd equation from: $c = 37$
Subtracting 4th equation from: $b = 22$
Subtracting 5th equation from: $a = 19$

Therefore, the solution is $\boxed{(19, 22, 37, 41, 52)}$. $\square$[/hide]


geometry
geometric transformation
reflection

[quote="shobber"]Can you please help me find something??
I want to know something about the USA high school education system.
I also need to know the "ridiculus" cases in the USA. (they give the lawyers price if they win it)(I don't know if i used the correct word. I hope you understand it.)[/quote]

shobber,

I am not sure that I understand your questions. What do you want to know about high school in the USA? and how is that related to "lawyers" and their cases?

[quote="shobber"]They are two seperated problems...
I am giving my classmates an introduction about the USA. So I think the education system can interest them, and the ridiculus cases shall make them enjoy it and concentrate more on my presentation.

Thanks in advance![/quote]

Okay.

1. High school in the USA is typically four years long. The years of education are called grades. There are twelve grades of compulsory (required) education altogether; high school is the last four grades (9-12). After graduation at the end of the twelefth grade students can either get jobs and become workers or enter university for more study.

Twelve years of education are required everywhere in the USA, but the national government does not make the rules for it. The rules about what will be taught and in what way are made by the governments of the individual states of which there are 50.  Usually what subjects are taught is pretty much the same from one state to another, but the way in which these subjects are taught varies widely from place to place. So it is quite possible to have two people both educated in the USA, having the same number of years of school, studying essentially the same material, and yet one will be better educated than the other because of where each went to school.

2. For your second question, I don't feel qualified to comment on what you call "rediculous" cases, I'm not a lawyer. Though there are many cases where someone has sought restitution in the American courts for what most people consider silly reasons. It's a problem. Some states have even made laws that say if you go to court for a silly reason and lose you have to pay a fine or go to jail.

As to the type of case you referred to in your original post, sometimes a lawyer will agree to represent a person on a "contingency" basis; this means if the lawyer does not win then you don't have to pay him. Of course if he does win, then his fee will be a percentage of the sum you are awarded, typically 1/3 to 1/2.

I don't suppose you have many lawsuits in China? Here if one person wrongs another but what they've done is not a crime then the one wronged can go to the court and ask that the person who wronged them compensate them for their pain and suffering. If the wronged person can prove his case, or rather if his lawyer can, then court orders the other to pay money to the wronged person. It is out of this money that a lawyer will take his percentage in a "contingency" case.

shobber,

There are many ways, and many ridiculous ways, in which the USA is unusual compared to other countries.
But I don't know if lawyers and the legal system are the best example or the first example to explain.

The USA has essentially the Anglo-Saxon law (the old "common law" in England) together with some modifications derived from the US constitution.   The "ridiculous" phenomenon you mention is that in cases involving large amounts of money, lawyers are often paid something like 0.30 to 0.50 of the amount (in case of victory).  The procedures for paying lawyers are not extremely different between the USA and other Western countries. What is different is that the amounts of money awarded by a court are often much higher in the USA, because:

(1) the amounts are usually decided by a [i]jury[/i] instead of a [i]judge[/i] (this is not always true, but is the basic difference in the USA compared to the rest of the world).  For most cases, the judge does not decide the result; the judge is a sort of manager of the legal process for the case, and the decision is made by a jury (a group of citizens selected to decide the case).

(2) the laws determining what a company/corporation/person have to pay, are different in the USA.  For example, one can bring a case against a doctor for bad treatment ("malpractice"), and there is no limit on the amount that one can claim in damages.  There are huge political controversies about this in the United States ("tort reform").

By the way, there is a big case recently, that illustrates the differences in USA and Chinese legal systems.
A Chinese electronics company is bringing a case against its American partner for something like $400 million.
This lawsuit is in the USA, because in China it would be hard to win large sums of money.  What is possible (and has happened) in China is simply that the Chinese government held the president of the company (an American Chinese) in jail when he visited China, without trial --- he "disappeared" for a few months.   I don't know if he has been released.

For a summary of US education system see

http://www.math.niu.edu/~rusin/papers/teaching-math/usa

i recall there was one case where a lady spilt mcdonalds coffee on her pants and she sued mcdonalds for making their coffee too hot (mcdonalds lost) she won

and there was another one where some guy put a laptop on his lap for like 24 hours and when he woke up his :censored: was very heated (u know what im talking about) and he sued the company and he won

both cases had very expensive lawyers. and btw there are 13 grades not 12 (k-12)

Well, kindergarten is not mandatory.  BTW the whole ridiculous cases thing is actually quite politically/historically important, as it reflects some of the bad sides of the US government policy.  Maybe you should talk about something else?

[quote="noneoftheabove"]... and btw there are 13 grades not 12 (k-12)[/quote]

Kindergarten (the first year of school, around age 5) is not compulsory.

[quote="Dr. No"]Kindergarten (the first year of school, around age 5) is not compulsory.[/quote]

This may vary from locality to locality.  Education is not compulsary through high school, by the way -- you can drop out at 18 (age of majority) or 16 (with parent's permission).  I also believe that in New York (and perhaps elsewhere), you are entitled to be educated until the age of 21 or graduation from high school.

American high schools vary very much from locality to locality -- there is no national school system.  Each state, and each school district within each state, is responsible to a large degree for all the details of the system: curriculum writing, funding, hiring and credentialling teachers, punishments, etc.  There are also private and parochial (religious, especially Roman Catholic) schools, and a small number of students are educated at home.


This should go in the Mathcounts Forum, where there is a whole topic with resources for practice and past competitions.

[quote="toadoncart"]Has anyone ever had MATHCOUNTS experience from the past years? cuz I need to know some stuff like available resources for studying MATHCOUNTS, how the test rooms look like, and what they provide for you. If you do, thanks a lot! Just post and I'll be grateful enough!  :lol:[/quote]

There are lots of people (usually), the tests are in packets (some can be found [url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47[/url]). They provide scrap paper and pencils.

AoPS vol 1 is a good start to MATHCOUNTS.


Has anyone ever bought stuff from A1books.com before and can say whether its reliable and credible or not? For some reason, many of its books are significantly cheaper than the same titles at amazon and barnes&nobles.

You might try campusi.com, which compiles a list of prices from 60 or so online bookstores -- it's very nice.

You might check to make sure that they aren't used copies.

Yes, that is actually where I found A1Books I think (or maybe another similar price-comparison site, I forget). I was taking Simon's advice and looking into buying [i]A Path to Combinatorics for Undergraduates[/i], but I found that A1Books sells it brand new for $27.57 while everyone else (including amazon and B\&N) sells it for $35-$40, a fact I found somewhat suspicious. So I was just seeing whether anyone else had bought books from A1Books before.

I found a book through a similar site selling also at $10-$15 less than anywhere else I could get it, and I ended up recieving it fine.  So while I can't speak to the specific store, I would say that you're probably all right.

credit cards make everything safe. hehe my parents got out paying a large sofa that was defective. because it needed to be repaired and the compnay was closing so they never reparied and my parents never payed hehe.

[quote="churchilljrhigh"]credit cards make everything safe.[/quote]

Generally, I hold the opposite to be true.

[quote="JBL"]Generally, I hold the opposite to be true.[/quote]
Well, the credit card companies do a lot to stop fraud. If you have charges on your bill that you don't think belong there, you just have to call them up and tell them and they'll investigate.

Yes, but if you pay in cash there's no way you can be defrauded.

Of course, someone could always steal your money.


How many of the first 30 positive integers can be expressed as a sum of two distinct prime numbers?

If you have knowledge of Goldbach's conjecture, you'll know that all even numbers greater than 4 and less than or equal to 30 will work, so that's already 14.

As for odd numbers, you can get $ 2\plus{}3$, $ 2\plus{}5$, $ 2\plus{}7$, $ 2\plus{}11$, $ 2\plus{}13$, $ 2\plus{}17$, $ 2\plus{}19$, and $ 2\plus{}23$, which contributes 9, so $ 14\plus{}9\equal{}\boxed{23}$

If you don't know Goldbach's conjecture, then you can just use casework based on the smaller of the two addends.

The problem with using Goldbach's conjecture is that it just states that all even integers greater than 2 can be expressed as two, not necessarily distinct, primes.

So we simply count the integers that can only be expressed as two of the same prime: 4=2+2, 6=3+3. So there are $14-2 = 12$ even integers that satisfy the conditions.

You are right for the odd numbers, except you miscounted: There are 8, not 9.
So the final answer is $8 + 12 = \boxed{20}$.


V=6
F=5
E=9

So answer is 20.

For n-sided polygon prism answer is 6n+2.


Sabet konid baraye har $n$ moadeleye zir dar adade tabiyi javab darad
$\sum_{i=1}^{n}x_{i}^{2}=y_{n}^{2}$
ke dar an $x_{i}$ ha adadhaye motemayezand .

Look for rational points on unit sphere in $\mathbb{R}^{n}$


I used to have a real account here. Now I am Yahoo! Slurp. I can't log out of this account, someone please help me log out.

EDIT: okay...I didn't think that this message would be posted. I just unsubscribed yahoo! slurp from the bot list. Before that, I wasn't able to post any messages at all.

I notice I have now posted in the wrong forum. Sorry to the administrator and can someone please kindly move this post to an appropriate sub-forum? Thanks.

firstly, you should hav cookies enabled (you should already have this)

secondly, you should be running internet explorer; sometimes other browsers do not let you log on

thirdly, install a firewall(if you did not already have one)

fourthly, scan your computer for viruses (if you have not already done so)

once you log out, use your 'real account'

I have done the first two. I'm not sure if I have a firewall or not.

With this account, I cannot send PMs...being a bot is so restrictive!

Um what's a bot if the bot doesn't do anything?
Just download several firewalls.
Usually cable/DSL providers and routers have a firewall or you can enable a firewall.
The Yahoo! and Google toolbars have firewalls (I think).
Haha yahoo firewall to block yahoo bot.

[quote="mysmartmouth"]You should probably PM [b]Valentin Vornicu[/b] about this problem, as it doesn't seem to be something we encounter daily.  He is the person who deals with a lot of the technical stuff.[/quote][quote="Yahoo! Slurp"]With this account, I cannot send PMs...being a bot is so restrictive![/quote]

A mod should move this to the "Support" forum.

Valentin Vornicu helped me out with my profile problem.
I'm sure he can help you too!

Massive necroposting  :huh: ?

http://en.wikipedia.org/wiki/Internet_bot

This is funny, however raises a nice issue about updating bot ips. Problem should be solved now. If you would tell us your username that would also be great Yahoo Slurp! :)


Encuentra todas las soluciones en numeros reales positivos $a,b,c,d$ para el siguiente sistema:

$a+b+c+d=12$
$abcd=27+ab+ac+ad+bc+bd+cd$

Primero, si $ab=k_{1}$, $ac=k_{2}$, $ad=k_{3}$, $cd=l_{1}$, $bd=l_{2}$, $bc=l_{3}$, $\sum_{i=1}^{3}k_{i}=x$, $\sum_{i=1}^{3}l_{i}=y$, entonces

$\sum (a-b)^{2}\geq 0\Leftrightarrow 3\sum a^{2}\geq 2\sum ab \Leftrightarrow 3\sum a^{2}+6\sum ab\geq 8\sum ab \Leftrightarrow$
$\Leftrightarrow 2\times 3^{3}=\frac{3(a+b+c+d)^{2}}{8}\geq x+y$
Por otra parte $a,b,c,d<9$, pues si, pongamos, $a\geq 9$, entonces $1\geq \left (\frac{b+c+d}{3}\right)^{3}\geq bcd$ y $a\geq abcd=27+\sum ab>27$, lo que es imposible, al ser $a+b+c+d=12$ y $b,c,d>0$.
Como $\forall i$, $k_{i}l_{i}=abcd$, las dos sucesiones estan dispuestas la una en orden de crecimiento inverso a la otra; de esto y de la primera observación se deriva:
$-81+3(x+y)\leq 27+x+y=abcd=\frac{1}{3}\sum_{i=1}^{3}(k_{i}l_{i})\leq$
$\leq \frac{1}{3}\frac{ (\sum_{i=1}^{3}k_{i})(\sum_{i=1}^{3}l_{i}) }{3}=\frac{xy}{9}$

$-3^{6}+3^{3}x+3^{3}y\leq xy \Leftrightarrow (x-27)(y-27)\geq 0$.
 $(x>27 \vee y>27) \rightarrow x+y>54$, por tanto, ambos factores serán no positivos y $x,y\leq 27$.

En particular $ab+ac+ad\leq 27 \Leftrightarrow a(12-a)\leq 27 \Leftrightarrow (a-6)^{2}\geq 9 \Leftrightarrow (a\leq 3 \vee a\geq 9)$.
 Por la observación del principio sólo puede ser $a\leq 3$ y, por simetría $a,b,c,d\leq 3\Rightarrow a=b=c=d=3$, que es la única solución.

la condicion es equivalente a:
$a+b+c+d=12$ y $2abcd=54+(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)$
esto es lo que necesitamos:
$(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)\ge12\sqrt{abcd}$...........(*)

$(\frac{a+b+c+d}{4})^{4}\ge abcd$..........................................(**)

$\sqrt[4]{abcd}\ge\frac{4}{\frac{abc+adc+abd+bdc}{abcd}}$...........................................(***)

y $(a+b+c+d)^{3}\ge16(abc+adc+abd+bdc)$..............................(****)

por (*) tenemos $2abcd\ge54+12\sqrt{abcd}$ por lo que $abcd\ge81$
y por (**) tenemos $81\ge abcd$ por lo tanto $abcd=81$
por(***) tenemos $abc+adc+abd+bdc\ge108$ y por
(****) tenemos $108\ge abc+adc+abd+bdc$ por lo que
$abc+adc+abd+bdc=108$
y debido a que $ab+ac+ad+bc+bd+cd=54$
entonces las soluciones a,b,c,d estan dadas por las raices del polinomio
$x^{4}-12x^{3}+54x^{2}-108x+81$ las cuales son $(3,3,3,3)$

[quote]$(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)\ge\sqrt{12abcd}...........(*)$
[/quote]
Como das esto por cierto???

[quote]$(a+b+c+d)^{3}\ge16(abc+adc+abd+bdc)..............................(****)$[/quote]

Y esta parte??

Creo que ninguna de las dos es obvia...

en la (*) me confundi al escribirlo pero ya lo corregi y tienes razon debi poner por que son ciertas:
bueno, (*) es por $MA \ge MG$
(**)por lo mismo
(***)por $MG \ge MH$
(****)una forma facil de demostrarlo es por muirhead si no pues con las desigualdades de las medias.
y ya quedan demostradas.


Hi,

I'm trying to solve this one with no result. Hope someone can give me a hint (not necessary a complete solution).

A prime power is an integer of the form $p^n$ where p is a prime and n > 1 is an integer. Find all pairs of prime powers that differ by 1. Prove your answer.

Thanks in advance,

Thanks Seamusoboyle for your hint. Actually I factored them, but I looked over them carelessly.

I think there's a solution, which is: $3^2 - 1 = 2^3$

And to prove that there's no other pair, I did like this:

$p^n - 1 = q^m$ <=> $(p-1)(p^{n-1} + p^{n-2} + ... + p + 1) = q^m$, since p is a prime, if p > 3, p is odd, then p-1 is even and is in the form 2k with k be any integer. Then $2 | q^m$ but that means $2 | q$ which is impossible since q is also a prime.

Any suggestion?

Use [url=http://mathworld.wolfram.com/CatalansConjecture.html]Catalan's Conjecture[/url] which has been proved recently.

I didnt know the catalan conjecture was proved! Thats so cool! Mathematicains everywhere breath a sigh of relief! Does anyone know how he went about proving it?

I posted a link to a prove some time ago: http://www.mathlinks.ro/Forum/viewtopic.php?t=32508 (post $5$)

[quote="ZetaX"][quote="t0rajir0u"]Denote the first prime $p$ and the second prime $q$.  We desire $p^n - q^m = 1$ for $n, m > 1$.  By FLT, for $n, m > 2$ there cannot be a solution (:D) so we only need to check $n, m = 2 \implies p^2 - q^2 = 1 \implies (p+q)(p-q) = 1$ which obviously has no solution.[/quote]
The use of FLT is simply wrong.[/quote]

Oops - $n \neq m$.   :blush:


Buenas gente,
solo para informarles de este link:

http://www.diamante.com.pe/file/carlosbr/

donde encuentran:
- eboooks
- exams
- instaladores
- etc.

PD: Esto solo estara por unos dias ..
arovecharlo  ;)

Carlos Bravo ;)
LiMa - PERU


category theory
advanced fields
advanced fields unsolved

Let $ X$ be a separated scheme over an affine scheme $ S$. Let $ U$ and $ V$ be open affine subsets of $ X$. Prove that $ U \cap V$ is also affine.

[hide="Hint"]Can you combine the affine (?) scheme $ U\times_SV$ with some other scheme to get $ U\cap V$ in terms of some diagram. First think of ordinary sets, then this will certainly translate to schemes. ;)[/hide]

[quote="Megus"]Third Edition: thank you for a hint, but I still can't do it  :blush: the problem is I don't know actually from what I can deduce affinity of a set (besides: closed subscheme of an affine scheme is affine but that doesn't help here), so could you help me further?[/quote]Good guess, since it does actually and is the essential point here: Can you see a morphism $ U\cap V\rightarrow U\times_SV$ somewhere at the horizon which turns out to be a closed immersion?[hide="Hint"]Start with the closed immersion $ \Delta: X\rightarrow X\times_SX$. ... and: Think of this thingies as of sets first. That makes it easier. ;)
[hide="Solution:"] The following diagram is Cartesian:
\[ \begin{array}{ccc}
U\cap V&\rightarrow&U\times_SV\\
\downarrow&&\downarrow\\
X&\rightarrow&X\times_SX
\end{array}\]To show this reduce to the category of sets, by Yoneda's lemma.
[/hide]
[/hide]

Upon request here a more detailed solution on this applying a technique using Yoneda's lemma. The technique allows to reduce to the case that that everything can be thought to be an ordinary set. Not to mention with that this is a benefit of [i]abstract nonsense[/i]. After running through this process in depth, it is pretty easy in mathematical every-day life to achieve results like in this post.

It is a bit lengthy, but it goes beyond things needed here. But nevertheless I hope it can be helpful for those you are interested.

[quote="Third Edition"]The following diagram is Cartesian:
\[ \begin{array}{ccc} U\cap V & \rightarrow & U\times_SV \\
\downarrow & & \downarrow \\
X & \rightarrow & X\times_SX \end{array}
\]
To show this reduce to the category of sets, by Yoneda's lemma.[/quote]
[hide="I. General remark:"]
Let $ \mathcal C$ be a category (I do NOT consider any set theoretical twists here).

Let $ X$ be an object in $ \mathcal C$. Then, for every object $ T$ in $ \mathcal C$, define $ X(T)$ to be the set of all morphisms from $ T$ to $ X$, i.e. $ X(T)\equal{}Mor(T,X)$. Every morphism $ v: T\rightarrow X$ in $ X(T)$ is called a $ T$-valued point in $ X$. Then $ X(.)$ defines a contravariant functor on $ \mathcal C$ with values in the category $ (Sets)$ of sets.

Every morphism $ u: X\rightarrow Y$ in $ \mathcal C$ induces a natural transformation of functors $ u(.): X(.)\rightarrow Y(.)$. Denote the set of natural transformations $ X(.)\rightarrow Y(.)$ by $ Mor(X(.),Y(.))$.[/hide]
[hide="II. Yoneda s lemma:"]
Now Yoneda's lemma guarantees that every natural transformation $ X(.)\rightarrow Y(.)$ of functors comes from a (unique) morphism $ X\rightarrow Y$. In more formal terms: The map \[ \left\{\begin{array}{ccc}
Mor(X,Y)&\rightarrow&Hom(X(.),Y(.))\\
u&\mapsto&u(.)
\end{array}\right .\]is a bijection.[/hide]
[hide="III. Applications of Yoneda s lemma - together with exercises ;) :"]

Let $ \mathcal C$ be a category.

1. Let $ F$ be a contravariant functor from $ \mathcal C$ to $ (Sets)$. Given two objects $ X$ and $ Y$ in $ \mathcal C$ together with natural equivalences between $ F$ and $ X(.)$, $ F$ and $ Y(.)$, then $ X$ and $ Y$ are isomorphic. Moreover there is exactly one isomorphism linking the two given natural equivalences together by composition.

This is the disguised version of the fact that a universal object is unique up to unique isomorphism. $ F$ is called [i]representable[/i]. [Exc.: Try to prove and give a list of examples.]

2. Let $ X,Y$ be objects in $ \mathcal C$. Show that the functor $ F(T)\equal{}X(T)\times Y(T)$ is contravariant. What would you call an object representing $ F$?

3. Let $ S$ be an object in $ \mathcal C$ and let $ \mathcal C_{/S}$ be the category (check) where objects are morphisms $ u: X\rightarrow S$ in $ \mathcal C$ and a morphism in $ \mathcal C_{/S}$ from $ u: X\rightarrow S$ to $ v: Y\rightarrow S$ is a morphism $ f: X\rightarrow Y$ with $ v\circ f\equal{}u$.
Given two objects $ X\rightarrow S$ and $ Y\rightarrow S$ in $ \mathcal C_{/S}$, what would you call an object representing - if it exists - the functor \[ F(T)\equal{}X(T)\times_{S(T)}Y(T)\,.\]
4. Let $ (Sch)$ be the category of schemes. Given a scheme $ X$ and open immersions $ i: U\rightarrow X$ and $ j: V\rightarrow X$ in $ (Sch)_{/X}$, show that $ U\cap V$ represents the functor $ F(T)\equal{}U(T)\times_{X(T)}V(T)$.[/hide]
[hide="IV. More detailed solution:"]
After the introductory remarks the result will drop out of the abstract nonsense:

That the commutative square is Cartesian means that the left upper object is the fiber product of $ X\rightarrow X\times_S X$ and $ U\times_S V\rightarrow X\times_S X$ in $ (Sch)_{X\times_SX}$. Since the fiber product is a representing object of a functor (see III.3), it suffices to check by "evaluating" the square at every $ T$-valued point. In other words, given an arbitrary object $ T$ in $ \mathcal C$, we need to show that the following diagram is Cartesian in $ (Sets)$:
\[ \begin{array}{ccc} (U\cap V)(T) & \rightarrow & (U\times_{S}V)(T) \\
\downarrow & & \downarrow \\
X(T) & \rightarrow & (X\times_{S}X)(T) \end{array}
\]
Plugging in the [u][i][b]natural[/b] [/i][/u] bijections \[ (U\cap V)(T)\equal{}U(T)\times_{X(T)}V(T)\] \[ (U\times_SV)(T)\equal{}U(T)\times_{S(T)}V(T)\] and \[ (X\times_SX)(T)\equal{}X(T)\times_{S(T)}X(T)\] the diagram reads
\[ \begin{array}{ccc} U(T)\times_{X(T)}V(T) & \rightarrow & U(T)\times_{S(T)}V(T) \\
\downarrow & & \downarrow \\
X(T) & \rightarrow & X(T)\times_{S(T)}X(T) \end{array}
\]But this diagram can easily be checked to be Cartesian.
[/hide]

Thank you very much, it's wonderful  :) Everything's clear now to me.

@third: I'm glad to see that you think the same about the yoneda lemma as me :-). but this seems to be common under algebraic geometers and everybody who wants to be it some day.

a generalization: let $ X \to S$, $ Y \to S$, $ S \to T$ morphisms in some category. then the canonical diagram

$ \begin{array}{ccc} X \times_S Y & \rightarrow & X \times_T Y \\
\downarrow & & \downarrow \\
S & \rightarrow & S \times_T S \end{array}$

is cartesian. this seems not to follow from simpler rules concerning fibred products, e.g. transitivity. the universal property can be directly verified: a commutative diagram

$ \begin{array}{ccc} P & \rightarrow & X \times_T Y \\
\downarrow & & \downarrow \\
S & \rightarrow & S \times_T S \end{array}$

corresponds, via the universal properties of the fibred products, to a triple of morphisms $ P \to S$, $ P \to X$, $ P \to Y$ such that $ P \to X \to T = P \to Y \to T$, $ P \to X \to S = P \to S$ and $ P \to Y \to S = P \to S$, i.e. to a pair of morphisms $ P \to X$, $ P \to Y$ such that $ P \to X \to S = P \to Y \to S$, i.e. to a morphism $ P \to X \times_S Y$. that's it. in this case, I would not use yoneda: the case of sets just uses the same arguments as above.

Here's "one line" solution from Mumford's red book: $ U \cap V \equal{} \Delta ^{ \minus{} 1} (U\times_SV)$,where $ \Delta$ is diagonal morphism coming from separatedness of $ X$ hence $ \Delta$ is closed immersion and $ U\times_SV$ affine so  $ U \cap V$ is affine.

this equation is nothing else than the cartesian diagram discussed above. so mumford hides many details: the equation is clear as sets, but we have to recoqnize this restriction of $ \Delta$ as a closed immersion, so you have to check that this comes from a base change of the original diagonal.

I totally agree with -oo-. In fact I had this equation in mind in my first post. But you need to get the categorical machinery - or something else -  moving to see that this simple equation makes sense in the category of schemes and yields an equality of [b]schemes[/b], i.e. as, e.g.,  locally ringed spaces. This needs some reasoning even if you know that $ U\times_SV$ is an open subscheme of $ X\times_SX$ and you take the preimage under the closed immersion $ \Delta$.


calculus
integration
logarithms
calculus computations

Evaluate $ \int_{0^{\plus{}}}^{1}\frac{x^{x}\left( x^{2x}\plus{}1\right) \left( \ln x\plus{}1\right) }{x^{4x}\plus{}1}dx$

Answer [hide]$ \equal{}0$[/hide]

I'm thinking that $ \int_{0}^\frac {1}{e} \equal{} \minus{} \int_\frac {1}{e}^{1}$. Gotta go. Have a good evening (or is it about $ 2: 30\,$a.m.in Singapore?).

Consider I =  $ \int \frac {x^{x}\left( x^{2x} \plus{} 1\right) \left( \ln x \plus{} 1\right) }{x^{4x} \plus{} 1}dx$

$ x^x \equal{} y \ \Rightarrow \ x^x(lnx\plus{}1)dx \equal{} dy$

I = $ \int \frac{y^2\plus{}1}{y^4\plus{}1} dy \equal{} \int \frac{1\plus{}\frac{1}{y^2}}{y^2\plus{}\frac{1}{y^2}}dy$

= $ \int \frac{1\plus{}\frac{1}{y^2}}{(y \minus{} \frac{1}{y})^2 \plus{} 2} dy$

= $ \frac{1}{\sqrt{2}} tan^{\minus{}1}(\frac{y \minus{} \frac{1}{y}}{\sqrt{2}})$

= $ \frac{1}{\sqrt{2}} tan^{\minus{}1}(\frac{x^x \minus{} \frac{1}{x^x}}{\sqrt{2}})$

Applying the limits :

$ \frac{1}{\sqrt{2}} [ tan^{\minus{}1}(\frac{x^x \minus{} \frac{1}{x^x}}{\sqrt{2}}) ]_{0^\plus{}}^1$

= $ \frac{1}{\sqrt{2}} \times 0$

= $ \boxed{0}$

I am kicking myself. Well done Nora-91. (That's one I won't forget in a hurry.)

:)

[b][color=blue]Thank You,[/color][/b] AndrewTom and IndoChina !!


trigonometry
function
LaTeX
ceiling function
floor function
algebra

Simplify:

$ Arcsin(sin(x))$

Write it in another way.

$ \arcsin (\sin x) \equal{} x, 0 \le x <2\pi$, right?

I believe it should be $ \minus{}\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$. You have to be careful with arcsines, because like sin(pi/12)=sin(11pi/12), and we want it to be well defined.

[quote="archimedes1"]$ \arcsin (\sin x) \equal{} x, 0 \le x < 2\pi$, right?[/quote]

$ \arcsin (\sin x) \equal{} x, \minus{}\frac{\pi}{2} \le x < \frac{\pi}{2}$

But I want a simplification in $ R$  :)

[quote="13375P34K43V312"]I believe it should be $ \minus{} \dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$. You have to be careful with arcsines, because like sin(pi/12)=sin(11pi/12), and we want it to be well defined.[/quote]

No, $ D(sin)\equal{}R$  :)

[quote="archimedes1"][quote="13375P34K43V312"]I believe it should be $ \minus{} \dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}$. You have to be careful with arcsines, because like sin(pi/12)=sin(11pi/12), and we want it to be well defined.[/quote]
oops

what is a simplification in $ R$?[/quote]

$ R$ is the set of REAL numbers I mean

For $ {\pi\over 2}\leqslant x\leqslant {3\pi\over 2}$ you have $ \arcsin(\sin x)\equal{}\pi\minus{}x$. Hence you need to reduce real $ x$ to interval $ \left(\minus{}{\pi\over 2},{3\pi\over 2}\right)$ and you can do it by using the argument $ z\equal{}x\minus{}2\pi\left[{x\over 2\pi}\plus{}{1\over 4}\right]$.

Therefore

$ \arcsin(\sin x)\equal{}\begin{cases}z & \minus{}\pi/2\leqslant z \leqslant \pi/2\\ \pi\minus{}z & \pi/2\leqslant z\leqslant 3\pi/2\end{cases}$

where $ z$ is defined as above.

Farenhajt , can you explain  your solution better?  :wink:

[quote="gauss1181"]This problem is wwaaayyy too easy... It should have been in the High School Basics forum.

Anyways, arcsin(sin x)=x iff x is in the interval [-pi/2, pi/2][/list][/hide][/code][/quote]

Please read posts before yours  :)

[quote="FOURRIER"][quote="Farenhajt"]using the argument $ z \equal{} x \minus{} 2\pi\left[{x\over 2\pi} \plus{} {1\over 4}\right]$.
[/quote]

What do you mean by this?

$ z$ is a complex number ?
Does the $ [$ means $ ($??[/quote]

$ z$ is just a new variable, needed for later calculations.

$ [\cdot ]$ is floor function ($ [x]$ is the largest integer not exceeding $ x$).

As for how to obtain the expression, we must find $ k$ such that

$ \minus{}{\pi\over 2}\plus{}2k\pi\leqslant x<{3\pi\over 2}\plus{}2k\pi$

$ 2k\pi\leqslant x\plus{}{\pi\over 2}<2\pi\plus{}2k\pi$

$ k\leqslant {x\over 2\pi}\plus{}{1\over 4}<k\plus{}1$

from where it follows

$ k\equal{}\left[{x\over 2\pi}\plus{}{1\over 4}\right]$, so $ z\equal{}x\minus{}2k\pi$ as defined before.

[quote="Farenhajt"][quote="FOURRIER"][quote="Farenhajt"]using the argument $ z \equal{} x \minus{} 2\pi\left[{x\over 2\pi} \plus{} {1\over 4}\right]$.
[/quote]

What do you mean by this?

$ z$ is a complex number ?
Does the $ [$ means $ ($??[/quote]

$ z$ is just a new variable, needed for later calculations.

$ [\cdot ]$ is floor function ($ [x]$ is the largest integer not exceeding $ x$).

As for how to obtain the expression, we must find $ k$ such that

$ \minus{} {\pi\over 2} \plus{} 2k\pi\leqslant x < {3\pi\over 2} \plus{} 2k\pi$

$ 2k\pi\leqslant x \plus{} {\pi\over 2} < 2\pi \plus{} 2k\pi$

$ k\leqslant {x\over 2\pi} \plus{} {1\over 4} < k \plus{} 1$

from where it follows

$ k \equal{} \left[{x\over 2\pi} \plus{} {1\over 4}\right]$, so $ z \equal{} x \minus{} 2k\pi$ as defined before.[/quote]

Now I see, Its actually a good idea.

But its not my simplification, Ill post mine later

Heres my simplification

$ arcsin(sin(x))\equal{}(\minus{}1)^k(x\plus{}k\pi)$

Where $ k\equal{}[\minus{}\frac{x}{\pi}\plus{}\frac{1}{2}]$

I "simplified" like this:

            $ \arcsin{\sin{x}} \equal{} x \minus{} [\frac {2x}{\pi}] \plus{} \frac {\pi}{4} \plus{} ( \minus{} 1)^{[\frac {2x}{\pi}]}\frac {\pi}{4}$

where $ [x]$ is the integer part of $ x$

 :)

By the way, is there a Latex symbol for the maximun integer function??

___________________________________________________

Thanks AstroPhys!

$ \left\lceil x \right\rceil$ \left\lceil x \right\rceil
$ \left\lfloor x \right\rfloor$ \left\lfloor x \right\rfloor.


1. Let $ r$ be a number such that $ r \plus{} 1/r$ is an integer. Prove that for every positive integer
$ n$, $ r^n \plus{} 1/r^n$ is an integer.

2. Show that for all $ n$, $ 2^{3^{n}} \plus{} 1$ is divisible by $ 3^{n \plus{} 1}$. [color=green]Edited by mod.[/color]

[quote="bigman"]

2. Show that for all $ n$, (2^3^{n + 1} is divisible by $ 3^{n \plus{} 1}$.[/quote]

NOTE:
The OP's post was edited by a moderator.

for the first one,
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=286311

There is something wrong with your question number 2!

[hide="2 in its edited form"]Use the factorization $ x^3 \plus{} 1 \equal{} (x \plus{} 1)(x^2 \minus{} x \plus{} 1)$ repeatedly to get $ n \plus{} 1$ factors of $ x^{3^n} \plus{} 1$, all of which are $ 0 \pmod{3}$ when $ x \equal{} 2$ is put in.[/hide]

[hide="The induction"]The base case is $ n\equal{}0$, where we have $ 3^1|2^{3^0}\plus{}1$, or $ 3|3$, which is true.

By the inductive hypothesis, $ 3^{n\plus{}1}|2^{3^n}\plus{}1$. For the inductive step, we shall now factor $ 2^{3^{n\plus{}1}}\plus{}1$. Since this is in the form $ x^3\plus{}1$, we can factor it as $ (x\plus{}1)(x^2\minus{}x\plus{}1)$. Doing this, we have $ 2^{3^{n\plus{}1}}\plus{}1\equal{}(2^{3^n}\plus{}1)(2^{2\cdot 3^n}\minus{}2^{3^n}\plus{}1)$. So we now at least know that $ 3^{n\plus{}1}|2^{3^{n\plus{}1}}\plus{}1$. Now all we have to do left is to prove that $ 3|2^{2\cdot 3^n}\minus{}2^{3^n}\plus{}1$. However, observe that by the inductive hypothesis, $ 2^{3^n}\equiv 2\pmod 3$. Therefore, $ 2^{2\cdot 3^n}\minus{}2^{3^n}\plus{}1\equiv 1\minus{}2\plus{}1\equiv 0\pmod 3$, so we have reclaimed an additional factor of 3 and we are done.[/hide]

Can you explain the part from "So we know..." to "However"

$ 3^{n\plus{}1}|2^{3^n}\plus{}1$ from the inductive hypothesis.

Also, we have shown that $ 2^{3^{n\plus{}1}}$ is the product of $ 2^{3^n}\plus{}1$ and $ (\text{other ugly expression})$, and so if we show that $ (\text{other ugly expression})$ is divisible by 3, then we've fulfilled the inductive step, because then we have $ 3^{n\plus{}2}|2^{3^{n\plus{}1}}$.


i need help with these pigeonhole questions...

14 integers are chosen from {1,2,3,...,28}. Prove that there exist 4 of the chosen integers that can be split into 2 groups of 2 with the same sum.

A child drinks at least 1 bottle of milk a day. Given that he has drunk 700 bottles of milk in a year of 365 days, prove that he has drunk exactly 29 bottles of milk in some consecutive days.

Let us first split it up into 14 groups,     1-28, 2-27, etc. This makes 14 groups. From these 14 groups we could at most have 15 number otherwise, we can add 2 pairs to be 29. Now let us pair then up so that we get 28-2, 27-3, etc. This makes 13 groups with 15 and 1 left over. From these groups we could at most take 1 from each and 2 from one of them. This means we could say that out of 1-28-2, we could at most chose 2. from 1-28-2-3-27, we could at most chose 3. etc. We get that out of all the number we could chose at most 13 such that no pair adds up to be the same number.
still working on second.


Tom, Dick and Harry started out on a $ 100$-mile journey. Tom and Harry went by automobile at the rate of $ 25$ mph, while Dick walked at the rate of $ 5$ mph. After a certain distance, Harry got off and walked on at $ 5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:

$ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ \text{none of these answers}$

[hide]
way too hard to explain with words...
Lets look at Harry, he travels for $ t_1$ hours at $ 25mph$ then for $ t_2$ hours at $ 5mph$, and his final displacement is $ 100$ miles

$ 25t_1 \plus{} 5t_2 \equal{} 100 \qquad(1)$

Now Tom walk at $ 5mph$ for $ t_1$...then tom and harry are $ 25t_1 \minus{} 5t_1 \equal{} 20t_1$ miles apart.
Then Dick drives back, and gets gloser to Tom at $ 30mph$,$ \Rightarrow$ it takes Dick $ t_3 \equal{} \frac {2t_1}{3}$ hours to reach Tom. In that time Tom has walked $ 5\cdot \frac{2t_1}{3}$ miles.

So now Tom and Dick are in the car at time $ \frac{5t_1}{3}$ and are $ 100 \minus{} \frac {25t_1}{3}$ miles from the finish, traveling at $ 25mph$...$ t_4 \equal{} \frac {100 \minus{} \frac {25t_1}{3}}{25} \equal{} 4 \minus{} \frac {t_1}{3}$

OKAY; $ t_3 \plus{} t_4 \equal{} t_2 \Longrightarrow \frac {2t_1}{3} \plus{} 4 \minus{} \frac {t_1}{3} \equal{} 20 \minus{} 5t_1$ from $ (1)$
Now $ t_1 \equal{} 3$
$ \Rightarrow 25(3) \plus{} 5t_2 \equal{} 100 \Longrightarrow t_2 \equal{} 5$

$ \Rightarrow t_1 \plus{} t_2 \equal{} 5 \plus{} 3 \equal{} 8 \longrightarrow \boxed{D}$!?!

Hope you understood that... :) [/hide]

Hmm... I think you're right, but I have a much less computational way of doing it.
[hide="Solution"]Consider Dick's journey. He walks for some distance, then goes by automobile for the rest of the way. Now consider Harry's journey. He goes by automobile for some distance, then walks for the rest of the way. They both take the same time to get to the end, so the time/distance they spend walking or going by automobile must match. (Think of it this way: If $ w$ is the time walked, and $ a$ is the time on automobile, and $ t$ is the total time, then both Dick and Harry have equations satisfying $ 100 \equal{} 5w \plus{} 25a \iff 20 \equal{} w \plus{} 5a, w \plus{} a \equal{} t$, which has a unique solution.)

Therefore, if Harry gets out of the car when he is $ 5w$ miles away from the destination, then when Tom meets Dick, they will be at the point that is $ 5w$ away from the beginning.
So, Tom's total distance is 100 [going all the way] + 2(100-10w) [overlaps the middle area two times to pick up Dick], and he has the rate of 25 the whole time, so $ 25t \equal{} 100 \plus{} 2(100 \minus{} 10w) \iff 5t \equal{} 60 \minus{} 4w$.

Now we have a nice simple system of 3 equations:
$ 20 \equal{} w \plus{} 5a 5t \equal{} 60 \minus{} 4w t \equal{} a \plus{} w$
Substitute the 3rd equation into the second, reducing it to a 2-equation system:
$ 20 \equal{} w \plus{} 5a 5(a \plus{} w) \equal{} 60 \minus{} 4w$
The second equation is equivalent to $ 5a \plus{} 9w \equal{} 60$, and subtracting $ 5a \plus{} w \equal{} 20$ from it, we have $ 8w \equal{} 40 \iff w \equal{} 5$. Now plugging this into $ 5t \equal{} 60 \minus{} 4w$ gives $ t \equal{} \boxed{8}$.[/hide]

Here's my solution:

Let $ t$ be the time when Harry got off; the car has traveled $ 25t$ while Dick traveled $ 5t$. The difference is $ 20t$. So it takes $ T \equal{} 20t/(15 \plus{} 5)$ for Tom and Dick to meet. Meanwhile Harry takes $ w \equal{} (100 \minus{} 25t)/5$ to walk to the destination.

So we have $ 5t \plus{} 5T \plus{} 25(w \minus{} T) \equal{} 100$.
Solving for $ t$ we get $ t \equal{} 3$. So the answer is $ 3 \plus{} w(3) \equal{} 3 \plus{} (100 \minus{} 25*3)/5 \equal{} 8.$

PS: I thought there are only 30 problems on the AHSME...

Wow, I spent forever wondering why there would be infinitely many solutions until I realized it was a 100 mile jouney  :ewpu:
The best way in my opinion is to synchronize the time.
After x hours, Tom and Harry have gone forward 25 x and Dick has gone forward 5x.
Now we need to find where Tom and Dick meet. We find they meet when Dick has gone forward (10/3)x and Tom has gone back (50/3)x, so Harry has also gone forward (10/3)x in this unit of time.
Now it is easy to see that Harry goes forward another 5x and Tom and Dick go forward another 25x in the same unit of time, and they are both at the same distance.
The total distance is 100=25x+(10/3)x+5x, so x=3.
But Dick went 5x and (10/3)x at the same rate, so it took him 3 and 2 hours, and then he went 25x at five times the rate, another 3 hours. So the answer is 8.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ be a prime integer, determine the number of $ p\minus{}Sylow$ groups in the permutation group on $ p$ letters.

The p-sylow subgroup is of order $ p$, so it suffices to find order $ p$ elements. They are just cycles of length $ p$. There are $ (p\minus{}1)!$ many of them. Notice that a cyclic group of order $ p$ has $ p\minus{}1$ generators. Therefore there are $ (p\minus{}2)!$ many of p-Sylow subgroups.


abstract algebra
vector
superior algebra
superior algebra unsolved

Are there any other relationships between $ Ker f$ and $ Im f$, for $ f$ a liniar aplication, exept that $ |G|\equal{}|ker f| |Im f|$  for $ f: G\minus{}>G$ morphism, $ G$ group?

this one doesn't hold either if your group is not finite :)

No it holds for arbitrary groups (but then, the statement is mostly uninteresting).
You also should better describe it as $ G/ \mathrm{ker} f \cong \mathrm{im} f$, as this additionally describes the group structure.

I'll add that the previous statement is also true for morphisms of rings, modules, vectors spaces, algebras...

I don't know the most general formulation of the theorem.  But it's quite general.


This is from Dummit and Foote's section on the group isomorphism theorems: If $ H \triangleleft G$ with prime index $ p$ and $ K \leq G$, then either (i) $ K \leq H$ or (ii) $ G \equal{} HK$ and $ |K : K \cap H| \equal{} p$.

[hide="hint"]
We will show that if $ K$ is not a subgroup of $ H$ then $ G\equal{}HK$ and $ |K: K \cap H|\equal{}p$
1. Note that $ HK/H$ is a subgroup of $ G/H$, so $ |HK: H|$ divides $ p$, but the only possible thing is $ |HK: H|\equal{}1$ or $ |HK: H|\equal{}p$, but $ |HK: H|\equal{}1$ implies $ HK\equal{}H$, contradict the fact that $ K$ is not a subgroup of $ H$, then $ |HK: H|\equal{}p$, $ HK/H$ isomorphic to $ G/H$, $ HK/H x H$ isomorphic to $ G/H x H$, finally $ G$ is isomorphic to $ HK$, but $ HK$ is a subgroup of $ G$, then $ G\equal{}HK$. Now $ HK/H$ is isomorphic to $ H/H \cap K$, so that $ |H: H \cap K|\equal{}p$. We finish our proof.
[/hide]

I don't think that $ G$ being isomorphic to its subgroup $ HK$ means $ G \equal{} HK$. After all $ \mathbb{Z}$ is isomorphic to $ 5\mathbb{Z}$ but of course $ \mathbb{Z} \neq 5\mathbb{Z}$. We didn't assume $ G$ is finite.

[b] Qruni [/b] got the idea. It is just the problem of communication.

Notice that if $ [HK : H] \equal{} p$ then because of a well-known identity:
\[ [G: HK] [HK: H] \equal{} [G: H] \equal{} p\]
we deduce $ [G: HK] \equal{} 1$ which is nothing but $ G\equal{}HK$.

[quote="mathangel"][b] Qruni [/b] got the idea. It is just the problem of communication.

Notice that if $ [HK : H] \equal{} p$ then because of a well-known identity:
\[ [G: HK] [HK: H] \equal{} [G: H] \equal{} p
\]
we deduce $ [G: HK] \equal{} 1$ which is nothing but $ G \equal{} HK$.[/quote]

or $ [HK: H] \equal{}1$...

[quote="vict85"]

or $ [HK: H] \equal{} 1$...

[/quote]

I think that mathangel omitted the analysis of that case because Qruni's treatment of [b]it[/b] is not that long-winded after all.

What mathangel [b]really[/b] left out of his argument was the verification of the fact that $ [K : K \cap H ] \equal{} p$.

For the sake of completion, I present below the proof of this fact.

[hide]
The second isomorphism theorem allows us to ascertain that $ K/ (K\cap H) \simeq KH / H \equal{} HK / H \equal{} G/H.$ Hence,

$ [K : K\cap H] \equal{} \left|K/ (K\cap H)\right| \equal{} \left| G/H \right| \equal{} [G: H] \equal{} p$

and we're done.
[/hide]


The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle.  Many triangles are formed when this is done.  What is the ratio, expressed as a common fraction, of the area of one the smallest triangles formed to the area of the original triangle? Express your answer as a common fraction.

after connecting the points there are 9 triangles since they are formed in a equilateral triangle, they are all congruent. small to big 1/9.

hmmm.... are all of them from the test we took? if so... i dont have to look at them...


Is anyone else going to Furman this weekend? My school actually has seven people going!

Walton will be there. Small group, but ought to be OK. Claus, Scott, Henry, and Chris.

Ah, I remember that Furman was on the same day as Lassiter last year...not so much the same this year.

Hmm, I did not, in fact, go...How'd you do?

Someone from Georgia won! Alas, it was not me. (Now wouldn't THAT be interesting! :-) It was Eric. He did a really good job.

What was the test like? Is it easy? Hard?

Could some1 pm me a copy?


:? -  I have looked everywhere for help with this problem-can you help show me how to set this up and solve it?

A ship left port and sailed east at the rate of 20 miles per hour.  One hour later a second ship lef the same port at the rate of 25 miles per hour, also traveling east.  In how many hours did the second ship overtake the first ship?

Thanks

[hide]After one hour, ship 1 went 20 miles (because time*rate=distance).  The second ship goes 5 miles per hour faster, so every hour, the second ship catches up 5 miles.  Therefore, it takes $\frac{20}{5}$ or 4 hours after ship 2 leaves to overtake ship 1.[/hide]


I am using internet explorer (how do you tell what version it is?), but I don't think it is related to that.

I posted several solutions to problems on week one, then I tried to post on problem 7 (posting a way to solve without a calculator).  I had written up my response, and clicked on preview (like I always do before posting), and it came up with a message that said
"General Error!
a bunch of stuff refering to where I was trying to post
cannot post, or something like that"

Put the link to the topic that you tried to reply to.

It worked this time, but I was wondering what the general error was about.

When I was posting on week 2 problem 10, this error came up:

General Error

SQL requests not achieved

DEBUG MODE

SQL Error: 2006 MySQL server has gone away

SQL Request: SELECT p.*, pt.post_subject, pt.post_sub_title, pt.post_text, pt.bbcode_uid FROM phpbb_posts p, phpbb_posts_text pt WHERE p.topic_id = 98274 AND pt.post_id = p.post_id ORDER BY p.post_time DESC, p.post_id DESC LIMIT 0, 20

Line : 383
File : class_posts.php

I will post the URL in my next message, becuase I had to copy that instead of the URL.

Here is the URL.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98274[/url]

Just refresh the page when that error appears (I don't know yet what's causing it, but it's not the forum code). There's probably a bottle-neck somewhere on the way.

[quote="ZetaX"]I just want to add that I got that error sometimes, too.[/quote]Yeah, but I have no cause right now. I assume there's something wrong with the my.cnf file on the server. I will check it by next week. Just refresh when you see that.

I've gotten that a couple of times - it doesn't seem to matter what forum or subforum - and each time, when I simply submitted the exact same thing again, it went through the second time. (Or didn't, if my dollar signs didn't balance, but in any case normal behavior.) So that would be my advice if you see it: ignore it and submit again.

I fiddled with some mySQL settings; let me know if you still see this.

Change to FireFox, then this problem vanishes ;)

Otherwise: just mark the text and save it somehow (either by Strg-C or better by putting it somewhere else).

[quote="b-flat"]How do you submit it again (without retyping it)?  On my computer, when I press back, the stuff I typed in was gone.  Does refreshing allow you to keep the things you typed?[/quote]

Usually, yes, hitting the refresh button will re-post form data.  (Doesn't work with some forms - I've never tried it in the forum.  Be careful to check that you haven't double-posted).  Do please post here if you're still getting the weird SQL error.

If you hit refresh within 10 seconds of the appearence of the error you will not be able to double post ;)

Not exactly the same but similar:
I tried to edit the first post of http://www.mathlinks.ro/Forum/viewtopic.php?t=5981 (bad headline), but after sending I got:
"WRONG_NOTIFICATION_MODE"
Nonetheless, the post seems to be successfully edited now...

Yeah, a minor glitch. When you write 1000 lines of code, this is bound to happen :D

I got the SQL error when I was previewing some $\LaTeX$ in the Test forum.

SQL requests not achieved

DEBUG MODE

SQL Error: 2006 MySQL server has gone away

SQL Request: SELECT c.class_id FROM classes c, enrollments e WHERE e.user_id = 8949 AND c.start_date <= DATE_ADD(NOW(),INTERVAL 7 DAY) AND c.end_date > DATE_ADD(NOW(),INTERVAL -50 DAY) AND e.dropped = 0

Line : 256
File : functions_valentin.ph

Nothing big, just hit refresh when that happens.

I just want to say that I got more or less the same when previewing today:

SQL requests not achieved

DEBUG MODE

SQL Error: 2006 MySQL server has gone away

SQL Request: SELECT p.*, pt.post_subject, pt.post_sub_title, pt.post_text, pt.bbcode_uid FROM phpbb_posts p, phpbb_posts_text pt WHERE p.topic_id = 100777 AND pt.post_id = p.post_id ORDER BY p.post_time DESC, p.post_id DESC LIMIT 0, 20

Line : 385
File : class_posts.php

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=away+server+mysql+2006+error+sql&t=97606

Nothing important, just a busy server. Hit refresh.


还 有 几 天 就 是 我 们 中 国 人 的 最 热 闹 的 时 间 ? 你 们 准 备 这 么 庆??? 哈 哈 ? 年 我 ? 能 看 看 晚 会 。 。 。

Ok,

? 喜 ? 财 & 身 体 ? 康 & 新 年 快 樂?

美国有什么晚会?

现在CCTV的晚会越?越差了...大家都?很爱看了...远?如90年代?的时候.

[quote="spang"] 你 们 准 备 这 么 庆??? 哈 哈 ? 年 我 ? 能 看 看 晚 会 。 。 。 [/quote]

我的事情比较多, 三??夜放烟?, 爆竹. 接下?7,8天全跑亲戚. ??玩?, 大鱼大肉应有尽有 :D . ?过从??看CCTV的所谓春节?欢晚会, 太无?了.

[quote="Alexander Lenin"]我的事情比较多, 三??夜放烟?, 爆竹. 接下?7,8天全跑亲戚[/quote]
亲戚这么多... :lol:

香港?能放爆竹...  :(

yeah..有香港人.. i speak cantonese.=)

大家在美国哪里啊?我在加州...
希望大家新年进步..希望我今年?以过AIME ?..唉好难啊..哈哈

[quote="yinjiaqiu"]大家在美国哪里啊?我在加州...
希望大家新年进步..希望我今年?以过AIME ?..唉好难啊..哈哈[/quote]

我也是一样，我在麻州（Massachusetts）。我今年最大的目标就是过 AIME.

[quote="beta"]我也是一样，我在麻州（Massachusetts）。我今年最大的目标就是过 AIME.[/quote]

?你好?..
我是没什么机会了.. 我sophmore year ?开始这方?的数学.. 今年目标?是想?对5-7题.. 希望能?功?..

[size=150] [color=red]这么多中国人哟??拜大年了。 :D  :D  :D  :D  :D[/color][/size]

T.T
第四次在美国过年了..都是一样的无?..唉..
好想回家..=[

大家春節好呀~~~~
很快便到年?一了~~

yinjiaqiu,
你是哪裡人?

大家春节好.
在此拜年 :)
(已??一了)

我比较幸?,我在自己家里过年
哈哈,我从没去过美国,
大家新年快?呀
 :?  :lol:  :P  :P  :blush:

[quote="yinjiaqiu"]i live in HK before, and im cantonese. 潮州人[/quote]

潮州..
我的?貫正是潮州呢

哇..唔係啊嘛..甘岩都得? 哈哈..man..we are homie. haha.. 新開心啦..
msn: yinjiaqiu@hotmail.com
得閑chat ?啦.


I'm unable to use the [b]large text[/b] tag of $ \text{\LaTeX}$ for equations and the complicated ones appear so small, you'll need a microscope to read that. Any idea anyone?

\displaystyle is the command.
If you use it in fractions, use \dfrac instead of \frac.
Move to LaTeX forum.

For continued fractions, use \cfrac.

Additionally, you can use \dbinom and a few others.


linear algebra
matrix
geometry
vector
superior algebra
superior algebra solved

let $A \in M_n(R)$ be a semi definite symetric matrix such that $A_{i,j} < 0$ for every $i \neq j$.
Prove that $Rank(A) \geq (n-1)$

the proof I read is not that natural: what is yours ?

[quote="Moubinool"]Post the not naatural proof Alek[/quote]
Yes, probably if you post it, we can make sence of it. I still don't have an idea!

[quote="alekk"]let $A \in M_n(R)$ be a semi definite symetric ...[\quote]

What do you mean by semi definite symetric ?

Maybe you mean something like

semi definite positive

or

semi definite negative

[quote="Moubinool"]
[quote="alekk"] let $A \in M_n(R)$ be a semi definite symetric ...[\quote]

What do you mean by semi definite symetric ?

Maybe you mean something like

semi definite positive

or

semi definite negative[/quote][/quote]

I would say he mean positive semidefinite [b]and[/b] symmetric.

I see noone posted a proof and since it happens to have solved a very useful exercise yesterday, I will post a solution (let's move some problems to solved section area!): since $A$ is semipositive definite (or positive, it doesn't really matter), $A$ is the Gramm matrix of a system of vectors $ x_1,...,x_n$. Moreover, it is not difficult to show that the rank of $A$ is equal to the rank of the family $ x_1,...,x_n$. So, it suffices to prove that if $ (x_i, x_j)<0$ for $i,j$ differents then the rank of the family $x_1,...,x_n$ is at least $n-1$. This follows from the following very nice exercise: if $x_1,...,x_s$ are such that $(x_i, x_j)<0$ for all $i,j$ differents then if $ a_1x_1+...+a_s x_s=0$ either all $a_i$ are 0 or none of the $a_i$ is 0. This means that any family with at most $ s-1$ elements is free (the vectors in this family are linearly independent). Thus $x_1,...,x_{n-1}$ are linearly independent and the rank is at least $n-1$.


Find the remainder of $ C^{2008}_{1234}$ divided by $ 7$.

[quote="azjps"][hide]See [url=http://en.wikipedia.org/wiki/Lucas%27_theorem]Lucas' theorem[/url]; $ {5562_7 \choose 3412_7} \equiv 6 \pmod{7}$.[/hide][/quote]

What else can we do without Lucas' theorem? :|


the second one is wrong! (a,b,c)=(1,2,2) is a counterexample I think.
For the first it is equivalent to:
$\sum_{cyc}c(a-b)^{2}\left( (a+b)(a^{2}+b^{2})+c^{2}(a+b)+c^{3}-7abc\right)$
So all we need to show is:
$(a+b)(a^{2}+b^{2})+c^{2}(a+b)+c^{3}\geq 7abc$ which is just AM-GM :)

[quote="Albanian Eagle"]For the first it is equivalent to:
$\sum_{cyc}c(a-b)^{2}\left( (a+b)(a^{2}+b^{2})+c^{2}(a+b)+c^{3}-7abc\right)$[/quote]
You have some mighty factoring skills! The solution is similar to mine. But for me, it is easy to see, because I make the inequality! :P


How far do I have to go in the sum

1 + 1/2 + 1/3 + 1/4 + 1/5 + . . .

until I have a total that is greater than $2\frac{1}{2}$?

(I mean, what is the last term I have to include in that sum to get a sum that is greater than $2\frac{1}{2}$)

This is a pretty interesting question..

what I woud do is change 2 and 1/2 to a mixed fraction which is 5/2

5/2 - 1    = 3/2
3/2 - 1/2 = 2/2
3/3 - 1/3 = 2/3
8/12 - 4/12 = 4/12
4/12 - 3/12 = 1/12
5/60 - 12/60  = wich becomes negative

so it's 12/60 = 1/5

Shinwoo, you made a tiny mistake there -- why don't you try it again?  (Your method is good!)

Ah!!!

I will fix it after the foot ball game :D

Go STEELERS!!

you know,i sadi I would fix my mistake after the football game, but I never fixed it :blush:

so, what did I do wrong? :(  :?

Look at the values you subtracted -- first you took out 1, then 1/2, then 1/3, then what?

i think JBL is right, i don't see where 4/12 came in

[quote="GoBraves"]Why are so many old topics coming back???[/quote]

Old topics are starting to come back because math92 is going back to all the old questions in this forum and posting on them even though they are old.


LaTeX
quadratics
number theory proposed
number theory

Let $ p$ be a prime, $ A = {p - n^2|n\in\mathbb N, n^2<p}$. Show that we can find integers $ a$ and $ b$ in $ A$ such that $ a > 1$ and $ b$ is divisible by $ a$.

[quote="t0rajir0u"]You mean $ A \equal{} \{ p \minus{} n^2 : n \in \mathbb{Z}, n^2 < p \}$?[/quote]
Yes. Sorry for using LaTeX incorrectly.  :blush:

Pardon me for introducing my own terminology here, but to save time, I call the set $ \{N \minus{} a^2 : a \in \mathbb{Z}, a^2 \leq N \}$ the quadratic walk of N. So your assertion is that for a prime number p, the quadratic walk of p contains one element that divides another element nontrivially.

I'm also guessing you want to add the restriction p > 5.

$ p \minus{} b^2 \equal{} M * (p \minus{} a^2)$

$ p \minus{} b^2 \equal{} Mp \minus{} Ma^2$

$ Ma^2 \minus{} b^2 \equal{} (M \minus{} 1)p$

M has to be a quadratic residue (mod p).

$ Ma^2 \equal{} (M \minus{} 1)p \plus{} b^2$

$ p \equiv b^2$ (mod M).

$ a^2 \equiv b^2$ (mod (M-1)).

If M is even, then b is odd.

Hmm.

http://www.mathlinks.ro/Forum/viewtopic.php?t=48452 has solutions (note that you need the condition $ p>5$ to satisfy $ a>1$).

  darij


inequalities
geometry
circumcircle
geometry unsolved

A problem to think, but better someone check it on the computer first.

Let $ ABC$ be a triangle and let $ P$ be a point inside $ ABC$. Denote be $ A_{1},B_{1},C_{1}$ the projections
of $ P$ onto triangle's sides. Let $ R,R_{1}$ be the circumcircles of triangle $ ABC$ and $ A_{1},B_{1},C_{1}$, respectively. Prove that

$ 9RR_{1}\geq (AP+BP+CP)(d_{a}+d_{b}+d_{c}).$

and what do $ d_{a},d_{b},d_{c}$ denote in this problem?is $ d_{a}$ the length of the bisector of $ \angle A$?  :maybe:

One more time:

Let $ ABC$ be a triangle and let $ P$ be a point inside the triangle $ ABC$.
Denote by $ A_{1},B_{1},C_{1}$ the projections of $ P$ onto triangle's sides.
Let $ R,R_{1}$ be the circumradii of triangles $ ABC$ and $ A_{1}B_{1}C_{1}$, respectively. Prove that

$ 9RR_{1}\geq (AP\plus{}BP\plus{}CP)(d_{a}\plus{}d_{b}\plus{}d_{c}),$

where $ d_{a},d_{b},d_{c}$  are the distances from $ P$ to triangle's sides.

Observe that for $ P\equiv H, I, O$ the inequality is correct !

Any ideas ?


[quote="mathnerd314"]How would you go about solving a problem like this:
Two circles have radii x and y, and their centers are q units apart. A rope is looped around the outside of the circles (that is, it doesn't cross itself). What is the length of the rope?

And another problem based on one in the GS marathon:
Two circles have radii x and y, and their centers are q units apar. A rope is looped around the circles such that it contains the internal tangents and parts of the circumference (that is, it crosses itself once). What is the length of the rope?[/quote]

Well you would make right triangles with right angles on the rope where the rope leaves the circle.. you can use that to find the length of the rope not touching the circle for both circles. Then you can use those right triangles to find the angles on the center of the circles and then use that to find the exterior angle on the other side of the circle, then use that and multiply it by the radius to find the length of the rope that touches the circumference. I think this works for both situations, you just have to plot different right triangles >.>
yeah i hope that helps

Thanks.

That solution seems to use trig, though, which is bad if you're doing a Mathcounts sprint round or something. Any other solutions?


Amber transferred her money from a simple-interest account to one that paid compound interest annually. Her initial deposit in the new account was $10,950. After 2 years at 5.5%, how much money would she have in the new account?

[hide]12188, i hate consumer math.[/hide]


[hide="Hint"]Just cross-multiply and combine like terms.[/hide]

Ah! It's so simple!
I was looking all over the identities, I forgot something basic.

Thanks!

[quote="Falkor"]I've been racking my brain at this! I need help to prove this.

$\frac{1+sin\theta+cos\theta}{1+sin\theta-cos\theta}=\frac{1+cos\theta}{sin\theta}$

Sorry if this is the wrong section.[/quote]

[hide="Solution"]
$\frac{1+\sin\theta+\cos\theta}{1+\sin\theta-\cos\theta}=\frac{1+\cos\theta}{\sin\theta}$

$(1+\sin\theta+\cos\theta)\sin\theta=(1+\cos\theta)(1+\sin\theta-\cos\theta)$

$\sin\theta+\sin^{2}\theta+\sin\theta\cos\theta=1+\sin\theta-\cos\theta+\cos\theta+\sin\theta\cos\theta-\cos^{2}\theta$

Cancel out:

$\sin^{2}\theta=1+\cos^{2}\theta$
$\sin^{2}\theta-\cos^{2}\theta=1$
This statement is true.
[/hide]

Yea it cancels out but are you sure about that identity?

[quote="anirudh"]
Cancel out:

$\sin^{2}\theta=1+\cos^{2}\theta$
$\sin^{2}\theta-\cos^{2}\theta=1$
This statement is true.
[/quote]

I think you meant:
$\sin^{2}\theta=1-\cos^{2}\theta$
$\sin^{2}\theta+\cos^{2}\theta=1$

Because I think that's what you get when you cancel...

Yiou can use the following remarkable identities :

$\boxed{\ \{\begin{array}{c}\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\\\ 1+\cos x=2\cos^{2}\frac{x}{2}\\\ 1-\cos x=2\sin^{2}\frac{x}{2}\\\ 1\pm \sin x=(sin\frac{x}{2}\pm\cos\frac{x}{2})^{2}\\\ \tan\frac{x}{2}=\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}\end{array}\ }$

You can use the formulas : $\boxed{\ \tan \frac{x}{2}=t\Longrightarrow \{\begin{array}{c}\sin x=\frac{2t}{1+t^{2}}\\\\ \cos x=\frac{1-t^{2}}{1+t^{2}}\end{array}\ }\ .$


How do you solve this problem without trial and error? :?:

What is the positive three-digit integer that is four times the product of its digits?

Thanks a lot !!

[hide]

hint:

100A + 10B + C = 4ABC

and no digit can be zero because then the product would be zero.[/hide]

and note that the number BC must be divisible by 4

A<7 because the largest value of ABC is 729.


The following more general inequality is proved in Algebraic Inequalities (page 341):
If $x,y,z$ are non-negative numbers, no two of which are zero, and $\frac{-3}{4}\leq r \leq \frac 7{8}$, then
$\sum \frac 1{x^{2}+rxy+y^{2}}\geq \frac{4(2r+5)}{(r+2)(x+y+z)^{2}}$.
Moreover, Marian Tetiva proved the inequality for $r=-1$.

[quote="mathboy_cntHDVN"]that is a siminal problem :
prove that :
$\sum\frac{1}{a^{3}+b^{3}}\geq \frac{20}{(a+b+c)^{3}}$[/quote]
Tou can see here! :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=123235


Let $ a$ and $ b$ be an integer numbers such that there are a consecutive integer $ c$ and $ d$ statisfying $ a\minus{}b\equal{}a^{2}c\minus{}b^{2}d$. Prove that $ |a\minus{}b|$ is a perfect square.

If $ p\parallel a-b|$, then $ p|(a^{2}-b^{2})c\pm b^{2}%Error. "Longritharrow" is a bad command.
p|a,p|b$.
Let $ p^{k}|(a,b)$, then$ p^{2k}|a-b$. Therefore $ |a-b|$ is perfect square.

An other method (Don't use exponent )
Case 1:
$ c\equal{}d\plus{}1$
$ a\minus{}b\equal{}d.(a^{2}\minus{}b^{2})\plus{}a^{2}$ (1)
So $ (a\minus{}b)|a^{2}$
Let $ gcd(a,b)\equal{}1$ then
$ a\equal{}km,b\equal{}kn$ which $ gcd(m,n)\equal{}1$ (2)
$ (m\minus{}n)(1\minus{}kd(m\plus{}n))\equal{}km^{2}$ (From (1) ,(2))
$ gcd(1\minus{}kd(m\plus{}n),k)\equal{}1$ then $ k|m\minus{}n$
Other easy to check $ gcd(m\minus{}n,1\minus{}kd(m\plus{}n))\equal{}1$ then $ |m\minus{}n|\equal{}kx^{2}$
So $ |a\minus{}b|$ is a perfect square.
Case 2 :$ d\equal{}c\plus{}1$ has same method.
An other result :There exist infinity solution  of this equation.


On a graph, a lattice point is an ordered pair (x,y) with integers x and y. Exactly 15 lattice point lie strictly in the interior of the triangular reion with vertices (0,0), (N,0) and (N,N), when N>0. What is the value of N? :wallbash_red:  :wallbash:

The points $ (0,0), (N,0),$ and $ (N,N)$ will always form an isosceles right triangle which is a $ 45^\circ \minus{} 45^\circ \minus{} 90^\circ$ triangle. There is a pattern of N and the number of lattice points in the interior.

$ [N,lattice points] \equal{} [3,1], [4,3], [5,6], [6,10], [7,15]$  Yahtzee! $ [7,15]$ works, so $ N\equal{}7$.
The pattern in the lattice points is a triangular sequence, meaning it's going $ , \plus{}2,\plus{}3,\plus{}4...$.

$ N\equal{}7$


[hide]13.50

For this problem its best if you work backwards
the price of the magazine is .75*5=3.75
3.75+0.75=4.50 which is 2/3 of half Sally's total money
sooo 6.75 is the half of Sally's money
so 6.75*2=13.50
[/hide]

[quote="MCrawford"]After spending one-half of her money and losing one-third of the remainder, Sally was left with just enough money to buy a magazine and a candy bar.  The price of the magazine is 5 times the price of the candy bar, which is 75 cents.  How many dollars did Sally have initially?  Express your answer as a decimal.[/quote]

[hide="solution"]
Magazine + Candy Bar = 4.50 dolars
Money before losing 1/3 of the remainder: 6.75 dollars
Original amount if money: [b]13.50 dollars[/b]
[/hide]

Working Backwards:
.75+5(.75)=4.50
That is the price of the candy bar and the magazine together.  4.50 is 2/3 of what number you need next, so you have to divide by 2. then times by 3.  6.75 is that answer.  That is only 1/2 of the final answer.  Double it to get 13.50 as the final answer.


hi, i'm ankit .can anyone please tell me what would
 be a reasonable score in kvpy written exam
 & what is standard of problems in biology
& chemistry

Hi,
I think you are referring to Kishore Vaigyanik Protsahan Yojana. I do not think this is right forum for this qeuestion. It should be in the India forum.The moderators please shift it. People there would be able to answer your question.


well... is 111...11 (91 1s) prime?

( we just need to find k where 10^91==1 (mod k) and k!=9,3)

but how?

sorry actually i mistyped the problem, it will be 93 instead of 91.

oh yeh...... how i could be so stupid...... :blush:


(a) Evaluate the infinite product
\[\prod_{n=2}^{\infty}\frac{n^{3}-1}{n^{3}+1}\]
(b) Show that an infinite product can be transformed into an infinite series by means of the identity $P=e^{\log P}$ and use this to solve (a) in another way.

If we look real hard, we see that the first term is zero, which should direct us to the correct answer.

Wow...I feel cheated. Why would she use such trivial problems? Pretend n starts at 2.


How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different? How many four-digit whole numbers have an even leftmost, an odd second digit, and four different digits?

[hide]For the first one, there are 5 possible first digits, 5 possible second digits, and  then 8 and 7 digits left for the last two respectively.  It is 5*5*8*7=1400.  The next one is the same except the leftmost can't be 0 so it is 4*5*8*7=1120[/hide]

[hide="The first question"]  5 * 5 * 8 * 7= 1400 [/hide]

[hide="The second question"] 4 * 5 * 8 * 7= 1120 [/hide]

[hide]5*5*8*7 = 1400, and 4*5*8*7 = 1120

be careful, 0 cant start the number
[/hide]


If only 4 individuals get to move on from States to Nationals, what is the point of a State Team test?  Do they also give out awards to high scoring teams/individuals?  How many?

in Indiana, they give team awards to top 6

Also, the top team's coach becomes the nat'l coach.

In my state we get awards for top 10 teams and top 20 individual.

Awards for top 6 teams and top 6 individuals.

In Wisconsin, top 4 teams, 10 individuals