I am sending in my solutions by email, but I realized that I have to send in new forms for this year. How should I do it? Should I scan the forms in, or mail them seperately, or what? Thanks

You can do it by mail, but given the deadline, fax would probably be better, although you might get a busy signal. I sent in my forms last year via fax.

So does that mean that the forms need to be sent by the deadline too?

The site says "You must submit the Entry Form with your next set of solutions you submit." It's not that much of a hassle, anyway.

Please just send them in sometime this week.  The deadline is firm for the solutions - the forms can wait a few days.

[quote="towersfreak2006"]That's fine. The implication was that the forms can be delayed if necessary.[/quote]
He said he also sent his solutions at that time...

[quote="Klebian"]Well, it matters on how you define "early this morning"
For example, I sent in MY solutions early this morning as well: 1 am (11 pm PST)[/quote]
Well, he lives in San Diego according to his profile, so early morning would in fact be past the deadline for him.

[quote="kevin gong"]wait, so if I sent both my entry/permission forms and my solutions early this morning, then it won't be accepted?  :([/quote]

I seriously doubt this would be a problem.  A friend of mine emailed his Round 3 Solutions at 5 AM CST last year, and the USAMTS graders accepted them anyways.  I really don't think Richard Rusczyk and Co. will stay up during each deadline and enforce a rigid 12 AM PST code - the USAMTS graders are pretty cool with this kind of stuff.  Nobody really discusses USAMTS solutions before the Math Jam anyways *cough* *cough*.  As long as you get your solutions in before the AoPs crew awakes from their slumber...

That is not to say though, that you should hand in your solutions at 6 AM in the morning.  Turn in your solutions before the official 12 AM EST, because you never know, the graders could be having a late night session one of these days... :roll:

EDIT:  Dude Samson, I totally didn't know you were a USAMTS answer guy!  Ownage...

[quote="lowest_reynolds"]
I seriously doubt this would be a problem.  A friend of mine emailed his Round 3 Solutions at 5 AM CST last year, and the USAMTS graders accepted them anyways.  I really don't think Richard Rusczyk and Co. will stay up during each deadline and enforce a rigid 12 AM PST code - the USAMTS graders are pretty cool with this kind of stuff.  [/quote]
Well, all you have to do is check the date and time on the fax or e-mail to figure out whether it's submitted past the deadline.  Perhaps the USAMTS admins are leniant about late submissions, but it wouldn't be because they don't notice when you've sent it in.

by early morning i meant around 6:00 am, and since i do live in san diego that means i passed the deadline by 6 hours  :(

but i think after it gets postmarked, the graders will only see that it was postmarked "10/11/06", so they'll probably reject it  :noo:

and through all this, i completely forgot that i could fax it last night... :oops_sign:

will they still accept the entry/permission forms though, or will they send the entire envolope back?  :(

We accept or reject late submissions on a case-by-case basis.  In some cases, we may ask for more information from you as to why your submission is late.  Just because your solutions are late doesn't mean that we will automatically reject them, but we're not obligated to accept them either.

In the future, avoid the uncertainly and turn in your solutions on time. :)


algebra
polynomial
number theory
relatively prime
algebra proposed

For [i]all[/i] nonzero integers $ n$?  For [i]some[/i] nonzero integers $ n$?

let's suppose there is rational number $ \dfrac{d}{c}$($ c$ and $ d$ are relatively prime) satisfying $ F(\dfrac{d^2}{c^2}) \equal{} 1$
we know that $ F(x) \equal{} (x \minus{} n^2)G(x)$($ G(x)$ is an integer coefficient polynomial). so $ F(\dfrac{d^2}{c^2}) \equal{} (\dfrac{d^2}{c^2} \minus{} n^2)G(\dfrac{d^2}{c^2})$
let's put $ k$ be a degree of $ F$,
then $ c^{2k} \equal{} c^{2k}F(\dfrac{d^2}{c^2}) \equal{} (d^2 \minus{} c^2 n^2)*c^{2k \minus{} 2}G(\dfrac{d^2}{c^2})$
a degree of $ G$ is $ k \minus{} 1$ so $ c^{2k \minus{} 2}G(\dfrac{d^2}{c^2})$ is an integer....
so $ d^2 \minus{} c^2 n^2$ divides $ c^{2k}$
but $ d^2 \minus{} c^2 n^2 \equal{} (d \minus{} cn)(d \plus{} cn)$, and $ d \minus{} cn$ and $ c$ are realtively prime. so do $ d \plus{} cn$ and $ c$
so $ d \minus{} cn \equal{} 1$,$ d \plus{} cn \equal{} \minus{} 1$ or $ d \minus{} cn \equal{} \minus{} 1$,$ d \plus{} cn \equal{} 1$.
anyway, then $ d \minus{} cn \plus{} d \plus{} cn \equal{} \minus{} 1 \plus{} 1 \equal{} 0$
so $ 2d \equal{} 0$ so $ d \equal{} 0$ which is contradiction.
therefore there is no rational number $ a$ satisfying $ F(a) \equal{} 1$

[quote="kunny"]Let $ P(x)$ be a polynomial with integer coefficients such that $ P(n^2) \equal{} 0$ for some non zero integers $ n.$ Prove that $ P(a^2)\neq 1$ for all non zero rational numbers $ a\neq 0.$[/quote]

Note that integers are also rational numbers and so, choose $ a \equal{} n$ for a contradiction where $ n$ is such that $ P(n^2) \equal{} 0, n\in \mathbb {N}\plus{}\{0\}$


Hey, i go to an academy which specializes on math and science, so i figured what cover could be better that one spelled out with the school's initials (GHAMAS) spelled out via mathematical (mainly parametric) functions.  a little online searching showed that it is possible, but some letters have proved nearly impossible to find.  So i come to all of you and set fourth a challenge- Spell out the entire alphabet using parametric functions.

they don't need to be perfect, but must be close enough to be recognizable.

also, they don't necessarily have to be all parametric, for instance, "A" could be a combination of an absolute function and a constant function.

i'll start- (A)
y= -2|x|+6
and
y= 3 (where -3/2 < x < 3/2)

For S I might try some of the inverse trig functions.


between.benzaldehyde and propanal

when the attacking $ Nu$ (a small one) is
a.strong and i mean very strong at  a.low temp. b.high temp

b.weak at a.low temp. b. at high temp

when the attacking  $ Nu$ (a big one) is

a.strong and i mean very strong at  a.low temp. b.high temp

b.weak at a.low temp. b. at high temp

[hide="Opinion"]On the basis of electronic and stereochemical considerations, propanal is more reactive than benzaldehyde, so for the moment I don't see how nucleophile size and strenght and temperature can change that. At least for the case "big nucleophile" propanal is certainly more reactive.[/hide]

[hide="my opinion"]in the first step the negative charge over o is more distribute than in propanal'sw first step.yes though the activation energy of bezaldehyde is higher but the TS relative to benzaldehyde
is more stable than the TS of propanal.so at high temp when reaction does ocuur in bot the TS may revert back to propanal due lower activation nenrgy than in case of benzaldehyde so.....[/hide]

[quote="Pardesi"]in the first step the negative charge over o is more distribute than in propanal'sw first step[/quote]

I don't understant what you mean by that.


Does anyone know around when the Irmo test will take place? It seems the first round test was given a lot later this year than it usually has, so I was just wondering when we have to go to Irmo...

[quote="joml88"]Email Mr. Rybarczyk about it (actually, I might email him since I was wondering myself).  When I saw him at USC Columbia he said February 18 (next Saturday) but apparently he has yet to notify qualifiers so I'm not so sure he'll stick to that date.  Btw, I think it might be at somewhere like AC Flora this year instead of Irmo.[/quote]

well, i don't think it will be the 18th considering when I talked to him, which was on Wednesday, he said he graded "most" of the tests, and considering we just took ours on Monday, he won't have all of them graded and sent out invitations to Irmo/Flora by the 18th.

Ah, well since you talked to him on Wednesday [i]you[/i] would probably be the best person to know :P  Anyway, I emailed him.  So when I get a response I'll keep everyone updated.

Ok, the response I got says that there will be a meeting on March 18 at Ben Lippen School for all of the people on the All State team.

so there wont be like. a second round qualifier like there was last year?

[quote="joml88"]This is the message I got.  It seems to me that there won't be a second round...

[quote]
The meeting will be on March 18th, at Ben Lippen (this needs to be confirmed
that it will be ok w/the school, but it should be fine).  It will be a
practice round for all those on the All-State team - of course, you're on
the team.[/quote][/quote]

Wow, then that must mean people did really bad on that test, since the cutoff was only 11...that's kind of strange... But yay, we don't have to waste our time on a second round!! :clap:

Do to the fact that many tests were sent in late, we will not have a qualifying round.  The prelimanary exam is the qualifying exam.  I feel that the students that are being invited would have been the students to qualify anyways.  Our first practice will be on March 18th (hopefully at Ben Lippen).


group theory
abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

Could anyone help me on these problems?

1. Let G be a finite group with P a Sylow p-subgroup. Show that $ N_G(P)$ is its own normalizer.
2. Show that there are no simple groups of order 148 or 56.

Thank you very much.
ftr

1. Let $ h \in N_{G}(N_{G}(P))$. Then $ hPh^{ \minus{} 1} \leq hN_{G}(P)h^{ \minus{} 1} \equal{} N_{G}(P)$. Therefore $ P$ and $ hPh^{ \minus{} 1}$ are Sylow p-subgroups of $ N_{G}(P)$. Since all Sylow p-subgroups of $ N_{G}(P)$ are conjugate and $ P\triangleleft N_{G}(P)$, there must be $ P \equal{} hPh^{ \minus{} 1}$. Hence $ h\in N_{G}(P)$.

2. Let $ n$ be the number of Sylow 7-subgroups of group $ G$ of order 56. From Sylow Theorem $ n\equiv 1\pmod{7}$ and $ n|56$, so $ n \equal{} 1$ or $ n \equal{} 8$. If $ n \equal{} 1$, then Sylow 7-subgroup is normal. If $ n \equal{} 8$, then there are 48 elements of order 7 (because intersection of different subgroups of prime order must be trivial), and it follows that there is only one Sylow 2-subgroup, which must be therefore normal.
If $ G$ is a group of order 148, again by Sylow Theorem we obtain, that there is only one Sylow 37-subgroup of $ G$.


Hi,

Is there any way to solve this problem without just using guess and check?

Three consecutive positive prime numbers have a sum that is a multiple of 7. What is the least possible sum?

Thanks!  :lol:

There's no non guess and check way for that. Sorry :\

Hmm i wouldnt be too sure.... you know that they have to be an odd multiple of 7 cuz 2,3,5 =12.

13,17,19 yields a solution 49....

like cricket said, the least prime cant be 2, since it would be even and 10 isnt divisible by 7. after that, i guess you really have to brute force it...since there isnt any steady formula for prime numbers. all you can really do is just go down the list, and yeah, 49 is the least sum, from that


There are three circles that are touching eachother and a line under them. The radius in the biggest circle is 2, the radius in the smaller circle is 1 and the radius in the smallest circle is unknown.

[url=http://img159.imageshack.us/my.php?image=tresirkler5xg.png]Picture here[/url]

What is the radius in the smallest circle?

You're using pythagoras theorem wrong. It's only true for right-angled triangles, and that is no right-angled triangle.

I think I have found a solution. Call the radius x. Then if you create a line with the distance $1 - x$ (call the endpoint of this P) from the midpoint of the smaller circle you can create a rightangled triangle, and a similar one with side $2 - x$ in the larger circle (call this endpoint G). Then call the distance between the midpoint of the smallest circle and P y. We know that PG has a total distance 3, so we can set up the following equations:

$(1 - x)^2 + y^2 = (1 + x)^2$ and $(2 - x)^2 + (3 - y)^2 = (2 + x)^2$. It's hard to explain without a figure, but I think it is right.

[quote="Masamune"]Did I even mention any triangles? I didn't say anything about pythagoras theorem either...

Anyway, thanks for the help, but I don't know what to do next.[/quote]

Just solve for $x$ by system of equations

[quote="Masamune"]Did I even mention any triangles? I didn't say anything about pythagoras theorem either...

Anyway, thanks for the help, but I don't know what to do next.[/quote]
Someone posted an incorrect solution before I posted mine. He then removed his post.

Kalle: Ok, then I understand.

Thanks for the help everyone.

EDIT: How did you get 2 * square root of 2?


floor function
email
number theory unsolved
number theory

Originally i posted the problem in algebra part. but nobody answered,maybe its too easy :maybe:  :blush:  but it could be number theoritic too. so please help:

Please solve the following problem.
For each +ve $ x \in \mathbb{R}$, define

$ E(x) = \{[nx]: n\in \mathbb{N}\}$
Find all irrational $ \alpha > 1$ with the following property:
 If a positive real $ \beta$ satisfies $ E(\beta)$ is a proper subset of $ E(\alpha)$ then $ \frac {\beta}{\alpha}$ is a Natural number.

Hint:
Use Beatty theorem:
If $ r,s$ are two positive irrational numbers then \[ \frac {1}{r} \plus{} \frac {1}{s} \equal{} 1\implies E(r)\cap E(s) \equal{} \phi\text{ and }E(r)\cup E(s) \equal{} \mathbb N\] :)

[quote="Akashnil"]Hint:
Use Beatty theorem:
If $ r,s$ are two positive irrational numbers then
\[ \frac {1}{r} \plus{} \frac {1}{s} \equal{} 1\implies E(r)\cap E(s) \equal{} \phi\text{ and }E(r)\cup E(s) \equal{} \mathbb N
\]
:)[/quote]
sorry but i didn't understand how to use that theorem here.can you explain :(
Note that you have to FIND $ \alpha$

[quote="ith_power"]Note that you have to FIND $ \alpha$[/quote]
I know. But don't take it too seriously. The answer can't be something like "the only $ \alpha$ is $ \pi$" or something :P

The real answer is: all $ \alpha > 2$ has the property

For some $ x > 1$ denote $ x'$ to be the positive real satisfying $ \frac {1}{x} + \frac {1}{x'} = 1$
And for all sets $ X$ denote $ X'$ to be the compliment set of $ X$ (Our universal set is $ \mathbb N$)
In other words, denote $ X'$ the set satisfying $ X\cap X' = \phi \text{ and }X\cup X' = \mathbb N$
By Beatty theorem $ (E(r))' = E(r')$ for all irrational $ r > 1$
Note that we trivially have $ E(x)\subset E(\frac {x}{n})$ for all real $ x > 1$ and integer $ n > 1$
If $ \alpha < 2$ then $ \alpha ' > 2$ and $ \frac {\alpha '}{2} > 1$ But
$ E(\alpha ')\subset E(\frac {\alpha '}{2})$
$ (E(\frac {\alpha '}{2}))'\subset (E(\alpha '))'$
$ E((\frac {\alpha '}{2})')\subset E(\alpha '') = E(\alpha )$
Calculating $ (\frac {\alpha '}{2})'$ gives that it isn't a multiple of $ \alpha$
So $ \alpha < 2$ doesn't have the property.

Assume that for some $ \alpha > 2$ there exists some $ \beta$ which isn't a multiple of $ \alpha$ and satisfying $ E(\beta )\subset E(\alpha )$

[b]Case 1[/b] $ \frac {\beta }{\alpha }$ is rational
Let $ \{\frac {\beta }{\alpha }\} = \frac {u}{v}$ where $ u,v$ are positive integers satisfying $ (u,v) = 1$ and $ u < v$
For all positive integers $ k$:
$ \lfloor (kv + 1)\beta \rfloor \in E(\beta )$
$ \lfloor (kv + 1)\beta \rfloor = \lfloor (kv + 1)\alpha \frac {\beta }{\alpha }\rfloor = \lfloor (kv + 1)\alpha (\lfloor \frac {\beta }{\alpha}\rfloor + \{\frac {\beta }{\alpha}\})\rfloor =$
$ = \lfloor (kv + 1)\alpha (\lfloor \frac {\beta }{\alpha}\rfloor + \frac {u}{v})\rfloor = \lfloor ((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \frac {\alpha }{v})\rfloor$
we shall prove that there exists $ k$ for which this $ \not \in E(\alpha )$. That is equivalent to:
$ \lfloor ((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha )\rfloor <$
$ \lfloor ((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \frac {\alpha }{v})\rfloor <$
$ \lfloor ((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \alpha )\rfloor$
Or equivalently:
$ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha )\} >$
$ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \frac {\alpha }{v})\} - \frac {\alpha }{v} >$
$ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \alpha )\} - \alpha\cdots (i)$
But for any irrational $ x$, the set $ \{\{mx\}: m\in \mathbb N\}$ is dense in $ (0,1)$
And $ (v\lfloor \frac {\beta }{\alpha }\rfloor + u)\alpha$ is irrational.
So we can find $ k$ satisfying $ \{k(v\lfloor \frac {\beta }{\alpha }\rfloor + u)\alpha \}\to \{ - \lfloor \frac {\beta }{\alpha }\rfloor \alpha \}^ -$.
In that case $ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha )\}=\{\{k(v\lfloor \frac {\beta }{\alpha }\rfloor + u)\alpha \}+\{ - \lfloor \frac {\beta }{\alpha }\rfloor \alpha \}\}\to 1$
and similarly
$ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \frac {\alpha }{v})\}\to \{\frac {\alpha }{v}\}$
and
$ \{((kv + 1)\lfloor \frac {\beta }{\alpha}\rfloor + ku)\alpha + \alpha )\}\to \{\alpha \}$
Now $ (i)$ is equivalent to $ 1 > - \lfloor \frac {\alpha }{v}\rfloor > - \lfloor \alpha \rfloor$
which is obviously true since $ \alpha > 2$ and $ v\ge 2$.
So in case 1 we have contradiction to existence of such $ \beta$

[b]Case 2[/b] $ \frac {\beta }{\alpha }$ is irrational
Again we have to find 2 positive integers $ k,l$ satisfying
$ \lfloor l\alpha \rfloor < \lfloor k\beta \rfloor < \lfloor (l + 1)\alpha \rfloor$
or $ k\beta - l\alpha \ge 1$ and $ (l + 1)\alpha - k\beta \ge 1\cdots (ii)$
But $ k\beta = \frac {k\beta }{\alpha }\alpha = (\lfloor \frac {k\beta }{\alpha }\rfloor + \{\frac {k\beta }{\alpha }\})\alpha = \lfloor \frac {k\beta }{\alpha }\rfloor \alpha + \{\frac {k\beta }{\alpha }\}\alpha$
So we take $ l = \lfloor \frac {k\beta }{\alpha }\rfloor$
$ (ii)$ is equivalent to $ \{\frac {k\beta }{\alpha }\}\alpha \ge 1$ and $ \alpha - \{\frac {k\beta }{\alpha }\}\alpha \ge 1$
$ \{\frac {k\beta }{\alpha }\}\in [\frac {1}{\alpha },1 - \frac {1}{\alpha }]\cdots (iii)$
But for any irrational $ x$, the set $ \{\{mx\}: m\in \mathbb N\}$ is dense in $ (0,1)$.
And $ \frac {\beta }{\alpha }$ is irrational. So there exists $ k$ satisfying $ (iii)$
In case 2 also we have contradiction to the existence of such $ \beta$

Hence $ \alpha$ has that property iff $ \alpha > 2$

[hide]
[hide]
Are you a Bangali?
[hide]
me am :P
[/hide]
[/hide]
[/hide]

[quote="Akashnil"][
[hide]
[hide]
Are you a Bangali?
[hide]
me am :P
[/hide]
[/hide]
[/hide][/quote]

[hide][hide][hide]check my email[/hide][/hide][/hide]


[quote="mathnerd314"]When I was in Private Messages, I clicked on "Find a Username" and it didn't display correctly (i.e. I couldn't enter anything into the text box). Please help. Thanks.[/quote]
See here, where I have the same question:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99298
Unfortunately, nobody ever replied.  :oops:  :roll:


Does anyone here watch numb3rs on cbs? If so, what do you think, particularly about the math in the show?

P.S. Last episode, Charlie mentioned something about Fourier analysis.... thing is, he pronounced it "furrier"... is this the right way, or is it "for-ee-ay"?

I like that show quite a bit. Although there are some real howlers if you pay attention to the math stuff, some of it is really interesting. I do think the second season is tuning out to be not quite as good as the first, though.

yeah, its beginning to be "charlie will figure this out for us!".......... "i calculated the probability and..." "lets talk about mom again!"

Obviously, the writers (or actors) of the show don't understand French pronunciation. I've never heard that version of Fourier.

Knowing too much math may detract from the enjoyment of a show like this- it's really much more about the characters.


consider the reaction

$RS(CH_{3})_{2}$  $\Rightarrow$  $R^{+}$  $+$  $S(CH_{3})_{2}$

why does the rate of the reaction decrease moderately when you increase the polarity of the solvent

[hide=" My answer"]

since this is a reaction where a cation is formed rate will depend on how easily the force between the cation and here the uncharged counterpart decreases . now no matter what the solvent is the carbocation doesn't solvate much due to charge distribution so we are here considering the solvation of the other part which tends to be neutral and probably would require a weakly polar solution :polar in order to seperate it from the carbocation . and bit non polarity to dissolve the non polar product.this is because the other part becomes increasingly non polar as the reaction proceeds.[/hide]

Your opinion on this ?

In a typical SN1 reaction, where a neutral reactant (for example, tert-butyl chloride) ionizes to a carbocation [(CH3)3C+] and a negative species (Cl-), the rate of reaction is greatly enhanced by using a polar protic solvent, because of its ability to stabilize positive and negative charges.

However, in your reaction, we have a positive species [RS+(CH3)2] that dissociates in another positive species (R+, a carbocation) and a neutral molecule (dimethylsulphide), and the more polar the solvent then the reaction will be moderatly slowered. Why?
Now, we know that if the charge on a transition state (t.s.) is bigger relative to reagent, then the use of polar solvents will stabilize and thus reduce the energy of the t.s. (decreasing the difference in energy between reagent and t.s. and so facilitating the reaction). However, if, as in this case, the reagent is already positively charged, than the t.s. will have a more dispersed charge (which is equivalent to smaller charge) relative to reagent, and so more polarity of the solvent will stabilize better the reagent than the t.s. - this will increase the difference in energy between reagent and t.s., and the reaction will be slowered.


If x,y,z are real positive numbers, prove that
$ (x\plus{}y\plus{}z)^2(xy\plus{}yz\plus{}zx)^2\leq3(y^2\plus{}yz\plus{}z^2)(z^2\plus{}zx\plus{}x^2)(x^2\plus{}xy\plus{}y^2)$

[hide="My very ugly proof..."]
i wrote
$ (x\plus{}y\plus{}z)^2(xy\plus{}yz\plus{}zx)^2\equal{}((x\plus{}y)(y\plus{}z)(z\plus{}x)\plus{}xyz)^2
\equal{}6\sum_{sym}x^3y^2z^1\plus{}12x^2y^2z^2\plus{}xyz(x^3\plus{}y^3\plus{}z^3)\plus{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3\plus{}(y^2\plus{}yz\plus{}z^2)(z^2\plus{}zx\plus{}x^2)(x^2\plus{}xy\plus{}y^2)$

We have to prove that
$ 6\sum_{sym}x^3y^2z^1\plus{}12x^2y^2z^2\plus{}xyz(x^3\plus{}y^3\plus{}z^3)\plus{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3\plus{}(y^2\plus{}yz\plus{}z^2)(z^2\plus{}zx\plus{}x^2)(x^2\plus{}xy\plus{}y^2)\leq3(y^2\plus{}yz\plus{}z^2)(z^2\plus{}zx\plus{}x^2)(x^2\plus{}xy\plus{}y^2)$

$ \Leftrightarrow$

$ 6\sum_{sym}x^3y^2z^1\plus{}6x^2y^2z^2\plus{}xyz(x^3\plus{}y^3\plus{}z^3)\plus{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3\leq2(\sum_{sym}x^4y^2z^0\plus{}2\sum_{sym}x^3y^2z^1\plus{}xyz(x^3\plus{}y^3\plus{}z^3)\plus{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3)$

$ \Leftrightarrow$

$ 2\sum_{sym}x^3y^2z^1\plus{}6x^2y^2z^2\leq2\sum_{sym}x^4y^2z^0\plus{}xyz(x^3\plus{}y^3\plus{}z^3)\plus{}x^3y^3\plus{}y^3z^3\plus{}z^3x^3$
which is true by Muirhead and AM-GM. :) [/hide]

Without loss of generality, we suppose that $ x \plus{} y \plus{} z \equal{} 1$. Notice that
\[ x^2 \plus{} xy \plus{} y^2 \equal{} (x \plus{} y)^2 \minus{} xy \equal{} (1 \minus{} z)^2 \minus{} xy \equal{} 1 \minus{} z \minus{} z(1 \minus{} z) \minus{} xy \equal{} 1 \minus{} z \minus{} (xy \plus{} yz \plus{} zx).
\]
Let $ xy \plus{} yz \plus{} zx \equal{} q$, and $ xyz \equal{} r$. We rewrite the desired inequality as
\[ q^2\leq 3[(1 \minus{} q)q^2 \minus{} r].
\]
That is,
\[ 3q^3 \minus{} 2q^2 \plus{} 3r\leq0.
\]
Notice that $ q^2\geq9r$. Then we need to show $ 3q^3 \minus{} 2q^2 \plus{} q^2/3\leq0$. This is true since
\[ q \equal{} xy \plus{} yz \plus{} zx\leq\frac13(x \plus{} y \plus{} z)^2 \equal{} \frac13.
\]
We are done.

[quote="hell_ever"]If x,y,z are real positive numbers, prove that
$ (x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} zx)^2\leq3(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)(x^2 \plus{} xy \plus{} y^2)$

[hide="My very ugly proof..."]
i wrote
$ (x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} zx)^2 \equal{} ((x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} xyz)^2 \equal{} 6\sum_{sym}x^3y^2z^1 \plus{} 12x^2y^2z^2 \plus{} xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} (y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)(x^2 \plus{} xy \plus{} y^2)$

We have to prove that
$ 6\sum_{sym}x^3y^2z^1 \plus{} 12x^2y^2z^2 \plus{} xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} (y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)(x^2 \plus{} xy \plus{} y^2)\leq3(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)(x^2 \plus{} xy \plus{} y^2)$

$ \Leftrightarrow$

$ 6\sum_{sym}x^3y^2z^1 \plus{} 6x^2y^2z^2 \plus{} xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3\leq2(\sum_{sym}x^4y^2z^0 \plus{} 2\sum_{sym}x^3y^2z^1 \plus{} xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)$

$ \Leftrightarrow$

$ 2\sum_{sym}x^3y^2z^1 \plus{} 6x^2y^2z^2\leq2\sum_{sym}x^4y^2z^0 \plus{} xyz(x^3 \plus{} y^3 \plus{} z^3) \plus{} x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3$
which is true by Muirhead and AM-GM. :) [/hide][/quote]

\[ x^2\plus{}xy\plus{}y^2\ge \frac{3}{4}(x\plus{}y)^2,\]
and similarly we have two other inequality. So it suffices to prove that
\[ (x\plus{}y)(y\plus{}z)(z\plus{}x)\ge \frac{8}{9}(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx),\]
which equivalent to
\[ xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)\ge 6xyz.\]

This true by AM-GM.

Thank you,both.
I realize the simplicity of the problem and how complicated i made it....


I feel so angry that everytime I take AMC-10 test, I get 104 to 116. I took all the AMC-10 test and I never reached the 120. Only the near its surface. :( It makes me mad that I can't do those problems and I'm only off by FEW POINTS!!! Can anyone tell me how I can increase my score and qualify for AIME this year for sure? I need practical suggestion other than saying "Take AMC-12" because that's impossible for me...

 :( Please help me.....

What advice do you want?  There's no magic formula for succeeding on the AMC 10.  Make sure you know all the math that is tested and do a ton of problems and really that's all you can do.  Seeing how you post frequently on AOPS I can tell you're doing a lot of problems so just keep solving problems and you should be fine.  AMC 10A isn't for another whole month anyway.

Thanks, that makes me feel more relief. :)

I guess I'll get better once I covered more AHSMEs.

Thanks again.

Although it is probably too late this year, when you study for next year's exam. Work on problems harder than AMC10, say AMC12 or AIME. That wat, when you go back and work on AMC10, all but the last few should be pretty simple.

here's an idea... like captcha000 said: work harder prblems now (from lik Jan 1 to jan 15 -20) then immediately start doing AMC10's... (until Jan 31... since AMC10 is on Feb. 1 (for some)). Also, a strategy I used at the Texas A&M contest, don't try to get an answer for al questions... focus on the ones you know... ie. don't do all 25 and have some mistakes, but do 20 or so, with no mistakes

With the scores you're getting, it sounds like you're trying too many problems. You can get more than 100 points by just doing 11/25 if you leave the other 14 blank, but I imagine you're not just doing that.

Once again, my advice is to not do so many problems, because it definitely seems like you are. Two blank problems is almost as good as one solved problem, and toward the end, that may not be a bad deal for you.

I think you're mistaking me.

I'm taking AMC-10. For AMC-10, you need to solve at least 17 problems in order to qualify.

I personally think taking AMC-12 is technically easier than taking AMC-10 when it comes to qualifying for AIME

You need at least 100 on AMC-12 to "pass," that's 11 answered correctly and 14 left blank minimum.

You need at least 120 on AMC-10 to "pass," that's 17 answered correctly and the other 8 left blank.

[quote="AMC website's FAQ"]
There are about a dozen problems in common between the AMC 10 A and the AMC 12 A, and about a dozen problems in common between the AMC 10 B and AMC 12 B.
[/quote]

If you can get most the problems in common, then you're probably set on the AMC-12

But...

...in order to prepare, like Nightflarer said, there's no magic formula. Try to experience as wide a range of problems as possible and remeber to take as little risk on the tests as you cananswer only those you are SURE about

[quote="Silverfalcon"]To Rep,

I usually solve from 17-18. It's usually stupid mistake that I make. :([/quote]

Yeah, that's the other thing, stupid mistakes kill :(

My friend missed the AIME last yr when on the first question of the AMC-12 he answered in cents, not dollars :(

to that i agree.
i used to make about 4 or 5 careless mistakes every test i took, but i cut it down to about 1 now. basically you just have to take more time with your work, do it fast enough that you have time, but not too fast that you just do really stupid things. also, check your work. i know i sound like a teacher, but it helps so much. especially like on the aime, when you have plenty of time, make sure you do.


Obviously [code] $ {...}$ [/code] doesn't work.

Use \{ and \} to get, for example, $\{1,2,3,\dots,n\}$

Symbols in LaTeX which have special meaning like & % # _ { } and of course the dollar sign, can be shown by putting \ in front of them: $\$ \& \% \# \_\{ \}$


symmetry
geometry
geometric transformation
rotation
factorial
videos

Can someone please explain counting with symmetries. Thanks

Counting with symmetry is basically counting the number of ways things can be arranged on a circle, that is, if two arrangements are "rotations" of each other, they are considered identical.  The key formula is that when arranging, for example, five distinct people around a round table, there are (n-1)! or 4! (factorial) = 24 successful outcomes.

For more details, use Alcumus.  If you have specific questions, you can post them here.


Problem:Find all real roots of equation:
$ 3x^2\plus{}11x\minus{}1\equal{}13\sqrt{2x^3\plus{}2x^2\plus{}x\minus{}1}$

Let's join!

Hi chuyentoan. Is there a typo here?

I get $ 9x^{4} \minus{}272x^{3} \minus{}223x^{2}\minus{}191x\plus{}170\equal{}0$, which doesn't seem to factorise nicely.


The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.

Solution: Let the 14 games be the pigeon holes. By the pigeon hole principle, we can stuff one player each to the 14 games. So, we have 6 players left. Now, we stuff the remaining 6 players to 6 pigeon holes. Thus, there are 6 pigeon holes with 2 players each, distinct. Notice that we have 8 slots left. These would be filled up by the remaining members. And thus, these represent the games with non distinct players.

Is this right? thanks!

Yes it is correct. This is, without argument, the most straightforward USAMO problem ever made in the history of mankind.


In how many ways can the integers from $1$ to $n$ be arranged such that, except for the first integer on the left, every integer differs by 1 from some integer to the left of it?

I am sorry I understood the problem incorrectly earlier; as result, I have deleted my previous solution.
[b]sky9073[/b], thanks for pointing it out.

I understood the problem a bit differently...
[hide]
Say we start with some integer $k (1 \leq k \leq n)$. Each new number must have [i]some [/i]integer [i]somewhere [/i]to its left that differs by 1, which means:

- the $k-1$ integers less than $k$ must come in decreasing order,
- the $n - k$ integers greater than $k$ must come in increasing order, but
- we are free to switch listing between "numbers greater than $k$" and "numbers less than $k$"

Essentially, we have a block-walking problem: $n-1$ "blocks" to walk (numbers to list), with $k-1$ "blocks" in one direction (numbers less than $k$) and $n-k$ in the other (numbers greater than $k$). For each $k$, that makes $\binom {n-1}{k-1}$ ways; summing up the cases for all $k$ gives $\binom{n-1}{0}+ \binom {n-1}{1} + \cdots +\binom{n-1}{n-2} + \binom{n-1}{n-1} = \boxed{2^{n-1}}$ ways.[/hide]

[hide="Bijective Proof"]
The sequences of length n match up exactly with the binary strings of length (n-1). Each zero/one tells us whether to move up or down to reach each next integer in the sequence, and the starting number in the path is simply given by n minus the number of ones in the binary string. By bijection, the number of sequences is 2^(n-1).

(I was more formal but wasn't logged in and lost the text of my post).
[/hide]

The problem is equivalent to this problem:

We have $n$ boxes arranged in a row, and will put a ball in each, with the following criteria:
1. We first pick a box to pick a ball in, and put a ball in it.
2. From there on, we can put balls only in boxes next to boxes that already have balls in them until each box is filled with a ball.

Here are my two solutions to this equivalent problem:

[hide=" Solution 1"]
Consider the number of ways to [i]remove [/i]the balls from the boxes without ruining the conditions of the previous statement. We now have $n$ full boxes, and can remove a ball from either the $n^{th}$ box, or the $1^{st}$ box. After each removal of a ball, the choices for the removal of the next ball are from the edges. So we will always will $2$ choices for the removal of a ball, and we will have $n-1$ removals. So there are $\boxed{2^{n-1}}$ ways to remove (or arrange) the balls.[/hide]
[hide=" Solution 2"]
Pick the $k^{th}$ box first, $1 \ge k \ge n$. We have $k-1$ boxes to the left of this box, and $n-k$ boxes to the right of this box. This problem is now equivalent to finding the number of ways we can write the numbers $1$ to $n$ except for some integer $k$ in order such that $k-1$ integers are in increasing order, and $n-k$ integers are in increasing order. There are clearly $\binom{n-1}{n-k}$ ways. However, we can pick any $k$ from $1$ to $n$, so the total number of ways is
$= \binom{n-1}{n-1}+\binom{n-1}{n-2}+\cdots+\binom{n-1}{n-(n-1)}+\binom{n-1}{n-n}$
$= \binom{n-1}{n-1}+\binom{n-1}{n-2}+\cdots+\binom{n-1}{1}+\binom{n-1}{0}$
$=\boxed{2^{n-1}}$ ways[/hide]


Let x_i (i=1,..,n) be n real positive numbers. Prove that

 sum(j=1,..,n)(A_j/B_j)  \leq n(A/B)

where

A=sum(i=1,..,n)(x_i)^2
B=sum(i=1,..,n)x_i

A_j =sum(i=1,..,n; i=/=j)(x_i)^2
B_j =sum(i=1,..,n; i=/=j)x_i

Very interesting and very difficult inequality.

when i posted the problem in the proposed problems section(created by me ),i remebered a similar inequality had appeared sumwhere,and this post was the one i found this inequality unsolved.so i'll try and give a proof
let A= \sum (X_i)^2  and B =  \sum (X_i)
then the inequality we want to prove is
 \sum  [A - (X_J)^2]/[B - X_j]  \leq  nA/B EQUIVALENT TO
 \sum  [1 - {(X_J)^2}/A]/[1 - {X_j}/B]   \leq  n
now expand 1/(1-{X_j)/B} as an infinite series and then truncate it after cancelling "n " on both sides,this inequality becomes the same as the one as I proposed and HARAZI solved it beautifully by setting B=1
I HOPE wot i have done is correct


Does anyone know the formula to find how many intersections there are in a plane with n lines?
I worded that badly, sorry. :(
Example:In a plane with 28 lines, how many intersections will there be?
I worded that even worse. :o  :(

For the maximum regions created by n intersecting lines it is:
$ \frac{n(n\plus{}1)}{2}\plus{}1$
I don't know if that is what you are looking for...

I believe he's asking for the number of intersections.    [hide="Something like this I think:"]
If we have $ 28$ lines in a plane, such that no $ 3$ lines intersect in the same place, what is the maximum number of intersections that can be formed?

First, we place one line down.  It can intersect a maximum of $ 0$ lines, since no other lines are down yet.  For the second one, we can intersect a maximum of $ 1$ line (the first line).  For the third, we can intersect a maximum of $ 2$ lines (once for each line).  This goes all the way until the 28th line, which intersects a maximum of $ 27$ lines.  So we have $ 0 \plus{} 1 \plus{} 2 \plus{} \dots \plus{} 27 \equal{} \binom{28}{2} \equal{} \frac {28\cdot27}2 \equal{} 378$.
Generally, if we have $ n$ lines in a plane, and we're looking for the maximum number of intersections possible, the equation is:
\[ \boxed{\binom{n}{2} \equal{} \frac {(n)(n\minus{}1)}2}\]
Hope this helps  :)[/hide]

[quote="gogators"]Does anyone know the formula to find how many intersections there in a plane with n lines?
I worded that badly, sorry. :(
Example:In a plane with 28 lines, how many intersections will there be?
I worded that even worse. :o  :([/quote]

Wouldn't that just be $ \binom{28}{2} \equal{} 3276$? There's an intersection with every 2 lines.

@jxl: I'm probably missing something, but where did you get that restriction from?

[quote="isabella2296"]
Wouldn't that just be ${ \binom{28}{2}=3276}$?  There's an intersection with every 2 lines.[/quote]
Isn't $ \binom{28}{2}=378$?
Also, I made a mistake with my equation; it is now edited  :)
[quote="isabella2296"]
@jxl: I'm probably missing something, but where did you get that restriction from?[/quote]
Are you referring to "Such that no 3 lines intersect in the same place"?
This part really doesn't matter... If you want to maximize the number of intersections, you wouldn't want more than 2 lines to intersect in the same place.  So I guess you can just disregard it?  :)

Man the problem as stated is actually pretty cool, as in, how many points of intersection CAN be obtained?

Clearly, 0 works, 1 works, 27 works, etc.

I wonder if there is a general formula for this sort of thing.

I think that the general formula is $ n(n\plus{}1)/2$, but I think $ \binom{28}{2}\equal{} 378$ is right.
Here is the link for the question and solution.(number 16)
http://www.floridamao.org/PublicPages/TestArchive.aspx
you can see the test and the solutions

You mean $ \frac{n(n\minus{}1)}{2}$  :wink:

I meant if you could figure out the number of POSSIBLE intersections.

[quote]You mean $ \frac{n(n\minus{}1)}{2}$  :wink:[/quote]
 er yes... :blush:

but anyway please help
im still not sure about the formula...


function
inequalities solved
inequalities

Here's an idea that might be useful: consider the triangle X, Y, Z where X, Y, Z are the points on arc BC, arc AC, arc AB respectively where the angle bisectors of ABC meet the circumcircle. This is well known (all the following letters mean angles): X=(B+C)/2, Y=(A+C)/2, Z=(A+B)/2. Then sinA/2=cosX, cosA/2=sinX and the others. The condition A, B, C <Pi/2 is equivalent to X, Y, Z>Pi/4 (simple angle calculations). So the new problems are:

If ABC is a triangle s.t. A, B, C>Pi/4 we have cosA+cosB+cosC >= 4/3*(1+cosA*cosB*cosC) and
sinA+sinB+sinC >= 4/3^1/2*(1+cosA*cosB*cosC)

This isn't even close to a soln, but I think it looks a bit friendlier like this.

P1. After some simple trigonometric manipulations we obtain
sin(A/2)sin(B/2)sin(C/2)=(cos A + cos B + cos C - 1)/4.
So initial enequlity is
sin(A/2)+sin(B/2)+sin(C/2) \geq (cos A + cos B + cos C)/3+1

Let A, B, C is an optimal set such that
f(A,B,C)=sin(A/2)+sin(B/2)+sin(C/2) - (cos A + cos B + cos C)/3 = min.

It is fact 0< A,B,C <pi/2.
Let us consider new set A+eps,B-eps,C. Due to optimal condition
0.5*cos(A/2) - 0.5*cos(B/2) + 1/3sin A - 1/3sin B = 0 ==>
0.5*cos(A/2) + 1/3sin A = 0.5*cos(B/2) + 1/3sin B.
Similarly
0.5*cos(A/2) + 1/3sin A = 0.5*cos(B/2) + 1/3sin B = 0.5*cos(C/2) + 1/3sin C.

Function 0.5*cos(t/2) + 1/3sin(t) has exactly one extremum on [0,pi/2] ==> set A,B,C contains two equal values.

Suppose pi/2>A=B>pi/4, C=pi-2A.
f(A,B,C)=g(A)=2sin(A/2)+1/3cos(A)+1/3cos(2A). Denote x=sin(A/2) then sin(pi/8)<x<sin(p/4).
g(A)=2/3(1+3x-5x2+4t4).
Denote h(x)=3x-5x2+4t4. h'(x)=3-10x+16x^3=0 ==>
x0=1/2, x1=(\sqrt 7 -1)/4, x2=(-\sqrt 7 -1)/4<0.
For x0 and x1 we obtain g(A)\geq 1 as with A=pi/4 and A=pi/2 ==> g(A)\geq 1 ==> min=1 (optimal set is (pi/3,pi/3,pi/3)).

P2. I think this inequlity can be proved by the same method as P1.

Are there more elegant proofs?


function
integration
real analysis
real analysis unsolved

Denote by $d_{n}$ the number of divisors of positive integer $n$ (including $1$ and $n$). Prove that $\mathop{\sum}_{n=1}^{\infty}\frac{d(n)}{n^{2}}<4.$
(G. Shevchenko)

More generally,
\[ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {d(n)}{n^{s}} \equal{} \sum_{a,b \equal{} 1}^{ \plus{} \infty}\frac {1}{a^{s}}\cdot\frac {1}{b^{s}} \equal{} \zeta(s)^{2},
\]
where $ a, b$ are divisors of $ n,$ $ \zeta$ is the Riemann zeta function and $ s$ is a complex number such that $ \text{Re}(s) > 1.$

Thus, here it yields that $ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {d(n)}{n^{2}} \equal{} \frac {\pi^{4}}{36} < 4.$

another solution:

We know that $ d(n) \leq2 \sqrt{n}$. Then $ \sum_{n\equal{}1}^{ \infty} \frac {d(n)}{n^2} \leq \sum_{n\equal{}1}^{\infty} \frac {2 \sqrt{n}}{n^2} <2 \int_1^{ \infty} x^{\minus{}\frac 32}dx\equal{}4$.


Show that the number $111...111$ with 1991 times the number 1, is not prime.

Only known primes of the type 11...11 are   (2 , 19, 23, 317, 1031) digits.
Larger ones of 49081  and 86453 digit long are most likely primes but there are no oter primes of that type  less than 100,000 digits llong.. these kind of primes are indeed rare.

$111\ldots11$ with $n$ 1's can be expressed as $D_{n} = \displaystyle\frac{10^{n}-1}{9}$.
Generally, for a composite $k = pq$ where $p$ and $q$ are greater than 1, we can say:
   $x^{p} \equiv 1 \bmod{x^{p} - 1}$
   $x^{pq} \equiv 1^{q} \equiv 1 \bmod{x^{p} - 1}$
   $x^{k} - 1 \equiv 0 \bmod{x^{p} - 1}$
Another proof of that fact, consider $x^{k}-1_{10}$ in base $x$. It becomes $(x-1)(x-1)(x-1)\ldots(x-1)(x-1)_{x}$ where the $(x-1)$'s are the digits and there are $k$ of them. Since $p$ divides $k$, the $k$ digits can be grouped into $q$ sets of $p$ $(x-1)$'s.
Both statements imply whether indirectly or directly that $x^{p} - 1 \mid x^{k} - 1$ if $p \mid k$. So, since we know $1<p<k$, we know $x - 1 < x^{p} - 1 < x^{k} - 1$. From which we can conclude $\displaystyle\frac{x^{k} - 1}{x^{p} - 1}>1$. If $x>2$, $x-1>1 \Rightarrow x^{p} - 1 > 1$. So, if $x>2$ and $k$ is composite, $x^{k}-1$ can't be prime because it is composed of two non-trivial factors. $\displaystyle\frac{D_{1991}}{D_{11}}=\displaystyle\frac{10^{1991}-1}{10^{11}-1}$, so, since $11 \mid 1991$, by our former reasoning, $D_{1991}$ must be composite.

P.S. For advanced people don't sweat the long explanation I intended it for Beginner's to understand it.

no use - beginners don't read such long proofs :D

$11\dots 11$ is called as [b]repunit number[/b].

[b]Lemma[/b]
$\underbrace{11\dots 11}_{1991 \ \text{times}}$ has at least two prime factors

[b]Proof[/b]
$\underbrace{11\dots 11}_{1991 \ \text{times}}=\frac{10^{1991}-1}{9}$.
Since $11\mid 1991,$then $10^{11}-1\mid 10^{1991}-1$.
By [b]Zsigmondy's theorem,[/b]$\exists p(\text{prime})$ s.t. $p\mid 10^{11}-1,p\nmid 10-1=9$.Thus $p\neq 3$.
Also by [b]Zsigmondy's theorem,[/b]$\exists q(\text{prime})$ s.t. $q\mid 10^{1991}-1,q\nmid 10^{11}-1,q\nmid 10-1=9$.Thus $q\neq 3,p$.Therefore $p,q\mid \underbrace{11\dots 11}_{1991 \ \text{times}}\blacksquare$ :coool:


We have $ 0.2\cdot 0.5\cdot 80\equal{}\boxed{8}$


Right now, my code works fine and produces the correct answer, but the problem is that this piece of code:
      while(gets(&Data[temp]))
      {
         temp+=strlen(&Data[temp]);
         Data[temp] = '\n';
         temp++;
      }

gets an error when it compiles because: "the gets method is dangerous" I did some research and rewrote it to this:

      while(fgets(&Data[temp], 2000,stdin))
      {
         temp+=strlen(&Data[temp]);
         Data[temp] = '\n';
         temp++;
      }

now the problem is that all the lines i print out have an extra end of line \n character because the fgets function dosen't trim it off. So if a line should be:
ASDF
GHHJ

I get:
ASDF

GHHJ

Can someone tell me how to solve this problem? Thanks.

That is because you are doing: Data[temp] = '\n';

!!


(a) prove that $\sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7}$ is irrational.
(b) suppose $a_i$, with $i=1,2, \cdots ,n$ are rationals s.t. $\sqrt{a_i}$ is irrational for at least one value of $i$. Prove that $\sqrt{a_1} + \sqrt{a_2} + \cdots + \sqrt{a_n}$ is irrational.

[url=https://artofproblemsolving.com/community/c6h27089]This[/url] should kill all such problems ;)


A circle is circumscribed about a triangle with sides $ 20$, $ 21$, and $ 29$, thus dividing the interior of the circle into four regions. Let $ A$, $ B$, and $ C$ be the areas of the non-triangular regions, with $ C$ being the largest. Then

$ \textbf{(A)}\ A \plus{} B \equal{} C\qquad
\textbf{(B)}\ A \plus{} B \plus{} 210 \equal{} C\qquad
\textbf{(C)}\ A^2 \plus{} B^2 \equal{} C^2\qquad \\
\textbf{(D)}\ 20A \plus{} 21B \equal{} 29C\qquad
\textbf{(E)}\ \frac{1}{A^2} \plus{} \frac{1}{B^2} \equal{} \frac{1}{C^2}$

[hide="Click for solution"]
Since $ 20^2\plus{}21^2\equal{}29^2$, the triangle has a right angle.  Thus its hypotenuse is a diameter of the circle so the region with area $ C$ is a semicircle and is congruent to the semicircle formed by the other three regions.  The area of the triangle is $ 210$, hence $ A\plus{}B\plus{}210\equal{}C$.  Thus, our answer is $ \boxed{\textbf{(B)}}$.
[/hide]


How many combinations of patterns can you make with 5 squares, so that each touch an edge, not a corner?

[][][][][]

^ for example, like that. except of course, they would be real squares.

Just count all 12.
They have a special name which I forget (possibly pentomino)
Just use casework :)

[quote="indianamath"][hide][hide][hide]4*3^3=108?[/hide][/hide][/hide][/quote]

u must have a [i]lot[/i] of time on ur hands and/or an evil sense of humor... but i don't get how u can use casework. i'm confused, cuz i'd probably be off listing somewhere...

Are you considering Five vetical squares different from five horizontal squares.
Just wondering :P

[quote="Necrowarrio0"].....I don't get this question....[/quote]

It is making any shape you want with 5 square piece of paper that are distinguishable.

From Wikipedia:
[hide][img]http://upload.wikimedia.org/wikipedia/commons/thumb/e/ed/Pentominos.svg/800px-Pentominos.svg.png[/img][/hide]

there was something about that in my one gifted class.

[quote="buzzer11"][quote="indianamath"][hide][hide][hide]4*3^3=108?[/hide][/hide][/hide][/quote]

Thats really annoying, just so you know.  :)[/quote]

with those I just quote them and with all of the hide tags just as type, I can just find the answer.

I think they are like tetris things. Pentrix?

[quote="adrenalinefix36"]Hello everyone, this is clearly my very first post. I am firstly hear to share with you all, my love for math, which i've recently discovered this past year. For this problem, i feel, with little qualification, that the answer is much higher than what others have posted. When i hear the words "how many combinations" i think permutations. To me it sees, starting with the first square, should have zero possible arrangements, as it is its relationship with another square that determines its possible arrangements(and w/o another square, blah blah, no comb.) So starting with the second block, can go on either of the 4 sides of a 2d square? After adding this square,there is an increasing number of sides that the 3rd square can inhabit(6). The pattern of N+2 seems to continue with the 4rth square having 8 and the fifth 10. I just thought the answer was then (1)(4)(6)(8)(10)=1920. Could someone explain my flawed method? I do understand the other ones, just not mine lol.. thanks[/quote]
I didn't understand this either, but this is what I think. The problem is not asking for permutations, but the patterns possible, both vertical and horizontal. Also, [][][][][] is the same as[]
                                []
                                []
                                []
                                []
Basically, all patterns are covered in pentominoes, and there are 12 different pentominoes, so the answer is 12


[quote="yair"]prove that if x=pi/2(n+1) then
sum of sec (jx)*sec (j-2)x form j=2 to n equals 3cos 2x csc^2 2x[/quote]

$ nx\equal{}\frac{\pi}{2}\minus{}x$ ---> $ tan(n\minus{}1)x\equal{}cotx, tan(nx)\equal{}cot2x$
$ sum of sec (jx)*sec (j\minus{}2)x\equal{}\sum_{j\equal{}2}^{n}\frac{1}{cox(jx)cos(j\minus{}2)x}\equal{}\frac{1}{sin2x}\sum_{j\equal{}2}^{n}\frac{sin2x}{cox(jx)cos(j\minus{}2)x}\equal{}\frac{1}{sin2x}\sum_{j\equal{}2}^{n}(tan(jx)\minus{}tan(j\minus{}2)x)\equal{}\frac{1}{sin2x}(tan(n\minus{}1)x\plus{}tan(nx)\minus{}tanx)\equal{}\frac{1}{sin2x}(\frac{cosx}{sinx}\minus{}\frac{sinx}{cosx}\plus{}\frac{cos2x}{sin2x})\equal{}\frac{1}{sin2x}(\frac{cos^{2}x\minus{}sin^{2}x}{sinxcosx}\plus{}\frac{cos2x}{sin2x})\equal{}\frac{1}{sin2x}(\frac{2cos2x}{sin2x}\plus{}\frac{cos2x}{sin2x})\equal{}\frac{3cos2x}{sin^{2}2x}$


superior algebra
superior algebra unsolved

Let $G$ be a finite group with order $p^{m}q$,$p,q$ are primes,then $G$ is solvable.

It's a particular case of a Burnside's theorem.
cf.          http://en.wikipedia.org/wiki/Burnside_theorem

This one is even stronger http://planetmath.org/?op=getobj&from=objects&id=8839 .

Strangely, it is contained in the unsolved section.

but do you have an easy solution for this one?


Ok I was practicing the 1999 USNCO multiple choice and got owned by these questions.  Any help would be greatly appreciated.  Thanks!

22. Which statements are true?
1. S° values for all elements in their standard states
are positive.
2. S° values for all aqueous ions are positive.
3. $\Delta$ S° values for all spontaneous reactions are
positive.
(A) 1 only
(B) 1 and 2 only
(C) 2 and 3 only
(D) 1, 2 and 3

[hide]Correct answer is A.  I see why 3 is wrong ($\Delta$G° determines spontanteity) and why 1 is right, but it is very surprising (as well as confusing) that two is false.  How can a solution of ions be more ordered than a perfect crystal at 0 K?[/hide]

For 3 (which I hope shows, it's supposed to be a continuous curve peaking around 34, can find the original [url=http://center.acs.org/education/olympiad/1999/OL99NI_keyincl.pdf]here[/url]): [hide]Correct answer is B.  Any thoughts?[/hide]

EDIT:  Oh and another random thing I was wondering about, what is the hybridization of sulfur in $SO_3$?  If I'm drawing it right, sulfur is surrounded by three double bonds for a trigonal planar structure.  I have no idea what kind of hybridization this implies.

EDIT 2:  I just did the free response, and the answer key says that $IF_3$ is a Lewis acid.  It seems it should be a Lewis base, as it has lone pairs... any ideas?

Once again, thank you for any help you can provide.

Some ions (such as some complex ions) have more order when in aqueous solution than when they're not in aqueous solution.
For that delta T problem, I thought it should be A, 10 degrees.
And yeah, I think IF3 should be a lewis base rather than a lewis acid.

And, I guess the S in SO3 would have sp3 hybridization.

For number 2, I also am surprised, shouldn't the entropy rise with the dissociation of the particles? hmmm...let me think about this...

The even more surprising thing is that it doesn't talk about $\Delta$S.  It talks about absolute entropy.  Can you have an ion with a negative value of entropy (i.e. more ordered than a perfect crystal at 0K)?  My guess is no, they just have an entropy of exactly 0, but honestly I have no idea.

Something like AlCl3 has a negative entropy in aqueous solution.


calculus
integration
real analysis
limit
real analysis theorems

For example, is

$lim_{h\to 0}\int_{a}^{b}f(x+h) dx \equiv \int_{a}^{b}lim_{h\to 0}f(x+h) dx$

The question of when the interchange of a limit and an integral is justified is one of the central motivating questions for the development of theories of integration.

Within the theory of the Riemann integral the classic theorem is this: If $f_{n}$ converges [b]uniformly[/b] to $f$ on the compact interval $[a,b],$ then $\int_{a}^{b}f_{n}$ converges to $\int_{a}^{b}f.$ (The theorem even implies that $f$ is Riemann integrable even if all that was previously known about integrability is that the $f_{n}$ were integrable.)

Using that theorem, we could say this: Suppose $f$ is continuous on a compact interval slightly larger than $[a,b].$ Since $f$ is uniformly continuous on that interval, it follows that $f(\cdot+h)$ tends uniformly to $f$ on $[a,b]$ as $h$ tends to zero. This then justifies the interchange of limit and integral.

Then there is the Lebesgue integral. My attitude is that the entire theory is built around convergence theorems, most prominently the Monotone Convergence Theorem and the Dominated Convergence Theorem - the integral is defined the way it is just so we'll have these theorems. I'm not going to explain those here - get a good textbook; these theorems will have a central place.

Within the Lebesgue theory, I can prove the following:

If $f\in L^{1}(\mathbb{R})$ (that is, $f$ is absolutely integrable on the line), then

$\lim_{h\to0}\int_{-\infty}^{\infty}|f(x+h)-f(x)|\,dx=0.$


A transposition of a vector is created by switching exactly two entries of the vector. For example, $(1,5,3,4,2,6,7)$ is a transposition of $(1,2,3,4,5,6,7).$  Find the vector $X$ if $S=(0,0,1,1,0,1,1)$, $T=(0,0,1,1,1,1,0),$ $U=(1,0,1,0,1,1,0),$ and $V=(1,1,0,1,0,1,0)$ are all transpositions of $X$. Describe your method for finding $X.$

U and V differ in four places, the 2,3,4,5 spot.  Its easy to show that the 1,6,7 spots of X must be 1,1,0 then.

Then examining S, it differs from what we know about X in two spots already.  So in total its 1011010.

Checking, it works.


$ |2\minus{}3x| < 5$

Solve.

EDIT U+OOPS MIStAKE

[quote="Bicameral"]$ |2\minus{}3 x| < 5$

Solve.[/quote]

[hide]If $ |a| < b$, then $ \minus{}b < a < b$. Try to prove this yourself.

With this observation, we get

$ \minus{}5 < 2\minus{}3x < 5$

$ \minus{}7 <\minus{}3x < 3$

$ \minus{}3 < 3x < 7$

$ \boxed{\minus{}1 < x <\frac{7}{3}}$[/hide]

Edited.

I had to double check to make sure I'm doing this right. thanks!  :lol:


Edward(Y)/Yang/Mark 2 0 0 0 2 2
Brady/Yuri/Barry 0 2 0 0 2 4
Taiho(A)/Tony(Y)/Yitian - - - - 0 0
There were a few changes in teams, and the third team came in very late.

The first two were from the 2004 ARML, the third was a bunch of questions I collected from my math team coach, the fourth and fifth were from the 7th Polya Mathematics competition, and the last was from the same competition of a different year.
Finally, I'd like to thank Maria(A) for being a great host.

$\text{R3-1. Compute the number of ordered 4-tuples }(x, y, z, w)\text{ of non-negative integers that satisfy: }\\x+y+z+w=7.$[hide]120[/hide]

$\text{R3-2. Let }T\text{=TNYWR. If }r_1, r_2, r_3,\ldots ,r_T\text{ are the roots of }x^T-x-2=0,\\ \text{compute }\displaystyle\sum_{k=1}^T (r_k)^T=(r_1)^T+(r_2)^T+\cdots +(r_T)^T.$[hide]240[/hide]

$\text{R3-3. Let }T\text{=TNYWR. Evaluate }\displaystyle\sum_{k=1}^T \frac{1}{k^2+k}.$ [hide]$\frac{240}{241}$[/hide]
Thanks stan

For R3, I think the summation you meant was 1/(k^2+k), otherwise the first term is undefined...

[quote="Bictor717"]Finally, I'd like to thank Maria(A) for being a great host.[/quote]

and geehoon for a mental support?

[quote="Geehoon"]and geehoon for a mental support?[/quote]
Yay Geehoon :)
mental and spiritual support in addition to providing some of the problems used  :D

Just wondering, is the A Team going to have any more online practice relays???

im thinking about whole new york city team relay practice on
[b]may 28th saturday[/b], from 12 pm to midnight  :)
well, first of all, its not going to be twelve hours straight practice.
I myself have a soccer game at 5pm, so im going to be in and out also.
but i hope there are enough people so that its ok for some people to be in and out.

I dont know.
we'll talk more about it at CCNY in few days.

I was hoping for another relay as well, but I wasn't sure when the best time would be. I'd be willing to cover for you during your out times. It would be cool to have a relay marathon-like thing.

And yes, Geehoon was with us on Saturday night in spirit and, for a brief moment, partial consciousness. Thanks for the moral support.

also, the A team practice need not be an only A team practice... maybe we should have it be open to anyone who wants to come and just offer a general open invitation. i'm not sure how hard it might be to run such a relay though--- just an idea!  :)

yes, thats what i was thinking about.
based on some experiences, now i think i can handle how to set up a relay with more than 30, 40 even 50 people.

It might help for the captains of each team to watch their teams, that way, Geehoon wouldn't need to handle 50 people...

rather, i would need three people to handle each room.


pigeonhole principle
number theory
relatively prime
advanced fields
advanced fields unsolved

:mad: Suppose that b is a positive integer which is divisible by neither 2 nor 5. Prove that  for each positive integer n, there is a power of b ending with n-1  0's and then a 1.

Example
i :) f n = 6, then we need an integer k s.t. b^k ends in 000001

We could even choose an exponent depending only on $n$. ($4$ if $n=1$, $20$ if $n=2$, $100$ if $n=3$, $5\cdot 10^{n-2}$ if $n\ge 4$)

The key (for your original question): If $a$ is relatively prime to 10 (not divisible by 10) it has an inverse $a^{-1}$ mod $10^{n}$ for each $n$ ($a\cdot a^{-1}\equiv 1\mod 10^{n}$. Combine this with your pigeonhole hint.


AMC
AIME
geometry
AMC 12
AMC 12 A
MATHCOUNTS
power of a point

Well, if your problem is that you can't figure out where to start, go through a mental list-try branching first geometry or algebra, now just as an example-there's a problem: here are your thought processes:
Aha! A geometry problem!
I can't use the parallel lines here...
Aha! An application for Power of a Point! Bingo!

Then proceed from there. I hope this helps. :D

[quote="keta"]I have studied the AMC and AIME for one and a half years. I now recognize a lot of the types of questions on the AIME, but reading numerous previous posts, many of you seem to be able to get 10+ easily. My question is, I have looked at all the previous questions and when doing them, I still don't know where to begin. Once I look at the solution though, its simply trivial or obvious. It's frustrating. How can I make the next jump to actually being able to solve 9 or 10 without looking at the solutions? Any advice is appreciated.[/quote]

Practice and experience are valuable on the AIME. The more problems you see, the easier they'll be. After a while, you'll start seeing your score go up. And many times, it'll go up faster than you thought.

That's exactly what I did to improve my score this year (12 points). I would read a question once, then re-read it to make sure I knew what they were asking. Then I would start thinking about what there was and what I could apply to the problem. For example, when I notice parallel lines the first thing I think of is similar triangles from the congruent angles. Similarly, when dealing with circles I often considering power of a point or drawing extra segments in order to form some kind of right angles (cyclic quadrilaterals are also useful). Just mentally list out possible theorems that might come in handy and try to apply them. If you seem to be getting nowhere, then move to the next one. After a while your thinking will become more coherent and you will begin to be able to solve the problems much more quickly.

First, you may try to find out if there is any superfluous information.  Many times, this will reveal a link in trying to figure out the problem.  Then, you just plain look for links between the pieces of information your given and the one youre trying to find.  For example, on AMC12A 2005 #23, you draw in the altitude of $\triangle DEC$, $h_E$, and it makes a right triangle.  You just linked two pieces of information to one that makes the problem trivial.

... and don't use any mathcounts time-managment strategy whatsoever unless it's as easy as number two...


modular arithmetic
quadratics
number theory proposed
number theory

Obviously x-1|p(x)
let q(x)=p'(x) r(x)=p''(x)
If q(1)=0 (or (x-1)^2|p(x)) we get sum m*(a/m)=0.
If p=4k+3 then sum(m)=(p-1)(p-2)/2 is odd, contradiction.
If p=4k+1 then (a/p)=1 iff (p-a/p)=1 and residues group in pairs of sum p and the sum is 0.
If p=8k+5 q(1)=0 or sum m^2*(m/p)=0.
Grouping m with p-m we get (sum 2m^2+sum(p^2-2m))(m/p)=0.
The latter sum is 0 by a) so we get sum (m/p)m^2=0 (m=1..(p-1)/2)
But sum m62 is odd, contradiction.

I didn't understand very well what Iura said, so I apologise if my solution is the same:
  Part a) is trivial. Let's prove part b). I will prove that $\sum_{i=1}^{p-1}{i^2\cdot(\frac{i}{p})} $ is 4 modulo 8. Note that $\sum_{i=1}^{\frac{p-1}{2}}{(2i)^2\cdot(\frac{2i}{p})}$ is congruent to 4 modulo 8 (just work modulo 2 with $ \sum_{i=1}{\frac{p-1}{2}}{i^2}$ mod 2). Thus, it is enough to prove that $\sum_{i=1}^{\frac{p-1}{2}}{(\frac{2i-1}{p})} $ is 0 mod 8. ( use the fact that $ x^2$ is 1 modulo 8 for any odd x). But this is easy, since we obviously have $\sum_{i=1}^{\frac{p-1}{2}}{(\frac{2i}{p})}+\sum_{i=1}^{\frac{p-1}{2}}{(\frac{2i-1}{p})}=0 $. Now, we also have $\sum_{i=1}^{\frac{p-1}{2}}{(\frac{2i}{p})}=\sum_{i=1}^{\frac{p-3}{2}}{(\frac{2\cdot(\frac{p-3}{2}-i)}{p})+1=1+\sum_{i=1}^{\frac{p-3}{2}}{(\frac{2i+1}{p})}=\sum_{i=1}^{\frac{p-1}{2}}{(\frac{2i-1}{p})} }$. The last two relations solve the problem.

what does 'trivial' mean HERE?I've spent 0.5 hours on it!
my solution for part 1:we only have to prove
$\sum_{i=1}^{p-1}(\frac{i}{p})=0,\sum_{i=1}^{p-1}i(\frac{i}{p})=0$ if and only if $p=2 or p=4k+1$
when$p=4k+3$,then$\sum_{i=1}^{p-1}i(\frac{i}{p})\ne 0$
(that's because $\sum_{i=1}^{p-1}(-1)^{f(i)}(\frac{i}{p})$ is an odd.)
when $p=4k+1$,we can pair the $\frac{p-1}{2}$ quadratic residues and non-quadratic ones,so adding up gets 0.

Part a is quite trivial :wink:

It suffices to examine $f(x) \cdot x$. Clearly this does not interfere with divisibility by $x-1$. Thus $f$ denoted below is just $f(x) \cdot x$.

$f(1) = \sum_{i=1}^{p-1} \left ( \frac{i}{p} \right ) =0$ clearly.

To show $(x-1)^2 \nmid f(x)$, it suffices to show $f'(1) =0$ iff $p \equiv 1 \pmod{4}$
The forward direction is trivial, just pair $i$ with $p-i$.
The reverse direction is also trivial, if $p \equiv 3 \pmod{4}$ then we need to show $\sum_{i=1}^{p-1} i \left ( \frac{i}{p} \right ) \not = 0$. The LHS is a sum of $(p-1)/2$ even terms and $(p-1)/2$ odd terms and hence is odd, so we're done.

Part b. can be dealt with in a similar fashion. We've already shown $(x-1)^2|P(x)$ so $(x-1)|P'(x)$. Hence we need to show $(x-1) \nmid (xP'(x))'$, or in other words show $\sum_{i=1}^{p-1} i^2 \left ( \frac{i}{p} \right ) \not = 0$.
From the fact that for $p=5$ we have the sum is $4$, we are inspired to show the sum is $4 \pmod{8}$.
Clearly if we take $\sum_{i=1}^{(p-1)/2} (2i)^2 \left ( \frac{2i}{p} \right )$ then it's $4 \pmod{8}$ because we simply have to show $\sum_{i=1}^{(p-1)/2} i^2 \left ( \frac{2i}{p} \right )$ is odd, which is trivial using above methods.

Then we need to show $\sum_{i=1}^{(p-1)/2} (2i-1)^2\left ( \frac{2i-1}{p} \right ) \equiv 0 \pmod{8}$. As $2i - 1$ is odd we know $(2i-1)^2 \equiv 1 \pmod{8}$, hence we have to show $\sum_{i=1}^{(p-1)/2} \left ( \frac{2i-1}{p} \right ) \equiv 0 \pmod{8}$.

However, this is easy...
$\sum_{i=1}^{(p-1)/2} \left ( \frac{2i-1}{p} \right ) = \sum_{i=1}^{(p-1)/2} \left ( \frac{p-2i+1}{p} \right ) = \sum_{i=1}^{(p-1)/2} \left ( \frac{2i}{p} \right )$

But $\sum_{i=1}^{(p-1)/2} \left ( \frac{2i-1}{p} \right ) + \sum_{i=1}^{(p-1)/2} \left ( \frac{2i}{p} \right ) = 0$, hence they are both 0 and therefore we are done as this implies $\sum_{i=1}^{(p-1)/2} \left ( \frac{2i-1}{p} \right ) \equiv 0 \pmod{8}$.

$a)$
We need to prove that $x^1||f(x+1)$ iff $p \equiv 3 \pmod 4$. $f(x+1) \equiv \sum_{i=1}^{p-1} i\left ( \frac{i}{p} \right )  (mod  x^2)$.
$p \equiv 1 \pmod 4$ then $\sum_{i=1}^{p-1} i\left ( \frac{i}{p} \right ) =0 $
$p \equiv 3 \pmod 4$ then $\sum_{i=1}^{p-1} i\left ( \frac{i}{p} \right ) \equiv \sum_{i=1}^{p-1} i= \frac{p(p-1)}{2} \equiv 1 \pmod 2$.
$b)$
We need to prove that $x^2||f(x+1)$,where $p \equiv 5 \pmod 8$.
$f(x+1) \equiv \sum_{i=1}^{p-1} i^2\left ( \frac{i}{p} \right )x^2 (mod  x^3)$.
$ \sum_{i=1}^{p-1}i^2\left ( \frac{i}{p} \right )= \sum_{0<i<p,2|i} i^2\left ( \frac{i}{p} \right )+\sum_{0<i<p,2 \nmid i} i^2\left ( \frac{i}{p} \right ) \equiv 4\sum_{0<i<p,2|i}\frac{i^2}{4}\left ( \frac{i}{p} \right )+\sum_{0<i<p,2 \nmid i}i^2\left ( \frac{i}{p} \right ) \equiv 4+\sum_{0<i<p,2 \nmid i}i^2\left ( \frac{i}{p} \right )\equiv4+\sum_{0<i<p,2 \nmid i} \left ( \frac{i}{p} \right )\equiv 4\pmod 8$.

[quote=Valentin Vornicu]Let $p$ be a prime number and $f\in \mathbb{Z}[X]$ given by
\[ f(x) = a_{p-1}x^{p-2} + a_{p-2}x^{p-3} + \cdots + a_2x+ a_1 , \]
where $a_i = \left( \tfrac ip\right)$ is the Legendre symbol of $i$ with respect to $p$ (i.e. $a_i=1$ if $ i^{\frac {p-1}2} \equiv 1 \pmod p$ and $a_i=-1$ otherwise, for all $i=1,2,\ldots,p-1$).

a) Prove that $f(x)$ is divisible with $(x-1)$, but not with $(x-1)^2$ iff $p \equiv 3 \pmod 4$;
b) Prove that if $p\equiv 5 \pmod 8$ then $f(x)$ is divisible with $(x-1)^2$ but not with $(x-1)^3$.

[i]Sugested by Calin Popescu.[/i][/quote]

[color=#f00]a[/color]
As $f(1)=\sum\limits_{i=1}^{p-1}\left(\frac{i}{p}\right)=0$, and
$$f'(1)
=\sum\limits_{i=1}^{p-1}(i-1)\left(\frac{i}{p}\right)
=\sum\limits_{i=1}^{p-1}i\left(\frac{i}{p}\right)
=\sum\limits_{i=1}^{p-1}(p-i)\left(\frac{p-i}{p}\right)
=(-1)^{\frac{p-1}{2}}\sum\limits_{i=1}^{p-1}(p-i)\left(\frac{i}{p}\right)
=-(-1)^{\frac{p-1}{2}}f'(1)$$
For $p\equiv 1\pmod 4$, we have $f'(1)=0$ and $(x-1)^2|f(x)$
For $p\equiv 3\pmod 4$, we have
$$f'(1)=\sum\limits_{i=1}^{p-1}i\left(\frac{i}{p}\right)\equiv\sum\limits_{i=1}^{p-1}i=\frac{p}{p-1}\equiv 1\pmod 2$$
so that $(x-1)||f(x)$


Hello,

I was just wondering, who use this forum?

In my case, I am David Jimenez, I am a grad student at GATech. What about you. I know Billy and Linda (but they can talk by themselves), but does anyone else use the forum (This Georgia forum?)

Best,

I moderate this forum. You can probably see that there isn't much stuff to moderate, but oh well. To introduce myself, I am Howard Tong, currently a freshman.  I go to Phillips Exeter, a boarding school in New Hampshire, but before that I went to Chamblee Middle. Any questions or anything, feel free to contact me through PM and such

Hi everyone! I'm Sam Brotherton, a freshman at the Rockdale Magnet School.


A man was pushing his car when he stoped outside a hotel, the man suddenly realised he was bankrupt. how is this possible?

The man had invested all of his money into the hotel and it was being shut down?

While these riddles are nice, the problem is that they frequently have many, may correct answers.  Whichever answer you have is just one of them.

sorry, there is only one answer to this riddle and that wasnt it

he was playing monopoly and he had the car piece
he stopped outside the hotel owned by another player and was bankrupt when he had to pay the other person

since i can't edit my post anymore, i'll ask on this post

did i get it right??????

yep the man was playing monopoly sorry it took so long to post answer but life got in the way

I got this in record time of 5 seconds. :(

this has already been solved (by me :P ), so plz don't post the answer anymore
it'll be considered spam
thanks


What is the greatest factor of 11! that is one greater than a multiple of 6?

Thank you! :P

Note that if the factor is $ 1$ greater than a mulitple of $ 6$, then it is also one greater than a mutiple of $ 3$ and a multiple of $ 2$. Hence for our number, it must be entirely composed of factors which are not multiples of $ 2$ or $ 3$. The only such factors of $ 11!$ are $ 5$, $ 7$ and $ 11$ (we actually have two fives). Trying out cases, we find that $ 5^2(7)\equal{}\boxed{175}$ is the largest.


abstract algebra
group theory
complex analysis
complex analysis unsolved

Show that for any point $ p\in\mathbb{C}/\Lambda$ (a complex torus) and any integer $ n\geqslant 1$, there is a point $ q\in\mathbb{C}/\Lambda$ such that $ nq \equal{} p$.  Furthermore, show that there are $ n^{2}$ such points $ q$.

[hide="Hint"]Show that the kernel of the group homomorphism on $ \mathbb C/\Lambda$ induced be the multiplication with $ n$ is isomorphic to $ \Lambda/n\Lambda$. Then derive $ |\Lambda/n\Lambda|\equal{}n^2$.[/hide]


Denote $n$ numbers among $1, 2, ..., n+1$ by $a_{1}, a_{2}, ..., a_{n}$ such that $|a_{1}-a_{2}|, |a_{2}-a_{3}|, ..., |a_{n}-a_{1}|$ are different numbers. Find all $n\geq 3$ for which, this is possible. (ie the necessary and sufficient condition.)

[mod.: made a canonical correction of an error]

Paying attention to the parity, and that the different numbers are exactly $1, ..., n$ leads us to the conclusion that $n=4k,\; n=4k-1$. Now one should prove that this condition is also sufficient. I didn't exactly get you Rust. :wink: Can anyone follow my proof?

[quote="Rust"]It is true for all n. My solution satisfied it for n=4k, and for 4k+1 .[/quote] As I told you $n=4k$ or $n=4k-1$ is a necessary condition. For, $c$ and $|c|$ has the same parity; hence $\Sigma_{i=1}^{n}|a_{i}-a_{i+1}|$ is even, where $a_{n+1}=a_{1}$. Thus $\frac{n(n+1)}{2}$ is even and the result follows. I just wanted you to show that this condition is sufficient as well.

Repetition is NOT allowed in choosing $n$ numbers, or to put it another way we throw away one natural number between $1$ and $n+1$ and consider the other ones.


induction
inequalities
number theory unsolved
number theory

Suppose there are n positive integers which have sum S and product P. Prove that there are no other n positive integers with this property.

$ 2\plus{}7\plus{}12 \equal{} 3 \plus{}4 \plus{}14 \equal{} 21$
$ 2*7*12 \equal{} 3*4*14 \equal{} 168$

If I understood it correctly, then you must have forgotten to add sth.

//69

... or find a counterexample!

But what I meant to say was, given a set of n positive integers, some pair of which differ by 1 and no pair of which differs by more than 1, can you find a different set of n positive integers which have the same sum and product as the first set? The original set is something like 6,6,6,7,7 or 12,12,13 or something like that.

No, then it is impossible.

Suppose we can find such a set. Let's say we got in our set $ k$ elements with value $ A$ and $ l$ elements with value $ A \plus{} 1$. Our "normal" set consists of $ t$ elements with value $ S$ and $ m$ elements with value $ S \plus{} 1$.

Then we obtain:
$ k*A \plus{} l*(A \plus{} 1) \equal{} t*S \plus{} m*(S \plus{} 1)$
$ A^k*(A \plus{} 1)^l \equal{} S^t*(S \plus{} 1)^m$
$ k \plus{} l \equal{} m \plus{} t$

It's very easy to find out that $ S \equal{} A$ (using second equation and third) . WLOG we may suppose that $ k > t$ (if they are equal then our set is not different from the first one). Then:

 $ A^k / A^t  \equal{} (A \plus{} 1)^m / (A \plus{} 1)^l$

But $ gcd(A,A \plus{} 1) \equal{} 1$ and considering it we obtain that $ k \equal{} t$ and $ m \equal{} l$. Anyway I didnt use the fact that  the sums must be equal. If I made sth wrong, then correct me, please.

/

You're not wrong, but that's not quite the problem I asked about.

"But what I meant to say was, given a set of n positive integers, some pair of which differ by 1 and no pair of which differs by more than 1, can you find a different set of n positive integers which have the same sum and product as the first set? The original set is something like 6,6,6,7,7 or 12,12,13 or something like that."

I never said the different set of n positive integers had to have no pair differing by more than 1.

You have an original set of n positive integers, with the property that it is of the form {x,x,...x,x+1,x+1,...,x+1}. Let's call such a set bunched. The question is, can you find a different set of n positive integers which has the same sum and product? The different set can be any set of n positive integers.

In fact, the problem in my last post is a subproblem of the following problem:

Let x1, x2,..., xN be N positive integers whose sum is KN+r, where r is a nonnegative integer less than N. Then the product of these N positive integers is less than or equal to ((K+1)^r)K^(N-r), with equality if and only if the integers x1, x2,..., xN are K+1,...,K+1,K,...K, where the K+1 appears r times and the K appears N-r times, in some order.

Can anyone prove that?

Actually, I think I've solved this problem now. It's by induction, surprisingly enough.

If K, N, r are positive integers such that 0<r<n and we have n positive integers x1, x2,..., xN such that x1+x2+...+xN=KN+r, then x1x2...xN=<((K+1)^r)K^(n-r) with equality if and only if x1, x2,..., xN = K+1,...,K+1,K,...K, where the K+1 appears r times and the K appears n-r times.
Demonstration:
Let K be an arbitrary positive integer.
Let r be an arbitrary positive integer.
Amongst the positive integers which are suitable candidates for n (the number of positive integers in our list with sum Kn+r), namely those which are strictly greater than r, let n be arbitrary.
Suppose we have that the theorem holds true for n numbers summing to Kn+r.
Consider the n+1 numbers x1, x2,..., x(N+1) who have sum K(N+1)+r.
If they are equal to K+1,...,K+1,K,...,K in some order when the K+1 appears r times and the K appears N+1-r times, then we have equality.
Suppose they do not have this form, i.e. not all the numbers are K+1 or K. Then there must be a number larger than K. Now we take two cases.
Case A: There is not a number smaller than K.
All numbers are >=K, hence there are clearly at least N+1=r K's amongst them. Pick N+1-r of these K's and cancel them to get (N+1)-(N+1-r)=r numbers whose sum is K(N+1)+r-K(N+1-r)=r+rK=(K+1)r. As these numbers are not all K+1, by the AM-GM inequality they have product <(K+1)^r. Hence adding the N+1-r K's back in, we have that the original N+1 numbers have product <((K+1)^r)k^(N+1-r).

Case B: There is a number smaller than K.
Then there is a pair of numbers, let's say they are x(N+1) and xN, such that x(N+1)>K and xN<K.
Hence 0<(x(N+1)-K)(K-xN) = K(x(N+1)+xN-K)-x(N+1)xN
Hence x(N+1)+xN-K>(xNx(N+1))/K.
Now consider the N numbers x1, x2,...,x(N-1), (x(N+1)+xN-K). They have sum K(N+1)+r-K=KN+r. Then by the theorem for K, r and N, they have product less than or equal to ((K+1)^r)K^(N-r).
Hence x1x2...x(N-1)(xNx(N+1))/K<((K+1)^r)K^(N-r).
Hence x1x2...x(N-1)xNx(N+1)<((K+1)^r)K^(N+1-r).

Hence given arbitrary K and r, and suitable N, the theorem for K, r and N implies the theorem for K, r and N+1.
Hence to prove the theorem in general all we have to do is prove that for arbitrary K and r, the theorem is true for the smallest possible N, namely r+1. Equivalently, we have to prove that the theorem is true in general for N numbers summing to KN+(N-1), where N is a positive integer.

To keep the notation as similar as possible, I will do it for N+1 numbers summing to K(N+1)+N. We want to prove that their product must be =<((K+1)^N)K, with equality iff the numbers are K+1,...,K+1,K.
Let's take a set of N+1 numbers with sum K(N+1)+N. Clearly we have equality if they are K+1,...,K+1,K. Otherwise, let us split the investigation into two cases.
Case A: There is not a K.
Then at least one of the numbers must be >K, and at least one of the numbers must be <K. Let's say x(N+1)>K and xN<K.
Hence 0<(x(N+1)-K)(K-xN) = K(x(N+1)+xN-K)-x(N+1)xN
Hence x(N+1)+xN-K>(xNx(N+1))/K.
Now consider the N numbers x1, x2,..., x(N-1), (x(N+1)+xN-K).
They have sum K(N+1)+N-K=KN+N=(K+1)N. By the AM-GM inequality, their product is =<(K+1)^N.
Hence x1x2...x(N-1)(xNx(N+1))/K<(K+1)^N.
Hence x1x2...x(N-1)(xNx(N+1))<((K+1)^N)K.

Case B: There is a K.
x1, x2,...,x(N+1) contains at least one K, hence we can cancel a K and get N numbers whose sum is K(N+1)+N-K=KN+N=(K+1)N. Hence by the AM-GM inequality, these numbers have product strictly less than (K+1)^N, because they are not all K+1. Putting the original K back in, the N+1 numbers have product <((K+1)^N)K. This completes the proof of the theorem. Q.E.D.


trigonometry
inequalities unsolved
inequalities

Let $ABCD$ be a quadrilateral inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  If $s$ is the semiperimeter we have

$ s \leq \sqrt{4R^2 +r^2} + r $

with equality if and only if at least one pair of opposite angles are right angles

[quote="socrates"][quote="manlio"]Let $ABCD$ be a quadrilateral inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  If $s$ is the semiperimeter we have

$s \leq \sqrt{4R^2 +r^2} + r$

with equality if and only if at least one pair of opposite angles are right angles[/quote]

 :D  :D[/quote]
Why need you in sending :D  :D 3 times :?:

hey socrates, enough spamming please :D  :spam:

This is not spam ;)
Have a look at the date of the first post and you'll understand why!


geometry
3D geometry
combinatorics proposed
combinatorics

There is an $8\times8\times1$ cube is tiled with 64, unit cubes. Could we tile the $4\times4\times4$ cube with these cubes that every two adjacent cubes in $8\times8\times1$ cube are adjacent in the $4\times4\times4$ cube?

the answer is no.
If we consider the arrangement as a mapping from an 8*8 square onto a 4*4*4 cube, then iIt is easy to see that the image of any 2*2 square is again a 2*2*1 block (= a 2*2 square), i.e. all four blocks are complanar.
This is already half of the proof.
By the adjacency conditions, it follows easily from there that the image of any 3*3 square is a 3*3 square again, i.e. all complanar.
Now by overlapping 3*3 squares, it follows that all the images are lying in the same layer. No cube. Contradiction.


linear algebra
matrix
Gauss
number theory
greatest common divisor
absolute value
linear algebra unsolved

True or false? Given a n x n matrix A with integer entries, there exists a n x n matrix B, having integer entries, such that det B = 1 and such that AB is upper triangular.

the statement is true, a hacked proof is attached

Could you please post the pdf version of this? It looks very weird when I open it with Microsoft Word or Open Office.

here is the pdf version, sorry for the inconvenience

Much better, thanks! However, I am confused. The official solution of problem 3 of IMC 2006 mentions this property of matrices with integer entries without a proof. Is it a known result, then? I didn't know it and apparently there is no trivial explanation for it, so I don't understand why they just mention it without proving it.

[quote="Arne"]Is it a known result, then? I didn't know it and apparently there is no trivial explanation for it, so I don't understand why they just mention it without proving it.[/quote]
I think so. It basically follows by Gauss-Jordan elimination. The rationale is pretty classic:

Multiplying by integer elementary matrices to the right leads to column operations.

Let's try to make the bottom row of the form $ (0,\ldots,0,k)$. First, we pick the integer with the smallest positive absolute value that lies on that bottom row, say it's $ |a|$. Now, for all other $ b$ on that bottom row we write $ b \equal{} aq\plus{}r$ (if $ b\neq 0$, then $ |q|$ is at least $ 1$). Now do the corresponding elementary operations (of determinant $ 1$), so that each $ b$ is replaced by $ r$. What happened? Well, the smallest positive absolute value has either decreased, or has remained the same, in which case we have only non-zero value on the bottom row, say it's $ \ell$. We move it on the last place, so that it looks like this: $ (0,0,\ldots,\ell)$. Mission accomplished.
(Note: We always reach this case, since we can't have a strictly decreasing infinite sequence of non-negative integers.)

Now say the row before the bottom one looks like this: $ \left( h_{1},\ldots,h_{n\minus{}1},h_{n}\right)$. Next, we do the same thing as we did above to $ \left( h_{1},\ldots, h_{n\minus{}1}\right)$ (don't touch $ h_{n}$). In the end, just move that non-zero value next to $ h_{n}$.

Continue like this and you'll get it.

However, I agree that they should have given some hints there.

PS: What I did is just what [i]hjbrasch[/i] says in the next post.

I guess it is a standard result in linear algebra based on a modified Gauss elimination carefully choosing your elementary transformations (which I actually applied in my hacked proof). I would be surprised if the proof of this standard result can avoid any "gcd-language" and "reduction language" in order to eliminate coefficients to finally get an upper triangular matrix.

[quote="perfect_radio"] in which case we have only non-zero value on the bottom row, say it's $ \ell$. We move it on the last place, so that it looks like this: $ (0,0,\ldots,\ell)$. Mission accomplished.[/quote]
Suppose we have only one nonzero entry on the last row. Then the elementary coloumn operation that sends it to the lower right corner has $ det\equal{}\minus{}1$

I admit that I didn't notice that, but it's easy to fix: if the resulting $ B$ has $ \det \equal{}\minus{}1$, then replace it with \[ B^{\prime}\equal{} B\cdot\begin{pmatrix}\minus{}1 & 0 & 0 &\ldots\\ 0 & 1 & 0 &\ldots\\ 0 & 0 & 1 &\ldots\\ \vdots &\vdots &\vdots &\ddots\end{pmatrix}.\]


Niech $( a_n )_{n = 1}^{\infty}$ bedzie rosnacym ciagiem zlozonym ze wszystkich ''nie kwadratow'' (liczb naturalnych $n = a^2 \cdot b$ z $( a, b ) = 1$ i $b \neq 1$).
Na przyklad $a_1 = 2$, $a_2 = 3$, $a_3 = 5$, $a_4 = 6$, $a_5 = 7$, $a_6 = 8$, $a_9 = 10$
Pokaz ze
\[ a_n = n + \left\lfloor \sqrt{n} + \frac{1}{2} \right\rfloor  \]


Hi everybody,
              Can anybody tell me what thoerem we can use without showing a proof ? If we don't show it, will they give us point ?

You can use basically any published theorem. Although it is preferred that you use some variant of its official name, if you can't remember it you might be able to get away with "well-known theorem" or something to that effect - the graders are familiar with more identities than we are. Of course, if you cite something patently obscure, you may want to reference where it can be found in mathematical literature. To be particularly safe you could outline a proof of the theorem in question.

Thank you, Mildorf. By the way, do you know when they release the lists of qualified people for the USAMO ?

Probably any theorem you know you can use without proof.  It's best if you know the name.

Everyone on the website's saying March 28. Is that date set in stone somewhere? (Just wondering...)

No...which is why I'm saying it could easily be before.


I have seen an interesting problem in a magazine, stating that if $A_1,...,A_k$ are $nxn$ matrices such that the sum of all $A_{\sigma(1)}\cdot...\cdot A_{\sigma(k)}$ over all permutations $\sigma$ is equal to the identity, then $n$ must be a multiple of $k!$. Yet, the solution is wrong (it was based on traces properties, but on wrong ones). Can you prove or disprove the statement?

Presumably the $A_i$ are matrices over the integers?


Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna.  In how many ways can we choose 6 starters if exactly two of the triplets are in the starting lineup?

I didn't see that the question required 6 people, but I did see that it required 2 of the triplets, and I accounted for that

We choose any two triplets. There are three choices, $\binom{3}{2}$=3. There can only be two triplets so there are still 11 players to fill in the rest of the 4 spaces. That makes it $\binom{11}{4}=330$. If we multiply them together, we get 330*3 which equals $\boxed{990}$.

Counted them by hand... JK.
[hide=Solution]
3choose2 times 11 choose 4
11 is 14-3 which is the triplets[/hide]


geometry
circumcircle
incenter
inequalities
inequalities unsolved

Let d_1 \leq d_2 \leq d_3 where d_1,d_2,d_3 are the distances of the circumcenter of a triangle to its excentres. Prove that

1)  R<d_1<2R

2) R<d_2<R \sqrt 5

3) 2R<d_3<3R

A part oif the solution. I want to prove that all d_i are >R. For this we should prove that Ia, Ib, Ic are outside the circumcircle. It's enough to prove it for Ia, since it's the same for the others:

Let D be the intersection between the bisector of angle BAC and the circumcenter of ABC and let I be the incenter of ABC. If Ia is inside the circumcenter, then it should be on [AD]. It's obvious that one of the angles ABD and ACD must be <=90. Let's assume ABD<=90. This means that IBD<90, but IBIa=ICIa=90, so Ia can't be inside the segment [AD]. This proves 2 of the 6 inequalities you've posted here:

d_1>R and d_2>R.

Hope I'll be back. :)

Remember that OI2 =R2 -2Rr? There's a similar formula for the exscribed circles:

OI_a2 =R2 +2R*r_a. I can prove it if you want me to.

Another part of the proof:

Let's assume OI_a<=OI_b<=OI_c. OI_c<3R iff OI_c2 < 9R2 iff r_a<4R iff S/(p-a)<abc/S, where p is the semiperimeter and S is the area of ABC.

We now must prove S2 <abc(p-a) iff p(p-a)(p-b)(p-c)<abc(p-a) iff p(p-b)(p-c)<abc. We put p-a=x, p-b=y, p-c=z, and we have to prove (x+y+z)yz<(x+y)(y+z)(z+x) for x,y,z>0, which is obvious.

Can you please post your proof of the formula

 OI_a^2 =R^2 +2R*r_a

Thank you very much Grobber.

I see this like so:

Because of the fact that the power of the point I_a refering to its circumscriebed circle (ABC's ) is : P_c( I_a)=2Rr_a it is equivalent to saying that:
OI_a^2=R^2+2Rr_a

cheers!


AMC AIME geometry algebra polynomial Vieta trigonometry

Hello,

Anyone know how to study/practises for the AIME/AMC 10/AMC 12 contests? I know that I can print off contests rom the website.  However, are there any tricks and topics that are usually on the math contests?

Thanks a lot

There really aren't any tricks that will drastically improve your performance on such contests. The most important thing is to do problems. If you don't have the requisite knowledge in a subject area, get a book or look up stuff online to learn it better.

Thanks.

However, are there any topics that have a lot of tricks? e.g. Geometry and some algebra. Where would I get such knowledge of the tricks because they don't teach the tricks at school and I don't really have many books on these topics (Where in this website or where outside of the site?)

Thanks again..

There aren't really any "tricks." You just learn the material. Eventually after studying and working with the material so much you become so much more comfortable with it that it seems simple. One could call these skills "tricks" but they really aren't.

There are a few areas where there are tricks (Vieta jumping is a prime example), but they are rare and not widely useful.

I signed up for the AMC 12 class for the summer. Hopefully it will help me win those extra points to get me into the AIME. I'm in 8th grade, so I can get in through AMC 10 and/or AMC 12.

I have written the AMC 12 and AIME twice. I believe that there are no tricks that determine one's performance on these contests. There are some main strategies which one can read in AOPS books or in any other problem book. "The Art and Craft of Problem Solving" by Paul Zeitz is a perfect book. Instead of searching for tricks one should learn some very useful theorems such as the law of sines, the law of cosines, Ptolemy's theorem, Pell's Equation and Vieta Jumping (as mentioned earlier).

In no universe does Vieta Jumping qualify as a "very useful theorem." It's a trick that's been useful in about 5 olympiad problems ever.

I think u should just work on what u feel is your weakest spot.

This is probably the wrong place to ask...sorry...but what is Vietas and Vieta Jumping?  If this can even sometimes help me on the AMCs I'd like to know (even if it isn't [b]usually [/b]that useful)

Vieta's [b]formulas[/b] are ways to express the coefficients of a polynomial in terms of symmetric sums of its roots. They are extremely useful and show up incredibly often. See [url=http://en.wikipedia.org/wiki/Vieta%27s_formulas]Wikipedia[/url] or [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas]AoPSWiki[/url].

Vieta jumping is a specialized and rarely useful number theory trick. It will never be useful on short-answer contests.


does anyone know what the cutoff score for AMC 10B?
it is best if it is the official cutoff socre.
i'm pretty sure i got 120.5, but last year the cutoff score was like 121, so i'm very worried
(i know i suk, but it is my first time taking it)

yep, 120 is the maximum qualifying this year. so no matter what the entire world got, 120 will make to AIME.

how can u be sure? i mean only top 1% get to take AIME, not necessary the people who get 120 or more, last year the cutoff for 10B was 121

They changed the rule -- now, it's the top 1% OR everyone who got over 120, whichever includes more people.

check the cover of this year's AMC 10 A or B, and you shall find the qualification modifications....


geometry
projective geometry
geometry solved

$M$ is the intersection between the polar of $A$ wrt the incircle, and the polar of $D$ wrt the incircle, which means that it is the pole of $AD$. In the same way we show that $N,P$ are the poles of $BE,CF$ respectively. This means that the line $\ell=MNP$ is the polar of $AD\cap BE\cap CF$, which is the Gergonne point $G$. Now, since the line connecting a point to the center of a circle is perpendicular to the polar of that point wrt the circle, it means that $IG\perp\ell$, as desired.

[quote="Socc"]Given the triangle $ABC$ . The incircle (I) of the triangle touchs $BC,CA,AB$ at $D,E,F$ respectively. $EF\cap BC$ at $M$;$FD\cap CA$ at $N$;$DE\cap AB$ at $P$
Show that $M,N,P$ lie on the same line and that line perpendicular to $IG$. With G is the Gergorne point of the triangle $ABC$[/quote]

This is a direct corollary of the Romanian TST problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=5485 : According to that Romanian TST problem, the lines IG and NM are perpendicular; in other words, the point M lies on the perpendicular to the line IG through the point N. Similarly, the point P lies on the perpendicular to the line IG through the point N. Thus, the three points M, N, P all lie on the same line which is perpendicular to the line IG.

  Darij

Something nonsense:
By Desargues Theorem, M,N,P are collinear.


invariant
geometry
3D geometry
combinatorics proposed
combinatorics

A cube $W$ with edge length 2 is composed of cubes $W_1, \ldots, W_8$ with edge length 1. Each of the six faces of a small cube either is colored white or black. Together there are $8 \cdot 6$ small cube faces of those 24 are colored white and the black each. Show that it is possible that $W$ can be composed of the cubes $W_1, \ldots, W_8$ such that on the outer faces of $W$ there are exactly as many white squares as black ones of length 1.

What we must do is pick a vertex from each of the small cubes, and select the faces containing that vertex. In this way, we select $24$ faces, $12$ of which have to be white.

We can see that if we select a vertex from each small cube as above, and then, for a certain cube, we select a vertex adjacent to the one we had selected before, then the number of white faces changes with $0$ or $1$. Call such an action a "good" move.

Select a vertex $v_i,\ i\in\overline{1,8}$ from each small cube, in any way that you want. Assume the number f white faces containing these is $n$. If now, for each cube $i$, we select the vertex opposite to $v_i$, we get prtecisely those faces which we ommitted the last time, so the number of white faces is $24-n$ this time. Since we can go from the previous position to this one through good moves, at some point we will have $12$ white faces (because $12$ is between $n$ and $24-n$ and we advance from $n$ to $24-n$ in steps of length $0$ or $\pm 1$).


inequalities
real analysis
real analysis unsolved

[quote="Michał"]Let $x_1<x_2<\ldots<x_n$ be positive integers. Prove that

$\sum_{i=1}^{n}\frac{\sqrt{x_i-x_{i-1}}}{x_i}<\sum_{i=1}^{n^2}\frac{1}{i}$[/quote]
Let's set $x_0 = 0$. For $n=1$, the estimate is obviously correct.
For $n=2$, we can estimate
$\sum_{i=1}^{2}\frac{\sqrt{x_i-x_{i-1}}}{x_i}< 1 + 1 < \frac{25}{12} = \sum_{i=1}^{4}\frac{1}{i}$.

Now the general case, $n > 2$:

Start with Cauchy-Schwarz.
$\sum_{i=1}^{n}\frac{\sqrt{x_i-x_{i-1}}}{x_i} = \sum_{i=1}^{n}\frac{\sqrt{i(x_i-x_{i-1})}}{x_i} \frac{1}{\sqrt{i}} \le (\sum_{i=1}^{n}\frac{1}{i})^{1/2} (\sum_{i=1}^{n}\frac{i(x_i-x_{i-1})}{x_i^2} )^{1/2}$.

Next work on the second factor: For $i > 1$, we have
${\frac{i(x_i - x_{i-1})}{x_i^2} < \frac{i(x_i - x_{i-1})}{x_i x_{i-1}} = i ( \frac{1}{x_{i-1}} - \frac{1}{x_i}})$
and hence by summation by parts
$\sum_{i=1}^{n}\frac{i(x_i-x_{i-1})}{x_i^2} < \frac{1}{x_1} + \sum_{i=2}^n i ( \frac{1}{x_{i-1}} - \frac{1}{x_i}) = \frac{2}{x_1} + \sum_{i=1}^{n-1} \frac{1}{x_i} - \frac{n}{x_n}$.
This is bounded by $2 + \sum_{i=1}^n \frac{1}{i}$, since clearly $x_i \ge i$ for all $i$. Therefore,
$\sum_{i=1}^{n}\frac{\sqrt{x_i-x_{i-1}}}{x_i}<\sqrt{\sum_{i=1}^{n}\frac{1}{i}(2+\sum_{i=1}^{n}\frac{1}{i})}$.

For $n > 2$, this is a sharper estimate, since
$\sum_{i=1}^{n^2} \frac{1}{i} = \sum_{i=1}^n \frac{1}{i} + \sum_{j=2}^n \sum_{k=1}^n \frac{1}{(j-1)n+k} > \sum_{i=1}^n \frac{1}{i} + \sum_{j=2}^n \sum_{k=1}^n \frac{1}{jn} = 2\sum_{i=1}^n \frac{1}{i} -1$
and $2A-1 \ge \sqrt{A(2+A)}$ whenever $A \ge \sum_{i=1}^3 \frac{1}{i}$.  :lol:

Nice moves, thanks. I see you must have some experience on that.


For a fixed point $ O$ and four points $ A_1,\ A_2,\ A_3$ and $ A_4$ in Euclid space, define the point $ G$ by $ \overrightarrow {OG} \equal{} \frac {1}{4}(\overrightarrow{OA_1} \plus{} \overrightarrow{OA_2} \plus{} \overrightarrow{OA_3} \plus{} \overrightarrow{OA_4})$. If a point $ P$ in the space satisfies $ |\overrightarrow{PG}| \equal{} 1$, then prove that $ \sum_{k \equal{} 1}^4 |\overrightarrow{PA_k}|^2 \minus{} \sum_{k \equal{} 1}^4 |\overrightarrow{GA_k}|^2 \equal{} 4$.

$ \sum_{k \equal{} 1}^4 |\overrightarrow{PA_k}|^2 \minus{} \sum_{k \equal{} 1}^4 |\overrightarrow{GA_k}|^2 \equal{} 4$ $ \iff$ $ 4(\overrightarrow{P}^2 \minus{} \overrightarrow{G}^2) \minus{} 2(\sum_{k \equal{} 1}^4\overrightarrow{A_k})(\overrightarrow{P} \minus{} \overrightarrow{G}) \equal{} 4$ we know: $ \overrightarrow{G} \minus{} \overrightarrow{O} \equal{} \frac {1}{4}(\sum_{k \equal{} 1}^4\overrightarrow{A_k} \minus{} 4\overrightarrow{O})\Rightarrow \overrightarrow{G} \equal{} \frac {1}{4}\sum_{k \equal{} 1}^4\overrightarrow{A_k}$ . What we want to prove is equivalent to: $ (\overrightarrow{P} \minus{} \overrightarrow{G})(2\overrightarrow{P} \plus{} 2\overrightarrow{G} \minus{} \sum_{k \equal{} 1}^4\overrightarrow{A_k}) \equal{} 2$ but $ \sum_{k \equal{} 1}^4\overrightarrow{A_k} \equal{} 4\overrightarrow{G}$, conclusion $ \iff$ $ (\overrightarrow{P} \minus{} \overrightarrow{G})(2\overrightarrow{P} \minus{} 2\overrightarrow{G}) \equal{} 2$ but that is clearly true because $ (\overrightarrow{P} \minus{} \overrightarrow{G})^2 \equal{} 1$


function
topology
real analysis
real analysis solved

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $a,b\in \mathbb{R}$, $a<b$, with $f(a)=f(b)$, there exists a $c\in(a,b)$ with $f(a)=f(b)=f(c)$. Prove that the function $f$ is monotonous on $\mathbb{R}$.

Assume we have, for instance, $f(x)<f(y),f(z)<f(y)$ for some $x<y<z$ We may also assume $f(x)=f(z)$. Now take $x'=\sup\{a\in[x,y]|f(a)=f(x)\}$ and $z'=\sup\{b\in[y,z]|f(b)=f(z)\}$. Then $x'<z',\ f(x')=f(z')$, but for all values $c\in(x',z')$ we have $f(c)>f(x')=f(z')$.

well, the previous isn't really a solution. the essence of this problem is that one can take sup of those sets. for this, one needs to observe that $\{f(x)\}$ is a closed set, so $f^{-1}(f(x))$ is a closed set. If we look at elements $a\in [x,y]$, then it is also bounded, so it is compact and a sup exists.

:what?: It was so obvious, that grobber didn't need to write it.

BTW, I believe he meant  $z'=\inf\{b\in[y,z]|f(b)=f(z)$.

A slight note on English: the general phraseology is 'monotonic' and not 'monotonous'; the last word contains connotations of boredom and drudgery  :)


A man walks into a hardware store. He buys 1 for a dollar, 50 for two dollars, and 200 for three dollars. What is the man buying?

[hide]House numbers - you pay a dollar per digit.[/hide]


Outline a synthesis of p-nitrobenzenesulphonic acid from benzene.

1). nitrate benzene
2). reduce nitro group to get aniline
3).aniline + Ac2O
4). sulphonation
5).H2O/H+
6). oxidation to get back nitro group.