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geometry
3D geometry

Here r some starters:-
[color=darkred]The world's costliestcoffee at $ 130\minus{}$300 a pound is called '[i]Kopi Luwak[/i]' which is the droppings of some marsupials that eat only the best coffee beans!![/color]
[color=green]There's a city called ROME in every continent!!(u might be knowing that)[/color]

because every1 is busy studying hard for cbse math exam  :rotfl:  :rotfl:

no pun intended

[color=darkred]WELCOME TO AOPS! INDIA
by [u]rituraj007[/u][hide]a.k.a.[u]readnow08[/u](only for now)[/hide]
[/color]
[color=green]Beware of [u]mads[/u] 'n' [u]mod[/u] or u will go to [u]hell[/u](for)[u]ever[/u]
there r [u]shre[/u] 'n' [u]shru[/u] whose vetti talks will [u]alter ur ego[/u]
there are [u]chem.rock[/u]ers & comp.sci. victims so don't forget to [u]include#[/u]
there is [u]euclidean geometer[/u]
we have [u]chappli[/u] from youngistan
along with [u]soumyashant nayak[/u] & [u]vihang[/u]
[u]Albert einstein[/u] is also here with his [u](nitish)express[/u]
[u](verizon)wires[/u] can calculate [u](velirium)maxima[/u] of functions all complex
also there r [u]varun[/u] & [u]apek[/u]
and [u]anitsahu[/u] who always tried to reach the apex

it doesn't rime,but it is true
sorry if i missed somebody[/color]

hey i wanted every one 2 see it at once
and that's triple posting

why havent u included yourself in it? :)

You missed sanjith, myself and a lot of others but that's okay :)

@varun--don't u see my name in intro???(by rituraj007)
@ashwath--apologies!!!!-i'll think of editting

we r not even frm same city
and u r talking of school!!!!!! :rotfl:  :rotfl:  :rotfl:

i thought u all know each other so well that u might be in the same skoll!!!!

that's the spirit of  AOPS  or for that matter  [size=150]JOI[/size]

i thoroughly like and would alsom want to join in ur friend circle if u all  dont mind

U r supposed to go to iit,right?
So why pune? :huh:

[quote="Aashay THE ROKR"]And I am coming to Pune for the next 5 yrs :help:[/quote]

You don't live in Pune?

And Scary, please don't change the topic, it's going really fine!!!

[hide]Gaurav's pic in newspapers today![/hide]

[quote="Rajiv"][quote="Aashay THE ROKR"]And I am coming to Pune for the next 5 yrs :help:[/quote][/quote]

You don't live in Pune?

And Scary, please don't change the topic, it's going really fine!!!

*Gaurav's pic in newspapers today!*
So is d pic of rida shiekh  :(  , bad time for gauravs pic to come in paper , it will go unnoticed  :P

[color=green]Do not quote entire posts if yours follows it. It is assumed you are replying to the one above. - [b]MODERATOR[/b][/color] said to me in Calculus Computations & Tutorials.

Do post the pic
And which paper by the way? :read:
The ones available here are Statesman and telegraph mainly :(
Besides whatz the ocassion? :o

@scary
I have joined IISER Pune for 5 yr MS program.
@Rajiv
I stay in Mumbai.

[quote="Rajiv"][hide]Gaurav's pic in newspapers today![/hide][/quote]
I went and checked today's TOI frantically for 5 minutes before realising what you meant :(

Not Gaurav [b]Shukla[/b], it is Gaurav [b]Patil[/b]. BTW, I'd really like to see what GS looks like. I guess you are 5 feet 5 (or more), fair, wear expensive watches, RBK shoes and own a Honda City (probably with a driver in white uniform)

[hide]Jokin', man! People around here take me way too seriously.... (bloody [hide="nerds"]Jokin' again![/hide])[/hide]

@Rajiv
I guess u own a BMW or limo or rolls royce(coz ur location's Monte Carlo and one can expect atleast that from a lucky guy living there  :rotfl: )

i thot i was luky to escape 4m pune but ive caught a bad cold here.
i kno its swine flu. i dont wanna die. i wanna njoy coll life.  :(
[hide]just kiddin...but read a toi after abt a week or so. situation in pune seems really bad[/hide]

@Sagar/franky
So what?
One of my classmates is havin off-season chic pox & one's been havin a bad throat infection for a week now  :(
One fell ill in Kol :(

Shutup dude! (literally , not in the insulting manner)  :P
Dude , you come here , 3 people dead , and anyone you see on the road is wearing a mask  :o
Its no joke. Its a National Threat now , please take it seriously.
To give an example of the seriousness , we got a weeks hollidays here at MPrakash , And Thats more than one could ask for , cause he doesnt give a single holiday for a period of two years!

By the way ,
[hide]People sneezing around me has acquired an exceedingly sinister angle.[/hide]

[quote=Sagar_Vaze]it was abhinav school and not khs which started the swineflu  :mad:
[hide]abhinav and khs are two best schools in pune and i was in khs  :)
we used to call them ABhiNaV=Adani Bhikari Nalstopche Vadari  :rotfl:  [/hide][/quote]

ohhh looking at this post after 13 years [hide=Note]PS : Im from abhinav and MPrakash  :oops: [/hide]


How many distinct,nondegenerate triangles with positive integer sides have perimeter equal to 100?

[quote="13375P34K43V312"][hide]$ a+b>c,b+c>a,a+c>b$. Let's consider the value of $ c$, which forces the value of $ a+b$, then from here we consider the possible values of $ a$ and $ b$.

Fix the value of $ c$. Then, $ a+b=100-c$. Since $ a+b>c$, $ 100-c>c$ and $ c<50$. Now, $ a+c=100-b>b$, and $ b+c=100-a>a$., Thus, $ a<50$ and $ b<50$. Hence, we're just looking for $ a+b+c=100$ such that all three are less than 50. However, note that the order does not matter, so eh this is going to be ugly casework.

I'm not really up to doing all of it right now, but one could use balls and urns to find all ordered triples (a,b,c), but you have to be careful, since they have to be positive, then subtract all the instances when a=b, b=c, or c=a. Note that a=b=c is not possible. Then, apply complimentary counting and subtract those with one of them greater than 50, wlog, let it be a. and then you'll be done.[/hide][/quote]

I can get the part where you subtract the iscoscles triangle ones.

Like you use balls and urns, to get 6a+3b+c=number of solutions using balls and urns whbere a are scalene triangles (counted 6times) and b isosceles triangles (counted 3 times using balls and urns) and c-equilaterals ones, but there arent any, but I dont know how to count the degerneate triangle cases.

[quote="13375P34K43V312"][hide]$ a+b>c,b+c>a,a+c>b$. Let's consider the value of $ c$, which forces the value of $ a+b$, then from here we consider the possible values of $ a$ and $ b$.

Fix the value of $ c$. Then, $ a+b=100-c$. Since $ a+b>c$, $ 100-c>c$ and $ c<50$. Now, $ a+c=100-b>b$, and $ b+c=100-a>a$., Thus, $ a<50$ and $ b<50$. Hence, we're just looking for $ a+b+c=100$ such that all three are less than 50. However, note that the order does not matter, so eh this is going to be ugly casework.

I'm not really up to doing all of it right now, but one could use balls and urns to find all ordered triples (a,b,c), but you have to be careful, since they have to be positive, then subtract all the instances when a=b, b=c, or c=a. Note that a=b=c is not possible. Then, apply complimentary counting and subtract those with one of them greater than 50, wlog, let it be a. and then you'll be done.[/hide][/quote]

well if $ a+b+c=100$ where all three are less than 50, you can actually start "guess and checking" but its logical because a pattern will arise.

start with a=1,no solution
a=2, (49,49) (1 soltuion)
a=3, (48,49) (1 solution)
a=4, (48,48) (47,49) (2 solution)
a=5  (47,48),(46,49) (2 solutions)
a=6 has three
a=7 has three

obviously its 1+1+2+2+3+3+4......until you get to a=33 (last term)
so when a=33, there are 16 terms.  So its 2(1+2+3+...16)=272

clealy what you do is you do $ {\frac{100-a}{2}}$, then count the numbers between that number after [b]celing it [/b](i dont know latex celing) and $ 49$ inclusive.  However, you need to stop at when $ a=33$ becaseu once you get to 34, you will already have counted that (since distinct tirangles)

i still think i may be overcounting...wait 1 sec..

ah wait dang i overcounted some very small cases like when a=30, a case is (30,31,32) but I counted that case when a=31 also as (31,30,32)...but thats what you are supposed to do, i just need to find when I first start overocunting but ithink the real answer is somewherelikearound 272(lessthan that forsure)

I would like to know the exact answer to this?

[quote="ordinary boy"]I think its 208..i semi-brute forcedit.[/quote]

Uh how did you get that?  Like what did you go up to, when a=33?


Farz konid $a,b,c>0$ sabet konid \[ \sum_{cyc}\frac{1+a}{1+b}\leq 3+a^2+b^2+c^2  \]

[quote="jensen"]fekr konam in masale kami sakht bood. :D[/quote]
etefaaghan masale aasoone; man bad saligheam :rotfl:


Given a $2007\times 2007$ array of numbers $1$ and $-1$, let $A_{i}$ denote the product of the entries in the $i$th row, and $B_{j}$ denote the product of the entries in the $j$th column. Show that
\[A_{1}+A_{2}+\cdots +A_{2007}+B_{1}+B_{2}+\cdots +B_{2007}\neq 0.\]

Supposse that we have any array and we change the number in the column $i$ and the row $j$. We have that the sum decreases (or increases) by 4 or remains equal (there are just three cases that can be checked manually). Thus the sum of the numbers is always the same modulo 4 (since each array can be obtained from any other one by changing some numbers). This means that the sum is equal modulo 4 to the sum of the array having 1 in each place. The sum in this case is $2007+2007 \equiv 2 (\bmod 4)$. Since $4\not|2$ we are done.

[quote="silouan"]Here is my solution .
Let $A_{1}+..+A_{2007}+B_{1}+...+B_{2007}=0$ .
Then there exist 2007 ,$-1$  between the numbers $A_{1},...A_{2007},B_{1},...,B_{2007}$

But $A_{1}A_{2}...A_{2007}B_{1}B_{2}...B_{2007}=1$  so between the numbers $A_{1},...A_{2007},B_{1},...,B_{2007}$  there exist an even number of $-1$ , contradiction[/quote]
@Silouan: Why should $A_{1}A_{2}...A_{2007}B_{1}B_{2}...B_{2007}=1$? I don't see how that arises from the given condition. Could you explain that to me again?
   AR

Because A_1*A_2*...A_2007 and B_1*B_2*...b_2007 are product of the entries so the product of them is positive.

Oh yeah... :blush: . I missed that out. Thanks.

[quote="barasawala"]Given a $2007$x$2007$ array of numbers $1$ and $-1$, let $A_{i}$ denote the product of the entries in the $i$th row, and $B_{j}$ denote the product of the entries in the $j$th column. Show that

$A_{1}+A_{2}+...+A_{2007}+B_{1}+B_{2}+...+B_{2007}\neq 0$[/quote]

[size=150][color=red] I have posted it also here:[/color][/size]

http://www.mathlinks.ro/Forum/viewtopic.php?t=135524

 There is also an other solution.

[u] Babis[/u]


[quote="euler_vn"]Find all the functions $f: \mathbb{R}\to\mathbb{R}$ such that:
$f(xf(y)+x)+f(yf(x)-x)=2xy,\ \ \forall x,y\in\mathbb{R}$[/quote]
Nobody?


trigonometry
algebra
polynomial
sum of roots

Let  tan $a$, tan $b$, tan $c$ be roots of $x^3-7x^2+56x-128=0$. Find $tan(a+b+c)$

There is a nice formula for $\tan{(a+b+c)}$. It sound like: $\tan{(a+b+c)}=\frac{\tan a+\tan b+\tan c - \tan{a}\tan{b}\tan{c}}{1-\tan{a}\tan{b}-\tan{b}\tan{c}-\tan{c}\tan{a}}$. Using Viete formulas we have ${\tan a+\tan b+\tan c=7, \tan{a}\tan{b}\tan{c}=128}$ and $\tan{a}\tan{b}+\tan{b}\tan{c}+\tan{c}\tan{a}=56$. So $\tan{(a+b+c)}=\frac{7-128}{1-56}=\frac{121}{55}$ :huh:

[quote="warut_suk"]For $x^2+ax+b=0$ let $c,d$ be the roots, then $c+d=-a$ and $cd=b$.
(can be extended to all degrees)[/quote]
This one is easy, but how do we extend it to all degrees?

[quote="warut_suk"]If $x^3+ax^2+bx+c=0$ and $d,e,f$ are roots, then
$d+e+f=-a$
$de+ef+fd=b$
$def=-c$
like this![/quote]
How do I prove it? I don't know anything about equations in 3 degrees...

Since $d,e,f$ are roots, $x^3+ax^2+bx+c=(x-d)(x-e)(x-f)$

Yes.  I didn't know they were known as Viete's Formulas, but I knew about them.

For $f(x)$ with degree $d$ written $a_1 x^d + a_2 x ^{d-1} + ... + a_{d-1} x + a_d$
sum of roots: $\frac{-a_2}{a_1}$
prod of roots $(-1)^d \frac{a_d}{a_1}$

[quote="shobber"]What is viete formula?[/quote]

[i]Theorem:[/i]
Let $f=a_{n}{X^n}+a_{n-1}{X^{n-1}}+...+a_{1}{X}+a_{0}\in C[X], a_{n}\ne 0$ a $n$ degree polynom. Numbers $x_{1},x_{2},...,x_{n}$ are roots of polynom $f$ if:

(1) $x_{1}+x_{2}+...+x_{n}=-\frac{a_{n-1}}{a_{n}}$

(2) $x_{1}x_{2}+x_{1}x_{3}+...+x_{n-1}x_{n}=\frac{a_{n-2}}{a_{n}}$

(3) $x_{1}x_{2}...x_{k}+...+x_{n-k+1}...x_{n}=(-1)^{k}\frac{a_{n-k}}{a_{n}}$

(4) $x_{1}x_{2}...x_{n}=(-1)^{n}\frac{a_{0}}{a_{n}}$

[i]Proof:[/i]
Using the degeneration in liniar factors we can rewrite $f$ like:
$a_{n}{X^n}+a_{n-1}{X^{n-1}}+...+a_{1}{X}+a_{0}=a_{n}(X-x_{1})(X-x_{2})...(X-x_{n})$. Calculating the $RHS$ and we write it like a polynom in $X$, then order it decreasing, after $X$ powers. Then we identify the coef. of the terms that have the powers of $X$ and we get the previously identities.

I'm not clearly with the proof either  :|

Slizzel, in (1), you mean $= \frac{-a_{n-1}}{a_n}$

I did not know (3).  That is a very useful formula.  Thanks


function
algebra
domain
algebra unsolved

Find all functions $ f: R\rightarrow \R$ satisfying $ f(x^2+f(y)-y)=(f(x))^2$ for all real numbers $ x,y$.
I read it from the VMEO contest III but I cant solve. Could you help me? Thank you very much!

Suppose $ f$ has a period $ p > 0$. We have $ f(x^2 \plus{} f(y) \minus{} y) \ge 0$, so $ f(a) \ge 0$ for $ a$ large enough (i.e., $ a \ge f(y) \minus{} y$ for a fixed $ y$). The periodicity of $ f$ implies $ f$ is always nonnegative.

$ y : \equal{} y \minus{} p \rightarrow f(x^2 \plus{} f(y) \minus{} y \plus{} p) \equal{} f(x)^2$
$ x : \equal{} x^2 \rightarrow f(x^2 \plus{} p \plus{} f(y) \minus{} y) \equal{} f(\sqrt {x^2 \plus{} p})^2$

$ Im f \subset R_ \plus{}$, therefore $ f(\sqrt {x^2 \plus{} p}) \equal{} f(x)$.

$ y : \equal{} \sqrt {y^2 \plus{} p} \rightarrow f(x^2 \plus{} f(y) \minus{} \sqrt {y^2 \plus{} p}) \equal{} f(x)^2 \equal{} f(x^2 \plus{} f(y) \minus{} y)$

Hence, for enoughly big $ a$, $ f(a \plus{} t) \equal{} f(a)$, where $ t \equal{} \sqrt {y^2 \plus{} p} \minus{} y$. Again, the periodicity of $ f$ completes the information: $ t$ is a period of $ f$. We also see that $ lim_{y \rightarrow \infty} t \equal{} 0$ and $ lim_{y \rightarrow \minus{} \infty} t \equal{} \infty$. Then all the real numbers are periods of $ f$ and $ f$ is constant. We get the solutions $ f(x) \equal{} 1$ and $ f(x) \equal{} 0$.

Now suppose $ f$ doesn't have periods (I am killing the grammar, right? What must I say? Has no periods? Hasn't periods?). Also, assume $ f(y) \equal{} f(z)$ for some $ y, z, y \not\equal{} z$. Then $ f(x^2 \plus{} f(y) \minus{} y) \equal{} f(x)^2 \equal{} f(x^2 \plus{} f(z) \minus{} z)$ for all $ x$. Then $ k \equal{} |y \minus{} z|$ is a period of $ f$ if we restrict the domain of $ f$ for some interval $ [a, \infty]$. Thus, taking $ m$ large enough and $ n \in \mathbb{N}$,

$ y : \equal{} m \rightarrow f(x^2 \plus{} f(m) \minus{} m) \equal{} f(x)^2$
$ y : \equal{} m \plus{} nk \rightarrow f(x^2 \plus{} f(m) \minus{} m \minus{} nk) \equal{} f(x)^2$

Every number can be represented as $ x^2 \plus{} f(m) \minus{} m \minus{} nk$ if we take a large $ n$ (and therefore a large $ x$). The above equations give $ f(x^2 \plus{} f(m) \minus{} m) \equal{} f(x^2 \plus{} f(m) \minus{} m \minus{} nk)$. With this relation, we link the negative numbers with the positive numbers, concluding that $ k$ is a period of the function. Contradiction.

The end: we've just proved $ f$ is injective. $ y : \equal{} 0$ gives $ f(x^2 \plus{} f(0)) \equal{} f(x)^2 \equal{} f(x^2 \plus{} f(y) \minus{} y)$. If $ x^2 \plus{} f(0) \equal{} x^2 \plus{} f(y) \minus{} y$, $ f$ is linear and it is trivial to verify $ f(x) \equal{} x$. The solutions are therefore $ f(x) \equal{} 0, 1$ and $ x$.

It's an online Mathematical competition of http://diendantoanhoc.net/forum/index.php (for some reasons, the site doesn't work at this time), and VMEO stands for Vietnamese Mathematical E-Olympiad.


superior algebra
superior algebra theorems

[quote="ZetaX"]What axioms do you use/allow¿
And what has this to do with this forum¿[/quote]

You can use contraposition. Please move this to proper forum.

kunny

Contraposition is just one out of many axioms. Thus again the same questions: what axioms (of set theory)¿

Sorry I don't know, because I'm not acquaintance with this field very well.
How do you prove the problem?


it shouldn't be unless the reader can't imagine a little bigger, beyond 10  :D

Not at all.  Most people in a middle school can look at them and see 21 10 times or just use brute force by adding it all up.

there are 20 numbers, the avergave of the numbers is 1+20  /2 so the sum is 210

didn't Pascal first think of this method of adding backward and dividing up in half?

this is the right level for classroom math.

I can think of some of my classmates doing this on a calculator.

[hide="the answer"]

20(21)is 420.
Half is 210. [/hide]

[quote="236factorial"]Is this too hard to classroom?

What is 1 + 2 + 3 + 4 +..... + 20?[/quote]
[hide]
Using the sum of an arithmetic series formula:

$\frac{20(1+20)}{2}$

$10*21$

$210$[/hide]

[quote="236factorial"]Is this too hard to classroom?

What is 1 + 2 + 3 + 4 +..... + 20?[/quote]

[hide="Answer"] $\text{20+1=21. 21*10=210. Thus, the answer is 210}$[/hide]


Ring Theory
superior algebra
superior algebra unsolved

Let A be a ring where every prime ideal is maximal. Show that any localization of A/N, where N is a nilradical,with respect to a maximal ideal m is a field.

Now, it is clear that the above localization will be a reduced ring (since A/N is reduced and localization and operation of taking nilradical commute).

...not sure how to proceed now....can we show somehow that A/N has also the property that every prime ideal is maximal? because that will solve the problem.

of course, every prime ideal of A/N is maximal, since (prime) (maximal) ideals in A/N correspond to (prime) (maximal) ideals in A containing N.

every localization of A/N is a reduced ring with only one prime ideal, thus a field.

sure. So, since every prime ideal in A/N is maximal + there is one-to-one correspondence between prime ideals of a localized ring and prime ideals containing in m, it follows that the only ideals of (A/N)_m are (0) and (1), so it's a field.


Hi, I have $ A,B\in M^n_m(\mathbb{R})$ and $ a,b,c\in\mathbb{R}$. How should I proof these simply equation?
1) $ c(aA\plus{}bB)\equal{}(ca)A\plus{}(cb)B$
2) $ \minus{}aA\equal{}(\minus{}a)A\equal{}a(\minus{}A)$
3) $ (aA)^T\equal{}aA^T$
This is my solution:
Let $ A(i,j)$ means $ a_{ij}$, ofc $ i\equal{}1,...,m$, $ j\equal{}1,...,n$ and now:
1) $ \Big(c(aA\plus{}bB)\Big)(i,j)\equal{}c\Big((aA\plus{}bB)\Big)(i,j)\equal{}c\Big(aA\Big)(i,j)\plus{}c\Big(bB\Big)(i,j)\equal{}$

$ \equal{}caA(i,j)\plus{}cbB(ij)$
And is it O.K.?

Yes. To show that two matrices are equal you need to show that all corresponding entries are equal. That is $ A\equal{}B$ if and only if $ A_{ij}\equal{}B_{ij}$ for all $ i,j$  (assuming $ A$ and $ B$ have the same dimension)
Then use definitions:

$ (cA)_{ij}\equal{}cA_{ij}$
$ (A\plus{}B)_{ij}\equal{}A_{ij}\plus{}B_{ij}$
$ (A^T)_{ij}\equal{}A_{ji}$


geometry
geometric transformation
rotation

ABCD-square.
Proof that the lines m,n,p and q cross into one point.

This is definately not High School Basics. Try Pre-Olympiad forum or Olympiad Geo forum.

Well, maybe, [hide="but"]does this not scream cyclic quadrilaterals...?[/hide]

[quote="modularmarc101"]I can barely make out the picture.. It looks really crowded.[/quote]That has to do with the lack of explanation from the OP.

Given a square $ ABCD$, let $ P$ be any point inside of the square. Lines $ q,m,p,n$ passes through the respective vertices $ A,B,C,D$ and are perpendicular to lines $ PB,PC,PD,PA$, respectively. Prove that lines $ q,m,p,n$ are concurrent.

[hide="Solution"]
Let $ AC\cap BD\equal{}O$
Let $ a$ be the perpendicular line from the point $ A$.
Let $ b$ be the perpendicular line from the point $ B$.
Let $ c$ be the perpendicular line from the point $ C$.
Let $ d$ be the perpendicular line from the point $ D$.

$ R_{o}^{90^\circ}: ABCD\to BCDA$

$ R_{o}^{90^\circ}:$
$ a\to l$
$ A\in a \to B\in l$
$ a\bot l \equal{}> l\equiv BP$

$ R_{o}^{90^\circ}:$
$ b\to n$
$ B\in b \to C\in n$
$ b\bot n \equal{}> n\equiv CP$

$ R_{o}^{90^\circ}:$
$ c\to m$
$ C\in c \to D\in m$
$ c\bot m \equal{}> m\equiv DP$

$ R_{o}^{90^\circ}:$
$ d\to k$
$ D\in d \to A\in k$
$ d\bot k \equal{}> k\equiv AP$

=> $ AP\cap BP\cap CP\cap DP\equal{}P$
=> $ R_{p}^{\minus{}1}\equal{}a\cap b\cap c\cap d$

[/hide]

[hide="Another Solution"]Let the points with right angles marked be $ Q,R,S,T$. Then it's clear that they all lie on a circle. None of lines $ m,n,p,q$ are tangent to the circle because they intersect at their right angle and at least one other point where they intersect with the other three lines. $ m,n,p,q$ can only have one intersection among them because otherwise there would be more than two intersections for some line.[/hide]


[quote="1234567890"][quote]New feature for AoPS students: All of your future assignments
and exam papers for AoPS Classes will now be archived in your
My Classes area in the form of your own personal solution library.[/quote]

So, the transcripts and finals and midterms from this summer's classes will never disappear?[/quote]

The transcripts, etc aren't permanent - your work and our comments are.

the only thing that will remain in your My Classes section is a copy of the paper you sent in to the instructors and the feedback that they gave to you.  the specific transcripts and forms such as the final/midterm for each class will not remain there.


vector
inequalities
real analysis
real analysis unsolved

Suppose that you have a vectorial subspace $F$ in $R^n$. Prove that either $F$ or its orthogonal contains a vector all of whose components are positive.

I guess you actually mean "a non-zero vector all of whose components are non-negative"? Because positive (strict inequality) doesn't work.

The "dual" problem would be stated like this: given several vectors $v_1,\ldots,v_n$ in $\mathbb R^k$, prove that either $0$ is a convex combination of $v_i$, or there is a $v\ne 0$ with $(v,v_i)\ge 0,\ \forall i$.

We can interpret $v_i$ as points in the affine space $\mathbb R^k$. If $0$ is not a convex combination of $v_i$, there is a hyperplane $\mathcal H$ separating $0$ and $v_i$. Now take $v$ to be $\overrightarrow{op}$, where $o$ is the origin, and $p$ is its orthogonal projection on $\mathcal H$.


Can someone provide me links to math contests which are intermediate to pre -olympiad  level .Not requiring lengthy knowledge based solutions but simple logical solutions(thats why intermediate to pre-olmpd). Also can someone provide names/links to multiple choice type math competitions.

Try the Michigan Math Prize Contest (MMPC) - http://gauss.math.oakland.edu/mmpc/previousexams.html

They have lots of multiple choice problems in part I, and once you've mastered those you can move to the part II problems that are free-response.

[quote="facis"]AMC: http://www.etsu.edu/coe/uschool/facstaff/wells/AMC.htm
[/quote]

Sometimes I really don't understand these bugs.

Thnx but I think I want harder level AIME looks good, but I cant find many past year questions or practice questions.only 3-4 years available.

AIME: [url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45[/url]

Any idea what happened to my last post???


[quote="akhrif"]let $a1,a2,a3,a4 .... an$ real numbers such that : $a1+a2+a3+...an=n$
prove that
$\sum\frac{1}{ai+1}\ge \sum\frac{1}{ai^{2}+1}$[/quote]
It's wrong. Try $n=2,$ $a_{1}=1.1$ and $a_{2}=0.9.$ :wink:

:D I dont think that it's wrong because I've solved it (it was proposed for moroccan contests)
moreover the example that you have done is true and nothing is wrong.
you can prove that the right part is inferior to $\frac{n}{2}$ only by using the remarquable identity $a^{2}+1\geq 2a$ and prove after that the left part is superior to $\frac{n}{2}$ by using AM-GM or ..

for you akhrif did you solve it?

Wait if the problem is this:
[quote="akhrif"]let $a_{1},a_{2},a_{3},a_{4},... ,a_{n}$ real numbers such that : $a_{1}+a_{2}+a_{3}+...+a_{n}=n$
prove that
$\sum\frac{1}{a_{i}+1}\ge \sum\frac{1}{a_{i}^{2}+1}$[/quote]

arqady is right about the counterexample.

But maybe there is some misunderstanding here.
Can you post the problem or is it like what I wrote above?

$a_{1},a_{2}.....a_{n}$ are  positive real numbers

The inequality is really wrong. Please check again.

i used (chebyshev) and i dont know that true or not :(

yes you are right  :rotfl:
the true inequality is
$a_{1},a_{2}....a_{n}$ are positive real numbers such that ${\sum_{i=1}^{n}a_{i}}=n$

prove that $\sum_{i=1}^{n}}{\frac{a_{i}}{(a_{i})^{2}+1}\leq \sum_{i=1}^{n}}{\frac{1}{a_{1}+1}$

when I was proving it 2 hours ago I ve said  that the right part is inferior to $\frac{n}{2}$  because of $a^{2}+1\geq 2a$ and I didn't see that there is no $a_{i}$ in the numerator.

[quote="vietnam2007"]
$a_{1},a_{2}....a_{n}$ are positive real numbers such that ${\sum_{i=1}^{n}a_{i}}=n$
prove that $\sum_{i=1}^{n}}{\frac{a_{i}}{(a_{i})^{2}+1}\leq \sum_{i=1}^{n}}{\frac{1}{a_{1}+1}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=74606


calculus
integration
calculus computations

Please help on the following question:

Prove that  \[ \int_0^1 \left(\frac{d^n}{dx^n}\left(x^2\minus{}1\right)^n \right)^2 dx\equal{} (2n!) \int_0^1\left(x^2\minus{}1\right)^n dx.\]

[hide="RHS"]$ (2n!)\int_0^1 (x^2\minus{}1)^n  dx\equal{} (2n!)\int_0^1 \sum_{k\equal{}0}^n \frac{n!}{k! n\minus{}k!}x^{2k} (\minus{}1)^{n\minus{}k}dx\equal{}$

$ (2n!)\sum_{k\equal{}0}^n \frac{n! (\minus{}1)^{n\minus{}k}}{k! n\minus{}k!} \int_0^1 x^{2k} dx\equal{}\sum_{k\equal{}0}^n \frac{n! 2n! (\minus{}1)^{n\minus{}k}}{k! n\minus{}k!(2k\plus{}1)}$
[/hide]
[hide="LHS"]
$ L(x)\equal{}\frac{d^n}{dx^n}\left(x^2\minus{}1\right)^n\equal{}\frac{d^n}{dx^n}\left(\sum_{k\equal{}0}^n \frac{n!}{k! n\minus{}k!}x^{2k} \right)\equal{}$

$ \sum_{k\equal{}0}^n \frac{n! (\minus{}1)^{n\minus{}k}}{k! n\minus{}k!}\frac{d^n}{dx^n}\left(x^{2k}\right)\equal{} \sum_{k\ge \frac{n}{2}}^n \frac{n! (\minus{}1)^{n\minus{}k}}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}x^{2k\minus{}n}$

$ [L(x)]^2\equal{}\sum_{i\equal{}n}^{2n} \sum_{k\plus{}j\equal{}i, \max(k,j)\ge \frac{n}{2}} \frac{n! (\minus{}1)^{n\minus{}k}}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}\frac{n! (\minus{}1)^{n\minus{}j}}{j! n\minus{}j!} \frac{2j!}{2j\minus{}n!}x^{2(i\minus{}n)}$

$ \int_0^1 [L(x)]^2 dx \equal{} \sum_{i\equal{}n}^{2n} \sum_{k\plus{}j\equal{}i, n\ge k,j\ge \frac{n}{2}} \frac{n! (\minus{}1)^{2n\minus{}i}}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}\frac{n! }{j! n\minus{}j!} \frac{2j!}{2j\minus{}n!}\frac{1}{2(i\minus{}n)\plus{}1}\equal{}$

$ \sum_{i\equal{}0}^{n} \sum_{k\plus{}j\equal{}i\plus{}n, n\ge k,j\ge \frac{n}{2}} \frac{n! (\minus{}1)^{n\minus{}i}}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}\frac{n! }{j! n\minus{}j!} \frac{2j!}{2j\minus{}n!}\frac{1}{2i\plus{}1}$
[/hide]

remains to show that

$ \frac{ 2n!}{i! n\minus{}i!}\equal{}\sum_{k\plus{}j\equal{}i\plus{}n, n\ge k,j\ge \frac{n}{2}} \frac{1}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}\frac{n! }{j! n\minus{}j!} \frac{2j!}{2j\minus{}n!}\equal{}$

$ \equal{}\sum_{k\ge \frac{n}{2}}^n  \frac{1}{k! n\minus{}k!} \frac{2k!}{2k\minus{}n!}\frac{n! }{n\plus{}i\minus{}k! k\minus{}i!} \frac{2n\plus{}2i\minus{}2k!}{n\plus{}2i\minus{}2k!}$

which may be .. impossible..


I have an assignment that I have completed and I just wanted to see if I'm on the right track... I've put my answers down, and the work... if you agree or disagree with any of my answers could you let me know... and I can try to find a way to fix my methods. Thanks

[b]PROBLEM A: An urn contains coloured balls: 5 red balls, 8 green balls, and 7 blue balls. Suppose four balls are drawn simultaneously from this urn.[/b]

1. What is the probability that all of the balls drawn are green?
0.014447884

2. What is the probability that balls of all three colours are obtained?
0.98245614

[b]PROBLEM B: Four male and three female students seat themselves randomly in a row of seven seats for a lecture.[/b]

3. What is the probability that the students are seated with the genders alternating?
0.028571428

4. What is the probability that Rachel and Roger are seated next to one another?
0.002380952381

[b]PROBLEM C: A bag contains five ping pong balls numbered from one to five. Suppose two balls are drawn at random from this bag with replacement.[/b]

5. What is the probability that the total of the two numbers on the selected balls is at least eight (8)?
0.24

6. Given that the total of the two numbers on the selected balls is at most four (4), what is the probability that the total is an odd number?
0.0666666

[b]PROBLEM D: The work area at a certain company is illuminated with a number of fluorescent fixtures requiring a total of 20 fluorescent tubes. Once each year the complete set of tubes is replaced. Suppose the company purchases 24 fluorescent tubes of which 3 are defective (this fact is unknown to the company), and then randomly chooses 20 tubes from this set of 24 new tubes to use in replacing the current tubes.[/b]
7. What is the probability that the 20 tubes selected include at most one defective tube?
0.061264822

8. What is the probability that the 20 tubes selected include more than one defective tube?
0.938735177

[b]PROBLEM E: A special deck of 30 cards consists of five cards numbered each of 1, 2, 3, 4, 5 and 6. The deck of cards is thoroughly shuffled putting the cards in random order.[/b]

9. If a “hand” of three cards is then dealt, what is the probability that the "hand" so dealt contains "one pair" (for example, two 5's plus another card that is not a 5)?
0.369458128

10. If a “hand” of four cards is dealt, what is the probability that it will contain “two pair” (for example, two 3’s and two 6’s)?
0.054734537

[b]PROBLEM F: A bag contains nine balls of which three are red, three are green and three are blue. The nine balls are randomly arranged in a row.[/b]

11. What is the probability that the arrangement will have all three balls of each colour together in groups (i.e. the three red balls side by side, the three green balls side by side, and the three blue balls side by side)?
0.003571428571

12. What is the probability that the row begins and ends with balls of the same colour?
0.083333

[b]PROBLEM G: Suppose that the letters of the ord “statistics” are randomly permuted.[/b]

13. How many of the possible resulting ‘words’ have all three t’s together in a group?
0.0083333

14. How many of these ‘words’ begin and end with t?
0.066666

[b]PROBLEM H: Do you wear your seatbelt when in a vehicle? Suppose that government information indicates that 88% of all people consistently wear their seatbelts. When accidents
occur, records show that 32% of people wearing seatbelts suffer injuries whereas 84% of people not wearing seatbelts suffer injuries.[/b]

15. What is the probability that a person involved in an accident suffers an injury?
0.3824

16. If a person involved in an accident is known to have suffered an injury, what is the probability that the person was wearing her/his seatbelt?
0.7364

1. combin (8,4)/combin (20,4)

2. combin (5,1)combin(8,1)combin(7,1)combin(17,1)/combin (20,4)

3. combin(4,1)combin(3,1)combin(3,1)combin(2,1)combin(2,1)combin(1,1)combin(1,1) / 7!

4. 2x6/ 7!

5. 3/25 + 2/25 + 1/25

6. (2/36) / (30/36) = 2/30

7. [combin(3,1)combin(21,19) + combin(3,0)combin(21,20)] / combin(24,20)

8. [combin(3,2)combin(21,18) + combin(3,3)combin(21,17)] / combin(24,20)

9.combin(6,1)combin(5,2)combin(5,1)combin(5,1) / combin(30,3)

10.combin(6,2)combin(5,2)combin(5,2) / combin(30,4)

11. combin(9,1)combin(2,1)combin(1,1)combin(6,1)combin(2,1)combin(1,1)combin(3,1)combin(2,1)combin(1,1) / 9!

12. 7!/3!3!1!  /  9!/3!3!3!

13. 7!/3!2!1!1!1!  /  10!/3!3!2!1!1!

14. 8!/3!2!1!1!1!  /  10!/3!3!2!1!1!

15. .88x.32 + .12x.84

16. .88x.32 / .88x.32 + .12x.84

[quote="brettburns83"]2. What is the probability that balls of all three colours are obtained?
0.98245614[/quote]
I got exactly half of that.  :maybe:

[quote="brettburns83"]4. What is the probability that Rachel and Roger are seated next to one another?
0.002380952381[/quote]
I got [hide] $\frac{2}{7}$ [/hide]


USAMTS search AMC AIME USA(J)MO USAMO probability

Personally, I thought this round 1 was much harder than last years round 1.  Last year, I solved all the problems of round 1 in one sitting, while this time I took much longer.

I thought it was much harder, but I don't know whether that's because the problems were generally more difficult or I'm just out of practice with these kinds of problems.  However, I have a lot more math experience than I did last year, and I still found these problems quite tough (can any of the USAMTS officials comment on the intent of this year's round in terms of difficulty?).

I really have no idea how hard last year's was... I thought it wasn't too difficult, but I just put the problems off too long. I should have been able to get all 5, but not in just three days or so... :P

These year's problems were a lot harder, at least for Round 1.  This was the most time I've ever spent on the first round of a USAMTS before.  Even my first year participating, when my math wasn't close to what it is now, I solved the problems in a decent amount of time.  So either I'm just slow, I'm bad with certain types of problems used here, or this round really was harder.  I'd say all three. :lol:

... But that is good if they're harder.  I like challenges.

[quote="chess64"][quote="bookaholic"]I found this round average-ish compared to previous rounds, but my perspective is messed up because I've seen both #2 and #4 (though not their solutions) before. I was surprised that two problems were already at least semi-well-known--I don't think that's ever happened before. Or maybe this is just the first time a known problem that I happened to know appeared. I still didn't get #4 though.[/quote]

Yup. On the last day I searched "can't be expressed as the sum of two primes" on google :ninja:

edit: look at the first result: http://www.google.com/search?hl=en&q=%22can%27t+be+expressed+as+the+sum+of+two+primes[/quote]

Wow, I'm surprised the problem was located elsewhere on the net.  Even though my solution was awful for #4, I'm glad I didn't know that.  I think looking into the problem yourself is much more beneficial, as far as your math skill goes.  Also, note #14 on the FAQ...

[quote="chess64"][quote="bookaholic"]I found this round average-ish compared to previous rounds, but my perspective is messed up because I've seen both #2 and #4 (though not their solutions) before. I was surprised that two problems were already at least semi-well-known--I don't think that's ever happened before. Or maybe this is just the first time a known problem that I happened to know appeared. I still didn't get #4 though.[/quote]

Yup. On the last day I searched "can't be expressed as the sum of two primes" on google :ninja:

edit: look at the first result: http://www.google.com/search?hl=en&q=%22can%27t+be+expressed+as+the+sum+of+two+primes[/quote]
I noticed that too - a Google search can actually turn up a compendium of different ways in which the problem is posed and solved.  They should have changed it somewhat so that you couldn't basically take your solution straight off the net (though there's really no personal satisfaction in doing that).

There's loads of sites with that problem. Try searching "Impossible problem" and I think you'll get a few more results.

I really didn't do that one purpose! I was just looking for a list of numbers that can't be expressed as the sum of two primes, to save some time, and I clicked the first link and there it was :blush:

I really didn't do it on purpose!

Edit: no_one_2000, if you're talking about, "Asking the question on a message board or a math help site would be asking another person for help," well I didn't post it on a message board, so... ;)

Did you ever find what you wanted? If Goldbach's Conjecture is true, then the only such numbers are two more than odd composites. And up to a few billion or trillion, the above is true.

Anyway, I really didn't think this round was all that hard, though I didn't participate last year. Really, I'm just a lot more interested in these problems than the "find all the possible sums of the numbers on each side of this hexagon" problems that popped up at least twice last year (I may not have participated last year, but I've certainly seen the problems). Besides, I didn't take that long. I did them all the last weekend (yeah, I got the mailing and decided to do it then) and got all 5.

[quote="joml88"][quote="zanttrang"] They should have changed it somewhat so that you couldn't basically take your solution straight off the net (though there's really no personal satisfaction in doing that).[/quote]
Perhaps.  But it is near impossible to come up with problems that haven't been posed and solved before somewhere else.  Even the AIME, USAMO, and IMO sometimes have problems that people have seen before.  It's even harder with the USAMTS where the competitors have an essentially unlimited supply of resources at their disposal.  I'm sure that the USAMTS stuff didn't realize that a simple google search could produce an answer...in fact I think the date of that post with the answer was after the USAMTS round 1 was realized...[/quote]
I have seen similar problems about guessing the ages of three children and guessing house numbers, in which the people have the conversation know some numbers but the reader doesn't, but I don't think I had seen this particular problem before. It is more complicated than the typical brain teaser.

A good portion of the problems on the USAMTS are borrowed from another source. George Berzsenyi had liked mining 15-year-past Hungarian math contests for problems. However, polishing up the problems for the USAMTS to emphasize the interesting mathematics usually changes the problem beyond recognition. For example, one problem I submitted started in this very forum, in a Mock USAMTS Problem set created by tetrahedr0n and jbl (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12857),

[quote="tetrahedr0n"]4. Find the positive integer n < 100 for which the number $A_n = 7^1 + 7^2 + ... + 7^n$ ends with the maximum number of equal digits.[/quote]
yet by the time it reached the USAMTS as problem 3/3/16 it had become

[quote]3/3/16. Define the recursive sequence 1, 4, 13, $\ldots$ by $s_1 = 1$ and $s_{n+1} = 3s_n + 1$ for all positive integers $n$. The element $s_{18} = 193710244$ ends in two identical digits. Prove that all the elements in the sequence that end in two or more identical digits come in groups of three consecutive elements that have the same number of identical digits at the end.[/quote]
Problem 4/1/17 could not be altered much without losing the essence of the problem. That some students may solve it through a good mathematical literature search finding the source of the problem rather than from scratch is a risk we were willing to take. As a grader, I would give them full credit (I don't know which problem in Round 1 I will grade. The NSA grading session is October 21.)

Erin Schram
USAMTS grader and occasional problem writer

[quote="Erin J. Schram"][quote="joml88"]Compared to last year's round 1 (the only USAMTS round I've done until now) this round was much harder.  Compared to other rounds I think it's about average or slightly harder.[/quote]

That's pretty much it. In past years, we put easy-to-grade problems on Round 1, because we get twice as many submissions for Round 1 as we do in the other rounds. The intent is not to make that round easier to solve  than the other rounds, but that is often the result. That was not the case this year.

Past years were written by George Berzsenyi. This year, I believe the USAMTS is written by the Art of Problem Solving people with advice from George Berzsenyi. That changes the style slightly.

Erin Schram
USAMTS grader[/quote]

I agree with both paragraphs. This round is definitely challenging than previous rounds (if you look at one of the past rounds' number one, one problem only required you to complete the square and you'll get the answer) but it wasn't necessarily evil. I got #2 wrong since I'm really poor at probability, counting. I didn't get to try #5 (I regret it now really! I should've found sometime to work on it!) but others were possible, not necessarily easy.

The style of problem was clearly different. I've solved many IMTS (That's USAMTS before it got name) problems and unlike those problems, this year's problems were clearly different in style. I don't know if any other people realized this but in some way, current USAMTS problems are LOT easier than previous ones (especially ones that you had to prove on dissection, drawing, etc.).

joml88 is exactly right about how hard it is to come up with a problem that is completely unknown. I've researched on many math competitions in many different nations, states, and there is always chance for you to find two problems that are similar. Moreover, some old math contests had EXACT problems in different years (Take one or two of early European olympiad problems). I do have to say that #4 wasn't clearly the best kind of problem to choose because I found this problem without trying (after the contest but the problem was posted before I think) in ONE OF THE SITES THAT AOPS HAD!

#3 in http://www.qbyte.org/puzzles/puzzle01.html (Nick's Mathematical Puzzles)

The solution to problem 5/4/15 is actually on the MathWorld website -- a fact I discovered by accident last year. :)

http://mathworld.wolfram.com/CalabisTriangle.html

http://www.usamts.org/Tests/us1504.pdf

The triangle in the problem is known as Calabi's Triangle: the only non-equilateral triangle for which the largest inscribed square can be inscribed three different ways. I thought it was kind of weird that they used that as a problem in the USAMTS.


Phil has some number of golf balls.  Now he gives half of his golf balls to Bob, and gives one quarter of the remaining golf balls to Sam. He now has 75 golf balls.  How many golf balls did he originally have?

*cough cough* this is a problem of the day. Not a problem of the tri-hour.

Meh.
[hide]
Call the number of golf balls he has $ g$.
After giving half of them away, he has $ \frac{g}{2}$ left. When he gives away one quarter of the remaining, he

keeps three quarters of the remaining. So after he gives away the golf balls, he has $ \frac{g}{2} \times \frac{3}{4}\equal{}\frac{3g}{8}$. Since he had 75 golf balls remaining, $ \frac{3g}{8}\equal{}75$
$ g\equal{}200$.
 [/hide]

Oops, thought it was like a marathon type of thing.
  :oops:
And that is correct mathcountsfreak.
 :)

If you're going to do a marathon, then I suggest creating [i]one[/i] topic.

[quote="jackdillon"][hide="The Answer"]$ \75*4*2 = 600$
 $ \ 600*1/2*1/4 = 75$
600 origanel golfballs[/hide][/quote]
No; he only gives away 1/4 of them, he keeps 3/4.

[hide]
he has 75% of the golf balls
Since 75%=75
Therefore the remaining golf balls are 25%
25%=25

75+25=100

He had half of the golf balls
100*2=200

ANSWER: 200 @))[/hide]

x/2 multiply 1/4= 75
x/8 =75
x = 75 multiply 8
x= 600

Let x be the total number of balls.

[hide="Solution"]Let $ x$ be the number of golf balls that Phil originally had.
\begin{align*} & x - \overbrace{\frac {1}{2}\cdot x}^{\text{Bob}} - \overbrace{\frac {1}{4} \cdot \frac {1}{2}\cdot x}^{\text{Sam}} = 75 \\
& \implies x - \frac {1}{2} \cdot x - \frac {1}{8} \cdot x = 75 \\
&\implies x\left(1-\frac{1}{2}-\frac{1}{8}\right) =75 \\
& \implies \frac {3}{8} \cdot x = 75 \\
& \implies x = 75 \cdot \frac {8}{3} \\
& \implies \boxed{x = 200} \end{align*}
[/hide]

I think there is something wrong with your solution shaun ong.
btw, this is the 10000th Classroom Math Post!
 :D


Let  $k,p$  are positive integer numbers. Prove that $\frac{1}{p+1}k^{p+1}<1^p+2^p+...+k^p<\frac{1}{p+1}{(k+1)^{p+1}}.$

This is obvious using calculus but I wonder if there is some elementary proof  :?:

[quote="silouan"]This is exactly application of the famous inequality of Arhimedes which says that if

$m,n$ are positive integers then

$1^m+2^m+.....+(n-1)^m<\frac{n^{m+1}}{m+1}<1^m+2^m+....+n^m$.

But it would be great if we have a proof for this.[/quote]
We can use induction on $n$ to prove $1^m+2^m+.....+(n-1)^m<\frac{n^{m+1}}{m+1}<1^m+2^m+....+n^m$.
When $n=2$ the inequality is obviously true.
Assume the inequality is true when $n=k$, i.e. $1^m+2^m+\cdots +(k-1)^m<\frac{k^{m+1}}{m+1}<1^m+2^m+....+k^m$
When $n=k+1$,
For the left side,
$1^m+2^m+\cdots +(k-1)^m+k^m<\frac{k^{m+1}}{m+1}+k^m=\frac{k^{m+1}+(m+1)k^m}{m+1}<\frac{(k+1)^{m+1}}{m+1}$
with the last inequality by binomial theorem.
For the right side,
$1^m+2^m+\cdots +k^m+(k+1)^m>\frac{k^{m+1}}{m+1}+(k+1)^m=\frac{k^{m+1}+(m+1)(k+1)^m}{m+1}$
So it suffices to prove
$k^{m+1}+(m+1)(k+1)^m>(k+1)^{m+1}$

$\Leftrightarrow (m+1)(k+1)^m>(k+1)^{m+1}-k^{m+1}$

$\Leftrightarrow \sum_{i=0}^m (m+1)\binom{m}{i}k^{m-i}>\sum_{i=0}^m \binom{m+1}{i+1}k^{m-i}$
This is true becasue $(m+1)\binom{m}{i}=\frac{(m+1)m(m-1)\cdots (m-i+1)}{i!}>\frac{(m+1)m(m-1)\cdots (m-i+1)}{(i+1)!}=\binom{m+1}{i+1}$


A license plate consists of two letters followed by two digits; for example: MP78. Neither the letters nor the digits may be repeated, and neither the letter “O” nor the number “0” may be used. When reading from left to right, the letters must be in alphabetical order and the digits must be in increasing order. How many different license plate combinations are possible?

Crazy casework :!:

[hide][b]For the letters, here's the casework[/b]:

If the first letter is A, then B-Z (not O) can be used so that's 24.
B: C-Z (not O) 23
22...

So that's $\frac{24 \times 25}{2}=300$.

[b]For the numbers[/b]:

If the first is 1, then 2-9 can be used so that's 8.
Going along the same pattern, 7, 6, ...

So that's $\frac{8 \times 9}{2}=36$.

Since the two are independent, $36 \times 300=\boxed{10800}$.[/hide]

[hide]
For the letters, there are 25 of them (no O)
If the first is A, then there are 24 choices for the second.
If the first is B, then there are 23.
So we have to sum 1-24.
$\frac{24(25)}{2}=12\cdot25=300$
For the numbers,
Same thing as before.
We have to sum 1-8
$\frac{8(9)}{2}=4\cdot9=36$
Multiplying together, we get
$10800$[/hide]

[hide]
Simple thought process:

Give me any two letters, I can arrange them in alphabetical order. Therefore, it is simply just $25C2$.

Give me any two numbers, I can arrange them in increasing order. Therefore, it is simply just $9C2$.

Multiply them together to get $10800$[/hide]


Prove that for all real number $x$ we have:
$(x^{3}-x+2)^{2}>4x^{2}(x^{2}+1)(x-2)$

Expand and move all to LHS. We want to prove

$x^{6}-4x^{5}+6x^{4}+9x^{2}-4x+4 > 0$

Subtract $12x^{2}$ from each side:

$x^{6}-4x^{5}+6x^{4}-3x^{2}-4x+4 >-12x^{2}$
$(x-1)^{2}(x^{4}-2x^{3}+x^{2}+4x+4) >-12x^{2}$

the left side is nonnegative and the right side nonpositive. And we know it's strict because they don't equal 0 at the same place.


$ 1)$ $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}0$

Tell the number of: a) complex roots, b) positive real roots possible c) negative real roots possible d) imaginary roots possible.

I know that there are b) 0 positive real roots because there are no sign changes, and c) 4, 2, or 0 negative roots, a) 4, 2, 0 complex roots, but I don't know how to find the number of imaginary roots possible?

[quote="Bicameral"]$ 1)$ $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$

Tell the number of: a) complex roots, b) positive real roots possible c) negative real roots possible d) imaginary roots possible.

I know that there are b) 0 positive real roots because there are no sign changes, and c) 4, 2, or 0 negative roots, a) 4, 2, 0 complex roots, but I don't know how to find the number of imaginary roots possible?[/quote]

Think of it as $ \frac {x^5 \minus{} 1}{x \minus{} 1}$

Does that lead anywhere?

EDIT:

[hide="Solution"]
Since $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} \frac {x^5 \minus{} 1}{x \minus{} 1}$

And $ x^n \minus{} 1 \equal{} (x \minus{} 1)(x \minus{} \omega)(x \minus{} \omega^2) \cdots (x \minus{} \omega^{n \minus{} 1})$

When $ \omega^m$ are the nth roots of unity.

When $ n \equal{} 5$, and we have divided out the root $ x \equal{} 1$, we are left with the question, which yields 4 complex roots, which are the 5th roots of unity.
[/hide]

[quote="SimonM"]Think of it as $ \frac {x^5 \minus{} 1}{x \minus{} 1}$

Does that lead anywhere?

EDIT:

[hide="Solution"]
Since $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} \frac {x^5 \minus{} 1}{x \minus{} 1}$

And $ x^n \minus{} 1 \equal{} (x \minus{} 1)(x \minus{} \omega)(x \minus{} \omega^2) \cdots (x \minus{} \omega^{n \minus{} 1})$

When $ \omega^m$ are the nth roots of unity.

When $ n \equal{} 5$, and we have divided out the root $ x \equal{} 1$, we are left with the question, which yields 4 complex roots, which are the 5th roots of unity.
[/hide][/quote]No, he is asking to get information about the roots of the equation by using Descartes Law of Signs, not solving the equation and then find out that they are 5th roots of unity...  :ninja:

I assumed in that context that imaginary referred to anything where the coefficient of the i value isn't zero. :oops:

[quote="10000th User"][quote="mountain.dew"]Well if the roots are not real, they're imaginary  :wink:[/quote]This claim is wrong. If the roots are not real, they're complex, not necessarily imaginary. A number is imaginary iff the real part of the complex number is zero.

Let $ P(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$. Using Descartes Law on $ P(x)$, it tells you that there are no positive roots, and from $ P( \minus{} x)$, there are 4 or 2 or 0 negative roots. So, correspondingly, there are 0 or 2 or 4 complex roots. Now look at the coefficient of $ x^3$. The sum of the roots is -1, a real number. This means that $ P(x)$ cannot have 4 imaginary roots because if it does, the sum of the roots would be zero or an imaginary number. Thus, there are either 2 or 0 imaginary roots. Any questions? ;)[/quote]

A few, but I can't give them words yet.  Imma go hit up AOPS II for a bit then come back and ask.

EDIT:  :)


What do you mean by the letters $A$,$B$,$C$?
beacuse if they are the angles then.......! :huh:

[quote="maokid7"]o yes sorry. $A,B,C$ are the angles of the triangle.[/quote]

[quote]If $a,b,c$ are the sides of a triangle and $A+B+C=\frac{\pi}{2}$ find the minimum value of

$\frac{aA+bB+cC}{a+b+c}$[/quote]

i'm a bit confused. isn't the sum of angles in triangle equal to $\pi$?

o man...this really isnt my day. yes the sum should be equal to $\pi$ sorry.  :blush:

It's not hard to show that $(a,b,c)$ and $(A,B,C)$ are similarly sorted (suppose otherwise and arrive at a contradiction). Then we can apply Chebyshev to get

$\frac{aA+bB+cC}{a+b+c}\ge \frac{(a+b+c)(A+B+C)}{3(a+b+c)}= \frac{A+B+C}{3}= \frac{\pi}{3}$.

Equality holds for any equilateral triangle.

[quote="maokid7"]what do u mean by "similarly sorted"?[/quote]

$a \ge b \ge c \Leftrightarrow A \ge B \ge C$.


$a,b,c$ are positive real numbers. Prove that :

$\frac{a^{2}+bc}{b+c}+\frac{b^{2}+ac}{a+c}+\frac{c^{2}+ab}{a+b}\geq a+b+c$

Double posting!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=148725[/url]

You should clarify the subject of problem.

You can delete that one I ve made a big mistake.I m so sorry.

:)

Regrettably, I'm a Japanese Communities Moderator, but I can't delete this.Because I don't have the right to delete the post other than Japanese Communities posting.

kunny

So I ask the moderator to delete this topic because there is another one with the same subject.


see here http://www.mathlinks.ro/Forum/viewtopic.php?t=131123  :wink:

\[(a^{2}-1)(b^{2}-1)(c^{2}-1)\leq 8\]
\[(a-\frac{1}{a})(b-\frac{1}{b})(c-\frac{1}{c})\leq \frac{8}{abc}\]
\[abc+\sum \frac{a}{bc}\leq \frac{9}{abc}+\sum \frac{ab}{c}\]
\[\left(\sum a\right)^{2}+\sum a^{2}\leq 9+\sum ab+\sum a^{2}(\frac{b}{c}+\frac{c}{b})\]
\[\sum ab \leq 9+\sum a^{2}(\frac{b}{c}+\frac{c}{b}-2)\]
\[\sum (\frac{b}{c}+\frac{c}{b}-2)\leq \sum a^{2}(\frac{b}{c}+\frac{c}{b}-2)\]
\[\sum (a^{2}-1)(\frac{b}{c}+\frac{c}{b}-2)\geq 0\]
and the last inequality is obvious, since $a^{2},b,^{2},c^{2}>1$ and $x+\frac{1}{x}\geq 2$ for $x>0$

[hide="for yosh"]saya iseng saja nulis lagi disini :mrgreen:[/hide]


geometry probability AMC USA(J)MO USAMO MATHCOUNTS trigonometry

Never had much experience with probability...
Even if I have no connection to MATHCOUNTS...

That is probably what I would've chosen if geometry was easier.

Hmm, I used to be a lot worse at geometry than counting, but now, I'm worse at counting :|

Probability and Geometry mainly. I haven't done Probability yet in C/P and I havent done Intro to geo yet.

Finally, somebody who agrees with me.  :)

Oh, just tough number theory problems in general. I can solve them, but I'm pretty slow and I have trouble, even after doing Introduction to NT. Not to say I don't like it, however!

I used to be pretty bad in geometry, but I'm getting better now that I got Intro to Geo. Into to C/P helped a lot too.

[quote="isabella2296"]I'm pretty weak in number theory, so that's what I put.[/quote]

NT is my favourite :(

I said I was bad at Number Theory, but I also said [b]that doesn't mean I don't like it[/b]. I think NT problems are interesting, I'm just terrible at them.

[quote="pythag011"]NT is my absolute favorite.

There are a lot of really nice NT problems on USAMO and IMO.

Hmm...probably probability is my weakest at MathCounts, though counting is my second strongest...[/quote]

That's the spirit!

I suck at probability...

my strength is with Algebra. If I could learn some more about Geometry, that would probably be a strong topic for me. As for number theory...I don't even know it.

It appears that Algebra is a strong subject for everybody, but may I remind you that Algebra includes the Distance and rate problems. Not that I think it [b]isn't[/b] easy, it is probably my strongest subject.

@qwerty: Do you know modular arithmetic and number bases? If so, that's number theory. I strongly recommend you get Introduction to Geometry. It really helps!

Algebra is my strength, for sure. It has got to be my best subject. I'm a little unsteady in counting and probability, and I'm getting a lot better in Geometry.

Number Theory pwns me... :(
But now I'm doing excessive practice in Intro to NT so I'm getting lots better! :D

Probability and Counting... I could do all the 08 nats sprint problems except for the last one... Counting xP.

I fail at number theory, but intend on getting better.  :)

So probability and geometry are the hardest topics according to AoPS users. I agree, those are my two weakest subjects too.

I said that my weakest was probability, but let's write them all:
Weakest: Probability
2.Counting
3.Geometry
4.Algebra
Strongest: Number Theory!!!

Wow, I'm not the only person weak in Number Theory. Here's my list:

Weakest: Number Theory
2: Counting
3: Probability
4: Geometry
Strongest: Algebra

[quote="Kaixi"]probablility >.>;;
[i]very[/i] problematic for me  :\
getting better though[/quote]
Same here! :P

1 (Weakest): Counting  }
2: Probability               } Improving
3: Algebra
4: Number Theory
5 (Strongest): Geometry

I guess that's my list

Favorites:

1 (least): Counting
2: Probability
3: Algebra
4: Number Theory
5 (most): Geometry

Hmm...

1: Geometry
2: Probability
3: Counting
4: Number Theory 2nd Strongest
5: Algebra Strongest

well, uh, it depends. At the MATHCOUNTS level, i dont really have a weakness. I just miss problems because of careless errors(a lot of them).

Above mathcounts, definitely number theory. Geometry is my best, then algebra. Im still working on intermediate counting and probability though.

Geometry is my best too, then algebra! Then counting/probability, and then last number theory.

The only reason why geometry is my weakest is because all the Nats. geometry problems are always so hard, and it looks like there are never any solutions, but then when the answer is explained to you, it always takes time for me to figure out how to do it again.

Number theory and geometry for clear concepts but algebra for knowledge

I don't really think that counting is really all that hard, you just need to know how to solve the problem.

Who chose other?
And why?

I chose Geometry. It is very error-prone for me.

Yeah, that's another reason why I chose it.

Who chose other?

I chose number theory, because I didn't even know what it was.

It is a little bit confusing to explain. You should look it up. There are usually many different topics in Number Theory. Or you should check out the stuff about the book on Number Theory on this website. They even have a few sample questions.

I'm terrible at Geometry. I find Algebra easier to reason through. But Trig is okay, so I'm fine with some Geo.

Number theory is just things like modular arithmetic and base numbers.

Yeah, that's pretty much what it is in general, but there are also many other types of problems in that category too, like the problems that use made up symbols as different operations like 'if a*b = 2a + 2b, then what is 5*6' and there are a few others, but I can't remember them all.

Yeah basically... but there are problems that I've seen in the Number Theory category that haven't been said though. You could just look at the Number Theory book by rruszcyk and see what topics are in it.

P.S. - nice new avatar gauss1181


induction
function
vector
number theory
totient function

Show that the $n$th triangular number is given by the formula

$T_{n}=\frac{1}{2}n(n+1)$

Show that the $n$th pentagonal number is given by

$P_{n}=\frac{1}{2}(3n^{2}-n)$

Show that the general expression for a $k$-gonal number is

$\frac{1}{2}k(n^{2}-n)-n^{2}+2n$

Find all odd $n<100$ for which a regular $n$-gon can be constructed.

[quote="brady00n"]Show that the $n$th triangular number is given by the formula

$T_{n}=\frac{1}{2}n(n+1)$[/quote]
[hide] Induction: $\frac{1}{2}1(1+1)=1$. Now, we must prove that if the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, the sum of the first $n+1$ natural numbers is $\frac{(n+1)(n+2)}{2}$. $\frac{(n+1)(n+2)}{2}=\frac{n^{2}+3n+2}{2}=\frac{n(n+1)}{2}+n+1$.
$QED$[/hide]

[quote="ZetaX"][quote="brady00n"]Find all odd $n<100$ for which a regular $n$-gon can be constructed.[/quote]
Hardly intermediate ;)

$n=1,3,5,15,17,51,85$[/quote]

how do you construct a regular 1-gon?

The last problem seems to have very little to do with the others  :rotfl:

Why can't an n-gon be constructed for any odd number? Like 7, 9, etc...

A regular $n$-gon can be constructed by ruler and compass iff $\varphi(n)$ (totient function) is a power of $2$.

The proof that $\varphi(n)$ being a power of $2$ is necessary requires some usage of vector spaces, especially their dimension, and of algebraic degrees.

The other direction uses stronger theorems, e.g. Galois theory.

Oh! You guys are talking about handmade.

Umm... what else¿ (assuming handmade=ruler and compass)


linear algebra
matrix
linear algebra unsolved

If a matrix A is in Jordan Normal Form, let A = U + V, where U is the diagonal matrix consisting of the diagonal entries of A, and V is the matrix consisting of the off-diagonal 1's, i.e. V = U - A. Is it necessary that UV = VU? Give a proof or a counterexample.

Yes.  $ U$ is block-diagonal and $ V$ has the same block structure, and each block of $ U$ is a scalar multiple of the identity.

$ U \equal{} \begin{bmatrix}\lambda_1 I & 0 & \cdots & 0 \\
0 & \lambda_2 I & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_k I\end{bmatrix}$ and $ V \equal{} \begin{bmatrix}N_1 & 0 & \cdots & 0 \\
0 & N_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & N_k\end{bmatrix}$. Write out that product in block form.


function
inequalities
inequalities proposed

To this problem, I suggest the following solution :

Consider the function $f$ defined by $f(x)=\frac{1+x}{1+x^{2}}$, then
$f'(x)=\frac{-x^{2}-2x+1}{(1+x^{2})^{2}}$, and so
$f"(x)=\frac{-2x^{3}-6x^{2}-10x+2}{(1+x^{2})^{3}}$. To check that $f$ is
concave it suffices to study the sign of the function
$g(x)=-2x^{3}-6x^{2}-10x+2$. Since $g$ is strictly decreasing on $R$ and
continuous, it is a bijection from $R$ to $R$. Consequently, there
is a unique real number $\alpha$ such that $g(\alpha)=0$. Hence, by
using the theorem of intermediate values, we obtain that
$0<\alpha<1/2$. Thus $g(x)<0$ for $x>\frac{1}{2})$; and so $f$ is
concave.

Now, by using Jensen theorem for concave functions, we get this
inequality: \[f({\frac{1}{5}}(a+b+c+d+e))\geq\frac{1}{5}({\frac{1+a}{1+a^{2}}+{\frac{1+b}{1+b^{2}}}+{\frac{1+c}{1+c^{2}}}+{\frac{1+d}{1+d^{2}}}+{\frac{1+e}{1+e^{2}}}})=A.\] On the other hand, \[\frac{1+a}{1+a^{2}}\leq\frac{1+a}{2a}\leq(\frac{1}{2}+\frac{1}{2a})\leq 1 \mbox{ if }a\geq 1.(\star\star)\] Moreover, AM-GM ensures that $a+b+c+d+e\geq 5$
$\Leftrightarrow\frac{ a+b+c+d+e}{5}\geq 1$. Hence, $(\star\star)$
is satisfied, and $f({\frac{1}{5}}(a+b+c+d+e))\leq 1$. This shows
that $5A\leq 5$.

Now the problem is solved for every positive numbers $a,b,c,d, e >\frac{1}{2}$.

For the second part, we may suppose that  $0\leq a\leq\frac{1}{2}$.

 it's so easy to prove the inequality by using order properties. for
 the
others suppositions it's so simple.

[quote="vietnam2007"]To this problem, I suggest the following solution :

Consider the function $f$ defined by $f(x)=\frac{1+x}{1+x^{2}}$, then
$f'(x)=\frac{-x^{2}-2x+1}{(1+x^{2})^{2}}$, and so
$f"(x)=\frac{-2x^{3}-6x^{2}-10x+2}{(1+x^{2})^{3}}$. To check that $f$ is
concave it suffices to study the sign of the function
$g(x)=-2x^{3}-6x^{2}-10x+2$. Since $g$ is strictly decreasing on $R$ and
continuous, it is a bijection from $R$ to $R$. Consequently, there
is a unique real number $\alpha$ such that $g(\alpha)=0$. Hence, by
using the theorem of intermediate values, we obtain that
$0<\alpha<1/2$. [/quote]

 I doubt your solution. B/c
$f(x)=\frac{1+x}{1+x^{2}}$
$f''(x)=\frac{2(x-1)(x^{2}+4x+1)}{(1+x^{2})^{3}}$

[quote="kunny"]For $0<x\leq 1,$ we have $\frac{1+x}{1+x^{2}}\leq \frac{3-x}{2}\Longleftrightarrow (x-1)^{3}\leq 0.$[/quote]
But It means nothing because $abcde=1$ implies that there exist 1 number greater than 1.  :lol:


[quote="mathematica"]Find all $f: \mathbb{R^{+}}\longrightarrow \mathbb{R^{+}}$ such that:

I.$f(x)+f(y)+f(z)+f(xyz)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$

II.$\forall$ $y>x\geq 1: f(y)>f(x)$[/quote]

This is a nice IMO shortlist problem see  [url]http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=17&year=2003&p=119999[/url]


Παρακάτω σας παραθέτω την άσκηση του test στα μαθηματικά κατεύθυνσης στο κεφάλαιο σχετικό με τις ευθείες. Μπορεί να μη σας φανεί δύσκολη, ωστόσο τη θεώρησα διδακτική. Σας τη παραδίδω μερικώς τροποποιημένη.

[b]ΑΣΚΗΣΗ[/b]

Δίνονται οι ευθείες: ($k\in\mathbf{R}$)

$\epsilon_{1}: \quad y=\left(1453\cdot k^{2006}+2\right)x+1821$

και

$\epsilon_{2}: \quad y=\left(\sin 2006\lambda-\sqrt{3}\cos 2006\lambda\right)x+1940$

Για ποιες τιμές του $k$ οι ευθείες είναι παράλληλες;

[quote="socrates"]...$k=0$....

 :)[/quote]

Πολύ ωραία Socrates :D

Ας κάνω όμως εγώ τα του $\LaTeX$, για να καταλάβουν όλοι το σκεπτικό του προβλήματος.

Κατ' αρχήν, από τη στιγμή που μας λέει για παραλληλία, θα πρέπει οι συντελεστές διευθύνσεως να είναι ίσοι. Για το συντελεστή διεύθυνσης της $\epsilon_{2}$, αν τον αναλύσουμε λίγο περαιτέρω θα πάρουμε: (Έστω $2006\lambda=x$)

$\sin x-\sqrt{3}\cos x=2\cdot\frac{1}{2}\sin x-2\cdot\frac{1}{2}\sqrt{3}\cos x = 2\left(\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\right)=2\left(\cos\frac{\pi}{3}\sin x-\sin\frac{\pi}{3}\cos x\right)=2\sin\left(\frac{\pi}{3}-x\right)$

Επίσης $\left|2\sin\left(\frac{\pi}{3}-x\right)\right|\leq2\quad\textsl{(1)}$
και για το συντελεστή διεύθυνσης της $\epsilon_{1}$ ισχύει:$1453\cdot k^{2006}+2\geq2\quad\textsl{(2)}$. Οι $(1)$, $(2)$ όμως παρουσιάζουν κοινή λύση μόνο για $2\sin\left(\frac{\pi}{3}-x\right)=1453\cdot k^{2006}+2=2\implies1453\cdot k^{2006}+2=2\implies k=0\ldots$


Find the $ 10^{th}$ term of the sequence $ \minus{}50$, $ \minus{}49$, $ \minus{}47$, $ \minus{}44$, $ \minus{}40$.

[quote="RunpengFAILS"]The numbers increase by triangular numbers so the 10th term is ${ -50+\frac{9\cdot{10}}{2}}=\boxed{-5}$[/quote]

What would be the formula for finding triangular numbers? Wouldn't it be n(n-1)/2?

Yeah, since we are finding the 10th term, 10 is our n. Therefore, it must be $ \frac{(10)(10-1)}{2} = \frac{10\cdot 9}{2} $.


AMC
AIME
geometry
rectangle
search
3D geometry
tetrahedron

Point P is inside a rectangular box. The distances from point P to four of the vertices of the box are 1, 2, 3, and 4. If the distance from P to another vertex is greater than 5, find that distance.

ABCD doesn't have to be a rectangle. The points could be non-coplanar since it's a box, not a rectangle.

[quote="matt276eagles"][quote="Elemennop"]Use the fact for any point P and a rectangle ABCD, then $PA^{2}+PC^{2}= PB^{2}+PD^{2}$. Then it's just plugging in to see which vertices are A,B,C,D on the cube.[/quote]

Can you show a proof of that?[/quote]
You could search it on google or MathWorld, I believe it is called the British Flag Theorem.

well, it's quite obvious if you draw lines parallel to both sets of sides through $P$.  the distance from $P$ to the rectangle is the same so you don't have to bother with that.

[quote="matt276eagles"][quote="Elemennop"]Use the fact for any point P and a rectangle ABCD, then $PA^{2}+PC^{2}= PB^{2}+PD^{2}$. Then it's just plugging in to see which vertices are A,B,C,D on the cube.[/quote]

Can you show a proof of that?[/quote]

I'm too lazy, but the brunt of the proof is to show it for P in the plane of ABCD by dropping perpendiculars, and then use the planar version to prove the spatial version by dropping perpendiculars again.

I'm kind of confused, I mean couldn't the other sides be anything? If you put the four vertex on one face, the and then P is on top of it, it would create a tetrahedron, which has nothing to do with the other side. Please clarify me if I thought/read something wrong.

[quote="matt276eagles"][quote="Elemennop"]Use the fact for any point P and a rectangle ABCD, then $PA^{2}+PC^{2}= PB^{2}+PD^{2}$. Then it's just plugging in to see which vertices are A,B,C,D on the cube.[/quote]

Can you show a proof of that?[/quote]

easily we can show that any rectangle can be represented on a cartesian grid with the points:

$A=(a,b)$, $B=(a,-b)$, $C=(-a,-b)$, and $D=(-a,b)$

with $a,b>0$. Define $P=(x,y)$. Then \[PC^{2}+PA^{2}=(x-a)^{2}+(x+a)^{2}+(y-b)^{2}+(y+b)^{2}\]  \[=2(x^{2}+y^{2}+a^{2}+b^{2})\] It can easily be shown that $PB^{2}+PD^{2}$ takes the same value, therefore they are equal as desired.

[quote="computer_nuke"]O, so P is coplanar with the points A, B, C, and D....[/quote]

not necessarily, i was just using that notation to prove the theorem


geometry
circumcircle
incenter
geometric transformation
angle bisector
geometry solved

1. The circles S1, S2 intersect at A and B. Let Q be a point on S1. The lines QA and QB meet S2 at C
and D, respectively, while the tangents to S1 at A and B meet at P. Assume that Q lies outside S2,
and that C and D lie outside S1. Prove that the line QP goes through the midpoint of CD.

2. A circle centered at O and inscribed in triangle ABC meets sides AC,AB,BC at K,M,N, respectively.
The median BB1 of the triangle meets MN at D. Show that O,D,K are collinear.

3. The incircle of triangle ABC touch sides AB,BC, CA at M,N,K, respectively. The line through A
parallel to NK meets MN at D. The line through A parallel to MN meets NK at E. Show that the
line DE bisects sides AB and AC of triangle ABC.

4. Let ABCDE be a convex pentagon, and letM,N, P,Q,R be the midpoints of sides AB,BC, CD,DE,EA,
respectively. If the segments AP,BQ, CR,DM have a common point, show that this point also lies on
EN.

5. In triangle ABC, AB = AC and the angle bisector of C meets AB at D. The perpendicular to CD
through the center of the circumcircle of ABC intersects BC at E. The parallel to CD through E meets
AB at F. Show that BE = FD.

6. In triangle ABC, AC = BC and O is the the circumcenter, I the incenter, and D lies on BC so that
lines OD and BI are perpendicular. Prove that ID and AC are parallel.

maybe not hard though...only that i dunno how to do them

do u have nathan altshieler court book?

all of these problems are there.

if not, where did u find these problems?

Number 4 was solved here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45945[/url]

My apologies, Valentin, for registering the second additional account for today, but I don't feel like logging in with any of my passwords on a computer infected by a trojan. Besjdes, after [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101347]this here[/url], I became rather mistrustful with respect to other people here.

Anyway, I wanted to say that four of your problems have been discussed:

[b]EDIT:[/b] 1. Now discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=101976 .

2. Theorem 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4, viewed from a slightly different direction. (I already told that in the other topic.)

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=33503 .

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=45945 .

[b]EDIT:[/b] 5. The problem is wrong, but a corrected version was solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=28336 .

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=24876 .

Please post each problem into one single thread, in future. As for now only problems 1 and 5 remain open, I propose you to open a new topic for problem 1 (it is actually quite easy) and in this thread solutions to problem 5 should be discussed.

  darij

[quote="darij grinberg"]Ashegh, are these problems really from Altshiller-Court? I remember seeing some of them in olympiads...

  darij[/quote]

i mean in his book, not in himself :D

[quote="Ashegh"]i mean in his book, not in himself :D[/quote]

What an answer  :D

No, actually I know that you are speaking of his book, I was asking whether you accidentally happen to know whether the problems are in the original edition (Nathan Altshiller Court, [i]College Geometry - An Introduction to the Modern Geometry of the Triangle and the Circle[/i], Barnes & Noble, 1952) or they were added in the Farsi translation in later years.

  Darij

oh, i dont know wether the book translated originally, or not.

but, i think there is nothing to add to the translated book. :D


Is there a way to prove AM-GM using Lagrange Multipliers? If so, how?


The sum of the heights of six people is 336 inches. No two people are the same height. What is the mean height of the third and fourth tallest people if the mean and the median height of all six people are equal?

[hide]The mean and median are clearly $\frac{336}{6}=56$.
This means that the mean of the third and fourth person is also $56$.
(When there are an even amount of people the median of the set is the mean of the two middle members.)[/hide]


[youtube]vLew1ggDfQQ[/youtube]

I'm speechless. This is the most important event in the history of StarCraft aside from the release of starcraft. I mean, what can I say, this game is sheer uber epicness. Seriously. I mean, WOW.

I thought YouTube had a 10-minute time limit? :huh:

Wow. I actually watched through the whole vid..... Nice! :D

[quote="(^_^)"]yeah....absolutly amazing that people would be willing to spend that much time on a game and people would be willign to watch that whole thing....[/quote]
Dude, like half the people in GFF spend about 2 hours here anyway. So really. So yeah, if you play starcraft you'll appreciate how insane that vid is.

i didn't know that MBC hosted SC games...

but seriously... woww...

I love that game. I know this is off-topic, but what do people around here think of SC II?

[quote="Nerd_of_the_Ages"][quote="kimmystar94"]i didn't know that MBC hosted SC games...[/quote]
Err, I think you read it wrong, look at it again. Anyway, if there's a pro SC of WC game, you should immediately be able to tell that it's in Korea.  :P[/quote]

i know. i'm Korean. smart one.

anyway, some of my friends are one of those SC freaks......

and read what wrong??

whatever... but seriously. this is crazy. i play crazyarcade (from nexon) and i can't play a single game for more than 5 mins.


modular arithmetic
number theory
relatively prime
number theory unsolved

There are no such digits, 'a'.
See since n>3, thus least value of n(n+1)/2 = 10, which becomes a number, since a is a digit, it must be a single digit-answer, thus proved...

soumik, you didn't understand the problem. Try $ n\equal{}36$

This is trickier than I thought at first. It's easy to find $ 10\cdot 11/2 \equal{} 55$ and $ 11 \cdot 12 / 2 \equal{} 66$, so $ a\equal{}5$ and $ a\equal{}6$ are possible. It's also quite easy to see that $ a$ cannot be any other digit [b]except possibly[/b] 1. I haven't been able to rule out 1 yet.

Oh! I'm damm idiotic to jump at conclusions so fast!!!! :(

Randomgraph, could you elaborate more as to why the other digits don't work?

Well, the reason in a nutshell is that $ f(n) \equiv 2 \pmod{10}$ has no solution (where $ f(n) \equal{} \frac{n(n\plus{}1)}{2}$), and the same goes for 4, 7 and 9. For 3 and 8 you just have to go one level deeper and notice that $ f(n) \equiv 33 \pmod{100}$ has no solution, and the same goes for 88. If you need more details on how to show those things, let me know.

The nuisance with numbers of the form 111...1 is that you can actually find $ n$s for which $ f(n)$ is congruent to 111....1 modulo an arbitrarily high power of 10. For instance, $ f(963541)\equal{}464206111111$. So a simple-minded argument modulo a fixed power of 10 is just not going to work. I'll see if I have time to look more closely into that.

will you explain me in brief please randomgraph ?

[quote="lajanugen"]will someone help me with this problem?
For which digits "a" do exist integers n>3 such that each digit of n(n+1)/2 equals a ?[/quote]
[hide="Solution"]
Notice that $ k \equal{} \frac {n(n \plus{} 1)}{2}$ is a perfect square if and only if $ 8k \plus{} 1$ is a perfect square, namely $ (2n \plus{} 1)^2$.  Now, $ k$ is a number with only digit $ a$.  If $ a$ is $ 2$, then $ 8k \plus{} 1$ ends in a seven, so it is not a perfect square.  If $ a$ is $ 3$, then $ 8k \plus{} 1$ ends with a $ 65$, so it is not a perfect square.  If $ a \equal{} 4$, then $ 8k \plus{} 1$ ends with a $ 3$, so it is not a perfect square.  If $ a \equal{} 5$ or $ 6$, then we can find such triangle numbers, namely $ 55$ and $ 66$. If $ a \equal{} 7$, then $ 8k \plus{} 1$ ends with a seven.  If $ a \equal{} 8$, then $ 8k \plus{} 1$ ends with a $ 05$, so it is not a perfect square.  If $ a \equal{} 9$, then $ 8k \plus{} 1$ ends with a three, so it is not a perfect square.  Thus, the only case to deal with is if $ a \equal{} 1$.

This gives that $ \frac {n(n \plus{} 1)}{2} \equal{} \frac {10^k \minus{} 1}{9}$ for some $ k\ge 2$, so\[ k \equal{} 8n \plus{} 1 \equal{} \frac {(10^k \minus{} 1)8}{9} \plus{} 1
\]which means that $ 9k \equal{} 8\cdot 10^k \plus{} 1$.  $ k$ is a perfect square, so $ 9k$ is a perfect square, say $ m^2$.  Then,\[ (m \plus{} 1)(m \minus{} 1) \equal{} 8\cdot 10^k \equal{} 2^{k \plus{} 3}\cdot 5^k\]Notice that the only factor that $ m \plus{} 1$ and $ m \minus{} 1$ can share (and they must share if $ k\ge 1$) is $ 2$.  Letting $ m \plus{} 1 \equal{} 2x$ and $ m \minus{} 1 \equal{} 2x \minus{} 2$, we get that $ x(x \minus{} 1) \equal{} 2^{k \plus{} 1}\cdot 5^k$.  $ x$ and $ x \minus{} 1$ are relatively prime, so either $ 5^k \equal{} x$ and $ 2^{k \plus{} 1} \equal{} x \minus{} 1$ or $ 2^{k \plus{} 1} \equal{} x$ and $ 5^k \equal{} x \minus{} 1$.  This means that $ |5^k \minus{} 2^{k \plus{} 1}| \equal{} 1$, which only happens when $ k \equal{} 1$ (or else $ 5^k > 2^{k \plus{} 1} \plus{} 1$.)  Yet, recall that $ k\ge 2$, which is a contradiction.  Thus, $ a\ne 1$, so $ a \equal{} 5, 6$ are the only solutions. [/hide]


algorithm
Support
calculus
college
linear algebra

I am a computer science and math student right now. I am aiming to do a lot of the learning on my own, (not in a college/school).

I plan on taking the CLEP Exam, info in the link. It is essentially the same as an AP Exam or a Credit by Exam for CALC1 in college.

http://www.collegeboard.com/student/testing/clep/ex_calc.html

I want to learn math to see if I enjoy it for its own sake, and for being generally more logical/smarter (mental exercise), and for complimenting work I may do in the sciences like physics and computer engineering/programming.

So, from what I understand, a lot of lower level Math, in classes like CALC1,2,3, Linear Algebra, and Differential Equations, is all about algorithms.

Anyway, what does it mean to learn by doing? How much of the learning is conceptual? How much is of it, is pretty much robotic rule following?

My main problem in learning Math, is getting stuck in cognitive loops trying to make sense of the conceptual/theoretical discussion presented prior to the exercises in every section.

But then there is this approach called "Learn by Doing", which says you basically do robotic exercises and drills and memorization, and then you learn math that way. Look at the examples before the exercises, and then just apply those algorithms to the exercises enough times so that you know the rules by heart.

I read a book on the topic, (The Equation for Excellence: How to Make Your Child Excel at Math)
[http://www.amazon.com/Equation-Excellence-Make-Child-Excel/dp/0980144604[/url]
. It basically argued that drilling, memorization, and other "cranking"/"grinding" learning behaviors are critical to developing math skill.

[edit: for the record I read another book on the topic as well (How Mathematicians Think)
http://www.amazon.com/exec/obidos/ASIN/0691127387/ref=nosim/librarythin08-20

He also said struggling through problems, (preferably with a mentor/tutor), is the best way to learn deeply, and develop your mind.

More importantly, and to the point, what is the best way to prepare for this CLEP CALC1 exam?

Also, the other notion I've been getting, is that only through struggling through problems do you really gain a deeper understanding. The memorization and drilling is important, but they are useful as supports for the problem solving which is kind of where the light bulb goes off, and the understanding clicks in your mind.

The critical subtext to this, which has really been bothering me, is that you also need someone who has seen the problem type you are working on, and can interface with you and let you struggle, make sure you are adequately struggling, and offer as limited assistance as is necessary for you to discover the solution on  your own.

Now, I don't know if this subtext was intended, but the analogy of doing math to doing a bench press lift in the weight room, really solidified this problem for me. And this is a problem, because I don't have access to someone who knows Math really well who can work with me one on one, and who knows me well enough to interface with my psychology to kind of train me and make sure I am working hard.

If the analogy didn't make sense, it goes like this. Most people don't use math in their work. However doing math makes you smarter and more generally intelligent. Similarly, few people become serious body builders, but many people play sports. Just as you do math to enhance your general intelligence for various non-explicitly mathematical applications, you do the bench press (or running (which doesn't need a spotter, weakness in the analogy)) to help your general athleticism in various sports, although in no sport, are you expected to actually lay down and do the exact bench press behavior.

The critical part of this analogy which is making me nervous is that, as Arnold Schwarzenger said in his movie Pumping Iron, paraphrased, "It is those last few reptitions that separate the great from the champions". And those last few reps are almost always ones which require spotting. Spotting means you need someone there, who can interface with you, and make you sweat.

Similarly, in math, you need someone who knows how to do the problems, and who is willing to interface with you, and make sure you work your hardest, because as the analogy/learning-theory goes, you learn the best by struggling through the problems.

I hope this hasn't been too much to read, but it is driving me nuts.

It's undeniable that if you can find a good mentor then this can help you a whole lot to get better at math.  Also, if you can do math with talented peers, hopefully some of whom are (for the time being) more talented than you, then that can help a whole lot.

Many people live in a place, or go to a school, where talented mentors and peers cannot be found.  Luckily, now that AoPS exists, they aren't completely out of luck.  Just doing a lot of problems on the AoPS forums, and reading solutions from other students, can go a long way.

Linear Algebra has the potential to be a very conceptual course, not just robotic manipulations.  It is often used as an "introduction to higher math", meaning proofs and more abstract concepts instead of just finger exercises.

Calculus can also be presented in a way that is wonderfully conceptual, not just memorization/manipulation.

I don't put much value in memorization and drilling.  I think understanding the derivations and proofs is the way to enjoy math and to be good at it.

Hello!  Welcome to the forum  :)

It seems you've thought about this quite a bit.  The answers to your questions depend a lot on how someone wants to go about learning math.  I'm sure that many people on this forum have different thoughts on this, but here's what I think:

To prepare for a standardized introductory test like the CLEP CALC I, you don't have to know much of anything conceptual.  It really is about learning algorithms.  This is something you can do from almost any exam-prep book.  But that's not why you're here.

[quote]I want to learn math to see if I enjoy it for its own sake, and for being generally more logical/smarter (mental exercise), and for complimenting work I may do in the sciences like physics and computer engineering/programming. [/quote]

There is a whole universe of math beyond the algorithms introduced in introductory level courses.  I like to describe higher mathematics as the study of everything that is verifiably true (doesn't that sound a little broad?  it is.)  Consequently, learning mathematics is a never ending journey.  This generally starts with an introduction to proof.  Based on the type of interest you've expressed in mathematics, I think this is what you want to study.  The type of conceptual and logical thinking that this develops is far more valuable than the algorithms you need for CLEP, and will undoubtedly help you in all of your academic endeavors (even humanities!).

You brought up an excellent point, one of the best ways to learn higher mathematics is by struggling through problems.  While the knowledgeable mentor certainly can help with this, these valuable people can be very hard to find (depending on where you live).  Another great way to struggle through problems is to work through a quality math book.  There are some books that provide a "gentle" introduction to higher mathematics.  However, even with a gentle introduction, you must be prepared to struggle through problems that you may not know how to do.  It takes quite a bit of effort to learn mathematics this way, but its payoff is very rewarding (not just academically; it's also incredibly exciting whenever you discover a nice solution to a tough problem).

A book you may want to consider: Liebeck, Martin [u]A Concise Introduction to Pure Mathematics[/u]

I haven't worked through this book myself, but I've heard it's good, and skimming through its contents makes me think it would be a good fit for you.  There are many other introductions to higher mathematics that are good too.  Just choose one and spend a lot of time working on problems.

A word of warning: It's very hard to generalize about class topics.  Some introductory linear algebra classes are entirely based on algorithms, while others go in-depth into mathematical proof.

Good luck!!

- Berkeley


Let $G$ be a graph with $n$ vertices and more than $\frac{(n-1)k}{2}$ edges where $k\geq 2$.show that $G$ contains an elelmentary cycle of length at least equal to $k+1$.

This is known as Erdös-Gallai theorem.

Pierre.


Hi everyone,

Could someone please help me with this?

According to the study, brain weights of Swedish men are normally distributed with a mean of 1.40 kg and a standard deviation of .11 kg.

Determine teh percentage of all samples of three Swedish men that have mean brain weights within .1 kg of the population mean brain weight of 1.4 kg. Interpret your answer in terms of sampling error.

I know that the mean=1.40 kg
n=3
standard deviation=.11 kg
standard deviation x bar=.064

z=-1.4/.064=-22.95
z=1.4/.064=24.22

Could someone please tell me if this is correct so far?

Thank you very much

I hope $ 0.064\equal{}0.11/\sqrt 3$. If so, your argument is correct up to that point :). After that it is a clear mess. What you should do is to compare the allowed deviation $ 0.1$ from the mean to the standard deviation $ 0.064$ you got. Comparing the mean itself to the standard deviation hardly makes any sense. Needless to say, out of the two lines
z=-1.4/.064=-22.95
z=1.4/.064=24.22
at most one can be correct, so you should be more accurate with your arithmetic as well.


How do you express the little dot that means multiplying?  :?

Do you mean \cdot as in $3!=1 \cdot 2 \cdot 3$ ? See http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

Yes, thats exactly what i meant. Thanks for your help.  :)


A log is in the shape of a right circular cylinder diameter 12. Two plane cuts are made, the first perpendicular to the axis of the log and the second at a $45^\circ$ angle to the first, so that the line of intersection of the two planes touches the log at a single point. The two cuts remove a wedge from the log. Find its volume. (http://www.kalva.demon.co.uk/aime/aime03b.html)

(BTW, does anyone have the real wording of this problem? I don't understand the Kalva version at all :()

[hide]
after drawing the 2D cross section of the side, we find that it is a 45-45-90 triangle with a leg of side 12, so the height of the wedge will be 12 too.  it is easy to see that the volume of the wedge is half of the cylinder, so $6^2 * \pi * 12 / 2 = 216\pi$
[/hide]

[quote="Farenhajt"]Does the picture help?[/quote]

Thanks! But one question: where are the plane cuts? I see one of them, the one that is at a 45 degree angle, but what about the other one that is perpendicular to the axis?

BTW, what did you use to make that picture? :)

The bottom red ellipse is the other cut (red lines represent the wedge). Perhaps I should've drawn the axis too, sorry.

EDIT: Picture is edited.

Aghh I misinterpreted the meaning of "axis" :mad:

The question is on the UNL site as one of the practice problems of 2004.

Yeah, I could never understand the question as worded by Kalva. I had to ask someone what it was describing.


function
modular arithmetic
number theory
least common multiple
totient function
relatively prime

How do you do questions like....

What is $2^{100}$ in mod 100?

Thanks.

A couple useful formulae:

Fermat's Little Theorem:
\[a^{p}\equiv a \bmod p \ \ \text{(for prime p)}\]
Euler's Generalization, aka Euler's Totient Theorem:
\[a^{\phi(n)}\equiv 1 \bmod n \]
where $\phi(n)$ is the totient function and $\phi(n)=a \prod_{i=1}^{m}\left( 1-\frac{1}{p_{i}}\right)$, if $a$ is composed of $m$ distinct primes $p_{i}$.

There's also the Chinese Remainder Theorem (look it up) which says that there is a bijection $\left( \mathbb{Z}_{r}\times \mathbb{Z}_{s}\right) \rightarrow \mathbb{Z}_{rs}$ for relatively prime $r$ and $s$ (i.e. that if you have two modular residues in mod $r$ and $s$, there is only one modular residue in mod $rs$ which satisfies both.) But I forget how to calculate it. Look it up on Mathworld or Google.

Carmichael's function is also incredibly useful:

$a^{\lambda(n)}\equiv 1 \pmod n$, with $\lambda(n)$ being the smallest exponent to satisfy this equation for all $a$.

$\lambda(n) = \text{lcm}(\lambda(p_{1}^{k_{1}}),\lambda(p_{2}^{k_{2}}),...,\lambda(p_{i}^{k_{i}}))$, for all prime powers in $n$.

$\lambda(p^{k}) = \varphi(p^{k})$, unless $p^{k}$ is a $2^{k},\ k \geqslant 3$, then $\lambda(2^{k}) = 2^{k-2}$


Jessica ran the 6.2-mile Bolder Boulder road race in 55 minutes. At the same pace, how many minutes will it take her to run the Horsetooth Half-Marathon (13.1 miles)? Express your answer as a decimal to the nearest tenth.

Set up a proportion-
6.2/55= 13.1/x

Cross multiplying,
6.2x= 720.5
x= approx. $ 116.2$ minutes


These angels are made of dihydrogen monoxide.

[hide="Colorless, odorless..."] Snow angels! [/hide]

do you mean angels or angles or poison i am so confused

[hide]SNOW/ICE/FROST ANGELS!!![/hide]
assuming that it was angELS and not anGLEs...
Tell me this....is dihydrogen monoxide the same thing as Oxygen DiHydride? :D

how about strange angles made of dihydrogen oxide?

[hide]hint think hybridization[/hide]

[hide="solution"]dihydride is $H_{2}$ and monoxide is $O$. So it is made of $H_{2}O$. It's made of water. the angels are made of snow, frost, or ice.[/hide]


[url]https://mathcounts.org/Page.aspx?pid=1336[/url]

so apparently IL is now down to 15th place team after the messup on GA... :(

Oh no Illinois fell down one spot from 14th to 15th, what a huge difference.

they also apparently managed to spell georgia wrong. pwned.


i'm very weak in inequalities.
can anyone post some basic problems in that topic?but not too elementary

thanks in advance

Hi krishkoushik,

There's lots of stuff here: [url]http://www.artofproblemsolving.com/Wiki/index.php/Algebra/Olympiad[/url].


function
calculus
integration
limit
real analysis
real analysis unsolved

Does there exist a function $ f\in L^2(\mathbb{R})$ such that the indefinite integral is discontinuous?

I'm almost certain the answer must be yes...

The answer is no.

If $ f\in L_{\text{loc}}^1(\mathbb{R}),$ (which is a weaker condition than $ L^2$), then the indefinite integral is continuous.

Since the property in question is local, not global, we may WLOG assume that $ f\in L^1.$ Define $ g(x)\equal{}\int_a^xf$ for some arbitrary fixed $ a.$

For $ h>0,\ g(x\plus{}h)\minus{}g(x)\equal{}\int_x^{x\plus{}h}f\equal{}\int_{\mathbb{R}}f\cdot \mathbf{1}_{[x,x\plus{}h]}.$

We're interested in what happens as $ h\to 0,$ so we replace any sequence $ h_k$ which tends to zero. The sequence of functions $ f\cdot\mathbf{1}_{[x,x\plus{}h_k]}$ tends to zero pointwise [i]a.e.[/i] and are dominated by the integrable function $ |f|.$ Hence by the Dominated Convergence Theorem, $ \lim_{h\to0^\plus{}}(g(x\plus{}h)\minus{}g(x))\equal{}0,$

A very similar argument can be made for the left hand limit, this time writing $ g(x)\minus{}g(x\minus{}h)\equal{}\int f\cdot\mathbf{1}_{[x\minus{}h,x]}.$

Hence $ g$ is continuous at each point.

------

Side lemma, justifying my first sentence: if $ f\in L^2(\mathbb{R}),$ then $ f\in L_{\text{loc}}^1(\mathbb{R}).$

By Cauchy-Schwarz, $ \int_a^b|f|\le \left(\int_a^b|f|^2\right)^{\frac12} \left(\int_a^b1\right)^{\frac12}\le\|f\|_2\cdot\sqrt{b\minus{}a},$

which is finite for any finite interval.


function
number theory proposed
number theory

[quote="lhvietbao"]Consider the function $f: [0;N] \to R$ which satisfies
$\mid f' \mid \leq A$ and $f''>0$ in its domain.
Let $k$ be the number of lattice points on the curve $y=f(x)$.
Prove that there exist a constant $c$ which depends only on $A$ such that
$k<cN^{\frac{2}{3}}$ for all $N$ and $f$ satisfying the above conditions.[/quote]

:)

[quote=lhvietbao]Consider the function f:[0;N] -> R which satisfy
$\mid f' \mid \leq A$ and f''>0 in its domain.
Let k be the number of lattice point on the curve y=f(x).
Prove that there exist a constant c which depend only on A such that
k<c$N^{\frac{2}{3}}$ for all N and f satisfying the above conditions.[/quote]


Decide whether there exist $10,000$ ten digit numbers divisible by $7$, all of which can be obtained from one another by a reordering of their digits.

woa...a digit slide...does there exist an integer, $n$, such that $7|(10^{n}-1)$...i think so!...n=6, but also n=12, and 6k for k\in Z...so yes


group theory
abstract algebra
number theory
least common multiple
superior algebra
superior algebra theorems

1.) What is the largest cyclic subroup of $Z_{6}$ x $Z_{8}$?
My answer is 30. Is it correct?

2.) Find all the index of <3> in $Z_{24}$

Your #1 is incorrect. There are no subgroups of order 30, as can be checked by basic divisibility criteria.

#2: That "all" shouldn't be there, but you can just count the cosets.

Sorry, the question in number 1 was suppose to be the largest order of the group. My idea is that, in Z_6, the largest possible order is 6 and in Z8, the largest order is 15. That is a group in the form of (a, b, c)(d, e, f, g, h). So the LCM of 6 and 15 is 30. Don't really know if I am right.

$S_{8}$ and $Z_{8}$ are not the same. You asked about cyclic groups, but your answer was based on symmetric groups. It's still wrong in that context, but for a subtler reason.

Also, "largest order" and "largest cyclic subgroup" are the same.

oops, sorry. I mixed up everything.  :blush:

You could begin by writing $G = \mathbb Z_{6}\times \mathbb Z_{8}\cong \mathbb Z_{3}\times \left( \mathbb Z_{2}\times \mathbb Z_{8}\right)$, since $\mathbb Z_{mn}\cong \mathbb Z_{m}\times \mathbb Z_{n}$ for $\gcd (m,n) = 1$.

Now use this lemma: If $\left( G,+\right)$ is commutative and $x,y \in G$ such that $\text{ord}\, x = m$ and $\text{ord}\, y = n$, where $\gcd \left( m, n \right) = 1$, then $\text{ord}\, (x+y) = mn$.

Take $x = (a,b,c) \in G$ and write it as $y+z$, where $y = (a,0,0)$ and $z = (0,b,c)$. The order of $y$ divides $3$ and the order of $z$ divides $8$ (since the exponent of $\mathbb Z_{2}\times \mathbb Z_{8}$ is $8$), so they are coprime.

I think this is sufficient to conclude that the answer to your question is $24$. We can obviously generalize.


find all $ x \in R$ such that :
$ \frac {logx^2}{(logx)^2}$ + $ \frac {logx^3}{(logx)^3}$+ $ \frac {logx^4}{(logx)^4}$ + ... = 8
 :lol:

[quote="stvs_f"]find all $ x \in R$ such that :
$ \frac {\log x^2}{(\log x)^2} \plus{} \frac {\log x^3}{(\log x)^3} \plus{} \frac {\log x^4}{(\log x)^4} \plus{} \ldots \equal{} 8$
 :lol:[/quote]
[hide="Hint"]Let $ y\equal{}\log x$, and think about arithmetic/geometric series. (I forget what the combination is called... arithmo-geometric?)[/hide]

[hide="Solution"]We have $ \sum_{k\equal{}2}^{\infty}\frac{\log x^k}{(\log x)^k}\equal{}\sum_{k\equal{}2}^{\infty}\frac{k}{(\log x)^{k\minus{}1}}$.

Let $ S\equal{}\frac{2}{\log x}\plus{}\frac{3}{(\log x)^2}\plus{}\cdots$

Then $ \frac{S}{\log x}\equal{}\frac{2}{(\log x)^2}\plus{}\frac{3}{(\log x)^3}\plus{}\cdots$.

So $ S\minus{}\frac{S}{\log x}\equal{}\frac{2}{\log x}\plus{}\frac{1}{(\log x)^2}\plus{}\frac{1}{(\log x)^3}\plus{}\cdots$.

Let $ S_2\equal{}S\minus{}\frac{S}{\log x}\minus{}\frac{2}{\log x}\equal{} \frac{1}{(\log x)^2}\plus{}\frac{1}{(\log x)^3}\plus{}\cdots$.

Then $ \frac{S_2}{\log x}\equal{}\frac{1}{(\log x)^3}\plus{}\frac{1}{(\log x)^4}\plus{}\cdots \equal{}S_2\minus{}\frac{1}{(\log x)^2}$

$ \frac{S_2}{\log x}\equal{}S_2\minus{}\frac{1}{(\log x)^2}\implies S_2(\log x\minus{}1)\equal{}\frac{1}{\log x}\implies S_2\equal{}\frac{1}{(\log x)(\log x\minus{}1)}$.

Therefore, $ S\equal{}S_2\plus{}\frac{2\plus{}S}{\log x}\implies S\equal{}\frac{1\plus{}(2\plus{}S)(\log x\minus{}1)}{\log x(\log x\minus{}1)}$. Solving for $ S$, we get $ S\equal{}\frac{2\log x\minus{}1}{(\log x\minus{}1)^2}$.

$ \frac{2\log x\minus{}1}{(\log x\minus{}1)^2}\equal{}8\implies 2\log x\minus{}1\equal{}8(\log x\minus{}1)^2$. Let $ y\equal{}\log x$ to get $ 2y\minus{}1\equal{}8y^2\minus{}16y\plus{}8\implies 8y^2\minus{}18y\plus{}9\equal{}0\implies y\equal{}\frac{18\pm6}{16}\equal{}\frac{3}{4},\ \frac{3}{2}$.

Therefore, $ x\equal{}10^\frac{3}{4},\ 10^\frac{3}{2}$.

However, when $ x\equal{}10^\frac{3}{4}$ the sum diverges, so the only real value of $ x$ is $ \boxed{x\equal{}10\sqrt{10}}$.[/hide]


Solve this: $(x+a)^4+(x+b)^4=0$. Am I blind?!  :? It might be easy....

$a$ must be equal to $b$ for this to have real roots. anything to the power of 4 is positive.

$\left( {x + a} \right)^4  + \left( {x + b} \right)^4  = 0 \Leftrightarrow \left( {x + a} \right)^4  =  - \left( {x + b} \right)^4 \mathop  \Leftrightarrow \limits_{\mathop {x \ne  - b}\limits^ \uparrow  } $

$ \Leftrightarrow \left( {\frac{{x + a}}{{x + b}}} \right)^4  =  - 1 \Leftrightarrow \left( {\frac{{x + a}}{{x + b}}} \right)^4  = \cos \pi  + i\sin \pi  \Leftrightarrow $

$ \Leftrightarrow \frac{{x + a}}{{x + b}} = \cos \frac{{\left( {2k + 1} \right)\pi }}{4} + i\sin \frac{{\left( {2k + 1} \right)\pi }}{4},k = 0,1,2,3 \Leftrightarrow $

$\mathop  \Leftrightarrow \limits_{\mathop {\theta _k  = \frac{{\left( {2k + 1} \right)\pi }}{4}}\limits^ \uparrow  } x + a = \left( {\cos \theta _k  + i\sin \theta _k } \right)x + b\left( {\cos \theta _k  + i\sin \theta _k } \right),k = 0,1,2,3 \Leftrightarrow $

$ \Leftrightarrow x\left( {1 - \cos \theta _k  - i\sin \theta _k } \right) = \left( {b\cos \theta _k  - a} \right) + i \cdot b \cdot \sin \theta _k ,k = 0,1,2,3 \Leftrightarrow $

$ \Leftrightarrow x \in \left\{ {\frac{{\left( {b\cos \theta _k  - a} \right) + i \cdot b \cdot \sin \theta _k }}{{1 - \cos \theta _k  - \sin \theta _k }}\left| {k = 0,1,2,3} \right.} \right\}$

Yeah  :D

hint
[hide]$x=y-\frac{a+b}{2}$[/hide]

[hide]
Since $(x+a)^4$ and $(x+b)^4$ are all greater or equal to $0$, so just solve for the real roots.[/hide]


here is one nice problem i found in harvard problem sets

you are given a 10.00g mixture of sodium sulfate and aluminum sulfate. you dissolve this mixture in water and add excess barium nitrate solution, forming a precipitate of barium sulfate. the total mass of barium sulfate formed is 19.46g. determine the mass of each compound in the original mixture.


The altitude from one of the vertices of an acute-angled triangle $ ABC$ meets the opposite side at $ D$. From $ D$ perpendiculars $ DE$ and $ DF$ are drawn to the other two sides. Prove that the length of $ EF$ is the same whichever vertex is chosen.


Prove that the number of ways to arrange n distinct charms on a bracelet that does not have a clasp is or is not equal to $ \frac{n!}{2n}$

[hide]prove, or just provide a counting argument?

For a counting argument, there are n! ways to order the charms. However, there are n rotations for each order.  Also, because you can flip all the charms and still get the same order, it is indeed n!/2n[/hide]


One-tenth is what part of three-fourths ?
(A) 3/40 (B) 1/8 (C) 2/15 (D) 15/2 (E) 40/3
.

[hide] $ \frac {1}{10}(\frac34) \equal{} \boxed{\frac {3}{40}}$. Thus, the answer is A.[/hide]

Wait, shouldn't it be $ \frac{\frac{1}{10}}{\frac{3}{4}}\equal{}\frac{1}{10}(\frac{4}{3})\equal{}\boxed{\frac{2}{15}}$

That's because it's asking for what part, not what $ \frac{1}{10}$ of $ \frac{3}{4}$ is. Therefore, we divide and not multiply.