My school system gets out much later than most others in the country. In fact, I will have to leave school a week early in order to attend MOP. This last week I'm missing is exam week, so in addition to the projects all the teachers cram in at the end of the year, I am also forced to take my exams early. Between studying and completing these final assignments, I am left with very little time.

How bad is it if I leave most of the MOP homework incomplete?

The graders/instructors will understand. Just try to at least think about every problem during whatever time you have. That shouldn't be too hard, since it looks like there are only 16 problems.

Actually blues have 32 I believe. 16 of which are shared with red.

I'm also in this situation, as are the other Virginians. Luckily, I have very few final exams, so I'm not too bad off. I hope you'll be able to work everything out.

So are we expected to be familiar with the problems, or to turn in write ups?

It says "At the beginning of the program, written solutions will be
submitted for review by MOSP graders, and you will present your solutions and ideas during the firrst few study sessions." so i guess yeah, you need to write them up.

Last year they never collected our writeups (this was for blue group). I don't think the graders really have time to grade 20-30 solutions from everyone at camp, especially since they'll also have to grade the MOP tests.

Our first two afternoon classes were spent reviewing the homework problems, someone who'd solved each problem came up and presented the solution on the board.

I think the black group had to collectively write up a set of solutions, but they never asked the blue MOPers to do that.

So basically, don't stress out about the problems, they don't really matter that much, just get whatever you can.

[quote="probability1.01"]This year might be different, though. I think I heard from someone that people would have to turn something in by the end of the first week; I'm not sure if it was the MOP hw or some other problem set.[/quote]

So what does Black group do?  Since it appears we have a LOT of homework problems...

zhai said it's not a good idea to do the hw problems


calculus
integration
trigonometry
calculus computations

$1 \int x \sin ^2 (6 x) dx$
$2 \int \frac{54}{\sqrt{21 -24 x -36 x^2}} dx$

These are the two problems I can't figure out on my own, If anyone can give a hand, I would be very grateful.  Thanks.

This is what I have done.
For the first one, I used integration by parts and got 4x^2sin(6x)cos(6x)
for the second one, I got  (27/36)*ln(-36x^2-24x+21)
But those answers are wrong according to the answers at the back of the book.

Got the second one, I am still working on the first one.


[quote="Wikipedian1337"]Prove that $ \sum_{sym} a^4b \minus{} a^3b^2 \minus{} a^3bc \plus{} a^2b^2c \ge 0$.[/quote]
$ a^4b \minus{} a^3b^2 \minus{} a^3bc \plus{} a^2b^2c \equal{} a^2b(a \minus{} b)(a \minus{} c)$, so we have to prove:
$ \sum_{cyc} a^2(b \plus{} c)(a \minus{} b)(a \minus{} c) \ge 0$
Using $ (a \minus{} b)(a \minus{} c) \equal{} \frac {1}{2} ( (a \minus{} b)^2 \plus{} (a \minus{} c)^2 \minus{} (b \minus{} c)^2 )$, we just have to prove:
$ \sum_{cyc} (a \minus{} b)^2 ( a^2(b \plus{} c) \plus{} b^2(a \plus{} c) \minus{} c^2(a \plus{} b) ) \ge 0$
Let $ S_c \equal{} a^2(b \plus{} c) \plus{} b^2(a \plus{} c) \minus{} c^2(a \plus{} b)$, $ S_b \equal{} a^2(b \plus{} c) \minus{} b^2(a \plus{} c) \plus{} c^2(a \plus{} b)$, $ S_a \equal{} \minus{} a^2(b \plus{} c) \plus{} b^2(a \plus{} c) \plus{} c^2(a \plus{} b)$, and assume wlog $ a \ge b \ge c$.
$ S_c \equal{} ac(a \minus{} c) \plus{} bc(b \minus{} c) \plus{} ab(a \plus{} b) \ge 0$
$ S_b \equal{} (a \minus{} b)(ab \plus{} bc \plus{} ca) \plus{} c^2(a \plus{} b) \ge 0$
We have to prove $ (a \minus{} b)^2S_c \plus{} (b \minus{} c)^2S_a \plus{} (c \minus{} a)^2S_b \ge 0$
But since $ (a \minus{} c)^2 \ge (b \minus{} c)^2$ and $ S_c,S_b \ge 0$, it is enough to prove:
$ (b \minus{} c)^2(S_a \plus{} S_b) \ge 0 \iff S_a \plus{} S_b \ge 0$.
But $ S_a \plus{} S_b \equal{} 2c^2(a \plus{} b) \ge 0$, and we are done :)

[color=red]Note: It only holds for $ a,b,c \ge 0$[/color]

After rewriting it as $ \sum_{cyc} a^2(b\plus{}c)(a\minus{}b)(a\minus{}c) \ge 0$ we can also use Vornicu-Schur.  :)

what is that Vornicu-Schur, I now that Schur is an inequality, and Vornicu is the administrator of mathlinks, but what is this Vornicu-Schur?  :?:


[quote="Farnak"]A rectangular room measures 30 feet in length and 12 feet in height and the ends are 12 feet in width. A fly with a broken wing rests on a point one foot down from the middle of the ceiling on one end of the room. A smudge of food is located one foot up from the middle of the floor on the other end of the room. The fly has just enough energy to walk 40 feet. Show that there is a path that will allow the fly to get to the food.[/quote]

Consider the net of the room.

Consider the row of the net which contains a side, and the two ends. The horizontal distance to be covered walking along that wall is $ 30\plus{}3\plus{}3$. The vertical distance to be covered is $ 10$. Therefore the total distance is

$ \sqrt{36^2\plus{}10^2} < 40$

I've heard of problems like this one! yeah, you just look at the net and it comes out nicely.

This works because when you lay out the room flat in a net, there's a diagonal straight between the fly and the food that you can measure using the vertical and horizontal lengths given to you in the problem.  However, I can't seem to find the exact length of such a path because the directions are sort of muddled...a picture would really help.

[quote="SimonM"]Consider the net of the room.

Consider the row of the net which contains a side, and the two ends. The horizontal distance to be covered walking along that wall is $ 30 \plus{} 3 \plus{} 3$. The vertical distance to be covered is $ 10$. Therefore the total distance is

$ \sqrt {36^2 \plus{} 10^2} < 40$[/quote] Where does the "+3+3" come from? I don't see it. When I look at a side wall with two ends attached as a row in the net, I see the fly walk a path with 30+6+6 feet horizontal displacement and 12-1-1 vertical displacement. That gives me  $ \sqrt {42^2 \plus{} 10^2} > 40$
We could use a different net, making a stack of floor, long wall, and ceiling, attached to each other, with one end wall attached to the ceiling (the one with the fly) and the other attached to the floor. A straight line there covers 30+1+1 feet horizontal distance and 12+6+6 vertical distance. The distance is:
$ \sqrt {32^2 \plus{} 24^2} \equal{} 40$

I agree with KMST. The horizontal distance is 42, and the vertical distance is 10.

But I don't really understand Farnak's wording. It's written 'one foot down from the middle of the ceiling' and 'one foot up from the middle of the floor'. Does 'middle' mean the center of the ceiling and floor?

Yeah, I messed up, those $ 3$s should be $ 6$s. I got confused about how many times I'd divided $ 12$ by $ 2$.

Still, you get the general idea


vector
function
linear algebra
linear algebra unsolved

Let $ V$ be a finite-dimensional vector space over $ F$. How does one exhibit an $ F$-algebra isomorphism of End($ V$) with End($ V^*$)? The natural choice of sending $ T$ to its dual map doesn't work.

Also: If $ B$ is a basis for the infinite-dimensional vector space $ V$ over $ F$, it is easy to see that $ V$ is the direct sum of copies of $ F$ indexed by $ B$. How does one show that the direct product of copies of $ F$ indexed by $ B$ has dimension strictly greater than $ V$? Try to go easy on the cardinal arithmetic: I know what countable and uncountable mean and that the cardinality of a set is strictly less than the cardinality of its power set, but that's about it.

For 1) There is not any canonical isomorphism of $ V$ onto $ V^*$ and then of $ end(V)$ onto $ end(V^*)$.
for 2) In the first case you obtain the functions $ f: B\rightarrow{F}$ s.t. $ f(x)\equal{}0$ except for a finite number of $ x$.
 In the second case you obtain all the functions $ f: B\rightarrow{F}$.
For example if $ B,F$ are countable sets then in the first case we obtain a countable set and in the second case we obtain a non countable set.


As you proceed from figure to figure, each shaded square is divided into four congruent squares and the upper left square of the four is painted white. By continuing the pattern, what fractional part of the tenth figure will be shaded? Express your answer as a common fraction in which the numerator and denominator are expressed in prime factored form using exponents.

[hide="Answer"]$ \frac {3^9}{2^{18}}$[/hide]

[hide]the first stage is 1, second is $ \frac{3}{4}$ third is $ \frac{9}{16}$

the nth stage is $ \frac{3^{n\minus{}1}}{4^{n\minus{}1}}$[/hide]

DonkeyKong, you screwed up your LaTeX. Something raised to the power of a # higher than ten has to be inputed like this:

base^{exponent} with the braces.
[hide]
The answer is simply

$ \left(\dfrac{3}{4}\right)^{9}$

$ \equal{}\dfrac{3^{9}}{2^{18}}$[/hide]

the first figure, $ \frac {3^0}{4^0}$ is shaded, the second $ \frac {3^1}{4^1}$, and so on, until the 10th it's $ frac{3^9}{4^9}$ ==> $ \frac {3^9}{(2^2)^18}$ which simplifies to $ \frac {3^9}{2^18}$

Lol Bob you made the same mistake in LaTeX that DonkeyKong made. Doesn't anyone read the posts before them? You can use
[code]
$base^exponent$[/code]
only for one-digit exponents!

Oh yeah, you should have \Rightarrow for your arrow and you should put a backslash before your frac :P


If $a,b>0$ and $n \in N $ we have:
$ \sqrt[n]{ab^{n-1}} \leq b+ \frac{a-b}{n}$. When does equality occur?

cheers! :D :D :cool:

? doesn't it comes from AM-GM directly? the equality occurs when a=b


geometry
incenter
function
geometry unsolved

Let I be the incenter of triangle ABC. Let the incircle of ABC touch the sides BC, CA, and AB at K, L and M, respectively. The line through B parallel to MK meets the lines LM and LK at R and S, respectively. Prove that angle RIS is acute.

A little lemma :) :

In acute triangle $\triangle ABC$, $CD^{2}\geq AD\cdot DB$, where $CD$ is a height of this triangle.

Proof is easy: let $\angle ACD=\alpha$ and $\angle BCD=\beta$, then $AD=CD\cdot tg\alpha$, $BD=CD\cdot tg\beta$, so the thesis is: $tg\alpha\cdot tg\beta\leq1$. Let's take $\alpha+\beta=\frac{\pi}{4}$, in that case $tg\alpha\cdot tg\beta$ is obviously 1. So, when $\alpha+\beta\leq90$ (our triangle is acute), $tg\alpha\cdot tg\beta\leq1$, because $tg\alpha$ is an increasing function (for $\alpha\in (0, \frac{\pi}{4}$)).

Now the problem. Let $\angle IAM=\alpha$ and $\angle KCI=\beta$. Then:
$\angle AML=90-\alpha$ => $\angle RMB=90-\alpha$. Doing the same, we have $\angle SKB=90-\beta$. $\angle KMB=\angle MKB=\alpha+\beta$. Because $KM$ is parallel to $RS$ we have $\angle RBM=\angle KMB=\alpha+\beta$. Angle $\angle SBK$ is also $\alpha+\beta$. $BI$ is a bisector of $\angle ABC$, so we have $\angle RBI=\frac{\pi}{4}$. As we can see now, triangle $\triangle RBM$ is similiar to triangle $\triangle KBS$, so we can write: \[RB\cdot BS=BM\cdot BK=BM^{2}\] .

$BI^{2}=BM^{2}+MI^{2}$, so $BI^{2}>BM^{2}=RB\cdot BS$. Now, using lemma, we have the thesis.

See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=37746[/url], etc.


logarithms
function
inequalities
algebra unsolved
algebra

Solve the equation in $R$
$ (x+1)^{x-1}=(x-1)^{x+1} $

[hide="My solution"]
Suppose x > 1. We have :

$ \displaystyle (x+1)^{x-1}=(x-1)^{x+1} <=> (x+1)^{1/(x+1)} = (x-1)^{1/(x-1)} $
$ <=> \log (x+1) / {(x+1)} = \log (x-1) / {(x-1)} = a $ for some a.

$ \log $ is a concave function, so if we take rays $ y=a x $ and we make a decrease from $ +\infty $ to $ -\infty $, we will get

1) at first no intersection with $ \log $ (so no $ x $ verify $ \log (x) = a x $ ) ;

2) then one intersection (so there is exactly one $ x $ verifying $ \log (x) = ax $) ;

3) then two intersections, whose distance strictly increase with(out) a (it's decreasing, so we can't say "with") ;

4) and then again one (when $ a $ is negative)

The only case where we can get $ x-1 $ and $ x+1 $ giving the same $ a $ is the third one, and we have that there can be only one solution because their distance is equal to a fixed number : $ 2 $.

There is in fact a solution, namely $ x=3 $ which gives $ 2^4 = 4^2 $.

We then have to study the case $ x \leq 1 $.

If $ x=1 $ we get $ 0^2 = 2^0 $, which is false. So $ x<1 $.
If $ x=0 $ we get $ 1^{-1} = (-1)^1 $, which is false.
If $ x=-1 $ we get $ 0^{-2} = (-2)^0 $, which is also false, since $ 0^{-2} $ in undefined.

Thus $ x-1 < 0 $, and because we are computing it's $ (x+1)$-th power, we need too have $ x+1 \in \Z $ , so $ x $ is a negative integer.

If $p$ is a prime number dividing $(x-1)$ (respectively $(x+1)$), then it also has to divide $(x+1)$ (respectively $(x-1)$), so $p=2$ and the two numbers are powers of 2 times $\pm 1$.

The only powers of 2 with a difference of 2 are 2 and 4, so we get $x-1=-4$, thus $x=-3$.

We have $(-2)^{-4}=(-4)^{-2}$, so $x=-3$ is solution.

We thus have [b]two[/b] solutions : $x=3$ and $x=-3$.[/hide]

--

Pierre.

You're right.  :blush:
I've modified my solution, the second part was completely wrong.

I used Bernoulli's Inequality to reduce it to proving the following:

(a+2)^A > a^B, where A=a^(a+2) and B=(a+2)^a, where real a > 1.

Can someone prove this?

I think it is difficult to prove that. It is a difficult problem.

The thoughts of Toumaf helped and I find a solution.
For x>1, equivalently we have that $ ln(x-1)^{x+1}=ln(x+1)^{x-1} $  <=>
(x+1)ln(x-1)=(x-1)ln(x+1)  <=>

$ \frac{ln(x-1)}{x-1}=\frac{ln(x+1)}{x+1} $   <=>

f(x-1)=f(x+1)  or  f(x)=f(x+2) where $  f(x)=\frac{lnx}{x} $ x>0

$ f'(x)=\frac{1-lnx}{x^2} $ x>0
We have also
f'(x)>0 at (0,e) and f'(x)<0 at $ (e,\propto) $
So f(x) increasing  at (0,e]   and f(x) decreasing at $ [e,\propto) $

Now for x=2 we have that f(2)=f(4)
So a solution to our problem is x=3

We will prove that this solution is unique.
Let x1 the solution above (x1=2).
Easily we find that the number x1+2>e

Now let's assume there is an another solution $ k $ with x1<k<e .
we have that k+e>e
Also x1<k so x1+2<k+2 and because  f is decreasing at $ [e,\propto) $
we have that f(x1+2)>f(k+2) => f(x1)> f(k) contradiction.
So there is one only solution.
In addition if we put where x the -x the equation which we have to solve is the same.
so the solutions to our problem are 3 and -3

Isn't this effectiely the same as the first solution...

Am I missiing sth???

Two solutions now and the second part of first solution was wrong as Toumaf said

[quote="toumaf"]

--

Pierre.[/quote]

 :huh:

Pierre.


Prove that for each prime $ p\le 100$, there always exists two integers x,y satisfying $ y^{37}\equiv x^3 \plus{} 11 (\mod p)$.

If $ (p\minus{}1,37)\equal{}1$ exist $ c$, suth that $ 37c\equal{}1\mod p\minus{}1$ and for any x we get solution $ y\equal{}(x^3\plus{}11)^c\mod p$.
If $ (p\minus{}1,3)\equal{}1$ exist b, suth that $ 3b\equal{}1\mod p\minus{}1$ and for any y we get solution $ x\equal{}(y^{37}\minus{}11)^b\mod b$.
Therefore for any p, suth that $ 222\not |p\minus{}1$ equation had solutions.


algorithm
induction
number theory proposed
number theory

$ \text{Prove that for any positif integers}$ $ a \leq n!$  $ \text{can be writeen as the sum of}$ $ k$ $ \text{distinct divisors of}$ $ n!$ $ \text{such that}$ $ k\leq n$

This is a problem from [i]Mathematical Olympiad Treasures[/i], first chapter.

Solution from the book:
[hide]The assertion is trivial for $ n \equal{} 1$, so suppose it holds true for some $ n > 1$. We have $ a\leq (n \plus{} 1)!$. By the division algorithm, there exist $ b$, $ c$ such that $ a \equal{} b(n \plus{} 1) \plus{} c$, with  $ 0\leq c\leq n$. Clearly, $ b\leq n!$; and thus, using the induction hypothesis, $ b$ can be represented as a sum of at most $ n$ distinct divisors of $ n!$, say
      $ b \equal{} d_1 \plus{} d_2 \plus{} ... \plus{} d_k$,
with $ k\leq n$. But then
      $ a \equal{} d_1(n \plus{} 1) \plus{} d_2(n \plus{} 1) \plus{} ... \plus{} d_k(n \plus{} 1) \plus{} c$,
 and the numbers $ d_1(n \plus{} 1), d_2(n \plus{} 1),...,d_k(n \plus{} 1)$ and $ c$ are distinct divisors of $ (n \plus{} 1)!$.[/hide]


number theory
greatest common divisor
number theory proposed

Find "All" triples of integers like (x,y,z) such that:

x^2 + y^2 = z^3

[hide]
There are infinitely many integers x,y,z which satisfy the equation; even with the condition that gcd of any two of them is 1:

let a,b be integers, then x = a^3 + ab^2, y = b^3 + ba^2, z = a^2 + b^2 are an infinite family of solutions.
the problem demands finding All the solutions.[/hide]

[hide="Hint"]Work in $ \mathbb{Z}[i]$, which is a unique factorization ring. You get $ z \equal{} \prod_{j\equal{}1}^k(a_j^2 \plus{} b_j^2)$, but expressing $ x$ and $ y$ is messy.[/hide]

also see PEN H41(but there is a condition for the question in PEN:gcd(x,y)=1)


integration
inequalities
calculus
analytic geometry
graphing lines
slope
function

Let n>=2 be an integer. For any division 0=x_0<x_1<..,x_(n-1)<x_n=1 of [0,1] we define A(x_0,x_1,..,x_n)={f:[0,1]->R|f derivable,  \int {0->1}|f(x)|dx=1, f(x_k)=0, k=0->n}.

Prove the existence of a division 0=a_0<a_1<..,a_(n-1)<a_n=1 s.t. for all f in A(a_0,a_1,.., a_n) we have sup {x in [0,1]} |f'(x)|>4n.

[i]Romanian NMO/ final round/ Brasov/ 2000/ problem 2/ 12'th grade[/i]

It's kind of hard for me to follow this, so I don't really know whether you've solved it or not :D , but here's a hint (in white :D ):

[color=white]

It's kind of natural to consider the equidistant division of [0,1]...

[/color]

This condition means the subdivision is equidistant
a(k+1)-a(k) = 1/(n \sqrt 2)

Since the length of the intervl is 1, shouldn't we have a(k+1)-a(k)=1/n? Maybe I'm seeing this wrong. Are your a(i)'s the same as my a_i's?

Before works with n points, let's try with two

Lemma: f:[a,b]->R derivable with f(a)=f(b)=0, then
 \int_[a,b] f(x)dx<(b-a)^2.M/4 where M = sup_[a,b]|f'|
Proof

There exist c, a<c<x
f(x)-f(a)=f(x)=f'(c)(x-a) \leq M(x-a) for a \leq x \leq (a+b)/2

There exist d, x<d<b
f(x)-f(b)=f(x)=f'(d)(x-b) \leq M(b-x) for (a+b)/2 \leq x \leq b

The derivability at (a+b)/2 will give <
 \int _[a,b]f(x)dx <M \int _{a to (a+b)/2}(x-a)dx + M \int _{(a+b)/2 to b}(b-x)dx =M/4.(b-a)^2

Now apply the lemma to the intervals [0,1/n], [1/n,2/n],..., [(n-1)/n,1]

\int_[0,1/n] f(x)dx< 1/n^2.M_1/4

\int_[1/n,2/n] f(x)dx<1/n^2.M_2/4

...............................................
\int_[(n-1)/n,1] f(x)dx< 1/n^2.M_n/4  with M_k = sup_[(k-1)/n, k/n]|f'|

We sum

 \int _[0,1] f(x)dx<1/(4n^2)(M_1+M_2+...+M_n)

but M_k  \leq sup_[0, 1]|f'| =M

so  \int _[0,1]f(x)dx<1/(4n^2)(M+M+...+M) = M/(4n)

M > 4n \int _[0,1] f(x)dx

Take a_k = k/n

Actually I don't see a proof for  \int_[0,1] |f|

I don't know if it's rigorous enough for you, but here it goes:

[color=white]

We have Int{0->1}|f(x)|dx=1, so Int{0->1/n}|f(x)|dx+Int{1/n->2/n}|f(x)|dx+..+Int{(n-1)/n->1}|f(x)|dx = 1, so the sum of these n integrals is 1, so the largest among them is >=1/n. Let's assume, without loss of generality, that Int{0->1/n}|f(x)|dx>=1/n (1).

We assume sup{x in [0,1]} |f'(x)|<=4n (*)

We take the cartesian plane. Take A(0,0), B(1/n,0). We now take 2 lines: d, which passes through A and has slope 4n, and d', which passes through B and has slope -4n. Let these 2 lines intersect at C. We have C(1/2n,2). It's easy to deduce from (*) that on the intervel [0,1/n] the graph of |f(x)| must lie inside the triangle ABC, so
Int{0->1/n}|f(x)|dx<=Area(ABC)=(2*1/n)/2=1/n (2). From (1) and (2) we find Int{0->1/n}|f(x)|dx = 1/n, but this can happen only when on the interval [0,1/n] the graph of |f(x)| consists exactly of the segments [AC] and [CB]. It's pretty obvious that a function satisfying this can't be derivable (it's not derivable at x=1/2n, which corresponds to point C).

[/color]

I'll post the official solution a bit later, Ok?


how to balance chemistry equations with maths, any method, or does anyone know a good programme for ti 89 titanium to balance equations ?

Well, there's a website with a JavaScript that does that for you :)

Other than that, I think it's just pretty much intuitive. (Guess and check, anyone?)

I came up with a pretty obvious way to balance equations using math, but I'm sure someone's thought of this before:

Take the simple example

_H(2) + _O(2) ~~~> _H(2)O

Call "x" the unknown multiplier in front of H on the left side of the formula. Now, since there are "x" hydrogen atoms on the left side there must be "x" hydrogen atoms on the right side, so "x" must go in front of H(2)O as well. So far we have:

xH(2) + _O(2) ~~~> xH(2)O

Next, we see there are x oxygen atoms on the right side of the formula, so there must be x oxygen atoms on the left side. To achieve this, the multiplier in front of O(2) must be x/2:

xH(2) + (x/2)O(2) ~~~> xH(2)O

To complete the balancing, just find the lowest possible whole number value of x so that all the multipliers are whole. In the above example, x=2 does the trick:

2H(2) + 1O(2) ~~~> 2H(2)O

Sometimes you will need two variables (x and y), in which case you would have to solve a system of equations to eliminate one variable.

If you want, try the above procedure on this:

_Al + _O(2) ~~~> _Al(2)O(3)

yea i came up with that when i did 1st year chem...too bad my teacher is like WTF LINEAR SYSTEM what???

so that failed

it is rarely useful...sometimes it is helpful for combustion reactions


Let$ 1\leq a;b;c \leq2$ Prove that
$ [4\minus{}(ab\plus{}bc\plus{}ca)][\sum\frac{1}{(a\plus{}b)c}] \leq \frac{3}{2}$

By$ \frac{4\minus{}ab}{(a\plus{}b)c} \leq\frac{4\minus{}ab}{a\plus{}b}\leq \frac{3}{2}$  -->Done :rotfl:


I will be doing a lesson on problem solving in my research class and will be giving some problems.  One is:
Call an integer $k$ a [i]palindrome[/i] if it is read the same way backward and forward.  How many [i]palindromes[/i] exist $0 \leq k \leq 1000000$?

I believe I have solved and I believe I have a strategy but perhaps my way of doing it is long and tedious.  My answer is $1989$.  Any confirmations?  I would also appreciate the methods by which you solved it.

[hide]
With palindromes, we only have to look at the possibilities for the first half of the digits in the number. The other digits just fall into place mirroring the first half

1 Digit numbers:
$10$

Two digit numbers
$9$ (since 0 cannot be the first digit)

Three digit numbers
$9 * 10$

Four digit:
$9 * 10$

Five:
$9 * 10 * 10$

Six:
$9 * 10 * 10$

Sum is:
$10+9+2(9)(10)+2(9)(10)(10)$
$19+2(990)$
$19+1980$
$1989$
[/hide]

Thanks.  Exactly how I did it.  I was thinking there was maybe some advanced way that I didn't know about.

[quote="chickendude"]$19+1980$
$1989$
[/quote]

ehh...

JRav, since you're 10 short, you probably forgot the [b]one digit [/b]palindromes.

[quote="chickendude"][hide]
$19+1980$
$1989$
[/hide][/quote]

Check your math again. $19+1980 \not= 1989$  :D

I'm not sure whether or not to count single digit palindromes.  I wouldn't think so.  After all, the word "I" isn't a palindrome (at least I don't think so).

read it backwards it is exactly the same thing.

do i detect some possible inherent discrimination here? hmmm...suspicious. i will have to keep an eye on you...

EDIT: actually if this is from a 1989 contest then no, because you want the answer to be 1989.  if this is from a 1999 contest yes, because you want the answer to be 1999.  otherwise flip a coin or put in both.  :lol:

I thought I made up the problem myself so I wasn't aiming for a particular answer.  If not, whatever.


Evaluate
\[\displaystyle\sum_{n=0}^\infty\sin^{-1}\left(\frac{1}{\sqrt{n^4+2n^3+3n^2+2n+2}}\right).\]

The result is very nice, but also pretty well hidden...

i think the experession will factorize to sum(cotinv(n^2+n+1)), after which it can be further rationalized

From here use the identity
\[\cot^{-1}{(x^2+x+1)}=\tan^{-1}{(x+1)}-\tan^{-1}{x}\]
Then the whole sum telescopes to
\[\displaystyle \tan^{-1}{\infty}-\tan^{-1}{0}=\frac{\pi}{2}\]


Sho that for any prime p and positive integer k, there exists a positive integer t such that pt has k consecutive zeros somewhere in its decimal representation.

- If p is not 2 or 5 :
Let N such that p^N >= 10^k (thus the decimal expansion of p^N uses at least k+1 digits). For n >= N, the string of the k+1 last digits of p^n varies over a finite set (not more than 10^(k+1) different possible values). Thus, from the pigeon-hole principle; there are n =< a < b such that p^a and p^b have the same last k+1 digits. It follows that p^a*(p^(b-a) - 1) = 0 mod[10^k].
Since p is a prime and coprime with 10, we deduce that p^(b-a) - 1 ends with k+1 digits '0'. Then p^(b-a) will do the job.

- If p = 2 :
As above, we prove that there exists b > 0 such that 2^b - 1 = 0mod[5^(2k)]. Then 2^(b+2k) - 2^(2k) = m*10^(2k) for some positive integer m. But, we have 2^(2k) = 4 ^4 < 10^k, then 2^(2k) uses not more than k digits. Thus 2^(b+2k) = m*10^(2k) + 2^(2k) has at least k digits '0'.

- If p  = 5 :
As above, there exists b > 0 such that 5^b - 1 = 0 mod[2^4k]. Then, 5^(b+4k) - 5^(4k) = m*10^(4k). Since 5^(4k) = 625^k < 10^2k, the conclusion follows in the same way as in the previous case.

Pierre.


Determine the 2 by 2 matrix $ A$ such that $ AX\equal{}XA$ for any 2 by 2 matrix $ X$.

$ \left[\begin{array}{cc} 1&0\\0&1\end{array}\right]$. This is a matrix that will surely work, since the identity matrix has properties $ XI\equal{}IX\equal{}X.$ Then, clearly, if $ A\equal{}I\equal{}\left[\begin{array}{cc} 1&0\\0&1\end{array}\right]$, then $ AX\equal{}XA$ for any 2 by 2 $ X$.

The answer is that $ A$ must be a "scalar" matrix, meaning $ A\equal{}\begin{bmatrix}a&0\\0&a\end{bmatrix}$ where $ a$ is any arbitrary number (*see footnote). This includes the possibility of the zero matrix.

This can be proved surprisingly quickly and efficiently. It's easy to see that all such matrices (scalar multiplies of the identity) do commute with everything. To prove the converse - that any matrix which commutes with all other matrices must be of this form - we need to take advantage of the "for all" quantifier in that statement by testing against certain specific $ X.$

Step 1. Assume $ A$ is the completely generic matrix $ \begin{bmatrix}a&b\\c&d\end{bmatrix}$ and let $ X\equal{}\begin{bmatrix}1&0\\0&0\end{bmatrix}.$

Then $ AX\equal{}\begin{bmatrix}a&0\\c&0\end{bmatrix}$ and $ XA\equal{}\begin{bmatrix}a&b\\0&0\end{bmatrix}.$

We must have $ AX\equal{}XA.$ For that to be true, we must have $ b\equal{}0$ and $ c\equal{}0.$

Step 2. By Step 1, we know know that $ A\equal{}\begin{bmatrix}a&0\\0&d\end{bmatrix}.$ Let $ X\equal{}\begin{bmatrix}0&1\\0&0\end{bmatrix}.$

Then $ AX\equal{}\begin{bmatrix}0&a\\0&0\end{bmatrix}$ and $ XA\equal{}\begin{bmatrix}0&d\\0&0\end{bmatrix}.$

We must have $ AX\equal{}XA.$ For that to be true, we must have $ a\equal{}d.$ We're done.

* By "arbitrary number" we mean any member of the field over which these matrices are taken, which kunny didn't specify. If he meant these to be real matrices, then $ a$ may be any real number. But any other field would work just as well - the argument used no specific field properties.

Thank you for your detail explanation as usual, Dr. Kent Merryfield.

Yes, the answer is scalar matrix with complex number elements.

In 1976~1997's age, the elements of any matrix was complex number, but at present, that is limited to real elements in Japanese university entrance exam.

To my disappointment, Matrix itself will be disappear from Japanese high school mathematics in 2012. What is on earth, what do governmemt officials think?

No way! In replace of Matrix, Complex plane is comming back in about ten years in 2012.


calculus
integration
calculus computations

$ \int_0^\infty \dfrac{x\lambda (\lambda x)^{n\minus{}1} exp({\lambda x})}{(n\minus{}1)!}dx$

thanks

If $ \lambda < 0$ and $ n \in \mathbb N_\plus{}$ than
$ \textbf I \equal{} \frac {1}{(n \minus{} 1)!} \cdot \frac {1}{\lambda} \int_0^\infty (x \lambda)^n e^{x \lambda} \, \textbf d (x \lambda ) \equal{} \frac {n!}{(n \minus{} 1)!} \cdot \frac {\lambda^{n}}{( \minus{} \lambda)^{n \plus{} 1}} \equal{} \boxed{( \minus{} 1)^{n \plus{} 1} \frac {n}{\lambda}}$


Let $ x,y,z\leq 0$such that $ x\plus{}y\plus{}z\equal{}2$.Prove that $ x^t(y\plus{}z)\plus{}y^t(z\plus{}x)\plus{}z^t(x\plus{}y)\leq 2$    $ t \in [2,3]$

[quote="DCTPKTCSPHN"]Let $ x,y,z\leq 0$such that $ x \plus{} y \plus{} z \equal{} 2$.Prove that $ x^t(y \plus{} z) \plus{} y^t(z \plus{} x) \plus{} z^t(x \plus{} y)\leq 2$    $ t \in [2,3]$[/quote]
We can find max of $ P\equal{}\sum \ x^t(y\plus{}z)$ with  $ t \in{R}$,but it 's very hard


abstract algebra
group theory
superior algebra

Let $ G$ be a finite group, and suppose that an automorphism $ T$ of $ G$ sends more than three-quarters of the elements of $ G$ onto their inverses. Prove that $ T\left(x\right)=x^{-1}$ for all $ x\in G$, and that $ G$ is abelian.

Proof that $G$ is abelian (using the statement: $f(x)=x^{-1}$ for all $x$):

Let $f$ be the mentioned automorphism. Consider $a, b\in G$. Then

\[ba=(a^{-1}b^{-1})^{-1}=f(a^{-1}b^{-1})=f(a^{-1})f(b^{-1})=ab.\]

Misha

Aww well will you please explicitly tell what you have done.You have ssumed that it is true for all x but on the contrary it is given taht the statement is true for 3/4 of the no of elements of G

I guess he left the pleasure of proving $f(x)=x^{-1} \forall x$ to us.

O dear thats what I have been trying to prove.And thats what I cant prove .Can any one help me in this??

Define $H:= \{ x | T(x)=x^{-1} \}$ and choose some $x \in H$.
Now choose $y$ so that $y,xy \in H \iff y \in H \cap (x^{-1}H)$.
This gives $x^{-1}y^{-1} = T(x)T(y)=T(xy)=y^{-1}x^{-1}$ so $xy=yx$ for all that $y$.
So $H \cap (x^{-1}H) \subset C_x$, the centalizer of $x$, and since $|C_x| \geq |H \cap (x^{-1}H)|>|G|/2$ the subgroup $C_x$ has index $1$ in $G$ and thus $C_x=G$, giving that $x$ commutates with all $y \in G$. This gives that $H \subset C_G$, the center of G, and the same way as before you see that $C_G=G$, so $G$ is abelian.
Now it's simple to conclude for every $x \in G$ by choosing an $y,xy \in H$ that $T(x)y^{-1}=T(xy)=y^{-1}x^{-1}$ giving $T(x)=x^{-1}$.

Here's why you need more than 3/4 of the elements: Take the dihedral group with 8 elements, and the identity automorphism. Since 6 of the elements have order one or two, those six (three fourths of the total) elements are sent to their inverses.

Heres another proof : Let $H$ be as defined by ZetaX, then for any $a$ in $G$ that is not in $H$, let $I = \{h \in H| ah \in H \}$, then $I$ has more than half the elements of $G$.  For any $h$ in $I$, we have $T(ah) = h^{-1}a^{-1}= T(a)T(h) = T(a)h^{-1}$,  which implies $T(a) = h^{-1}a^{-1}h$. Hence for any two elements $h_{1}$ and $h_{2}$ in $I$, we have $T(a) = h_{1}^{-1}a^{-1}h_{1}= h_{2}^{-1}a^{-1}h_{2}$,  or $h_{1}h_{2}^{-1}$ commutes with $a$. Now let $H' = \{h_{1}h_{j}^{-1}| h_{1}\in I , h_{j}\in I \}$. This set also has greater than half the number of elements of $G$. $H' \cap I$ is non-empty, but that would mean that $T(a) = a^{-1}$. That $G$ is abelian follows immediately.

[Moderator edit: Initially this topic contained a discussion of the following problem as well:
[quote]If $ f$ is an automorphism of a finte group such that $ f(x) \equal{} x \iff x \equal{} e$ and $ f(f(x)) \equal{} x$ for all elements $ x$ of the group, prove that it is the unique $ f(x) \equal{} x^{ \minus{} 1}$.[/quote]
I have split this discussion into a new topic ( http://www.mathlinks.ro/Forum/viewtopic.php?t=200576 ) because I don't see how it is related to the original problem. - Darij]


find minimal of expression P
P= $\sum\frac{x^{20}}{y^{11}}$   with $x,y,z >0$and$x+y+z =2001$

[quote="tranthanhnam"]Find minimal of expression P:
$P=\frac{x^{20}}{y^{11}}+\frac{y^{20}}{z^{11}}+\frac{z^{20}}{x^{11}}$
with $x,y,z >0$ and $x+y+z =2001$.[/quote]
That's easy.
\[ \frac{x^{20}}{y^{11}}+\underbrace{y+y+\cdots+y}_{11\ \text{times}} \ge 12\sqrt[12]{x^{20}}=12x^{\frac53} \newline \cdots  \]
Take sume, we get:
\[ P+11(x+y+z) \ge 12\left(x^{\frac53}+y^{\frac53}+z^{\frac53}\right)  \]
By Jensen's:
\[ \frac{x^{\frac53}+y^{\frac53}+z^{\frac53}}{3} \ge \left(\frac{x+y+z}{3}\right)^{\frac53} \newline \cdots  \]
That's our claim.

Sorry. I forgot the equality in my solution. But it could be fixed easily.


[hide="To Tarquin"]Tarquin, it's 22. Think of it this way. Draw yourself a Venn Diagram, and put 8 in the intersection of the algebra and geometry. You have 17 total taking algebra, but 8 are already taking geometry, so you have 9 that are just taking algebra. With the geometry, you have 8 of the geometry students also taking algebra, and there are 13 total geometry students, so there are 5 that are just taking geometry. 9 + 8 + 5 = 22. [/hide]

[quote="MCrawford"]Every member of a math club is taking algebra or geometry and 8 are taking both.  If there are 17 taking algebra and 13 taking geometry, how many members are in the club?[/quote]

[hide]
22. 17+13-8.[/hide]

elements if you look at my work I got the same answer, I just added incorrectly.

[quote="MCrawford"]Every member of a math club is taking algebra or geometry and 8 are taking both.  If there are 17 taking algebra and 13 taking geometry, how many members are in the club?[/quote]

In general, the answer is $\text {Group A pop. } + \text { Group B pop. } - \text {total number people in both } = \text {total people}$, or $17+13-8=22$.

[hide]Ill to a mental VEnn diagram

17+13 = 30... subtract 8 because we're double counting, and we have.... [b]22 ppl! There's your asnwer.[/b][/hide]

sry.. i just wanted to practice using latex even there really isnt a need in this question :D
[hide]
$ 17 + 13 - 8 = 22 $[/hide]
so latex really is no differnt for the person typing?? you just add dollar signs and it looks cool??? cool.. lol


çeşitli olimpiyat problemlerin tartışıldığı bir forum

http://www.geomania.org/portal/Forum/default.asp

Osman Hocam, siteye üye olamıyorum. Yeni üye kabul etmiyoruz şeklinde bir yazı var. Ve forumlar üyelik istiyor. Yardımcı olabilir misiniz?

Bir site adresi daha verelim.

http://www.egitimokulu.com

eren hocam şu anda üye alımında bir kısıtlama yoktur.forumlar istifadenize açıktır.iyi çalışmalar...  http://www.geomania.org/portal/forum/

ziyaret etmek isteyen arkadaşlar için sitenin güncellenmiş linki: http://www.geomania.org


how to calculate limit of such an infinite sum?

$ \sum_{n \equal{} 1}^{\infty} \frac {1}{n^2}$


Here is my conjecture, I haven't solve
We well known $ \frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\ge\sqrt{3}$ and $ 4R+r\ge\sqrt{3}s$ with $ P$ is an arbitrary point in plane and $ R,r,s$ are circumradii, inradii, semiperimeter of $ ABC$. Prove that the shaper inequality but for acute triangle, only
\[ \frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\ge\frac{4R+r}{s}\ge\sqrt{3}. \]
:)

this problem is not true,it has been discussed in a Chinese mathematics forum which belongs to the China Inequality Rescarch Group(http://www.irgoc.org)(but it is in Chinese language),In China Inequality Rescarch Group,there are many expert on Geometric inequality.

Really? Can you show me an counter-example?, I repeat inequality but for [u]acute[/u] triangle!


Prove that $\sum_{k=1}^{n}{\frac{|sink|}{k}=\frac{2}{\pi}\cdot lnn+O(\sqrt{{ln n}})}$.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25735[/url] for a stronger result.


Two congruent $30^\circ - 60^\circ - 90^\circ$ triangles are placed so that they overlap partly and their hypotenuses coincide.  If the hypotenuse of each triangle is 12, the area common to both triangles is
\[ \text{(A)}\ 6\sqrt{3} \qquad \text{(B)}\ 8\sqrt{3} \qquad \text{(C)}\ 9\sqrt{3} \qquad \text{(D)}\ 12\sqrt{3} \qquad \text{(E)}\ 24 \]

[hide]
This one is hard to explain.

After a [b]proper[/b] diagram has been drawn, it's really pretty simple, but I'm too lazy to write a solution.
The answer is $\displaystyle D$.[/hide]

[hide="Hint"]
Ok. I'm going to attempt to describe the diagram. Draw your first 30-60-90 triangle. Then on the hypothenuse, construct a 30 degree angle on the 60-degree side of the original and construct a 60-degree angle on the 30-degree side of the original. Extend the two lines, and we now have our other congruent triangle.
[/hide]

[hide="More hints"]
Find other 30-60-90 triangles. Then use the formula $[area]=\frac{1}{2}ab\sin C$. And yes, the answer is D.
[/hide]

edited out because i was angry when i posted this

oh wait! i figured it out

thanks tarquin, but i actually didnt need that.  I just needed to think about it some more instead of getting angry that I couldn't get it lmao.  Im glad i got it without looking at your diagram because I was pissed a minute ago that i couldn't.

Anyway here is what it should be:
$6^2 + x^2 = 6\sqrt{3}^2$
$x = \sqrt{72}$

The area of the overlapping region is the area of the whole triangle minus the area of the right triangle shown below.
$\frac{6\sqrt{3}(6)}{2} - \frac{6\sqrt{72}}{2}$
Which is
$12\sqrt{3}$

P.S.  Sorry about all of that up there.  I should really think about a problem more before thinking that I cant find the answer.


trigonometry
number theory proposed
number theory

Find all $x$ and $y$ which are both rational multiples of $\pi$ with
$0<x<y\leq \frac{\pi}{2}$ and have the property that
$\tan x+\tan y=2$


Is there really such thing as an irregular triangle? It doesn't seem possible, yet it does... Are equilateral triangles "regular" triangles?   :huh:  :ewpu: suddenly im more confused...

A regular polygon has all sides and all angles equal, so equilateral triangles are the only regular triangles.

I guess, using the definition of "regular" being all angles and side different, "irregular" can be defined as a scalene triangle.  Such as:
[asy]pair A=(0,0), B=(6,3), C=(2,4);

draw(A--B--C--cycle);[/asy]

Actually, regular means all angles are congruent and all sides are congruent, like

[asy]draw(polygon(5));[/asy]

And

[asy]draw(polygon(8));[/asy]

But not

[asy]draw((0,0)--(4,8)--(3,2)--cycle);[/asy]

A polygon is regular IF AND ONLY IF all of its angle measures and sides are congruent.

i was showing an "irregular" triangle, or scalene.

Isosceles triangles are also irregular, so perhaps that example is somewhat misleading.


Isoceles triangles $AC_{1}B$ and $AB_{1}C$ with an angle $\varphi$ at the vertex are constructed outwards on the unequal sides $AB$ and $AC$ of a scalene triangle $\triangle ABC$.
$\text{(a)}$ Let $M$ be a point on median $AA_{1}$ (or on its extension), let $M$ be equidistant from $B_{1}$ and $C_{1}$. Prove that $\angle B_{1}MC_{1}= \varphi$.
$\text{(b)}$ Let $O$ be a point of the midperpendicular to segment $BC$, let $O$ be equidistant from $B_{1}$ and $C_{1}$. Prove that $\angle B_{1}OC = 180-\varphi$.
Problems in Plane and Solid Geometry, Viktor Prasolov, #1.50

Is there any way you can provide a picture?

:blush: sorry it took so long.. here's for part (a)
[img]http://img49.imageshack.us/img49/4971/prasolov150ajm9.png[/img]
for part (b):
[img]http://img118.imageshack.us/img118/9798/prasolov150bdg8.png[/img]


Prove that a finite graph is bipartite if and only if it contains no cycle of odd length.


limit
function
real analysis
real analysis unsolved

A bounded sequence $ \{x_n\}$ of real numbers satisfies $ \lim_{n->\infty} (x_n-2x_{n+1}+x_{n+2}) = 0$.
Prove that $ \lim_{n->\infty}{(x_n-x_{n+1}) = 0}$.

$ \lim_{n \minus{} > \infty}(x_{n} \minus{} 2x_{n \plus{} 1} \plus{} x_{n \plus{} 2}) \equal{} 0$ implies $ \left|(x_{n} \minus{} x_{n \plus{} 1}) \minus{} (x_{n \plus{} 1} \minus{} x_{n \plus{} 2})\right| < \epsilon$ for each $ \epsilon > 0$ and $ n$ great enough.

$ \epsilon > \left|(x_{n} \minus{} x_{n \plus{} 1}) \minus{} (x_{n \plus{} 1} \minus{} x_{n \plus{} 2})\right|\geq \left|x_{n} \minus{} x_{n \plus{} 1}\right| \minus{} \left|x_{n \plus{} 1} \minus{} x_{n \plus{} 2}\right| \Rightarrow$

$ \lim_{n \to \infty}|x_{n} \minus{} x_{n \plus{} 1}|$ exists. Let it be $ L$. If $ L \equal{} 0$, we are done. Otherwise $ L > 0$. But if
$ \lim_{n \to \infty}(x_{n} \minus{} x_{n \plus{} 1})$ would not exist, that means there would exist an infinite subsequence $ a_{n_{i}} \equal{} x_{n_{i}} \minus{} x_{n_{i}\plus{}1}$, for which $ \lim_{i \to \infty}{a_{n_{i}}} \equal{} L$ and an infinite subsequnce $ b_{n_{j}} \equal{} x_{n_{j}} \minus{} x_{n_{j}\plus{}1}$ for which $ \lim_{j \to \infty}{b_{n_{j}}} \equal{} \minus{} L$.

Let $ I \equal{} \left\{n; n\; \mathrm{index\; of}\; \{a_{n}\} \right\}$ and $ J \equal{} \left\{n; n\; \mathrm{index\; of}\; \{b_{n}\} \right\}$ .

But then there would exist infinitely many adjacent pairs $ n_{k}, n_{k\plus{}1}$, for which $ n_k \in I$, $ n_{k\plus{}1} \in J$ and vice versa. Therefore we would have an infinite subsequence $ c_{n_l}\equal{}x_{n_l} \minus{} 2x_{n_l \plus{} 1} \plus{} x_{n_l \plus{} 2}$ converging to $ 2L \neq 0$. Contradiction.

Hence $ \lim_{n \to \infty}(x_{n} \minus{} x_{n \plus{} 1}) \equal{} L$, where $ L \neq 0$ and $ L \in \mathrm{R}$.

WLOG $ L > 0$. Therefore for $ n_0$ great enough $ x_{n} \minus{} \frac {L}{2} > x_{n \plus{} 1}$ for all $ n > n_0$. Now sending $ n$ toward $ \infty: x_{n} < x_{n_0} \minus{} \frac {L}{2}(n \minus{} n_0).$ So $ x_{n}$ is not bounded. Contradiction. Hence $ L \equal{} 0$.

[quote="tmrfea"]
$ \epsilon > \left|(x_{n} \minus{} x_{n \plus{} 1}) \minus{} (x_{n \plus{} 1} \minus{} x_{n \plus{} 2})\right|\geq \left|x_{n} \minus{} x_{n \plus{} 1}\right| \minus{} \left|x_{n \plus{} 1} \minus{} x_{n \plus{} 2}\right| \Rightarrow$
$ \lim_{n \to \infty}|x_{n} \minus{} x_{n \plus{} 1}|$ exists. [/quote]
Why?

It is sadly to see that so nice problem still didn't get a solution. First time I saw it in Sadovnichii, Konyagin "Problems from student's math olympiads." Maybe Fedor knows the origin... :)

Suppose there is $ a > 0$ that $ |x_{n \plus{} 1} \minus{} x_n|\geq a$ for infinitely many $ n$. Let's choose arbitrary small $ \varepsilon > 0$ ($ \varepsilon < < a$) and $ N \equal{} N(\varepsilon)$ that
\[ |(x_{n \plus{} 2} \minus{} x_{n \plus{} 1}) \minus{} (x_{n \plus{} 1} \minus{} x_n)| \equal{} |x_{n \plus{} 2} \minus{} 2x_{n \plus{} 1} \plus{} x_n| < \varepsilon
\]
if $ n\geq N$. Note that
\[ x_{n} \minus{} x_{n \plus{} p \plus{} 1} \equal{} (x_n \minus{} x_{n \plus{} 1}) \plus{} (x_{n \plus{} 1} \minus{} x_{n \plus{} 2}) \plus{} ... \plus{} (x_{n \plus{} p} \minus{} x_{n \plus{} p \plus{} 1}).
\]
Denote $ s \equal{} |x_n \minus{} x_{n \plus{} 1}|$, then
\[ |x_{n} \minus{} x_{n \plus{} p \plus{} 1}| > s \plus{} (s \minus{} \varepsilon) \plus{} (s \minus{} 2\varepsilon) \plus{} ... \plus{} (s \minus{} p\varepsilon) \equal{} (p \plus{} 1)s \minus{} \frac {p(p \plus{} 1)}{2}\varepsilon
\]
if $ s \minus{} p\varepsilon > 0$. But $ |x_n \minus{} x_{n \plus{} p\plus{}1}| < M$ since $ (x_n)$ is bounded. Choose $ n$ such that $ s\geq a$ and $ p\approx \frac {a}{\varepsilon}$. Then we obtain $ M > (\approx \frac {a^2}{2\varepsilon})$ for arbitrary small $ \varepsilon$. Contradiction.

How about the continuous analogue?

Suppose $ f$ is a bounded $ C^2$ function on $ \mathbb{R}$ and that $ f''(x)\to 0$ as $ x\to\infty.$ Prove that $ f'(x)\to 0$ as $ x\to\infty.$

(This might be a little easier, but it's certainly related.)


superior algebra
superior algebra unsolved

I have a question about the FIT of rings.. I don't really understand how to use the theorem. Could anyone give me an example (hopefully not a trivial one)?

Thanks!!

the isomorphism theorems are so fundamental dealing with quotient structures; if you work with them, you will see enough applications. ok, three very simple examples:
$\mathbb{R}[x]/(x^{2}+1) \cong \mathbb{C}$.
$\mathbb{Z}[i]/(1-i) \cong \mathbb{Z}/2$
$\mathbb{C}[x,y] / (x+y) \cong \mathbb{C}[x]$
these isomorphisms are also useful to decide whether an ideal is prime/maximal.


modular arithmetic
algebra
system of equations
algebra unsolved

If $ P_n(x) \equal{} x^n \plus{} x \plus{} 1$, find n if

$ P_2(x)|P_n(x)$

Let $ (x^2 \plus{} x \plus{} 1)(a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_1x \plus{} a_0) \equal{} x^{n \plus{} 2} \plus{} x \plus{} 1$.

Let $ n \geq 3$ (you can easy to check that $ n \equal{} 0$ and $ n \equal{} 3$ are solutions).

By multiplying in left side, we will have system equations:
$ a_n \equal{} 1$;
$ a_{n \minus{} 1} \plus{} a_n \equal{} 0$;
$ a_{n \minus{} 2} \plus{} a_{n \minus{} 1} \plus{} a_n \equal{} 0$;
...
$ a_2 \plus{} a_1 \plus{} a_0 \equal{} 0$;
$ a_1 \plus{} a_0 \equal{} 1$;
$ a_0 \equal{} 1$.

Note that there is exist three-element summ in this system of equations (because $ n \geq 3$).

You can find solutions of it: $ a_0 \equal{} a_3 \equal{} a_6 \equal{} ... \equal{} ... \equal{} 1$, but $ a_n \equal{} 1$, $ a_{n \minus{} 1} \equal{} \minus{} 1$, $ a_{n \minus{} 2} \equal{} 0$.
So, it must be $ n$ divisible by 3. And for all $ n$ (if $ n$ is divisible by 3) this system equations has solutions. So, $ x^{n \plus{} 2} \plus{} x \plus{} 1$ is divisible by $ x^2 \plus{} x \plus{} 1$ if and only if $ n$ is divisible by 3.

Answer: all $ n$ such that $ n \plus{} 1$ is divisible by 3.

If you don't feel like doing all the coefficient expansion, notice that $ x^2 \plus{} x \plus{} 1 | x^3 \minus{} 1$, so

$ P_n(x) \equiv x^{n'} \plus{} x \plus{} 1 \pmod {x^3 \minus{} 1} \implies P_n(x) \equiv x^{n'} \plus{} x \plus{} 1 \pmod{P_2(x)}$,

where $ n'$ is $ n$ reduced mod 3. Clearly then $ P_2(x) | P_n(x)$ iff $ n \equiv 2 \pmod{3}$.

Let $ \omega, \omega^2$ we the two roots of $ x^2\plus{}x\plus{}1$. Then they must both be roots of $ x^n\plus{}x\plus{}1$, so $ \omega^n\equal{}x^2$. Thus, $ n\equiv 2\mod{3}$.

Now, each of these positive n work because \[ x^{3m\plus{}2}\plus{}x\plus{}1\equal{}x^{3m\plus{}2}\minus{}x^{2}\plus{}x^2\plus{}x\plus{}1\\\equal{}x^2(x\minus{}1)(x^2\plus{}x\plus{}1)(x^{3(m\minus{}1)}\plus{}...\plus{}1)\plus{}x^2\plus{}x\plus{}1\]

[quote="me@home"]
Now, each of these positive n work because
[/quote]
I think, there is no reason to explain, why every $ n$ in form $ n \equal{} 3m \plus{} 1$ works, because you had find that $ \omega^n \equal{} x^2$ if and only if $ n\equiv 2\mod{3}$ (with property $ x^3 \equal{} 1$).


$ \forall n \in \mathbb{Z}$
find $ x$ such that $ x|n \plus{} 1$

What?

If you meant $ x \mid n\plus{}1$ for all $ n$, make the substitution $ m \equal{} n \plus{} 1$, so we need $ x \mid m$ for all $ m$, thus $ x$ needs to divide all numbers, thus $ x \equal{} \pm 1$.

If you meant $ x \mid n\plus{}1$ for a given $ n$, make the substitution $ m \equal{} n \plus{} 1$, so then $ x$ is just any divisor of $ m$.

 :huh:


trigonometry
geometry
circumcircle
geometric transformation
reflection
angle bisector
geometry unsolved

The median, height, and bisector of triangle $ ABC$ from vertice $ B$ divide angle $ A\hat{B}C$ in four equal angles. Find the angles of this triangle.

The solution I found finds the relations between the sides and then I have to use trigonometry so finish the problem. I was wondering if there was a simple geometry only solution to this problem. If it helps the answer [hide="are"]$ 22,5^\circ$ , $ 90^\circ$ , $ 67,5^\circ$[/hide]

If E, Y, L in CA are feet of the B-altitude, internal angle bisector of <ABC and the B-median, respectively, and if O is the triangle circumcenter, then BO is reflection of BE in BY, so that the circumcenter O is on the B-median BL, which implies that O = L and the angle <ABC is right. To find the remaining angles <CAB, <BCA without trigonometry, see [url]http://www.mathlinks.ro/viewtopic.php?t=128688[/url].

[quote]and if O is the triangle circumcenter, then BO is reflection of BE in BY, so that the circumcenter O is on the B-median BL, which implies that O=L and the angle <ABC is right[/quote]

Sorry you lost me there. We don't know that BO belongs to BL, so we can't say that its the reflection. Anything I am missing?

In any triangle ABC with B-altitude BE, bisector BY of <ABC, and circumcenter O, <EBC = 90 - <BCA = <ABO => <YBE = <OBY => BO is reflection of BE in BY. In the given triangle ABC, the B-median BL is also reflection of BE in BY. And sorry, you do not have to follow any link, <CAB = <LAB = <ABL = 90/4 = 22.5 and <BCA = 67.5.

Took me a while to figure it out but I got it, thank you for this solution


inequalities
function
algebra
polynomial
geometry
logarithms
rational function

I have struggled with this one for a whole night. I think the best way for this is only use Jensen. And I thank you very much to Soarer for  his very nice solution. It is so beautiful !

Ok. Here is his:

Set $a=e^{x}, b= e^{y}, c= e^{z}$
 So $abc=1 \Leftrightarrow x+y+z= 0$
$f(x)= 3e^{x}-\sqrt{8e^{2x}+1}$
  and $f(x)$ is a convex and because $f''(x)=3e^{x}+\frac{64e^{4x}}{(8e^{2x}+1)^{\frac{3}{2}}}-\frac{16e^{2x}}{(8e^{2x}+1)^{\frac{1}{2}}}$
$f''(x)\geq 0$
so we have $f(x)+f(y)+f(z)\geq3 f(\frac{x+y+z}{3})$
   $\Leftrightarrow LHS\geq3 f(0)$
   $\Leftrightarrow LHS\geq 0$
  So we are done.
   :lol:

Gibbenergy, put $e^x= \frac{1}{2}$.

we get $f'' <0$.

That function is not convex.

Well, you are right. Sorry for my careless. I will do it again.
   Thank you so much Chang Woo-JIn

I think what silversunboy posted is a nice way to prove the original problem as well as the generalization :)

Do there exist a solution without considerin 3(or more) situations?

[quote="Ji Chen"][quote="harazi"]Is it true that if $ a,b,c > 0$ verify $ abc \equal{} 1$, then
$ \sqrt {1 \plus{} 8a^{2}} \plus{} \sqrt {1 \plus{} 8b^{2}} \plus{} \sqrt {1 \plus{} 8c^{2}}\leq3(a \plus{} b \plus{} c)$?
[/quote]
$ \sum\sqrt {1 \plus{} 8a^{2}} \equal{} \sum\sqrt {\sqrt [3]{a^{2}b^{2}c^{2}} \plus{} 8a^{2}}\equiv\sum\sqrt {x^{2}y^{2}z^{2} \plus{} 8x^{6}}$

$ \leq\sum\frac {48x^{4} \plus{} 45x^{3}(y \plus{} z) \plus{} 3x(y^{3} \plus{} z^{3}) \minus{} 5x^{2}(y^{2} \plus{} z^{2}) \plus{} 10y^{2}z^{2}}{16(x \plus{} y \plus{} z)}$

$ \equal{} 3(x^{3} \plus{} y^{3} \plus{} z^{3})\equiv3(a \plus{} b \plus{} c)$.
[hide="Generalization."]
If $ x,y,z$ be nonnegative real numbers such that $ xyz \equal{} 1$, and $ 0\leq k\leq k_{0} \equal{} 11.607577302867926253141\cdots$, then

$ \sqrt {1 \plus{} kx^{2}} \plus{} \sqrt {1 \plus{} ky^{2}} \plus{} \sqrt {1 \plus{} kz^{2}}\leq(x \plus{} y \plus{} z)\sqrt {1 \plus{} k}$,

where the $ k_{0}$ be a root of the following irreducible polynomial

$ 6912k^{9} \minus{} 31232k^{8} \minus{} 535456k^{7} \minus{} 368177k^{6} \minus{} 147406k^{5} \minus{} 711839k^{4} \minus{} 754408k^{3}$
$ \minus{} 192235k^{2} \minus{} 26456k \minus{} 1456$,

with equality if $ k \equal{} k_{0}$,

$ x \equal{} 0.33849484707370319678177\cdots$ be a root of the following irreducible polynomial

$ 9s^{9} \minus{} 28s^{8} \minus{} 82s^{7} \plus{} 116s^{6} \plus{} 545s^{5} \minus{} 400s^{4} \plus{} 32s^{3} \minus{} 400s^{2} \plus{} 896s \minus{} 256$,

$ y \equal{} z \equal{} 1.7187945546210849737333\cdots$ be a root of the following irreducible polynomial

$ 16t^{9} \minus{} 16t^{8} \minus{} 20t^{7} \plus{} 16t^{6} \minus{} 16t^{5} \minus{} 23t^{4} \plus{} 10t^{3} \minus{} 4t^{2} \minus{} 2t \plus{} 3$.[/hide][/quote]

Dear Ji Chen,

I must ask, how did you come up with the upper bound for the square root? What is your thinking in this area? Obviously, you want something that sums up to the desired expression and has equality where the equation has equality. But I would still like to get a better idea as to what you are thinking. What types of functions to do you try to use? Is there a better method than just considering an arbitrary rational function?

--alex

[quote="harazi"]Assume that $a$, $b$, $c$ are positive real numbers satisfying the condition $abc=1$.
Prove that $\sqrt{1+8a^{2}}+\sqrt{1+8b^{2}}+\sqrt{1+8c^{2}}\leq 3\left(a+b+c\right)$.[/quote]
Tangent Line method does not help here:
$\sqrt{1+8a^{2}}+\sqrt{1+8b^{2}}+\sqrt{1+8c^{2}}\leq 3\left(a+b+c\right)\Leftrightarrow \sum_{cyc}f(a)\geq0$, where
$f(a)=3a-\sqrt{1+8a^2}-\frac{1}{3}\ln{a}$.
$f'(a)=\frac{9a-1}{3a}-\frac{8a}{\sqrt{1+8a^2}}<0$ for $0<a\leq\frac{1}{9}$.
But for $a>\frac{1}{9}$ we obtain $f(0.225)<0$.

Note that we can wlog $a \ge b \ge c$
Then it is easy to prove that $ \frac{a^2+b^2+c^2}{3}+8c^2,\frac{a^2+b^2+c^2}{3}+8b^2,\frac{a^2+b^2+c^2}{3}+8a^2 \succ{9c^2,9b^2,9a^2} $
But since $f(x)=\sqrt{x}$ is concave inequality is obvious from Karmata's Majorisation inequality:
$ \sqrt{1+8c^2}+\sqrt{1+8b^2}+\sqrt{1+8a^2}\le$
$ \sqrt{\frac{a^2+b^2+c^2}{3}+8c^2}+\sqrt{\frac{a^2+b^2+c^2}{3}+8b^2}+\sqrt{\frac{a^2+b^2+c^2}{3}+8a^2} \le \sqrt{9c^2}+\sqrt{9b^2}+\sqrt{9a^2}=3a+3b+3c$
The proof is also respective with unlimited number of variables :roll:

[quote="hipernova"]Note that we can wlog $a \ge b \ge c$
Then it is easy to prove that $ \frac{a^2+b^2+c^2}{3}+8c^2,\frac{a^2+b^2+c^2}{3}+8b^2,\frac{a^2+b^2+c^2}{3}+8a^2 \succ{9c^2,9b^2,9a^2} $
But since $f(x)=\sqrt{x}$ is concave inequality is obvious from Karmata's Majorisation inequality:
$ \sqrt{1+8c^2}+\sqrt{1+8b^2}+\sqrt{1+8a^2}\le$
$ \sqrt{\frac{a^2+b^2+c^2}{3}+8c^2}+\sqrt{\frac{a^2+b^2+c^2}{3}+8b^2}+\sqrt{\frac{a^2+b^2+c^2}{3}+8a^2} \le \sqrt{9c^2}+\sqrt{9b^2}+\sqrt{9a^2}=3a+3b+3c$
[color=#FF4000][b]The proof is also respective with unlimited number of variables[/b][/color] :roll:[/quote]

This proof is also wrong. :)

[url=http://en.wikipedia.org/wiki/Majorization]Majorization[/url]

[quote="silversunboy"][quote="Sung-yoon Kim"]Well, it can be proved by force. (24 degrees) But I proved by Jensen.

(Not simple : we must consider 3 cases. )[/quote]
(this is not my solution)
We use Jensen.
a>=b>=c
*a>=1>=b>=c
It's not very complicated,you can use B.C.S to prove it
Hint: a+b+c>=ab+bc+ca
*a>=b>=1>=c (using Jensen)
You must prove that f(a,b,c)>=f( \sqrt ab, \sqrt ab,c)
then you prove that  f(a,a,c)>=0
Hope that you could find the solution your self! :)  :)  :)[/quote]

who can give a detailed proof with the obove mentioned method?

Harazi, did you create it by hand or by computer?

[quote="MathUniverse"][quote="arqady"]
Thank you, quykhtn-qa1! You are right.
The MV method helps here by beautiful Can's idea: $\sqrt{1+8b^2}+\sqrt{1+8c^2}\leq(b+c)\sqrt{8+a}$.[/quote]

Can you please post proof after using MV?[/quote]
Do you mean the following: $\sqrt{1+\frac{8}{b^4}}+2\sqrt{1+8b^2}\leq3\left(\frac{1}{b^2}+2b\right)$?

[quote="MathUniverse"][quote="arqady"]
Thank you, quykhtn-qa1! You are right.
The MV method helps here by beautiful Can's idea: $\sqrt{1+8b^2}+\sqrt{1+8c^2}\leq(b+c)\sqrt{8+a}$.[/quote]

Can you please post proof after using MV?[/quote]
[quote="zdyzhj"][quote="arqady"][quote="quykhtn-qa1"]
It is wrong.
We have: $ f'(a)=\frac{9a-5}{3a}-\frac{8a}{\sqrt{1+8a^2}} $ .[/quote]
Thank you, quykhtn-qa1! You are right.
The MV method helps here by beautiful Can's idea: $\sqrt{1+8b^2}+\sqrt{1+8c^2}\leq(b+c)\sqrt{8+a}$.[/quote]

dear arqady, what is the following?[/quote]
let me write full solution of [b]can_hang2007[/b]:
assume $c=min(a,b,c)\Longrightarrow c\le 1$
use cauchy inequality we have:
$\sqrt{8a^2+1}+\sqrt{8b^2+1}\le \sqrt{(a+b)(\frac{8a^2+1}{a}+\frac{8b^2+1}{b})}=(a+b)\sqrt{c+8}$
so we need prove:
$(a+b)(3-\sqrt{c+8})+3c\ge \sqrt{8c^2+1}$
but we have: $3-\sqrt{c+8}\ge 0,a+b\ge 2\sqrt{ab}$
so we need to prove:
$2\sqrt{ab}(3-\sqrt{c+8})+3c\ge \sqrt{8c^2+1}\iff 6\sqrt{ab}+3c\ge 2\sqrt{8ab+1}+\sqrt{8c^2+1}$
from AM-GM inequality we have:
$\sqrt{8ab+1} \le \frac{8\sqrt{ab}+1+\frac{9(8ab+1)}{8\sqrt{ab}+1}}{6}$
$\sqrt{8c^2+1}\le \frac{8c+1+\frac{9(8c^2+1)}{8c+1}}{6}$
so we need to prove:
$10(2\sqrt{ab}+c)\ge 3+\frac{18(8ab+1)}{8\sqrt{ab}+1}+\frac{9(8c^2+1)}{8c+1} (1)$
$\sqrt{ab}=t\Longrightarrow c=\frac{1}{t^2},t\ge 1$
$\Longrightarrow (1) \iff 10(2t+\frac{1}{t^2})\ge 3+\frac{18(8t^2+1)}{8t+1}+\frac{9(t^4+8)}{t^2(t^2+8)}$
$\iff 2(2t+\frac{1}{t^2}-3)\ge 2(\frac{9(8t^2+1)}{8t+1}-8t-1)+(\frac{9(t^4+8)}{t^2(t^2+8)}-\frac{t^2+8}{t^2})$
$\iff \frac{2(2t+1)(t-1)^2}{t^2}\ge \frac{16(t-1)^2}{8t+1}+\frac{8(t^2-1)^2}{t^2(t^2+8)}$
$\iff 2t+1\ge \frac{8t^2}{8t+1}+\frac{4(t+1)^2}{t^2+8}$
it is true because:
$t>\frac{8t^2}{8t+1}$ and $t+1-\frac{4(t+1)^2}{t^2+8}=\frac{(t+1)(t-2)^2}{t^2+8}\ge 0$
complete prove.the inequality holds when $a=b=c=1$


geometry
3D geometry
AMC
USA(J)MO
USAMO
number theory unsolved
number theory

I reduced a practice USAMO I was taking to showing that no cubes of the form $ x(x \minus{} 2)$ existed, but couldn't satisfactorily convince myself of it... So..

1. Would I have gotten a 6 if I stated it as if it was obvious on the USAMO ("We know no cubes of this form exist and therefore...")? Or more like a 2? (Note that this was like a sub-case of one of three cases in the problem, so I still made a very large amount of significant progress. This was not a major tenet of the problem)

2. What's the answer? Are there known results? It seems like there should be a nice proof...

For the record, the problem I ended up with this on was [hide]2005-2[/hide] Also, the numbers I was particularly interested in were actually any integers of the form $ x(x \minus{} 2)$, but rather of the special case [hide]$ (3^r*7^{157} \plus{} 1)(3^r*7^{157} \minus{} 1)$[/hide]. So it would suffice to have shown that only those aren't cubes.

I haven't actually read the solution (just in case the answer to this question shows up in it) to the problem in question because if this is a significant flaw and/or not really provable I'd like to keep working on the problem.

Oops, sorry, I did know about that case but forgot to mention it... with the assumption that n > 4 is it statable as I said?

I actually meant to say x > 4, since as you can see from the numbers I had (in the spoiler-text), x was quite certainly larger than 4.

However, for case 2, why exactly must $ \frac{x}{2}$ and $ \frac{x}{2}\minus{}1$ be cubes? Can't we have $ x$ (or $ x\minus{}2$) is $ 2*a^3$ for some a and the remaining one is $ 2^{3k\plus{}2}*b^3$ for some k,b? That case was my concern in stating it as obvious/not obvious.

[quote="t0rajir0u"]As stated, it's not obvious (edit: but the proof I just had was incorrect).  Try http://en.wikipedia.org/wiki/Mih%C4%83ilescu's_theorem .[/quote]

Oooh...

So we reduce the problem to
$ x^2\minus{}a^3\equal{}1$ which by the cited theorem is clearly unsolvable for the given x? Sweet!

Any idea if the inability to have cited that theorem would have netted me a 6 vs a 2 though?

Mihailescu's Theorem is a little stronger than you need.  The curve $ y^2 \equal{} x^3 \plus{} 1$ is a [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] - see the discussion of the $ n \equal{} 1$ case.  Alternately, there may be a very easy way to prove the particular case you're after (mods and so forth).


geometry
geometric transformation
dilation

Write net ionic reactions:
1.A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide
2.A solution of ammonia is added to a solution of ferric chloride

[quote="adchia"]1. Cl_2 + OH -> Cl- + ClO_3 - + H_2O (disproportionation)

2. In water, NH3 + H2O -> NH4+ + OH-.

Iron and hydroxide would merge to form a precipitate, so the precipitation reaction dominates the solution.

H2O + NH3 + FeCl_3 -> Fe(OH)_3 + NH_4+[/quote]

cool, thanks


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

let $A \in M_n(C)$ be an arbitrary matrix.
$C_A= \{M \in M_n(C) | MA=AM\}$, $E_A=\{P(A) | P \in C[X] \}$.
Find the best constants $d_1$ and $d_2$ such that:
$dim(C_A) \leq d_1$ and $dim(E_A) \leq d_2$ for all $A \in M_n(C)$.

Can you correct your post so that you're asking two different questions?

All answers here are independent of the field. If $A$ is a multiple of the identity, $C_A$ is the whole space, which has dimension $n^2$.
The dimension of $E_A$ is the degree of the minimal polynomial of $A$, which is at most $n$. Examples of matrices with any degree $n$ minimal polynomial can be easily constructed.

sorry I forgot to say that $A$ is not $kI$ because otherwise it is to easy for the first one  :)

It is not difficult to prove that the dimension of the commutant is at least $n$ (just trigonalize A and choose matrices that commute to it as triangular). Much more difficult is to find all matrices for which the dimension is equal to $n$. About the superiour bound, I think that it is $ n^2-2n+2$. I found a problem given in an oral examination to Centrale Paris that says that there always exist $d_1,...,d_r$ integers which add up to $n$ and such that the imension of the commutant of the matrix is equal to the sum of the $r$ numbers. With this, one can prove the upper bound. The problem is that I don't know how to solve the problem from Centrale. Any idea?

yes I did like this to find the upper bound. To prove the result remember that if $u$ and $v$ commutes then the proper spaces of $u$ are stable under $v$. Then use Jordan blocks for example  :)

Alekk, could you please be more explicit? Thanks a lot.

I've just realized my solution was false  :blush:  i don't know why I thought in my stupid head that $A$ was diagonalisable   :(
In any case it is a good pb  :D


Gauss
logarithms
limit
calculus
calculus computations

If $ \{ a_{n}\}$ is a sequence of positive terms, prove that if there exists $ N \geq 1$, $ s > 1$ such that

$ \frac{a_{n+1}}{a_{n}}= 1-\frac{1}{n}+\frac{f(n)}{n^{s}}$ for all $ n \geq N$,

where $ f$ is bounded on $ \mathbb R^{+}$, then $ \sum a_{n}$ diverges.

This problem is a particular case of the so called Gauss test .

Gauss test:

Let $ \sum a_{n}$ be a positive infinite series and suppose that

$ \frac{a_{n+1}}{a_{n}}=1-\frac{\beta}{n}+\frac{\theta_{n}}{n^{1+k}}$,

where $ \theta_{n}$ is a bounded sequence and $ k>0$.

Then

a) If $ \beta>1$, then  the series converges.

b) If $ \beta \leq 1$, then  the series diverges.

In your case $ \beta=1$ and hence the series diverges.

Didi, that's right.  Here's the proof I have of the part for $ A=1$:

First suppose $ \{ a_{n}\}$ and $ \{ b_{n}\}$ are two sequences of positive terms for $ n \geq N$, and put $ c_{n}= b_{n}-b_{n+1}a_{n+1}/a_{n}$.  Then, if $ c_{n}\leq 0$ for $ n \geq N$ and if $ \sum 1/b_{n}$ diverges, so does $ \sum a_{n}$, for if $ n-1 \geq N$ we have
$ \sum_{k = N}^{n-1}b_{k}a_{k}-b_{k+1}a_{k+1}= b_{N}a_{N}-a_{n}b_{n}\leq 0$, so $ \sum a_{n}$ dominates $ \sum 1/b_{n}$.

Now suppose $ \{ a_{n}\}$ is a sequence of positive terms, and there exists $ N \geq 1$, $ s > 1$ such that
$ \frac{a_{n+1}}{a_{n}}= 1-\frac{1}{n}+\frac{f(n)}{n^{s}}$ for all $ n \geq N$,
with $ |f| < M$ on $ \mathbb R^{+}$.  Put $ b_{n+1}= n \log n$ and $ c_{n}= b_{n}-b_{n+1}a_{n+1}/a_{n}$.  We have
$ c_{n}= (n-1) \log (n-1)-n \log n \left( 1-\frac{1}{n}+\frac{f(n)}{n^{s}}\right) \\ = (n-1) \log(1-1/n)-\frac{ \log n f(n) }{n^{s-1}}$

Since $ s-1 > 0$ and $ f$ is bounded, $ \lim_{n}\frac{ \log n f(n) }{n^{s-1}}= 0$.  Also, $ (n-1) \log(1-1/n) = (n-1) \left(-\frac{1}{n}+o(1/n) \right) =-1+1/n+o(1)$ as $ n \rightarrow \infty$, so $ \lim_{n}(n-1) \log(1-1/n) =-1$.  Thus, $ \lim_{n}c_{n}=-1$, so for large enough $ n$, $ c_{n}\leq 0$.  Since $ \sum 1/b_{n}$ diverges, so does $ \sum a_{n}$ by the what was shown first.

For $ A \neq 1$ the proof is similar (and shorter) and uses that first lemma on $ c_{n}$.  My question is: do you have a shorter or nicer proof?


[quote="GameBot"]A broken toaster toasts two slices of bread at once but toasts only one side, so the bread must be flipped over to toast the other side. It takes 45 seconds to toast one side. What is the least number of minutes required to toast five slices of bread on both sides? Express your answer as a mixed number.[/quote]

Let the slices be A, B, C, D, and E.

First, the toaster can toast one side of both A and B, which will take 45 seconds, and then the second sides, which will take another 45 seconds. Then, it can toast one side of both C and D, which takes 45 seconds, and then the second sides, which takes another 45 seconds. Repeat this one more time with just E, and then our answer is $ 45 \plus{} 45 \plus{} 45 \plus{} 45 \plus{} 45 \plus{} 45 \equal{} 4\frac{1}{2}$.

I don't think so...
Perhaps I'm wrong... but...:

Let's use a point value system:
0 means not toasted at all, 1 means 1 side toasted, 2 means both sides toasted

Step 0: $ 0 0 0 0 0$
On each step, let us toast 2 pieces of bread at once:
Step 1: $ 1 1 0 0 0$
Step 2: $ 2 1 1 0 0$
Step 3: $ 2 2 2 0 0$
Step 4: $ 2 2 2 1 1$
Step 5: $ 2 2 2 2 2$

So it takes 5 steps to toast 5 pieces of bread.
Each step takes 45 seconds = 3/4 of a minute, so the answer is $ 5 * \frac{3}{4} \equal{} \frac{15}{4} \equal{} \boxed{3\frac{3}{4}}$

[quote="nonumbered"]on ftw it says 33/4[/quote]

That's $3\frac{3}{4}$.


inequalities
calculus
integration
function
trigonometry
complex analysis
complex analysis unsolved

Use the mean value inequality (or "theorem", or whatever it is called in English). You may as well use some basic integration (I remember Kent Merryfield giving me such a solution on the forum).

q: show /e^(ia) - e^(ib)/ <= /a - b/ for a, b real numbers and /x/ stands for the modulus of a complex number x.
a: we show a small fact:
fact: for y >=  0 one has: y - siny >= 0 and y + siny >= 0
check: f(y) = y - siny has f'(y) = 1 - cosy >= 0 thus f is increasing and
f(y) >= f(0) = 0 - sin0 = 0. For the second inequality use a similar function g(y) = y + siny and have g'(y) = 1 + cosy >= 0 and thus
g is also increasing so g(y) >= g(0) = 0 + sin0 = 0. but both inequalities we just proved yield the following as a corollary:
corollary: for x > 0, x^2 >= (sin(x))^2.
we now get back to the problem. we can assume a > b and thus : a - b > 0.
now using Euler's formula e^(ia) = cosa + i.sina we have:
/e^(ia) - e^(ib)/^2 = /(cosa - cosb) + i.(sina - sinb)/^2 = (cosa - cosb)^2 + (sina - sinb)^2 = 2 - 2.cos(a - b) = 2.(1 - cos(a - b)) = 2.2.(sin((a - b)/2))^2 <= 4.(sin((a - b)/2))^2 <= 4.((a - b)/2)^2 = (a - b)^2 using
the corollary. we are done.
___________________________________________________________

[quote="oldman"]Let $a,b\in\mathbb{R}$. Show that \[|e^{ia}-e^{ib}|\leq |a-b|\] [/quote]

Eum...
$|e^{ia}-e^{ib}|=\sqrt{2-2\cos(b-a)}\le|a-b|$ since $2-2\cos x\le x^{2}$? (the latter by taylor series of $\cos$)

suppose a<b

$|e^{ib}-e^{ia}|=|\int_{a}^{b}ie^{it}dt|\leq \int_{a}^{b}|ie^{it}|dt=b-a= |b-a|$


Let $\mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $f: \mathbb{N} \mapsto \mathbb{N}$ such that $\forall n \in \mathbb{N},$ $f^{1989}(n) = 2 \cdot n$ ?

Let $n_1,n_2,\ldots$ be the increasing sequence of odd numbers $\ge 1$, put, for all $k\ge 0$, $f(n_{1989k+i})=n_{1989k+i+1},\ \forall i\in\overline{1,1988}$, $f(n_{1989k+1989})=2n_{1989k+1}$, and then extend the function to even numbers by using $f(2n)=2f(n),\ \forall n\ge 1$.

[quote="orl"]Let $ \mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $ f: \mathbb{N} \mapsto \mathbb{N}$ such that $ \forall n \in \mathbb{N},$ $ f^{1989}(n) = 2 \cdot n$ ?[/quote]
Define for all $ n \in \mathbb{N}$: $ m_n = \frac {\frac {n}{2^{e_2(n)}} + 1}{2}$.
Then this one works:
$ f(n) =  \{\begin{array}{cc} t \mid m_n & (2(m_n - t) + 1)2^{e_2(n) + 1} \\
t \nmid m_n & (2m_n + 1)2^{e_2(n)} \\
\end{array}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose $ %Error. "display" is a bad command.
\ F$ is a feild.define $ %Error. "display" is a bad command.
\ U T_n (F)$={upper triangle matrices $ %Error. "display" is a bad command.
\ [a_{ij}]_{n\times n}\mid \forall i,j a_{ij}\in F,a_{11} = a_{22} = ... = a_{nn} = 1\}$.proove $ %Error. "display" is a bad command.
\ U T_n (F)$ is a nilpotent group and the class of nilpotency of $ %Error. "display" is a bad command.
\ U T_n (F)$ is n_1.as you know as definition we have $ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z(G)})$=$ %Error. "display" is a bad command.
\dfrac{\ Z_2(G)}{Z(G)}$ so that $ %Error. "display" is a bad command.
\ Z_2(G)$  is a normal subgroup of G that include $ %Error. "display" is a bad command.
\ Z(G)$ .if we continue we have$ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z_i (G)})$ =$ %Error. "display" is a bad command.
\dfrac{\ Z_{i+1} (G)}{Z_{i}(G)}$ and ${ %Error. "display" is a bad command.
\ Z_1(G)}$=$ %Error. "display" is a bad command.
\ Z(G)$.if we have m$ %Error. "display" is a bad command.
\in\mathbb{N} that Z_m (G)$=G then we say that  G is a nilpotent group and if m be the smallest natural number then we define m as the class of nilpotency of G.

pls use any reasonable group theory text book

1) Use this definition:
http://en.wikipedia.org/wiki/Nilpotent_group
2) The key is as follows:
Let $ T_k\equal{}\{U\in\mathcal{M}_n: j\minus{}i\leq{k}\rightarrow{u_{i,j}\equal{}0}\}$.
Prove that if $ U,V\in{T_k}$ then $ (I\plus{}U)(I\plus{}V)\minus{}(I\plus{}V)(I\plus{}U)\in{T_{k\plus{}1}}$.

Correction; I have written too quickly.
You must prove this condition:
if $ U\in{T_0},V\in{T_k}$ then $ (I\plus{}U)^{\minus{}1}(I\plus{}V)^{\minus{}1}(I\plus{}U)(I\plus{}V)\minus{}I\in{T_{k\plus{}1}}$.
I hope that it's correct.


trigonometry
function
calculus
derivative
real analysis
real analysis unsolved

Let f(x) = series from k=1 to infinity of (sin(kx) over k^3).
Show that f(x) is differntiable.  Can we determine if f is twice-differentiable?

Please help!

$f(x) = \sum_{k=1}^{\infty}\frac{ \sin (kx) }{k^{3}}$.  This function satisfies

$| f(x) | \le \sum_{k=1}^{\infty}\frac{1}{k^{3}}$

so it is defined.  Its derivative satisfies

$| f'(x) | \le \sum_{k=1}^{\infty}\frac{1}{k^{2}}$

So it is also defined.  But the second derivative is indeterminate.

That makes sense, but why are we unable to determine if the second derivative is differentiable?  Why couldn't we repeat this same process again?

Thanks!

$f''(x) = \sum_{k=1}^{\infty}\frac{-\sin (kx) }{k}$

We cannot guarantee that this converges since we can only bound it below the harmonic series.

The series converges for all x. For $x\in(0;2\pi)$ it converges to $\frac{x-\pi}2$. For $x\in(\varepsilon;2\pi-\varepsilon)$ the convergence is uniform.

Ah.  It reminded me of a familiar Fourier series, but I wasn't sure.


A man is at a base of a tree 10 miles due south of a water well. His home is located 20 miles southeast of the tree. What is the total distance he must travel from his starting location to go the shortest distance to the water well and then the shortest distance to return home? Assume all paths are coplanar. Round to the nearest hundreth of a mile.

[hide] To get to the tree, he needs to travel 10 miles (since the tree is directly south of the well).  Now, his home is 20 miles southwest of the tree.  Therefore, if he were to walk directly south (from the tree) for $10\sqrt{2}$ miles, then directly east for $10\sqrt{2}$ miles, he would reach is house.  Now, this imples that if the man were to walk $10+10\sqrt{2}$ miles south then $10\sqrt{2}$ miles east, from the well, then he would reach home.  Applying the Pythogorean Thereom on these two lengths, we have the shortest length is $\sqrt{(10+10\sqrt{2})^{2}+10\sqrt{2}^{2}}=\sqrt{500+200\sqrt{2}}.$  This makes 27.98 miles if we round it to the nearest hundredth.  [/hide]

[quote="The QuattoMaster 6000"][hide] To get to the tree, he needs to travel 10 miles (since the tree is directly south of the well).  Now, his home is 20 miles southwest of the tree.  Therefore, if he were to walk directly south (from the tree) for $10\sqrt{2}$ miles, then directly east for $10\sqrt{2}$ miles, he would reach is house.  Now, this imples that if the man were to walk $10+10\sqrt{2}$ miles south then $10\sqrt{2}$ miles east, from the well, then he would reach home.  Applying the Pythogorean Thereom on these two lengths, we have the shortest length is $\sqrt{(10+10\sqrt{2})^{2}+10\sqrt{2}^{2}}=\sqrt{500+200\sqrt{2}}.$  This makes 28 miles if we round it to the nearest tenth.  [/hide][/quote]

Last time I checked something to the nearest tenth had a decimal. I meant nearest hundreth. My bad. Edited.


[quote="darthfader"]I hope I posted this on the right forum..

$ If$

$ \frac {{(2x^5 \minus{} x^4 \plus{} 3x^3 \minus{} 4x^2 \plus{} 5x \minus{} 1)}}{{(3x^5 \plus{} 2x^3 \plus{} 7x^2 \minus{} x \minus{} 10)}} \equal{} \frac {2008}{2009}$

$ Find$ $ the$ $ exact$ $ value$ $ of$

$ \frac {{(3x^4 \minus{} 5x^3 \plus{} 26x^2 \minus{} 17x \minus{} 17)}}{{(12x^5 \minus{} 3x^4 \plus{} 13x^3 \plus{} 2x^2 \plus{} 13x \minus{} 23)}}$

Can someone help me to work around this? Can anyone show me how? Any help is appreciated. Thanks. :)[/quote]

Let $ 2x^5 \minus{} x^4 \plus{} 3x^3 \minus{} 4x^2 \plus{} 5x \minus{} 1\equal{}2008k$,   and $ 3x^5 \plus{} 2x^3 \plus{} 7x^2 \minus{} x \minus{} 10\equal{}2009k$.

Then $ \frac {{(3x^4 \minus{} 5x^3 \plus{} 26x^2 \minus{} 17x \minus{} 17)}}{{(12x^5 \minus{} 3x^4 \plus{} 13x^3 \plus{} 2x^2 \plus{} 13x \minus{} 23)}}$
$ \equal{}\frac{(2)(2009k)\minus{}(3)(2008k)}{(2)(2009k)\plus{}(3)(2008k)}$
$ \equal{}\frac{\minus{}2006k}{10042k}$
$ \equal{}\minus{}\frac{1003}{5021}$


function
topology
algebra
domain
advanced fields
advanced fields unsolved

According to the book General Topology by Stephen Willard, there's a theorem and a consequence of this (corollary) which is:

$\emph{Theorem}$: If $X$ is a first countable space and $E \subset X$, then $x \in \overline{E}$ iff there is a sequence $(x_{n})$ contained in $E$ which converges to $x$.

$\emph{Corollary}$. Let $X$ and $Y$ be first countable spaces.  Then

$f: X \to Y$ is continuous iff whenever $x_{n}\to x$ in $X$, then $f(x_{n}) \to f(x)$ in $Y$.

Well the problem is that in my proof I don't use the fact that $Y$ is first countable, so i don't know if I am wrong or what, so I would like you could check it:

$\emph{Proof}$  Suppose $f$ is not continuous, i.e. there is $x_{0}\in X$ such that given $V$ nhood of $f(x_{0})$, $f(U) \nsubseteq V$ , for any nhood $U$ of $x_{0}$.  $X$ is first countable, so there's a countable nhood base $\{U_{n}\ / \ n=1,2,...\}$ at $x_{0}$ in $X$. We can suppose $U_{1}\supset U_{2}\supset \cdots$ (just replacing $U_{n}$ by $\bigcap_{k=1}^{n}U_{k}$ where necessary) , then pick $x_{n}\in U_{n}$ such that $f(x_{n}) \in f(U_{n})-V$ (if there not exists that $x_{n}$ then $f(U_{n})-V= \varnothing$ , but that contradicts the fact ,  $f(U) \nsubseteq V$ , for any nhood $U$ of $x_{0}$) , thus clearly $x_{n}\to x_{0}$, but $f(x_{n}) \nrightarrow f(x_{0})$ (there is no point of the sequence $f(x_{n})$ in the nhood $V$  :) ), so it is a contradiction with the hypothesis; therefore $f$ is continuous.

The converse is true even if X or Y are not first countable (homework for u :P )

am I wrong? If I am i cannot see the mistake.  A exercise says :

"Find spaces $X$ and $Y$  and a function $F: X \to Y$ which is not continuous , but which has the property that $F(x_{n}) \to F(x)$ in $Y$ whenever $x_{n}\to x$ in $X$"

well i could answer it , actually it is here  http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=adriana&t=112962

but as you can see I just use the fact that $X$ was not first countable, so if anyone could help me i would be really thankful

Adriana

You're right. Only first countability of $X$ is required.

cool how fast the answer was ! thank you I will write to Stephen telling he has to see my proof :P

I found a topology textbook on the shelf, and opened it up looking for this result. The first page I tried had it (pure luck), and the result was stated with only the domain first countable.


Let´s try to solve the next problem:

Let A be a matrix with complex eigenvalues. Show that exist a real invertible matrix P such that $ P^{\minus{}1}AP\equal{}J$, where J is the 2x2 matrix (0,-1;1,0)

You want $ \mathbf{A} \in \mathcal{M}_2(\mathbb{Z})$ in order for the result to even be plausible, and even then the result isn't true (take $ \mathbf{A} \equal{} \left[ \begin{array}{cc} \minus{}1 & \minus{}1 \\ 1 & 0 \end{array} \right]$.)

Perhaps you mean this? There exists invertible $ P$ and real numbers $ x,y$ such that $ P^{\minus{}1}AP\equal{}xI\plus{}y\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}.$

If the eigenvalues of $ A$ are pure imaginary, then $ P^{\minus{}1}AP\equal{}y\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}.$

Note that t0rajir0u's example doesn't have pure imaginary eigenvalues.

(Side note: I refuse to call that  particular matrix $ J$ because in my mind the name $ J$ is already claimed by a different matrix.)


if $ x^2 \plus{} 2y^2$ is an odd prime then it has the form 8n +1 or 8n+3
my solution:
x must be odd.so $ x^2$=1mod 8
$ x^2 \plus{}2y^2$=1 mod8 if y is even.
if y is odd then $ y^2$=1mod8 and $ x^2 \plus{}2y^2$=3 mod 8

Can you prove the converse: if $ p$ a prime is of the form $ 1\plus{}8n$ or $ 3\plus{}8n$ then $ p\equal{}x^2\plus{}2y^2$ for $ x,y,n$ integers?

Disclaimer: the only proof I know of requires a certain theorem which may or may not be very well known.


modular arithmetic
function
number theory

What is the sum of the tens and units digits of $ 9^{99}$?

Tens and units digits for digits for $ 9^n$ are: $ 09,81,29,61,49,41,69,21,89,01,09$, so it repeats every 10.

Therefore, the answer is $ 8\plus{}9\equal{}17$

[hide="Method 1: theoretically"] $ \lambda(100) \equal{} 20$ (see [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url]), so $ 9^{99} \equiv 9^{\minus{}1} \bmod 100$.  Moreover, since $ 9 \cdot 11 \equiv \minus{}1 \bmod 100$ we know that $ 9 \cdot 89 \equiv 1 \bmod 100$, so $ 9^{\minus{}1} \equiv 89 \bmod 100$. [/hide]
[hide="Method 2: algebraically"] $ 9^{99} \equal{} (10 \minus{} 1)^{99} \equiv (\minus{}1)^{99} \plus{} 99 (\minus{}1)^{98} 10 \equiv 90 \minus{} 1 \equiv 89 \bmod 100$. [/hide]

[quote="SnipedYou"]The last two digits of $ 9^{x}$ repeats itself every 10 repetitions.

So $ 9^{99} \equiv 9^{9} \pmod{100}$
If the exponent is odd in $ 9^{x}$, then the units digit is $ 9$, and the tens digit will be one less, then the units digit of the power its raised to (This applies to odd exponents). So the tens digit is $ 8$ and the sum is $ 8 \plus{} 9 \equal{} 17$[/quote]

hmmm....
i have been using this technique for a year without knowing that i was using modular arithmetic  :D


trigonometry
trig identities
Law of Cosines

[quote="archimedes1"]If $ \frac{a^{3}+b^{3}+c^{3}}{a+b+c}=c^{2}$ in $ \triangle ABC$, then find $ \tan C$.[/quote]

[hide]$ a^{3}+b^{3}+c^{3}= (a+b)c^{2}+c^{3}$
$ a^{3}+b^{3}= (a+b)c^{2}$
$ (a+b)(a^{2}-ab+b^{2}) = (a+b)c^{2}$
$ a^{2}-ab+b^{2}= c^{2}$

By the law of cosines,
$ a^{2}-2ab\cos C+b^{2}= c^{2}$

This implies that
$ \cos C =\frac{1}{2}$

From the Pythagorean Identity,
$ \sin C = \frac{\sqrt{3}}{2}$

Thus,
$ \tan C = \frac{\sin C}{\cos C}= \boxed{\sqrt{3}}$.[/hide]

EDIT: Oops, messed up the signs.


On the official USAMTS site, the 'past problems' archive only goes back to year 10. Is there any way to obtain older problems, free or paid?

IMTS is something you can do.

If you scroll down, there is IMTS Solution Thread where I collect solutions for the IMTS problems.

If you can help, that would be great. (You don't have to post on this forum if you want to)


I just bought Geometer Sketchpad, and I was wondering how to import the images created into Latex. I think I need to convert the images to PNG format or PDF first. Is there a way to export Geometer images to PNG or do I just need to print the image to a PDF file? Thanks.

You have a couple of options.  One is that you can use a virtual PDF printer (I like [url=http://www.cutepdf.com/Products/CutePDF/writer.asp]CutePDF Writer[/url], but there are several) to create a PDF document and then just include it in your LaTeX file using \includegraphics.  (If you do this, make sure that you resize the output page within your virtual printer, so that you don't have to mess around with a bounding box when you try to include it into LaTeX.)  Another is to cut-and-paste from Sketchpad over to a drawing program, or to use the PrintScrn button.

I don't believe that Sketchpad will export images directly into another format (which is by far its biggest flaw).


Given a circumscribed quadrilateral $ABCD$. $AD$ is the longest side and $AB \cap CD=M$, $AC \cap BD = N$.
Prove that $\frac {\pi}2 \leq \angle {AND} \leq \frac {\pi}2+\frac {\angle {AMD}}2$

[i]Edited by Myth[/i]

You mentioned Sedrakyan book in the source of this problem.Which Sedrakyan do you mean?Do you mean the Armenian IMO team deputy leader?

The following solution was invented by my students.

Step 1. The figure for this problem looks like on the picture. Indeed, we have $AB\leq AD$, thus $YB\leq XD$. It implies that $\angle XID\geq \angle YIB$. And in the same way $\angle XIA\geq \angle ZIC$. It immediately gives $\angle AID\geq \pi/2$, so $\angle BAD+\angle CDA=2\pi-2\angle AID\leq \pi$.

Step 2. The upper bound. Construct $AB'=AB$. Then $C'\in [CD]$, so $\angle CAD\geq \angle C'AD=\frac{1}{2}\angle A$. Analogously, $\angle BDA\geq \frac{1}{2}\angle D$. It follows that $I$ lies inside of $\triangle AMD$. Therefore $\pi/2+\frac{1}{2}\angle AND=\angle AID\geq \angle AMD$.

Step 3. The lower bound. Suppose the contrary. Then we obtain $AD^2< AM^2+MD^2$ and $BC^2<MB^2+MC^2$, on the other hand $AB^2>AM^2+MB^2$ and $CD^2>MC^2+MD^2$. It follows that $AD^2+BC^2<AB^2+CD^2$, but $AD+BC=AB+CD$ and $AD$ is the maximal from numbers $AB,BC, CD$ and $AD$. So it is impossible to have $AD^2+BC^2<AB^2+CD^2$. Contradiction.


projective geometry
geometry proposed
geometry

[b][size=100][color=DarkBlue]A quadrilateral $ ABCD$ is given and let be the points $ E\equiv AB\cap CD$ and $ F\equiv AD\cap BC.$ The line segments $ PA,\ PB,\ PC,\ PD,$ where $ P$ is an arbitrary point in the plane, intersect $ EF,$ at points $ A',\ B',\ C',\ D',$ respectively. Prove that the line segments $ AC',\ A'C,\ BD',\ B'D,$ are concurrent at one point.[/color][/size][/b]

Kostas Vittas.

[quote="mr.danh"]Denote O=AC ∩ BD, I=PO∩EF. Then, AC',CA',BD',DB' are concurrent at K which is the point satisfying (OIPK)=-1.[/quote]
Thank you dear [b][size=100]mr.danh[/size][/b] for your interest.

It is true and easy to show the result you mentioned, after the concurrency as the problem states.

But how we can prove at first, that each of the line segments $ AC',\ A'C,\ BD',\ B'D,$ intersects the line segment $ EF,$ at the same point $ Q$ ( instead of $ K$ in your wording ), as the harmonic conjugate of $ P,$ with respect to the points $ O\equiv AC\cap BD$ and $ I\equiv EF\cap PO$ ?

Could you please, share with us your idea, giving some more details?

Best regards, Kostas Vittas.

Dear vittasko,
I think it is not difficult. We know that if M=BD∩EF, then (BDOM)=-1, the pencil D'(BODM) is also harmonic. So by intersecting it with the line (POI), it follows that (OIPQ)=-1 where Q=BD'∩(POI). Similarly, AC',CA',BD',DB' all pass through Q as the harmonic conjugate of P wrt O and I.
Note : When OP is parallel to EF, M=∞ and Q is symmetric to P with respect to O.

Really, it is very easy with the argument of harmonic conjugation you mentioned, dear [b][size=100]mr.danh[/size][/b].

The solution I have in mind, it is based on the [b][size=100]Desarques theorem[/size][/b] and although it is not a better approach, however it is useful ( I think ) to be presented here, as an alternative one.

$ \bullet$ So $ ($ schema [b][size=100]t=210603(a)[/size][/b] $ ),$ we can say that the triangles $ \bigtriangleup OBC,\ \bigtriangleup QB'C',$ where $ O\equiv AC\cap BD$ and $ Q\equiv AC'\cap B'D,$ are perspective, because of the points $ A\equiv OC\cap QC',\ D\equiv OB\cap QB'$ and $ F\equiv BC\cap B'C',$ are collinear.

Hence, we conclude that the line segments $ BB',\ OQ,\ CC',$ are concurrent at one point and because of $ P\equiv BB'\cap CC',$ we have that $ AC'\cap B'D\equiv Q\in OP$ $ ,(1)$

$ \bullet$ Similarly, the triangles $ \bigtriangleup OAB,\ \bigtriangleup Q'A'B',$ where $ Q'\equiv A'C\cap B'D,$ are perspective, because of the points $ C\equiv OA\cap Q'A',\ D\equiv OB\cap Q'B'$ and $ E\equiv AB\cap A'B',$ are collinear and so, we have that the line segments $ AA',\ BB',\ OQ',$ are concurrent at one point and because of $ P\equiv AA'\cap BB',$ $ \Longrightarrow$ $ AC'\cap B'D\equiv Q'\in OP$ $ ,(2)$

From $ (1),$ $ (2),$ $ \Longrightarrow$ $ Q'\equiv Q$ and so, we have that $ AC'\cap A'C\cap B'D\equiv Q\in OP.$

By the same way, we can prove that $ AC'\cap A'C\cap BD'\equiv Q,$ and the proof is completed.

Best regards, Kostas Vittas.


are there 24? I'm not sure.. u can use that middle line to count the the number on one side, then multiply by two, then count the triangles that go across the middle line

This has been posted somewhere in this forum recently.

[quote="slamchess"][quote="Quevvy"][quote="slamchess"]how do you count these? And figure it out?[/quote]
I would suggest counting the number of triangles made of 1 triangles, then 2 and so on.

[hide]Using this, there are 8 triangles made up of 1.
5 triangles made of 2.
6 triangles made of 3.
2 triangles made of 4.
2 triangles made of 5.
0 triangles made of 6.
0 triangles made of 7.
1 triangle made of 8.

I get 24....[/hide]

$\lfloor\left(\sqrt{\lceil5^{2}\rceil}*\lfloor(\sqrt{10})^{4}\rfloor\right)-1\rfloor^{th}\text{post}!$[/quote]

ur right, but is there an easier way?[/quote]
There's probably not a formula because the triangles are always different so I would recommend doing it that way so that you don't miss any.


geometry
parallelogram
angle bisector
geometry unsolved

ABCD is a parallelogram which x is the interior angle bisector of angle BAD. The line x intersect DC at F and intersect BC at G. If E is the circumcentre of triangle CFG, prove that BECD is a cyclic quadrilateral.

Thanks.


Princeton college Duke AMC AIME USA(J)MO USAMO

Let either me or mysmartmouth know if there are any that should be added to this list. [b]I mainly just copied this from last year so a lot of it's tentative.[/b]

(T = Tentative)

[b][u]November 2008[/u][/b]
November 8th (Saturday) - [url=http://www.math.uga.edu/mathmeet.html/]UGA Math Tournament[/url]
November 18th (Tuesday) - [url=http://www.unl.edu/amc/]AMC 8[/url]

? - [url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State Part I[/url] (L)
? - [url=http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]Princeton University MathContest[/url]
? - [url=http://www.math.duke.edu/dumu/contests/2006/index.html]Duke Math Meet[/url] (T)

[b][u]December 2008[/u][/b]

[b][u]January 2009[/u][/b]
? - Rockdale Mathematics Competition (T)

[b][u]February 2009[/u][/b]
February 10th (Tuesday) - [url=http://www.unl.edu/amc/]AMC 10/12 A[/url]
February 21st (Saturday) - [url=http://math.cofc.edu/MATHMEET/]College of Charleston Math Meet[/url]
February 25th (Wednesday) - [url=http://www.unl.edu/amc/]AMC 10/12 B[/url]

? - [url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State Part II[/url] (L)

[b][u]March 2009[/u][/b]
March 17th (Tuesday) - [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml]AIME[/url]

[b][u]April 2009[/u][/b]
April 1st (Wednesday) - [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml]Alternate AIME[/url]
April 28th - [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml]USAMO Day 1[/url]
April 29th - [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml]USAMO Day 2[/url]

? - Walton MathFest (T)

[b][u]May 2009[/u][/b]
End of May/Early June? - [url=http://www.arml.com/]ARML[/url]

[u][b]Unknowns:[/b][/u]
[url=http://www.math.sc.edu/contest/]USC Columbia[/url]
USC Upstate
[url=http://math.furman.edu/tournament/tournament.html]Furman[/url]
[url=http://www.coastal.edu/math/mathcontest/]Coastal Carolina[/url]
USC Aiken?
Winthrop?

[u][b]Potential Tournaments:[/b][/u]
[url=http://www.gacs.pvt.k12.ga.us/facstaff/sbaggett/math_tournament.htm]GACS[/url]
[url=http://www.woodward.edu/academics/us/math/mathteam_sjv_tournament.asp]Woodward JV[/url]
[url=http://home.comcast.net/%7Edrgarnerjr/RMC.htm]Rockdale[/url]
[url=http://www2.fultonschools.org/teacher/pricea2/MathVarsity/MVarsityMilTourn.html]Milton Math Tournament[/url]
[url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State[/url]
[url=http://www.math.gatech.edu/%7Epme/hsmc/]Georgia Tech[/url]
[url=http://web.mit.edu/hmmt/www/]Harvard MIT Math Tournament[/url]

The dates of local Math Contests for 2008-2009:

Sat, Oct. 25th - Columbia College Math Day for Girls (Columbia)
Sat, Nov. 15th - USC Upstate Math Contest (Spartanburg)
Sat, Dec. 6th - USC Math Contest (Columbia)
Thurs, Jan. 22nd - SCISA Math Meet (Orangeburg)
Tues, Feb. 10th - AMC-10/12A
Sat, Feb. 21st - College of Charleston Math Meet (Charleston)
Wed, Feb 25th - AMC-10/12B
Tue, Mar. 17th - AIME I
Sat, Mar. 21st - Furman Math Tournament (Greenville)
Wed, April 1st - AIME II
TBD (April) - Coastal Carolina Math Contest (Conway)
Fri, April 17th - Clemson Calculus Challenge (Clemson)
Tue & Wed, April 28th & 29th - USAMO

[u]Columbia Math Circle Dates:[/u]
Sept. 27th, Oct. 18th, Nov. 8th, Dec. 13th

[u]Teams being organized by the Columbia Math Circle:[/u]
Sat, Nov 8th	– ARML Power Contest
Sat, Nov. 15th 	– Duke University Math Meet
Sat, Jan. 31st 	– Princeton University Math Contest
February 7-22	– ARML Power Contest
Sat, May 30th 	– ARML (All-State Math) Competition

And then there are contests at the Citadel and Wando High School, as well as the USAMTS, Mandelbrot, I-Test, and Math League, none of which I know the dates for.

The results of each of these contests will be used to select members of the SC ARML Team. If you'd like to be considered for the team, it is highly advised that you participate in at least 2 or 3 contests (especially the AMC).

USAMTS

Round 1 due date: October 14, 2008
Round 2 due date: November 24, 2008
Round 3 due date: January 12, 2009
Round 4 due date: March 9, 2009

Just a reminder that the first round of the [url=http://www.usamts.org/Problems/U_Problems.php]USAMTS[/url] is due Tuesday, October 14th (same day as the PSAT by the way), and the next session of the Columbia Area Math Circle will be Saturday, October 18th.

Is the Columbia Math Circle sending a team to Princeton?

Not to my knowledge, mainly because of the cost. If that issue can be overcome, and there is strong student interest, then it is a possibility.


trigonometry
function
limit
calculus
calculus computations

$ f \equal{} \sin(\frac{1}{x})$

Is f continuous at x = 0?

No.

For a function to be continuous at a point $ a$, it must satisfy $ \lim_{x \to a} f(x) \equal{} f(a)$. In this case, $ f(a)$ doesn't even exist.

Can you find a sequence, say $ x_{n}$, that converges to 0 with $ f(x_{n})$ not converging to $ f(0)$? If your answer is yes, then f is not continuous.

Following akech suggestion, you can try

$ a_n\equal{}\frac{1}{2n\pi}$ and $ b_n\equal{}\frac{1}{2n\pi\plus{}\frac{\pi}{2}}$

where both $ a_n\rightarrow 0$ and $ b_n\rightarrow 0$

but what can you say about $ f(a_n)$ and $ f(b_n)$ ? :)

Or the one that is commonly used: $ \frac{1}{n}$


search
number theory unsolved
number theory

Let 2n-1 integers be given.
Show that you can choose n of them whose sum is divisible by n !

Lemma: if 2p-1 integers are given, with p prime, then there is a p-element subset with sum divisible by p.

To prove the final result, it is sufficient to prove the following assertion: If the result holds for two numbers m and n, it holds for their product mn.

We have a set of 2mn - 1 = (m-1) + m(2n-1) integers. We can find 2n-1 disjoint sets of m elements, the sum of each set divisible by m. Let us call their sums [tex]mA_1, mA_2, ... mA_{2n-1}[/tex]. By our lemma, we can find n of the numbers [tex]A_1, A_2, ... A_{2n-1}[/tex] whose sum is divisible by n. Therefore, we have found a set of mn elements whose sum is divisible by mn.

I haven't figured out how to prove the lemma yet, maybe it's the harder part of the problem  :P

There are proofs of that lemma here on the forum. Search for "Erdos-Ginzburg-Zif".

[quote="pbornsztein"]Or look at the polycopié on Animath's page Olivier.

Pierre.[/quote]

Could you give a link please.

Here :
http://www.animath.fr/cours/arithm.pdf

pb #157.

Pierre.


the equation is ( d^2 + d = 2t )

i have to get an equation in which i plug in t and get d. i need d to be all by itself, any suggestions??

[img]http://img199.imageshack.us/img199/1035/untitled16xo.png[/img]

By completing the square we get that $d^2 + d = (d + \frac{1}{2})^2 - \frac{1}{4}$. I think that answers your question at least partly.

edit: to clarify you get that either $d = -\frac{1}{2} + (2t + \frac{1}{4})$ OR  $d = -\frac{1}{2} - (2t + \frac{1}{4})$. There is no way to explicitly state the whole equation with d as the 'subject' in only one equation.

Or you could set up a quadratic equation
$d^2+d-2t=0$
then solve. :)

or you could just use the quadratic formula


calculus
integration
algebra
polynomial
algebra unsolved

hello,substituting $ y\equal{}x^3\minus{}3x$ in the second equation and $ z\equal{}(x^3\minus{}3x)^3\minus{}3(x^3\minus{}3x)$ in the third equaton we get $ ((x^3\minus{}3x)^3\minus{}3(x^3\minus{}3x))^3\minus{}3((x^3\minus{}3x)^2\minus{}3(x^3\minus{}3x))\equal{}x$,simplifying and factoring we have to solve
$ x(x\minus{}2)(x\plus{}2)(x^6\minus{}7x^4\plus{}14x^2\minus{}7)(x^3\minus{}x^2\minus{}2x\plus{}1)(x^3\plus{}x^2\minus{}2x\minus{}1)(x^6\plus{}x^5\minus{}5x^4\minus{}4x^3\plus{}6x^2\plus{}3x\minus{}1)(x^6\minus{}x^5\minus{}5x^4\plus{}4x^3\plus{}6x^2\minus{}3x\minus{}1)\equal{}0$
Sonnhard.

i guess what sonnhard has done is not useful

what i had done is find out the integral solns and then look for rationals and irrationals

here is what i did..............

first i wud like to find out integral solns if any other than zero

now we get x^2 - 3 = y/x
                 y^2 - 3 = z/y
                 z^2 - 3 = x/z

so we conclude that (x^2 - 3)(y^2 - 3)(z^2 - 3)=1            .......i.

from here since x,y and z are integers then so r x^2 - 3 and so on

now since product of three integers =1

then all of the three have to be 1

so we have x^2 - 3 =1 and so on so from here we conclude that x=y=z=2
or
x=y=z=(-2)     are the only integral solns possible

now we take the case when some of them can be zero or all of them zero

now let x=0 then we have y=0 and similarly z=0
and so on

so from here we conclude that no single of them can be zero at a time
and other two integers or rationals so all have to be zero when one of
them is zero

hence we have x=y=z=0 as one more soln

now we r left to finding all the rational and irrational solns of this
eqn............

so from here on

i have tried to get

the value of a,b,c

where

cyclic [x]= a
cyclic [xy]=b
xyz=c

so that i can make a polynomial having x,y and z as its roots

and from the relation .i.

i have

found that

(c+3a+1)(c+3a-1)=3[(b+3)^2]

and using the fact that cyclic {[x]^3} = 4 cyclic{x}

we have

a(a+2)(a-2)=3(ab-c)

but not able to find how to proceed as the questions asks for all the REAL SOLNS AND NOT INTEGRAL SOLNS
 :)

[quote="Jaswinder"]Find all real solutions to the following system of equations
x^3 - 3x = y
y^3 - 3y = z
z^3 - 3z = x

 :)[/quote]
Uhm
You can see $ x,y,z \in [\minus{}2;2]$
Thus you can substitute $ x\equal{}2cos a$ hence $ y\equal{}2cos (2a)$ and $ z\equal{}2cos (4a)$
The last is using the third equation to see $ cos a\equal{}cos (16a)$
Yoou can finish This solution ,I think so

t0rajir0u=10000 posts, Loup Blanc=600 posts.
I feel very small.

Hi Allnames,
good idea but $ y\equal{}2cos(3a),z\equal{}2cos(9a),x\equal{}2cos(27a)$.

[quote="loup blanc"]t0rajir0u=10000 posts, Loup Blanc=600 posts.
I feel very small.

Hi Allnames,
good idea but $ y \equal{} 2cos(3a),z \equal{} 2cos(9a),x \equal{} 2cos(27a)$.[/quote]
VERY SORRY  :blush:
so we get $ a\equal{}\frac{k\pi}{13}$ or $ a\equal{}\frac{k\pi}{14}$
Thus we have the values of $ x,y,z$


$ a^2\plus{}b^2\plus{}c^2\equal{}3$ and $ a,b,c >0$.Prove that !
$ 9 \ge a^3\plus{}b^3\plus{}c^3\plus{}2ab\plus{}2bc\plus{}2ac$

[quote="bigbang195"]$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ and $ a,b,c > 0$.Prove that !
$ 9 \ge a^3 \plus{} b^3 \plus{} c^3 \plus{} 2ab \plus{} 2bc \plus{} 2ac$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, your inequality is equivalent to $ w^3\leq f(u,v^2).$
But, $ w^3$ gets a maximal value, when two numbers from $ \{a,b,c\}$ are equal.
In this case after homogenization and after substitution $ b\equal{}c\equal{}1$ we obtain:
$ (a\minus{}1)^2(3a^4\minus{}6a^3\plus{}14a^2\minus{}12a\plus{}10)\geq0,$ which is true.

[quote="bigbang195"]$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ and $ a,b,c > 0$.Prove that !
$ 9 \ge a^3 \plus{} b^3 \plus{} c^3 \plus{} 2ab \plus{} 2bc \plus{} 2ac$[/quote]

My solution  :)
 $ 9 \ge {a^3} \plus{} {b^3} \plus{} {c^3} \plus{} 2ab \plus{} 2bc \plus{} 2ac$
$ \Leftrightarrow 18 \ge \left( {2{a^3} \plus{} 2{b^3} \plus{} 2{c^3}} \right) \plus{} 2\left( {2ab \plus{} 2bc \plus{} 2ac} \right)$

From AM - GM inequality
$ {a^4}\plus{}{a^2} \ge 2{a^3},{b^4} \plus{} {b^2} \ge 2{b^3},{c^4} \plus{}{c^2} \ge 2{c^3}$
$ {a^2}{b^2} \plus{} 1 \ge 2ab,{b^2}{c^2} \plus{} 1 \ge 2bc,{c^2}{a^2} \plus{} 1 \ge 2ca$

Therefore,
$ \left( {{a^4}\plus{}{b^4} \plus{} {c^4}} \right)\plus{}\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)\plus{}2\left( {{a^2}{b^2}\plus{}{b^2}{c^2}\plus{}{c^2}{a^2} \plus{} 3} \right) \ge \left( {2{a^3}\plus{}2{b^3} \plus{} 2{c^3}} \right)\plus{}2\left( {2ab \plus{} 2bc \plus{} 2ac} \right)$
And,
$ \left( {{a^4} \plus{} {b^4} \plus{} {c^4}} \right) \plus{} \left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right) \plus{} 2\left( {{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2} \plus{} 3} \right)$
$ \equal{} {\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)^2} \plus{} {a^2} \plus{} {b^2} \plus{} {c^2} \plus{} 6 \equal{} 9 \plus{} 3 \plus{} 6 \equal{} 18$

The proof is completed. The equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$