Let m,n be positive integer and S in {1,2,..,2 m .n}
      |S|=(2 m -1)n+1
Prove that :exist a_0<a_1<...<a_m  \in S s.t :
        a_i is divisible by a_(i-1) for all 1 \leq i \leq m

Each number in {1,2,...,2 m n} has the form 2 a b (1), where a,b  \geq  0 and b is odd.

Let's count how many times a number b may occur in the decomposition of the elements of S.
For p = 1,2,...,m-1 there are exactly 2 p-1 n odd numbers in {2 p n + 1, 2 p n + 2,...,2 p+1 n}. Each of them may occur in S at most m-p times (in the form b, 2b, ..., 2 m-p-1 b ).
Then there are at most N =  \sum {1 \leq p \leq m-1} (m-p)n2 p-1 numbers for which in (1) b is an odd number  \geq  2n.

But N = N_1 - N_2 where :
N_1 = \sum {1 \leq p \leq m-1} mn2 p-1  = mn(2 m-1  - 1).
and N_2 = \sum {1 \leq p \leq m-1} np2 p-1.
Let f(x) = 1 + x + .... + x m-1  = (x m  - 1)/(x-1).
Then : f'(x) =  \sum {1 \leq p \leq m-1} px p-1  = ( (m-1)x m  - mx m-1  + 1)/(x-1) 2 .
Thus N_2 = nf'(2) = n(m2 m-1  - 2 m  + 1).

It follows that N = n2 m  - mn - n.
Then, there are at least |S| - N = mn + 1 elements in S for which the decomposition (1) use odd numbers b which are not greater than 2n. By the pigeon-hole principle, there are at least m+1 elements of S which are sharing the same odd number b, say x_i = 2 a_i b where a_0 < a_1 < ... < a_m. Then x_0 < x_1 < ... < x_m and x_i divides x_(i+1) for i = 0,...,m-1, and we are done.

Pierre.


Here are two vector space questions that seem intuitively obvious but I want a more conceptual understanding.

1. Is 4x - 3y = 7 a vector space? Answer: no.
2. Is 2x + y = 0 a vector space? Answer: yes.

My book explains that the answer to #1 is no because you can find two x- and y-values s.t.
$\\ 4x_{1}-3y_{1}= 7 \\ 4x_{2}-3y_{2}= 7$

And since the sum of these two equations $= 14 \not=7$, 4x - 3y = 7 is not a vector space since it is not closed under addition.

My book explains that the answer to #2 is yes because it is closed under addition, scalar multiplication, and satisfies the other properties.

WHAT I DON'T GET:

When one asks "does 4x - 3y = 7 constitute a vector space", I assume that one means "does V, the set of vectors (x, y) |  4x - 3y = 7 constitute a vector space". So when adding 4x - 3y = 7 to itself as an example, I don't see how you are proving that it cannot be a vector space. I would think that the proof would look more like:

Since the zero vector is not contained in V, as $4(0)-3(0) \not=7$, V is not a vector space.

Can someone tell me if my proof is correct and what the author meant when adding 4x - 3y = 7 to itself to disprove closure under addition? I guess the reason I don't fully understand adding 4x - 3y = 7 to itself is because you're adding 7 to 7, which is not the sum of two vectors, but the sum of two scalars.

[quote="oaklandsun"]Here are two vector space questions that seem intuitively obvious but I want a more conceptual understanding.

1. Is 4x - 3y = 7 a vector space? Answer: no.
2. Is 2x + y = 0 a vector space? Answer: yes.
[/quote]
What is sum of two vector in your problem?

The book gives the sum as

$4(x_{1}+x_{2})-3(y_{1}+y_{2}) = 14 \not= 7$

and concludes that $4x-3y = 7$ does not constitute a vector space since it is not closed under addition.

Suppose $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are in the set; that is, $4x_{1}-3y_{1}=7$ and $4x_{2}-3y_{2}=7$. Is $(x_{1}+x_{2},y_{1}+y_{2})$ in the set? That's what adding the equations is about; we find that this sum is never in the set.


function
real analysis
real analysis unsolved

let $ \lambda\in\mathbb{R}$.
find all sequences $ (a_n)$ of nonnegative reals
such that for every  $ n\in\mathbb{N}$ holds
\[ n^{\lambda}\sum_{k\equal{}1}^{\infty} a_{kn} \equal{}\sum_{k\equal{}1}^{\infty}a_k\]

I assume that we are talking about converging series only. When $ \lambda\le 0$, this is clearly impossible. When $ \lambda>1$, it can be easily reduced to a rather well-known problem. Just consider $ b_k\equal{}a_k\minus{}\frac c{k^\lambda}$ with $ c>0$ chosen so that $ \sum_k b_k\equal{}0$. Then we get $ \sum_kb_{nk}\equal{}0$ for all $ n$ and the series converges absolutely whence (this is not trivial but has been discussed many times) all $ b_k\equal{}0$.

Thus, it remains only to consider the case $ 0<\lambda\le 1$.

It turns out that if we look not at just the result but at the solution of the problem I refered to, we will get
\[ a_1 \equal{} S\prod_{p < n, p\text{ prime}}(1 \minus{} p^{ \minus{} \gamma}) \plus{} o(1) \text{ as }n\to\infty\]
where $ S \equal{} \sum_k a_k$. Since that product tends to $ 0$, we get $ a_1 \equal{} 0$. Passing to the subseries $ \sum_{k\ge 1}a_{kn}$, we conclude that all $ a_n \equal{} 0$.

i don't understand the $ 0<\lambda\le 1$ case ... could you explain this?

Do you know the solution of the $ \sum_k b_{kn}\equal{}0$ problem I referred to?

I understand what you have written in your first post - the case $ \lambda>1$ where you used numbers $ b_{kn}$.
But harmonic series for $ 0\le\lambda<1$ are divergent and I can't use the same idea for this case

I'll take this as "No", so let's go over the full thing.

Let $ \mu(n)$ be the Mobius function ($ 0$ if $ n$ is not square-free and $ \minus{}1$ to the power of the number of prime divisors of $ n$ otherwise). Recall that $ \sum_{d|n}\mu(d)\equal{}\begin{cases}1,&n\equal{}1\\0,&n>1\end{cases}$. Now let $ D$ be the product of the first $ m$ primes $ p_1,\dots,p_m$. Write
\[ \begin{aligned}
S\prod_{k\equal{}1}^m(1\minus{}p_k^{\minus{}\gamma})& \equal{}\sum_{d|D}\mu(d)d^{\minus{}\gamma}S
\equal{}\sum_{d|D}\mu(d)\sum_{n}a_{nd}\\ & \equal{}\sum_n a_n\sum_{d|(n,D)}\mu(d)
\equal{}\sum_{n: (n,D)\equal{}1}a_n\equal{}a_1\plus{}o(1)
\end{aligned}\]
as $ m\to\infty$. The rest should be clear.


inequalities
quadratics
function
inequalities proposed

$ x_1 \le x_2 \le x_3 \le ..x_n \le y_1 \le y_2 \le ...y_n$

$ Prove$   $ that:$
$ (\sum_{i\equal{}1}^n x_i\plus{}y_i)^2\ge4n(\sum_{i\equal{}1}^n xiyi)$

I don't think it is correct inequality.
Let $ n\equal{}2, x_1\equal{}y_1\equal{} 0$,
$ (x_2\plus{}y_2)^2 \geq 8x_2y_2$. but $ x_2\equal{}y_2\equal{}k$, $ 0\geq k^2$.

If $ x_{1}\equal{}y_{1}$ then we have $ (x_{2}\plus{}y_{2})^2\ge0$ which is true obvious.

[quote="NextPeace"]I don't think it is correct inequality.
Let $ n \equal{} 2, x_1 \equal{} y_1 \equal{} 0$,
$ (x_2 \plus{} y_2)^2 \geq 8x_2y_2$. but $ x_2 \equal{} y_2 \equal{} k$, $ 0\geq k^2$.[/quote]
If n=2 and $ x_1 \equal{} y_1 \equal{} 0$ then $ x_2 \equal{} 0$ because $ x_1 \le x_2 \le y_1$ then the inequality becomes $ y_2^2\ge0$

Let $ x_i \equal{} X_1 \plus{} \cdots \plus{} X_i$ and $ x_n \equal{} X$, and let $ y_i \equal{} Y_1 \plus{} \cdots \plus{} Y_i \plus{} X$. All the $ X_i,Y_i$ are nonnegative by assumption. In these terms the inequality can be rewritten
\[ \left(nY \plus{} \sum_i (i \minus{} 1)(X_i \minus{} Y_i)\right)^2 \minus{} 4 \sum_{i < j} (X_i \plus{} \cdots \plus{} X_j)(Y_i \plus{} \cdots \plus{} Y_j) \geq 0.
\]
For all $ i$, this as a quadratic function of $ X_i$ is minimized when
\[ (i \minus{} 1)X \equal{} 4(n \minus{} 1)(y_i \plus{} \cdots \plus{} y_n) \minus{} 3(i \minus{} 1)Y
\]
(*) or when $ X_i \equal{} 0$.

Since all the variables walk along the nonnegatives, a compact set, by Weierstrass's theorem at some point there is global minimum. But at this point all the equalities (*) must hold or otherwise there are zeros, for if that was not do we could substitute another nonnegative $ X_i$ (the solution of (*) or zero) that makes an even smaller value, contradicting the global minimaness of the point considered. So we only need to consider it all when [b]for each $ X_i$ either equality (*) is true or $ X_i \equal{} 0$.[/b] But (*) cannot be satisfied by more than one $ X_i$ and unless all the $ X_i$ are zero (in which the solution follows) there is exactly one nonzero $ X_i$. In this case we can express $ X_i$ through the $ Y_i$ by (*) and finish the problem by any means.

Let the $ x_i$ be fixed and the sum $ y_1 \plus{} \cdots \plus{} y_n$ be fixed. Consider $ y_1$.

If $ y_1 > x_n$, then $ y_1^{\ast} \equal{} x_n$ and $ y_n^{\ast} \equal{} y_n \plus{} y_1 \minus{} x_n$ [color=olive]and the rest the same[/color] is a new sequence satisfying all [color=olive]that is required[/color] and for which the left side is the same and the right side is greater. Apply the same to $ y_2$, etc until you get $ x_n \equal{} y_1 \equal{} \cdots \equal{} y_{n \minus{} 1}$. Now do the same to get $ x_{n \minus{} 1} \equal{} x_n$ (substract from $ x_1$ and then from $ x_2$ if needed, etc) until you get
\[ 0 \equal{} x_1 \equal{} \cdots \equal{} x_i < x_{i \plus{} 1}\leq x_{i \plus{} 2} \equal{} \cdots \equal{} y_{n \minus{} 1}\leq y_n
\]
($ i$ may be zero or $ n$). Now the inequality is $ (a \plus{} kb \plus{} c)^2 \geq 4n(ab \plus{} (k \minus{} n)b^2 \plus{} bc)$ which is equivalent to $ (a \plus{} c \plus{} b (k \minus{} 2 n))^2\geq 0$.

[color=darkred][b]EDIT:[/b][/color] The variables were assumed to be nonnegative.

Nice, but the numbers can be negative as well. This seems to be a really hard problem.

[hide="Solution"]
Let $ f(t) \equal{} \sum_{i \equal{} 1}^n (t \minus{} x_i)(t \minus{} y_i)$
because of $ x_1 \le x_2 \le x_3 \le ..x_n \le y_1 \le y_2 \le ...y_n \equal{} > f(x_n) \le 0, f(y_1) \le 0$ and $ f(x_1) \ge 0, f(y_n)\ge0$ so there's such real number $ t_0$ that $ f(t_0) \equal{} 0$
$ f(t_0) \equal{} \sum_{i \equal{} 1}^n (t_0 \minus{} x_i)(t_0 \minus{} y_i) \equal{} nt^2 \minus{} (\sum_{i \equal{} 1}^n x_i \plus{} y_i)t \plus{} \sum_{i \equal{} 1}^n x_iy_i \equal{} 0$
This must have at least one real solution so $ D \ge 0$
$ (\sum_{i \equal{} 1}^n x_i \plus{} y_i)^2 \minus{} 4n(\sum_{i \equal{} 1}^n x_iy_i) \ge 0$
[/hide]


In how many different ways can four students stand in a straight line if two of the students refuse to stand next to each other?

[hide]There are $ 24$ different ways to put them, but we must subtract out the unwanted results. There are $3!(2)=6(2)=12$ unwanted results, so $ \boxed{12}$ is our answer.[/hide]

the answer is actually 12

your solution was great, but you miscounted the number of unwanted results

there are three places where the two kids can be placed so that they are next to each other:
11xx, x11x, xx11
(where 1's are the kids in question and x's are the other kids)

there are two ways to arrange the two kids who do not want to stand next to each other

and there are two ways to arrange the other kids

so the total number of unwanted results is $ 3*2*2\equal{}12$

$ 24\minus{}12\equal{}12$

Would doing 4!/2! still be correct? It gives the right answer, but is it a correct way of solving the problem?

[hide=Solution]There are only a three ways this could happen
\begin{align*}
oxox\\oxxo\\xoxo
\end{align*}
In each of these cases, there are $2!\cdot2!=4$ ways to arrange these students.

The answer is $3\cdot4=\boxed{12}$.

[hide=Solution]If there were no restrictions, the total number of ways is $4! = 24$ ways. We subtract the number of arrangements we don't want which is $3!\cdot2! = 6\cdot2 = 12$. So the answer is $24-12=\boxed{12}$.[/hide]

[quote=vallon22]
there are two ways to arrange the two kids who do not want to stand next to each other

and there are two ways to arrange the other kids
[/quote]
I don't get this part of everyone's solutions. Can someone help?
[hide=Here was my incorrect sol btw]24 ways for the kids to stand non restricted, and there are 3 ways for these kids that don't like each other to stand together:
00xx, x00x, xx00 (0 being the kids in question and x being other kids)
Order matters though right? So 3!=6 ways. 24-6=18 which is wrong [/hide]

[quote=AlwaysOnce][hide=Here was my incorrect sol btw]24 ways for the kids to stand non restricted, and there are 3 ways for these kids that don't like each other to stand together:
00xx, x00x, xx00 (0 being the kids in question and x being other kids)
Order matters though right? So 3!=6 ways. 24-6=18 which is wrong [/hide][/quote]

[hide]There are 24 ways for the kids to stand non-restricted, and there are 3 ways for the kids that don't like each other to stand together, but if order matters, there would be 4 ways for the kids to fill in the positions 00 and xx (2 each, $2\cdot 2=4$), so $3\cdot 4 = 12$ and $24-12=12$


I am a student that goes to Sharon High School in Sharon, MA. This year, my school will be offering the AMC A but not the AMC B. I would really like to take the B. If I understand correctly, one may take the test at other schools.

Is there any school out there in the Norfolk region (or surround regions... i am willing to travel) that offer the AMC B and would have space for an extra student.

I prefer to take the AMC 10, but I will also be delighted to take the AMC 12.

Thanks a lot in advance
-Nandubetax

An update, I am willing to travel to high schools that are not in the Norfolk region.

Thanks in advanced

Well, many institutions of higher learning (colleges) offer the AMC B tests. [url=http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml]Here[/url] is the list. Look's like there's one college in Massachusetts which offers the tests.


If you can currently get around 3-5 on the AIME, the AIME problem series is probably just right for you.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76662

It says here that the 4-8 range is optimal...I guess 3-5 somewhat falls under that.

[quote="Karth"]I saw that for most of the classes in AoPS, there are "Are you ready?" tests, or pre-tests. What would the pre-test for AIME Problem Series be? I currently keep getting around 3 to 5 on the AIMEs from the Resources List. Will this be sufficient for me to benefit from the class? Thanks.[/quote]

If you're getting in the 3-5 range and have a strong handle on the essentials of high school math up through what's normally called Algebra II and Trigonometry, then the AIME problem series is likely a good one for you.

[quote="Karth"]But as filletwho said, would the 4-8 range be the BEST for AIME Problem Series? Or should I wait before taking the course? Which would be the best plan to maximize my AIME score with regards to AIME problem series?

Also, randomly, while taking the class, for "homework," is the homework previous AIMEs, or does the instructor put problems on the classroom forum after class?[/quote]

Think of that range as a general guide.  Perhaps that general guide includes some students every familiar with contest-taking, so the AIME Problem Series is best for them when they're getting 6-8 problems.  Perhaps there are students with all the curriculum and smarts, but far less contest-taking experience -- such students might be at essentially the same level of mathematical maturity while only getting 3-5 problems right (and for them, the course might do more to increase their scores by helping them overcome the experience gap).

Past AIME and AIME-level problems are posted on a private message board after each class.

[quote="PenguinIntegral"]Is there some information on structure, how many problems per class, etc? Will you group them by subject? Are there tests? What are the sources of the problems? How is a class done? A few examples, then problems? Are there lots of hard, really super difficult problems to have fun with?[/quote]

Each class covers 5-10 problems depending on the difficulty of the problems and concepts on any one day.  The problems are either very hard AMC problems, AIME problems, or problems at a similar difficulty level.  Around a dozen problems are posted on the message board after each class.

A 90 minute class focuses on two or three topics within a subject area.  Within each topic, problems are discusses that are progressively more difficult.

None of our problem series classes involve tests.

[quote="Karth"]Oh so as we do problems, will the instructor point out:

It is a good method to often do ___ when approaching a problem like this.

?[/quote]

Yes.

[quote="Karth"]Oh so as we do problems, will the instructor point out:

It is a good method to often do ___ when approaching a problem like this.

?[/quote]

The entire course is written with explanations as to how to set up each problem and how to look for clues as to how to follow through once you've identified all the important information in a problem.

Maybe you should go to a math jam and see what it's like.


search
inequalities unsolved
inequalities

[quote="nkht-tk14"][u][i]Problem 1[/i][/u] [i]Let a,b,c be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2$ =3
    Prove that
       $ 5(a \plus{} b \plus{} c) \plus{} \frac {3}{abc} \geq 18$[/i]
[u] Problem 2
   [/u]
  [i]  If a,b,c be positive real numbers such that [/i]$ abc \equal{} 1$
[i]prove that[/i] $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 7 \geq 5(a \plus{} b \plus{} c)$
        [i]  Argady , Nayel .... Can you help me?[/i]
    :roll:[/quote]
The first see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904
For the second I have not found the link. Try the Mixing Variables method. :wink:

Look [url=http://www.mathlinks.ro/viewtopic.php?search_id=824528242&t=128529]here[/url] for the second one  :wink:

These two problem could be proved easily using p q r method. :wink:


Without using the quadratic formula, show that the sum of the roots of ax^2 + bx +c is -b/a  and the product of the roots is c/a.

let the roots be $r_1$ and $r_2$
by the factor theorem, for all x,
$x^2+\frac{b}{a}x+\frac{c}{a}=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$

the only way these can be equal for all x is if the coefficients are the same, so

$r_1+r_2=-\frac{b}{a}$
and
$r_1r_2=\frac{c}{a}$

by the way, you should not post proofs in getting started...

[quote="Altheman"]by the way, you should not post proofs in getting started...[/quote] This is OK because it is simple proof problem and no special trick in particular.


geometry
circumcircle
incenter
geometry proposed

This problem gave me Sailor I think:

Let  $ABC$ be a triangle with circumcircle $C(O)$.
Let $F$ be a circle, which is tangent to $AB,AC$ and $C(O)$
The point of tangency to $C(O)$ is $D$.

Prove that $DI$ bisects angle $BDC$ (where $I$ is incenter of $ABC$).

http://www.mathlinks.ro/Forum/viewtopic.php?t=22925 .

  Darij

It follows from the property stated in the first post on this topic. A simple angle chase will show that $\angle DBI+\angle DCI=\frac{\pi-\angle BDC}2$. At the same time, we know, that $\angle NDI=\angle CDI$, so we get $\angle DCI=\angle DIB$, which means that $BDI,DCI$ are similar (which proves the assertion).


Let $ p$ be a prime number congruent to 3 modulo 4. Suppose that $ p|x^2 \plus{} y^2$ for some integer $ x$ and $ y$. Prove that $ p|x$ and $ p|y$.

[hide="Stuff"]So then $ x^2 \plus{} y^2 \equiv 0 \mod p$, $ (x \plus{} yi)(x \minus{} yi)\equiv 0 \mod p$, only when $ x\equiv y\equiv 0 \mod p$? Anywhere close? D= doesn't use $ p\equiv 3\mod 4$...and it doesn't make much sense to me in the first place...[/hide]

You would have to justify the use of complex numbers. The ring $ \mathbb{Z}[i]$ is usable, but not all primes stay prime in $ \mathbb{Z}[i]$: for example, $ 13\equal{}(3\plus{}2i)(3\minus{}2i)$. And of course 13 divides $ 3^2\plus{}2^2$ but neither 2 nor 3.

The simplest approach here would be to reduce the problem to showing $ t^2\equiv \minus{}1\pmod{p}$ has no solution when $ p\equiv 3\pmod{4}$.


If eight apples cost the same as four bananas, and two bananas cost the same as three cucumbers, how many cucumbers can Tyler buy for the price of 16 apples?

16 apples is the same as eight bananas, and eight bananas are the same as $ 4 \times 3 \equal{} 12$ cucumbers.


Generally the MATHCOUNTS Challenge problem posted here have been of the harder variety of MATHCOUNTS problems and a few topics have been discussed that are definitely beyond the scope of MATHCOUNTS problem solving.

I want to make this clear because I would like to encourage students who find these problems very challenging to sit in on the MATHCOUNTS Math Jams.  During the Math Jams the problems are more closely taylored to the actual MATHCOUNTS competition.  Usually we just take problems from the actual competition and alter them slightly.

Well you can take Advanced AMC8/MATHCOUNTS courses for success in states and nationals!!

It is a very good class ;)

^It is a good class I agree, but if you want further depth of a subject, take the intro courses

ok it's just that my parents don't like to pay for stuff like these classes


geometry
trapezoid
geometric transformation
vector
dilation
perpendicular bisector
geometry proposed

We have a trapezoid $ABCD$, and two squares $DAEF$ and $CBGH$. Construct the perpendicular bisector of $CD$. $M$ is the intersection of this perpendicular bisector and $FH$.
Prove $MF=MH$.

N ice problem.
Draw perpendicular  $FK$ to the line $CD$($K$ is on $CD$) and draw the perpendicular $DL$ to the line $BA$($L$ is on $AB$).We now prove that the triangles $FDK$ and $DAL$ are congruent. $AD=FD, \angle{FDK}=\angle{ADL}, \angle{DKF}=\angle{DLA}$ thus the triangles $FDK$ and $DAL$ are congruent. So $KD=DL$. In the same way if we draw perpendicular $HR$ to $DC$($R$ is on $DC$) we'll get that $CR=DL$. Denote the midpoint of $DC$ by $N$.We have that $KN=KD+DN=DL+DN=CR+CN=RN$, thus $N$ is the midpoint of the segment $KR$.On the other hand  $FK$ is parallel to $MN$ and parallel to $HR$ so $M$ is the midpoint of $FH$.
QED

Let $A(0,0)$,$B(0,1)$,$C(c,b)$,$D(a,b)$.We can easily obtain $F(a-b,a+b)$ and $H(c+b,b+2-c)$,so the midpoint of $FH$ is $M((a+c)/2,b+1+(a-c)/2 )$.The equation of the bisector is $x=(a+c)/2$,so M is a point of the bisector,and the proof is complete.

Through the translation of vector $\vec{DC}$ the square AEFD becomes A'E'F'C and through the translation of vector $\vec{CD}$ the square BGHC becomes the square B'G'H'D (EE'=FF'=GG'=HH'=CD, EE'||FF'||GG'||H'||CD and the quadrilateral FF'HH' is parallelogram). From a old problem results that the perpendicular lines on CD from the points C, D meet the lines FH', F'H respectively in them middlepoints U, V respectively (UV=FF'=HH', UV||FF'||HH')Thus, the lines FH, F'H', UV have the same middlepoint. a.s.o.
Remark. If C=D then the problem is wellknown. The author is resourcefull man, even lousy man (excuse me for the word "lousy") because he commited a dilate of the point C (or D) to the segment CD parallel with AB ! It is great (if such it is !). I commited the reversetrip. The problem is nicely, indeed.I was enraptured and I congretulate you,.


The increasing sequence 1,3,4,9,10,12,13,... consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Of the first googol (10100) terms in the sequence, how many are powers of 3?

Is the answer 333?

Please lock! This is from the MathCamp Qualifying Test!

Problems from an ongoing qualifying test should not be posted on the forum.  Topic locked.


Simple the following expression:
$T=\sin1^\circ.\sin2^\circ.sin3^\circ \dots\sin89^\circ\sin90^\circ$


AMC
USA(J)MO
USAMO
induction
number theory
divisibility tests

This is a problem from Art of Problem Solving Volume 1. It is probem #105 in chapter 5.   "For each of n=84 and n=88, find the smallest integer multiple of n whose base 10 representation consists entirely of 6s and 7s."  My teacher couldnt figure out he solution to the problem.  Icouldnt either.

Well, let me tell you how I did it for 84.  I don't know if there's a better way.  [hide]The answer is 76776.  The way I did it is this:  first, the units digit has to match.  We see that this means our number has the form 84*4, 84*9, 84*14, ..., 84*(5n - 1), ...

So, we plug the first few values for n into that formula to get 336, 756, 1176, 1596, 2016, 2436, 2856, 3276, 3696, 4116 and we see that the tens digit goes {3, 5, 7, 9, 1, 3, 5, ...}.  So, we need our tens digit to be 7.  This means that n can be 3, or 8, or 13, or (5m - 2).  So our number has to have the form 84*(5*(5m - 2) - 1) = 84*(25m - 11).  Now, we plug in the first few values for m to get 1176, 3276, 5376, 7476, 9576, 11676, 13776, 15876, 17976, 20076, and we see that the hundreds digit goes {1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, ...}.  This means that m can be either (10j - 4) or (10k - 3).  This means our original number has the form 84*(25*(10j - 4) - 11) = 84*(250j - 111) or 84*(25*(10k - 3) - 11) = 84*(250k - 86).  Plugging in the first few values for j gives us 11676, 32676, 53676, 74676, 95676, and plugging in the first few values for k gives us 13776, 34776, 55776, 76776, 97776, so 76776 is our answer.[/hide]  The same process works just fine for 88.  Maybe even a little bit faster.

Hah.. that was one of the questions on the Bergen Academies (AAST) entrance exam last year...

Okay, I'll admit that I peeked at the AoPS solution for the number 84, and it was nice.  (This problem came from USAMTS.)  Here's a hint in spoiler: [hide]Use divisibility by 4 to determine the last two digits.  Use divisibility by 3 to figure out how many 7's there probably are.  Test the resulting handful of combinations for divisibility by 7.[/hide]

Hmm ... I like my method better :)  Although using divisibility rules would have sped it up, I suppose.  And not having Mathematica to do the actual arithmetic would have slowed it down :)

Okay, hotshots, here's a problem that I just made up.  Let n be a positive integer that is not a multiple of 5.   Prove that there exists a multiple of n whose base 10 representation consists entirely of 6's and 7's.

I've seen things much like this (but usually easier) before, but we'll pretend that's not the case:

[hide]First, numbers also not divisible by 2.  Take a number b divisible by neither 2 nor 5.  Consider the sequence 1 mod b, 11 mod b, 111 mod b, 1111 mod b, ...

eventually by the pigeon-hole principle, two of these numbers are the same.  Take their difference and we get 11...1100...00 is divisible by b.  b has no 2's or 5's, so that means 11...11 is divisible by b, which means 66...66 is divisible by b.

I can reduce the even case to powers of 2:  assume we can do it for powers of 2.  For the same reasoning as above, any odd number divides some number of the form 100...00100...00100... ... ...001, where we have the same number of zeroes in each case.  We just take enough zeroes so that when me multiply this number by the multiple of the power of 2 that is all 7's and 6's, we don't get carrying and we don't have any leftover zeroes.  Refactoring this new number gives us a multiple of our power of two times some odd number, which corresponds to any even number which implies every non-multiple of 5 has a multiple composed of 6s and 7s.  So now I just need to do it for powers of 2 [/hide]

Looks good so far.  Your method of reducing the general case to powers of 2 is related to, but simpler than, the method I came up with.

Big hint for powers of 2:

[hide]

This is pretty much just like last year's USAMO #1: you can prove by induction that there is exactly one n-digit integer with all digits 6's and 7's that is divisible by 2^n.

[/hide]

Yep.  I hadn't seen that USAMO problem before, but you're right that it is practically the same problem.  One question:  As you said, induction shows that there is a unique n-digit number using 6's and 7's that is a multiple of 2n.  Is there a nice pattern for the 6's and 7's?  For example, if I were to try 250, is there an easy way to construct the 50 6's and 7's?  Or should I be satisfied with the inductive definition?  I tried the powers of 2 up to about 1024, but I didn't see a nice pattern.

Yeah, I had basically figured out what Alison just said but hadn't gotten the time to write it down.  It doesn't look to me like there is a nice pattern, but I don't honestly know.


calculus
integration
trigonometry
calculus computations

In the future, it would be nice if you gave your questions more enlightening titles.  If I come back here trying to find a particular one of the integrals you just posted, I'm going to have to look through as many as 7 posts to find it.  Thanks.

[quote="prithwijit"]Evaluate $\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^{2}}}{(1+x^{2})\sqrt{2+x^{2}}}dx$.[/quote]

Let $J: =\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^{2}}}{(1+x^{2})\sqrt{2+x^{2}}}dx$

And let $K: =\int_{0}^{1}\frac{x\tan^{-1}\left( \frac{x}{\sqrt{2+x^{2}}}\right) }{(3+x^{2})\sqrt{2+x^{2}}}dx$

Integrating J by parts with $u=\tan^{-1}\sqrt{2+x^{2}}$ and du=blah, dv=blah, dv=blah gives $J=\frac{\pi^{2}}{18}-K$ and I have shown to 60 decimal places that J/K=15.000000... hence I get $J=\frac{5\pi^{2}}{96}$ which agrees with out-right numerical approx. of J.

Would someone prove that we indeed have $J=15k$ ? then the proof is solid.

Hi prithwijit. How did you get from $ I_{2}\equal{}$ a double integral to the products of integrals?

I just split the integrand in $ I_2$ into partial fractions.

Thanks prithwijit. (Go raibh mile maith agat.)

i think that fubini's theorem has also been applied there. no?


In each of three boxes are four straws with integer lengths 1cm through 4cm, inclusive. A straw is randomly selected from each box. What is the probability that the three straws chosen can make a triangle?

Any better way than listing?

[quote="chess64"]You [i]are[/i] using the triangle inequality, correct?[/quote]

Yeah.  :D

The best you can do is list them based on whatever restrictions you can find.

[hide="I would say..."]First side is 1: can have 1,1; can have 2,2; can have 3,3; can have 4,4.
First side is 2: can have 2,2 or 2,3; can have 3,3 or 3,4; can have 4,4.
First side is 3: can have 3,3 or 3,4; can have 4,4.
First side is 4: can have 4,4.

That would be 13 distinct triangles, four of which are equilateral, meaning one permutation; eight are isosceles, meaning three permutations; one is scalene, meaning six permutations. Therefore, $4\cdot 1+8\cdot 3+1\cdot 6=34$, so we have $\frac{34}{4^3}=\frac{17}{32}$.[/hide]

I'm not sure which is faster, though.

[quote="dakyru"]List the ones that won't work:

$(1,1,4) (1,2,4) (1,3,4) (2,2,4) (1,1,3) (1,2,3)(1,1,2)$

Add up the number of permutations of each:

$3 + 6 + 6 + 3 + 3 + 6 + 3 = 30$

The probability of it not forming a triangle is then $\frac{30}{64}$ and therefore the probability of a triangle being formed is $\frac{34}{64}$ or $\frac{17}{32}$.[/quote]

I tried both and definetly found the finding the ones that won't work faster. Jesusfreak197's way is interesting, but I'm not sure if it will actually be faster. (will you list and recognize which ones are isosceles and equilateral and add them up in 80 seconds?). Still, the listing isnt' too speedy.

[quote="236factorial"](will you list and recognize which ones are isosceles and equilateral and add them up in 80 seconds?)[/quote]

Quite possibly, but it depends on how you look at things. It took me longer to write it out, but I could have done it pretty quickly if I was just thinking it out in my head. Maybe the other way would have been even faster; I don't really know.


Please choose the choice that represents the state you are from.

Wrong section. This probably belongs in FF.

Um, as I said in the subtitle, I meant for the survey to be among FTWers...

obviously the person who made this put in a ftw forum, but it got moved

because FTW forum is for serious discussion of FTW like glitches or bugs.
I would think it would get moved to global though.


Hey I have a question about system requirements. This game says that it requires 256 Mb of RAM to run/install. But I just realized this computer I'm on only has 248 MB of RAM.

Is there any way I could still play the game (maybe just going a little slower), or is 256 Mb a hard, unavoidable requirement?

Machines that do not have enough RAM can experience problems as they use virtual memory and continuously write to disk. Be sure your PC meets the recommended system requirements.

-jorian

If your RAM is that close it shouldn't matter. I'm surprised that you have 248 MB RAM, as 256 is much more common. But 256 isn't a hard requirement, it's just a recommendation.

the RAM doesn't matter. It might make the game run a little bit slower, but heck, just set priority for it all the way up to max. That might mess up your other programs, but the game's worth more.


hmmm

[hide]

Multiply everything by 19!

Then you get  :Sigma: n = 2 - 9 of 19Cn

19C10 = 19C9 and so forth

so its 2^19/2 - 1 - 19 = 2^18 - 20

Divide that by 19! and you get something ugly

[/hide]

Yep, that's how I did it too.

Also using Maple I got it to equal:

[hide]

14563/6758061133824000

[/hide]


calculus
integration
inequalities
real analysis
real analysis unsolved

Find all continuous maps $f: [0,+\infty[ \rightarrow [0,+\infty[$ such that $\forall x\geq 0: f\left( x\right) \leq \int_{0}^{x}f\left( t\right) dt$

Note that we must have $f(0)=0.$

Let $g(x)=e^{-x}\int_{0}^{x}f(t)\,dt.$

Then $g(x)=0,\,g(x)\ge0$ for $x\ge0,$ and

$g'(x)=e^{-x}f(x)-e^{-x}\int_{0}^{x}f(t)\,dt$

$=e^{-x}\left[f(x)-\int_{0}^{x}f(t)\,dt\right]\le0.$

If $g(0)=0$ and $g'(x)\le0$ for $x\ge0,$ then $g(x)\le 0.$

Since we also have $g(x)\ge0,$ then we have that $g(x)=0$ for all $x\ge 0.$ It's a few short steps from there to $f(x)=0$ for all $x\ge0.$

Very nice.

@Dr kent - how did you think of bringing the exponential? that is the tool that solves the problem.

When you're looking at a differential or integral inequality, the solution to the corresponding differential or integral equation is usually very useful. In this case, the corresponding differential equation is $f'(x)=f(x)$, which has multiples of $e^{x}$ as solutions.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=proof&t=83256]Here[/url]'s a previous topic discussing the same problem.


$ 0 < x_0 < y_0$
$ x_{n \plus{} 1} \equal{} \sqrt {x_ny_n}$
$ y_{n \plus{} 1} \equal{} \frac {x_n \plus{} y_n}{2}$

why do the rows $ x_n$ and $ y_n$ have the same limit?....please help

Assume limit exists for both sequences (that too can be proved easily if you require). We need to show they are equal.
By definition of limit, when $ n\to \infty\implies y_n \equal{} y_{n \plus{} 1}$. We also have $ y_{n \plus{} 1} \equal{} \frac {x_n \plus{} y_n}{2}\implies y_{n} \equal{} \frac {x_n \plus{} y_n}{2}\implies x_n \equal{} y_n$
Hence when $ n\to \infty\implies x_n \equal{} y_n$

For reals $ 0 < b \le a$, show that

$ \frac {1}{8} \frac {(a \minus{} b)^2}{a} \le \frac {a \plus{} b}{2} \minus{} \sqrt {ab} \le \frac {1}{8} \frac {(a \minus{} b)^2}{b}$.

What does this tell you about $ \lim_{n \to \infty} |x_n \minus{} y_n|$?  What is the rate of convergence?


Three distinct numbers are chosen at random from the set [4,8,13,16,24,26,30,33,38,44,46,47,51,64,83]. What is the probability that none of the chosen numbers is odd?

Well first we can find the probability that you choose 3 even numbers. Since there are 11 even numbers and a total of 15 numbers the probability. 11!/(8!*3!) would be the total number of ways that you pick out 3 even numbers while 15!/(3!*12!) would be the number of ways to choose all the numbers. Then you divide the even by the total.

I used the combination formula to find the total number of ways.

You wrote the question wrong. The set should be: [4,8,13,16,24,26,30,33,38,44,46,47,51,64,[b]83[/b]] not 84.

There are 15 total numbers. Of those, 10 are even and 5 are odd. What is the probability that you get all even (aka dont get any odd)? Well, on your first pick, you want an even:
Prob:10/15
on you second pick you want an even as well
Prob:9/14
Sam with your third pick; you want an even
Prob:8/13

Multiply the 3 together and you get the probability that you never get an odd. The answer should be 24/91.

Wait how do you know the question is written wrong?

[quote="AIME15"]I believe the problem is from MMPC, a contest that they will both attend, or ICAE, a school they both attend :lol:[/quote]

Yeah, this question is from the Michigan Mathematics Prize Competition, which is like the pinnacle of michigan math competitions (it also happens to be the only math competition exclusive to michigan...)

[quote="AIME15"]I believe the problem is from MMPC, a contest that they will both attend, or ICAE, a school they both attend :lol:[/quote]

Er, yeah, its ICAE. I think that's what you were referring to, at least.

I typo way too much, which explains my epic failure rating on FTW.


algebra
function
domain
trigonometry
calculus
calculus computations

Find the eqt. of curve to which the length of each tangent   between two axes is equal to constant $ a$.

How to set up a differential equation for finding out the the eqt. of the curve?

A bash-approach to getting a diff. eqn. is as follows:

The equation of the tangent line to a curve $ y \equal{} f(x)$ at $ x \equal{} a$ is $ y \equal{} f(a) \plus{} f'(a)(x\minus{}a)$. The y-intercept of the line is $ f(a) \minus{} af'(a)$, and the x-intercept is $ a \minus{} \frac{f(a)}{f'(a)}$.

If you want the distance between them to be a constant (say $ k$) for all tangents to the curve, then $ \left(x \minus{} \frac{f(x)}{f'(x)}\right)^2 \plus{} \left(f(x) \minus{} xf'(x)\right)^2 \equal{} k^2$ for all $ x$ in the domain of $ f'$.

One (perhaps unhelpful) way to write this differential equation is as $ x^2(y')^4 \minus{} 2xy(y')^3 \plus{} (x^2\plus{}y^2\minus{}k^2)(y')^2 \minus{} 2xyy' \plus{} y^2 \equal{} 0$.

[quote="jmerry"]Well, I might be able to say something... if I had any idea what that was supposed to look like. Length between two axes?[/quote]

Intuitively, the curve is Asteroid.

[quote="J.Y.Choi"]You can solve this problem by using Clairaut's ODE. The curve equation you want is the astroid.  See the book "Mathematical Methods for Physicists" by Arfken.[/quote]

What page or chapter is this in?

[quote="mathchallenger"]Asteriod?... You mean this?...
http://en.wikipedia.org/wiki/Asteroid

 :rotfl:[/quote]

The spelling is pronunciation of Japanese.

The right is [color=red]Astroid[/color], [url=http://www.gap-system.org/~history/Curves/Astroid.html]this[/url]. :lol:

$ Let A\left(t \minus{} \frac {f(t)}{f'(t)}\right),\ B(0,\ f(t) \minus{} tf'(t))$ for the point of the tangency,

$ AB \equal{} a\Longleftrightarrow \sqrt {1 \plus{} f'(t)^2}\left\{t \minus{} \frac {f(t)}{f'(t)}\right\} \equal{} a$

$ \Longleftrightarrow \{tf'(t) \minus{} f(t)\} \sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$

$ \Longleftrightarrow t^2\left\{\frac {f(t)}{t}\right\}'\sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$

[color=red]Last Edited[/color]

[quote="kunny"][quote="mathchallenger"]Asteriod?... You mean this?...
http://en.wikipedia.org/wiki/Asteroid

 :rotfl:[/quote]

The spelling is pronunciation of Japanese.

The right is [color=red]Astroid[/color], [url=http://www.gap-system.org/~history/Curves/Astroid.html]this[/url]. :lol:

$ Let A\left(t \minus{} \frac {f(t)}{f'(t)}\right),\ B(0,\ f(t) \minus{} tf'(t))$ for the point of the tangency,

$ AB \equal{} a\Longleftrightarrow \sqrt {1 \plus{} f'(t)^2}\left\{t \minus{} \frac {f(t)}{f'(t)}\right\} \equal{} a$

$ \Longleftrightarrow \{tf'(t) \minus{} f(t)\}' \sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$

$ \Longleftrightarrow t^2\left\{\frac {f(t)}{t}\right\}\sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$[/quote]

Hello
Kunny

How to solve your suggested D.E. ?

[quote="kunny"][quote="mathchallenger"]Asteriod?... You mean this?...
http://en.wikipedia.org/wiki/Asteroid

 :rotfl:[/quote]

The spelling is pronunciation of Japanese.

The right is [color=red]Astroid[/color], [url=http://www.gap-system.org/~history/Curves/Astroid.html]this[/url]. :lol:

$ Let A\left(t \minus{} \frac {f(t)}{f'(t)}\right),\ B(0,\ f(t) \minus{} tf'(t))$ for the point of the tangency,

$ \overline{AB} \equal{} a\Longleftrightarrow \sqrt {1 \plus{} f'(t)^2}\left\{t \minus{} \frac {f(t)}{f'(t)}\right\} \equal{} a$

$ \Longleftrightarrow \{tf'(t) \minus{} f(t)\} \sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$

$ \Longleftrightarrow t^2\left\{\frac {f(t)}{t}\right\}'\sqrt {1 \plus{} f'(t)^2} \equal{} af'(t)$

[color=red]Last Edited[/color][/quote]

Sorry, I made a typo. I have just fixed it as this.

Regrettably, I have no clue to solve the differential equation now, let's wait the solution by Experts. :lol:

I think pkMathew's idea is to find the equation of curve by finding the D.E. and then eliminating the parameters.

[hide="Here is my idea"]
Suppose
$ P(x,y)$ is the tangential point of the curve

$ \{xf'(x) \minus{} f(x)\} \sqrt {1 \plus{} f'(x)^2} \equal{} af'(x)$, where $ y \equal{} f(x)$  and  $ \frac {dy}{dx} \equal{} f'(x)$

Let
$ p \equal{} \frac {dy}{dx}$

Hence,

$ \{xp \minus{} f(x)\} \sqrt {1 \plus{} p^2} \equal{} ap$,

differentiate w.r.t.x,

$ p \plus{} x\frac {dp}{dx} \minus{} p \equal{} \frac {d}{dx}\frac {ap}{\sqrt {1 \plus{} p^2}}$

$ \implies x\frac {dp}{dx} \equal{} a\cdot \frac {\sqrt {1 \plus{} p^2} \minus{} \frac {p}{2}\cdot \frac {2p}{\sqrt {1 \plus{} p^2}}}{1 \plus{} p^2}\frac {dp}{dx}$

$ \implies x \equal{} a\cdot \frac {(1 \plus{} p^2) \minus{} p^2} {(1 \plus{} p^2)^{3/2}}$

$ \implies \frac {x}{a} \equal{} \frac {1} {(1 \plus{} p^2)^{3/2}}$

$ (\frac {x}{a})^{2/3} \equal{} \frac {1}{1 \plus{} p^2}$

Suppose the inclination of tangent is $ \theta$,i.e. $ p \equal{} \frac {dy}{dx} \equal{} \pm\tan\theta$

Then $ 1 \plus{} p^2 \equal{} 1 \plus{} \tan^2\theta \equal{} \sec^2\theta$

$ \implies (\frac {x}{a})^{2/3} \equal{} \cos^2\theta$  or  $ \frac {x}{a} \equal{} \cos^3\theta$

The intercept form of the tangent is

$ \frac {x}{a\cos\theta} \plus{} \frac {y}{a\sin\theta} \equal{} 1\cdots(*)$

Put $ \frac {x}{a} \equal{} \cos^3\theta$ into (*) and get

$ (\frac {y}{a})^{2/3} \equal{} \sin ^2\theta$

$ \implies (\frac {x}{a})^{2/3} \plus{} (\frac {y}{a})^{2/3} \equal{} 1$[/hide]

While this starts out with a differential equation, it isn't really worthwhile to keep going with it. That square root is very inconvenient.

Let's try something else: a parametrization. Since I'd rather use the letter $ a$ for something else, I'll rename the fixed length of that tangent segment $ L$.

Let $ a$ be the $ x$-intercept and $ b$ be the $ y$-intercept of the tangent line at the point $ (x,y)=(x(t),y(t))$. We have
\begin{align}a^2+b^2 & = L^2 \\
\frac{x}{a}+\frac{y}{b} & = 1 \\
\frac{x'}{a}+\frac{y'}{b} & = 0\end{align}
from the length of the segment, its equation, and the tangency. We're also left with complete flexibility in how fast the point moves, and we might choose that later.
What's the trick? $ a$ and $ b$ are also variables. Differentiate (1) to get
\begin{align*}aa'+bb' & = 0 \tag{4}\end{align*}
and (2) to get
\begin{align*}0 & = \frac{x'}{a}+\frac{y'}{b}-\frac{a'x}{a^2}-\frac{b'y}{b^2}\\
& = -\frac{aa'x}{a^3}-\frac{b'y}{b^2}\\
& = \frac{bb'x}{a^3}-\frac{bb'y}{b^3}\\
\frac{x}{a^3} &=\frac{y}{b^3} \tag{5}\end{align*}
Now we've eliminated the derivatives. Let's solve the system $ (1),(2),(5)$.
$ 1=\frac{x}{a}+\frac{y}{b}=a^2\frac{x}{a^3}+b^2\frac{y}{b^3}= (a^2+b^2)\frac{x}{a^3}$
$ 1=L^2\frac{x}{a^3}=L^2\frac{y}{b^3}$
$ x^{\frac23}=\left(L^{-2}a^3\right)^{\frac23}=L^{-\frac43}a^2$ and similarly $ y^{\frac23}=L^{-\frac43}a^2$ Add them, and
$ x^{\frac23}+y^{\frac23}=L^{-\frac43}(a^2+b^2)=L^{-\frac43}L^2=L^{\frac23}$
That's the equation of an astroid, and we're done.

Note the connection to the classic "pipe around the corner" optimization problem.

Let two corridors meet at right angles, and let the axes be the outer walls of the respective corridors. A thin pipe of length $ L$ can be carried horizontally around the corner precisely when the junction of the two inner walls lies outside the astroid.

That brings in all the methods you might use on that calculus problem.

Note also that this curve is rectifiable in closed form - that is, you can explicitly compute the arclength. Can you connect that in to the rest of it?

[quote="gauss202"]What page or chapter is this in?[/quote]
Weber & Arfken, [i]Esssential mathematical methods for physicists[/i], pp. 419-420.

It introduces a method of solving ODE like $y=xy'+f(y')$, which you have been tried to solve. Your equation 'looks' like a nonlinear ODE involving $f'(x) ^4$, but a simple observation leads to the form
\[ f(x)=xf'(x) \pm \frac{kf'(x)}{\sqrt{(f'(x))^2+1}}\]
from $()^2+()^2=k^2$.


[quote="pkothari13"]$ABC$ is a triangle and the points $A',B',C'$ are on sides $BC,CA,AB$ and $AA',BB',CC'$ meet at $O.$ Also, $\displaystyle\dfrac{AO}{AO'} + \displaystyle\dfrac{BO}{BO'} + \displaystyle\dfrac{CO}{CO'} = 92.$  Find $(\displaystyle\dfrac{AO}{AO'}) (\displaystyle\dfrac{BO}{BO'}) (\displaystyle\dfrac{CO}{CO'}).$

Does anyone know how to do this using mass point geometry?[/quote]
Hey, what is the point $O'$ :?. By any chance, did you mean this: If

$\dfrac{AO}{A'O} + \dfrac{BO}{B'O} + \dfrac{CO}{C'O} = 92$,

find

$\dfrac{AO}{A'O} \cdot \dfrac{BO}{B'O} \cdot \dfrac{CO}{C'O} =\ ?$

If this is the case, then assign (unknown) masses $m_A, m_B, m_C$ to the vertices A, B, C, such that the point O is the center of mass of the system. By the rule of moments,

$m_A \cdot AC' = m_B \cdot BC',\ \ m_B \cdot BA' = m_C \cdot CA',\ \ m_C \cdot CB' = m_A \cdot AB'$

The center of mass O does not change, if:
(a) the mass $m_A$ stays at A and the masses $m_B, m_C$ are piled up at A', or
(b) the mass $m_B$ stays at B and the masses $m_C, m_A$ are piled up at B', or
(c) the mass $m_C$ stays at C and the masses $m_A, m_B$ are piled up at C'.

Using the rule of moments sequentially for all 3 above situations,

$(a)\ \ m_A \cdot AO = (m_B + m_C) \cdot A'O$
$(b)\ \ m_B \cdot BO = (m_C + m_A) \cdot B'O$
$(c)\ \ m_C \cdot CO = (m_A + m_B) \cdot C'O$

$(a)\ \ \dfrac{AO}{A'O} = \dfrac{m_B + m_C}{m_A}$

$(b)\ \ \dfrac{BO}{C'O} = \dfrac{m_C + m_A}{m_B}$

$(c)\ \ \dfrac{CO}{C'O} = \dfrac{m_A + m_B}{m_C}$

Substituting this to the given equation:

$\dfrac{m_B + m_C}{m_A} + \dfrac{m_C + m_A}{m_B} + \dfrac{m_A + m_B}{m_C} = 92$

$\dfrac{(m_B + m_C)m_Bm_C + (m_C + m_A)m_Cm_A + (m_A + m_B)m_Am_B}{m_Am_Bm_C} = 92$

and to the desired expression

$\dfrac{m_B + m_C}{m_A} \cdot \dfrac{m_C + m_A}{m_B} \cdot \dfrac{m_A + m_B}{m_C} =$

$= \dfrac{(m_B + m_C)(m_C + m_A)(m_A + m_B)}{m_Am_Bm_C} =$

$= \dfrac{[(m_B + m_C)m_C + (m_B + m_C)m_A](m_A + m_B)}{m_Am_Bm_C} =$

$= \dfrac{(m_B + m_C)m_Cm_A + (m_B + m_C)m_A^2 + (m_B + m_C)m_Bm_C + (m_B + m_C)m_Am_B}{m_Am_Bm_C} =$

We want to keep the term $(m_B + m_C)m_Bm_C$ in the numerator, because it is also in the given expression,  and multiply out the others:

$= \dfrac{m_Am_Bm_C + m_C^2m_A + m_Bm_A^2 + m_Cm_A^2 + (m_B + m_C)m_Bm_C + m_Am_B^2 + m_Am_Bm_C}{m_Am_Bm_C} =$

then collect the appropriate terms together:

$= \dfrac{2m_Am_Bm_C + (m_C + m_A)m_Cm_A + (m_B + m_C)m_Bm_C + (m_A + m_B)m_Am_B}{m_Am_Bm_C} =$

$= 2 + 92 = 94$

...[size=150]WOW[/size] thank you so much! You made it look so easy! And sorry for the typo, you interpreted it the correct way.

'Wow'  is an appropriate emotion in this case,

     provided that  the difference

     AM / MA'  *   BM / MB'  *  CM / MC'  - ( AM / MA'  +   BM / MB'  +  CM / MC'  )

    has been  =  2    since Euler time  (and probably even for some time before that ...)

    Thank you.

    M.T.


Compute

\[ \sum_{d|m}\frac{m}{d}\varphi(d), m\in\mathbb{N}^*\]

where $ \varphi$ is the Euler's indicator with $ \varphi(1)\equal{}1$

we take $ f(m)\equal{}\sum_{d|m}\frac{m}{d}\varphi(d)$
if $ pgcd(m,n)\equal{}1$ then
$ {f(nm)\equal{}\sum_{d|mn}\frac{m}{d}\varphi(d)
\equal{}\sum_{d|m,d'|n}\frac{mn}{dd'}\varphi(dd')
\equal{}(\sum_{d|m}\frac{m}{d}\varphi(d))(\sum_{d'|n}\frac{n}{d'}\varphi(d'))
\equal{}f(n)f(m)}$
now if we suppose that $ m\equal{}\prod_{i\equal{}1}^{h}p_i^{a_i}$ where $ p_i$ are $ \neq$ primes we have $ f(m)\equal{}\prod_{i\equal{}1}^hf(p_i^{a_i})$
we have :
$ f(p_i^{a_i})\equal{}\sum_{d|p_i^{a_i}}\frac{m}{d}\varphi(d)\equal{}\sum_{k\equal{}0}^{a_i}p^{a_i\minus{}k}\varphi(p^{k})\sum_{k\equal{}0}^{a_i}p^{a_i\minus{}1}(p_i\minus{}1)\equal{}(a_i\plus{}1)(p_i\minus{}1)p^{a_i\minus{}1}$
so $ f(m)\equal{}d(n)\varphi(n)$ where $ d(n)\equal{}|\{d\in\mathbb{N}: d|n\}|$


calculus
integration
real analysis
real analysis unsolved

If $ f$ is Lebesgue integrable on $ (0,a)$ and $ g(x)\equal{}\int_x^a t^{\minus{}1}f(t) dt$, then $ g$ is integrable on $ (0, a)$ and  $ \int_0^a g(x) dx\equal{}\int_0^a f(x)dx$.

Just change the order of integration :wink:


Compute the smallest positive integer $n$, greater than $200$, with the property that $n$ is the sum of $5$ consequetivie integers, $6$ consequetivie integers, and $7$ consequetive integers.

[hide]n must be divisible by 5
n must be divisible by 3 and be odd
n must be divisible by 7

LCM(5,3,7)=105. n is an odd multiple of 105 greater than 200.

n=315[/hide]

[quote="lkryptonicl"][quote="JavaMan"][hide]
We see that what they're asking for is a number $n$ such that:
$n \equiv 0 (\mod 5)$
$n \equiv 0 (\mod 7)$
$n \equiv 3 (\mod 6)$
That number must be in the form 210n + 105. Smallest number above 200 would be when n = 1, 210+105 = 315.
[/hide][/quote]

But why does n have to be congruent to 0 in mod5 and 7 but not in mod6. Maybe im misssing something????[/quote]

If the starting number was $x$, the sum of 5 consecutive would be $5x+15=5(x+3)$
If there were 6 consecutive, it would be $6x+21$ and $21\equiv 3\pmod 6$
If there were 7, then the sum would be $7x+28=7(x+4)$

Or if you start counting with the middle terms before dividing by $5$ or $7$, then the $5$ consecutive terms would be $n-2,n-1,n,n+1,n+2$ which has a sum clearly divisible by $5$, and the seven consecutive terms can be similarly added, so the number is also divisible by $7$.  The six terms, just do it however cuz you'll still get the same answer (this way of numbering is useful sometimes).  If you want you could say $n-2,n-1,n,n+1,n+2,n+3$ so the sum is $\equiv 3\pmod6$


calculus
integration
trigonometry
complex analysis
real analysis
real analysis unsolved

for $ |\alpha|<\infty$, solve :D

$ \textbf{(i)}\;\;\int_{0}^\infty\;\;\frac{\cos\; (\alpha x)}{1+x^{2}}\;\;\textbf dx$

$ \textbf{(ii)}\;\;\int_{0}^\infty\;\;\frac{\cos\; (\alpha x)}{1-x^{2}}\;\;\textbf dx$

$ \textbf{(iii)}\;\;\int_{0}^\infty\;\;\frac{\sin\; (\alpha x)}{1+x^{2}}\;\;\textbf dx$

$ \textbf{(iv)}\;\;\int_{0}^\infty\;\;\frac{\sin\; (\alpha x)}{1-x^{2}}\;\;\textbf dx$

[ [b]PS:[/b] check out my [i]new [/i]avatar, fox mulder must be getting
very jealous right about now...  :rotfl: ]

(1) Without using complex analysis, you may solve it like this:

\begin{eqnarray*}\int_{0}^{\infty}\frac{\cos (\alpha x)}{1+x^{2}}\; dx & = & \int_{0}^{\infty}\cos (\alpha x) \left( \int_{0}^{\infty}\sin t e^{-xt}\; dt \right) \; dx \\ & = & \int_{0}^{\infty}\int_{0}^{\infty}\cos (\alpha x) \sin t e^{-xt}\; dx \; dt \\ & = & \int_{0}^{\infty}\frac{t \sin t}{t^{2}+\alpha^{2}}\; dt \\ \end{eqnarray*}

Now let $ I(\alpha) = \int_{0}^{\infty}\frac{\cos (\alpha x)}{1+x^{2}}\; dx$. For $ \alpha > 0$,

\begin{eqnarray*}\frac{d}{d\alpha}I(\alpha) & = &-\int_{0}^{\infty}\frac{x \sin (\alpha x)}{1+x^{2}}\; dx\\ & = &-\int_{0}^{\infty}\frac{t \sin t}{\alpha^{2}+t^{2}}\; dt \quad \quad (\text{where }t = \alpha x)\\ & = &-I(\alpha). \end{eqnarray*}

So $ I(\alpha) = I(0)e^{-\alpha}= \frac{\pi}{2}e^{-\alpha}$ for $ \alpha \geq 0$. Then it follows from $ I(-\alpha) = I(\alpha)$ that $ I(\alpha) = \frac{\pi}{2}e^{-|\alpha|}$.

For (3), I introduced $ I(\alpha) = \int_{0}^{\infty}\frac{\sin (\alpha x)}{1+x^{2}}\; dx$ and found that it satisfies $ I''-I+\frac{1}{\alpha}= 0$ with $ I(0) = 0, I'(\infty) = 0$.

I was not able to solve it, so I used mathematica and finally I conjectured that $ I(\alpha) = \frac{e^{-\alpha}\text{Ei}(\alpha)-e^{\alpha}\text{Ei}(-\alpha)}{2}$, where $ \text{Ei}(x)$ is the [url=http://mathworld.wolfram.com/ExponentialIntegral.html]exponential integral[/url]. (But I know nearly nothing about this function...)

[quote="sos440"](1) Without using complex analysis, you may solve it like this:

\begin{eqnarray*}\int_{0}^{\infty}\frac{\cos (\alpha x)}{1+x^{2}}\; dx & = & \int_{0}^{\infty}\cos (\alpha x) \left( \int_{0}^{\infty}\sin t e^{-xt}\; dt \right) \; dx \\ & = & \int_{0}^{\infty}\int_{0}^{\infty}\cos (\alpha x) \sin t e^{-xt}\; dx \; dt \\ & = & \int_{0}^{\infty}\frac{t \sin t}{t^{2}+\alpha^{2}}\; dt \\ \end{eqnarray*}

Now let $ I(\alpha) = \int_{0}^{\infty}\frac{\cos (\alpha x)}{1+x^{2}}\; dx$. For $ \alpha > 0$,

\begin{eqnarray*}\frac{d}{d\alpha}I(\alpha) & = &-\int_{0}^{\infty}\frac{x \sin (\alpha x)}{1+x^{2}}\; dx\\ & = &-\int_{0}^{\infty}\frac{t \sin t}{\alpha^{2}+t^{2}}\; dt \quad \quad (\text{where }t = \alpha x)\\ & = &-I(\alpha). \end{eqnarray*}

So $ I(\alpha) = I(0)e^{-\alpha}= \frac{\pi}{2}e^{-\alpha}$ for $ \alpha \geq 0$. Then it follows from $ I(-\alpha) = I(\alpha)$ that $ I(\alpha) = \frac{\pi}{2}e^{-|\alpha|}$.
[/quote]

nice!  :coolspeak:

unlike yours, my approach actually led me to a [b]2nd [/b]order ODE... :|

let $ I\;=\;\int_{0}^\infty\;\frac{\sin \;(\alpha x)}{x\;(1+x^{2})}\;dx$

now, $ \frac{dI}{d\alpha}\;=\;\int_{0}^{\infty}\frac{\cos \;(\alpha x)}{1+x^{2}}\; dx$ (this is what we are interested in)

and, $ \frac{d\;^{2}I}{d\alpha\;^{2}}\;=\;-\int_{0}^\infty\;\frac{x\;\sin \;(\alpha x)}{(1+x^{2})}\;dx\;=\;\alpha^{2}\;I-\frac{\pi}{2}$

so, $ \frac{d\;^{2}I}{d\alpha\;^{2}}-\alpha^{2}\;I\;=\;-\frac{\pi}{2}$

solving for $ I$ and then taking $ \frac{dI}{d\alpha}$ will give us the answer.


Let $ f: Z \to Z$ be such that $ f(1) \equal{} 1, f(2) \equal{} 20, f(\minus{}4) \equal{} \minus{}4$ and $ f(x\plus{}y) \equal{} f(x) \plus{}f(y)\plus{}axy(x\plus{}y)\plus{}bxy\plus{}c(x\plus{}y)\plus{}4 \forall x,y \in Z$, where $ a,b,c$ are constants.

(a) Find a formula for $ f(x)$, where $ x$ is any integer.
(b) If $ f(x) \geq mx^2\plus{}(5m\plus{}1)x\plus{}4m$ for all non-negative integers $ x$, find the greatest possible value of $ m$.

[b]
put $ x \equal{} y \equal{} 0$,then we have
$ f(0) \equal{} \minus{} 4$
put $ x \equal{} 0$ ,then we have $ c \equal{} 0$,
put $ x \equal{} 1, y \equal{} \minus{} 1$, we have
$ f(0) \equal{} f( \minus{} 1) \plus{} f(1) \minus{} b \plus{} 4 ,f( \minus{} 1) \equal{} b \minus{} 9$
put $ x \equal{} y \equal{} \minus{} 1$,
$ f( \minus{} 2) \equal{} 2f( \minus{} 1) \minus{} 2a \plus{} b \plus{} 4 \equal{} \minus{} 2a \plus{} 3b \minus{} 14$
put $ x \equal{} y \equal{} \minus{} 2$,
$ f( \minus{} 4) \equal{} 2f( \minus{} 2) \minus{} 16a \plus{} 4b \plus{} 4, \minus{} 4 \equal{} \minus{} 20a \plus{} 10b \minus{} 24, \minus{} 2 \equal{} 2a \minus{} b .....(1)$
put $ x \equal{} y \equal{} 1$, we have,
$ f(2) \equal{} 2f(1) \plus{} 2a \plus{} b \plus{} 4$
$ 14 \equal{} 2a \plus{} b .......(2)$
use $ (1), (2)$, we have
$ a \equal{} 3, b \equal{} 8$
so $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} 3xy(x \plus{} y) \plus{} 8xy \plus{} 4$
put y=1, we have
$ f(x \plus{} 1) \minus{} f(x) \equal{} 3x^2 \plus{} 11x \plus{} 5$
$ \sum_{x \equal{} 1}^n[f(x \plus{} 1) \minus{} f(x)] \equal{} \sum_{x \equal{} 1}^n[3x^2 \plus{} 11x \plus{} 5]$
hence, $ f(x) \equal{} x^3 \plus{} 4x^2 \minus{} 4$
(b)$ f(x) > \equal{} mx^2 \plus{} (5m \plus{} 1)x \plus{} 4m$
     $ x^3 \plus{} 4x^2 \minus{} 4 > \equal{} mx^2 \plus{} 5mx \plus{} 4m \plus{} x$
    $ (x \plus{} 4)(x \plus{} 1)(x \minus{} 1) > \equal{} m(x \plus{} 1)(x \plus{} 4)$
   $ x > \equal{} m \plus{} 1$ for all non-negative  integer x
 hence, the greatest possible value of m is $ \minus{} 1$


What is a free program that can generate chemical structural diagrams when you enter names of chemicals?

I would like also to know, since I didn't know that such programs exist.

Look around wikipedia and see how the contributors made the SVGs?

[quote="n0vad3m0n"]http://www.acdlabs.com/download/

It's similar to an MS Word document, where you can type regular documents, but you can also structural formulas. It also includes a compound-namer, molecule templates, 3D structure viewer, and more.

I highly recommend it.[/quote]

Thanks  :D

[quote="n0vad3m0n"]http://www.acdlabs.com/download/

It's similar to an MS Word document, where you can type regular documents, but you can also structural formulas. It also includes a compound-namer, molecule templates, 3D structure viewer, and more.

I highly recommend it.[/quote]

dudeman!!!!!! thanx bunches!!


is it true that every integer can be written as $ x^2\plus{}y^2\minus{}z^3$ for some $ x,y,z\in\mathbb{Z}$ ?


Right, but then you must make one more step, namely splitting $2y^2$ into $y^2+y^2$ in order to get a workable factorization.

As far as I can see, you convert every term into two, and then cancel just one out of the two, so you don't get very far (without further factorization or some manual computing).

the sophie german is just completing the square, and using difference of squares...

[quote="chess64"]Well here's one way using Sophie-Germain: http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3988[/quote]

Well LOOK at it carefully, chess64 :D

They did just what I said - they split $2b^2$ terms in both pairs of parentheses into $b^2+b^2$ (see the last line of the first formula sequence)

Please compare post #7 and your reply in post #8.

Well, it isn't completely necessary to do that (I didn't do it that way when I did the problem, and I still got the answer easily), but agreed, it would be better to split up the $2b^2$. :)

[hide=Solution] The answer is $373$. We proceed by using the sofie-german identity on each of the sums. Note that $324=4*3^{4}$. Notice that the majority of the fraction cancels when using the identity, so the answer is just $\frac{61^2+9}{1^2+9}=\frac{3730}{10}=373$.[/hide]

As soon as we see the $10^4$ and the other powers of $4$, we think of the [I]Sophie-Germain Identity[/I]. We also see that $324$ can be factored as $4(3^4)$, which basically confirms  the usage of Sophie-Germain in this problem. We repeatedly use the identity on all the terms, and we then cancel like terms on the numerator and on the denominator

Note that $p^4 + 324 = ((p+3)^2+9)((p-3)^2+9)$. Then plugging this in and cancelling terms, we obtain the fraction $\frac{61^2+9}{1^2+9} = \boxed{373}$.

[quote=Math4Life2020][quote=OlympusHero]Note that $p^4 = ((p+3)^2+9)((p-3)^2+9)$. Then plugging this in and cancelling terms, we obtain the fraction $\frac{61^2+9}{1^2+9} = \boxed{373}$.[/quote]

Uhhh... the identity in your post isn't true? The $p^0$ coefficients on the LHS (0) and the RHS (324) are different.[/quote]

there are no $p^0$ coefficients and there is no RHS

also this is from intermediate algebra book and they use no sophie germain

[quote=Math4Life2020]in the "identity" $p^4=((p+3)^2+9)((p-3)^2+9)$
the RHS stands for the Right Hand Side
and in case you didn't know the $p^0$ term is the same thing as a constant term[/quote]

((p+3)^2+9)((p-3)^2+9)=p^4+324

i think he forgot the 324 but he still used it as p^4+324 instead of p^4

[hide=Solution]Note that each term being multiplied is in the form of $n^4+324$. In order to begin to cancel terms, we want to write $n^4+324$ as the product of at least two terms. Since, $324=18^2$, we have that $n^4+324=n^4+18^2+36n^2-36n^2=(n^2+18)^2-36n^2$ (completing the square). Since both terms are squares, this is the difference of two squares, which we can rewrite as $(n^2+18-6n)(n^2+18+6n)=(n(n-6)+18)(n(n+6)+18)$. Since each of the values of $n$ differ by multiples of $6$, all the terms cancel except for $\frac{58(58+6)+18}{4(4-6)+18}=\frac{3730}{10}=\boxed{373}$.[/hide]


calculus
derivative
function
vector
limit
analytic geometry
graphing lines

What are partial derivatives used for? Ordinary derivatives can be used for finding minima and maxima of functions, among other things; can you do something analogous with partial derivatives?

If you have a function in several variables, and you hold all but one variable constant and take the derivative, what does that tell you about the "picture" of the function itself? Could you provide an example?

I'm in high school and studying calculus casually (i.e. not as a class), and this is one question I've always wondered about.

Thanks.

The mere existence of partial derivatives is somewhat problematical, as that doesn't guarantee even that a function is continuous. You want a little more in a "nice" function.

One way to look at it is that a partial derivative is a rate of change in a specific direction - and that makes partial derivatives a special case of [b]directional derivatives.[/b]

If $\mathbf{b}$ is a unit vector (a vector of length one), then the directional derivative of $f$ at $\mathbf{x}$ in the direction $\mathbf{u}$ is given by

$D_{\mathbf{u}}f(\mathbf{x})=\lim_{t\to0}\frac{f(\mathbf{x}+t\mathbf{u})-f(\mathbf{x})}{t}.$

Now, the thing is that for a "nice" function (a function that is differentiable when we define that the way we want to), the directional derivatives exist in all directions and moreover, we can use the partial derivatives as a basis for all directional derivatives. Here's the formula. Let $\mathbf{x}=(x,y,z)$ and $\mathbf{u}=(a,b,c)$ and use the notation $f_{x}$ for $\frac{\partial f}{\partial x}.$ Then

$D_{\mathbf{u}}f(\mathbb{x})= f_{x}(\mathbf{x})a+f_{y}(\mathbf{x})b+f_{z}(\mathbf{x})c =\nabla f\cdot \mathbf{u}.$

--

You ask about maxima and minima. For a smooth function to have either a maximum or minimum at a point, the directional derivative must be zero in all directions. From the formula above for directional derivatives, and imagining all possible $\mathbf{u},$ that can only happen if all of the partial derivatives are zero simultaneously. (In vector terms, $\nabla f(\mathbf{x})=\mathbf{0}.$) As a practical matter, finding such critical points requires us to solve $n$ simultaneous equations for $n$ unknowns, and in many of the interesting cases, these are nonlinear equations. That can be a challenge.

More interesting is the issue of [b]constrained[/b] optimization problems - can we find a maximum or minimum of a function of several variables, subject to a condition that restricts those variables to lie on a particular surface or curve. The key to finding critical points is now to require that the directional derivative be zero in all [i]allowed[/i] directions. Following up on that insight leads to the technique known as Lagrange Multipliers - try looking that up in whatever textbook or other resource you have.

So partial derivatives are just directional derivatives in directions parallel to the axes?

What's $\nabla$?

$\nabla$ is a convenient notation for something that is both a differential operator and a vector.

For a scalar-valued function $f$, $\nabla f$ is a vector-valued function called the [i]gradient[/i] of $f$. For the sake of the argument, say $f=f(x,y,z)$ is a function on (some portion of) $\mathbb{R}^{3}.$ Then the gradient of $f$ is

$\nabla f=\left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle = \frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}.$

A different notation with essentially the same mathematical content is the [i]differential[/i], also known as the [i]total differential[/i] of $f:$

$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy+\frac{\partial f}{\partial z}\,dz$

There are other uses of the $\nabla$ notation, but for where you are right now, it's probably best to stop with the notion of the gradient.

In your first post, you said something about the "picture" of a function. Useful pictures can be elusive. For functions that map $\mathbb{R}$ into $\mathbb{R}$ (simple functions of one variable), you have been conditioned to think of the graph of the function. Technically, the graph is the set ${\{(x,y}: y=f(x)\}.$ In general, for a function from $A$ into $B,$ the graph of the function is a subset of the Cartesian product $A\times B.$ For a function from $\mathbb{R}^{2}$ into $\mathbb{R},$ we may consider its graph to be a subset of $\mathbb{R}^{3}.$ For anything larger, the graph lives in a space with too many dimensions for us to be able to draw useful pictures.

But in that one case, $f: \mathbb{R}^{2}\mapsto\mathbb{R},$ we can visuallize. Consider $z=f(x,y)$ to be a surface of height $z$ above the $xy$ plane. Then the vector field $\nabla f$ always points "uphill" in the direction of steepest slope, and the magnitude of the gradient is the actual slope in that steepest direction. At the "top of the hill" - at a maximum (or a minimum) - $\nabla f$ is the zero vector and has no direction. But another kind of point at which $\nabla f=\mathbf{0}$ is a "saddle point" or "minimax." Think of a pass through a ridge. Consider how the function $f(x,y)=x^{2}-y^{2}$ behaves near the origin.


[quote="sludgethrower"]Now this is really difficult for me to write...but:

A is a 3 by 3 matrix. let RnCm = the entry in the nth row and mth column:

R1C1 = 1, R1C2 = 0, R1C3 = -1
R2C1 = 0, R2C2 = 1, R2C3 = -1
R3C1 = 0, R3C2 = 1, R3C3 =  2

Similarly, the 3 by 3 matrix B has entries:

R1C1 = x + 2, R1C2 = x - 2, R1C3 = x + 3
R2C1 = -4,      R2C2 = 4,      R2C3 = 2
R3C1 = 2,       R3C2 = -2,     R3C3 = 3

(i) Find the product AB.
(ii) Obtain the determinants of A and of AB
(iii) Hence obtain an expression for det B[/quote]

[hide="(i)"]So multiply the first row of A and the first column of B to get the first entry. Then multiply the first row of A and the second column of B to get the second entry. and so on.... Doing this....
                    AB
              |x   x   x|
              |-6  6  -1|
              |0   0   8|[/hide]

[hide="(ii)"]Det A=3, Det AB=96x
Just to let people who don't know how to do this, we have a 3x3 matrice. Let's call the entries $a_{1}$, $a_{2}$, $a_{3}$... so on. So a determinant of a 3x3 is $(a_{1}a_{5}a_{9}+a_{2}a_{6}a_{7}+a_{3}a_{4}a_{8})-(a_{3}a_{5}a_{7}+a_{2}a_{4}a_{9}+a_{1}a_{6}a_{8})$[/hide]

[hide="(iii)"]So now, the Det B=34x+6[/hide]

Part (iii) of your solution is incorrect. Hint you have detA and detAB, from simple determinant algebra you can obtain detB.


During his weekend, John wants to cross a mountain, but the local forester has imposed him some rules:
1. his track has to be 2*n-2 meters, where the value n is given by the forester. He has to leave from the base of the mountain, and he has to get to the oder side of the mountain, at the same altitude.
2. he isn't allowed to descend under the starting altitude
3. he can walk using 3 kind of steps: a horizontal step of length of 2 (such as (2,0) ), a up step of length of 1 (such as (1,1) ), a down step of length of 1 (such as (1,-1) )
4. his track isn't allowed to have a top at the altitude of 1. (this means that at any point in his track, John,from the starting altitude, isn't allowed to climb one step and then descend)

Enuntiation

Given the number n, determine in how many ways John can cross the mountain, respecting the forester's rules.

INPUT

On the first line there is the number n.

OUTPUT

The output has to contain a single number, resembling in how many ways John can cross the mountain.

Restrictions

1<=n<=100

SAMPLE INPUT

3

SAMPLE OUTPUT

3

Time limit: 0.1 seconds

No information regarding the altitude of the mountain is given?? no altitude nothing? are u sure of the problem statement?
One more thing where can I find all the problems of the Romanian olympiad preparation

thanks
best regards

the altitude varies, it can be 0, or maximum n-2, because the track has to have exactly 2*(n-1) steps

I can post the other problems (the ones that aren't too big) :D

I don't understand the last restriction (number 4), could you explain a little bit?

[hide]This is just a simple DP. Let f[x][y] = total number of paths from (x,y) to (2n-2,0).
Then we get f[2n-2][0] = 1, f[2n-2][y] = 0 for y>0.

f[x][0] = f[x+2][0] + f[x+2][2]
f[x][y] = f[x+2][y] + f[x+1][y+1] + f[x+1][y-1] for y>0.

(ignoring cases where x+2>2n-2 of course, when that part of it is just 0.)[/hide]

Restriction number 4 means that if John is in a position such as (k,0), he isn't allowed to climb to (k+1,1) and then descend to (k+2,0).

OK, I get it now. Uhm, so TripleM, how do you check condition number 4?

He does it with the line

f[x][0] = f[x+2][0]+f[x+2][2].

If he did not consider condition 4, the line would look like

f[x][0] = f[x+2][0]+f[x+1][1],

since those are the two positions that you can go to from $(x, 0)$. However, since we do not want to consider the path $(x, 0) \rightarrow (x+1, 1) \rightarrow (x+2, 0)$, we must assume that if we have $(x, 0) \rightarrow (x+1, 1)$ it must go to $(x+2, 2)$. So by using f[x+2][2], he excludes the case in which we have a peak of height 1.

But then we should also consider the path $(x,0) \rightarrow (x+1,1) \rightarrow (x+3,1)$, right?

Hmm, so perhaps the line should be

f[x][0] = f[x+2][0]+f[x+2][2]+f[x+3][1].

I believe this means

f[x][0] = f[x+1][1]

which makes sense because we can find a sort of bijection between the two sets of (reverse) paths:

$(x+2, 2) \rightarrow (x+1, 1)$ goes to $(x+2, 2) \rightarrow (x+1, 1) \rightarrow (x, 0)$

$(x+3, 1) \rightarrow (x+1, 1)$ goes to $(x+3, 1) \rightarrow (x+1, 1) \rightarrow (x, 0)$

$(x+2, 0) \rightarrow (x+1, 1)$ goes to $(x+2, 0) \rightarrow (x, 0)$.

Whoops, yeah, I missed that one case. f[x][0] = f[x+1][1] is surprisingly simple :)


trigonometry
inequalities
geometry
circumcircle
inequalities unsolved

$P$ is a point in tri-angle $ABC$,
note $\angle PAB=u_1$, $\angle PAC=u_2$,$\angle PBC=v_1$,$\angle PBA=v_2$,
$\angle PCA=w_1$,$\angle PCB=w_2$,
PROVE:$\sin (u_1+v_1+w_2)\sin (u_1+v_2+w_1)\sin (u_2+v_1+w_1) \geq 8 \sin u_1 \sin v_1 \sin w_1$

Peter Scholze and me have just solved this problem; the solution is surprisingly simple. Here is my version of this solution:

[color=blue][b]Problem.[/b] Let P be a point inside a triangle ABC. Prove the inequality

$\sin\left(\measuredangle PAB+\measuredangle PBC+\measuredangle PCB\right)\cdot\sin\left(\measuredangle PBC+\measuredangle PCA+\measuredangle PAC\right)\cdot\sin\left(\measuredangle PCA+\measuredangle PAB+\measuredangle PBA\right)$
$\geq 8\cdot\sin\measuredangle PAB\cdot\sin\measuredangle PBC\cdot\sin\measuredangle PCA$.[/color]

[i]Solution.[/i] Let the lines AP, BP, CP meet the circumcircle of triangle ABC at the points X, Y, Z (apart from A, B, C), respectively. Let the lines AP, BP, CP meet the lines YZ, ZX, XY at the points X', Y', Z', respectively.

Since the points A, B, C, X, Y, Z lie on one circle, we have < XAB = < XYB; in other words, < PAB = < PYX. Similarly, < PBC = < PZY and < PCB = < PYZ. Hence,

< PAB + < PBC + < PCB = < PYX + < PZY + < PYZ = < PZY + (< PYX + < PYZ)
= < Z'ZY + < ZYZ' = 180° - < ZZ'Y

(by the sum of angles in triangle YZ'Z). Hence, $\sin\left(\measuredangle PAB+\measuredangle PBC+\measuredangle PCB\right)=\sin\measuredangle ZZ^{\prime}Y$. On the other hand, similarly to < XAB = < XYB, we obtain < YBC = < YZC, or, equivalently, < PBC = < YZZ'. Hence,

$\frac{\sin\left(\measuredangle PAB+\measuredangle PBC+\measuredangle PCB\right)}{\sin\measuredangle PBC}=\frac{\sin\measuredangle ZZ^{\prime}Y}{\sin\measuredangle YZZ^{\prime}}=\frac{YZ}{YZ^{\prime}}$

(by the sine law in triangle YZ'Z). Similarly,

$\frac{\sin\left(\measuredangle PBC+\measuredangle PCA+\measuredangle PAC\right)}{\sin\measuredangle PCA}=\frac{ZX}{ZX^{\prime}}$

and

$\frac{\sin\left(\measuredangle PCA+\measuredangle PAB+\measuredangle PBA\right)}{\sin\measuredangle PAB}=\frac{XY}{XY^{\prime}}$.

Hence, the inequality in question,

$\sin\left(\measuredangle PAB+\measuredangle PBC+\measuredangle PCB\right)\cdot\sin\left(\measuredangle PBC+\measuredangle PCA+\measuredangle PAC\right)\cdot\sin\left(\measuredangle PCA+\measuredangle PAB+\measuredangle PBA\right)$
$\geq 8\cdot\sin\measuredangle PAB\cdot\sin\measuredangle PBC\cdot\sin\measuredangle PCA$,

being equivalent to

$\frac{\sin\left(\measuredangle PAB+\measuredangle PBC+\measuredangle PCB\right)}{\sin\measuredangle PBC}\cdot\frac{\sin\left(\measuredangle PBC+\measuredangle PCA+\measuredangle PAC\right)}{\sin\measuredangle PCA}$
$\cdot\frac{\sin\left(\measuredangle PCA+\measuredangle PAB+\measuredangle PBA\right)}{\sin\measuredangle PAB}\geq 8$,

becomes

$\frac{YZ}{YZ^{\prime}}\cdot\frac{ZX}{ZX^{\prime}}\cdot\frac{XY}{XY^{\prime}}\geq 8$.

But this is easy to prove: Using the AM-GM inequality, we have

$YZ=YX^{\prime}+ZX^{\prime}\geq 2\sqrt{YX^{\prime}\cdot ZX^{\prime}}$;
$ZX=ZY^{\prime}+XY^{\prime}\geq 2\sqrt{ZY^{\prime}\cdot XY^{\prime}}$;
$XY=XZ^{\prime}+YZ^{\prime}\geq 2\sqrt{XZ^{\prime}\cdot YZ^{\prime}}$.

Hence,

$\frac{YZ}{YZ^{\prime}}\cdot\frac{ZX}{ZX^{\prime}}\cdot\frac{XY}{XY^{\prime}}\geq\frac{2\sqrt{YX^{\prime}\cdot ZX^{\prime}}}{YZ^{\prime}}\cdot\frac{2\sqrt{ZY^{\prime}\cdot XY^{\prime}}}{ZX^{\prime}}\cdot\frac{2\sqrt{XZ^{\prime}\cdot YZ^{\prime}}}{XY^{\prime}}$
$=8\cdot\sqrt{\frac{YX^{\prime}}{ZX^{\prime}}\cdot\frac{ZY^{\prime}}{XY^{\prime}}\cdot\frac{XZ^{\prime}}{YZ^{\prime}}}$.

But the lines XX', YY', ZZ' concur at one point (namely, at the point P); thus, by the Ceva theorem, applied to the triangle XYZ, we have $\frac{YX^{\prime}}{ZX^{\prime}}\cdot\frac{ZY^{\prime}}{XY^{\prime}}\cdot\frac{XZ^{\prime}}{YZ^{\prime}}=1$, and thus this becomes

$\frac{YZ}{YZ^{\prime}}\cdot\frac{ZX}{ZX^{\prime}}\cdot\frac{XY}{XY^{\prime}}\geq 8$.

And the problem is solved.

Note that, as this solution shows, equality holds if and only if the point P is the centroid of triangle XYZ. This holds if and only if the point P is one of the two focii of the Steiner inellipse of triangle ABC (see [url=http://groups.yahoo.com/group/Hyacinthos/message/3928?expand=1]Hyacinthos message #3928[/url]).

  Darij

darij and Peter Scholze .thanks sooooooooooooooooooooooo much.
 :)


function
analytic geometry
advanced fields
advanced fields unsolved

Let $ u(s,t)$ be a bounded continuous real function, $ s\ge0$, $ t\ge0$, and $ u(0,0)\equal{}0$. Prove that there exists a bouded continuous function $ w(s,t)$ such that
$ u\le w$
$ w(0,0)\equal{}0$
$ w(s,t)$ is increasing in $ s$ for fix $ t$
$ w(s,t)$ is concave in $ t$ for fix $ s$
Extra question: Could we assume more that $ w(s,t)$ is increasing in $ t$ for fix $ s$?
This is a claim in a paper. It's quiet strange to me, and I have no idea in proving it.  :oops:

Let $ \cal F$ be the family of continuous functions $ f: D\equal{} R^2_{\ge 0}\to R$ with $ u\le f$ fulfilling the monotonicity and concavity condition. $ \cal F$ in non-void since it contains some constant function. $ \cal F$ is closed under taking the minimum of finite sub-families.

Define $ w: \equal{}\inf \{f|f\in \cal F\}$. Then $ w$ is contained in $ \cal F$:
- It is easily checked that $ w$ fulfils the monotonicity and concavity condition.
- $ w$ is continuous: On every compact subset $ K$ of $ D$, $ w$ is the decreasing limit of functions in $ \cal F$, i.e. $ f_n\downarrow w$, hence $ w$ is continuous on $ K$, by Dini's theorem. Exhausting $ D$ by compact subsets yields that $ w$ is continuous on $ D$.

Futhermore $ w(0,0)\equal{}0$: Otherwise $ w(0,0)>0$. Then there is $ \delta>0$ such that $ |u(s,t)|<\frac{w(0,0)}{4}$ for $ 0\le s,t,\le \delta$. Now define a new function $ w^*$ by drawing a straight line from $ w(s,t)$ to $ \frac{1}{2}w(0,0)$ for any point $ (\delta,t)$ or $ (s,\delta)$ with $ 0\le s,t\le\delta$ and $ w\equal{}w^*$ if one of the coordinates $ s,t$ is greater than $ \delta$. Then $ w^*$ is in $ \cal F$ and $ w^*\le w$, hence $ w^*\equal{}w$. Contradiction by evaluating the functions at (0,0).

I cannot see any reason why this method can't be applied to impose the other monotonicity condition.

[quote="MegaRockman"]Why on every compact subset $ K$ of $ D$, $ w$  is the decreasing limit of functions in $ \cal F$?
Could you explain it more explicit? :maybe:[/quote]
If $ g_n$ is a sequence in $ \mathcal F$ approximating $ w$, take $ f_n\equal{}\min(g_1,\ldots,g_n)\in \mathcal F$. Then $ f_n\downarrow w$.

[quote="Third Edition"][quote="MegaRockman"]Why on every compact subset $ K$ of $ D$, $ w$  is the decreasing limit of functions in $ \cal F$?
Could you explain it more explicit? :maybe:[/quote]
If $ g_n$ is a sequence in $ \mathcal F$ approximating $ w$, take $ f_n \equal{} \min(g_1,\ldots,g_n)\in \mathcal F$. Then $ f_n\downarrow w$.[/quote]
but when we put $ w \equal{} \inf\limits_{f\in\cal F}f$, it means that $ w(s,t) \equal{} \inf\limits_{f\in\cal F}f(s,t)$, pointwise. Hence, for each $ (s,t)$, there exists a sequence $ \{f_n(s,t)\}$ decrreasing to $ w(s,t)$. I don't know what to do next

Note that for the sequence $ f_n$ of functions I constructed it holds $ f_{n\plus{}1}\le f_n$. It is not sufficient to look  for monotonicity at a single point (s,t).

ah, I mean the convergence of $ f_n(s',t')$ to $ w(s',t')$, where $ (s',t')$ is in a neighborhood of $ (s,t)$. Indeed, we did fix some $ (s,t)$, then choose a sequence $ f_n(s,t)$, which decreasing to $ w(s,t)$. But how about the other points in a neighborhood of $ (s,t)$. Does such sequence $ \{f_n\}$ convergent pointwise to $ w$ in a neighborhood of $ (s,t)$

[quote="Third Edition"]- $ w$ is continuous: On every compact subset $ K$ of $ D$, $ w$ is the decreasing limit of functions in $ \cal F$, i.e. $ f_n\downarrow w$, ... .[/quote]
Thinking about MegaRockman's concerns the innocent line above turned out to be a really nasty problem. Since there is no universal condition on the family $ \mathcal F$ the proof breaks down here - for the time being.

Can someone amend the proof? Comments are welcome.  :welcome:

this problem names "concave envelope". But, I have not found any reference yet.

I constructed w out of the blue sky without having any reference at hand I am afraid.  :(


AMC
AMC 10
geometry
probability
quadratics

I'm taking the AMC10 in a couple days, and what should I expect on it?  Is there an emphasis on a specific branch of math?  And are there any official old tests?  (everything here seems to be mostly on the AMC12s).  Geometry, I understand well, but the probability and the number theory's killing me...

It varies. You can't certain kinds of math to show up on the test, or to be weighted one way or the other. However, you can expect it to not include the more advanced math from the AMC12. If you know what most of the problems on the 12 are talking about, don't worry about your ability to do stuff on the 10.

first couple problems are usually related to exponents and substitutions of values

around the bottom, there is usually a circle problem....

when it gets to the middle, a little bit of probability and clever algebra
you should be able to know AT LEAST, quadratics and how to set up variables

last parts of the test is always geometry. you should be used to proving triangles

good luck,, although, it's probably over, good luck with AMC10 B

Check out the chart of problem topics on our website at:
http://www.unl.edu/amc/mathclub/quizes/04,0-questpercent.html

This same list is in the AMC 10/12 Math Club Package, your teacher may have a copy.

Steve Dunbar
AMC Director


If I choose four cards from a standard $ 52$-card deck, with replacement, what is the probability that I will end up with one card from each suit?

Multiplying the probabilities, we get $ 24\cdot\frac{13}{52}^4\equal{}\frac{3}{32}$. I think. :maybe:

[quote=kungfugirl]Why multiply by 24?[/quote]

Because there are $4!$ ways to choose which one goes first.

Alcumus solution: [hide]Because we are replacing the cards, at each draw, there is a $\frac{13}{52} = \frac{1}{4}$ probability of ending up with a card from any given suit. Because we are looking for one card from each of the four suits, it doesn't matter what suit the first card drawn represents. After one card is drawn and replaced, the probability that the second card drawn wil $\textit{not}$ be from the same suit as the first card is $\frac{3}{4}$. Similarly, after two cards have been drawn and replaced, the probability that the third card drawn will not be from either of the suits of the first two cards is $\frac{2}{4} = \frac{1}{2}$. Finally, the probability that the fourth card drawn will not be from any of the same suits of the first three cards that were drawn and replaced is $\frac{1}{4}$. Thus, our final probability is $\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \boxed{\frac{3}{32}}$.
[/hide]

This is how I did it:

[hide]First draw: 1/1 chance of new suit. Second draw: 3/4 chance of new suit. Third draw: 1/2 chance of new suit. Fourth draw: 1/4 chance of new suit. Multiply them, 1/1*3/4*1/2*1/4 = 3/32.[/hide]


You just find the circumference of each one. The number of revolutions is the total distance divided by the circumference.

Question #2
[hide]
[quote]A truck has tires each with a diameter of 3 feet, 6 inches. The tires are exchanged for larger tires, each with a diameter of 4 feet, 9 inches. There are 5280 feet in 1 mile. How many fewer revolutions than the smaller tire does the larger tire make in traveling 5000 miles? Express your answer to the nearest thousand.[/quote]
1.Find the circumference of each tire. 3 feet, 6 inches = 3.6 pie, 4 feet 9 inches = 4.9 pie.
2. We multiply 5000 and 5280 then divide by 3.6 pie and separatly 4.9 pie.
3. That gives us 2334272.499 and 1714975.713. We subtract the two numbers giving us a difference of 619296.786.
Inform me if I'm wrong.
Question #1 is hard. Still working. I can't get any Smilies in. Sort of had the same method as the previous person.  [/hide]

[quote="13375P34K43V312"]#1 is easily brute-forceable. It won't take more than 5 minutes.[/quote]
Brute force is the only way (I think)...
[hide="2"]Count how many perfect powers there are. There are 31 squares. There are are 6 cubes that aren't squares (1, 64, 729 have already been counted). The fours are all squares. The only fifth is 32. Sixth's have all been counted (2/3). Sevenths is only 128. No 8s (2s), 9s (3s), 10s(2/5). All elevenths (other than 1) are too big. So, 39 are perfects. 1000-39=961[/hide]

[quote="nutz_for2.718281828"]Here are two workout problems. I don't know the solutions.

A “perfect power” is a number that can be written as a positive integer raised to an integer power greater than one. For example, 125 is a perfect power because it is equal to $5^{3}$. The list 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, … contains every positive integer less than 1000 that is not a perfect power. How many integers are in the list?

A truck has tires each with a diameter of 3 feet, 6 inches. The tires are
exchanged for larger tires, each with a diameter of 4 feet, 9 inches. There are
5280 feet in 1 mile. How many fewer revolutions than the smaller tire does the
larger tire make in traveling 5000 miles? Express your answer to the nearest
thousand.

Thanks. :)[/quote]
for first problem, since the highest exponent below 1000 of an integer other than 1 is 9 ($2^{9}$), all the numbers without a perfect Square root, Cube Root, Fifth Root, or Seventh Root would be in the list.
other than 1, the cubes start with 8, so no need to check for cubes until 8.
other than 1, the fifth powers start with 32, so no need to check for fifth powers until 32. 243 is the only other fifth power.
other than 1, the seventh powers start with 128, and is the only 7th power.
So basicly other than 1, 32, 243, and 128, only squares and cubes need to be worried about.
There are 30 squares between 1 and 1000, and 9 cubes (does not include 1). two of those numbers are repeats.
37+4=41, and 1000-41=$959$.

Thanks, ProtestanT. I just checked the answers, and you got it right.

On number two, here's what I did.[hide="Look in here."]I just took the circumferences in feet: 3.5pi, 4.75pi. Then I Found out how many feet were in 5000 miles. Then I divided that y each of the circumfrences seperately, and subtracted, took the absolute value, and floored the result. I got it wrong. :huh: [/hide]

[quote="nutz_for2.718281828"]I tried that. It didn't work.[/quote]
How does it not work?

[quote="nutz_for2.718281828"]Here are two workout problems. I don't know the solutions.

A “perfect power” is a number that can be written as a positive integer raised to an integer power greater than one. For example, 125 is a perfect power because it is equal to $5^{3}$. The list 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, … contains every positive integer less than 1000 that is not a perfect power. How many integers are in the list?

A truck has tires each with a diameter of 3 feet, 6 inches. The tires are
exchanged for larger tires, each with a diameter of 4 feet, 9 inches. There are
5280 feet in 1 mile. How many fewer revolutions than the smaller tire does the
larger tire make in traveling 5000 miles? Express your answer to the nearest
thousand.

Thanks. :)[/quote]

this might still be in my calculator as we did it last week in mcounts...i cant get to it now, but i got some crazy decimal (for the second one)

as for the first one i tried subtracting fourths, fifths, etc. but something was messed up

-jorian

sixth powers are also third and second powers

(i know, i couldn't figure out what to do then...i don't remember what i did, but i think i got ~970)

-jorian

nuts_for [i]e[/i], could you tell me which year
and which workout # the truck tire problem came from?

if it's this year i have the solutions :thumbup:
I have them under my bed that is.... :sleeping:


Given $ f(x) \equal{} \frac{\sqrt{x\minus{}1}}{x\minus{}2}$, what is the smallest possible integer value for $ x$ such that $ f(x)$ has a real number value?

The radical needs to be nonnegative, so x=0 is minimum.

eh... \[ x\minus{}1\geq 0 \Rightarrow x\geq \boxed{1}\]


function
limit
integration
complex analysis
complex analysis unsolved

Hi,

I was wondering whether the function $ F(s): \equal{}s$ has Dirichlet-series expansion on some half-plane. So I naively tried to compute the coefficients using the formula

$ f(n)\equal{}\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{\minus{}T}^T F(\sigma \plus{} it)n^{\sigma\plus{}it}dt$,

but I got a term that is indefinite as $ T\rightarrow\infty$. Am I missing something here (resp. I need to 'debug' my calculation), or the identiy function does in fact not have an expansion as [i]absolute-converging[/i] Dirichlet series on any half-plane? If so, does the identity function have any expansion as Dirichlet series (without the condition for absolute convergence) at all?

Thanks in advance,

efq

PS: This might be some dumb question as it is late here and I´m somewhat irritated at the moment :)

Of course, it turned out to be a dumb question. Any convergent Dirichlet-series is at least $ o(|t|)$, and I was thinking all the time of $ O(|t|)$ for some unknown reasons :blush: Thanks go to [b]outermeasure[/b] at sosmath.com for bringing it to my attention!


[quote="jarno"]solve: $ \sqrt {2 \minus{} \sqrt {3}cosx \plus{} sinx} > 1$[/quote]

<=>$ \sqrt {2sin\left(x\minus{}\frac {\pi}{3}\right)\plus{}2} > 1$
<=>$ 2sin\left(x\minus{}\frac {\pi}{3}\right)\plus{}2 > 1$
<=>$ 2sin\left(x\minus{}\frac {\pi}{3}\right) > \minus{}1$
<=>$ sin\left(x\minus{}\frac {\pi}{3}\right) > \minus{}\frac{1}{2}$
<=>$ sin\left(x\minus{}\frac {\pi}{3}\right) > sin\left(\minus{}\frac{\pi}{6}\right)$
<=>$ \minus{}\frac{\pi}{6}\plus{}k2\pi<x\minus{}\frac{\pi}{3}<\frac{7\pi}{6}\plus{}k2\pi$
<=>$ \frac{\pi}{6}\plus{}k2\pi<x<\frac{3\pi}{2}\plus{}k2\pi$, where k is an integer.


[quote="mathemagician1729"]Is it possible to conceal the css file that someone custom made on aops blogs to prevent theft of css?[/quote]

I changed it BTW.  And I doubt you can.  I only did it cause I thought you hacked my account, but now I understood....

[quote="mathemagician1729"]Is it possible to conceal the css file that someone custom made on aops blogs to prevent [b]theft[/b] of css?[/quote]

Did you copyright it?

Anyway, I believe there isn't anyway to conceal the code, but if anyone "steals" your code, its just their lack of orginality.

If you want people to see your blog, then they automatically download the css. There is no way to "conceal" it :)


algebra
polynomial
modular arithmetic
trigonometry
complex numbers
algebra solved

[hide="First part"]Take $r$ such that \[ r^k+r+1=0, \] \[ r+1=-r^k. \] Then \[ r^{2k+1}+r+1=0, \]\[ r^{2k+1}-r^k=0, \] \[ r^k(r^{k+1}-1)=0. \] Clearly, $r\neq0$ and $r\neq1$, so \[ r^k+\ldots+r+1=0. \] That is, $r$ is a $(k+1)$th root of unity. But subtract the first equation from this and divide by $r^2$ to get \[ r^{k-3}+\ldots+r+1=0. \] That is, $r$ is a $(k-2)$th root of unity too for every choice of $r$. Thus, $r$ is a third root of unity, which implies $k=2$.[/hide]

oo nice use of roots of unity haha

here's how i approached it:

[hide]
For the first part, using synthetic division we see that:

\[ \frac{x^{2k+1} + x + 1}{x^k + x + 1} = x^{k+1} - x^2 - x + \frac{x^3 + 2x^2 + 2x + 1}{x^k + x + 1}  \]

Doing a quick check, we see that the only solution is when $k = 2$.

To solve the second part, we can use long division again, using $k=2$ and $n$. Hence we have:

\[ \frac{x^n+x+1}{x^2 + x + 1} = x^{n-2} - x^{n-3} + x^{n-5} - x^{n-6} + \ldots  \]

Though, the important steps are to note that during the long division process, you have tentative remainders of $x^{n-3} + x + 1$, and then after two more steps you reach $x^{n-6}+x+1$. We also note that in our previous synthetic division, we saw that if $n = 2k+1 = 5$, then there will also be no remainder, and this is obviously true when you have $n=2$. From these examples, this suggest something in mod 3. Hence, as long as $n \equiv 2 \mod 3$, we know that $x^n + x + 1$ will be divisible by $x^2 + x + 1$.

This can also quickly be shown using induction. For base case $n=2$, it is obviously true. Let's assume a case of $n$, such that $n \equiv 2 \bmod 3$ is true. Then we have $n+3$ as our next case, preserving the modulus. Then, using synthetic division, we arrive at:

\[ \frac{x^{n+3} + x + 1}{x^2 + x + 1} = x^{n+1} - x^{n} + \frac{x^n + x+ 1}{x^2 + x+1}  \]

 From our inductive hypothesis, we know that the fractional part above gives no remainder, and hence $x^{n+3} + x+ 1$ is also divisible by $x^2 + x+1$, and we are done. $\blacksquare$

[/hide]

i think its right..

Why did you only consider $k=2$ for the second part?

[hide="Another way to solve k=2 using roots of unity"]
If $\omega_1,\omega_2\neq1$ be distinct third roots of unity, so that $\omega_1^2+\omega_1+1=0$ and $\omega_2^2+\omega_2+1=0, \omega_1^3=\omega_2^3=1$.

For any $n=3m+2$, $m\in\mathbb{Z}$, $x^n+x+1=0$ for $x=\omega_1,\omega_2$. Since all the roots of $x^2+x+1=0$ are roots of $x^n+x+1$, then $(x^2+x+1)|(x^n+x+1)$ for all $n\equiv2\pmod{3}$. On the other hand, if $n\not\equiv2\pmod{3}$, $x^n+x+1\neq0$...[/hide]

[quote="gotztahbeazn"]Find all positive integers $k$ such that the polynomial $x^{2k+1} + x+ 1$ is divisible by the polynomial $x^k + x+ 1$. For each $k$, specifiy the integers $n$ such that $x^n + x + 1$ is divisible by $x^k + x+1$.[/quote]

part one:
let $Q(x)=x^k+x+1$ and $P(x)=x^{2k+1} + x+ 1=x^{2k+1} -x^k+x^k+ x+ 1$
$=x^k(x-1)(\sum_{i=0}^{k} x^i)+(1)(x^k+ x+ 1)$

evidently, if $Q(r)=0$, then we must have that $P(r)=0$ and the reverse, note that the second terms is already divisible, so we ignore that, also, we cannot have that $Q(0)=0$, or $Q(1)=0$, so those factors cannot introduce divisibility, so we just look at $Q(x)|Z(x)$ where $Z(x)=\sum_{i=0}^k x^i$. They have the same degree, so they must have all roots in common, and only be a constant multiple different using the factor theorem. However, we have that they have the same lead coefficient, so $Q(x)=Z(x)$. Then $Q(x)$ has only 3 terms, so $Z(x)$ has 3 terms, giving $\boxed{k=2}$

part 2:
define $Q(x)=x^k + x+1$, and $P(x)=x^n + x + 1=x^n -x^k+x^k+ x + 1=x^k(x-1)(\sum_{i=0}^{n-k-1}x^i)+(1)(x^k+ x + 1)$

Define similarly, $Z(x)=\sum_{i=0}^{n-k-1}x^i$, clearly, $Z(r)=0$ for $r=e^{\frac{2i\pi m}{n-k-1}}$ for $m\in \overline{1,n-k-1}$, the exact logic from before gives us $Q(x)|Z(x) \leftrightarrow Q(x)|P(x)$. So $Q(e^{\frac{2i\pi m}{n-k-1}})=1+e^{\frac{2i\pi m}{n-k-1}}+e^{\frac{2i\pi m*k}{n-k-1}}=0$
Looking at the imaginary part, we get $\cos\frac{2\pi m}{n-k-1}=\cos\frac{2\pi m*k}{n-k-1}$
or $\pm\frac{2\pi m}{n-k-1}=2p\pi+\frac{2\pi m*k}{n-k-1}$
$\frac{m}{n-k-1}*(k\pm1)=p$...stuck....is the question asking for a value of n in terms of k??

can anyone finish the second part please? why is the answer ONLY considered for k=2? thanks!!


what was the point of the french revolution (ie what were the people trying to acheive)?

i mean, i know the causes and effects, etc., but i don't see what the whole thing was about...

What was the point of the American revolution?

The French peasants were all super poor and had nothing to eat and I think they had to pay more taxes than all the people in the super high estate. So they wanted equality and stuff and so they destroyed the absolutist government thing.

GOVERNMENT. THIS IS ALWAYS THE CASE.

And The American Revolution, isn't quite justified, the colonists could've been like the Canadians.

i think it went like this:

peasants: we're hungry.
french royalty: go eat cake.
peasants: yum. now we hate you.
french royalty: what can you do about it? chop our heads off or something?
all: *hearty laugh*

at least, that's what i'm told.

I think not seeing the point only implies some big gap in the knowledge of the causes. (Of course you may not agree to the revolutionaries, that's a different matter.)

[quote="junggi"][quote="funcia"]the colonists could've been like the Canadians.[/quote]
i thought britain gave canada independence because they were afraid that they would be like US and start a revolution?
dunno I'm bad at history[/quote]
unless i am just not paying attention in history(which could be but eh i love canada.) i don't know wiki it but it's late so i am not able to do that at the moment...

Actually, Captain Sugar, there's no evidence that Marie Antoinette actually said "Go eat cake" when the peasants said there was no bread. It's a common misconception.  :wink:

Now for the real story.

It all started with Louis the fourteenth. He started waging all sorts of expensive wars against various countries, with little results. On top of that, he basically drained the royal treasury by building the palace of Versailles. His great grandson Louis XV continued his legacy. Finally Louis XVI ascends the throne. Throughout the past hundred years, the royals had increased taxes in order to pay for their lavish lifestyles, with peasants starving in the villages. Louis XVI then decides to help the US with their little Revolution, further draining the treasury and increasing taxes. Those who dared protest were sent to the Bastille. France finally fell into disorder, and in 1791, the French Revolution started.

So basically the French Revolution was about having enough food to eat, not being thrown in jail, etc.

[quote="junggi"][quote="funcia"]the colonists could've been like the Canadians.[/quote]
i thought britain gave canada independence because they were afraid that they would be like US and start a revolution?
dunno I'm bad at history[/quote]
Uhm, no I don't think so, I'd say Canada became independent in 1931.  And it has always been on good terms with the UK, officially their monarch is still Queen Elizabeth II. Remember that after the American Revolution, there have been wars with the UK in which Canada was on the same side as the UK

What I'd like to say about the French revolution is:
- it took a really time before French really stopped having kings and emperors.  When Napoleon was finally defeated in 1815, France had three more kings until 1848, and after that Napoleon's nephew (well some say he wasn't) became emperor, until the third republic started in 1870 (now they have the Fifth Republic)
-  I wouldn't portray the revolution as a really miraculous change for the people.  I mean, those wars under Napoleon with pretty much everyone else in Europe severely bankrupted the country.
- I wouldn't exactly say the idea of bringing "Equality, fraternity, liberty" to everyone was that serious.  Remember, some sources go as far as saying that even in 1870, only 30% of all people in France spoke French.  What makes the French identity is the common language, but -ironically- it's not the original language in most of the constituent regions.


Let $ AB$ be a chord of a circle. Points $ P$ and $ Q$ are on the same side of $ AB$; $ P$ is outside the circle; $ Q$ is inside the circle. Prove that $ angle APB  <  angle AQB$.

\[\theta  = \frac{1}{2}\left( {arc.AB + arc.CD} \right) > \frac{1}{2}arc.AB = \omega  > \varphi \]


Ok, I'll continue with the inequalities ...

This is too easy but I'll post it anyway :

If a, b, c :ge: 0 then a/bc + b/ca + c/ab :ge: 2/a + 2/b - 2/c.

This is a one - line argument !


I submitted the solutions two days ago, but apparently the site is down. can i re-submit it ?

 :P

Not sure what you mean about 'the site is down'.  If we have already received your solutions, you cannot re-submit them.

[quote="dhuang614"]is there a reason for this? what if I accidentally submitted an empty email or the wrong solutions before the deadline?[/quote]

If you send in a blank email, of course you can re-send.  We have many reasons for not allowing re-submissions, not least of which is that it is a major headache to receive multiple solutions and have to sort out which is 'official'.

I resubmitet mine today because I didn't get a reply so I didn't know if you recieved it is that okay and did you recieve it?

It typically takes us several days to go through all the submissions and get them marked as received - we'll take your first submission as the official one.


Have yourself a nice old fashioned mole frying!

Oh snap I can't find moles.  :( [img]http://www.megatonik.com/wp-content/uploads/2007/09/moley-mole.jpg[/img]

10/23? And a happy mole day to you. Here it was 6:02 just a few minutes ago,

Because of this, I decided to look up what a mole is.
mole

1. any of various small insectivorous mammals, esp. of the family Talpidae, living chiefly underground, and having velvety fur, very small eyes, and strong forefeet.
2. a spy who becomes part of and works from within the ranks of an enemy governmental staff or intelligence agency. Compare double agent.
3. Machinery. a large, powerful machine for boring through earth or rock, used in the construction of tunnels.
4. A value of 6.02*10^23,used in chemistry.
I guess you are talking about #4 right?

[quote="Jongao"]Because of this, I decided to look up what a mole is.
mole
.
.
.
2. a spy who becomes part of and works from within the ranks of an enemy governmental staff or intelligence agency. Compare double agent.
[/quote]
Aaahhh... I finally got where do "The Mole" come from" :D

WHen I think of mole, i think of someone who is untrustworthy...

I think of the Giant Mole that lives underneath Falador (if you understood this post please seek medical help IMMEDIATELY!)

[quote="mewto55555"]if you understood this post please seek medical help IMMEDIATELY![/quote]

Phew!
@AIME15: We don't wanna know.

OMG HAHAHAHAHHA!

googled it...

[size=150]MEWTO55555 PLAYZ RUNESCAPE!?[/size]

seriously...mew you used to play runescape?

thats super funny...

back on topic, why is a mole such a random number like 6.002*10^23?

Maybe something scientifically pertinent?

$ \pi\equal{}3.1415...$ why not a mole??

Because there were no moles when $ \pi$ was invented :D

A mole of a substance is as many kilograms of that substance as its atomic weight. The number is he number of molecules of the substance in the mole.

Actually, it's grams, not kilograms.  For example, since carbon's atomic weight is about 12.01, 1 mole of carbon would weigh about 12.01g.

$ 6.02 \times 10^{23}$ isn't random.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1048072513&t=214946