An unfair coin is tossed 10 times. The probability of heads on each toss is $0.4$. Let $a_n$ be the number of heads in the first $n$ tosses. Find the probability that $a_n/n \le 0.4$ for $n = 1, 2, \ldots ,9$ and $a_{10}/10 = 0.4$. (http://www.kalva.demon.co.uk/aime/aime02b.html)

I feel too lazy to do this now, but I think that brute force (i.e. a graph, will work well here.

[hide="Answer"]We have $4$ heads in $10$ tosses. This means that the first two tosses cannot be heads and the last toss must be heads. So, we have T T _ _ _ _ _ _ _ H, and we must place three H in those spaces and fulfill the conditions. From here, we list out the earliest places where each consecutive heads can appear:

T T H T H T T H T H

We must now find the number of ways to arrange the first three heads in the seven available spots. Listing:

HTHTTHT
HTHTTTH
HTTHTHT
HTTHTTH
HTTTHHT
HTTTHTH
HTTTTHH
THHTTHT
THHTTTH
THTHTHT
THTHTTH
THTTHHT
THTTHTH
THTTTHH
TTHHTHT
TTHHTTH
TTHTHHT
TTHTHTH
TTHTTHH
TTTHHHT
TTTHHTH
TTTHTHH
TTTTHHH

This gives $23$. Looking at kalva's answer, that's the right number, and it should be prime factorized. The probability of ending with 4 heads is $\boxed{\frac{2^4(3^6)(23)}{5^{10}}}$ after simplification.[/hide]

hmm wats the most efficient way to get all those cases tho? cuz on a real AIME under time restraint and everything ud have to be a lil worried that u missed something...

Probably not. That only took me about 5 minutes to do. It's fairly simple, since you realize that the third can only be in one of the last two spots, and the second can only be in one of the last five. The first can technically be in any spot, although in practice it only has the first five as options, since the second and third must be in the last two. You just need to work through the possibilities. I guess the one thing you can't be sure of with my method is whether $23$ is actually the right number, but you should have time to double-check and make sure you didn't miss anything. Also, I'm guessing that the original question had something like $\frac{a^pb^qc}{d^r}$, find $a+b+c+d+p+q+r$. When you see that there's one prime factor, you can guess that $23$ is probably right.

However, if anyone has a more definite solution, I'd like to see it. :)

[quote="JesusFreak197"]Probably not. That only took me about 5 minutes to do. It's fairly simple, since you realize that the third can only be in one of the last two spots, and the second can only be in one of the last five. The first can technically be in any spot, although in practice it only has the first five as options, since the second and third must be in the last two. You just need to work through the possibilities. I guess the one thing you can't be sure of with my method is whether $ 23$ is actually the right number, but you should have time to double-check and make sure you didn't miss anything. Also, I'm guessing that the original question had something like $ \frac {a^pb^qc}{d^r}$, find $ a \plus{} b \plus{} c \plus{} d \plus{} p \plus{} q \plus{} r$. When you see that there's one prime factor, you can guess that $ 23$ is probably right.

However, if anyone has a more definite solution, I'd like to see it. :)[/quote] I think I have one.  :)

[hide]First, we make a list of the maximum number of heads you can have once you are at a certain number of flips.  1-0, 2-0, 3-1, 4-1, 5-2, 6-2, 7-2, 8-3, 9-3, 10-4. Obviously, you must land tails on the first two flips and you must land heads on the last flip. So we have to arrange the remaining $ 3$ heads among the $ 7$ spots. To make it easier, we will put the 3rd and 4th flips in group A, the 5th, 6th, and 7th flips in group B, and the 8th and 9th in group C. We have two cases.
Case 1: Heads is flipped on only one of the turns in group C.
There are two ways to choose on which turn heads was flipped. Then, to choose the remaining two heads, we can have two heads in group B or one in group B and one in group A. The total is $ 2*(\binom{3}{2} \plus{} 3*2) \equal{} 18$.
Case 2: Heads are flipped for both of the turns in group C.
Since the other head can be in any of the remaining $ 5$ flips, the total is just $ 5$.

$ 18 \plus{} 5 \equal{} 23$. Now, because there are $ 4$ heads, there are $ 6$ tails. So the total probability of the requirements of the problem being filled out is $ 23*(\frac {2}{5})^4(\frac {3}{5})^6 \equal{} \frac {23*2^4*3^6}{5^10}$.
[/hide]

Note the original problem is [url=http://www.artofproblemsolving.com/Wiki/index.php/2002_AIME_II_Problems/Problem_12]here.[/url]


Find the fiftieth decimal place in the decimal representation of the sum of 1/6 and 1/7.

\begin{align*}
\frac{1}{6}+\frac{1}{7} &= \frac{13}{42} \\
&= 0.3\overline{095238} \\
\end{align*}

The 50th digit after the decimal is the 49th digit of the repeating block.  $ 49\equiv 1\pmod{6}$, so we want the first digit, which is $ \boxed{0}$.

[quote="bassamali01"]what is $mod$[/quote]
it's modular arithemtic, it's covered in intro to NT

[quote="bassamali01"]long form of nt???[/quote]

Look it up...

where.
Why don't you say it youself

Don't confuse me plz

Thanks
Is it in Counting and Probability

No. It is in introduction to number theory.

Is there any information or video available about it in this website


calculus
function
analytic geometry
graphing lines
slope
calculus computations

[quote="socialette"]Determine the values of a and b so that the function is continuous and differentiable for all x € R where  f(x) is defined as follows:
f(x) =  x^2  + a , x < 2; bx – 3, x ≥ 2

  :D[/quote]

At x=2, the slope and y-value have to be the same:

$2x = b$
$b = 4$

$x^{2}+a = bx-3$
$4+a = 2b-3$
$4+a = 8-3$
$a = 1$

$\boxed{a=1,b=4}$


trigonometry
function
logarithms
complex analysis
calculus
calculus computations

I'm posting this here because I suspect that the answer involves calculus.  Does anyone know of a proof to the above identity?

this just follows from the power series representations of $\sin z$, $\cos z$, $e^{z}$, for complex $z$:

$\sin z = \frac{z}{1!}-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-...$

$\cos z = \frac{1}{0!}-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-...$

$e^{z}= \frac{1}{0!}+\frac{z}{1!}+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+...$

get any book on complex analysis...

I always thought that Euler's formula for the complex case followed from the way $e^{iz}, \sin z$ and $\cos z$ were defined for complex values. What I mean is the above Taylor series expansions of $e^{iz}, \sin z$ and $\cos z$ are really definitions: they are not formulas in the sense we normally understand.

A key point: consider the function defined by the power series

$f(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}.$

We know that this converges for all complex $z.$ (The power series has an infinite radius of convergence; the complex variables name for this is that it's an [i]entire function[/i].)

We know that for $x\in\mathbb{R},\ f(x)=e^{x},$ the function we know (and think we're familiar with) from calculus. The series converges for $z$ not on the real axis, as well. Is this a sufficient reason for us to accept the notation $e^{z}$ for $f(z)?$

Being an exponential carries many other properties as well, most prominently the collection of algebraic identities known as the "laws of exponents." This leads to the following claim:

[b]Theorem:[/b] If $f(z)$ is defined as $\sum_{n=0}^{\infty}\frac{z^{n}}{n!}$ for any $z\in\mathbb{C},$ then $f(z+w)=f(z)f(w)$ for all $z,w\in\mathbb{C}.$

[I won't post the proof. Why spoil the fun? If you don't know it already, it's worth trying to prove this.]

Corollaries of this would include that $f(z)\ne0$ and $f(-z)=\frac1{f(z)}.$ Since we can replicate the laws of logarithms, we seize the opportunity to make the [i]definition[/i] $e^{z}=f(z)$ equals this power series.

Now, where does that leave us? Suppose $z=x+iy,$ with $x,y\in\mathbb{R}.$ Then:

$e^{z}=e^{x+iy}=e^{x}e^{iy},$ by the theorem.

$e^{x}$ is our familiar real exponential. Now expand out the power series for $e^{iy}$ and compare it to what you know about cosine and sine.

---

I'm saying that FieryHydra is essentially correct in that we're talking about definitions, but I'm adding the point that the "laws of exponents" theorem above needs to be true or else the alleged definitions would be hollow and unconvincing.


I have the solution to this problem, but I don't really understand it. Could somebody give me a solution with as much detail as possible? Here it is;

$ arccsc \equal{} \minus{}1$

I understand what $ arc$ is, and that $ csc \equal{} \frac{1}{sin}$, but I am still having trouble understanding the solution. Thanks.

[hide="solution"]
Take $ csc$ of both sides:
$ csc(arccsc(x))\equal{}csc(\minus{}1)$
Since $ csc(arccsc(x))\equal{}x$, we get
$ x\equal{}csc(\minus{}1)$

This is the answer to the problem.
[/hide]

Hope that helped  :)

Just as a general tip. (This tip is from AoPS Vol. 2), Anytime you are given something along the lines of inverse trig functions, always try to rewrite that as a normal trig function, and then solve it from there. So, if you have something like:

Find $ \arcsin 1.$
Just convert this into something of the form: $ \sin x \equal{} 1,$ and then solve this, it usually helps.

Okay, thanks a lot, that really helps.  :lol:


algebra polynomial calculus integration AMC USA(J)MO USAMO

Let $ a,b,c$ be three distinct integers, and let P be a polynomial with integer coefficients. Show that in this case the conditions:

$ P(a) \equal{} b$, $ P(b) \equal{} c$, $ P(c) \equal{} a$

cannot be satisfied simultaneously.

[quote="1234567890"]Looks good to me.  :). But you might need some more work at the end. I don't think the fact that a=b=c, and k_1=k_2=k_3 are immediately obvious. But that's easy to fix.[/quote]
Here's how we can fix that part.

We have the following results:
$(a-b)| P(a)-P(b) \Rightarrow (a-b)| (b-c)$
$(b-c)| P(b)-P(c) \Rightarrow (b-c)| (c-a),$ and
$(c-a)| P(c)-P(a) \Rightarrow (c-a)| (a-b)$.

Now, we know that for any two non-zero integers $x$ and $y,$ $x|y$ and $y|x \Rightarrow x=y.$ So, from the above statements we conclude $a-b=b-c=c-a =k(\text{say}).$ Adding them up, we get $3k=0 \Rightarrow k=0 \Rightarrow a=b=c,$ which is clearly a contradiction!

Suppose a-b=0, so a=b. Then P(a)-P(b) also is equal to zero, so by the Factor Theorem, a-b is a divisor of P(a)-P(b). :)

It's not a difficult lemma to prove.

[hide="Proof 1"] Of course $a-b | (a^{n}-b^{n})$.  Hence $a-b | c_{n}(a^{n}-b^{n}) \implies a-b | \sum_{k=1}^{n}c_{k}(a^{k}-b^{k})$ [/hide] [hide="Proof 2"] Consider the ring of polynomials of $a, b$ taken $\bmod a-b$.  Of course, $P(a) \equiv P(b)$. [/hide]

[quote="maokid7"]but what if $a-b\neq0$?[/quote]

It doesn't matter. The factor theorem states that if g(x)=0 implies f(x)=0, then g(x) is a factor of f(x).

[quote="PenguinIntegral"][quote="drunner2007"][quote="maokid7"]but what if $a-b\neq0$?[/quote]

I think that you can look at the general form of $P(x)$, plug things in, and then simplify.[/quote]
How so?[/quote]

I give this proof in my post.  You might better understand it from [hide="the other direction"] Let $P(x) = \sum_{k=1}^{n}c_{k}x^{k}$ be a polynomial.  Then

$P(a)-P(b) = \sum_{k=1}^{n}c_{k}\left( a^{k}-b^{k}\right)$

And $a-b | a^{k}-b^{k}$.  QED. [/hide]

Could someone please comment on my proof?
[hide]Note that substituting $ c \equal{} P(b),b \equal{} P(a)$ to $ a \equal{} P(c)$ we get $ P(P(P(a))) \equal{} a$ for integral $ a$.

Then let $ deg(P) \equal{} n$ and $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \plus{} a_0$, it follows (from doing the substitution) that the term with the highest power of $ x$ in $ P(P(P(x)))$ is $ {a_n}^{n^2 \plus{} n \plus{} 1}x^{n^3}$. In order to have $ P(P(P(x))) \equal{} x$, it is only possible to have either $ n^2 \plus{} n \plus{} 1 \equal{} 0,n^3 \equal{} 1$ or $ a_n \equal{} 1,n^3 \equal{} 1$.

The first case does not yield any positive integral solution, while the second one yield $ a_n \equal{} 1,n \equal{} 1$. So $ P(x) \equal{} x$.
However, If $ P(x) \equal{} x$, then $ P(b) \equal{} b$, but this must also be equal to $ c$ according to the problem statement, contradicting the condition $ b\neq c$.[/hide]
By the way, are you allowed to leave out the small details like the substitutions and the nonexistence of integral solution?
Also, why don't we have 1988 and earlier years usamo problems in http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27 ?
I had to search for usamo 1974 in order to get to this page (also i don't think the search yields all the 6 problems... so some haven't been posted yet)... so is there another archive where I can find the past years (<1989) problems? Thanks.


Let $ n\in N^*$
$ f(n)\equal{}(1\minus{}\frac{1}{n})^n$
$ g(n)\equal{}(1\plus{}\frac{1}{n})^n$
Let $ A\equal{}f(N^*), B\equal{}g(N^*)$
$ f: N*\to A$, $ g: N^*\to B$.
Prove that
a) $ f,g$ have inverse map
b) $ f$ increasing, $ g$ decreasing

not exactly sure what you are looking for...

anyway the relationship from $ f$ to $ g$: $ f(\minus{}x) \equal{} (1\plus{}\frac{1}{x})^{\minus{}x} \equal{} g(x)^{\minus{}1}$ hence $ g(x) \equal{} \frac{1}{f(\minus{}x)}$


In convex pentagon $ABCDE$, the sides $AE,BC$ are parallel and $\angle ADE=\angle BDC$. The diagonals $AC$ and $BE$ intersect at $P$. Prove that $\angle EAD=\angle BDP$ and $\angle CBD=\angle ADP$.

Could you clarify what you mean with diameters? is this pentagon cyclic?

Tiger100's confusion was due to your usage of the word 'diameter', instead of diagonal:

[quote="behdad.math.math"]In convex pentagon ABCDE, two sides AE, BC are parallel. <ADE  =  <BDC two diameters AC, BE intersect each other on P.
prove that <EAD=<BDP and <CBD=<ADP.[/quote]

Let the line passing through $B$ parallel to $DE$ intersect $DP$ at $F$. Then $FBCP\sim DEAP$, so $\angle CFB=\angle ADE=\angle BDC$, so $FBCD$ is cyclic. Thus $\angle DBC=\angle DFC=\angle FDA$. Similarly the other one.


search
algebra
polynomial
Ramsey Theory
graph theory
combinatorics open

Question:

What simple facts about complete graphs can be used to obtain bounds for
Multicolor Ramsey numbers?

I have thrown together an attempt at doing this.
(Has anyone got any other good ones!)

Try this....

Take the muticolor Ramsey number R2(3,3,3,3)
[defined for the sake of this argument as :
What is the number of vertices of the smallest complete graph such that
however its edges are colored in four colors one cannot avoid a monochromatic triangle?]

Here is a really bad lower bound for R2(3,3,3,3) obtained using a simple method....

The thickness of a graph G is defined by the minimum number of subgraphs
of G that are planar, but whose union is G.

Using the Four Color theorem (any planar graph can be vertex colored in 4 colors
such that no vertex of the same color is adjacent) we can immediately see that
any complete graph that R2(3,3,3,3) vertices must have thickness at least 3.

This is because we can create a graph coloring of a graph of thickness 2 with no
monochromatic triangles from this by coloring each of the planar layers in 2 of
the 4 colors.

This is easily done by 4 vertex coloring (according to the Four Color Theorem)
each planar layer then coloring the resulting triangles as follows:

make our vertex colors Red, Green, Blue and Yellow, and our two edge colors
Pink and White.

Triangle   edge

RGB        RG  : P   RB : P   GB : W
RGY        RG  : P   RY : W  GY  : W
RBY        RB  : P   RY : W   BY : W
GBY        GB : W   GY : P   BY : W

How bad is this? Bad!
There is a formula for thickness in complete graphs, I believe it is |_ 1/6(n+7) _|
Compare the estimates for R2(3,3,3,3) (around 60) with this and you see why!

However, I think there is much scope for improvement.
Anybody know of any other such methods?

I haven't read the whole your post, but the simpliest estimates, indeed, give 61 for $R_2(3,3,3,3)$.

I think there are a plenty of articles and books devoted to this theme. It is better to search for them.

I have read various papers on the subject of finding bounds for such numbers.

Let us take the first known non-trivial such number R(3,3,3).

Let me quote from a paper by Fettes, Kramer and Radziszowski entitled
'An Upper Bound of 62 on the Classical Ramsey Number R(3,3,3,3)'.

This paper gives a brief overview of the history for the solution of R(3,3,3).
The seminal article on this problem (by Greenwood and Gleason) is quoted as follows:

"R(3,3,3) >= 17 was shown by giving a good 3-coloring of K16.
 The construction relies on finite field theory using the Galois Field of
 order 16 and considers elements of the field to be vertices of a graph..."

Kalbfleisch and Stanton (1968) showed that in a 3-edge coloring of K16
without monochromatic triangles, the subgraph induced by the edges of any one
color is isomorphic to the Greenwood-Gleason graph (see mathworld.wolfram.com,
an excellent website, for a drawing).

I could not translate any of the ideas in papers about R(3,3,3)
into the concept below that I found interesting.
Does anybody else think this is intriguing?

K16 should have a thickness of 3 (whereas K17 is the first complete graph
that should have thickness 4 as is my understanding).

If one could find the 3 planar graphs representing the thickness of K16 they
would have to be pretty much planar complete. Total number of edges in
K16 is 16*15/2 = 120, any planar graph on 16 vertices can have at most
3*16 - 6 = 42 edges.

Now 4-color the vertices of each planar graph and the corresponding edges in
3 colors using the following:
(vertices Red, Green, Blue, Yellow : edges Pink, White, Orange)

triangle   edge

RGB        RG pink RB white GB orange
RGY        RG pink RY orange GY white
RBY        RB white RY orange BY pink
GBY        GB orange GY white BY pink.

I would imagine we now have three planar graphs that have a fairly even spread
of edge colors.

The only problem left to overcome occurs when putting the layers together again:

Layer 1: 2 vertices of the same color (WLOG) red can be connected to another
         vertex, say blue. Therefore, in this layer, we have two edges of the
         same color, white, attached to a vertex.

Layer 2: WLOG we can say that the edge joining the two red vertices in layer 1
         is in this layer. We need to guarantee when we 4-color this layer that
         the resulting edge is NOT white in color. This would require us to
         guarantee that the two vertices here are NOT colored green and yellow
         or red and blue!

I think this is a problem that cannot be overcome in a general sense.
One would have to have the 3 planar graphs to test at hand, and using their
respective chromatic polynomials see if it is possible to do!

I was toying with the idea of trying to come up with such graphs if they
are possible, as we know the Greenwood-Gleason graph pretty much defines
the possible solutions, but I failed at the first hurdle : trying to find
the 3 planar graphs that combine to form K16!!

Anyway, one last point, it seemed no more than a coincidence to me that
the number R(3,3,3,3), as yet unfound, has lower and upper bounds
converging close to the complete graphs of thickness 10 ?!


The number of students in Teresa's graduating class is more than 50 and fewer than 100 and is 1 less than a multiple of 3, 2 less than a multiple of 4, and 3 less than a multiple of 5. How many students are in Teresa's graduating class?

If a number is 3 less than a multiple of 5, then it ends in a 2 or a 7. We are given the number is 2 less than a multiple of 4, so it can't end it seven (because mutliples of 4 end in even digits so 2 less than a multiple of 4 also ends in an even number). So were are down to 5 possibilities: 52,62,72,82,92. Of these only 62 and 82 are 2 less than a multiple of 4. And only $ \boxed{62}$ is 1 less than a multiple of 3.

Alternatively, the number is

2mod3
2mod4
2mod5

Combining the first two parts with the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese Remainder Theorem[/url] we see the number is 2mod12.

Thus the number is 2mod12 and 2mod5. Using the Chinese Remainder Theorem again, the number is 2mod60.

We now know the number is ...-58,2,62,122...

The only number greater than 50 and less than 100 of these is $ \boxed{62}$

Sorry for the bump but I have a question. When you have $x mod a$, $x mod b$, and $x mod c$, and if a, b, and c are relatively prime, is it always equal to x mod (a * b * c)?

[quote=tenebrine]Not just x mod (a*b*c) but x mod LCM(a, b, c)[or a multiple of the LCM].
This is because for a number to be x mod a, it is also x mod [any multiple of a].
This is the same for b, and c, and therefore a number n will be x mod a, b, and c when it is x above a multiple of a, b, and c.
By definition, that is their LCM.[/quote]

woah rare post of tene :omighty:

ok yeah i was wondering a similar thing cuz im trying to be way better at modulo stuff thanks

[quote=hansenhe][quote=sushi1233][quote=tenebrine]Not just x mod (a*b*c) but x mod LCM(a, b, c)[or a multiple of the LCM].
This is because for a number to be x mod a, it is also x mod [any multiple of a].
This is the same for b, and c, and therefore a number n will be x mod a, b, and c when it is x above a multiple of a, b, and c.
By definition, that is their LCM.[/quote]

Thanks![/quote]

the upvote function exists -_-[/quote]

I'm sorry, I'm kind of new to this, but is there something wrong when people post "Thanks!"?

not really, but a better way is to click the thumb up button next to a post


Isaac is taking a Math Olympiad test with 6 questions. The scores fore each question range form 0 through 10 inclusive. If Isaac never scores lower on a later problem than an earlier one, how many ways can Isaac score on the 6-question test?

I already know how to answer this question (I did it for a 1,2,3 problem test with up to 5,6,7 points respectively and found a pattern).
Obviously, thats not much of a proof. I am extremely certain this problem can be solved by recursion-that is, if we know the whole list of an $ f_n$ question test with score from 0-10, we can let the first $ n$ questions of $ f_{n \plus{} 1}$ be any arrangement of the list of $ f_n$ (since that will be a valid arrangement). So all we need to do is worry about the last number of $ f_{n \plus{} 1}$, which is somewhat dependent on the last number of $ f_n$. But from there I dont really know how to continue. If anyone could give me [b]hints[/b] for a recursive solution, that would be appreciated.

Hi;

Can you please provide what the pattern you discovered is?

I don't understand your question. Are you asking:

1. How many possible overall scores are there?
[hide="Hint"]The answer is all of them. These scores can be realized through score patterns of the form 0 ... 0, something, 10 ... 10.[/hide]

2. How many possible score patterns are there (lists of scores for individual questions)?
[hide="Hint"]Suppose there are actually 8 questions, of which the questions we're interested in comprise the middle 6. Suppose we force the score on the first question to be 0 and the score on the last question to be 10, so that the new string of 8 questions satisfies the same non-decreasing pattern as the middle 6. Each set of the middle 6 is associated with exactly one set of 8 scores (just tack a 0 and a 10 on the beginning and end, respectively).

[hide="Further Hint"]What must be true of the seven improvements?[/hide]

The above leads to a direct solution. You can form a recursive solution of the same spirit using the Hockey Stick Identity[/hide]

Thanks for the hint, TZF. I see how to do the problem

The official solution for this problem is interesting. They form a cool bijection which trivializes the problem. Consider a 6-by-10 grid. The 11 lines running horizontally are labeled 0,1,2,3,4,5,6,7,8,9,10 respectively.The 6 columns are labeled Question 1, Question 2, Question 3,..., Question 6. Then the number of ways to get from the bottom left point of the grid to the top right point of the grid using the shortest route possible (16 steps in this case) will be our answer. This is true because each time we make a move, it must be either horizontally and cannot be downwards, so effectively we are limiting a later question from being "lower" than an earlier question. Since its a 10x6 grid, our answer is $ \binom{10 \plus{} 6}{10} \equal{} 8008$

It pretty cool if you think about it.

We could also consider the number of times each of 0,1,2,...,10 can appear, which is between 0 and 6 inclusive, but sum to 6. Then by the balls and urns argument the answer is $ \binom{11 \plus{} 6 \minus{} 1}{6}$.
TZF, unfortunately I don't see how your hint is going to lead to a recursive solution... please elaborate a little more (yes I'm really bad at identifying recurrence) :(
Also Thunder365 may I ask where you got this problem? (very few problems have official solutions, especially those clever ones)

[quote="mathwizarddude"]We could also consider the number of times each of 0,1,2,...,10 can appear, which is between 0 and 6 inclusive, but sum to 6. Then by the balls and urns argument the answer is $ \binom{11 \plus{} 6 \minus{} 1}{6}$.
[/quote]
Thats a really nice solution. It works because any way you choose the 6 numbers, there will always be 1 and only 1 way to arrange them validly, right?
[quote="mathwizarddude"]
Also Thunder365 may I ask where you got this problem? (very few problems have official solutions, especially those clever ones)[/quote]
It was a BMO problem from this year


cf. $\frac{\varphi(2^{n}-1)}{n}$ is number of [url=http://mathworld.wolfram.com/PrimitivePolynomial.html]primitive polynomials[/url] of degree n over GF(2)

[hide]
Let $m=2^{n}-1$. Then $n$ is the smallest integer $d$ such that $2^{d}\equiv 1\pmod{m}$, and $2^{\varphi (2^{n}-1)}\equiv 1\pmod{m}$, so $n$ divides $\varphi (2^{n}-1)$.
[/hide]

I just typed an entire solution, but my browser closed.  Typing it again...

[b]Solution.[/b] Consider the integer $a^{n}-1$.  Clearly, $a^{n}\equiv 1\pmod{a^{n}-1}$ and the order of $a$ is $n$ modulo $a^{n}-1$.  Since $\gcd(a^{n}-1,a)=1$, by Euler's Theorem $a^{\varphi(a^{n}-1)}\equiv 1\pmod{a^{n}-1}$.  The order $a$ modulo $a^{n}-1$ must divide $\varphi(a^{n}-1)$, i.e. $n|\varphi(a^{n}-1)$. $\Box$


How many revolutions of a wheel with a fixed center and an outside diameter of six feet are required so that a point on  its rim will travel one mile?  Answer in terms of pi.

[quote="mysmartmouth"]The answer is [hide="possibly"]880[/hide]  That may be wrong, but I don't think it would be expressed in terms of $pi$.[/quote]

Why would it not be expressed in $pi$?[hide="The Answer"] 880/:pi: [/hide]

EDIT: oops sorry I forgot to put the "/" sign

[quote="GoBraves"][quote="mysmartmouth"]The answer is [hide="possibly"]880[/hide]  That may be wrong, but I don't think it would be expressed in terms of $pi$.[/quote]

Why would it not be expressed in $pi$?[hide="The Answer"] 880 :pi: [/hide][/quote]

Oops, I missed something there.  Your answer is probably right.

[quote="elementsofmathkid"][quote="DarkKnight"][hide]
$\frac {5280}{6\pi}$. I don't know if this is right though.[/hide][/quote]
Simplify your answer.[/quote]

[hide]
Oh, so $280.1126998$. or $\frac {880}{\pi}$[/hide]

[quote="math92"]dark knight i think that your answer is correct
can u tell me how u did that though?[/quote]

5280 feet in a mile, divide that by 6 and get 880 (times :pi: ).  That simplifies (with a calculator), and you get your answer!

Yeah so we have 5280 ft. in a mile and the wheel has circumference of $6\pi$ ft. Then we have $\frac {5280}{6\pi}$. That simplifies to $\frac {880}{\pi}$.

[hide]In my last post, I forgot to say that my answer is 880/:pi:[/hide]

[quote="mathgeek2006"][quote="mysmartmouth"][quote="math92"]dark knight i think that your answer is correct
can u tell me how u did that though?[/quote]

5280 feet in a mile, divide that by 6 and get 880 ([b]times [/b]:pi: ).  That simplifies (with a calculator), and you get your answer![/quote]
I bolded "times" because wouldn't it be "divided by"[/quote]

It would be. Sorry.


At a round table are $1994$ girls playing a deck of $n$ cards. Initially, one girl holds all of the cards. In each turn, if at least one girl holds at least two cards, one of these girls must pass a card to each of her two neighbors. The game ends when and only when each girl is holding at most one card.
a) Prove that if $n > 1993$, then the game cannot end.
b) Prove that if $n < 1994$, then the game must end.


$ \frac{\tan x \plus{}\tan y}{1\minus{}\tan x \tan y}\equal{}\frac{\frac{\sin x}{\cos x}\plus{}\frac{\sin y}{\cos y}}{1\minus{}\frac{\sin x \sin y}{\cos x \cos y}}\equal{}\frac{\frac{\sin x}{\cos x}\plus{}\frac{\sin y}{\cos y}}{\frac{\cos x \cos y \minus{} \sin x \sin y}{\cos x \cos y}}$$ \equal{}\frac{\sin x}{\cos x}\plus{}\frac{\sin y}{\cos y} \cdot  \frac{\cos x \cos y}{\cos x \cos y \minus{} \sin x \sin y}\equal{}\frac{\sin x \cos y \plus{} \cos x \sin y}{\cos x \cos y \minus{} \sin x \sin y}$

[quote="mr. math"]Verify this identity:

(sinx cosy + cosx siny) / (cosx cosy - sinx siny) = (tanx + tany)/ (1-tanx tany)

without using sum and difference formulas.[/quote]

sinxcosy + cosxsiny is sin(x+y)
and cosxcosy-sinxsiny is cos(x+y)
Hence LHS is tan(x+y) , which is same as RHS on being expanded.
These are identities , most basic.

Dividing the LHS by cosxcosy, we get (tanx + tany)/(1 - tanxtany)

This process is how we find the identity tan(x+y) = (tanx + tany)/(1 - tanxtany), as the RHS is just sin(x+y)/cos(x+y) = tan(x+y)


integration
trigonometry
calculus
function
derivative
calculus computations

[quote="zeus_three"]I wasn't sure whether this I was just not seeing something obvious in this problem, because every other question on the paper was not very difficult...

solve:

$\frac{d}{dx}(x^{2}\int_{0}^{x^{2}}\frac{sinu}{u}du)$

I tried using the product rule, that is: $\frac{d}{dx}uv=u'v+uv'$ but that still leaves the annoying sine integral, which of course can't be represented by elementary functions...[/quote]

You have two function here,
$f(x) = x^{2}$
$g(x) = \int_{0}^{x^{2}}\frac{\sin u}{u}du$

Note, you can think of $g(x)$ as $h(x^{2})$ where $h(x)=\int_{0}^{x}\frac{\sin u}{u}du$.

Now, $g'(x)=h'(x^{2}) = 2x h'(x^{2}) = 2x \cdot \frac{\sin x}{x}$. (FTC)

And $f'(x)=2x$

Now just use product rule to finish this off.

uhh.. by using product rule, doesn't that still leave the sine integral unevaluated?

i.e.
$2x \int_{0}^{x^{2}}\frac{\sin u}{u}du+2xsin(x^{2})$?

It does. Just leave it like that- that's the best you can do.

As jmerry said $\int \frac{\sin u}{u}du$ doesn't have an known (in terms of known functions) anti-derivative. Sometimes we write $\mbox{Si}(u)$ for this anti-derivative because it does appear at times in mathematical physics. So it has some importance.


linear algebra
matrix
algebra
polynomial
calculus
integration
vector

Let $ A,B$ be two square integer matrices of size $ n\times n$ such that $ AB$ equals the matrix with 1's in all its entrances except for the diagonal which is filled with 0's. Prove that $ AB\equal{}BA$.

This appears to be false.

Let $ A\equal{}\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}\minus{}1&1\\1&0\end{bmatrix}$

Then $ AB\equal{}\begin{bmatrix}0&1\\1&0\end{bmatrix}$ but $ BA\equal{}\begin{bmatrix}\minus{}1&0\\1&1\end{bmatrix}$

Is there some other restriction on $ A$ and $ B$ besides what you've already stated?

I am sorry! I Wrote what I remembered about the problem. The correct version from the original test requires an additional condition: $ A$ and $ B$ must have non-negative entries!
Now it becomes interesting!

Is there a graph-theoretic way of formulating the problem? $ J-I$ is the matrix of a pretty important graph, after all.

Sure.  Suppose that the edges of a directed graph $ G$ are colored either red or blue in such a way that there is exactly one way to get from a vertex $ v_1$ to any other vertex $ v_2$ by following a red edge, then a blue edge.  We want to show that we can also do this by instead following a blue edge, then a red edge, which seems plausible.

Combinatorics seems to be a red herring here. The condition of the problem says that $ \mathrm{rank}\left(AB \plus{} I_n\right) \equal{} 1$, what means that the characteristic polynomial of $ AB$ is divisible by $ \left( \minus{} X\right)^{n \minus{} 1}$ but not by $ \left( \minus{} X\right)^n$, what in turn means that the characteristic polynomial of $ BA$ is divisible by $ \left( \minus{} X\right)^{n \minus{} 1}$ but not by $ \left( \minus{} X\right)^n$ (since the characteristic polynomials of $ AB$ and $ BA$ are the same, as proven in [url=http://www.mathlinks.ro/viewtopic.php?p=1491761#1491761]http://www.mathlinks.ro/viewtopic.php?t=275576 post #9[/url], Corollary 2), what yields $ \mathrm{rank}\left(BA \plus{} I_n\right) \equal{} 1$. On the other hand, the matrix $ BA \plus{} I_n$ must have its main diagonal full of $ 1$'s (because $ \mathrm{Tr}\left(BA\right) \equal{} \mathrm{Tr}\left(AB\right) \equal{} 0$, while all entries of $ BA$ are $ \geq 0$, shows that the matrix $ BA$ has its main diagonal full of $ 0$'s). Now, knowing that all entries of $ BA \plus{} I_n$ are nonnegative integers and that $ \mathrm{rank}\left(BA \plus{} I_n\right) \equal{} 1$ (what means that every $ 2\times 2$ minor of $ BA \plus{} I_n$ has determinant $ 0$), it is clear what $ BA\plus{}I_n$ must be...

  darij

Searching for examples of this:

For one, you could let $ P$ be any permumation and then let $ A\equal{}(J\minus{}I)P\equal{}J\minus{}P$ and $ B\equal{}P^T\equal{}P^{\minus{}1}.$ Call those the trivial examples.

For something nontrivial, try this: $ C\equal{}\begin{bmatrix}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\1&0&0&0&0\end{bmatrix}.$

Let $ A\equal{}C\plus{}C^4$ and $ B\equal{}C^2\plus{}C^3.$ (The pentagon and the pentagram.)

From a 1969 paper in the Journal of Algebra, by Bridges and Ryser, entitled [i]Combinatorial Designs and Related Systems[/i]:

[b]Theorem 1.2.[/b] Let $ X$ and $ Y$ be nonnegative integral matrices of order $ n>1$ be such that $ XY\equal{}(k\minus{}a)I\plus{}aJ.$ Let $ k\ne a,$ and let $ k$ and $ a$ be relatively prime. Then there exist positive integers $ r$ and $ s$ such that $ XJ\equal{}JX\equal{}rJ$ and $ YJ\equal{}JY\equal{}sJ,$ in particular $ YX\equal{}XY.$

And finally, a proof. It's not mine - I'm relaying it from my matrix expert friend, and it's similar to the argument in the reference I just gave.

I never did follow Darij's proof, and at this point I'm not sure I believe he had one.

Notation: $ J$ will stand for the matrix of all $ 1$s and $ \mathbf{1}$ for the column vector of all $ 1$s, so that $ \mathbf{1}\,\mathbf{1}^T = J.$

Note that $ J - I$ is invertible and $ (J - I)^{ - 1} = \frac1{n - 1}J - I.$ This can be verified just by multiplying it out.

We have that $ AB$ is a nonnegative integer matrix such that $ AB = J - I.$ Note that since $ AB$ is invertible, both $ A$ and $ B$ are also invertible, which means that neither $ A$ nor $ B$ can have any zero columns. We'll use that.

Multiply by the inverse to get $ AB\left(\frac1{n - 1}J - I\right) = I.$ But inverses commute, so $ B\left(\frac1{n - 1}J - I\right)A = I,$ or
\begin{align}\frac1{n - 1}BJA - BA & = I\end{align}
Now define $ \mathbf{1}^TA = e^T = \begin{bmatrix}e_1 & \cdots & e_n\end{bmatrix}$ and $ B\mathbf{1} = f = \begin{bmatrix}f_1 \\
\vdots \\
f_n\end{bmatrix}.$ That is, $ e$ is the matrix of column sums of $ A$ and $ f$ is the matrix of row sums of $ B.$ Since we observed that $ A$ and $ B$ cannot have columns or rows of zeros and since the entries are nonnegative, we have that each $ e_i$ and each $ f_i$ is strictly positive.

$ BJA = B\mathbf{1}\,\mathbf{1}^TA = fe^t$ is the matrix with entries $ f_ie_j.$

Look at equation (1). On the right, we have $ I$ with integer entries, and in the middle, we have $ BA$ with integer entries. That means that $ \frac1{n - 1}BJA$ must have integer entries, which in turn implies that $ n - 1$ divides $ f_ie_j$ for all $ i,j.$ But these are strictly positive, hence $ f_ie_j\ge n - 1$ for all $ i,j.$

Take the trace in equation (1). We note that $ \text{trace}\,(AB) = 0,$ so $ \text{trace}\,(BA) = 0$ and thus $ \frac1{n - 1}\,\text{trace}\,(BJA) = n,$ which we rewrite as

$ \frac1{n - 1}\sum_if_ie_i = n$ or $ \frac1n\sum_if_ie_i = n - 1.$

So the average value of the entries $ f_ie_i$ is $ n - 1.$ But we had already established that each $ f_ie_i\ge n - 1.$ The only way for the average to equal the minimum is for all of the entries to be the same, so $ f_ie_i = n - 1$ for all $ i.$

But now $ f_ie_j\ge n - 1$ and $ f_ie_i = n - 1,$ so we can divide by $ f_i$ (which is positive) to get that $ e_j\ge e_i.$ Since this is true for all $ i$ and $ j,$ then $ e_i = e_0$ is constant for all $ i.$ Similarly, $ f_i = f_0$ is constant, with $ e_0f_0 = n - 1.$

By direct computation, we now have $ \frac1{n - 1}BJA = \frac1{n - 1}fe^t = \frac1{n - 1}(n - 1)J = J.$

Finishing: $ J - BA = I,$ so $ BA = J - I$ and we are done.

========

We have also proved some kind of graph-theoretic regularity. Each of $ A$ and $ B$ must have both constant row sums and constant column sums. $ e_0$ is the row/column sums of $ A$ and $ f_0$ is the row/column sums of $ B,$ with $ e_0f_0 = n - 1.$ Above, I talked about the trivial examples we get by letting one or the other of $ A$ and $ B$ be a permutation. To get a nontrivial example, we must have both $ e_0 > 1$ and $ f_0 > 1,$ and we can only do that if $ n - 1$ is composite. Hence the $ 5\times 5$ example above is the smallest possible nontrivial example.

A generic example. Let $ C$ be the standard generator of the circulants. See post #9 to see what I mean (that's the $ 5\times 5$ version there.)

Suppose $ n\minus{}1\equal{}ab.$ Let $ A\equal{}I\plus{}C\plus{}C^2\plus{}\cdots\plus{}c^{a\minus{}1}$ and $ B\equal{}C\plus{}C^{a\plus{}1}\plus{}C^{2a\plus{}1}\plus{}\cdots\plus{}C^{(b\minus{}1)a\plus{}1}.$

[quote="Kent Merryfield"]I never did follow Darij's proof, and at this point I'm not sure I believe he had one.[/quote]

You are right, there was a flaw. Basically the argument that $ \mathrm{rank}\left(BA \plus{} I_n\right) \equal{} 1$ is wrong. But it can be rescued: Show (as in your proof) that $ A$ is invertible. Now, $ BA \plus{} I_n \equal{} A^{ \minus{} 1}\left(AB \plus{} I_n\right)A$, so that $ \mathrm{rank}\left(BA \plus{} I_n\right) \equal{} \mathrm{rank}\left(AB \plus{} I_n\right) \equal{} 1$.

I have not really read your post yet, but your argument seems to be more generalizable than mine. Then again, the end of my proof can be modified using the following nice (yet trivial) lemma:

If $ U\in\mathbb{Z}^{n\times k}$ is a matrix of rank $ 1$ (when seen as a matrix over $ \mathbb{Q}$), then there exist two matrices $ V\in\mathbb{Z}^{n\times 1}$ and $ W\in\mathbb{Z}^{1\times k}$ such that $ U \equal{} VW$. (This is a known result if $ \mathbb{Z}$ is replaced by a field, but it also holds over $ \mathbb{Z}$ and every unique factorization domain or Euclidean domain. It follows from Smith's normal form, but this is actually overkill.)

  darij


A rocket is travelling with a speed $ 20\frac{km}{s}$ in a non-gravity space. To fix the direction of motion, it turns on an engine, which pushes the gasses with constant speed $ 3\frac{km}{s}$ w.r.t. the rocket perpendicularly to the direction of its motion. The engine is on till the mass of the rocket declines for $ \frac{1}{4}$ of the initial.

For what angle has the rocket changed the direction of motion and what is its final speed?

Could you provide a diagram to support your question?


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Suppose $ G$ is the Galois Group over $ \mathbb{Q}$ of one of the polynomials:
i) $ x^{11}\minus{}1$
ii)$ x^4\minus{}2$
For your choice of polynomial, find the generators of the splitting field $ E$ over $ \mathbb{Q}$ and describe the elements of G by their effect on these generators. Finally, describe the group by generators and relations.

I just need a good head start on how to solve this problem. Thanks.

[quote="calc rulz"]What are the roots of each of these polynomials?[/quote]

[quote="selenne431"]Suppose $ G$ is the Galois Group over $ \mathbb{Q}$ of one of the polynomials:
i) $ x^{11} \minus{} 1$
ii)$ x^4 \minus{} 2$
For your choice of polynomial, find the generators of the splitting field $ E$ over $ \mathbb{Q}$ and describe the elements of G by their effect on these generators. Finally, describe the group by generators and relations.

I just need a good head start on how to solve this problem. Thanks.[/quote]

i) $ X^{11} \minus{} 1 \equal{} (x \minus{} 1)(x^{10} \plus{} x^9 \plus{} ... \plus{} x \plus{} 1) \equal{} (x \minus{} 1)\Phi_{11}(X)$
so the roots are $ 1, \zeta_{10} , \zeta_{10}^2 , ... , \zeta_{10}^{10}$
ii) $ X^4 \minus{} 2$
roots are $ \pm 2^{\frac {1}{3}} , \pm i 2^{\frac {1}{3}}$

For i) I believe the splitting field is $ \mathbb{Q}(\zeta_{10})$
For the second ii) the splitting field is $ \mathbb{Q}(2^{\frac {1}{4}},i)$

please confirm so far

I'm still lacking theory on how to proceed to compute the Galois group


search
factorial
algebra
binomial theorem

Is there any type of formula for the expansion of powers of trinomials, like how you have the Binomial Theorem for binomials?

I'll do you infinite better.  The Multinomial Theorem:

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

True, and I have a given a sketch of a proof for this theorem on this forum, you can search for it. :)

[quote] Wow.  That's pretty cool, but I think it will be a little while before I'm able to use it effectively...  Too many things to remember in that theorem... [/quote]

Yeah, for most AMC/AIME problems, the Binomial Theorem will suffice, such as #24 on the 2006 AMC 12, even though it involved three terms.[/quote]

[quote="DanK"]Wow.  That's pretty cool, but I think it will be a little while before I'm able to use it effectively...  Too many things to remember in that theorem...[/quote]

If you [u]understand[/u] the binomial theorem, the multinomial theorem should follow naturally.

So try this example: Expand $(x+2y+3z)^4$.

can someone translate the multi nomial theorem into englsih please?

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=58166 .

Well, all posts lead back to the same theorem...

The factorial fraction thing comes from the number of ways to arrange the powers. It's like the # of ways to rearrange words, like MISSISSIPPI for example.

yes, I basically understand it all now. :)


[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73716[/url]
Don't revive that thread( :P ).


[quote="Identity"][i]A and B are two towns, 120 km apart. A car starts from A to travel to B at the same time as a second car, whose speed is 20 km/h faster than the first, starts from B to travel to A. The slower car reaches B 1 hour and 48 minutes after it passes the other car. Find their speeds.[/i][/quote]

I bet there is a better method, but here is my solution:

[hide]Let $ x$ be the slow car's speed. So $ x+20$ is the fast car's speed (in kph). We know that together, they travel at a speed of $ 2x+20$ kph, so after $ 120/(2x+20)=60/(x+10)$ hours, they meet. Now, the slower car take another $ 9/5$ hours (1 hour and 48 minutes) to reach B. So, the total time the slower car takes to get to B is
$ \frac{60}{x+10}+\frac{9}{5}$.
But by the distance-time-rate formula, we also know that this time must equal
$ \frac{120}{x}$.
Set these two quantities and solve:
[hide="Click here if you want to see the mess"]$ \frac{60}{x+10}+\frac{9}{5}= \frac{120}{x}$
$ \frac{12}{x+10}+\frac{9}{25}= \frac{24}{x}$
$ 300x+9x(x+10) = 600(x+10)$
$ 100x+3x(x+10)=200(x+10)$
$ 100x+3x^{2}+30x = 200x+2000$
$ 3x^{2}-70x-2000=0$
$ (x-40)(3x+50)=0$[/hide]
$ x=40 \text{ or }\frac{-50}{3}$

Since $ x>0$ (it's a speed), $ x=40$. So the desired speeds are $ 40 \text{ and }60 \text{ kph}$.[/hide]

is this as easy as it looks. i got the same thing as vishalarul. or am i missing something

Thanks, it's correct  :wink:. The problem I had was finding an expression for the time they meet. I've never really thought of adding the speeds and dividing the distance...


Can we Prove this for $a,b>0$: $\frac{2a}{a+b}+\frac{b}{2a}\ge \frac{1}{2}\left(\frac{a-b}{a+b}\right)^{2}$

It's true, but is not at all tight... is there a typo?

Sorry, there really is a typo. It must be: $\frac{2a}{a+b}+\frac{b}{2a}\ge\frac{1}{2}(3+\left(\frac{a-b}{a+b}\right)^{2})$

$\frac{4a}{a+b}+\frac{b}{a}\ge 3+\left(\frac{a-b}{a+b}\right)^{2}$.
This Inequality is homogenious. So we can set $a+b=1(a,b<1)$. We get:
$4a+\frac{b}{a}\ge 3+(a-b)^{2}\Leftrightarrow$
$4a+\frac{1}{a}-1\ge 3+4a^{2}-4a+1 \Leftrightarrow$
$4a^{3}-8a^{2}+5a-1\le 0 \Leftrightarrow$
$(a-1)(2a-1)^{2}\le 0$ This is right for all $a<1$. Equality holds, if $a=b$ or if $b=0$.

I have got another solution.
Put $x=\frac{b}{a}\geq 0$ . So we have to prove this inequality:
$\frac{2}{1+x}+\frac{x}{2}\geq \frac{1}{2}(3+(\frac{1-x}{1+x})^{2})$
but it `s easy to see that this inequality is equivalent to
$x(x-1)^{2}\geq 0$ which is right for all x$\geq$0.

[quote="me@home"]
$\frac{4a}{a+b}+\frac{b}{2a}\geq 2+\frac{2b}{(a+b)^{2}}$
Tomorrow I'll try to prove this... :P[/quote]

isn't it $\frac{2a+2b}{a+b}+\frac{b}{2a}\geq 2+\frac{2b}{(a+b)^{2}}$?

i think you switched the a and the b

There is another little typo: It must be: $a>0,b\ge 0$, becouse there is equality when $a=1,b=0$

you guys are both right... :oops:
so without typos it is much easier... \[\frac{2a+2b}{a+b}+\frac{b}{2a}\geq 2+\frac{2b}{(a+b)^{2}}\] is what we want to prove
$\equiv \frac{b}{2a}\geq \frac{2b}{(a+b)^{2}}$
$(a+b)^{2}\geq (a+a)$
but in our range this is same as:
$(a+b)\leq a+a$ which is true by definition...

[quote="me@home"]
$\equiv \frac{b}{2a}\geq \frac{2b}{(a+b)^{2}}$
$(a+b)^{2}\geq (a+a)$
[/quote]

 :huh:

wow i'm all over the place...
i'll just give up sorry to have massacred such an easy problem... maybe i'll go back to the beginning and work out a real proof...


Evo i zadaci sa Federalnog takmicenja BiH Jedan zadatak za 4. je bio pogresno formulisan tako da ga nisam htio objaviti u onoj formi (moglo se naravno doci do kontradikcije u samom uslovu sto je donosilo 7 bodova)

Prvi razred:

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(Znam da spamam, ali nisam mogao izdrzati)

Neki se ne bi slozili da se radi o kontradikciji (sifra: "Ø->Ø")  :lol:

Bilo je bas fino takmicenje :D
Posebno sto sam bio prvi u prvim razredima :D  :)


Does anyone know what day the Mathcounts State is for Northern California? :|

i'll probably do bad because it's so competitive...and i definitely won't get into the nationals like [i]some[/i] of you people.  :omighty:

thanks

March 17@UC Davis.

How dare it be on St. Patrick's day...

It's okay, at least you're going to pwnify me :( :( :(

Which chapter are you from?

My goal is to not get last.
You will "pwnify" me.

yeah, as if i was going to pwn anybody...

I was in the diablo chapter, it was today.

No,but seriously, my goal for the state is not to have a heart attack around all those smart people (unlike me).  :stretcher:

[quote="#H34N1"]Lucky. At Ohio, we have it on Mar. 10.[/quote]

I know... I wish we had more time to prepare :(

Our Chapter was late and our state is early.

Us too! But, I actually prefer less prep time, as it means others have less prep tme. I'd rather it be difficult for all instead of easy.

[quote="#H34N1"]Lucky. At Ohio, we have it on Mar. 10.[/quote]

Florida is March 30th  :D

[quote="Ubemaya"]If you get beat by me I will laugh at you cause that will be seriously sad.[/quote]

Uh,about that....no, I am seriously not smart, and will get pwned by everyone...

wait...ubemaya, which chapter did you come from?

Hm... can we keep this to the California forum?  Thanks!  Dates for MC competitions are on the website.


[quote="6666deikhoong"]Without directly evaluating, show that

|3 c a+b|
|3 a b+c| = 0
|3 b a+c|[/quote]

I suppose that you mean :

Show that, $ \forall a,b,c\in\mathbb{R}$, determinant $ \begin{vmatrix}3 & c &a\plus{}b\\3 & a & b\plus{}c \\3 & b & a\plus{}c \end{vmatrix}\equal{}0$

Replace third column of the matrix by the sum of second and third and get your result.

But direst computation of the determinant is nearly as quick.

... and it's certainly not an olympiad problem! :(

.


geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry proposed

Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ and circumradius $R$. If $M=AB \cap CD, N=AD \cap BC$ show that:

$(1) \ \ \   MN^2 = MO^2 + NO^2 - 2R^2$

$(2) \ \ \   MN = LP + LQ$, where $L$ the midpoint of $MN$ and $LP, LQ$ are the two tangent conducted from $L$ to the circle $(O)$

$(3)$ The circle with diameter $MN$ is orthogonal to $(O)$

[b]Remark.[/b] A possible solution of (1) is given in
http://www.mathlinks.ro/Forum/viewtopic.php?t=25075

We can deduce both $(2)$ and $(3)$ (which are clearly equivalent :)) from $(1)$.

I'll assume it well known that $T=AC\cap BD$ is the orthocenter of $OMN$. Let $M'$ be the projection of $N$ on $OM$. If we show that $R^2=OM\cdot OM'$, we're done, because this means that the inversion wrt the circle $(ABCD)$ turns $M$ into $M'$, which also lies on the circle $(MN)$, so that inversion invariates the circle $(MN)$, which is equivalent to the two circles being orthogonal.

Let's thus prove $OM\cdot OM'=R^2$ from problem $(1)$:

We have $2R^2=OM^2+ON^2-MN^2=2\cdot OM\cdot ON\cdot \cos (\angle MON)=2\cdot OM\cdot OM'$, Q.E.D.

I think this problem (or a very similar one) has been discussed before. If I have the time, I'll post a self-contained proof a bit later.

Actually, here's a stand-alone proof for $OM\cdot OM'=R^2$, where the definitions we use are the ones above.

I'll also assume that $MM'\cdot MO=MA\cdot MB=$the power of $M$ wrt the circle $(ABCD)$. This means that $MM'\cdot MO=MS^2$, where $MS$ is tangent to the circle $(ABCD)$. We thus get $OM\cdot OM'+MM'\cdot MO=OM^2=R^2+MS^2=OM\cdot OM'+MS^2$, and from here we get the desired result: $OM\cdot OM'=R^2$.

[quote="grobber"]
I'll assume it well known that $T=AC\cap BD$ is the orthocenter of $OMN$. [/quote]

I do not understand, Grobber, where this fact in your first proof is used.

Leon

Looks like I didn't use it then :). I was thinking about other things which may be said about this figure. It's so rich in properties that I got lost in it :). Sorry about that.


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

Show that for all real positive numbers $ x$, $ y$ and $ z$ the following inequality is true: $ \frac{(2x\plus{}y\plus{}z)^2}{2x^2\plus{}(y\plus{}z)^2 }\plus{} \frac{(2y\plus{}z\plus{}x)^2}{2y^2\plus{}(z\plus{}x)^2 }\plus{}\frac{(2z\plus{}x\plus{}y)^2}{2z^2\plus{}(x\plus{}y)^2 }\le 8$

WLOG $ x\plus{}y\plus{}z\equal{}1$, we have to prove: $ \sum \frac{(1\plus{}x)^2}{2x^2\plus{}(1\minus{}x)^2}\equal{}\sum \frac{x^2\plus{}2x\plus{}1}{3x^2\minus{}2x\plus{}1} \le 8$, but:

$ 3 \cdot LHS\equal{} \sum \frac{3x^2\plus{}6x\plus{}3}{3x^2\minus{}2x\plus{}1} \equal{} \sum (1\plus{}\frac{8x\plus{}2}{3x^2\minus{}2x\plus{}1}) \le \sum (1 \plus{} \frac{8x\plus{}2}{\frac{2}{3}}) \equal{} \\ \equal{} \sum (1\plus{}12x\plus{}3) \equal{} (4\plus{}4\plus{}4)\plus{}12(x\plus{}y\plus{}z)\equal{}24 \equal{} 3 \cdot RHS$

May I know what is source of this inequality?

Old USAMO problem, posted several times
[url]http://www.mathlinks.ro/viewtopic.php?p=336208#p336208[/url]

How is assuming $ x\plus{}y\plus{}z\equal{}1$ WLOG?

[quote="triplebig"]How is assuming $ x \plus{} y \plus{} z \equal{} 1$ WLOG?[/quote]
It is homogeneous :wink:


calculus
derivative
integration
real analysis
real analysis unsolved

How would you determine the first derivative of the following expression with respect to Q?  (Q is also the limit of the integral).  I am solving for the optimal Q for this expression and i was having trouble with the first order conditions.

[b] $\int\limits_{0}^{Q}x f(x)\; dx + Q\int\limits_{Q}^{infinity} f(x)\; dx .$ [/b]

thanks

The Fundemental Theorem of Calculus has two sides- one of them is the derivative of an integral. Note that $\int_Q^\infty f(x)\,dx=\int_{\infty}^Q -f(x)\,dx$

By the Fundamental Theorem of Calculus (assuming some conditions involving the continuity of $f$ and the convergece of the integral)

$\frac{d}{dQ}\int_0^Qxf(x)\,dx=Qf(Q)$ and $\frac{d}{dQ}\int_Q^{\infty}f(x)\,dx=-f(Q).$

So we put this together with the product rule to get

$\frac{d}{dQ}\left[\int_0^Qxf(x)\,dx+Q\int_Q^{\infty}f(x)\,dx\right] =Qf(Q)-Qf(Q)+\int_Q^{\infty}f(x)\,dx$

Incidentally, the $\text{\TeX}$ code for the infinity symbol is \infty.


real analysis
function
topology
calculus
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
I was working on a separate proof when a question arose about sequences and sets.
Can a sequence be an element of a set? If so or if not, why?

Consider the following sequence and set
Let, $ {s}$ be the sequence,
$ {s} \equal{} {2}, {4}, {6},\dots, \infty$

Let, $ {S}$ be the set,
$ {S} \equal{} \{{0}, {1}, {2},\dots, \infty\}$

Then is the following statement correct?,
$ {s} \in {S}$

[b]2. Relevant equations.[/b]
Knowledge of sequences and sets.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I reason yes, a sequence can be an element of a set. I believe this is true if all the elements in a sequence are also elements in a set. However, I note that sequences and sets are not the same (even if all the elements and the order is) since fundamentally sequences need an explicitly defined order and sets do not.
Am I correct in this reasoning?
Lastly, can a set be an element of a sequence?
[/hide]
Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

1) There is no "$ \infty$" at the "end" of the sequence $ s \equal{} (2, 4, 6, \ldots)$, nor in the set $ S \equal{} \{0, 1, \ldots\} \equal{} {\bf Z}_{\geq 0}$.

2) Certainly, a set can contain a sequence as an element.  However, the set you've written does not contain the sequence you've written as an element.  Rather, it happens that $ S$ contains as elements all the numbers which appear as elements of $ s$.  (In other words, the [i]underlying set[/i] of the sequence $ s$, i.e., the set of values that appear in $ s$, is contained in $ S$.  Containment here is as a subset, not an element.)  So, there is a sense in which "the sequence $ s$ is contained in the set $ S$" could be made precise, but "as an element of the set" is not that sense.

Consider the [url=http://en.wikipedia.org/wiki/Lp_space#.E2.84.93p_spaces]$ \ell^{p}$ spaces[/url], which are sequence spaces.

Hey,

[quote="JBL"]
1) There is no "$ \infty$" at the "end" of the sequence $ s \equal{} (2, 4, 6, \ldots)$, nor in the set $ S \equal{} \{0, 1, \ldots\} \equal{} {\bf Z}_{\geq 0}$.
[/quote]
Ah I see. If I may ask, why is this the convention (in other words why don't we write "$ \infty$" at the end)?

[quote="JBL"]
2) Certainly, a set can contain a sequence as an element.  However, the set you've written does not contain the sequence you've written as an element.  Rather, it happens that $ S$ contains as elements all the numbers which appear as elements of $ s$.  (In other words, the [i]underlying set[/i] of the sequence $ s$, i.e., the set of values that appear in $ s$, is contained in $ S$.  Containment here is as a subset, not an element.)  So, there is a sense in which "the sequence $ s$ is contained in the set $ S$" could be made precise, but "as an element of the set" is not that sense.
[/quote]
I see, ok that makes sense. So it is incorrect to say that,
$ {s} \in {S}$

However, how would I mathematically write, "the sequence $ s$ is contained in the set $ S$"?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

It's kind of meaningless: $ \infty$ isn't actually a number so you can't include it as an element in your sequence.

As you've defined it, $ S$ isn't a sequence space so it's elements aren't actually sequences.  However, the elements of $ s$ are in $ S$.  We can say that $ (s_{n})\subset S$, which means the sequence $ (s_{n})$, whose elements come from $ S$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=263075]This[/url] is the topic that got APSStudent thinking this way. He's pretty far off track, and in particular, pretty far off track from what I was actually doing there. I'm not quite sure how to explain things to him. (I also just fixed a typo I had there - don't know if that will help or not.)

One thing I'll confess to: while I may not have been using all of the vocabulary (for instance, the word "measure" did not explicitly occur in my post), I was clearly using the [i]thought patterns[/i] native to a course on Lebesgue measure and integral.

I think the best way to put it is that it's purely notation.

Rhetorical question:  what is the difference between the set $ \{ 0 \}$ and the set $ \{ \{ 0 \} \}$?

The answer is that the first set contains a number and the second set contains [b]a set.[/b]  (In fact, it contains the first set.)  These aren't the same set, nor do they have the same elements.

Second rhetorical question:  what is the difference between the sets $ \{ 1, 2 \}, \{ \{ 1 \}, 2 \}$, and $ \{ \{ 1, 2 \} \}$?

Third rhetorical question:  what is the difference between the sets $ \{ 1, 2, 3, ... \}$ and $ \{ \{ 1, 3, 5, ... \}, \{ 2, 4, 6, ... \} \}$?

Now let's talk about sequences.  A sequence doesn't have to have a pattern in it, and it doesn't have to consist of numbers.  For example, the sequence $ a_1 \equal{} \text{Happy}, a_2 \equal{} \text{Sad}, a_3 \equal{} \text{Angry}$ is a sequence of emotions.

[b]Definition:[/b]  A sequence with elements in a set $ S$ is a function from $ \mathbb{N}$ to $ S$.

If $ S$ is the set of sets, it is meaningful to talk about a sequence of sets and this is not a fundamentally different concept from any other type of sequence.  This definition should hopefully make it clearer that sequences are not the same thing as subsets of $ S$, nor are they the same thing as objects in the set $ S$.

Your rhetorical questions are pretty important for topology.  If I define $ X\equal{}\{a,b,c\}$, then $ \{X,\emptyset,\{a,b\},b,\{b,c\}\}$ is a topology on $ X$, but $ \{X,\emptyset,\{a,b\},\{b,c\}\}$ is not.  As a result, depending how you define a topology, you can actually make sequences do really weird things.


A fence is to be built to enclose a rectangular area. The fence along three sides is to be made of material that costs 5 dollars per foot. The material for the fourth side costs 15 dollars per foot.

(A) If the area is 5000 square feet, find the dimensions of the rectangle that will allow the most economical fence to be built.

(B) If 3,000 dollars is available for the fencing, find the dimensions of the rectangle that will enclose the most area.

Id appreciate assistance. :D

[hide]Let two of the sides of the rectangle be long x, and the other two y. Then $\text{Cost}=5(x+2y)+15x=10(2x+y)$ and $\text{Area}=xy$
a) We need $\text{min(2x+y)}$ when $xy=5000$.
by AM-GM $\frac{2x+y}{2}\ge \sqrt{2xy}=100$ and equality holds when $2x=y$. So $y=100$, $x=50$ and $\text{Cost}=2000$
Try to do b) alone
[/hide]
[hide]b) We need $\text{max(xy)}$ when $2x+y \le 300$.
Again by AM-GM $\sqrt{2xy}\le \frac{2x+y}{2}=150$. Now $y=150$, $x=75$ and $\text{Area}=11250$.
[/hide]


ratio
geometric series
geometric sequence

If five geometric means are inserted between 8 and 5832, the fifth term in the geometric series is:
\[ \text{(A)}\ 648 \qquad \text{(B)}\ 832 \qquad \text{(C)}\ 1168 \qquad \text{(D)}\ 1944 \qquad \text{(E) none of these} \]

I'm unsure about geometric means, but,
[hide]there are seven terms including 8 and 5832, and these 7 terms are a geometric sequence, so we can use the formula a(n) = a(1) * dn - 1, when d is the multiplying factor, a(n) is the term we want, and a(1) is the first term.
5832 = 8 * d6.
d = 3 (multiplying factor).
the first term is 8, the fifth term in the five terms we added must be 1944, or D.    [/hide]

Oh well, it the question meant to ask for the 5th term from the number 8, then the answer would be:
[hide]Since we already know the 5th number we added between the 8 and the 5832 is 1944, then the sequence beginning from the 8 would be 1944 / 3, or 648 [color=orange]A[/color].[/hide]


An investement of 5000$ , earning interest compounded semi annually grew to 9000$ in 15 years. What was the nomimal rate?

Let me rewrite the question:
An investment of $ \$$5,000, earning interest compounded semi-annually grew to $ \$$9,000 in 15 years. What was the nominal rate?
Is this correct?

[hide="Solution?"]
[list]$ 9000 \equal{} 5000\left(1 \plus{} \frac {r}{2}\right)^{15 \times 2}$?

$ \frac {9}{5} \equal{} \left(\frac {2 \plus{} r}{2}\right)^{30} \implies \left(\frac {9}{5}\right)2^{30} \equal{} (2 \plus{} r)^{30}$

$ \boxed{2\sqrt [30]{\left(\frac {9}{5}\right)} \minus{} 2 \equal{} r}$

$ \boxed{r\approx3.957\%}$.[/list]
Thank you bertram for the code.[/hide]

[quote="Patterns_34"]Let me rewrite the question:
An investment of $ \$$5,000, earning interest compounded semi-annually grew to $ \$$9,000 in 15 years. What was the nominal rate?
Is this correct?

[hide="Solution?"]
[list]$ 9000 \equal{} 5000\left(1 \plus{} \frac {r}{2}\right)^{15 \times 2}$?

$ \frac {9}{5} \equal{} \left(\frac {2 \plus{} r}{2}\right)^{30} \implies \left(\frac {9}{5}\right)2^{30} \equal{} (2 \plus{} r)^{30}$

$ \boxed{2\sqrt [30]{\left(\frac {9}{5}\right)} \minus{} 2 \equal{} r}$

$ \boxed{r\approx3.957}$ percent.[/list]
Where is the code for the percent symbol?[/hide][/quote]

Try $ \%$ ($ \$$\%$ \$$) ;)


algebra
polynomial
trigonometry
algebra unsolved

Prove that the polynomial
$ P(x)=x^{n+1}\sin (n-1)\theta - x^n \sin n\theta  +x \sin \theta$
is divisible by
$ x^2-2x \cos\theta +1$

[b]Remark:[/b] If $ x_1\neq x_2$, then: $ (x - x_1).(x - x_2)|P(x)\Leftrightarrow P(x_1) = P(x_2) = 0.$
$ x^2 - 2x.cos{\theta} + 1 = [x - (cos{\theta} - i.sin{\theta})]. [x - (cos{\theta} + i.sin{\theta})]$
and
$ P(cos{\theta} + i.sin{\theta}) =$
${ = (cos{\theta} + i.sin{\theta})^{n + 1}.sin(n - 1).{\theta}} - (cos{\theta} + i.sin{\theta})^n.sin \ n.{\theta} + (cos{\theta} + i.sin{\theta}).sin{\theta} =$
${ = [cos (n + 1).{\theta} + i.sin (n + 1).{\theta}].sin(n - 1).{\theta}} - (cos n.{\theta} + i.sin n.{\theta}).sin \ n.{\theta} + \\
+ (cos{\theta} + i.sin{\theta}).sin{\theta} =$
$ = [cos(n + 1).{\theta}.sin(n - 1).{\theta} - cos n.{\theta}.sin \ n.{\theta} + cos{\theta}.sin{\theta}] +$
$ + i.[sin(n + 1).{\theta}.sin(n - 1).{\theta} - sin^2 n.{\theta} + sin^2{\theta}] =$
$ = \frac {1}{2}.\{[(sin {2n.\theta} - sin {2.\theta}) - sin {2n.\theta} + sin {2.\theta}] + i.[(cos {2.\theta} - cos{2n.\theta}) - (1 - cos{2n.\theta}) + \\
+ (1 - cos {2.\theta})]\} = 0 = P(cos{\theta} - i.sin{\theta}).$


limit
quadratics
induction
algebra
polynomial
number theory proposed
number theory

1)Let k is a natural number
Find all n
$ 5^{k}|7^{n}\plus{}1$
2)Find all n
$ 47^{\frac{n}{2}}|48^{n}\minus{}1$
More general
$ f:\ \mathbb{N}\to\mathbb{N}$
$ \lim_{ n\to\infty}\frac{f(n)}{n}<\infty$
For each a odd prime p and a natural x,Prove that there finite n such that
$ p^{f(n)}|x^{n}\minus{}1$

Find all $ P(x)\in Z[x]$ such that
$ P(n)|2^{n}\plus{}3^{n}$ for all natural number n.

Let$ (a,b)\equal{}1$
find all $ P(x)\in Z[x]$
$ P(n)|a^{n}\plus{}b^{n}$ for all n

This problem from my teacher
Find all $ P(x)\in Z[x]$
$ P(n)|n^{n\minus{}1}\minus{}1$

[quote="TTsphn"]1)Let k is a natural number
Find all n
$ 5^{k}|7^{n}\plus{}1$[/quote]
[hide]$ 7^{n}\plus{}1\equiv0\bmod5$, and therefore $ 7^{n}\equiv4\bmod5$

No powers of $ 7$ end in $ 4$, but $ 7^{n}$ ends in $ 9$ for all $ n\equiv2\bmod4$.[/hide]

for each k find all n satisfy .You solved when$ k\equal{}1$ .

[quote="TTsphn"]for each k find all n satisfy .You solved when$ k \equal{} 1$ .[/quote]

It's a matter of finding the order of $ 7$ modulo $ 5^{k}$, and dividing that by $ 2$ (because clearly $ 7$ is a quadratic nonresidue modulo $ 5^{k}$).  Clearly, we must get something which is a power of $ 5$ multiplied either by 1 or by 2, by a group order argument.  Then also prove the following:

$ a^{5}\equiv\minus{}1\mod{5^{k\plus{}1}}\implies a\equiv\minus{}1\mod{5^{k}}$

This we prove by induction on $ k$ (ie, a solution $ a$ to the first congruence above must satisfy $ a^{5}\equiv\minus{}1\mod{5^{k}}$, and so $ a\equiv\minus{}1\mod{5^{k\minus{}1}}$.  So take $ a\plus{}5^{k\minus{}1}b$ where $ a <5^{k\minus{}1}$ and $ b<5$ and raise this to the fifth power using binomial expansion.  Take this modulo $ 5^{k\plus{}1}$ and find that we will have a unique solution for $ b$.)

[quote="TTsphn"]No we can find order of 7 mod $ 5^{k}$
This solution I will post later.[/quote]

Er...yes we can, and I showed how above...
Was my explanation not clear enough?

These problem very interesting.
The first I take from Mathnfriend.org and general it .
So can you solve it?

$ n|2^{n}\plus{}3^{n}$
a)Prove that $ 5|n$
b)Prove that there exist infinite n satisfy .
c)Find all $ P(x)\in Z[x]$ satisfy
$ P(n)|2^{n}\plus{}3^{n},\forall n\in N$
2)Does there exist $ P(x,y)$ such that
$ P(3[2t],2[3t]) \equal{} 0,\forall t\in R$

This problem I take from a Olympiad competition
Find all $ n\in N$ such that
$ n|3^{n}\minus{}2^{n}$
More general
$ n|(a\plus{}1)^{n}\minus{}a^{n}$
 2)Very easy
Find all $ k\in N$ such that there exist infinite n such that
$ n|(a\plus{}k)^{n}\minus{}a^{n}$

1)Find all n such that
$ 2^{n}\minus{}1|3^{n}\minus{}1$

We know that n is odd. So 2^n -1 =7 mod 12.
So it has a prime divisor q such that (3/q)=-1. So the only solution is n=1.

A problem from the first
Find all n such that
$ 9^{n}\equal{} 1(\mod 7^{n})$
Let $ f(n) \equal{}\frac{n}{2}$
Find all n
$ 9^{n}\equal{} 1(\mod 7^{f(n)})$
In the special case we can find all solution of this equation.

Find all m such that for any $ n\in N$ then
$ \gcd(2^{m}\minus{}1,2^{n}\plus{}1)\equal{}1$
Hint
$ \gcd(x^{m}\minus{}1,x^{n}\minus{}1)\equal{}x^{\gcd(m,n)}\minus{}1$

Yes all odd natural number m
It is quite easy .
But can you post solution.?

If q divides 2^m -1 and 2^k|q-1, then there is a c such that
c^2^k=2(mod q).
Hence, q cannot divide 2^x +1 for any integer x.

I have a other method
$ d\equal{}\gcd(m,n)$
$ \gcd(2^{m}\minus{}1,2^{2n}\minus{}1)\equal{}2^{d}\minus{}1$(because m odd)
Other
$ 2^{2n}\minus{}1\equal{}(2^{n}\minus{}1)(2^{m}\plus{}1)$ and $ \gcd(2^{n}\plus{}1,2^{d}\minus{}1)\equal{}1$

Some problem in this topic I think quite well but now very was be solved.
Can you post solution .
I will post if no one post them.

I post solution for the problem
$ P(n)|3^{n}\plus{}2^{n}$
Case 1:$ \deg P > 0$
Lemma 1: A polynomial has degree>0 has infinite prime divisor.
Call is p
$ p|P(n_{0})$
then
$ p|3^{n_{0}}\plus{}2^{n_{0}}$(1)
But
$ p|P(n_{0}\plus{}p)$ then
$ p|3^{n_{0}\plus{}p}\plus{}2^{n_{0}\plus{}p}$(2)
From (1) and  (2) then
$ p|3^{p}\minus{}2^{p}$ tradition.
so deg P=0
$ P(n) \equal{} c,\forall n$
then
$ c|5$
Easy to check
$ c\in\{1,\minus{}1\}$


Problem: Give $ a,b,c>0$ and $ x\plus{}y\plus{}z\equal{}xyz$. Prove that:

$ (x^2\plus{}y^2\plus{}z^2)^2 \geq 3(x\plus{}y\plus{}z)(\frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{xz}{y})$

$ : P$

[quote="vipCD_A1"]Problem: Give $ a,b,c > 0$ and $ x \plus{} y \plus{} z \equal{} xyz$. Prove that:

$ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq 3(x \plus{} y \plus{} z)(\frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {xz}{y})$

$ : P$[/quote]

It's easy. Here is my solution.

Substitute $ x\plus{}y\plus{}z$ with $ xyz$ in the left side.

So we have to prove that:

$ x^4\plus{}y^4\plus{}z^4\plus{}2x^2y^2\plus{}2y^2z^2\plus{}2z^2x^2\geq3(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)$

Or:

$ x^4\plus{}y^4\plus{}z^4\geq x^2y^2\plus{}y^2z^2\plus{}z^2x^2$

which is true from AM-GM:

$ \dfrac{x^4\plus{}y^4}2\geq x^2y^2$
$ \dfrac{y^4\plus{}z^4}2\geq y^2z^2$
$ \dfrac{x^4\plus{}z^4}2\geq z^2x^2$


function
absolute value
algebra unsolved
algebra

let $ f$ is a function such that
$ f(x)\equal{}ax^{2}\plus{}bx\plus{}c$
such that $ a$, $ b$, $ c$ are real numbers such that:
$ la\minus{}b\plus{}cl\equal{}<1$ and $ la\plus{}b\plus{}cl\equal{}<1$ and $ lal\equal{}<1$

prove that
$ (\forall x \in [\minus{}1,1])$   $ lf(x)l\equal{}<2$

$ lxl$ is the absolute value of $ x$

The first we has $ |c|\leq\frac {3}{2},|b|\leq 1$
We prove for $ a > 0$ the case $ a < 0$ has same method.
$ f'(x) \equal{} 2ax \plus{} b$
[u]Case 1[/u]:$ \frac {b}{2a}\in[ \minus{} 1,1]$
Use deriverty we has :
$ 1\geq f(1)\geq f(\frac { \minus{} b}{2a})$
$ f(\frac { \minus{} b}{2a})\leq f(1)\leq 1$
So enough to prove that $ f(\frac { \minus{} b}{2a})\geq \minus{}2$
$ \leftrightarrow 8a \plus{} 4ac \minus{} b^2 \geq 0$
Because $ 6a \plus{} 4ac\geq 0$ and $ 2a \minus{} b^2\geq 0$ so it true.
[u]Case 2[/u]
$ \frac { \minus{} b}{2a} > 1$ so $ f'(x) < 0,\forall x\in [ \minus{} 1,1]$
But $ |f(1)|\leq 1,|f( \minus{} 1)|\leq 1$ so $ |f(x)|\leq 2$
[u]Case 3[/u]
$ \frac { \minus{} b}{2a} < \minus{} 1$ then $ f'(x) > 0,\forall x\in [ \minus{} 1,1]$
$ |f(1)|\leq 1,|f( \minus{} 1)|\leq 1$
So we has the inequalities.
And 2 is the best real such that $ |f(x)|\leq c$
For example:$ x^2 \minus{} \frac {3}{2}x \plus{} 1$


Hi, everyone.

Prove that orthocenters of four triangles formed by taking three out of four given lines are collinear.
I feel like having seen it before on forum, but can't find.
Thanks in advance. :blush:

hello 77ant , this is the Steiner's line  the four-line quadrangle

Thanks, tiger100

Your advice is great help for me. I will search "steiner line"


linear algebra
matrix
linear algebra open

It is easy to see  the matrix A = (a(i,j)) in Mn(R)

where  a(i,j) = (-1)^max(i,j)  is diagonalisable.

What can you say about the eigenvalues ? eingenvectors ?

[quote="fredbel6"]Interesting
Well all eigenvalues will be real, and all eigenvectors of different eigenvalues are orthogonal

But  there is a chance you already knew that?[/quote]

I guess this is too general -- any hermitian matrices possess these properties.


Let $ V$ be a topological vector space over $ \mathbb{C}$. Let $ U \subseteq V$ be an open set containing $ 0$.

Is it true that \[ \bigcap_{\left|\alpha\right|\equal{} 1} \alpha U\] is also an open set containing $ 0$?

Prove/disprove!

The preimage of $ U$ under the map $ (\alpha, v)\to \alpha v$ is an open subset of $ \mathbb C\times V$ containing $ (0,0)$. Therefore, there exist $ \epsilon > 0$ and a neighborhood $ W$ of $ 0\in V$ such that $ \alpha w\in U$ for all $ |\alpha| < \epsilon$ and $ w\in W$. The set $ (\epsilon /2)W$ is contained in $ \bigcap_{\left|\alpha\right| \equal{} 1} \alpha U$

But that does not imply that the intersection is open again, right?

You've only showed that the intersection contains some open set containing $ 0$, not that it is an open set containing $ 0$ itself.

I got a hint for the problem: the statement is true, and I should luse the compactness of $ \left\{\alpha \in \mathbb{C}: \left|\alpha\right| \equal{} 1\right\}$.

(But I don't know what to do with it.)

Indeed, let's use the compactness of $ \mathbb T\equal{}\left\{\alpha \in \mathbb{C}: \left|\alpha\right| \equal{} 1\right\}$. Fix $ a\in \bigcap_{\alpha\in\mathbb T} \alpha U$. We would like to prove that each $ \alpha\in\mathbb T$ is contained in an open arc $ I_{\alpha}\subset \mathbb T$ such that $ a$ is contained in the interior of the set $ \bigcap_{\beta\in I_{\alpha}} \beta U$. [If we can do this, then the proof can be finished by compactness.]

We have $ a\in \alpha U$ and $ \alpha U$ is open. Pick a balanced neighborhood of zero $ W$ such that $ a\plus{}W\plus{}W\subset \alpha U$. Or something like that...

Sorry, I do not understand. Could you elaborate?

I'm afraid no. I lost interest in the problem.

That line of reasoning should work. I'll take a look at it when I get a chance :).

Let $ S^1 \equal{} \{\alpha \in \mathbb{C} : |\alpha| \equal{} 1\}$ denote the standard unit circle.

Multiplication gives a continuous map $ \phi : \mathbb{C} \times V \to V$. Let $ W \equal{} \bigcap_{|\alpha| \equal{} 1} \alpha U$, and suppose $ x \in W$. Then for each $ |\alpha| \equal{} 1$, we know $ \alpha x \in U$. Since $ \alpha x \in U$, then $ (\alpha, x) \in \phi^{\minus{}1}(U)$; therefore there are open sets $ A_\alpha \subset \mathbb{C}$ and $ B_\alpha \subset V$ such that $ (\alpha,x) \in A_\alpha \times B_\alpha$ and $ \phi(A_\alpha \times B_\alpha) \subset U$. By compactness of $ S^1$, we know that there exist $ \alpha_1, \dots, \alpha_n$ such that $ S^1 \subset \bigcup_i A_{\alpha_i}$. Letting $ B \equal{} \bigcap_i B_{\alpha_i}$, it follows that $ x \in B$ and $ \phi(S^1 \times B) \subset U$. Therefore $ x \in B \subset W$; this shows that $ W$ is open. Trivially, $ 0 \in W$ as well.

Interestingly enough, you can figure out the above argument by looking at $ V \equal{} \mathbb{C}$.


Find the largest interval $ M \subseteq \mathbb{R^ \plus{} }$, such that for all $ a$, $ b$, $ c$, $ d \in M$ the inequality
\[ \sqrt {ab} \plus{} \sqrt {cd} \ge \sqrt {a \plus{} b} \plus{} \sqrt {c \plus{} d}\]
holds. Does the inequality
\[ \sqrt {ab} \plus{} \sqrt {cd} \ge \sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} d}\]
hold too for all $ a$, $ b$, $ c$, $ d \in M$?
($ \mathbb{R^ \plus{} }$ denotes the set of positive reals.)

[hide="my solution"]
Let $ a \equal{} b \equal{} c \equal{} d$ then the inequality $ 2a \ge 2\sqrt {2a}$ must hold, so $ M \subseteq [2, \infty)$
We'll show that $ M \equal{} [2,\infty)$
Squaring given ineq we have
\[ ab \plus{} cd \plus{} 2\sqrt {abcd} \ge a \plus{} b \plus{} c \plus{} d \plus{} 2\sqrt {(a \plus{} b)(c \plus{} d)}\]
but $ ab \minus{} a \minus{} b \equal{} (a \minus{} 1)(b \minus{} 1) \minus{} 1\ge 0$ and $ cd \ge c \plus{} d$
Let $ a \equal{} \max \{ a,b,c,d \}$ then $ (a \plus{} b)(c \plus{} d) \le 4a^2 \le a^4$ because $ a \in M \equal{} [2,\infty)$

Similarly we prove second inequality.
[/hide]

Is it OK?

[quote="micf"]Let $ a \equal{} \max \{ a,b,c,d \}$ then $ (a \plus{} b)(c \plus{} d) \le 4a^2 \le a^4$ because $ a \in M \equal{} [2,\infty)$[/quote]
I don't understand why this line leads to the solution. But you've already shown that $ ab\ge a\plus{}b$ and $ cd \ge c\plus{}d$ which proves the inequality. So yes.

I don't know what I was thinking about while I was writing the reply, sorry  :rotfl: This line obviously shouldn't be there...
Even squaring is not necessary  :D

Squaring the 2nd one you get $ ab\plus{}cd\plus{}2\sqrt{abcd}\ge a\plus{}b\plus{}c\plus{}d\plus{}2\sqrt{(a\plus{}c)(b\plus{}d)}$ and this follows from $ (a\minus{}1)(b\minus{}1)\ge1$ which gives $ ab\ge a\plus{}b$ and from $ 2(\frac{ab}{2}\minus{}1)(\frac{cd}{2}\minus{}1)\plus{}2(\frac{ad}{2}\minus{}1)(\frac{bc}{2}\minus{}1)\ge 4$ which gives $ abcd\ge (a\plus{}c)(b\plus{}d)$


hello, i think the searched limit is $ \frac{1}{2}$.
Sonnhard.

For all $ m,N\in \mathbb{N}$,
\[ \sum_{k \equal{} 1}^N( \minus{} 1)^kk^m \equal{} \frac {1}{2}( \minus{} 1)^N N^m \plus{} \mathcal{O}\left(N^{m \minus{} 1}\right)
\]
since in this case $ N\equal{}2n$ is even the limit is indeed $ \frac {1}{2}$.


:clap2: Anirudh actually solved a problem without spamming!!! yay!

My actual question was, the problem is cited as if its from a contest in tajikstan, yet it is elementary school material??!?!
Can you give mroe info about the source, this doesn't really make sense as it is such a trivial problem

[hide]multiplying 2nd by 8 and soliving gives $x=-y$ then plugging in to first equation gives $(x,y)=(2,-2)$[/hide]

klutzyzherry14, please hide your answers. this is to be fair to the people who are about to post a solution.

sorry mtms5467... i 4got...will be careful next time :oops:


function
real analysis
advanced fields
advanced fields unsolved

Let $[a,b]$ be a bounded real interval.
(a) Show that is $f: [a,b] \rightarrow \mathbb{R}$ is absolutely continuous, then $f(E)$ has Lebesgue measure 0 whenever E is a subset of $[a,b]$ with Lebesgue measure 0.
(b) Show that if $f: [a,b] \rightarrow \mathbb{R}$ is absolutely continuous, then $f(E)$ is Lebesgue measurable for every Lebesgue measurable subset $E$ of $[a,b]$.

(a) $k$ is fixed. For every $\epsilon >0$, you can cover $E$ by a countable disjoint union of open (in $[a,b]$) intervals $I_{n}$ such that $\sum {m(I_{n})}< \epsilon$. But then $m^{*}(f(\cup {I_{n}})=\cup {f(I_{n}}) )< 2^{-k}$ if $\epsilon$ is small enough, by absolute continuity of $f$. Whence $f(E)$ is measurable and $m(f(E))=0$.

(b) $\epsilon >0$. You can cover $E$ by a countable disjoint union ${U_{k}=\cup_{n} {I_{n,k}}}$ of open (in $[a,b]$) intervals $I_{n}$ such that $m(U_{k} - E)<\epsilon$. Now by the same proof as (a) $m^{*}(f(U_{k}-E))<\frac{1}{2^{k}}$ if $\epsilon$ is small enough.  $f(U_{k})$ is measurable since $f(U_{k}) = \cup_{n} f(I_{n,k})$ and ${f(I_{n,k})= \cup_{a_{i},b_{i} \in %Error. "mathdd" is a bad command.
{Q} \cap I_{n,k}}{f([a_{i},b_{i}])}}$ where $f([a_{i},b_{i}])$ is compact. But then $f(U_{k})=f(U_{k}-E) \cup f(E)$ and $liminf \ f(U_{k}) = N \cup f(E)$, where $N$ is null (this comes from the fact that $m^{*}(N = liminf \ f(U_{k}-E)) \leq \sum {m^{*}(\cap_{n=k \dots +\infty}f(U_{k}-E))} \leq \sum_{k} 2^{-m*k}$ for every $m$; we used the subadditivity and the monotony of $m^{*}$). Since $liminf \ f(U_{k})$ is measurable,  $f(E)$ is also measurable.

For (b), if I use the fact that any Lebesgue measurable set E is $H \cup N$ where $H$ is a countable union of closed sets and $N$ is a null set, can the argument be simplified?

Yeah of course. Actually I was not really aware of this fact ...


limit
trigonometry
real analysis
real analysis unsolved

I'm new to limits of functions.  :blush:  Find $\lim_{x\rightarrow0}{\frac{x-\sin{x}}{x^2}}$ without using derivates or l'Hospital :)

$\boxed {\ f: \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow R\ ,\ f(x)=\frac{x-\sin x}{x^2}\ }\ ;\ |x|<\frac{\pi}{2}\Longrightarrow f(-x)=-f(x)\ .$
$\blacktriangleleft\ 0<x<\frac{\pi}{2}\Longrightarrow \sin x<x<\tan x\Longrightarrow 0<x-\sin x<\tan x-\sin x\Longrightarrow$
$0<f(x)<\frac{\sin x\cdot (1-\cos x)}{x^2\cos x}\Longrightarrow$$0\le\lim_{x\searrow 0}f(x)\le \lim_{x\searrow 0} \frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{x}{\cos x}\Longrightarrow$
$\boxed {\ \lim_{x\searrow 0}f(x)=0\ }\ .$ I used the well-known limits $\lim_{x\to 0}\frac{\sin x}{x}=0\ ,\ \lim_{x\to 0} \frac{1-\cos x}{x^2}=\frac 12\ .$
$\blacktriangleright\ \lim_{x\nearrow 0} f(x)=\lim_{x\searrow 0}f(-x)=-\lim_ {x\searrow 0} f(x)=0\Longrightarrow\boxed {\lim_{x\nearrow 0} f(x)=0\ }\ .$
$\blacktriangleleft\blacktriangleright\ \lim_{x\nearrow 0}f(x)=0\ ,\ \lim_{x\searrow 0}f(x)=0\Longrightarrow \boxed {\ \lim_{x\to 0} f(x)=0\ }\ .$

[quote="Slizzel"]whoa!  :what?: I couldn't imagine that form. How did you transform $\sin{x}$?[/quote]

let $y=\sin x=a_0+a_1x+a_2x^2+a_3x^3+...$

so $y'=\cos x=a_1+2a_2x+3a_3x^2+...$

$y''=-\sin x=2a_2+6a_3x+...$

$y'''=-\cos x=6a_3+...$

so when $x=0$ , $y=0=a_0$ , $y'=1=a_1$ , $y''=0=2a_2$ , $y'''=-1=6a_3$

So $\sin x=x-\frac{1}{6}x^3+...$

*Virgil Nicula method is nice without using power series  :coolspeak:

Yeah shyong, but you still used derivates  :P.

Thanks, Virgil Necula! There is a typo there $\lim_{x\rightarrow 0}{\frac{\sin x}{x}}=1$ :P


A triangle has side lengths measuring 30, 40 and 50 units. What is the length of its shortest altitude, in units?

Obviously, $ 2$ of the altitudes are $ 30$ and $ 40$.  However, the third altitude from the hypotenuse is shorter.  The area of this triangle is $ 600$ sq. units.  If we turn the triangle, we can see that the base becomes $ 50$.  Thus, the altitude's length is $ 2 \times \frac{600}{50}$, or $ \boxed{24}$.

But why is the altitude from the hypotenuse shorter?

Note that $ A\equal{}\frac{1}{2}bh$.  Thus, $ b$ and $ h$ are inversely proportional; so as $ b$ increases, $ h$ decreases and vice versa.  We want $ h$ minimized; thus $ b$ is maximized, and $ b\equal{}50$ and $ h\equal{}24$.

Because a 3-4-5 triangle is a right triangle, a 30-40-50 triangle must be a right triangle as well.

The area of a 30-40-50 right triangle is (1/2) x (30) x (40) = 600 square units.

Two of the altitudes are 30 units and 40 units. Now we must find the other altitude, or the one that extends from the hypotenuse to the opposite vertex. We take 50 units as the base.

We shall introduce a variable. (1/2) x (50) x (h) = 600.

h = [color=red]24[/color] units :rotfl:

If we were to construct a 30-40-50 right triangle and draw an altitude from the vertex to the hypotenuse, we can see that all three triangles created are similar and thus we can set up the proportion $\frac {50}{30}=\frac {40}{x}$.
Solving for $x$, we get $\boxed {24}$.


calculus
integration
function
trigonometry
symmetry
real analysis
complex numbers

Let $ \Phi$ be an even, rapidly decreasing function, and all variables real.  Show:
\begin{align*} \frac {1}{8}\int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty} \Phi(s)\Phi(t)e^{i(s + t)x}(s - t)^2ds dt & \\
& \hspace{ - 125pt} = \left(\int_0^{\infty}\Phi(t)\cos(xt)dt\right)\left(\int_0^{\infty}\Phi(t)t^2\cos(xt)dt\right) + \left(\int_0^{\infty}t\Phi(t)\sin(xt) dt\right)^2 \end{align*}
Or if you'd prefer to stick with real analysis, just do the real part:
\begin{align*} \frac {1}{8}\int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty} \Phi(s)\Phi(t)\cos\big[(s + t)x\big](s - t)^2ds dt & \\
& \hspace{ - 150pt} = \left(\int_0^{\infty}\Phi(t)\cos(xt)dt\right)\left(\int_0^{\infty}\Phi(t)t^2\cos(xt)dt\right) + \left(\int_0^{\infty}t\Phi(t)\sin(xt) dt\right)^2 \end{align*}
Apparently the imaginary part is zero but I haven't checked that part.

$ \frac {1}{8}\int_{ \minus{} \infty}^{\infty}\int_{ \minus{} \infty}^{\infty} \Phi(s)\Phi(t) \cos ( (s \plus{} t)x)(s \minus{} t)^2ds dt  \equal{}$

$ \frac {1}{8}\int_{ \minus{} \infty}^{\infty}\int_{ \minus{} \infty}^{\infty} \Phi(s)\Phi(t) [\cos (sx)\cos (tx) \minus{}\sin (sx) \sin(tx) ](s \minus{} t)^2ds dt  \equal{}$

$ \frac {1}{8}\int_{ \minus{} \infty}^{\infty} \Phi(t) \left( \cos (tx) \int_{ \minus{} \infty}^{\infty} \Phi(s) \cos (sx)(s \minus{} t)^2ds  \minus{}
 \sin(tx) \int_{ \minus{} \infty}^{\infty} \Phi(s) \sin (sx) (s \minus{} t)^2ds \right) dt  \equal{}$

$ \frac {1}{8}\int_{ \minus{} \infty}^{\infty} \Phi(t) \left( \cos (tx) \int_{ \minus{} \infty}^{\infty} \Phi(s) \cos (sx)(s^2 \plus{}t^2) ds  \plus{}
 2t \sin(tx) \int_{ \minus{} \infty}^{\infty} s\Phi(s) \sin (sx) ds \right) dt  \equal{}$

$ \frac {1}{4}\int_{ \minus{} \infty}^{\infty} \Phi(t) \left( \cos (tx) \int_{ 0}^{\infty} \Phi(s) \cos (sx)(s^2 \plus{}t^2) ds  \plus{}
 2t \sin(tx) \int_{ 0}^{\infty} s\Phi(s) \sin (sx) ds \right) dt  \equal{}$

$ \frac {1}{4}\int_{ \minus{} \infty}^{\infty} \Phi(t) \left( \cos (tx) \int_{ 0}^{\infty} s^2 \Phi(s) \cos (sx)ds\plus{}t^2\cos (tx) \int_{ 0}^{\infty}\Phi(s) \cos (sx)ds   \plus{}
 2t \sin(tx) \int_{ 0}^{\infty} s\Phi(s) \sin (sx) ds \right) dt  \equal{}$

$ \frac {1}{2}\int_{0}^{\infty} \Phi(t) \left( \cos (tx) \int_{ 0}^{\infty} s^2 \Phi(s) \cos (sx)ds\plus{}t^2\cos (tx) \int_{ 0}^{\infty}\Phi(s) \cos (sx)ds   \plus{}
 2t \sin(tx) \int_{ 0}^{\infty} s\Phi(s) \sin (sx) ds \right) dt  \equal{}$

$ \frac {1}{2}\int_{0}^{\infty} \Phi(t) \cos (tx) \int_{ 0}^{\infty} s^2 \Phi(s) \cos (sx)ds dt \plus{}\frac {1}{2}\int_{0}^{\infty} \Phi(t) t^2\cos (tx) \int_{ 0}^{\infty}\Phi(s) \cos (sx)ds dt  \plus{}
 \int_{0}^{\infty} \Phi(t) t \sin(tx) \int_{ 0}^{\infty} s\Phi(s) \sin (sx) ds dt  \equal{}$

$ \left(\int_0^{\infty}\Phi(t)\cos(xt)dt\right)\left(\int_0^{\infty}\Phi(t)t^2\cos(xt)dt\right) \plus{} \left(\int_0^{\infty}t\Phi(t)\sin(xt) dt\right)^2$

That's very nice Ken; more structured than what I did:  I took advantage of all the symmetry encountered along the way and combined the integrals accordingly.  I'll post my work if you or anyone else wishes to see it.  How about the general case now:
\begin{align*} \int_{ - \infty}^{\infty}\int_{ - \infty}^{\infty} \Phi(s)\Phi(t)\cos\big[(s + t)x\big](s - t)^{2n}ds dt & \\
& \hspace{ - 150pt} = 8\sum_{k = 0}^{n - 1}\binom{2n}{k}\left(\int_0^{\infty} \Phi(t)f_k(xt)t^k dt\right)\left(\int_0^{\infty} \Phi(t)f_k(xt)t^{2n - k}dt\right) \\
& \hspace{ - 125pt} + 4\binom{2n}{n}\left(\int_0^{\infty}\Phi(t)f_n(xt)t^n dt\right)^2 \end{align*}
Where:
\[ \hspace{ - 35pt}f_k(u) = \begin{cases}\cos(u) & k \text{ 0 or Even} \\
\sin(u) & k \text{ Odd} \end{cases}
\]
I believe it's correct and never published before so unless someone here objects, I'll tentatively propose Shaw's Criteria for the Riemann Hypothesis to hold (you guys mind?):
\begin{align*} \mathcal{S}(x,n) & = 8\sum_{k = 0}^{n - 1}\binom{2n}{k}\left(\int_0^{\infty} \Phi(t)f_k(xt)t^k dt\right)\left(\int_0^{\infty} \Phi(t)f_k(xt)t^{2n - k}dt\right) \\
& \hspace{35pt} + 4\binom{2n}{n}\left(\int_0^{\infty}\Phi(t)f_n(xt)t^n dt\right)^2\geq 0;\hspace{50pt} x\in\mathbb{R}, n = 0,1,2,\cdots \end{align*}
with:
\[ \hspace{ - 180pt}\Phi(u) = 2\sum_{n = 1}^{\infty}(2n^4\pi^2e^{9/2 u} - 3n^2\pi e^{5/2 u})e^{ - n^2\pi e^{2u}
}\]
thereby perhaps offering another route to this problem via a  careful study of sums and products of two relatively simple real expressions:
\begin{align*} C_k(x) & = \int_0^{\infty}\Phi(t)\cos(xt)t^{k}dt \\
S_k(x) & = \int_0^{\infty}\Phi(t)\sin(xt)t^{k}dt \end{align*}
Interested readers can see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208424&sid=518429ee91121b582fd4d6432c439404]AoPS link[/url] for background.

[edit] corrected formula

Also see G. Gasper's paper: [url=http://arxiv.org/abs/math.CA/9307210]Sum of Squares[/url], equation 1.3 for Jensen's criteria regarding the double integral.

Also, I forgot that Myth a few weeks ago showed me how to easy show the imaginary part is zero: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208772&sid=d9d826ef9d098f052716a2f147a534d5]Integral is Zero[/url]


Anyone have any advice for the upcoming AMC 12A/10A? How do you spend the weekend before the AMC? What do you bring to the AMC? What food/drinks keeps your brain thinking fastest? Or anything else?

Eh, I believe this has been asked several times, but see here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255666[/url]


Let $S=\{1,2,3,\ldots,100\}$.

a) Find the probability of choosing two distinct numbers of $S$ such that the addition of them is an even number.

b) Find the probability of choosing two distinct numbers of $S$ such that the addition of them is divisible by 3.

[hide="Solution to Part A"]Replace all numbers in the set with their value $\pmod{2}$.  Then the set contains 50 0's and 50 1's.  We must choose 2 0's or 2 1's.  Therefore, the probability is $\frac{1}{2}\times\frac{49}{99}+\frac{1}{2}\times\frac{49}{99}=\boxed{\frac{49}{99}}$.[/hide]

[hide="Solution to Part B"]Replace all numbers in the set with their value $\pmod{3}$.  Then the set contains 34 0's, 33 1's, and 33 2's.  We must choose two 0's or a 1 and 2.  Regardless of which number we choose first, there are 99 total numbers remaining and 33 of them give us a multiple of 3.  Therefore, the probability is $\boxed{\displaystyle\frac{1}{3}}$.[/hide]

[hide="number 1"]Since two evens or two odds added together make an even number, we have:

${\frac{\binom{50}{2}+\binom{50}{2}}{\binom{100}{2}}}=\frac{49}{99}$[/hide]

[hide="number 2"]We know that the remainders when the numbers are divided by $3$ are in the order:

$1,2,0,1,2,0....0,1$, so we have a total of $34$ ones, $33$ zeroes, and $33$ twos. So we need the two numbers we choose to add up to $0$, so we have:

${{\frac{\binom{34}{1}\binom{33}{1}}{\binom{100}{1}\binom{99}{1}}}+\frac{\binom{33}{2}}{\binom{100}{2}}}$[/hide]

Now that I think about it, Matt276eagles' way of doing number 2 is a lot better

[quote="LoboSolitario"]Very nice [b]matt276eagles[/b] and [b]Iversonfan2005[/b]. But i would like to know if remainder argument is the unique way to do part [b]b[/b].[/quote]

It' s certainly the easiest way to do it..but ofcourse you can do it by breaking it down to cases using the fact that the digits adding up to a multiple of $3$ for the two numbers, but that would be a LOT more complicated


if we were to take a square(2 dimensional cube) of side length $ x$,
the sum of its area, perimeter and  number of vertices  would be

$ x^2 \plus{} 4x \plus{} 4$
we notice this factors to $ (x\plus{}2)^2$

with a cube, of length $ x$, the sum of its volume,surface area,  edges and number of points would be

$ x^3 \plus{} 6x^2 \plus{} 12x \plus{} 8$
we notice this factors to $ (x\plus{}2)^3$

with a line(1 dimensional cube)

the sum of its length and number of endpoints is

$ x\plus{}2$
or $ (x\plus{}2)^1$

we can extend this pattern to find the number of edges ,vertices, faces, volume, and hypervolume(?)  of a 4dimensional cube.

in general, we can find the number of 1-faces 2-faces 3-faces, 4-faces......n-faces of an $ n$dimensional cube by expanding the following $ (x\plus{}2)^n$

and then looking at the coefficient of $ x^j$ where $ 1\le{j}\le{n}$

the questions i had was

is there a way to prove this relationship? do we need to prove this relationship?

i was trying to find more about this online, but there was no such reference to a formula for finding the number of j-dimensional faces.

does anyone know of a link?

[quote="iYOA"]is there a way to prove this relationship? do we need to prove this relationship?[/quote]
Yes.  Probably the easiest way to do it is to define the $ n$-cube abstractly as the set of ordered $ n$-tuplets of zeroes and ones.  Can you define faces, edges, etc. in this abstract context?  $ n$-volume and points being $ x^n$ and $ 2^n$ are more or less obvious; how does this extend to the other cases?

I was very pleased with myself once upon a time when I noticed this same pattern :).  It may also help you to view the $ n$-cube as the product of a line segment by the $ (n \minus{} 1)$-cube.  Then the $ x$ part of $ x \plus{} 2$ is the interior of the segment, while the $ 2$ part is the two endpoints.  (You may have to think a while to understand exactly what that means.)


Consider the number line, from 0 to m*n where m and n are odd and coprime.
Mark the multiples of m and the multiples of n. Let these m+n-1 numbers be
X(1),X(2) ... X(m+n-1) in order. Prove:
X(0)X(1) - X(1)X(2) + X(2)X(3) - X(3)X(4) ... + X(m+n-2)X(m+n-1) = 1

hey ppl, no-one interested in my problem?  :(  :(
please try this problem, the solution ive seen isnt very elegant and im very interested in a nicer soln.

this was discussed before i think iura gave a sol :maybe:


If a triangle has two sides of lengths 5 and 7 units, then how
many different integer lengths can the third side be?

Ok, use the triangle inequality.

So, the third side is called x. the inequality is this:  2<x<12. So x ranges from 3 to 11 and there are 9 values from 3 to 11 inclusive.

The inequality states that a+b>c, where a, b, and c are different sides of the triangle. So x+5>7 gives x>2, and 5+7>x gives x<12.


geometry
ratio
3D geometry
tetrahedron
analytic geometry
function
linear algebra

Haha... It gets REALLY messy; I gave up on it, but I guess if you're willing to deal with the algebra, it's possible.

It would be a little easier because one of the faces is a right triangle (I think they purposely made it that way).

[quote="13375P34K43V312"]Is there a possible analytic geometry solution to #5 on round 1? It's probably really messy, but I'm curious to find out whether anyone actually got the correct answer that way.[/quote]
Yes, there is - check out my solutions post in the Solutions thread.  It requires a lot of working with plane equations, so it's a long solution, but it's actually relatively clean.

This was the first way I tried: find D, then pretty much everything else with ratios, but I ended up finding a minor but important error in my initial work to find D and realized I would have to redo all of the rest of it. So I decided to look simpler and found the similar tetrahedra that prove it in like two steps.

There is a way; in fact, I was pursuing it for awhile.  A set the right angle of the tetrahedron (I forget which one) at the origin and along the x- and y-axes.
It is possible and doesn't involve really messy algebra, but there are just a lot of steps and equations.

It was pretty messy, got something like $\frac{4\sqrt{47}}{3}$ for the z-value of point D (used AB and BC as x & y axes) and was like, wow.

Disgusting.

I got some ultra-simple answer though, don't know if it's right.

Yea it works:
$\\A = (0,0,0)\\ B =(6,0,0) \\ C=(6,8,0) \\ D=(\frac{-2}{3}, 4,\frac{4\sqrt{47}}{3})$

For the plane parallel to $ABC$, you get: $z=\frac{4\sqrt{47}}{3}-\frac{2\sqrt{47}\cdot2^\frac{2}{3}}{3}$

For the plane parallel to $ABD$, you get: $32\sqrt{47}x+160z=96\sqrt{47}\cdot2^\frac{2}{3}$

For the line of intersection between the two planes you get: $x=\frac{19\cdot2^\frac{2}{3}}{3}-\frac{20}{3},z=\frac{4\sqrt{47}}{3}-\frac{2\sqrt{47}\cdot2^\frac{2}{3}}{3}$

For the intersection of this line and $ABC$, you get:$(32\sqrt{47}x+160z=96\sqrt{47}\cdot2^\frac{2}{3},4-2\cdot2^\frac{2}{3},\frac{4\sqrt{47}}{3}-\frac{2\sqrt{47}\cdot2^\frac{2}{3}}{3})$

For the intersection of this line and $ACD$, you get:$(32\sqrt{47}x+160z=96\sqrt{47}\cdot2^\frac{2}{3},6\cdot2^\frac{2}{3}-4,\frac{4\sqrt{47}}{3}-\frac{2\sqrt{47}\cdot2^\frac{2}{3}}{3})$

The distance between these points is $8\cdot2^\frac{2}{3}-8$ or $8(\sqrt[3]{4}-1)$

... ewwww

I started off looking for an analytic solution, then I got stuck.  I looked at the problem for a while, and I wondered why I didn't get the non-analytical solution earlier. =\

[quote="renatzu"]I started off using some analytical geometry to find the height of D and the coordinates of some of the points, but I don't know how to find the equations of planes or use linear algebra, so I had to do something different.  I ended up noticing that if you move plane $Q$ perpindicular to one of the faces, the length of $RS$ scaled as a linear function.  Using the points I had already found,  I was able to find two points of this linear function, and therefore set up an equation.  I then just evaluated the function at the appropriate point.[/quote]
That's exactly what I did. :D


find the line that is tangent to the equation $ y \equal{} x^4 \plus{} 2x^3 \minus{} 3x^2 \minus{} 2x \plus{} 1$ at two points.

[hide=":|"]There is someone new and strange on AoPS. The Invisible One ... his face is not as invisible as you think.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90497]this thread[/url] for a nearly identical problem and a detailed method of solving it  :)


Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E. What is the length of CE?

I don't know how to draw those awesome pictures, but you just draw a square, and then find that the line is sqrt5.


Just to tell everyone, there's a free program called "POVRay" and it's a raytracing program designed to make really good 3D images.  All you need to do is learn the programming.

NOTE: Even though the programming looks complicated.

[quote="SplashD"]where do you get it?[/quote]

http://www.povray.org/download/
:juggle: