Hi Anderson, and other people who are reading this! First things first, I have something to say to our team and other Pennsylvanians only!! (you non-PA people who are reading this-DO NOT READ ON!!)

 [hide]OK, what did I say? [quote](you non-PA people who are reading this-DO NOT READ ON!!)
[/quote]
[hide] Please stop. It is getting old
[hide]You know what, really, it's getting annoying.  :mad:
Alright, here it is: Can you believe Hillary Clinton won PA? I mean, come on!
[/hide]
[/hide]
[/hide]

So, some of you know who I am. One of you know after a supposed leak, right, Ray? Anyway, I'd rather stay somewhat anonymous, but whatever. So... I guess here are a few problems. Linkie to site:
http://www.mathcountscentral.com/practice/set1.htm
You could do a few, if you want to. Make us all proud (and beat the MO team!!  :wink:)

Hi guys, does anyone happen to know the 8th grader from Winchester Thurston who's good at math?

i still dont know who you are....

probability- i do happen to know the 8th grader from WT who is good at math.

i dont like hillary clinton, she would ruin our country if she ever became prez.

Evil Republicans  :mad:
Just joking.  :)


inequalities
topology
function
real analysis
real analysis unsolved

I'm doing a presentation in class on Monday on Young's Inequality and The Hausdorff-Young Inequality. My professor suggested that I should give the statement of the Stein-Weiss Interpolation theorem and possibly a sketch of the proof.

I've searched online and I've found many statements of theorems with this name, and I've also checked out two books (Bergh-Lofstrom "Interpolation Spaces" and Stein-Weiss "Introduction to Fourier Analysis on Euclidean Spaces") and I'm finding so many different statements of the theorem, and some don't even make sense to me with the knowledge that I have.

I am using Folland's book for my Real Analysis class.

Any suggestions/help?

Thanks so much

I recommend that you try wikipedia (hey, why not?).  After that, I suggest you try the Mathworld site, which is right here: http://mathworld.wolfram.com/

Try looking for "Reisz-Thorin Interpolation Theorem" or "complex method of interpolation." That's what is needed for Hausdorff-Young and Young.

But neither the Hausdorff-Young Inequality nor Young's Inequality are optimal on $ \mathbb{R}^n.$ That would be  Beckner's Inequality.

Here's a version of the statement of Riesz-Thorin:
Suppose $ T$ is a linear operator that maps the integrable measurable simple functions over one measure space into measurable functions over another measure space, and suppose there are indices $ p_0,p_1,q_0,q_1\in[1,\infty]$ and constants $ M_0$ and $ M_1$ such that $ \|Tf\|_{q_0}\le M_0\|f\|_{p_0}$ and $ \|Tf\|_{q_1}\le M_1\|f\|_{p_1}$ for all simple functions $ f.$

Then $ T$ can be extended to a linear operator from $ L^p$ to $ L^q$ for any $ t\in (0,1),$ $ \frac1p\equal{}\frac{1\minus{}t}{p_0}\plus{}\frac{t}{p_1}$ and $ \frac1q\equal{}\frac{1\minus{}t}{q_0}\plus{}\frac{t}{q_1}.$ The resulting operator satisfies $ \|Tf\|_q\le M_0^{1\minus{}t}M_1^t\|f\|_p.$

I already tried wikipedia, mathworld, and planet math.  Also, someone else is presenting the Riesz-Thorin Interpolation Theorem.  I already have proofs for Young and Hausdorff-Young.  The problem is that I don't even know what the correct statement of Stein-Weiss is, so I'm not sure how to go about proving it.

Thanks for any more advice.


Let m_a,m_b,m_c be the medians of a triangle of sides a,b,c. Prove that

 m_a(bc-a^2)+m_b(ca-b^2)+m_c(ab-c^2) \geq 0

[size=100][color=red]The solution is short (not mine).Suppose M,N are the midpoints of AB,AC and G is the centroid of the triangle ABC.We will prove this inequality by Ptolemee for the quadrilateral AMGN:m_a*a \leq m_b*c+m_c*b.Then,we get m-a*a^2 \leq m_b*ca+m_c*ab.Finally,we have the answer.
We can easily prove this inequality:0 \leq m_a^2*(bc-a^2)+m_b^2*(ac-b^2)+m_c^2*(ab-c^2).How about  replacing  1,2 by a positive integer n? :D  :D[/color][/size]


Given any square matrix $ A$. Let $ \sigma(A)$ be the set of all eigen value of $ A$, then $ \sigma\left(\left[\begin{matrix} A & A \\ A & A \end{matrix} \right] \right) \equal{} ...$

Note that an eigenvector is non-zero per definitionem !!!

[hide="Solution"]$ \ldots\equal{}(2\sigma(A))\cup\{0\}$[/hide]


AIME II answer was just posted on the http://www.unl.edu/amc

Specifically, the answers are at http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/06answers-aime1+2.html .

please avoid creating new threads that discusses the same things as already existing threads...

Mod should lock/delete this ;)


A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 6 chocolates, 6 cinnamon, and 3 plain. If a box contains 12 doughnuts, how many different options are there for a box of doughnuts?

Hint: inclusion-exclusion principle

[hide]
Casework seems simplest....
If it has 0 plain, there's just 1 way.
If it has 1 plain, there's 2 ways: 5 and 6 or 6 and 5.
If it has 2 plain, then there's 3 ways: 4 and 6, 5 and 5, or 6 and 4.
If it has 3 plain, then there's 4 ways: 4 and 5, 5 and 4, 3 and 6, 6 and 3.
1+2+3+4=10.[/hide]

[hide="Or..."]
Pick 3 donuts not to buy. Then there are
3 ways for one type only.
$ \binom32\cdot2\equal{}6$ ways for two types.
1 way for all three types.
Which still adds to 10.
[/hide]


Define \[q(n) = \left\lfloor\frac{n}{\lfloor\sqrt{n}\rfloor}\right\rfloor (n = 1, 2, ...).\]

Determine (with proof) all positive integers $n$ for which $q(n) > q(n+1)$.

A little experimentation does the trick.

[hide="Solution"]I claim that $q(n)>q(n+1)$ if and only if $n=x^{2}-1,$ where $x$ is a positive integer.

If statement: Let $n=x^{2}-1$
$q(n)=\left\lfloor\frac{x^{2}-1}{\lfloor\sqrt{x^{2}-1}\rfloor}\right\rfloor=\left\lfloor\frac{x^{2}-1}{x-1}\right\rfloor=x+1$
$q(n+1)=\left\lfloor\frac{x^{2}}{\lfloor\sqrt{x^{2}}\rfloor}\right\rfloor=\left\lfloor\frac{x^{2}}{x}\right\rfloor=x$
Hence, $q(n)>q(n+1).$

Only if statement: Let $n=x^{2}+m,$ where $0\le m \le 2x-1.$
$q(n)=\left\lfloor\frac{x^{2}+m}{\lfloor\sqrt{x^{2}+m}\rfloor}\right\rfloor=x+\left\lfloor\frac{m}{x}\right\rfloor$
$q(n+1)=\left\lfloor\frac{x^{2}+m+1}{\lfloor\sqrt{x^{2}+m+1}\rfloor}\right\rfloor=x+\left\lfloor\frac{m+1}{x}\right\rfloor$
Clearly, $q(n+1)\ge q(n).$
[/hide]


If log 2 = .3010, and log 3 = .4771, the value of x when 3^(x+3) = 135 is approximately ? (no idea how to get log5/log3 by hand ><)

[hide="Hint"] $ \log 2 \plus{} \log 5 \equal{} \log 10$ [/hide]


At the forum overview ( http://www.mathlinks.ro/Forum/index.php ), it says there is a post (post-ID is 1011981) by t0rajir0u at 21:01 (time zone: UTC+1) in "Lateral: Did he want it at all¿" (Forum - High School - Other Problem Solving Topics - Puzzles & Brainteasers; Thread-ID is 183774). But if I click on it (normally I would get that thread with that post focused), I get the thread only and no post focused.
If I view all posts by t0rajir0u, I see this post, too (and trying to open it has the same effect). So this post exists, but not in this thread...

Here's a quote:
[quote="t0rajir0u"]Is the item useful in everyday life?

Does he keep it in his house?[/quote]

I've the impression something went wrong  :wink:

Well if it's really important: I'm using FireFox 2.0.0.11 (with no relevant extensions) on Windows XP.

Edit: I see it in the window below the typing box if typing a new post there, too.

Edit 2: After posting an answer, I can see it normally now. It's the first post of page 2, but I'm absolutely sure there was no second page before (I checked twice).

Edit 3: If I delete my answer, the thread again reduces to one page and that post gets invisible again.

this has happened
a post would have been deleted and it messes up a bit


Let $ f: Q\rightarrow R$ be a function such that $ |f(x)\minus{}f(y)|\leq (x\minus{}y)^{2}$ for all $ x,y\in Q$.Prove that f is constant


For $n\in\mathbb{N}$, let $a_{n}$ denote the closest integer to $\sqrt{n}$. Evaluate \[\sum_{n=1}^\infty{\frac{1}{a_{n}^{3}}}.\]

You used the next smaller integer, not the closest one.


induction
algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let A be a ring and let $ A[[x]]$ be the ring of formal power series $ f\equal{}\Sigma_{n\equal{}0}^\infty a_{n}x^n$ with coefficients in A. Show that
if $ f$ is nilpotent, then $ a_{n}$ is nilpotent for all $ n\geqslant 0$. Give an example that converse is not true.

denote $ p : A[[x]] \to A[[x]]/x \equal{} A$ the projection. if $ f \equal{} \sum_{n \equal{} 0}^{\infty} a_n x^n$ is nilpotent, then $ a_0 \equal{} p(f)$ is nilpotent. if $ a_0,...,a_n$ are already nilpotent, so is $ a_{n \plus{} 1} \equal{} p(x^{ \minus{} (n \plus{} 1)}(f \minus{} a_0 x^0 \minus{} ... \minus{} a_n x^n))$. thus by induction, every $ a_n$ is nilpotent.

the converse does not hold. the idea is to bound the sequence of nilpotency exponents $ N(a_n)$: from the proof above, we get $ N(a_{n \plus{} 1}) \leq N(f) \plus{} N(a_0) \plus{} .... \plus{} N(a_n)$. thus if the $ N(a_n)$ are increasing, we get $ N(a_{n\plus{}1}) \leq N(f) \plus{} (n\plus{}1) N(a_n)$. this is not satisfied if $ N_n \equal{} n! * n$. observe that every sequence of nilpotency exponents arises in some ring, e.g. if we divide out nilpotency equations from a polynomial ring of a sequence of variables.

what is a sufficient condition for $ f$ to be nilpotent?

[quote="-oo-"]what is a sufficient condition for $ f$ to be nilpotent?[/quote]Considering a completion of the polynomial ring, it is more appropriate and natural to deal with [i]topologically nilpotent[/i] elements I believe.

show me this induction $ a_{n\plus{}1}\equal{} p(x^{\minus{}(n\plus{}1)}(f\minus{}a_{0}x^{0}\minus{}...\minus{}a_{n}x^{n}))$

-oo- already did... what is unclear about it?

[quote="-oo-"]

what is a sufficient condition for $ f$ to be nilpotent?[/quote]

It holds when $ A$ is Noetherian.

Steve

@3rd: interesting! indeed, it's not hard to show that $ f$ is topologically nilpotent if and only if all coefficients are nilpotent.
@steve: clear, if $ (a_0,a_1,...)$ is finitely generated the nilpotency exponents are bounded.

any ideas for a general characterization? e.g. we conjecture that $ f$ is nilpotent if and only if there is some constant $ N$ such that $ N(a_{n\plus{}1}) \leq N \plus{} (n\plus{}1)(N(a_n)\minus{}1)$.


[b](1) If a>0,b>0 and x3+ax2+bx+2=0 has all real rots. Then prove that
a+b>=3(2^1/3+4^1/3)[/b]

Let $ p$, $ q$, and $ r$ represent the roots of $ x^3\plus{}ax^2\plus{}bx\plus{}2\equal{}0$.
Then $ a\equal{}\minus{}(p\plus{}q\plus{}r)$, $ b\equal{}pq\plus{}qr\plus{}rp$, and $ pqr\equal{}\minus{}2$.
By Descartes' Rule of Signs, there are no positive roots of the equation $ x^3\plus{}ax^2\plus{}bx\plus{}2\equal{}0$, so $ p, q, r \le 0$.
$ pqr\equal{}\minus{}2$ implies that $ p, q, r \ne 0$, so $ p, q, r<0$, and $ pq, qr, rp >0$.
By AM-GM, $ \dfrac{a}{3}\equal{}\dfrac{\minus{}p\minus{}q\minus{}r}{3} \equal{} \dfrac{|p|\plus{}|q|\plus{}|r|}{3} \ge (|pqr|)^\frac{1}{3}\equal{}2^\frac{1}{3}$, so
$ a \ge 3\cdot 2^\frac{1}{3}$.
By AM-GM, $ \dfrac{b}{3}\equal{}\dfrac{pq\plus{}qr\plus{}rp}{3} \ge (p^2 q^2 r^2)^\frac{1}{3} \equal{} (pqr)^\frac{2}{3}\equal{}(\minus{}2)^\frac{2}{3}\equal{}4^\frac{1}{3}$, so
$ b \ge 3 \cdot 4^\frac{1}{3}$.
Adding the two inequalities we get
$ a\plus{}b \ge 3(2^\frac{1}{3}\plus{}4^\frac{1}{3})$.


calculus
integration
function
ratio
geometric series
calculus computations

I have an integration test tomorrow morning which will consist of all integration, including improper integration. Because of this, I will definitely need to pull out some divergent and convergent functions to show that certain integrals diverge using the Comparison Theorem. Since we are allowed to use a sheet of paper for tomorrow's test, I feel that I ought to bring an inventory--or at least a few--integrals that--while easy to show--diverge. I greatly appreciate any feedback. Alright, nvm. That was an irrational wish of mine.

Geometric series are important for comparison tests. In fact, comparing a geometric series to a general series $ a_{n}$ that has undergone a certain operation is precisely how you derive the root and ratio test.
The power series (often called p-series) is another powerful comparison tool. Some other things to remember for a basic test on the subject matter is that sinusoidal functions are bounded between -1 and 1 and that $ 0 < e^{\minus{}kn} < 1$ (assuming $ k > 0$ and $ n \in \mathbb{N}$)

So, a sample problem would be to determine if $ a_{n} \equal{} \sum_{i\equal{}1}^{n} \frac{\sin(n)}{n^{3}}$ converges or diverges as $ n \rightarrow \infty$. Can you tell what the outcome would be?


There are $n\geq 3$ pupils standing in a circle, and always facing the teacher that stands at the centre of the circle. Each time the teacher whistles, two arbitrary pupils that stand next to each other switch their seats, while the others stands still. Find the least number $M$ such that after $M$ times of whistling, by appropriate switchings, the pupils stand in such a way that any two pupils, initially standing beside each other, will finally also stand beside each other; call these two pupils $ A$ and $ B$, and if $ A$ initially stands on the left side of $ B$ then $ A$ will finally stand on the right side of $ B$.

In the end, the permutation we obtain is some mirror flip along a line. Every student must cross this line at least once, and must do so at one of two places. Furthermore, if two students cross the line at the same place, they must cross each other. It follows that the number of transpositions is at least $\min_a {a \choose 2}+{n -a \choose 2}$. This bound is tight.

As an aside, we can prove a lower bound for the less restricted problem in which we may transpose arbitrary pairs of students. Note that for every triple of students, at least one pair of these students must be transposed at some point in time. Each time we transpose a pair of students, we satisfy this condition for $n-2$ triples. Thus, we need at least ${n \choose 3}/(n-2) = \frac{n(n-1)}{6}$ transpositions. When is this bound tight?

For the follow-up question, something is clearly wrong: with $2n + 1$ students, $n$ switches suffice: switch students $n + 1 + i$ and $n + 1 - i$ for every $i \in \{1, \ldots, n\}$.

Oops...

Perhaps the follow-up question should be the following: Given two adjacent blocks of students with $A$ and $B$ people respectively, we may swap the blocks at the cost of $AB(n-A-B)$ moves. What about when these swaps have the same cost?

Can everyone solve it clearly? Thanks very much!
I spent a lot of time to find the number of transformations but it's so hard to make sure that whether this number is true or not.


let three sets $A_{1}\cup A_{2}\cup A_{3}$={1,2,3,...3n} so that $a_{k}$ have n elements .prove that there
 exist a permutation $\sigma$ so that $(A _{\sigma(1)}+A _{\sigma(2)})\cap A _{\sigma(3)} \not= \phi$


Given the equation;

(a+b)/2=4ab/(a+b)

Determine every value a/b can take.

I'll appreciate if someone reposts the equation in LaTeX, this is my very first post and i don't know how to use it  :)  Thanks.

hello, your equation is equivalent to
$ \frac{(a\plus{}b)^2}{2}\equal{}4ab$
$ a^2\plus{}b^2\plus{}2ab\equal{}8ab$
$ a^2\plus{}b^2\minus{}6ab\equal{}0$
$ \frac{a}{b}\plus{}\frac{b}{a}\minus{}6\equal{}0$
$ t\plus{}\frac{1}{t}\minus{}6\equal{}0$ with $ t\equal{}\frac{a}{b}$
it gives
$ \frac{a}{b}\equal{}3\pm 2\sqrt{2}$
Sonnhard.


AMC
AIME
geometry
3D geometry
floor function
LaTeX
AIME I

So, I'm going to start an AIME Marathon. We just post an AIME question, or an AIME level question and the next person answers it.

So, I'll begin:

[size=200]Problem 1[/size]

Each angle of a regular $r$-gon is $59/58$ times larger than each angle of a regular $s$-gon. What is the largest possible value of $s$?

[quote="mathgeniuse^ln(x)"][quote="rnwang2"][hide="Solution"]
Each angle of the $r$-gon is $\frac{180(r-2)}{r}$.

Each angle of the $s$-gon is $\frac{180(s-2)}{s}$.

Thus, $\frac{180(r-2)}{r}=\frac{59}{58}\cdot\frac{180(s-2)}{s}$.

This simplifies to $r=\frac{116s}{118-s}$.

Clearly, $s<118$.

So the largest possible value of $s$ is $\boxed{117}$.[/hide]
If this is right someone else can post the next question.[/quote]

Wow that is just like that from the pamphlet.[/quote]

What pamphlet?

NVM.

[hide]Anyways, for #2, it is just, you have to figure out the number of perfect squares and cubes less than or equal to 500.  This is 27 after checking.

Adding 27 we get 527.

But 512, is a perfect cube so we have to add one more to get 528

$528$[/hide]

[quote="anirudh"]Yes. You are right. :)

Problem 2

The sequence 2, 3, 5, 6, 7, 10, 11, ... consists of all positive integers that are not a square or a cube. Find the 500th term.[/quote]

[hide] First take away the perfect squares. From 1 to 500, there are 22. Thus, we now have to go up to 522. Now we note that there are no perfect squares between 500 and 522, so we're still good. The next perfect square is 529. Now let's take away the perfect cubes. There are 8 between 1 and 522. However, $1^{3}$ and $4^{3}$ have already been taken out, since they're perfect squares. Thus, we take out only 6 and the 500th number is $\boxed{528}$ [\hide]

If I'm right, here's problem 3.

Let the product of the non-real roots of $x^{4}-4x^{3}+6x^{2}-4x=2005$ be $K$. Find $\lfloor \sqrt{K}\rfloor$.[/hide]

gah mathgeniuse^ln(x) beat me

What pamphlet?

Problem 2 looks like some MC problem I saw earlier this year.
[hide="probably calculated wrong"]
This shouldn't be the best way of doing this.

So, the answer would be 500, except for all of those annoying squares and cubes.

Well, there are 22 squares less than 500 and 7 cubes.

This would add 29 to our answer, but $1^{3}=1^{2}$ and $4^{3}=8^{2}$ mean that 1 and 64 have been counted twice so our answer should be 527.  But, there may be squares or cubes between 500 and 527.  Wow, what a surprise: $8^{3}=512$ so 528 is the final answer.[/hide] he beat me too. whatever.

The solutions manual.

And the next question.

Problem 4.

What is the value of
$\sqrt{(28)(29)(30)(31)+1}$?

Problem 3 from mathgeniuse:
[hide]
If we have an expression of the form $n(n+1)(n+2)(n+3)+1$, we can expand to get $n^{4}+6n^{3}+11n^{2}+6n+1$  which factors into $(n^{2}+3n+1)^{2}$.  So, $\sqrt{(28)(29)(30)(31)+1}=28^{2}+3(28)+1=869$.
[/hide]

Someone else can post a problem.

[quote="SM4RT"]@mathgeniuse^lnx: umm..somebody already posted a problem three..[/quote]
Sorry, but usually people don't hide the questions.

But yeah, I put problem 4, and you put problem 5.

[quote="rnwang2"][quote="13375P34K43V312"]

If I'm right, here's problem 3.

Let the product of the non-real roots of $x^{4}-4x^{3}+6x^{2}-4x=2005$ be $K$. Find $\lfloor \sqrt{K}\rfloor$.[/hide]

gah mathgeniuse^ln(x) beat me[/quote]
[hide="Solution"]$x^{4}-4x^{3}+6x^{2}-4x+1=2006$.

$(x-1)^{4}=2006$.

$(x-1)^{2}=\pm\sqrt{2006}$.

$-\sqrt{2006}$ gives the nonreal roots.

$x^{2}-2x+1=-\sqrt{2006}$.

$x^{2}-2x+1+\sqrt{2006}=0$.

The product of the roots is $1+\sqrt{2006}$.

Since $6^{4}=1296$ and $7^{4}=2401$, $\lfloor{\sqrt{1+\sqrt{2006}}}\rfloor=\boxed{6}$.

[/hide][/quote]

Yeah, though the original (2005 AIME I) problem asked for $\lfloor K \rfloor$, not the square root.

[quote="PenguinIntegral"][quote="rnwang2"]A set of building blocks contains a number of wooden cubes. The six faces of each cube are painted, each with a single color, in such a way that no two adjacent faces have the same color. Given that only five different colors have been used and that no two of the blocks are identical in their colorings, what is the maximum number of blocks there can be in the set?[/quote]

That's way too easy for an AIME problem.[/quote]

Oh really? Ignore it then.  :oops:

EDIT: The first problem was pretty easy too...

I don't know if this is right (seems too easy, even with what PenguinIntegral said), but...

[hide]
Let the five colors be A, B, C, D, E.  we color the top face A, then the four faces adjacent to it have to be B, C, D, E.  There are 3 ways to do that.  Then the last face has to be A.  So, for each color we duplicate (here A), there are 3 possibilities.  So the answer is 5 * 3 = [b]15.[/b]
[/hide]

Next problem:

7. Let m = (10^4 + 324)(22^4 + 324)(34^4 + 324)(46^4 + 32^4)(58^4 + 324) and n = (4^4 + 324)(16^4 + 324)(28^4 + 324)(40^4 + 324)(52^4 + 324). Find m/n.

^^I love this problem because it's the first 14 I ever did.  Still haven't solved a 15.

A set of positive numbers has the triangle property if it has three elements which are the side lengths of a non-degenerate triangle. Find the largest n such that every 10-element subset of {4, 5, 6, ... , n} has the triangle property.

[quote="rnwang2"][quote="vvr1590"]I don't know if this is right (seems too easy, even with what PenguinIntegral said), but...

Let the five colors be A, B, C, D, E.  we color the top face A, then the four faces adjacent to it have to be B, C, D, E.  There are 3 ways to do that.  Then the last face has to be A.  So, for each color we duplicate (here A), there are 3 possibilities.  So the answer is 5 * 3 = [b]15.[/b]
[/quote]
Actually, that's not right. But whatever. Just forget about the problem.

EDIT: [hide="Answer"]The answer is 55 by the way[/hide][/quote]

to what question (ur original one) or the one vvr1590 posted??

EDIT: @ mdk this question i think was already posted on this very forum isnt the answer
[hide]
253[/hide]


What major product will be obtained when 1,2-divinylcyclohexan-1-ol is treated with potassium hydride?

perhaps cyclodec-5-eneone?
see edited post. how stupid of me there's obviously no compound like dat.dunno why i wrote oct, i meant dec.
it seems like oxy cope rearrangement to me.

Obviously that answer is wrong, since there are carbons missing. Are you going to keep throwing away answers like that, with no justification?  :whistling:


Let $n$ be a positive integer with $k\ge22$ divisors $1=d_{1}< d_{2}< \cdots < d_{k}=n$, all different. Determine all $n$ such that \[{d_{7}}^{2}+{d_{10}}^{2}= \left( \frac{n}{d_{22}}\right)^{2}.\]


Prove or disprove that there exists four prime numbers $p_1, p_2, p_3, p_4$ such that any consecutive pair are twin primes.

If by "twin primes" you mean primes whose difference is 2, then here is [hide="the solution"]Those numbers can be written as $p_1, p_1+2, p_1+4, p_1+6$. Any prime number greater than $3$ divided by $3$ gives the remainder $1$ or $2$. If $p_1$ gives remainder $1$, then $p_1+2$ gives remainder $0$ and can't be prime. If $p_1$ gives remainder $2$, then $p_1+4$ gives remainder $0$ and can't be prime. And if $p_1=3$ we get $p_4=9$ which is composite. For $p_1=2$ all the other numbers are even, therefore composite.

Conclusion: Such numbers don't exist.[/hide]


Yeah, Calculus from BC is enough. In fact, our school used to have the Physics C-Calculus BC classes blocked (single class, two periods).

The physics B exam is algebra based, and covers a wide variety of topics, but not as much in depth. For example, it goes over fluid mechanics, atomic and nuclear physics, optics, waves, and thermal physics, which isn't touched by our C class.   The C exam is calculus based, and goes over a few topics intensively. Our semesters are separated into mechanics for fall, and electricity and magnetism for spring. That's pretty much what the C exams cover (they're named for a reason ;)). If you're planning to major in engineering, you probably want to take the C exam.

i am taking physics contest course now. And i find that the most widly used calculus in physics is Derivative and Integral. They can be really useful sometimes.

I heard that there are two C tests: One on Mechanics and One on E&M. is that true??

Yep. You're not required to take them together, so if you think you're definitely set to go on only one of them, it might be better just to hold off on the other, especially if you're loaded with AP tests in other subjects. They're somewhat related (force is always going to be ma), so usually people take them both together (at my school, anyways).

i'm taking physics c right now as well.  our class went sort of slowly and we didnt finish mechanics until march.  my teacher is nevertheless attempting to rush through electricity and magnetism until the ap exam on may 9.  i don't want to take both.  i'd rather just study mechanics intensely so that I can get a 5 on its ap.  i know that the collegeboard allows you to take one or the other, but how would I be able to get that to work in my school?  everyone has always just taken both, and most people do very poorly on electricity and magnetism.  but i'm only a freshman, so i'd rather just do mechanics now and then maybe take electricity and magnetism through EPGY.

DOnt worry too much about the AP Physics C. Even if one gets 60% of the questions right, is is  guranteed a 5 on both. I speak from experience (though i think defiantely scored muchmore than 50% on both) But genereally, AP teachers believe that 50% is all thats needed. Also they dont test too much on concepts, ie yo u wont find very hard problems. I think the point is just to test your ability to recognize and use basic formulas.

[quote="qweretyq"]I know that you can take one of the Physics C exams E&M or mechanics.
But can you take both the same year?
And can you take it with Physics B?
Like in the same year take AP physics B, Physics C mechanics, and physics C E&M tests?[/quote]

You can take both the C's in the same year, but I do  not know about taking it all together with B. I am assuming yes, because if you can take both C's, then why not all togehter?? :)

[quote="towersfreak2006"]There is usually a conflict though, so you would have to talk to your counselor about ordering re-scheduled tests.[/quote]

does that mean the times overlap?? because i looked at this years AP dates and hours in our school, and neither one of C-M, C-E&M, or B were overlapping or happening at the same time.

Are you sure? The two C exams never conflict, since people usually take those two together, but B and C are in the same time slot.

The conflicts are:
Calculus AB/Calculus BC
French Language/Human Geography
Computer Science A/Computer Science AB/Spanish Language
European History/Studio Art
French Literature/German Language
Biology/Music Theory
Physics B/Physics C
Chemistry/Environmental Science
Latin Literature/Latin: Vergil
Government & Politics (Comparative)/Art History

[quote="ShadowLord"]from what  know..it involve calculus..but how much?is the stuff in BC enough?
besides,is it really worth to take the c exam instead of B?[/quote]

Although Physics C does require that you know calculus, it will never ask a complicated integral or derivative, so BC calc is much more than enough.

C exam covers some calculus, AB material should be enough but I STRONGLY recommend taking BC concurrently

This thread is 8 years old. What was the use of your comment?

This kid loves to make troll posts, so I do not see any point.

On a tangent, Physics has a "s" in it.

I am taking AP Physics B this coming school year, and it does not use calculus, but only a thorough amount of algebra and trigonometry.
Both parts of AP Physics C (Mechanics and E & M) involve calculus to some extent. What is provided in the AP Calculus BC course should be plenty.


search
inequalities
inequalities proposed

$ a,b,c,x,y,z>0$ and $ xy\plus{}xz\plus{}yz\equal{}3$.Prove that !
$ \frac{a}{b\plus{}c}(x\plus{}y)\plus{}\frac{b}{a\plus{}c}(y\plus{}z)\plus{}\frac{c}{a\plus{}b}(y\plus{}z) \ge 3$

Search for it. It's a famous inequality of Gabriel Dospinescu.

[quote="bigbang195"]$ a,b,c,x,y,z > 0$ and $ xy \plus{} xz \plus{} yz \equal{} 3$.Prove that !
$ \frac {a}{b \plus{} c}(x \plus{} y) \plus{} \frac {b}{a \plus{} c}(y \plus{} z) \plus{} \frac {c}{a \plus{} b}(y \plus{} z) \ge 3$[/quote]
See here:
http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf


limit
integration
function
calculus
calculus computations

Let $ k, m$ be natural numbers. Find the limit

$ \lim\limits_{n\rightarrow \infty} n\int\limits_{0}^{1}\frac{x^{kn}}{1\plus{}x^{m}}dx.$

that is the correct answer... and a generalization, (where $ f$ is a continuous function):

$ \lim\limits_{n\rightarrow \infty}n\int\limits_{0}^{1}\frac{f(x^{n})}{1\plus{}x^{m}}dx$.


[hide="Answer"] [b]X = 63 [/b][/hide]
[hide="Solution"]This problem is easy to solve if you know basic algebra; if you factor 216 into 76*36 (Sorry I haven't installed Tex yet). Then all you have to do is divide both sides by 34*75, leaving you with x=32*7, which when sinplified equals 63.

[size=75][color=darkblue]<Note from moderator rcv:  You don't have to install TeX.  You just need to read the LaTeX guide, available from the menu bar on the left of your screen (AoPS users).  For example, $21^6 = 7^6 \times 3^6$.>[/color][/size] [/hide]

[quote="MCrawford"]Find the integer $n$ such that $n \times 3^4 \times 7^5 = 21^6$.[/quote]

[hide]$n\cdot 21^4\cdot 7=21^6$

$7n=7^2\cdot 3^2$

$n=63$[/hide]


In a triangle $ ABC$ such that $ \angle B \equal{} 2\angle A$ ,its lengths of three sides make a arithmetic progression with common difference $ 1$ in increasing order.
Find the lengths of the three sides.

[hide]Ans:
                $ (1)\text{ the length of sides are } 4,5 \text{ and  }6 .A\equal{}41.41^o ,B\equal{}82.82^o , \text{ and } C\equal{}55.77^o$

                 $ (2) \text{the length of sides are } \frac{ 1\plus{}\sqrt 13}{2} , \frac{3\plus{}\sqrt 13}{2}  ,\frac {5\plus{}\sqrt 13}{2}$
[/hide]

[quote="vinskman"][hide]Ans:
                $ (1)\text{ the length of sides are } 4,5 \text{ and }6 .A \equal{} 41.41^o ,B \equal{} 82.82^o , \text{ and } C \equal{} 55.77^o$

                 $ (2) \text{the length of sides are } \frac { 1 \plus{} \sqrt 13}{2} , \frac {3 \plus{} \sqrt 13}{2} ,\frac {5 \plus{} \sqrt 13}{2}$
[/hide][/quote]

That's right,vinskman!
Please write how to solve it if possible.

$ \text {Thanks,since the details of proof are quite long . I am now busy for preparing my documents.}$

$ \text {I will come back to complete the proof.}$

[hide]$ \text {Since B=2A, the possible arrangement of three angles : }$
$ (1) B>A>C ,(2) B>C>A ,(3) C>B>A$
$ (1)\text {when }B>A>C \text{  ; their corresponding angles are a+1, a,a-1 , where a is some positive real.}$
$ \text  { Use Sine Formula again,}$
$ \frac{\sin2A}{a+1} = \frac{\sin A}{a} = \frac{\sin(180^o-A-2A)}{a-1}$

$ \implies \frac{\sin2A}{a+1} = \frac{\sin A}{a} = \frac{\sin3A}{a-1}$

$ \implies \frac{\sin2A}{a+1} = \frac{\sin A}{a} = \frac{\sin3A}{a-1}=\frac{\sin 2A+\sin A+\sin 3A}{a+1+a+a-1}
        =\frac{\sin 2A+\sin A+\sin 3A}{3a}$

$ \implies \frac{\sin2A}{a+1} = \frac{\sin A}{a} = \frac{\sin3A}{a-1}=\frac{\sin 2A+2\sin 2A+\cos A}{3a}=\frac{\sin 2A(1+2\cos A)}{3a}$

$ \text{consider : } \frac{\sin2A}{a+1} = \frac{\sin A}{a} \text{ and } \frac{\sin2A}{a+1} =\frac{\sin 2A(1+2\cos A)}{3a}$

$ \implies \frac{\sin2A}{\sin A} = \frac{a+1}{a}  \text{ and } \frac{{3a} }{a+1} =\frac{\sin 2A(1+2\cos A)} {sin2A}$

$ \implies \cos A = \frac{a+1}{2a} \text{ and } 1+2\cos A =\frac{{3a} }{a+1}$

$ \implies  \frac{a+1}{2a} = \frac{2a-1}{2(a+1)}\implies a^2-3a-1=0$

$ \implies a=\frac{3+\sqrt 13} {2}$

$ \text{Thus , the three lengths a-1,a,a+1 are : } \frac{1+\sqrt 13} {2},\frac{3+\sqrt 13} {2} , \frac{5+\sqrt 13} {2}$
----------------------
$ (2)\text {when }B>C>A\text{  ; their corresponding angles are a+1, a,a-1 , where a is some positive real.}$

$ \text  {Also use Sine Formula again,}$
 $ \frac{\sin2A}{a+1} = \frac{\sin3A}{a} = \frac{\sin A}{a-1}$

$ \implies \frac{\sin2A}{a+1} = \frac{\sin 3A}{a} = \frac{\sin A}{a-1}=\frac{\sin 2A+\sin 3A+\sin A}{a+1+a+a-1}
 =\frac{\sin 2A+\sin 3A+\sin A}{3a}$

$ \implies \frac{\sin2A}{a+1} = \frac{\sin 3A}{a} = \frac{\sin A}{a-1}=\frac{\sin 2A+2\sin 2A+\cos A}{3a}=\frac{\sin 2A(1+2\cos A)}{3a}$

$ \text{consider : } \frac{\sin2A}{a+1} = \frac{\sin A}{a-1} \text{ and } \frac{\sin2A}{a+1} =\frac{\sin 2A(1+2\cos A)}{3a}$

$ \implies \frac{\sin2A}{\sin A} = \frac{a+1}{a-1}  \text{ and } \frac{{3a} }{a+1} =\frac{\sin 2A(1+2\cos A)} {sin2A}$

$ \implies \cos A = \frac{a+1}{2(a-1)} \text{ and } 1+2\cos A =\frac{{3a} }{a+1}$

${ \implies  \frac{a+1}{2(a-1)} = \frac{2a-1}{a+1}}\implies a^2-5a=0$

$ \implies a=5$

$ \text{Thus , the three lengths a-1,a,a+1 are : } 4,5,6$
----------------------------
$ (3)\text {when }C>B>A\text{, Similary , it cannot get  possible solution to a .}$
[/hide]

This is an old IMO problem.

[hide]Denote the length of $ AB,BC,CA$ by $ c,a,b$, respectively. We claim that $ b^2\equal{}a(a\plus{}c)$. Take the point $ D$ on the extension of $ AB$ beyond $ B$ such that $ BD\equal{}a$. Do some angle chasing, and note that $ \triangle CBD\sim\triangle ACD$. The claim follows easily.

Since $ \angle B>\angle A$, we must have $ b>a$. Now, consider three cases $ b>a>c$, $ b>c>a$ and $ c>b>a$. The rest is now easy. We should get the answer $ 4,5,6$.[/hide]

thank you vinskman!

@joh
Oh,really?I didn't know this is a problem of IMO.
yes,using similarity is the most clear solution :)


This is not the place to put a question like this, but... I need to know how to figure this out quickly.

I am using mathematica.
I used NDsolve on a differential equation.
I can plot it over a certain region.
I can also make a Table of it, but unfortunately the table would just print out R[1], R[1.5], etc... as opposed to numeric points.
I want to export the values to a data table, but I can not figure out how to do so.

Sorry mods for putting this in the wrong place, but I need a lot of people to see it, so they can help me out quickly.

Any help will be appreciated.

After 6 hours of learning, I was able to accomplish my goal.  Yay to me.

I think I'll borrow this thread a little bit. Do you use Maple? I have Maple 9, and it is totally screwed up. It doesn't give correct taylor expansions for $\sqrt x$ around any even square.  For example taylor(sqrt(x), x=4, 3); gives

$(2+{\frac{1}{4}} \left( x-4 \right) -{\frac{1}{64}} \left( x-4 \right) ^{2}+O \left( \left( x-4 \right) ^{3} \right) )$.

This is obviously a nutty answer as it is not equal to 2 when x = 4. Why does it do this? Do you have any idea?

edit: the crazy part is that I got this answer now when I chose to convert to latex, but it does NOT give the correct answer in the window, and when doing more complicated expressions, it's totally wrong. Do you have this problem? The part where it screws up is that it answers $(1/4)x - 4$ instead of what you want. Similarly, taylor(x^(1/2) + cos(Pi*x), x=1, 2); gives $(1/2)x - 1$. Unless I am insane this is just wrong. Is this a known bug?

I'm sorry, I haven't used Maple much.  The only thing I've done with maple is graph some things.


Is the AMC itest certified? I'm just wondering.

Thanks,
bpms

[quote="chess64"]lol... is the AMC iTest certified?

let's see, there have been 2 iTests so far and 57 AMC's...[/quote]

I thought there have beem 3 iTests.

[quote="krustyteklown"]well, the AMC is a lot more prestigious and taken seriously than the itest, so i dunno what that would really mean[/quote]
[quote="chess64"]
lol... is the AMC iTest certified?

let's see, there have been 2 iTests so far and 57 AMC's...[/quote]
I know all this, I'm just wondering if the itest decided to certify the AMC (not that it makes too much of a difference).
EDIT:Fixed quote tags

[quote="Boy Soprano II"]Then shouldn't this go in the itest forum instead of the AMC forum?[/quote]
I'm not sure   :maybe:
Anyways, is it true that the itest director called the AMC a "failure"?

There was a huge flamewar in the itest forum once, when people quoted stuff that the itest director allegedly put on his blog (I say "allegedly" because I never bothered to verify it), and it got rather nasty.  But it seems to have subsided for some time now, so why don't we try to keep it that way by not talking about it? :wink:


algebra
polynomial
number theory unsolved
number theory

Let $ \{P_1(x),P_2(x),...,P_n(x)\}$ be a set of nonconstant polynomial with integer coefficients.Prove that for some integer positive $ m$, the integer $ P_1(m),P_2(m),...,P_n(m)$ are all composite.

We'll prove that for some [b]large enough[/b] positive integer $ m$, the integers $ P_1 (m), P_2 (m) , \cdots , P_n (m)$ are all composite.

First, if any of the $ P_i$'s are composite for every large enough integer $ m$, that polynomial may be ignored.
(You'll see why as you read the solution)

Now, assume that each of the polynomials can be prime for infinitely many positive integer $ m$.

Then, we can choose $ n$ [b]distinct[/b] primes $ p_1 , p_2 , \cdots , p_n$, and integers $ q_1 , q_2 , \cdots , q_n$ such that $ P_i (q_i ) \equal{} p_i$ for all $ i\equal{}1,2, \cdots , n$.

Claim. There exists an integer $ n_i$ such that for all integers $ k>n_i$, $ P_i (q_i \plus{} kp_i )$ is composite.
Proof. This is trivial; observe that $ P_i (q_i \plus{} kp_i )$ is always divisible by $ p_i$, and only a finite number of them can be equal to $ p_i$.

Due to the claim and by CRT, if we choose a sufficiently large integer $ m$ such that $ m \equiv q_i (mod p_i )$ for all $ i$, the problem is solved immediately.

[quote="qwerty414"]We'll prove that for some [b]large enough[/b] positive integer $ m$, the integers $ P_1 (m), P_2 (m) , \cdots , P_n (m)$ are all composite.

First, if any of the $ P_i$'s are composite for every large enough integer $ m$, that polynomial may be ignored.
(You'll see why as you read the solution)

Now, assume that each of the polynomials can be prime for infinitely many positive integer $ m$.

Then, we can choose $ n$ [b]distinct[/b] primes $ p_1 , p_2 , \cdots , p_n$, and integers $ q_1 , q_2 , \cdots , q_n$ such that $ P_i (q_i ) \equal{} p_i$ for all $ i \equal{} 1,2, \cdots , n$.

Claim. There exists an integer $ n_i$ such that for all integers $ k > n_i$, $ P_i (q_i \plus{} kp_i )$ is composite.
Proof. This is trivial; observe that $ P_i (q_i \plus{} kp_i )$ is always divisible by $ p_i$, and only a finite number of them can be equal to $ p_i$.

Due to the claim and by CRT, if we choose a sufficiently large integer $ m$ such that $ m \equiv q_i (mod p_i )$ for all $ i$, the problem is solved immediately.[/quote]
nice  :D


we can suppose, obvious, that $gcd(a,b) = gcd(b,c) = gcd(c,a) = 1$.

if the the two expressions are integers, we have that they difference are one integer.

so,

$\frac{(a-b)(b-c)(c-a)}{abc} \in Z$.

but, $gcd(c-a,c) = gcd(c-b,c) = 1$ (and the analogous).

so, $c|a-b$, $b|c-a$ and $a|b-c$.

now, is easy.  :)

[quote="e.lopes"]we can suppose, obvious, that $gcd(a,b) = gcd(b,c) = gcd(c,a) = 1$.[/quote]
I don't think this assumption is valid. We can assume $\gcd(a,b,c)=1$, but that is much weaker than the hypothesis $\gcd(a,b) = \gcd(b,c) = \gcd(c,a) = 1$.

--------------------------

A hint for the second part of spider_boy's question:
[hide]Consider the polynomial $(x-\tfrac{a}{b})(x-\tfrac{b}{c})(x-\tfrac{c}{a})$, and use the Rational Root Theorem.[/hide]

[quote="spider_boy"]Prove that there exists natural numbers $a,b,c$ such that only one of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ is an integer.
Prove that if both of the above expressions are integers then $a=b=c$.[/quote]

$a=1,b=2,c=4$

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} = \frac{1}{2}+\frac{2}{4}+\frac{4}{1} = 5$
$\frac{b}{a}+\frac{c}{b}+\frac{a}{c} = \frac{2}{1}+\frac{4}{2}+\frac{1}{4} = 4.25$

For the second one, just consider $f(x)=(x-a/b)(x-b/c)(x-c/a) \in {\mathbb Z}[X]$ and since $x-a/b$, $x-b/c$ and $x-c/a$ divide $f$, they are from ${\mathbb Z}[X]$, so $a/b,b/c,c/a \in {\mathbb Z}$, so $a=b=c$.


search
function
geometry unsolved
geometry

Problem 5:
Let $ ABC$ be a triangle in which $ AB \equal{} AC$ . Let $ D$ be the midpoint of $ BC$ and $ P$ be a point on $ AD$. Suppose $ E$ is the foot of the perpendicular from $ P$ on $ AC$. If $ \frac{AP}{PD}\equal{}\frac{BP}{PE}\equal{} \lambda$ , $ \frac{BD}{AD}\equal{} m$ and $ z \equal{} m^{2}(1 \plus{} \lambda)$, prove that $ z^{2}\minus{}(\lambda^{3} \minus{} \lambda^{3}\minus{} 2)z \plus{} 1 \equal{} 0$.
Hence show that $ \lambda\geq2$ and $ \lambda \equal{} 2$ if and only if $ ABC$ is equilateral.

[quote="makar"]Problem 5:
Let $ ABC$ be a triangle in which $ AB \equal{} AC$ . Let $ D$ be the midpoint of $ BC$ and $ P$ be a point on $ AD$. Suppose $ E$ is the foot of the perpendicular from $ P$ on $ AC$. If $ \frac {AP}{PD} \equal{} \frac {BP}{PE} \equal{} \lambda$ , $ \frac {BD}{AD} \equal{} m$ and $ z \equal{} m^{2}(1 \plus{} \lambda)$, prove that $ z^{2} \minus{} (\lambda^{3} \minus{} \lambda^{3} \minus{} 2)z \plus{} 1 \equal{} 0$.
Hence show that $ \lambda\geq2$ and $ \lambda \equal{} 2$ if and only if $ ABC$ is equilateral.[/quote]

$ z^{2}\minus{}(\lambda^{3}\minus{}\lambda^{3}\minus{}2)z\plus{}1 \equal{} 0$
[size=150][color=blue][b]the form will be
z^2 +2z+1 and it will never equale zero  because  z > 0
or i missed something :stink: [/b][/color][/size]

[quote="almonkez"][quote="makar"]Problem 5:
Let $ ABC$ be a triangle in which $ AB \equal{} AC$ . Let $ D$ be the midpoint of $ BC$ and $ P$ be a point on $ AD$. Suppose $ E$ is the foot of the perpendicular from $ P$ on $ AC$. If $ \frac {AP}{PD} \equal{} \frac {BP}{PE} \equal{} \lambda$ , $ \frac {BD}{AD} \equal{} m$ and $ z \equal{} m^{2}(1 \plus{} \lambda)$, prove that $ z^{2} \minus{} (\lambda^{3} \minus{} \lambda^{3} \minus{} 2)z \plus{} 1 \equal{} 0$.
Hence show that $ \lambda\geq2$ and $ \lambda \equal{} 2$ if and only if $ ABC$ is equilateral.[/quote]

$ z^{2} \minus{} (\lambda^{3} \minus{} \lambda^{3} \minus{} 2)z \plus{} 1 \equal{} 0$
[size=150][color=blue][b]the form will be
z^2 +2z+1 and it will never equale zero  because  z > 0
or i missed something :stink: [/b][/color][/size][/quote]

is there any body here!!!
what is the wrong with this?? :huh:

[quote="apratimdefermat"]The problem has been stated wrong.
Besides this problem has already been discussed extensively.Please use the "search" function before posting. :wink:

Follow this link : [url]http://www.mathlinks.ro/viewtopic.php?t=132002[/url][/quote]
thanks :apratimdefermat
but the first topic was sent :
[quote]Posted: Sun Feb 04, 2007 11:08 pm  [/quote]

the scond topic was sent :
[quote]Posted: Tue Sep 01, 2009 9:53 am[/quote]
i found it while searching unsolvid problems
so that i asked about it didnt see  the first topic
thanks again for help
mr mohamed : almonkez


My club has 15 members.  In how many ways can we choose a president, vice-president, secretary, and treasurer, if either the secretary or the treasurer must be elected vice-president and no other member can hold more than one office?

Count the number of ways to elect 3 different people to president, secretary and treasurer.  For each one of those ways, you have two choices of who becomes vice president, so you'd double your first count.

15 choices for pres.
14 choices for secretary
13 choices for treasurer

2*15*14*13=5460


Find all real numbers $ x$ such that there exist $ n\in Z_{+}$ such that :
                  $ 15([x])^{2}+10[x]x+18[x]+2x+6-4^{n}=0$
(where $ [x]$ denote the largest integer not more than $ x$ )


Find the necessary condition for $ \frac{p}{q} \equal{} a \in \mathbb{Q}$ to have a terminating decimal representation.
For instance, $ \frac{1}{2}$ terminates but $ \frac{1}{3}$ does not. Why?

[hide="Hint"]
The denominator has a factor of the form $ 2^x5^y$
[/hide]

Let (p,q)=1

$ 10^k(p/q)\equal{}r$ so $ 10^kp\equal{}qr$ Now, (p,q)=1, so $ q|10^k$. Let q=$ 2^m5^n$. Then the problem is easy.

This isn't really either Olympiad-level or algebra.  More interesting is the question of computing the periods of the periodic representations, something Gauss spent a lot of time doing in prime bases as a child (that is, he liked to compute the period of the terminating representation of $ \frac {1}{p}$ in base $ q$ for different primes $ p$ and $ q$).


Do you know how to solve this iby using differential?

Please show me.

[quote]
Find $ f$:R$ \rightarrow$R
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$[/quote]

[quote="N.T.TUAN"]We have $ f'_x(x \plus{} y) \equal{} f'_x(x)$, therefore $ f_x'(x)\equiv C$, and we're done![/quote]

Thanks!

It's elegant,sorry.


Let $P$ a point inside a equilateral triangle such that distances from $P$ to the three vertex are $3,4,5$. Find the area of the triangle.

A direct method would be:

1. Use the know formula $3(a^4+b^4+c^4+s^4)=(a^2+b^2+c^2+s^2)^2$ to find the side $s$ of the equilateral triangle.

2. The needed area would be $A=\frac{\sqrt3}{4}\cdot s^2$.

However, I am 100% sure that there is a more beautiful way to do it :)

[quote="Ezbakhe Yassin"]Use the know formula $3(a^4+b^4+c^4+s^4)=(a^2+b^2+c^2+s^2)^2$ [/quote]

Sorry, but I don't know this formula. Can you give a proof or a link?

a really similar problem was posted, just with 6,8,10 and finding a side...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67031[/url]

Thanks but this gives me no proof for the formula  :(

Construct circles with radii 3,4, and 5 on each vertex, respectively. Each pair of circles will have 2 intersection points, one of them being P. Then for each pair of circles (same thing as each pair of distances from P,) we have a quadrilateral. I have to run, but I will add to this later.

haha ok I've seen this question quite a few times. It's in this book called Mathematical Olympiad Challenges, by Titu Andreescu and Razvan Gelca.

Try rotating the triangle by 60 degrees.

Yup.  I learned that method from that book too...great book. :)


Hello AoPS users. I am just making a post for people to post you-tube videos that they really like and enjoy to watch.

Here is a video that I think is pretty cool;

Frozen Grand Central ( a bunch of people stand still in public)

It won't let me watch it, but I think that I have seen that video before. I just watched it about two weeks ago. Big coincidence. Is this the Obama vs. McCain video, where the polling machine doesn't work?

it's where the polling machine is rigged and kills homer.

I didn't think it ended up killing him, it just spit him out.

Anymore videos?

[youtube]adc3MSS5Ydc[/youtube]

And

[youtube]rw2nkoGLhrE[/youtube]

Parody of Shut up and Drive, made by two of my friends

[youtube]7ewV6YMJbJk[/youtube]

Better than alcohol, drugs, caffeine, or money:

[youtube]fP_BNP7NWNs[/youtube]

This song is AWESOME.
[youtube]LDiDK_yBCw0[/youtube]

Geese!!! This got posted on alot in the four days I was offline. I saw this really cool video with Jackie Chan running down the side of a building. It was really cool.

[quote="The past tense of write is"][youtube]P9dpTTpjymE[/youtube][youtube]V1fJDIXEq9w[/youtube][/quote]

Was wondering if a video could go in a quote.

[size=200] :D [/size]

Hmm, I never thought about that. Has anybody seen the video I said at the top?

[quote="eggylv999"][youtube]itkB3dA6eRY[/youtube]
It doesn't work, so [url]http://www.youtube.com/watch?v=itkB3dA6eRY[/url][/quote]

Is that you eggy?

yes... my teammate recorded it at a math practice after we finished.

:rotfl:  :rotfl:  :rotfl:  :rotfl:
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahahahahaha
hahaha
hahahahahaha
hahahahahahahahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha
hahaha

ive posted this all over aops but:

[youtube]k4aHL12vtEM[/youtube]

@Splatyango - At the beginning I was watching and I was wondering what was going on with alll the polar bear pictures.  :)

[color=blue][b]The Mathlete Rap![/b][/color]

[youtube]cR-41A3spQs[/youtube]

You [b]must[/b] watch this the whole way through the first time.

[youtube]LXA6G16_IBY[/youtube]

I  watched about 6 minutes of it, but I didn't finish. I'll finish it later.  :)

It was really random.  :D

THIS IS SCARY!
[youtube]CdTEO4PslSg[/youtube]

dude someone needs to post more vids

[youtube]F9irG_uXsII[/youtube]

it is awesome
[youtube]d5XrmSBBOtk[/youtube]

I know a good video:
search: stick figures on crack 3
  :rotfl:  :D  I cant stop laughing XD

[youtube]xwfsXBKQx4Q[/youtube]

serously..
this might not make sense for some people..
but amsp 08-ers..
this was the higlight of that night.

That was the best two-and-a-half minute experience of my lifetime. XD

Awww that's in the past! DX

[youtube]BNjXgS0_CWE[/youtube]

If you don't wanna watch the entire thing, skip to the end.

Hm...watch "how abe lincoln really died" by the whitest kids you know.  I feel like I shouldn't post the video here, though, so go watch it on youtube.

[youtube]ACqM_bFzmAA[/youtube]

EPIC fail!

Rotfl
Rotfl
ROTFL
ROTFL
THAT WAS SOOOOOOOOOOOOOOOOOOOO FUNNY!!!

go to youtube and search up "counterstrike for kids"


geometry
geometric transformation
reflection
Pythagorean Theorem

The rules for a race require that all runners start at A, touch any part of a 1200-meter wall, and stop at B. What is the number of meters in the minimum distance a participant must run? Express your answer to the nearst meter.

Yeah, I thought I remembered this one, but I couldn't remember what the diagram looked like.

[hide="Answer"]The 1200 m wall needs to be split up proportionally to the other two walls. So, the point the runner should touch on the 1200 m wall is $\frac{3}{8}$ of the way along the wall, or 450 m. Using Pythagorean Theorem, we have $\sqrt{300^2+450^2}+\sqrt{500^2+750^2}=150\sqrt{13}+250\sqrt{13}=400\sqrt{13}\approx \boxed{1442}$ m.[/hide]

I hope I did that right...

that's right (i think). Reflect the top of A across the 1200m line and apply pythagoras. I just thought of that a couple of minutes ago..

[quote="anirudh"]what would be the fastest way for this?[/quote]

I don't know. But a quick way is the one I mentioned above  :roll:


ratio
trigonometry
geometry
cyclic quadrilateral
similar triangles
trig identities
Law of Cosines

Hint: use law of cosines to find one diagonal.  Now use Ptolemy's theorem to find the other.

[quote="WarpedKlown1335"]Couldn't you use similar triangles for this too?[/quote]
I don't think we can.

Hmm...yeah you're right, never mind.

I thought it'd be somewhat like the proof for Ptolemy's, but that's until I tried it.

[quote="warut_suk"]A cyclic quadrilateral has sides of length $2, 3, 4$ and $6$, in this order. Determine the two ratios in which the diagonals divide each other.[/quote]

I just did a problem like this for IS.

Are you sure the numbers are correct? I think something's wrong :huh:

How'd you go about solving that problem?

Are you sure you can't use similar triangles?  You can find all four partial lengths of diagonals in terms of one.... then use Ptolemy if you want to know the actual lengths and not just ratios.

I don't think these side lengths are right because by Ptolemy's:
$2(4) + 3(6) = 26$
(If I have interpreted the side lengths correctly)
$26$ is the product of the diagonals, so the lengths of the diagonals are then either 13 and 2 or 26 and 1, both of which will not work because of the triangle inequality.

By Law of Cosines, we have

$2^2+3^2-2(2)(3)\cos {\theta} = 4^2+6^2-2(4)(6)\cos {(180^\circ -\theta)}$

$\cos (180^\circ -\theta) = -\cos{\theta}$

$13-12\cos {\theta} = 52+48\cos {\theta}$
$0= 39+60\cos {\theta}$
$-13/20= \cos {\theta}$
$d_1^2=13+12\left( \frac{13}{20} \right)\\ d_1=\sqrt{104/5}$

Ptolemy yields $(2)(4)+(6)(3)=(d_1)(d_2) \implies d_2=\frac{26}{\sqrt{104/5}}$

The ratio is $\frac{d_1}{\frac{26}{d_1}}=\frac{d_1^2}{26}=\frac{104}{130}=\frac{52}{65}$ and the other ratio ($d_2/d_1$) is $\frac{65}{52}$.

Even if the answer is wrong, you can see how to do the problem. ;) I might have a made a (few) arithmetic mistake(s). :D

i believe that he is looking for the ratio that the diagonals divide each other, not the ratio of the diagonals

and for the ratios, i got 1:4 and 1:1 (draw the diagonals and think of the similar triangles that are formed by the fact that if 2 inscribed angles subtend the same arc, they are congruent) there is no need to find the lengths of the diagonals, nor use the law of cosines or ptlomy's theorem

Right, I misinterpreted the problem. In fact, I don't even see how you [i]could[/i] use Law of Cosines or Ptolemy's for the real problem. :?


[quote="t0rajir0u"][hide="Quick"] $(a-1)(b-1) > 0 \Leftrightarrow$
$ab+1 > a+b$ [/hide][/quote]

Nice.

there needs to be more...

a>b does not imply that |a|>|b|

You caught me.   :P  I was waiting for someone to say that.

[hide="As for the rest..."]$(a+1)(b+1) > 0 \Leftrightarrow$
$ab+1 >-a-b$ [/hide]

Well done, t0rajir0u! The following is my stupid way...  :blush:
\[\left|\frac{a+b}{1+ab}\right|<1\Leftrightarrow\]
\[\left(\frac{a+b}{1+ab}\right)^{2}<1\Leftrightarrow\]
\[(a+b)^{2}<(1+ab)^{2}\Leftrightarrow\]
\[a^{2}+b^{2}<1+a^{2}b^{2}\Leftrightarrow\]
\[(a^{2}-1)(b^{2}-1)>0\]
As $|a|<1, |b|<1$, we get $a^{2}<1, b^{2}<1$. Hence
\[(a^{2}-1)(b^{2}-1)>0\] is true. Then the inequaity \[\left|\frac{a+b}{1+ab}\right|<1\] is also true.

@2above if a squared plus b squared equals one then a and b must still have a positive value less than 1 so its just the same proof

[quote=dajeff][quote=harry1234]
This is not true. If $a=0, b=1,$ then we get $1<1.$
[/quote]

Note that $|b|=1$, which is not less than $1$.[/quote]

sqing posted a different problem

[quote=ifai]Let $a, b$ be real numbers that $|a|<1$ and $|b|<1$. Prove that \[\left|\frac{a+b}{1+ab}\right|<1.\][/quote]
[b][size=150](Moscow Maths Olympiad 1952 , Grade 9, P10)[/size][/b]

Let $a, b$ be real numbers that $|a|\le 1$ and $|b|\le 1$. Prove that \[\left|\frac{a+b}{1+ab}\right|\le 1.\]

[quote=sqing]Let $a, b$ be real numbers such that $a^2+b^2=1.$  Prove that $$\left|\frac{a+b}{1+ab}\right|\leq 1.$$[/quote]

[quote=duck_master][hide= Modified version of @ifai's proof for the second problem]
\[\left|\frac{a+b}{1+ab}\right|\leq 1\Leftrightarrow\]
\[\left(\frac{a+b}{1+ab}\right)^{2}\leq 1\Leftrightarrow\]
\[(a+b)^{2}\leq (1+ab)^{2}\Leftrightarrow\]
\[a^{2}+b^{2}\leq 1+a^{2}b^{2}\Leftrightarrow\]
\[(a^{2}-1)(b^{2}-1)\leq 0\]
As $a^2+b^2=1$, we get $a^{2}\leq1, b^{2}\leq1$. Hence
\[(a^{2}-1)(b^{2}-1)\leq 0\] is true. Then the inequality \[\left|\frac{a+b}{1+ab}\right|\leq 1\] is also true.[/hide]

Anyways there's already an inequality marathon going on in this forum so I don't want this to become another one.[/quote]
Thanks.

[hide=Solution]
let p = a+b and q = ab. Then we have $p^2-2q = 1$ and we want to prove that $\frac{p}{1+q} \ge \frac{2\sqrt{2}}{3}$ or $3p \ge 2\sqrt{2} + 2q\sqrt{2}$.
The given equality is also equivalent to $(p^2-1)\sqrt{2} = 2q\sqrt{2}$. Substituting it in gives $3p \ge 2\sqrt{2} +p^2\sqrt{2} -\sqrt{2}$.

Factoring this gives $(p-\sqrt{2})(p\sqrt{2}-1) \le 0$. Applying AM-GM on $a^2+b^2=1$ gives $1 \ge 2ab$. Since a and b are positive reals, $p \le \sqrt{2}$ since $p^2 = a^2+ 2ab + b^2 \le 2$. Also, we have $p^2 > a^2+b^2 = 1$, so $p > 1$. Thus $1 < p \le \sqrt{2}$, so $(p-\sqrt{2})(p\sqrt{2}-1) \le 0$ is satisfied. Equality occurs at $a = b = \frac{\sqrt{2}}{2}$[/hide]

[quote=sqing]Let $a, b$ be positive real numbers such that $a^2+b^2=1. $ Prove that$$\frac{a+b}{1+ab}\geq \frac{2\sqrt{2}}{3}.$$
[size=50](MathematicalReflections  3(2019))
[/size][/quote]

Next time, create your own threads if you want to add a new problem. You also did this three months ago, which resulted in a twelve year bump.

[quote=NikoIsLife][quote=ifai]Let $a, b$ be real numbers that $|a|<1$ and $|b|<1$. Prove that \[\left|\frac{a+b}{1+ab}\right|<1.\][/quote]

[hide=Solution]Since $|a|<1$ and $|b|<1$, we let $a=\tanh x$ and $b=\tanh y$ for some real numbers $x$ and $y$.

Then, by the Hyperbolic Tangent Addition Formula,
$$\frac{a+b}{1+ab}=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}=\tanh(x+y)$$
And, since $|\tanh(x+y)|<1$, this implies that $\left|\frac{a+b}{1+ab}\right|<1$.[/quote]

So extra...

[hide=*][quote=sqing]
Let $a, b$ be real numbers that $|a|\le 1$ and $|b|\le 1$. Prove that \[\left|\frac{a+b}{1+ab}\right|\le 1.\][/quote]
[quote=sqing]Let $a, b$ be positive real numbers such that $a^2+b^2=1. $ Prove that$$\frac{a+b}{1+ab}\geq \frac{2\sqrt{2}}{3}.$$
[/quote][/hide]
Let $a, b\ge 0$ and  $a^2+b^2=1. $ Prove that$$1\geq \frac{a+b}{1+ab}\geq \frac{2\sqrt{2}}{3}.$$

[quote=greenturtle3141][quote=nikbot05]Next time, create your own threads if you want to add a new problem. You also did this three months ago, which resulted in a twelve year bump.[/quote]

Sqing posted a new, very relevant problem, so it's completely on topic.

Also, sqing is a very prominent member of the community and has contributed much to the study of inequalities.  Please show them some respect.[/quote]
Thank you very much.


Find the smallest k such that the following ineq. takes place for all real a,b,c lie on [0,1]
ab + bc + ca + 1  >= a k + b k + c k

Really very easy...
 :)
For $a=1,b=0$ then:

$c \ge c^k \leftrightarrow c^{k-1}\le 1 \rightarrow k\ge 1$
(Because $c$ lied in $[0,1]$)
With $k=1$,we have:
$(1-a)(1-b)(1-c)\ge 0 \leftrightarrow 1+ab+bc+ca \ge a+b+c+abc$

We are done!


There is a right triangle. If a and b are the catetes, c is the hypothenuses and h is the perpendicular from the hypothenuses to the 90 degree angle . Prove that the equality $ \frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\equal{}\frac{1}{h^{2}}$ is true.  :)

$ \frac{1}{a^{2}}\plus{}\frac{1}{b^{2}} \equal{} \frac{a^{2}\plus{}b^{2}}{a^{2}b^{2}} \equal{} \frac{c^{2}}{a^{2}b^{2}} \equal{}(\frac{1}{\frac{ab}{c}})^{2}$

However, notice that the two triangles h divides the original triangle are similar, and thus $ \frac{a}{c} \equal{} \frac{h}{b}$, or $ \frac{ab}{c} \equal{} h$

Thus $ \frac{1}{a^{2}}\plus{}\frac{1}{b^{2}} \equal{} \frac{1}{h^{2}}$

This problem can done by some one. It's easy problem.

actually its a known formula:
another prove is that $ (ABC)\equal{}\frac{ab}{2}\equal{}\frac{hc}{2} < \equal{} >\frac{1}{h^2}\equal{}\frac{c^2}{a^2b^2}$
so we get $ \frac{1}{a^2} \plus{} \frac{1}{b^2}\equal{}\frac{c^2}{a^2b^2} < \equal{} > a^2\plus{}b^2\equal{}c^2$,which is true.


set theory
real analysis
real analysis theorems

I had a bit of a random idea and I want to hear some criticisms of it.

Would it be valid to say that the decimals in a real number must be countable?

So for example, pi would be:

{1, 4, 1, 5, 9, 2, 6, 5...}

Which is of course countable because you can name any position a digit appears.

However this would mean that:

0.000...1 = {0, 0, 0, ..., 1}

In such a set, if it were countable, 1 would never be in it.  So 0.000...1 = 0.0000... = 0

Which makes sense with the reals.

Actually, the thing you're poking at here is not so much the cardinality [$\{0, 0, 0, \ldots, 1\}$ is a countable multiset] as the ordinal structure.  When we represent most real numbers as decimals, we have a countable multi-set with a particular ordering on it [that particular ordering being the straight-forward ordering $\omega$ of the integers].  You haven't changed the cardinality of the set, but you have given a different ordinal structure, and in standard analysis we don't consider any such ordering to correspond to a real number.

Thanks for the reply, that helped quite a bit.

I'm trying to find places where I can study more on ordering and ordinal structure.  I tried google and wikipedia, but came up empty.  Any suggestions?

Well, I would think that any book on set theory should cover the ordinals and well-orderings.  My set theory class didn't use a book, so I don't have a particular suggestion, but those are the topics that are relevant.  You might also try [url=http://mathworld.wolfram.com]Mathworld[/url] to see what it has -- the page on ordinals seems pretty good, but you probably have to already know what it's talking about for it to be of much use.

Best of luck!


Find all positive integers $ x,y$ satisfying $ y^{x^2}\equal{}x^{y\plus{}2}$

[quote="SUPERMAN2"]Find all positive integers $ x,y$ satisfying $ y^{x^2} \equal{} x^{y \plus{} 2}$[/quote]

$ (x,y)\equal{}(1,1)$ is a solution.

For $ x,y \neq 1$, let $ x\equal{}z^a, y\equal{}z^b$, where $ a,b \ge 1$, $ z \ge 2$ are integers.

Then the equation becomes $ z^{bz^{2a}}\equal{}z^{a(z^b\plus{}2)}$
$ \Leftrightarrow bz^{2a}\equal{}a(z^b\plus{}2)$

[b]Case 1:[/b] $ z$ odd, then $ \gcd(z,z^b\plus{}2)\equal{}\gcd(z,2)\equal{}1$

$ z^{2a}|a<z^{2a}$, impossible.

[b]Case 2:[/b] $ z$ even, then let $ z\equal{}2k$ we get

$ b(2^{2a})(k^{2a})\equal{}2a(2^{b\minus{}1}k^b\plus{}1)$

[b](a)[/b] $ b \ge 2$, then $ \gcd(2^{2a},2^{b\minus{}1}k^b\plus{}1)\equal{}1 \Rightarrow 2^{2a}|2a<2^{2a}$ , contradiction.
[b](b)[/b] $ b\equal{}1$, then the equation becomes $ 2^{2a}k^{2a}\equal{}2a(k\plus{}1)$
$ k^{2a}, k\plus{}1$ coprime $ \Rightarrow k^{2a}|2a$, impossible if $ k>1$
So $ k\equal{}1 \Rightarrow 2^{2a}\equal{}4a \Rightarrow a\equal{}1$
$ \Rightarrow (x,y)\equal{}(2,2)$ are solutions for this case.

So $ (x,y)\equal{}(1,1),(2,2)$ are the only solutions.


modular arithmetic
counting
distinguishability

i)Prove that $ a!b!\ |\ (a\plus{}b)!\ ,\ (\forall) (a,b)\in \mathbb{N}\times \mathbb{N}$

ii)Show that $ p$ divides $ \frac{(p\minus{}1)!}{k!(p\minus{}k\minus{}1)!}\plus{}(\minus{}1)^{k\plus{}1}\ ,\ (\forall)k\in \{0,1,...,p\minus{}1\}$ , $ p$ - prime , $ p>2$ .

ii) Rewrite the term as $ a_k\equal{}\binom{p\minus{}1}{k}\plus{}(\minus{}1)^{k\plus{}1}$.

Now $ (x\plus{}1)^{p\minus{}1}\minus{}1\plus{}x\minus{}x^2\plus{}...\plus{}x^{p\minus{}1}(\minus{}1)^{p}\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_{p\minus{}1}x^{p\minus{}1}$.

$ (\minus{}1)(1\minus{}x\plus{}x^2\minus{}...\plus{}(\minus{}x)^{p\minus{}1})\equal{}(\minus{}1)\frac{1\minus{}(\minus{}x)^p}{1\minus{}(\minus{}x)}\equal{}\minus{}\frac{1\plus{}x^p}{1\plus{}x}$ since $ p$ is odd.

Now $ (x\plus{}1)^{p\minus{}1}\minus{}\frac{1\plus{}x^p}{1\plus{}x}\equal{}\frac{ (x\plus{}1)^p\minus{}x^p\minus{}1}{1\plus{}x}$

It is easy to see that  $ (x\plus{}1)^p\minus{}x^p\minus{}1$ has all coefficients divisible by $ p$ since $ p|\binom{p}{k}$ $ k\equal{}1,2,...,p\minus{}1$ and it checks for $ k\equal{}0,p$ as well. Now $ \frac{1}{x\plus{}1}\equal{}(1\minus{}x\plus{}x^2\minus{}x^3...$ which does not change the fact that the coefficients are already zero in mod $ p$ so each $ a_k\equiv 0\mod p$ so we're done.

[hide="2 alternative"]
We have to show $ \binom{p\minus{}1}{k} \equal{} (\minus{}1)^k \pmod{p}$

It is clearly true for k = 0, otherwise:

$ \binom{p\minus{}1}{k} \equal{} (p\minus{}1)(p\minus{}2)\cdots(p\minus{}k) \cdot 1^{\minus{}1} 2^{\minus{}1} \cdots k^{\minus{}1}$
$ \equal{} (0\minus{}1)(0\minus{}2)\cdots(0\minus{}k)\cdot 1^{\minus{}1} 2^{\minus{}1} \cdots k^{\minus{}1} \equal{} (\minus{}1)^k \pmod{p}$

[/hide]

[hide="1 simple"]Combinatorially, $ \frac{(a\plus{}b)!}{a!b!}$ is the number of ways to arrange $ a$ indistinguishable objects of one type and $ b$ indistinguishable objects of another type in a row, so it must be an integer. That is, $ a!b!|(a\plus{}b)!$.[/hide]


Seeding will be based on highest rating.  Post here to sign up.  The tourney will be double-elimination.  I will start when I get 32 people.  [i]Don't sign up unless you are sure you will be active.[/i]

Sign-ups:

AIME15 (2030)
Mewto55555 (2015)
FantasyLover (1917)
ernie (1891)
nt811581147 (1883)
Jongao (1806)
isabella2296 (1775)
mousy (1754)
jjx1 (1725)
nikeballa96 (1701)
ZachAttack (1601)
r15s11z55y89w21 (1573)
ReseesPieces (1488)
CubeX (1467)
jonathanchou711 (1408)
Erasable (1321)
Eugene Kevin (1252)
007math (????)
firestar58 (????)
SonyWii (????)

Also, please sign up with [b][size=200]highest rating, as written above in smaller font.[/size][/b]  This would be greatly appreciated.

12 spots remain!

gah omg ReseesPieces(1488)....i know pathetic right.........if it goes up(cough cough no cough cough), ill tell u

Erasable! :lol: 1203  :( keeps dropping down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down and down

I will join the countdown cup. I hope this may be a little bit ez. My rating cannot afford many blowbacks. Please inform me on FTW if you recieve my message. :maybe:

Eugene Kevin (1252)    :censored:  :spam: yay! ill sign up.

Hm, I suppose I will [b]join[/b] another tourney destined to fail :)

Find my highest on the Leaderboard.

Cubex. in these cds you can cd people with guests. 32 people seem like a lot of people.....

My highest rating is 1725

Me join....    :blush:   I stink though...  Rating: 1408   Going to GO DOWN!!!!

[quote="CubeX"]When will we begin next round of competition?

P.S. I answered your PM, ERNIE![/quote]

When you get an opponent.  Refer to the bracket.

Oh yes, and nikeballa96 and scai1 are both active.

[b]ZachAttack defeats r15s11z55y89w21 3-2.[/b]

Oh, good grief! I'm playing SonyWii next. I'm going to be like the first eliminated from this tournament.

[quote="CubeX"]I do not think andrewnycpops is regularly active. I can never find him! I just saw him once a few month ago, I think?[/quote]

He's active; I've seen him.

I can never find nikeballa96? I think he's not active on FTW anymore... :(

[quote="scai1"]I think he's not active on FTW anymore... :([/quote]

He's not as active as he used to be. ;)

I know nike in real life and he kinda gets paranoid in this time of year with all the contests lol

he goes on every now and then lol

andrewnycpops is not acitve these days. Once I saw him, but he wouldn't cd me......

Try to get him to be active on a specific date or bye him..

WOW.....
dang.....
i wish i was playing....
 :ninja:  :ninja:  :ninja:

yeah, i've been studying more for nats and aime, so, yeah...
pm me.

Cubex u never beat me 3-2 plus u never contacted me.

ya...
plz don't revive old topics
this is obviously dead


integration
AMC
AIME
MATHCOUNTS
email
geometry
AIME I

All right, I'm going to try and set up a game, and hope it works out well.

Okay so here's the story:

In this game, you are trying to be "The Top Nation". The level of "Topness" will vay based on your economy, # of citizens, etc. etc.

After I get an initial start-up list of people (12) I will use random.org to decide the size/resources/population of your nation.

After your nation has started, you may do/automatically have the following:
1) Advance your nation in terms of millitary.
2) Advance your nation in terms of commerce.
3) Advance your nation in terms of science.

The outcomes and benefits of 1-3 are known only by me, and the level of increase will base on your nation's resources, and random.org.

Automatic:
1) Military - Land size. Based on the size of your nation, your power here will increase.
Your chances of winning wars is higher, and your population increases.

2) Commerce - Wealth. Depending on the resources and number of allies, your power here will increase. You will gain more resources, and your population increase.

3) Science - Knowledge. Depending on the population and knowledge of your nation, your power here will increase, and you will have more advanced technology. (Which the outcome is only known by me)

What you can do:
1) Military - You can wage wars against other nations to gain territory. The outcome of your war will be based on the population (and random.org).

2) Commerce - You can barter with other nations on resources you have. Also, gaining peace with other nations also helps.

3) Science - You can use your wealth to build educational buildings. The more you have, the more knowledge you gain.

As the game continues, you may encounter freak happenings, which may be beneficial to you, or non-beneficial.

Once the game has started: (You can join after the initial start up)

You may choose to start your own nation, which then you will have to ask me to plot your land, or you may join one nation.

OR

You can join a civilization, of which then the leader of your nation (one of the orginal 12) will set you to their guidelines.

This thread is only for global peace, war and economy.
If you wish to gain power in military, commerce or science, pm me.

Map will be up soon.

Map shown.

Each color has been assigned to a number 1-12. You can pick any number that is left on the list:

1,2,3,4,5,6,7,8,9,10,11,12.

Any space that is white may be claimed. You can only claim it if if is within one space of your territory.

I will pm you your color as you choose a color.

Here begins Year 1.

[quote="Meta_Knight"]i pick the best one ^^[/quote]

Okay seriously dude... I'll give you... 6, since thats your post number.

[quote="1=2"]I pick $ \int_{1}^{2\sqrt [3]{2}} x^2 dx$[/quote]

Comeon. I phail at whatever that is.

List remaining (1 of these is taken by 1=2)

2,3,4,5,7,9,10,11

Zank ye ffao.

6 spots remain!!!
2,3,4,7,10,11

EDIT: New commerce rules

The more cities you have (which will be explained as you gain progress) the higher your commerce level.

EDIT2: More rules coming as soon as all 12 register.

here's the map so far:

EDIT: new map
EDIT 2: BAH new map

[quote="Dojo"]Comeon. I phail at whatever that is.[/quote]

Calculus.

And my first move is to attack meta knight... i think...

Offer peace treaty with 007math, hedgehog12, ffao, and fishy.

Okay thanks.

Sorry but everything from post #175 to now is INVALID.

From now on, per year, you can only advance your account twice a year.

[b]claim land, attack mathgeek88 blindly with [u]all[/u] men, after buying all possible troops[/b]

If possible, I'll help mod too if needed.

ANNOUNCEMENT!
my nation has undergone an election and a new leader has been chosen.
kdange is now the leader of my nation.
all my resources, land, and people now belong to kdange.
just informing the rest of the world.

Eh, can you post the spreadsheet one more time before giving away your mod seat?

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO THE WORLD IS ENDING

[b]trade .01 uranium for nuke[/b]

[b]nuke center of map[/b]

EVERYBODY DIES! :D

Eh, I'll create a more "logical" and less "chaotic" sequel...

Well, make sure it doesn't end up dead like all the other ones related to this one

I will take any land that is still open, if any, or be hired as a political advisor by anyone!

(Or any kind of advisor, general, or helper.)

Of course, this has been mod-abandoned :roll:

My army of cockroaches takes over the world!!!!!!!!!!!!!!!!!


modular arithmetic
algebra
polynomial
inequalities
number theory unsolved
number theory

Let $m,n\in N$ and $p$ is prime number and $n<p<\frac{2m}{2m-1}n$ prove that
$p\mid \sum^n_{i=0}\left(\begin{array}{c}n\\i\end {array}\right)^{2m}$ :P

wow looks hard....the only case i can handle is $m=1$ because the sum becomes $\binom{2n}{n}$ wich is clearly a multiple of $p$. I think a combinatorial interpretation of the sum to get a factor of it or things like that would be  the better idea

The following solution was given by [b]RIP[/b].

Let $r=p-n$. It follows from
\[(-1)^{n-k}(n-k)!\equiv(k+r)(k+r+1)\ldots(p-1)\equiv-\frac{1}{(k+r-1)!}\pmod p,\]
that
\[\sum_{k=0}^{n}\binom nk^{2m}\equiv(n!)^{2m}\sum_{k=0}^{n}\left(\frac{(k+r-1)!}{k!}\right)^{2m}\equiv(n!)^{2m}\sum_{k=0}^{p-1}Q(k)\pmod p,\]
where $Q(k)=\left(\frac{(k+r-1)!}{k!}\right)^{2m}=\left((k+r-1)(k+r-2)\dots(k+1)\right)^{2m}$ is a polynomial of $k.$
The inequality $p<\frac{2m}{2m-1}n=\frac{2m}{2m-1}(p-r)$ implies that $\deg(P(k)) = 2m(r-1) < 2mr < p$ and hence $\deg(P(k)) < p-1.$
It is easy to see (or to use [url=http://mathworld.wolfram.com/BernoulliNumber.html]formula (41)[/url]) that $\sum_{k=0}^{p-1}k^{d}\equiv 0\pmod p$ for every $d=0, 1, \dots, p-2,$ implying that $\sum_{k=0}^{p-1}Q(k)\equiv 0\pmod p.$

Just for the sake of completeness, this was also proven by me at http://problem-solving.be/pen/viewtopic.php?t=173 (generalizing PEN problem E 16 - note that my inequality condition is [i]slightly[/i] weaker), but it seems that my solution is identical to RIP's one.

  Darij


if $ f(z) \equal{} z^4 \minus{} 2z^3 \plus{} z^2 \minus{} 12z \plus{} 20$ and C is the circle $ |z| \equal{} 5$ evaluate $ \oint _c\frac {z f'(z)}{f(z)} dz$

Note that $ f(z) \equal{} (z\minus{}2)^2(z\plus{}1\minus{}2i)(z\plus{}1\plus{}2i)$, and the winding number of the contour about each zero is one. Since $ f(z) \equal{} z$ is entire, we may apply a generalization of the Argument Theorem to obtain

\[ \int_C \frac{z f'(z)}{f(z)} \ dz \equal{} 2\pi i \left(2(1) \plus{} 2(1) \plus{} (\minus{}1\plus{}2i)(1) \plus{} (\minus{}1\minus{}2i)(1)\right) \equal{} 4\pi i.\]

can you explain me the generalization of argument theorem and winding number.

Take a look at this article:

http://en.wikipedia.org/wiki/Argument_principle

The generalization is given under the "Consequences" section.


trigonometry
analytic geometry
graphing lines
slope
inequalities

Q1  A line through A (-5,-4) meets the line X+3Y+2=0, 2X+Y+4=0 and X-Y-5=0 at B, C, D respectively. If (15/AB) ^2+ (10/AC) ^2+= (6/AD) ^2 then equation of line is ---.

Q2 how many numbers are between 1 and 100000 such that sum of digits are 12.

Q3 if a ,b c, d are positive  such that abcd =1
then what us the minimum value of (1+a) (1+b) (1+c) (1+d).

1.
Let the point of intersection be $ B(ABcos(\theta) \minus{} 5,ABsin(\theta) \minus{} 4)$
$ ,C(ACcos(\theta) \minus{} 5,ACsin(\theta) \minus{} 4)$
$ ,D(ADos(\theta) \minus{} 5,ADsin(\theta) \minus{} 4)$ respectively .
satisfying these points  in respective eq.
and get d value of AB,AC,AD in terms of $ \sin(\theta),\cos(theta))$
put the value of AB,AC,AD in the given cond.
get the value of $ tan(\theta)$.
u have a point (-5,-4) and slope by which u can write d eq. of line .
Rest in next posts.
thank u

Q3) Applying AM - GM inequality to (1+a) , (1+b) , (1+c) and (1+d) separately we get the min value to be 2 X 2 X 2 X 2 = 16

No 2
Hint
[hide]Now let us 1st begin by trying to find 2 digited numbers whicch meet the given criteria and let us depict the numbers as 10a + b
where a + b = 12
now apply IEP  to find the number of values of (a,b) such that both r less than 10
similarly proceed for a + b + c = 12,to get 3 digited numbers whose digits add up to 12
and go on till
a + b + c + d + e + f = 12
Sry for not giving detailed solution.......I really cant type in latex...despite how hard I try :oops: [/hide]

[quote="Pocha"]IEP = inclusion exclusion principle
I din work it out entirely..I'll PM u the answer when I get it.[/quote]
thanks


[quote="karthik_k"]Two and a half work crews take two and a half days to clear two and a half hectares of forest.How many work crews are required to clear four hectares of forest in 2 days ??  :D[/quote]

[hide]

It's a proportionality problem.

$\frac{2.5\text{ crews}}{2.5\text{ days}\times 2.5\text{ hectares}}=\frac{x}{2\text{ days}\times 4\text{ hectares}}$

$\therefore\, x=3.2\text{ crews}$.

[/hide]

Dr.No's answer is right...BTW , math92 , how did you get that 8 ?
 :?  A wild guess ? ;)

yes, i just looked at it and i guessed 8


calculus
integration
trigonometry
calculus computations

\[ \int {\frac{{\sqrt {\cot x}  \plus{} \sqrt {\tan x} }}
{{4 \plus{} 3\sin 2x}}} dx\]

Set $ t\equal{}\sqrt{\tan x}$

$ \int \frac{\sqrt{\cot x}\plus{}\sqrt{\tan x}}{4\plus{}3\sin 2x}dx$

$ \equal{}\int \frac{\left( t^{2}\plus{}1\right) ^{2}}{\left( 5t^{2}\plus{}2\right) \left( 1\plus{}t^{4}\right) }dt$

$ \equal{}\frac{1}{\sqrt{14}}\tan ^{\minus{}1}\frac{\sqrt{14\tan x}}{2(1\minus{}\tan x)}\plus{}C$


calculus
derivative
real analysis
limit
function
real analysis unsolved

Let $ E\subset\mathbb{R}^n$ be open
Let $ \mathbf{f}: E\rightarrow\mathbb{R}^m$

The definition given by Rudin is
"$ \mathbf{f}$ is differentiable at $ \mathbf{x}\in\mathbb{R}^n$ if and only if $ \exists A\in L(\mathbb{R}^n,\mathbb{R}^m)$ such that
$ \lim_{\mathbf{h}\rightarrow\mathbf{0}}\frac{|\mathbf{f}(\mathbf{x}\plus{}\mathbf{h})\minus{}\mathbf{f}(\mathbf{x})\minus{}A\mathbf{h}|}{|\mathbf{h}|}\equal{}0$

However, if we follow strict analogy with single-variable case, we can give the definition
"$ \mathbf{f}$ is differentiable at $ \mathbf{x}\in\mathbb{R}^n$ if and only if $ \exists \mathbf{r}\in \mathbb{R}^m$ such that
$ \lim_{\mathbf{h}\rightarrow\mathbf{0}}\left|\frac{\mathbf{f}(\mathbf{x}\plus{}\mathbf{h})\minus{}\mathbf{f}(\mathbf{x})}{|\mathbf{h}|}\minus{}\mathbf{r}\right|\equal{}0$

What is the advantage of defining derivative as a linear transformation rather than as a similar function?

The advantage is that you get a meaningful and useful definition rather than some formal nonsense. Indeed, try to apply your definition to the identity map in $ \mathbb R^2$. You'll see that the derivative in your sense just does not exist.

Or: for $ f$ to be differentiable we need $ f(x \plus{} h) \minus{} f(x)$ to be small when $ h$ is small, but we don't need it to be in some fixed direction (i.e. a function can change at different rates in different directions and still be differentiable).  The standard definition captures this, but yours does not.

thanks for the help

The direction was seemingly my main problem with the original definition. So, with my definition, you take $ \mathbf{h}\rightarrow\mathbf{0}$ along a fixed direction and get the derivative along that direction, which I can see to be quite lame.


what type of numbers get reversed when they are multiplied by 4? (reversed means for example abbc, equals to cbba when reversed.)

your question is not really clear. Maybe explain in detail?

Umm... something very similar to this I believe is on the 2005 CPSC; it could be a coincidence, but I don't think we should discuss it (correct me if I'm wrong, I don't have a copy of the test with me).

it can be easily proven that if you have a three digit number, then the sum of the first and third digit of the number with it's reverce is always equal to 9. and that the second digit of of the difference of a three digit number with it's reverce is also 9!

but i'm not sure if this holds true for other numbers as well,

Colloborative Problem Solving Contest hosted by http://www.natassessment.com -- theres a forum on it here.

The number 2178 is the smallest one there is.
All other numbers are 219999.....9999978 for however many 9's you want.

hmmm, there was a question like this on last year's mathcamp qualifying quiz, it was 'for each n, how many n-digit numbers are there such that 9*x=x reversed'.  it turns out that for each x in that group, 2*x is in the set of numbers 4*number=number reversed (i dont have a proof yet though)

for 9*x=x reversed, there are $F_{[n/2]-1}$ n-digit numbers, where $F_k$ is the kth fibonacci #.  i proved that before i think, i can dig it up if anyone wants it.

ya i wouldnt mind seeing a proof about that...it'd be cool :D


Suppose $ x \notin \{\frac {k \pi}{2} : k \in \mathbb{Z} \}$. Find, with proof, all $ x$ such that
\[ \sqrt {2} \equal{} (\sin x)^{2 \sin^2 x} \plus{} (\cos x)^{2 \cos^2 x}.
\]

Note $ \sin x, \cos x > 0$. Let $ f(z) \equal{} z^z$, then for $ z > 0$, $ f''(z) \equal{} z^{z \minus{} 1} \plus{} (\ln z \plus{} 1)^2 z^z > 0$   :| , and by Jensen's, $ f(\sin^2 x) \plus{} f(\cos^2 x) \ge 2f\left(\frac {\sin^2 x \plus{} \cos^2 x}{2}\right) \equal{} \sqrt {2}$, equality when $ \sin^2 x \equal{} \cos^2 x \Longrightarrow x \equal{} \frac {(8n \plus{} 1)\pi}{4}, \forall n \in \mathbb{Z}$.

.. is there a better way?

I'm not sure, that's how I did it.

Is there any easier (e.g., non-calc) way to show the convexity of $ f(z) \equal{} z^z$?


algebra
polynomial
linear algebra
Set systems
combinatorics

Let $n>1$ be an integer. A set $S \subset \{ 0,1,2, \ldots, 4n-1\}$ is called [i]rare[/i] if, for any $k\in\{0,1,\ldots,n-1\}$, the following two conditions take place at the same time

(1) the set $S\cap \{4k-2,4k-1,4k, 4k+1, 4k+2 \}$ has at most two elements;

(2) the set $S\cap \{4k+1,4k+2,4k+3\}$ has at most one element.

Prove that the set $\{0,1,2,\ldots,4n-1\}$ has exactly $8 \cdot 7^{n-1}$ rare subsets.

yes, you are correct
the number is indeed $8\cdot 7^{n-1}$.

[quote=Taha1381][quote=maxal]I believe that the formula $7^n$ is incorrect.

Let $f(n)$ be the number of rare subsets of $\{0,1,\dots,4n-1\}$ and $g(n)$ be the number of rare subsets such that they do not contain $4n-1$ or $4n-2$.
Considering all cases of (not) containing elements $4n-4$, $4n-3$, $4n-2$, $4n-1$ we get
\[ \begin{array}{l}f(n) = 6 f(n-1) + 2 g(n-1)\\g(n) = 3 f(n-1) + g(n-1)\end{array}  \]
Since the characteristic polynomial of the matrix $\left(\begin{matrix} 6 & 2\\ 3 & 1\end{matrix}\right)$ is $x^2-7x$, for $n>1$ we have $f(n) = 7 f(n-1)$ that together with $f(1)=8$ implies $f(n)=8\cdot 7^{n-1}$.[/quote]

Can anyone explain more on how did we get to recurrent relations?[/quote]


calculus
integration
function
algebra solved
algebra

Let $\{a_n\}_{n = 0}^{\infty}$ be a non-decreasing sequence of real numbers such that \[ a_0 = 1,\;(n - 1)a_n = na_{n - 2}\;(n \geq 2).\] Compute $a_1$.

I've found $a_1 = \pi$ using Stirling's formula. Am I right because writing all the computations will be really boring...

Pierre.

You can use the Wallis product - you hardly have to do any computations.

Ok, Grobber, I propose an agreement for $ \frac {3 \pi} {4}$  :D
So, Arne what is the good answer (I think it is Grobber's one, according to Wallis' integral) ?

Pierre.

:(  In fact, I checked my computations and it is $ \frac { \pi} {2}$ as Grobber found.
Note that the result cannot be $ \pi$ since $a_0=1$ and $a_2 = 2$ and the sequence is non-decreasing....(aargh).

Let $a_1 = \alpha$.
From the recurrence relation, it is easy to find that, for all $p \geq 0$, we have :
$a_{2p} = \frac {4^p(p!)^2} {(2p)!}$ and $a_{2p+1} = \frac {(2p+1)!} {4^p(p!)^2} \alpha$.

We want that, for all $p$ : $a_{2p-1} \leq a_{2p}\leq a_{2p+1}$, which turns to be equivalent to :
\[ C_2(p) = \frac {4^{2p}(p!)^4 }{(2p)!(2p+1)!} \leq \alpha \leq \frac {4^{2p-1} (p!)^2(p-1)!^2} {(2p)!(2p-1)!} = C_1(p)\]

Now it is easy to verify that the function $C_1(p)$ is non-increasing (it is equivalent to  $(2p+1)^2 \geq 4p(p+1)$, which is true) and that the function $C_2(p)$ is non-decreasing (it is equivalent to $(2p+1)(2p+3) \leq 4(p+1)^2$ which is true again).
Then, to get the claimed result it is necessary and sufficient to prove that both $C_1$ and $C_2$ tends to $ \frac { \pi} {2}$ as $p$ tends to $+ \infty$.

But, according to Stirling's formula :
${C_1(p) \sim 4^{p-1} \frac {2pe \pi (p-1)^{2p-1} } { \sqrt {2p(2p-1)} (2p-1)^{2p-1} } \sim 4^{p-1}e \pi ( \frac {p-1 }{ 2p-1} )^{2p-1} }$

Moreover $ ( \frac {p-1} {2p-1} )^{2p-1} = \frac {1} {2^{2p-1}} \times (1 - \frac {1} {2p-1} )^{2p-1}$
and $(1 - \frac {1} {2p-1} )^{2p-1} = e^{ (2p-1)ln(1 - \frac {1} {2p-1}) }$ which tends to $e^-1$.
Thus $C_1(p) \sim \frac {\pi } {2}$.

In another hand, as above :
$C_2(p) \sim 4^pe \pi ( \frac { p} {2p+1} )^{2p+1}$
And as above we have
$\frac { p} {2p+1} )^{2p+1} \sim \frac {1} {e \times 2^{2p+1} }$
Thus $C_2(p) \sim \frac { \pi} {2}$, and we are done.

Pierre.


If a semicircle is constructed on the side of an equilateral triangle, as pictured (Attachment), what percent of the area of the semicircle lies outside the triangle?  Write your answer rounded to the nearest whole percent.

ANS: [hide]12[/hide]

not easy, but not hard.

we have triangle $ ABC$ with midpoints D, E, and F on the lines AB, AC, and BC respectivly. let the radius be r. we have the base of the larger triangle is also 2r. the area of the semicircle is $ \pi r^{2}/2$. now we have a 60 degree sector CEF. if we subtract the area of triangle CEF from that sectors area, we end with half of the area of the semicircle outside the triangle.so we have the area of the sector is $ \pi r^{2}/6$. so the area outside the triangle is...wait lets say r is one because were looking for the percent....so we have the area of the part outside the triangle is $ (2\pi\minus{}3\sqrt{3})/12$. multiply that by two get your answer. im in a hurry ill edit my post later.


geometry
3D geometry
algebra solved
algebra

From (little) cubes ABCD with AB=1 we construct a cube A'B'C'D' with A'B'=45
(it contains 45^3 little cubes).
In exactly 1998 little cubes are viruses.Viruses can pass in another little cube only if this cube has at least 3 common sides with virused cubes.
Can the viruses occupy al the cube A'B'C'D'?

The problem was given at the National Olympiad of Moldova republic in 1998.(Does anyone have the solutions form that olympiad?)

No, the virus cannot occupy the whole cube :

For convenience, we assume that infection propagates minute by minute.

Color the little infected cubes in black. Let S(t) be the total (exterior) area of the region of the space formed by the black squares at time t (for exemple if there were only two neighbour black squares then S(t) = 10*1 since there are only 10 exterior black faces).

Suppose that a not infected cube becomes infected :
If it had 6,5,4 or 3 infected cubes among its neighbours then S(t+1) = S(t) - x, where x = 6,4,2 or 0 respectively.
It follows that S(t+1)  \leq S(t) that is the total black area cannot increase.

At the beginning, we have S(0)  \leq 1998*6
But, if we want the whole cube to be infected then we must obtain S(t) = 6*48 2  at some t \geq 0.
Since 6*48 2  > 6*1998, it cannot be done...

Pierre.


A drawner contains 2 brown and 3 gray socks. The socks are taken out of the drawer one at a time. What is the probability that the fourth sock removed is gray? Express your answer as a common fraction.

It's just $ \frac{3}{5}$. All socks have a $ \frac{3}{5}$ probability of gray and $ \frac{2}{5}$ probability of brown.

However the problem seems to be saying that the socks are not replaced, so the answer is not just 3/5... the probabilities change based on how the contents of the drawer change.

For example if the first three socks you draw are grey, then the probability of the fourth sock being gray is 0.

What are the posibilities then?  Well, to draw a grey sock on the 4th step, you'd need to draw either one or two browns:

Sequences with 1 brown, 4th grey
BGGG GBGG GGBG each occuring with probability $ \frac{2}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\cdot\frac{1}{5}\equal{}\frac{12}{625}$

Sequences with 2 brown, 4th grey
BBGG BGBG GBBG each occuring with probability $ \frac{2}{5}\cdot\frac{1}{5}\cdot\frac{3}{5}\cdot\frac{2}{5}\equal{}\frac{12}{625}$

Then the total probability of these events is $ 3\cdot\frac{12}{625}\plus{}3\cdot\frac{12}{625}\equal{}\frac{72}{625}$

Actually, you forgot to decrease the number of socks you're choosing from.

It'd be $ \frac25*\frac34*\frac23*\frac12$ three ways and
$ \frac25*\frac14*\frac33*\frac22$ three ways.

Simplify and it's $ \frac1{10}*3 + \frac1{10}*3 = \boxed{\frac35}$

But if you've done these problems before, you'll realize that the probability of getting a specific outcome on the $ nth$ selection without replacement is the same as getting it with replacement.

I DID!  And after all that talk about non-replacement, I went and forgot to decrease the denominators　：）

So, surprisingly it comes out to be 3/5, but that is NOT the result of this logic : 3 gray socks in a set of 5, therefore 3/5.  It's just a coincidence.

Unfortunately this allows someone naively putting 3/5 to get full credit :/

Sorry again about the ridiculous mistake... I'll try to make them as infrequently as possible!

Well, I thought about this problem a little more... and it appears that the naive answer is always right!  To me it seemed counterintuitive at first... but apparently it's provable.

THM:  If T is a collection of objects with N blue objects and T-N nonblue objects in a box, if the objects are removed one at a time, then the probability that the k'th object is blue is N/T.

Proof:  Of course, the probability that the first ball picked is blue is N/T.

Imagine that you wrote out every possible pattern of length T that could occur.  Put it into a large matrix which has T columns.  Swap the 1st and the kth columns of the matrix.  Notice that after you swap, you still have a complete list of all patterns that could occur!  The new list is the same as the old list, except the rows are probably in a different order.

This basically shows that all columns of the matrix contain the same number of blue objects.  Then the probability the kth object selected is blue is the same probability as if the first object selected was blue, which is N/T.\qed


[hide="Click for solution"]
This is simply the number of ways to choose $ 2$ points from $ 12$, or $ \binom{12}{2}\equal{}66$, or $ \boxed{\textbf{(D)}}$.
[/hide]

[quote="Brut3Forc3"]Given $ 12$ points in a plane no three of which are collinear, the number of lines they determine is:
$ \textbf{(A)}\ 24 \qquad\textbf{(B)}\ 54 \qquad\textbf{(C)}\ 120 \qquad\textbf{(D)}\ 66 \qquad\textbf{(E)}\ \text{none of these}$[/quote]

[hide]

When you have a line, it must go through two or more points.  Therefore, to get the number of lines $ 12$ points will give is simply $ \binom{12}{2} \equal{} \boxed{\textbf{(D)}\ 66 }$[/hide]


limit
integration
real analysis
trigonometry
search
calculus
real analysis unsolved

let $ f$ be Lebegues integrable on $ [a,b]$
prove that $ \lim_{n\to \infty}\int_{a}^{b}f(x)|cosnx|dx\equal{}\int_{a}^{b}f(x)dx$

thanks in advanced

If you think just a little bit, you'll realize that the factor $ \frac 2\pi$ is desperately missing on the right hand side ;). Once corrected, the problem can be directly derived from the Riemann-Lebesgue lemma by decomposing $ |\cos x|$ into its Fourier series. You can also just mimic the proof. It can be repeated almost verbatim.

I believe that a more general form of this problem has already been discussed (with Kent as one of the main contributors) but I cannot find the link now :(.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=534005051&t=21933]Here[/url] is one link from 2005. There are others, but I haven't found them, either.

My search found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=146991]two[/url] threads. I prefer using integration by parts, which works with an arbitrary integrable periodic function.


D,E,F are midpoints of $ \overline{AB}, \overline{BC}$, and $ \overline{CA}$. Also G, H, and I are midpoints of $ \overline{FD}, \overline{DE}$, and $ \overline{FE}$ respectively. If area of $ \triangle{GHI}$ = 12, what is the area of $ \triangle{ABC}$?

A. 192
B. 180
C. 154
D. 384

[hide]$ S_{\Delta ABC} \equal{} 16S_{\Delta GHI} \equal{} 192$[/hide]

Yep i got A too becuase GHI = 1\16 $ \times$ ABC so 12 $ \times$ 16 = 192 = $ \boxed{A}$

[quote="xxxxx"]that's the answer I got, but the correct answer was this though:
[hide]C. 154[/hide][/quote]
Where is this problem from?

The answer is definitely 192, not 154.


limit
ratio
real analysis
real analysis unsolved

For $ n =1,2,..., $  let
\[
x_n=(4n)^n\frac{1^n+2^n + ...+ n^n}{ 1^{2n}+2^{2n}+3^{2n}+...+(2n-1)^{2n}}\; .
\]
[b]Questions:[/b]
[i]
a)  Investigate if exists  $  L: = \lim\limits_{n\to \infty}x_n $   .
b) If $  L $ exists, find    $\lim\limits_{n\to \infty} n\left( x_n-L \right) $ .
[/i]

$1^{n}+...+(n-1)^{n}$ is equivalent to  $\frac{1}{e-1}n^n$
yes flip 2004  the limit for a) is e, thx.

My opinion is that the answer to first question is e=Napier's constant.

For a)

$\sum_{k=1}^{n}k^n \sim \frac{e}{e-1}n^n$

$\sum_{k=1}^{2n-1}k^{2n} \sim \frac{(2n)^{2n}}{e-1}$

$x_n \sim (4n)^n \frac{\frac{e}{e-1}n^n}{\frac{(2n)^{2n}}{e-1}}$

After simplification $x_n\sim e$

For b) I think if the limit exist it is a rationnal expression in term of the Napier constant.

2Moubinool: could you explain prove of you equivalence:
$\sum_{k=1}^{n}k^n \sim \frac{e}{e-1}n^n$  As far as i know there are lot of ways to prove this one, i tryied to use for examlpe Stolz theorem- and i have got the ratio between $\sum_{k=1}^{n}k^n$ and $\frac{e}{e-1}n^n$ tends to zero=so if i did it right there mustn't be equivalence,but o(...)

[quote="Moubinool"]This problem posted by flip2004 is similar to problem in

http://rgmia.vu.edu.au/authors/ALupas.htm

7/ A. Lupas, A Sequence Convergent to Napier's Constant[/quote]

Hmm. It seems you revealed the true identity of our famous and active forum member.  ;)


For positive integer $ n$, consider the subset $ M$ of the set $ \{1,\ 2,\ \cdots ,\ n\}$ such that if $ m\in{M}$, then  $ 2m\notin{M}$.
For the set $ M$, let $ g(M)$ the number of  the subset of $ M$, denote the possible maximum value of $ g(M)$ by $ f(n)$. Answer the following questions.

(1) If $ n$ is the multiple of four, then prove that $ f(n)\geq \frac {n}{2} \plus{} f\left(\frac {n}{4}\right)$.

(2)  If $ n$ is the multiple of four, then prove that $ f(n)\leq \frac {n}{2} \plus{} f\left(\frac {n}{4}\right)$.

(3) Find $ f(3\cdot 2^{125})$.

[b]Kunny[/b], can you please clarify this?

$ M$ is a subset of $ \{1,2,3...,n\}$ with the restriction; if $ m\in M$ then $ 2m\not\in M$

$ g(M)$ is the number of possible subsets of $ M$, (i.e. if $ M$ contains $ i$ elements then $ g(M)\equal{}2^i$)

$ f(n)$ is the maximum possible value of $ g(M)$, in terms of $ n$...?


If $a,b,c$ are distinct positive numbers, prove \[a^{a}b^{b}c^{c}>\left(a^{pc+qb}b^{pa+qc}c^{pb+qa}\right)^{{}^{\frac{1}{p+q}}}\; ,\; \forall p,q\in (0,\infty).\]

[hide]
Note: All summations are cyclic.

Take the $\log$ of both sides; it remains to show that

$p\sum a \log{a}+q\sum a \log{a}> p\sum c \log{a}+q\sum b \log{a}$.

But $(a,b,c)$ and $(\log{a}, \log{b}, \log{c})$ are similarly ordered, by rearrangement we have $\sum a \log{a}> \sum c \log{a}$ and $\sum a \log{a}> \sum b \log{a}$ (strict because they are distinct) so we just add up $p$ times the first and $q$ times the second to get the desired inequality.
[/hide]

[quote="paladin8"]
Note: All summations are cyclic. Take the $\log$ of both sides; it remains to show that
$p\sum a \log{a}+q\sum a \log{a}> p\sum c \log{a}+q\sum b \log{a}$.
But $(a,b,c)$ and $(\log{a}, \log{b}, \log{c})$ are similarly ordered, by rearrangement we have $\sum a \log{a}> \sum c \log{a}$ and $\sum a \log{a}> \sum b \log{a}$ (strict because they are distinct) so we just add up $p$ times the first and $q$ times the second to get the desired inequality.
[/quote]Thanks, very nice/proposer


advanced fields
advanced fields unsolved

Build a non deterministic Turing machine that accepts the Language $ L\equal{}\{ xyx | x,y\in \{a,b\}^* , |x|\geq 1, |y|\geq 0\}$

So let your input be on the tape, where the end of the input has a $ \$$ as a marker. Now read symbol-by-symbol along the input tape, placing each one after $ \$$ (reject if any of these are not an $ a$ or a $ b$). Do this until the end of $ x$, which is possible to decided nondeterministically. Then scan through $ y$ (if there is anything there) and verify that it is part of the correct alphabet. Again you know when you've reached the end of $ y$ nondeterministically. Then scan back and forth along the remaining input symbols and the ones you've placed after $ \$$ and make sure they match. If they do, then accept, otherwise reject.

The question was more in the terms of writing down the table for the transition function.


trigonometry
calculus
calculus computations

Given that x = sec (y/2) , o≤y≤pi, show that dy/dx = 2/(x√(x^2 - 1))

Unfortunately, this computer has a backslash key which doesn't work.

dx/dy = (1/2)sec(y/2)tan(y/2)
dy/dx = 2/(sec(y/2)(tan(y/2))
         = 2/(x(sqrt((sec(y/2)^2) -1)) )
         = 2/(x(sqrt((x^2) -1)))

You can always find one easily on this site and copy+paste it =)

Thanks Fermat's Little Turtle, using CTRL C and CTRL V, I've got:

$ \frac {dy} { dx} \equal{}  \frac {1} {2} \ sec (\frac {y} { 2})\ tan (\frac {y} {2})$

$ \frac {dy} {dx} \equal{} \frac {2} { \sec(\frac {y} {2})\tan(\frac {y} {2}) }$

$ \equal{} \frac{2} {x \sqrt {\sec ^{2} (\frac {y} {2}) \minus{} 1}}$

$ \equal{} \frac {2} {x \sqrt{x^{2} \minus{} 1}}$


Let S be a subset of {1, 2, 3, ..., 2046} with |S| = 21.
Show that there exist distinct a, b, c in S such that bc < 2a < 4bc.

[hide]We will show that if b<a<c and c<=2b then bc<2(a)(a)<4bc.  Then we will show three numbers of satisfying those conditions exist in every 21-element subset of the first 2046 natural numbers.

a>b, so 2(a)(a)>2(b)b>=bc because 2b>=c, so 2(a)(a)>bc.  2b>c>a, so 2bc>(a)(a) and 4bc>2(a)(a).  So bc<2(a)(a)<4bc if b<a<c and c<=2b.

If a 3 number sub-subset of the 21-element subset such that b<a<c and c<=2b is not to exist, then the third smallest element of the subset needs to be more than twice the smallest element, or c>2b.  Similarly, the fifth smallest element of the subset needs to be more than twice the third smallest element, the seventh more than twice the fifth, etc.

Because the smallest element is at least 1, the third smallest is at least 3, the fifth smallest is at least 7.  We can see that this sequence with a1=1 and an+2=2an+1 will result it a21=2^11-1=2047.  So the largest element of the 21-element must be at least 2047, which is a contradiction because only the first 2046 natural numbers are being dealt with.

So in every 21-element subset of 2046, there exist three distinct numbers such that b<a<c and c<=2b, which implies bc<2(a)(a)<4bc.[/hide]

After solving that problem, I feel like I need some waffles.  Or some overpriced chocolate bars.  Maybe I should inform NATO of my victory over a side dish of some sort of sprout.


Does anyone know the Top 10 people that placed at CSULB's Math Day at the Beach? Just Curious.

Sean: I'll post some results in the Southern California ARML forum.