algorithm
induction
ceiling function
USAMTS
trigonometry

How'd ya'll do? I got 52413, total 15, final score 83.

Round 4: 53553, total 85. Why don't you make a poll?

Oof, 5 [b]5[/b] 5 [b]5[/b] 4, total score 99...

"Good work, but your proof has a few gaps"

Honestly, looking back at the proof, I think it's both clear and watertight...I can't see why they'd take off a point there, but give me a 5 on something like, say, round 3 problem 1.

5 [b]5[/b] 5 [b]5[/b] 5 = 100.  I definitely wasn't expecting this...

I don't have my scores yet, but I'm very, very satisfied with USAMTS.

I'm definitely going to do it again next year :)

ugh

5 2 2 1 5, worst I've done.
but I think they screwed up on #4, and possibly #3, so I appealed both, I might be able to squeeze that up to a 5 2 4 4 5, giving me a 76 for the year, but as of now, I have a 71.

[quote="K81o7"]Oof, 5 [b]5[/b] 5 [b]5[/b] 4, total score 99...

"Good work, but your proof has a few gaps"

Honestly, looking back at the proof, I think it's both clear and watertight.[/quote]
You posted your solutions, so I looked at your answer for problem 5. I consider your answer to be watertight, but I am mentally adding in the content that you left unmentioned. Your solution is not clear.

For example, consider the first few lines for your inductive step:
[quote]We assume $A_{2n+1}\geq n+1$. Then
\[\sum_{i=1}^{2n+1}x_{i}\geq (2n+1)(n+1).\]
But
\[(2n+2)n<(2n+1)(n+1)<(2n+2)(n+1)\]
\[\Longrightarrow\quad \sum_{i=1}^{2n+2}x_{i}\geq (2n+2)(n+1)\]
because that value has to be a multiple of $2n+2$, and must be greater than or equal to $(2n+2)(n+1)$.[/quote]
The math is valid, the reasoning is valid, but the reader has no clue that you are talking about $A_{2n+2}$. I would fill in the missing details as:
[quote]We assume $A_{2n+1}\geq n+1$. Then
\[(2n+1)A_{2n+1}= \sum_{i=1}^{2n+1}x_{i}\geq (2n+1)(n+1).\]
Consider $(2n+2)A_{2n+2}$. Because
\[(2n+2)n<(2n+1)(n+1),\]
we have
\[(2n+2)A_{2n+2}=\sum_{i=1}^{2n+2}x_{i}> \sum_{i=1}^{2n+1}x_{i}> (2n+2)n.\]
Thus, $A_{2n+2}>n$, and since it is an integer, $A_{2n+2}\geq n+1$.[/quote]
Same calculations, but the purpose of the calculations is much more evident. Your proof by induction is establishing properties of $A_{2n+2}$ and $A_{2n+3}$, therefore, referring to those values directly helps the clarity.

Furthermore, when you gave the sequence $A_{n}=\lceil\frac{n}{2}\rceil$ at the end of your solution, you proved abstractly by contradiction that no two consecutive $x_{n}$ are equal. Most people instead explicitly gave a formula for the $x_{n}$. Not only is the explicit method clearer, it is also shorter.

[quote="K81o7"]I can't see why they'd take off a point there, but give me a 5 on something like, say, round 3 problem 1.[/quote]
A different person graded your solution to problem 3/1/18 than graded your solution to problem 5/4/18. We graders vary. An NSA grading session typically puts eight graders on a problem. Problem 5 was a tough problem to grade, so we put a dozen graders on it. I even helped out after I was done grading problem 1. We establish rubrics to keep the grading consistent, but even with a rubric, individual judgment is required.

Erin Schram
USAMTS grader

5 5 5 5 3

For a total score of 98.  I was hoping for 100, but I knew my proof on #5 was a little shady.

[quote="calc rulz"]5 5 5 5 3

So 94 total =\

I thought my proof was rigorous, so this means I can challenge it! :)
(He said I didn't explain why I used the greedy algorithm, but I believed I justified everything so I'm going to look back...)[/quote]
The solution you posted here in the forum (and reposted just recently, but I had read the first one, too), looks complete and rigorous to me, though a trifle unpolished.

Your proof by induction does have many steps in common with a greedy algorithm, but since you are working by the rules of induction, it is not a greedy algorithm. You never said you were using a greedy algorithm, so for the grader to mention it means that he or she misunderstood your solution.

I have no authority to change grades, so you have to post in the Grading Issues subforum to get a change.

Erin Schram
USAMTS grader

225--

Good score IMO for a half hour, 10% effort!

uh... i got a 25 on round 4... absolutely no clue how

unfortunately i started on round 3 so my total is only 37 for the year.

54554 = 23
72+23=95

My proof for #3 definitly doesn't deserve a 5, but I think my proof for #2 does. Fortunatly, It all cancels out, (in my favor :D ), so I don't have to rant.

87, 3 commended.

I pretty much lost my motivation on round 4 and only solved 3 of the problems.  I did think they all deserved 5s instead of 4s, though.  Oh well, 87 or 90 its still a silver and I can still be proud of my 75/75 on the first 3 rounds.

Somehow my #1 was commended.  I don't know why. It was just like everyone else's solution.

Probably because it was the only solution I had that was right.

[quote="ehehheehee"]96 (barely made gold)

Do you think they should make more levels or change the levels so people don't lose motivation on the last round?  The range from 75-95 is pretty big.[/quote]

Heh I happen to agree with you on that too.

19/ 25/ 25/ 55[b]5[/b]55 = 94.

A lot higher than I expected for my 1st year here.

I wish they would subdivide the silver level too, but that’s just me a bit mad that I missed gold because I didn’t know what I was doing 1st round.

5 5 5 5 5

Previously: 25, 25, 21.

Thus, 96 total --> gold.  :D

Yay! My score is now 5 5 5 5 5 for 96 total :)

[b]5 5 5[/b] 5 3, though I'm appealing the last one. That's a total of 97 as-is, and 99 if I win the appeal, my only error coming the only time I used a calculator to check my answers to the trig problem (Round 3 Problem 2). Let that be a lesson...

i only took part 4, and the grader didnt understand one of my proofs (my train if thought is pretty eccentric)

still im happy with a 21

With my appeals I am now up to 52335, but I can't do any better, so I'm stuck with 74. I blame this on round 3 problem 2 (tan7x+cot7x or whatever it was), because I forgot to check 5pi/4, which just happened to be a solution, so I got a 3.

grrr, so close to silver.

btw, would a 75 get you a silver, or do you need 76?

[quote="Xaero"][quote="13375P34K43V312"]btw, would a 75 get you a silver, or do you need 76?[/quote]

I think you need a 76.[/quote]

No, you definitely need a 75. [url]http://www.usamts.org/About/U_AbPrizes.php[/url]

how does one go about getting an appeal?

i think one of my solutions was right but slightly hard to understand

also, does anyone know how to view the solutions i submitted from the website?

To get an appeal, post a new topic in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=213]Grading Issues Forum[/url]. Make sure that your reason for protesting is valid.

You can't view the solutions that you submitted from the website. You should keep a copy of them.


function
number theory
relatively prime
totient function

1) What is the remainder when $ 6^{555}$ is divided by $ 13$?
2) What is the remainder when $ n^{555}$ is divided by $ 13$?

[hide="1"]By [url=http://en.wikipedia.org/wiki/Euler%27s_totient_theorem]Euler's Totient Theorem[/url], since 6 and 13 are relatively prime, we have $ 6^{12}\equiv 1 \mod 13$.

$ 555\equiv 3\mod 12$

$ 6^{555} \equal{} (6^{12})^{46}\cdot6^3 \implies 6^{555}\equiv 1\cdot 6^3\mod 13$.

Hence, $ 6^{555}\equiv 8 \mod 13$[/hide]

[hide="2"]We can divide this problem by 2 cases:

1) Where $ n\equiv 0 \mod 13$
Obviously 0...

2) Other cases i.e. $ n$ has remainders other than 0 when divided by 13
Likewise as Number 1, $ n^{12}\equiv 1 \mod 13$.

Again, $ 555\equiv 3\mod 12$.

$ n^{555} \equal{} (n^{12})^{46}\cdot n^3\implies n^{555}\equiv n^3\mod 13$[/hide]

For #1, isn't that Fermat's little theorem?

$ a^{p-1} \equiv 1 %Error. "modp" is a bad command.
$

for p being any prime?

[quote="FantasyLover"][quote="modularmarc101"]For #1, isn't that Fermat's little theorem?

$ a^{p - 1} \equiv 1 %Error. "modp" is a bad command.
$

for p being any prime?[/quote]I believe Euler's Totient Function is generalized form of Fermat's Little Theorem.[/quote]
Correct


function
calculus
algebra theorems
algebra

How can I find if the function f(n)=n*tan(180/n) is increasing or decreasing for n natural number greater than 2?

I think in this case f(n) is decreasing.
U can use calculas to solve it.


Wow, repost when the same problem is exactly in the topic now below this one ;)
But this ones title is much better *changes old title*


Let $ABCDE$ be a regular pentagon. The star $ACEBD$ has area 1. $AC$ and $BE$ meet at $P$, while $BD$ and $CE$ meet at $Q$. Find the area of $APQD$.

Let $R = BE \cap AD$ and $S_{1} = \left[APR\right]$, $S_{2} = \left[AER\right]$.
Then $\left[ACEBD\right] = 4S_{1} + 2S_{2}$ and $\left[APQD\right] = 2S_{1} + S_{2} = \frac{1}{2}$.


in a particualr sequence, we have:
$a_0=0$ and $|a_n|=|a_{n-1}+3|$ for $n\geq 1$.
what is the minimum value for $|a_1+a_2+a_3+\cdots a_{2006}|?$

for all k $3|a_k,$ and $2|a_k$ if and only if 2|k. Therefore $|a_1+\dots +a_{2006}|=3(m0d \ 6)$. Sequense $a_{2k}=0, a_{2k+1}=3(-1)^k$ satisfied recurent equation with sum=3. Therefore minimum =3.


floor function
combinatorics unsolved
combinatorics

Let $ S$ be the set of all odd positive integers less than $ 30m$ which are not multiples of $ 5$, where $ m$ is a given positive integer. Find the smallest positive integer $ k$ such that each $ k$-element subset of $ S$ contains two distinct numbers, one of which divides the other.

2m+1

[hide="hint"]consider chains a 3a 9a ...[/hide]

I'm pretty sure it can't be 2m+1... if m=1 just pick $ {3,7,11}$ or any other set of three primes.

1. $ k\geq$ Number of primes in $ S$.
2. Let $ S(n)$ denote the set of all odd numbers less than $ n$ which are not the multiples of 5. We observe $ 30m\minus{}1$ is not divisible by any number less than 7(other than 1).
$ k\geq |S|\minus{}\lfloor |S\big (\frac{30m\minus{}1}{7}\big )|\rfloor$

first find how many numbers are not divisible by 2, 3 or 5.
after that consider the chains $ C_a\equal{}\{a,3a, 9a, ...\}$ defined by these numbers. show that our set should not have two elements from the same chain. so what can we say about k in terms of the number of such chains?


Here is a cute simple problem I proposed for juniors TST this year (of course, it wasn't given):
  For any $M>0$ and any positive integer $a$ there exist infinitely many positive integers $b$ such that $|a^m-b^n|>M$ for any positive integers m,n.

I like it :).

For a prime factor $p|a$, we can take $p|b$, and $b$ large enough so that for any solution of $|a^m-b^n|=k\le M$ $m,n$ must be so large that the largest power of $p$ which divides both $a^m,b^n$ is larger than $M$. If we also take $b$ to be a multiple of some prime which does not divide $a$, we can also eliminate the only possibility left, namely $a^m=b^n$ for some $m,n$.

I hope it's correct.

Why did they not give it? It is quite simple. :?

Take very large $k$ and $b=a^k(a+1)$, then $|b^m-a^n|>a^k$ for $n\leq k$ and $a^k\mid b^m-a^n\neq 0$ for $n>k$.

Great!! Indeed, it is a trivial problem, but I like it... it sounds nice...  :)


Let the triangle ABC and H your ortocenter. Such that AB<>AC.
Let B', C' the intersection of (BH and AC),  (CH and AB).
Let M be the midpoint of BC and D the intersection of  B'C' and BC. Prove that DH and AM are perpendicular.
I use analitic geometry. I want to now if someone have another solution....

see at http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.


geometry
rectangle
inradius
geometry unsolved

Let $ABCD$ be a rectangle, and let $\omega$ be a circular arc passing through the points $A$ and $C$.
Let $\omega_{1}$ be the circle tangent to the lines $CD$ and $DA$ and to the circle $\omega$, and lying completely inside the rectangle $ABCD$.
Similiarly let $\omega_{2}$ be the circle tangent to the lines $AB$ and $BC$ and to the circle $\omega$, and lying completely inside the rectangle $ABCD$.
Denote by $r_{1}$ and $r_{2}$ the radii of the circles $\omega_{1}$ and $\omega_{2}$, respectively, and by $r$ the inradius of triangle $ABC$.
[b](a)[/b] Prove that $r_{1}+r_{2}=2r$.
[b](b)[/b] Prove that one of the two common internal tangents of the two circles $\omega_{1}$ and $\omega_{2}$ is parallel to the line $AC$ and has the length $\left|AB-AC\right|$.

[quote="andrew"]Lemma.Let $ ABC$ is a triangle and $ w$ is a circle passing through $ A,C,v$ is a circle,which is tangent to $ AB,BC$ at points $ X,Y$ and $ w$ at point $ P.I$ is incenter of $ ABC$,then $ A,I,P,X$ lie on the same circle and $ C,I,P,Y$ lie on the same circle.(it is true in two cases on picture).
[/quote]
I'm eager to know solution. Can anyone solved?

It is based on Thebault theorem. See [url]http://www.mathlinks.ro/viewtopic.php?t=41667[/url].


function
logarithms
algebra unsolved
algebra
Functional Equations

Given is a non-constant function $f: \mathbb{R} \longrightarrow \mathbb{R}$ such that for any $x$ holds:

$f(x+1) \leq f(2x+1)$
$(3x+1) \geq f(6x+1)$
$f(3)=2$

a) prove that $f(x)$ takes value $2$ arbitrary many times
b) give an example of such function

a) is quite easy but I have no idea for b) [I found this problem in some Khrabrov's paper and there was given an example of such function but I'm rather curious how to obtain a result

[quote="Myth"]$f(x)\equiv 2$.
What do you actually mean?[/quote]

That doesn't hold for negative x values.

[quote="Megus"]Maybe I translated/misunderstood something - Myth can you look here:

http://www.math.spbu.ru/user/aih/papers/selprobl.zip

problem 26 if I remember correctly
[this problem above is a modification of problem 26][/quote]
They ask for non-trivial example. Moreover, they stated that only continuous such function is $f(x)=\equiv 2$ (easy task ;) ).

Actually I thought about non-trivial case :) - what can I think other about after I saw an example they gave  :D ?

So any ideas of finding such a non-trivial example ?

I even though that it is a typo: $f(3x+1)\geq f(6x+1)$, since it is equivalent to $f(x+1)\geq f(2x+1)$, and a) is more than trivial. It is strange...

Hm, indeed  :D  so from where they have taken the function that were shown? I'm nothing good at Russian but I guess this function is shown as an example of the function which would suffice given conditions, or am I wrong?  :?

[quote="Myth"]I don't know :? I haven't found the true source of this problem.[/quote]

Problem is correct and trivial.
Of course, $f(x+1)=f(2x+1)$, but for noncontinuous function it is not imply
that f is conctant. If f is continuous at point $x=1$ then f is constant.
It's standard fact about Cauchy functional equation. Equation
$f(x+1)=f(2x+1)$ is redused to Cauchy equation after shifting, taking the
logarithm etc.


$\bigstar$ Let $a,b,c,d$ be positive real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}+2abcd=6$. Prove that:
\[a+b+c+d\le 4\]

[quote="zaizai-hoang"]$ \bigstar$ Let $ a,b,c,d$ be positive real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}+2abcd=6$. Prove that:
\[ a+b+c+d\le 4 \]
[/quote]
I think, it's wrong. :(
Try $ a=b=c=1.4$ and $ d=\sqrt{7.649536}-2.744.$ :wink:
Maybe you have meant that
$ \bigstar$ Let $ a,b,c,d$ be positive real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}+abcd=5$. Prove that:
\[ a+b+c+d\le 4 ? \]
It's true. :wink:

If $ x_{1},x_{2},...,x_{n}$ are non-negative numbers such that $ x_{1}+x_{2}+...+x_{n}=n$, then

$ (n-1)(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})+nx_{1}x_{2}...x_{n}\ge n^{2}$.

This is a particular case of the problem 18, chapter 5, in Algebraic Inequalities (page 243).
See also Proposition 3.5 in my article The Equal Variable Method from Journal of Inequalities in Pure and Applied Mathematics (JIPAM):
http://jipam.vu.edu.au/images/059_06_JIPAM/059_06_www.pdf
and
http://jipam.vu.edu.au/article.php?sid=828

[quote="zaizai-hoang"]hix, above inequality of hungkhtn :P maybe He have made mistake :P I found it in his book (Vietnamese verision  :roll: ) Sorry everybody ...[/quote]

Yes, actually a friend of mine gives this problem to me and I have not time to check right (anyway, it is in practical problem set). Also the right problem of VASC and Arqady is not hard after using the contradiction method.  :blush:  Anyway, the English book won't make this mistake, since it is checked very carefully by Gil and me. Thank you, Hoang and Arqady, VASC for finding this one.


I am trying to register an ICQ number ，but fail。
Who can help me to get an ICQ number？

please send it to my Email：thatax@sina.com

thanks a lot. :)


integration
calculus
trigonometry
calculus computations

Find the solution of the following equation

\[ {\left( {\int\limits_0^x {\frac{{3x \minus{} {x^3}}}{{1 \minus{} 2{x^2} \minus{} 3{x^4}}}dx} } \right)^\prime } \equal{} \frac{{\varepsilon \varphi x}}{{1 \plus{} {x^2}}},x \in \left( { \minus{} \frac{\pi }{6},\frac{\pi }{6}} \right)\]

hello, it is better to write
$ \left(\int_{0}^{x}\frac{3t-t^3}{1-2t^2-3t^4}\,dt\right)^'$.
Sonnhard.

Yes it was a typo! Thank you
Right one
\[ {\left({\int\limits_{0}^{x}{\frac{{3t\minus{}{t^{3}}}}{{1\minus{}2{t^{2}}\minus{}3{t^{4}}}}dx}}\right)^\prime }\equal{}\frac{{\varepsilon\varphi x}}{{1\plus{}{x^{2}}}},x\in\left({\minus{}\frac{\pi }{6},\frac{\pi }{6}}\right)\]

What does $ {\varepsilon\varphi x}$ mean?

[quote="JoeBlow"]By the fundamental theorem of calculus, $ \frac {3x \minus{} x^3}{1 \minus{} 2x^2 \minus{} 3x^4} \equal{} \frac {\epsilon \phi x}{1 \plus{} x^2}$, so $ \boxed{\epsilon \phi \equal{} \frac {x^2 \minus{} 3}{3x^2 \minus{} 1}\Leftrightarrow x \equal{} \pm \sqrt {\frac {3 \minus{} \epsilon \phi}{1 \minus{} 3\epsilon \phi}}}$.[/quote]

I think you havent found solution, since you found x in terms of x. For that i am responcible, again...meh...i kept tanx in greek language..So εφx=tanx. Sorry guys again.
So i think right one..is
\[ {\left({\int\limits_{0}^{x}{\frac{{3t\minus{}{t^{3}}}}{{1\minus{}2{t^{2}}\minus{}3{t^{4}}}}dx}}\right)^\prime }\equal{}\frac{{\tan x}}{{1\plus{}{x^{2}}}},x\in\left({\minus{}\frac{\pi }{6},\frac{\pi }{6}}\right)\]

$ \tan x \equal{} \frac {x(3\minus{}x^2 )}{1\minus{}3x^2}$, so $ x \equal{} 0$ is a solution by inspection.  We can verify that no other solutions exist on $ \left( \minus{} \frac {\pi}{6},\frac {\pi}{6}\right)$ by noting that $ \frac {d}{dx}\tan x \equal{} \sec ^2 x < \frac {4}{3}$ and $ \frac {d}{dx}\frac {x(3\minus{}x^2 )}{1\minus{}3x^2 } \equal{} \frac {3(1 \plus{} x^2)^2}{(1 \minus{} 3x^2)^2}\geq 3$ (where we've made use of the fact that  $ \frac {\pi}{6} < \sqrt {\frac {1}{3}}$).

\[ \boxed{\boxed{\tan 3\theta \equal{} \frac {3\tan \theta \minus{} \tan ^ 3 \theta}{1 \minus{} 3\tan ^ 2 \theta}}}\]

Thank you both. My solution is based on notation of Kunny. Im changin variable at calculation of I, t=tanu and then im using the identity that kunny wrote. After this i use the equation for derive the integral and find solution.


Let x, y, z be real numbers in interval (-1,1) such that $ xy\plus{}yz\plus{}zx\equal{}1$. Prove that
\[ 6 \sqrt[3]{(1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)} \le 1 \plus{} (x\plus{}y\plus{}z)^2
\]

[quote="STARS"]Let x, y, z be real numbers in interval (-1,1) such that $ xy \plus{} yz \plus{} zx \equal{} 1$. Prove that
\[ 6 \sqrt [3]{(1 \minus{} x^2)(1 \minus{} y^2)(1 \minus{} z^2)} \le 1 \plus{} (x \plus{} y \plus{} z)^2
\]
[/quote]
[hide=" my solution"]
from AM-GM
$ 6 \sqrt [3]{(1 \minus{} x^2)(1 \minus{} y^2)(1 \minus{} z^2)} \le 6 \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)$ so we have to prove
$ 6 \minus{} 2(x^2 \plus{} y^2 \plus{} z^2) \le 1\plus{} (x\plus{}y\plus{}z)^2$
$ 5 \le 3(x^2 \plus{} y^2 \plus{} z^2) \plus{} 2$
$ 1 \le x^2 \plus{} y^2 \plus{} z^2$ and this is true, because we have
$ x^2 \plus{} y^2 \plus{} z^2 \ge xy\plus{}yz\plus{}zy \equal{} 1$
[/hide]
Hope it's good :)


algebra
polynomial
complex numbers
number theory proposed
number theory

Let $ a_1,..,a_n$ be n complex numbers which satisfy that :
$ \sum_{i\equal{}1}^{n} a_1^k \in Z \forall  k \in N^*$
Prove that :  It exits $ n$ numbers $ d_1,...,d_n \in Z$ which satisfy that:
$ a_1,a_2,...a_n$ are complex roots of this equation
$ x_n\plus{}\sum_{i\equal{}1}^{n}  d_ix^{n\minus{}i} \equal{}0$

For $ k\equal{}0,1,\ldots$ let $ p_k\equal{}\sum_{i\equal{}1}^ka_i^k$ be the $ k$-th power sum and $ e_k\equal{}\sum_{I\subseteq\{1,\ldots,n\}\atop |I|\equal{}k}\prod_{i\in I}a_i$ with the convention $ \sum_{\emptyset}\ldots \equal{}0$ and $ \prod_{\emptyset}\ldots \equal{}1$. These are the elementary symmetric polynomials evaluated for $ a_1,\ldots,a_n$. The polynomial $ P(x)\equal{}\prod_{i\equal{}1}^n(x\minus{}a_i)$ has exactly the $ a_i$ as roots and can be transformed into $ P(x)\equal{}\sum_{k\equal{}0}^n(\minus{}1)^ke_kx^{n\minus{}k}$, in particular, the leading coefficient is $ 1$. Using Newton's identities we can relate the $ e_k$s with the $ p_k$s via the equations (*) $ \;ke_k\equal{}\sum_{i\equal{}1}^k(\minus{}1)^{i\minus{}1}e_{k\minus{}i}p_i$ holds. Since the $ p_i$ were assumed to be integers we must have $ k!e_k\in\mathbb{Z}$. More generally, we have for the greatest common divisors the identity $ \forall k\in\mathbb{N}: \,(p_1,p_2,\ldots,p_k)\equal{}(1!e_1,2!e_2,\ldots, k!e_k)$. Note, that for higher order power sums $ m>n$ we have $ \;0\equal{}\sum_{k\equal{}0}^n(\minus{}1)^{k\minus{}1}e_kp_{m\minus{}k}$. Multiplication of both sides with $ n!$ implies the identity $ 0\equiv n!e_np_{m\minus{}n}\bmod{n}$, i.e. if $ T\equal{}(n!e_n,n)$ we have $ \forall i\in\mathbb{N}: \,\frac{n}{T}|p_i$ which implies $ \forall k\in\mathbb{N}: \,\frac{n}{T}|k!e_k$, in particular for $ k\equal{}n$ we get $ n|T^2$ ???? now I am getting tired  :(  any hints ?

:)  . I use this lemma :
If $ n$ complex numbers $ a_1,...,a_n$  sastify $ \sum p(\sum_{i\equal{}1}^n a_i^k ) \in Z \forall k \in N^*$.($ p$ is prime) then
$ \sum_{i\equal{}1}^n a_i \in Z$

I do not understand your comment. The power sums $ \sum_{i\equal{}1}^na_i^k$ are already integers for any $ k$, so why multiplying it with primes to show that $ \sum_{i\equal{}1}^na_i$ is an integer, which we already knew before ??

Let $ E_{k,i}$ is $ k$-th power sum of $ n$ numbers $ a_1^i,...,a_n^i$
You proved  $ k! E_{k,1} \in Z$  ,  so  it's easy for you to prove :$ k!E_{k,i} \in Z \forall k \in N^* ,k \le n , i \in N^*$(*) ,isn't it ?
After you prove (*) , you can see my lemma useful :)


probability
expected value
real analysis
real analysis unsolved

It is well known that if $d \in \{1, 2\}$ then a simple symmetric random walk in $R^{d}$ will visit every point infinitely many times.

Suppose that such a random walk starts at the origin at time $t=0$. I wonder if there is any reasonable way to estimate the expected value $D(t)$, the minimum distance from the origin of all points that have not yet been visited at time $t$.

I have tried to think about this problem but honestly it is very daunting and I have not had any clever ideas. I will try to return to it soon, but I thought I would post it here so anyone who is interested can post their ideas here and enjoy the problem. :)

The case $n=2$ appears untractable. You can compare it to
[quote]the famous problem of the self-avoiding random walk. Let $x_{n}$ be simple random walk in the plane, and let $T$ denote the first time that $x_{n}$ visits a point which was occupied at some time less than $n$... the distribution of $x_{T}$ presents a problem that seems far out of reach;... the event "$x_{T}=x$" depends on the entire history of the random walk up to time $T$, and $T$ may be arbitrarily large. Even a seemingly more modest problem, namely that of finding $\lim_{n\to\infty}(P[T>n])^{1/n}$, lies beyond the horizon of the present state of the art.[/quote]
This is from "Principles of random walk" by Frank Spitzer. Granted, the book was written in the 1960s, but the self-avoiding random walk is still [url=http://mathworld.wolfram.com/Self-AvoidingWalkConnectiveConstant.html]far from being understood.[/url]

The case $n=1$ is entirely different, because it only involves $\max_{1\le n\le N}x_{n}$ and $\min_{1\le n\le N}x_{n}$ instead of the entire history of the walk.


[quote=dragon96]a+c=10 (because if it sums to 0, then both are 0)
b+c=9 (from carrying)
a+d=9 (from carrying)

a=b+1 (subtract 2nd eq from 1st)
b+1+d=9 (plug above into 3rd eq)
c=d+1 (subtract above from 2nd eq)
a+d=b+c=9 (balances)
a+b+c+d=9*2=18[/quote]

Your method is correct but there is an easier way

First we start with this just like in your strategy:
$a+c=10$ (because if it sums to 0, then both are 0)
$b+c=9$ (from carrying)
$a+d=9$ (from carrying)

Then we realize something:
$a+b+c+d=(a+d)+(b+c)=9+9=\boxed{18}$


Suppose that $n \ge 2$. Prove that \[\sum_{k=2}^{n}\left\lfloor \frac{n^{2}}{k}\right\rfloor = \sum_{k=n+1}^{n^{2}}\left\lfloor \frac{n^{2}}{k}\right\rfloor.\]

I prove a lemma (I take from a book )
Consider an inject function $ f[a,b]\to [c,d]$ and decrease.
$ \sum_{a\leq k\leq b}f(k) \minus{} \sum_{c\leq k\leq d}f^{ \minus{} 1}(k) \equal{} [b]S(c) \minus{} [d]S(a)$
$ S(x) \equal{} [x]$ if $ x\not\in Z$
$ S(x) \equal{} 0$ if $ x \equal{} 0$
$ S(x) \equal{} x \minus{} 1$ if $ x\in Z$ and different from 0.
Now consider the function :
$ f[1,n]\to\ [n,n^2]$
$ f(x) \equal{} \frac {n^2}{x}$
then $ f(x)$ is inject and decrease.
So $ \sum_{k \equal{} 1}^n[\frac {n^2}{k} \minus{} \sum_{k \equal{} n}^{n^2}[\frac {n^2}{k}] \equal{} n(n \minus{} 1)$
$ \Rightarrow \sum_{k \equal{} 2}^n[\frac {n^2}{k}] \equal{} \sum_{k \equal{} n \plus{} 1}^{n^2}[\frac {n^2}{k}]$


A bicycle wheel with a radius of 13 inches makes 6 revolutions per second. What is the speed of the bicycle in inches per second?

If $ r\equal{}13$, then $ d\equal{}26$ and $ C\equal{}\pi (26)\equal{}26\pi$ inches.  Six times this is the total distance traveled in a second so $ \omega\equal{} 6\cdot 26\pi\equal{}156\pi\mathrm{{in}/{s}}$.


Consider operator [tex]A(B,C):R^n\rightarrow R^n[/tex], defined as follows:
Take matrices [tex]B:R^n\rightarrow R^m[/tex] and [tex]C:R^m\rightarrow R^n[/tex] . For each vector [tex]x\in R^n [/tex] take [tex]y=Bx\in R^m[/tex], then rearrange its components in increasing order (vector [tex]y^*[/tex]) and then multiply by  [tex]C[/tex] to obtain [tex]z=Cy^*=A(B,C)x[/tex]
The question is: how to compute eigenvalues and eigenvectors of [tex]A[/tex] given matrices [tex]B[/tex] and [tex]C[/tex]?
Warning: I don't know solution of this problem and even not sure there is some.  I'm interested in any ideas.

Well, one thing I noticed about your * operator is that it is usually linear.  Everywhere except for a negligible set, it is locally linear, looking like a matrix with mostly 1's and 1 zero in each row and column.  It only isn't linear when two of the components of y are the same.  If we ignore B for a second, then, we can break it down into finitely many regions and compute the eigenvectors in each region through normal methods (tossing out those which don't actually lie in the region being considered).  Then we should be able to do the same thing for those subspaces where two or more components of y are equal.

The problem with this, as far as I see it, is that B makes everything more complicated.  In particular, the negligible spaces of [tex]\mathbb{R}^m[/tex] when two components are equal might come from very non-negligible subsets of [tex]\mathbb{R}^n[/tex].  An obvious example of this would be when B = 0.  If we assume that this doesn't happen, what I described above should work.  If it does happen, I'm not positive.  I'm not even sure if there is a nice way to test when it does and doesn't happen, which might also be a problem.

I don't think that subspaces on which two components of y are equal make any difference. For each y, there is matrix S(y) s.t. y*=S(y)y. There are m! different matrices S(y) corresponding to m! transpositions of m components of y. So we can take m! matrices A(y)=CS(y)B, find their eigenvectors and take only those for which matrix S(y) was actually the one we've used. If some components of y are equal, there is more than one matrix S(y), but    S(y)y is the same for all of them, so we can take any of them.
I think the problem with this approach is that you have to find eigenvectors for m! matrices, which requires too much work (and m can be much greater than n - the dimension of space in which A operates)
There is a more special question about this operator: for given m and n, what is the maximum number of eigenvalues of A? It is less then m!n, because each of m! matrices A(y) has no more than n eigenvalues. But I think the actual maximum must be smaller (at least when m is much greater than n)

The point I was making is that on certain large sets, it is linear and we can treat it as locally linear as a result.  At pointss where two components are equal, it is not locally linear, so we need to deal with them seperately.

I don't think this operator lends itself at all to any other method, because it doesn't act nicely on spaces other than the ones I described (or smaller ones than that).  I don't even know how you would begin to approach it in such a way.

How'd you get interested in it?


Explain the abnormal stability of the benzene molecule.

Note simply saying because it has a conjugated $ \pi$ system is not good enough.

Well, I don't know whether we can draw the resonating structures using Latex, so anyways.

Benzene is aromatic, there 3 pairs of electrons in alternate positions to each other. So the electrons are fully delocalized over the benzene ring. So there are 2 resonating structures which contribute equally. And there is a factor called resonance energy, which provides the extra stability.

[quote=Wikipedia]
Resonance hybrids are always more stable than any of the canonical structures would be, if they existed[1]. The delocalization of the electrons lowers the orbital energies, imparting this stability. The resonance in benzene gives rise to the property of aromaticity. The gain in stability of the resonance hybrid over the most stable of the (non-existent) canonical structures is called the resonance energy. A canonical structure that is lower in energy makes a relatively greater contribution to the resonance hybrid, or the actual picture of the molecule. In fact, resonance energy, and consequently stability, increase with the number of canonical structures possible, especially when these (non-existent) structures are equal in energy. Resonance energy of a conjugated system can be 'measured' by heat of hydrogenation of the molecule. Consider the example of benzene. The energy required to hydrogenate an isolated pi-bond is around 28.6 kcal/mol (120 kJ/mol). Thus, according to the VB picture of benzene (having three pi-bonds), the complete hydrogenation of benzene should require 85.8 kcal/mol (360 kJ/mol). However, the experimental heat of hydrogenation of benzene is around 49.8 kcal/mol (210 kJ/mol). The difference of 36 kcal/mol (150 kJ/mol) can be looked upon as a measure of resonance energy. One must bear in mind again that resonance structures have no physical existence. So, even though the term 'resonance energy' is quite meaningless, it offers an insight into how different the VB picture of a molecule is from the actual molecule itself. [/quote]

Also many a time we associate stability and reactivity. Benzene is highly immune to addition reactions under normal conditions because of the delocalization.

What about explaining this in terms of MO Theory? Also explaining the instability of Cyclobutadiene using MO Theory.

P.S. How can we tell whether molecular orbitals are degenerate or not. Also which ones are higher in energy than others.

You have six molecular orbitals: three bonding and three anti-bonding. These orbitals are to be filled with six electrons. Therefore, all electrons go to bonding orbitals, and all bonding orbitals end up filled: we thus have a closed shell configuration, similar to noble gases, which is very stable. For cyclobutadiene, there are four orbitals and four electrons: one bonding, two non-bonding, and one anti-bonding. Two electrons will go the bonding orbital, but each of the non-bonding orbitals will get one electron. Thus we would get a di-radical, a not stable species.

Has the number of nodes got anything to do with degeneracy? Or is it something else?

1) What is a node?

2) What are the degenerate orbitals?

Nodes are regions of zero electron density.

Degenerate orbitals have the same energy.

Very good.

3) What can you say about p-orbitals?

By the way this is just a general question now (not specifically referring to benzene).

$ p$ orbitals have a nodal plane (an angular node).

All three p orbitals have one node and are degenerate.

4) What about d orbitals? f orbitals?

$ 2$ and $ 3$ nodal planes respectively I think (can't remember too much about $ f$-orbitals).

Therefore it seems that orbitals with the same number of nodal planes are degenarate.

This is true of both molecular/atomic orbitals?

As an exercise, consult an Organic Chemistry book on the Conjugated Dienes chapter and see the molecular orbitals for butadiene.

Actually do you mind if I pm you? I want to ask some stuff about that problem I posted about the geometry of $ \ce{H3\plus{}}$ but I don't want to discuss it openly (not until others have tried anyway).

Be my guest. But don't forget its (almost) always best to have several opinions about a subject.


A finite set of positive integers is called [i]isolated [/i]if the sum of the numbers in any given proper subset is co-prime with the sum of the elements of the set.

a) Prove that the set $A=\{4,9,16,25,36,49\}$ is isolated;

b) Determine the composite numbers $n$ for which there exist the positive integers  $a,b$ such that the set
\[ A=\{(a+b)^2, (a+2b)^2,\ldots, (a+nb)^2\}\] is isolated.

Yet another very easy one we chose to ignore :).

The sum of all the numbers at point (a) is a prime, so that's that.

For (b), if $n>6$ or if $n=4$ (the only composite below $6$), then there is a divisor $d>1$ of $n$ which divides the sum of all the elements in the set, which is $S=na^2+n(n+1)ab+\frac{n(n+1)(2n+1)}6b^2$. On the other hand, at least one of $a+b,a+2b,\ldots,a+nb$ is divisible by $n$ and thus by $d$, meaning that it's not coprime with the sum $S$. The answer must be $n=6$.

Not that the problem is not trivial, but could I know dear Grobber why at least one of the n numbers must be a multiple of n?

Sorry about that, I rushed into it, and that's just yet another stupid mistake of mine. :)

If $b$ is not coprime with $\frac n{(n,6)}$, then take $d=(b,\frac n{(n,6)})$. We then have $d|ka^2+k(k+1)ab+\frac{k(k+1)(2k+1)}6b^2=(a+b)^2+\ldots+(a+db)^2$, and we also have $d|$ the sum of all the numbers.

On the other hand, if $b$ is coprime with $k=\frac n{(n,6)}$, one of $a+b,a+2b,\ldots,a+kb$ is divisible by $k$.

Is it Ok now?

Ah, Grobber! You are so quick. I was writing my correction to you post :( Thus my work is vain...

Yes, grobber, but you see... obvious things always produce bad things!

Don't blame Grobber to much, Harazi!  ;)  :D

Hmm.. I don't know what that means.. I'm sure I'm supposed to learn a lesson from all of this though :D. Sorry again  :blush: .


how can I change row width in a table? i tried putting \vspace{0.25 in} with each entry, but it spaced the entire table out.

Use p{0.25in} where you would normally put l,c or r eg \begin{tabular}{p{0.25in}cc}

sorry, but that made the columns wider.
i meant to ask row height.
without adding anything extra, the table comes out with small rows. how do I make them taller?

Before the table environment use \renewcommand{{\arraystretch}{[i]factor[/i]}.
[i]factor[/i] is the multiplying factor for the height. So 1 means keep it as it is, 2 means double the height, 1.5 means multiply the height by 1.5 etc.


Not complete but one of the conditions [hide="is"]$k$ has to be divisible by $3$[/hide] if solutions are possible

[quote="José"]Find all numbers $k$ such that $2^k=k-1$[/quote]
[hide]
Graphing:
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=762[/img]
We see that there will be "no solution" because the exponential outruns the line very quickly.
[/hide]

[hide]I'm working now with integers k and a.

Its clearly that k=2a+1

2^(2a)=a

Its easy to see that it hasnt any solutions.[/hide]

umm sry mods but this solution uses calc a bit...

[hide]

let $f(k) = 2^k - k + 1$, then $f'(k) = \ln(2) \cdot 2^{k} - 1$

we want to find the minimum of this function, so we set $f'(k) = 0$ (it's easy to check that this is indeed a minimum and not a maximum)

hence $\frac{1}{\ln (2)} = 2^{k}$

which gives ${k = -\log_2 {\ln 2}}$

substituting this back in we get:

${{2^{-\log_2 {\ln 2}}} + \log_2 {\ln 2}} + 1$ as our $y$ value

now, we know that ${2^{-\log_2 {\ln 2}}} >0$, and that ${1 > -\log_2 {\ln 2}}$, so the $y$ value is positive

since $f(k) > 0$ for all real $k$, there are no real solutions in $k$ to our original equation

[/hide]

Clearly $k$ is odd, so let $k=2k_1+1$. The equation becomes $2^{2k_1+1}=2k_1$ or $4^{k_1}=k_1$. Done.

[quote="gotztahbeazn"]ummm and wats a good way to show that $4^k = k$ has no solutions?[/quote]

this problemwas practically on our last arml homework  :P

$4^k=k \implies 4=k^{\frac{1}{k}}$ but $f(k)=k^{\frac{1}{k}}$ has it's max value at $(k,f(k))=(e,e^{\frac{1}{e}})$ so no solution

[quote]ummm and wats a good way to show that $4^{k_1} = k_1$ has no solutions?[/quote]

$k_1$ must be a 4th power, so let $k_1=4^{k_2}$. Then we get ${4^{4^{k_2}}}=4^{k_2}$. Take log_4 of both sides to get $4^{k_2} = k_2$. So if we let $k_2=4^{k_3}$ we get $4^{k_3}=k_3$. Infinite ascent? :D

[quote="José"]and with non-integrers?[/quote]In your original post, you ask for "numbers" that satisfy the equation, and then you ask for "non-integers."  gotztahbeazn proved that there are no real solutions, but note that there are an infinite number of nonreal solutions.  Try these, for example (they're approximations, by the way):
$3.49681121009787-11.00914826965604i$
$4.3529549037675-20.1578268537269i$
$4.8839423109013-29.2700164531399i$
The complete (exact) solution is
$k=1-\frac{W_n(-2\ln{2})}{\ln{2}}$, where $n$ can be any integer and $W_n(z)$ is the generalized Lambert function.  Here's a good link about the generalized Lambert function: [url]http://functions.wolfram.com/ElementaryFunctions/ProductLog2[/url].


[quote="Interval"]If the circumference of a circle is doubled, the diameter of the circle

(1) remains the same
(2) increases by 2
(3) is multiplied by 4
(4) is doubled[/quote]
[hide]The circumference is $\pi d$ so then the new circle's circumference would be $2\pi d$ which is doubled (4)[/hide]

This is WAY to easy for HSB. put stuff like this in classroom math.

Way to "easy" for you but math note-taking for me.

yeah, but it still belngs in classroom math.

it probably should but mr. ruszyk told my to keep all the non- physics related stuff in here so oh well...

the circcumference is directly porportional to the radius and the diameter.

[quote="jli"]it probably should but mr. ruszyk told my to keep all the non- physics related stuff in here so oh well...[/quote]
Not that I'm questioning his authority or anything, but do these questions seriously belong in HSB? :maybe:

[quote="Quevvy"][quote="#H34N1"][hide=Classroom Math]
$d=\pi C$
$C=\frac d\pi$
$2c=2\frac d\pi\implies 4$[/quote]
doesn't $C=d\pi$?[/quote]

Yeah it does


[b]Problem.[/b] Let $f(x)=x^{4}+x^{3}+x^{2}+x+1$.  Find the remainder on dividing $f(x^{5})$ by $f(x)$.

Are you sure it's $f(x^{5})$? If it was $f(x^{6})$ then it would be a nice problem; as it is right now I think you end up with a very messy remainder.

There is a very clear solution with a very clean remainder.  [i]Clean[/i] being the antonym of [i]messy[/i] defined by you. :P

boxedexe wrote:There is a very clear solution with a very clean remainder.  Clean being the antonym of messy defined by you.

Well now you editted f(x) (at least from what it was at lunchtime)

anyways

[hide]

, deg r < 4

roots of f(x) are , , , so we have , but then r(x) - 5 has four roots and thus it's not degree four GASP

so r(x) - 5 is zero identically aka r(x) = 5 is our remainder.

[/hide]

Sorry about that.  I edited it a few minutes after I posted it. Also, your answer is correct.

[hide="Polynomial Modulus, One More Time"] $x^{5}\equiv 1 \bmod f(x) \implies f(x^{5}) \equiv f(1) = 5 \bmod f(x)$ [/hide]

[quote="me@home"]Does that idea work in general? the only reason you can use it is that $f(x)$ factors as $\frac{x^{5}-1}{x-1}$, unless I am missing something.[/quote]

The only reason the other solutions turns out so slick is cause f(x) is roots of unity too

[quote="me@home"]Does that idea work in general? the only reason you can use it is that $f(x)$ factors as $\frac{x^{5}-1}{x-1}$, unless I am missing something.[/quote]

As I state in the WOOT topic, in general if $A(x) \equiv B(x) \bmod P(x)$ then $Q(A(x)) \equiv Q(B(x)) \bmod P(x)$ for arbitrary polynomials $A, B, P, Q$.


$ (a_{n})_{n\geq 1}$ be a sequence of positive integers such that $ s(a_{n})\geq n\forall n\geq 1$. Prove that $ \frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}}<3.2\forall n$. Can we replace $ 3.2$ by $ 3$? Where $ s(k)$ is sum of the digits of $ k$ in base $ 10$.

[quote="N.T.TUAN"]$ (a_{n})_{n\geq 1}$ be a sequence of positive integers such that $ s(a_{n})\geq n\forall n\geq 1$. Prove that $ \frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}}< 3.2\forall n$. Can we replace $ 3.2$ by $ 3$? Where $ s(k)$ is sum of the digits of $ k$ in base $ 10$.[/quote]

It is not too hard to see that if $ S_{n}\equal{}\frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}}$ then $ S$ is maximized for the sequence $ (a_{i})_{i}\equal{} 1,2,3,4,5,6,7,8,9,19,29,\ldots,99,199,299,\ldots,999,\ldots$.  I.e.
\[ S\le\sum_{1\le k\le 9}\frac{1}{k}\plus{}\frac{1}{10k\plus{}9}\plus{}\frac{1}{10^{2}k\plus{}99}\plus{}\frac{1}{10^{3}k\plus{}999}\plus{}\ldots\]
\[ \le\sum_{1\le k\le 9}\bigg(\frac{1}{k}\plus{}\frac{1}{10k\plus{}9}\plus{}\frac{1}{10^{2}k\plus{}99}\bigg)\plus{}\sum_{1\le k\le 9}\frac{1}{k}\bigg(\frac{1}{10^{3}}\plus{}\frac{1}{10^{4}}\plus{}\ldots\bigg).\]
In the last line, the first sum is $ < 3.1$ (and also $ >3$), and $ \sum_{1\le k\le 9}\frac{1}{k}< 2.83$ yields the second is $ <\frac{2.83}{900}$ when summing the geometric series.  Thus, $ S < 3.2$.  Moreover, we have shown that there exists a sequence such that $ S > 3$.


I was looking through my school calender and I noticed a "Terra Nova test" will be given to all juniors in February. What exactly is this? Is it something like the PLAN test? I've never heard of it  :?

Google didn't really provide any speicifc information about it.

I took a Terra Nova test in 4th grade, but I think it's been replaced by the NJASK in New Jersey...

According to [url=http://www.ctd.northwestern.edu/mats/eligibility.html]here[/url], it's a [quote]nationally normed, standardized achievement, aptitude, or abilities test[/quote]  If it's like the tests I've taken before (which I'm pretty sure it is), it's basically a test where you answer a bunch of questions on different subjects, and it tells you how smart you are compared to other people in the same grade.


The FAQ states transcripts of prior classes are available. How do I get that. I missed one of my class last week, due to a game!

At the left of your screen, if you use the AoPS skin, is a list of links, like Bookstore, etc. Click "My Classes". This should direct you to a list of the classes you're currently taking. Click the class that you wish to find the transcript for. You should then be able to find the transcript.


What percent of all three-digit numbers contain the digit 5 at least once

[hide="Hint"] How many 3-digit numbers [b]don't[/b] contain 5? [/hide]

These are'nt doubts. I know the answers to them

[hide]
There are 999-100+1=900 3-digit numbers.
In each 100 set, there are 19 integers with at least one 5 in them.
But, in the 500 set, all of the integers have at least one 5.
$\frac{100+19+19+19+19+19+19+19+19}{900}=\frac{252}{900}$

So, the answer is $28$ percent.[/hide]

[quote="Klebian"][quote="bpms"][quote="pascal_1623"][hide]
There are 999-100+1=900 3-digit numbers.
In each 100 set, there are 19 integers with at least one 5 in them.
But, in the 500 set, all of the integers have at least one 5.
$\frac{100+19+19+19+19+19+19+19+19}{900}=\frac{652}{900}$

So, the answer is $72.\overline{4}$ percent.[/hide][/quote]
[hide]$8*9*9=648$ that don't contain a five...[/hide][/quote]

Well, the only mistake he made was that he put 652 instead of 252. (Adding error?)[/quote]
I was just pointing out a better way...


A lucky number is a positive integer which is 19 times the sum of its digits. How many different lucky numbers are there?

 I badly need help.. Thanks in advance i really dont have any idea how to solve this.

[hide="Hint"]Can such a number have $ 1$ digit? $ 2$ digits? $ 3$ digits? $ \ge4$ digits? If it contains two digits, call them $ a$ and $ b$ and obtain a Diophantine in two variables. Does this equation have solutions? (Consider the number modulo $ 9$.) If it contains three digits, we can obtain a Diophantine in three variables. We can do casework on the value of the leftmost digit, for instance.[/hide]

[quote="rahul1993"][hide]CASE 1 assuming it is a two digit number
so for positive integers a,b we have
19a+19b=10a+b
-9a=18b LHS is positive but RHS is negative
CASE 2 assuming it is a 3 digit number
so for positive a,b,c we have
19a+19b+19c=100a+10b+c
9a=b+2c ,            $ 0 \le a,b,c \le 9$
so only solutions we get are
(a,b,c)=(1,1,4) , (1,3,3) , (1,5,2) , (1,7,1) , (2,2,9) , (2,4,7) , (2,6,5) , (2,8,3)
CASE 3 considering it is a 4 digit number
so for positive integers a,b,c,d we have
19a+19b+19c+19d=1000a+100b+10c+d
109a+9b=c+2b ,           $ 0 \le a,b,c,d \le 9$
we can see for a $ \ge 1$ LHS > RHS so we know a=0 for which we get same solution as CASE 2
now you can see for any number with $ \ge 3$digits LHS > RHS
so we have 8 lucky numbers[/hide][/quote]

[hide="Correction"]The reasoning here is correct, but you fail to consider $ b\equal{}0$ or $ c\equal{}0$. Indeed, you missed the solution $ (1,9,0)$ where $ a\equal{}1$ and your solutions with $ a\equal{}2$ are incorrect: the solution triplets should be $ (2,0,9)$, $ (2,2,8)$, $ (2,4,7)$, $ (2,6,6)$, $ (2,8,5)$. You also missed the case $ a\equal{}3$ where the only solution is $ (3,9,9)$. Therefore, we have the complete solution set:

$ (a,b,c) \in \{(1,1,4), (1,3,3), (1,5,2), (1,7,1), (1,9,0), (2,0,9), (2,2,8), (2,4,7), (2,6,6), (2,8,5), (3,9,9)\}$ There are $ 11$ such numbers.[/hide]


At 10:03 p.m., megan said:

it is 8 hours and 12 minutes after 6 hours and 54 minutes before the end of my 2 hour and 31 minute long appointment.

at what time did her appointment begin?

explain your answer.

Sorry, you get x + 4 hours - 11 mins = 10.03 pm. Not sure what i did there.


inequalities
trigonometry
inequalities proposed

Giving $3$ positive real $x,y,z$ satisfy:$xy+yz+xz=3$.Prove that:
              $\sum \frac{1}{\sqrt{x^2+3}+x} \leq 1$  :)  :)

Sign reversed.
Subsitute $x=\sqrt{3}\cot A,y=\sqrt{3}\cot B,z=\sqrt{3}\cot C$, then the desired inequality is transformed into\[ \tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\ge\sqrt{3}  \]Consider $f(x)=\tan\frac{x}{2}$, $f''(x)=\frac{1}{2}\sec^2\frac{x}{2}\tan\frac{x}{2}\ge0$. By Jensen's Inequality, we are done.
Alternatively, we may consider $\sum\tan\frac{A}{2}\tan\frac{B}{2}=1$. Then the desired inequality follows from $(x+y+z)^2\ge3(xy+yz+zx)$.


inequalities
geometry
inradius
trigonometry
geometry unsolved

(R.Pirkuliev) Prove the inequality
\[ \frac1{\sqrt {2\sin A}} \plus{} \frac1{\sqrt {2\sin B}} \plus{} \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {p}{r}},
\]
where $ p$ and $ r$ are the semiperimeter and the inradius of triangle $ ABC$.

[hide="partial solution"]We simplify the LHS:

$ \dfrac{1}{\sqrt {2\sin{A}}} \plus{} \dfrac{1}{\sqrt {2\sin{B}}} \plus{} \dfrac{1}{\sqrt {2\sin{C}}} \equal{} \dfrac{1}{\sqrt {2}}(\dfrac{1}{\sqrt {\sin{A}}} \plus{} \dfrac{1}{\sqrt {\sin{B}}} \plus{} \dfrac{1}{\sqrt {\sin{C}}})$

$ \equal{} \dfrac{1}{\sqrt {2}}\cdot \dfrac{\sqrt {\sin{A}}\sin{B}\sin{C} \plus{} \sqrt {\sin{B}}\sin{C}\sin{A} \plus{} \sqrt {\sin{C}}\sin{A}\sin{B}}{\sin{A}\sin{B}\sin{C}}$

$ \equal{} \dfrac{1}{\sqrt {2}}\cdot \dfrac{\sqrt {\sin{B}\sin{C}} \plus{} \sqrt {\sin{C}\sin{A}} \plus{} \sqrt {\sin{A}\sin{B}}}{\sqrt {\sin{A}\sin{B}\sin{C}}}$

Now we must prove that

$ \dfrac{1}{\sqrt {2}}\cdot \dfrac{\sqrt {\sin{B}\sin{C}} \plus{} \sqrt {\sin{C}\sin{A}} \plus{} \sqrt {\sin{A}\sin{B}}}{\sqrt {\sin{A}\sin{B}\sin{C}}}\leq \sqrt {\dfrac{p}{r}}$

Since everything is always positive, we can cross-multiply to get

$ \sqrt {r}(\sqrt {\sin{B}\sin{C}} \plus{} \sqrt {\sin{C}\sin{A}} \plus{} \sqrt {\sin{A}\sin{B}})\leq \sqrt {2p\sin{A}\sin{B}\sin{C}}$

We multiply both sides by $ \sqrt {s}$ where $ s$ is the semiperimeter to get

$ \sqrt {[ABC]}(\sqrt {\sin{B}\sin{C}} \plus{} \sqrt {\sin{C}\sin{A}} \plus{} \sqrt {\sin{A}\sin{B}})\leq p\sqrt {\sin{A}\sin{B}\sin{C}}$[/hide]

I think I recall something about [ABC] and the product of the sines of the angles, but I don't remember what it was. Feel free to correct me on that matter.

this problem is much easier than that, the inequality is equivalent to $ \sum\dfrac{1}{\sqrt a}\leq \sqrt {\dfrac{s}{Rr}} \equal{} \dfrac{2s}{\sqrt {abc}}$.

after multiplying it follows that the inequality is equivalent to $ \sum\sqrt {bc}\leq 2s$, which is well-known... (i mean, it follows from the well-known fact $ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx$)

$ \sum_{} {\frac {1}{\sqrt {2sinA}} \leq {\sqrt \frac {p}{r}}}$ $ \Leftrightarrow$ $ \sum_{} {\sqrt {\frac {r}{2sinA}}} \leq \sqrt p$ $ \Leftrightarrow$ $ \sum_{} {\sqrt {\frac {bc}{2(a + b + c)}}} \leq \sqrt {\frac {(a + b + c)}{2}}$
$ \Leftrightarrow$ $ \sum_{} {\sqrt {bc}} \leq a + b + c$ $ \Leftrightarrow$ $ \sum_{} {{(\sqrt a - \sqrt b)}^2} \geq 0$ [b]QED[/b]

We know that $ S = \frac {1}{2}ab\sin C \Longrightarrow 2\sin C = \frac{4S}{ab} $ from here follows that:
$ \sqrt {\frac{ab}{4S}} + \sqrt{\frac{bc}{4S}} + \sqrt{\frac{ac}{4S}}$ $=$ $\frac{\sqrt{ab} + \sqrt{bc} +\sqrt{ac}}{\sqrt{4p\cdot r}}$
$\sqrt{ab} \leq \frac{a+b}{2}$ ,$\sqrt{bc} \leq \frac{b+c}{2}$ and $\sqrt{ac} \leq \frac{a+c}{2}$ , because $ a, b, c $ are the sides of triangle.
Now adding these 3 relations we see:
$\sqrt{ab} + \sqrt{bc} + \sqrt{ac} \leq a+b+c = 2p $ , but $ \frac{2p}{\sqrt{4p\cdot r}} = \sqrt{\frac{p}{r}} $

Ok I am using $s$ as semiperimeter because $p$ is commonly used for perimeter and that confuses me :P. Our desired is $$\frac1{\sqrt {2\sin A}} + \frac1{\sqrt {2\sin B}} + \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {s}{r}}$$ Recall that Extended Law of Sines tells us that $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$ which means that $$\frac{1}{2\sin A}=\frac{R}{a}$$ and similarly. So it follows that our desired inequality is $$\sum_{cyc} \sqrt{\frac{R}{a}}\le \sqrt{\frac{s}{r}}$$ so $$\sqrt{R}(\sum_{cyc}\frac{1}{\sqrt{a}})\le\sqrt{\frac{s}{r}}$$ which tells us $$\sum_{cyc} \frac{1}{\sqrt{a}}\le\sqrt{\frac{s}{Rr}}$$ We want to relate the circumradius, inradius, semiperiemter and side lengths...aha! We can use the fact that $$sr=[ABC]=\frac{abc}{4R}$$ (this equality statement is easy to prove, I can prove it if anyone asks for it). Thus $Rr=\frac{abc}{4s}$, we can plug this into our inequality we just got to get $$\sum_{cyc} \frac{1}{\sqrt{a}}\le\sqrt{\frac{s}{Rr}}=\frac{2s}{\sqrt{abc}}$$ So it suffices to show that $$\sum_{cyc} \sqrt{ab}\le 2s$$ We know that $s=\frac{a+b+c}{2}$ so we can substitute this in to get that our desired is now $$\sum_{cyc} \sqrt{ab}\le a+b+c$$ and this is true by trivial inequality after rearranging to $\frac{(\sqrt{a}-\sqrt{b})^2+(\sqrt{a}-\sqrt{c})^2+(\sqrt{c}-\sqrt{b})^2}{2}\ge0$ so we are done.


advanced fields
advanced fields unsolved

Let f be usc and less than infty on [a,b]. Show that there exists continuous f_k on [a,b] s.t. f_k converging decreasingly to f.

Thanks again for your help.

The question was already answered in your previous almost identical post.


The sum of infinite terms of the series

$\frac{1}{(1+a)(2+a)}+\frac{1}{(2+a)(3+a)}+\frac{1}{(3+a)(4+a)}+.....$
where a is a constant is.....
(Sorry if this type of sum's become too old :roll: )

[hide]$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

change each term in sum into this form and every thing cancels out except for

$\frac{1}{1+a}$[/hide]

first way: telescoping

second way: first factorize the first two terms in $\frac{2}{(1+a)(3+a)}$, then factorize this last one and the third into $\frac{3}{(a+1)(a+4)}$...at least you will obtain $\frac{n}{(a+n+1)(a+1)}$ which tends to $(1+a)^{-1}$

(i did the first solution when i first saw the exercise, years ago...looking to it better, then, on the first step i noticed the sum of fraction could be an easier way, three months later :P )


μια πολυωνυμική  για θαλή γ λυκείου...

Βρείτε όλα τα πολυώνυμα $ p(x)$ ώστε $ x\cdot p(x\minus{}1)\equal{}(x\minus{}26)\cdot p(x)$ για κάθε πραγματικό $ x$.

Tο θέμα έχει ξαναμπει εδώ:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1179691412&t=177624


geometry
circumcircle
conics
geometric transformation
reflection
homothety
ellipse

Let $ ABCDEF$ be a cyclic hexagon and point $ P$. Let $ A',B',C',D',E',F'$ be circumcenter of triangles $ PAB,PBC,PCD,PDE,PEF,PFA$. Prove that $ A'D',B'E',C'F'$ are concurrent.

[b]Lemma:[/b] Let $ P$ be a point on the plane of a circle $ (O,r).$ $ Q$ is a variable point on $ (O)$ and $ \mathcal{L}$ is the perpendicular line to $ PQ$ through $ Q.$ Then the envelope of $ \mathcal{L}$ is a conic $ \mathcal{H}.$

[b]Proof:[/b] Assume that $ P$ lies inside $ (O)$  and let $ P'$ be the reflection of $ P$ about $ Q.$ $P'$ lies on the homothetic circle $ (O')$ of $ (O)$ under the homothety with center $ P$ and coefficient $ 2.$ $ \mathcal{L}$ becomes the perpendicular bisector of $ PP'$ and let $ K\equiv \mathcal{L} \cap O'P'.$ We have $ PK \plus{} KO' \equal{} P'K \plus{} KO' \equal{} 2r \equal{} \text{const}.$ Thus, $ K$ moves on the ellipse $ \mathcal{H}$ with foci $ P,O'$ and center $ O.$ Since $ \mathcal{L}$ is the external bisector of $ \angle PKO',$ it follows that $ \mathcal{L}$ is tangent to $ \mathcal{H}$ at $ K.$
 ______________________________________________

Hexagon $ A'B'C'D'E'F'$ is homothetic to the hexagon formed by the perpendicular lines $ \mathcal{L}_1 , \mathcal{L}_2 , \mathcal{L}_3 , \mathcal{L}_4 , \mathcal{L}_5 , \mathcal{L}_6$ to $PA,PB,PC,PD,PE,PF$ through its vertices under the homothety with center $ P$ and coefficient $ \frac {_1}{^2}.$ From the the above lemma, the hexagon bounded by $ \mathcal{L}_1 , \mathcal{L}_2 , \mathcal{L}_3 , \mathcal{L}_4 , \mathcal{L}_5 , \mathcal{L}_6$ is circumscribed in a conic $ \mathcal{H},$ thus $ A'B'C'D'E'F'$ is also circumscribed in a conic $ \mathcal{H}'.$ By Brianchon's Theorem, the diagonals $ A'D',B'E'$ and $ C'F'$ of the tangential hexagon $ A'B'C'D'E'F'$ concur.

Consider the inversion $I$ with center $P$, power $\omega$.
$I$: $A\mapsto A'', B\mapsto B'', C\mapsto C'', D\mapsto D'', E\mapsto E'', F\mapsto F''.$
$(PAB)\mapsto A''B'', (PBC)\mapsto B''C'', (PCD)\mapsto C''D''$, $(PDE)\mapsto D''E'', (PEF)\mapsto E''F'', (PFA)\mapsto F''A''.$
Let $A_1, B_1, C_1, D_1, E_1, F_1$ be the reflections of $P$ wrt $A''B'', B''C'', C''D'', D''E'', E''F'', F''A''$, respectively.
Then $A'\mapsto A_1, B'\mapsto B_1, C'\mapsto C_1, D'\mapsto D_1, E'\mapsto E_1, F'\mapsto F_1.$
So $A'D', B'E', C'F'$ are concurrent iff 3 circles $(PA_1D_1), (PB_1E_1), (PC_1F_1)$ are coaxal. $(*)$
Since $A, B, C, D, E, F$ are concyclic we get $A'', B'', C'', D'', E'', F''$ are concyclic. By Pascal's theorem, the intersections of $A''B''$ and $D''E''; B''C''$ and $E''F''; A''F''$ and $C''D''$ are collinear, which follows that the centers of 3 circles $(PA_1D_1), (PB_1E_1), (PC_1F_1)$ are collinear. Therefore $(*)$ is true. We are done.

[quote=daothanhoai]A generalization in file attachment[/quote]

See the topic [url=http://www.artofproblemsolving.com/community/c6h626275]Six-points circle and one point problem[/url] (Part [b] B[/b]) for [b]1[/b] and my post at the same topic for [b]2[/b] :)


I hope that $A,B,C$ are angles of a triangle :)
In what follows "all" the sums are cyclic.

$\sum sinA =\frac{s}{R}$. Moreover $\sum cosAcosB=\frac{r^2+s^2-4R^2}{4R^2}$ so it suffices to prove that $2Rs \geq \sqrt{3}(r^2+s^2-4R^2)$.
If $r^2+s^2-4R^2 \leq 0$ the ineq follows instantly. So suppose $r^2+s^2-4R^2 \geq 0$. Squaring the above inequality we get $4R^{2}s^{2} \geq 3(r^2+s^2-4R^2)^{2}$. Now using the well known result $s^2 \leq 4R^2+4Rr+3r^2$ the ineq transforms into $R^2s^2 \geq 12r^2(R+r)^2$. The rest follows by the easy to prove inequalities : $s^2 \geq 27r^2$ and $3R \geq 2R+2r$

Here is my solution:Put [tex]A=\frac{C'+B'}{2},B=\frac{A'+C'}{2},C=\frac{A'+B'}{2} [/tex]then [tex]A'+B'+C'=180[/tex]
And the giveven ineq becomes:
[tex]\sum sin(\frac{A'+B'}{2})\geq 2\sqrt{3}(\sum sin\frac{A'}{2}sin\frac{B'}{2})[/tex]
Applying the following resuilt:
[tex]\sum sin(\frac{A+B}{2})\geq \sum sinA\geq 2\sqrt{3}(\sum sin\frac{A}{2}sin\frac{B}{2})[/tex]

Yes, but I don't understand ineq:[tex]s^2\le 4R^2+4Rr+3r^2[/tex].
Thank you!

what exactly don't u understand?why it holds?

[quote="keira_khtn"]Here is my solution:Put $ A\equal{}\frac{C'\plus{}B'}{2},B\equal{}\frac{A'\plus{}C'}{2},C\equal{}\frac{A'\plus{}B'}{2}$then $ A'\plus{}B'\plus{}C'\equal{}180$
And the giveven ineq becomes:
$ \sum sin(\frac{A'\plus{}B'}{2})\geq 2\sqrt{3}(\sum sin\frac{A'}{2}sin\frac{B'}{2})$
Applying the following resuilt:
$ \sum sin(\frac{A\plus{}B}{2})\geq \sum sinA\geq 2\sqrt{3}(\sum sin\frac{A}{2}sin\frac{B}{2})$[/quote]

I think this proof is false as A',B' and C' can be negative.

Am I right?

Thank you very much.


I have scored 2250 on the SAT reasoning test (Reading 760, writing 690). I live in India. But might I be excused from taking the TOEFL?( I want to apply to the UCs ,Caltech and perhaps CSULB)

How do the colleges you plan to apply to define the group of applicants who must take the TOEFL?

Caltech says that they encourage but do not require students from other countries to take the ToEFL. I am not sure about the UCs. But I think they require us to take the ToEFL. I hope an American citizenship and a good SAT score are reasons good enough for me to skip the ToEFL. The $150 fee is not a small amount, especially when your family doesn't earn in dollars.

with a 760 in reading, you definetely don't have to take the TOEFL. I think a 550 or something like that exempts you, so you're okay. Good job by the way.

Does a 550 really exempt you ?

I got 1900 on the SAT 1 (Math 800, Reading 580, Writing 520)
I think I can do better.
I am wondering if I should retake the SAT or go for TOEFL instead. I am trying for MIT, Duke, Caltech and some others.

do yourself a favor and take the SAT again. If improve your verbal and writing scores get the Barrons study guide.


Given are a triangle $ABC$, the its incircle $w$ and the points: $D\in BC\cap w$, $E\in CA\cap w$, $F\in AB\cap w$; $L\in AD,\ FL\parallel BC$; the midpoints $M$, $N$ of the segments $[FE]$, $[FD]$. Prove that $L\in MN$.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=74386

Hello Virgil !

  Who told you that it was a classic problem with level = ** ?

   T.Y.

   M.T.

It is my opinion. I have the right to it (by the way - the rights of the man).


among 12 people,in every 9 people there are always 5 people who pairwise know.prove that there are at least 6 people who pairwise know.
I have read it somewhere before...

[quote="a_vakilian"]i think it is Ramsy[/quote]
yes i think with ramsey,we can prove it easily. :D

[quote="Katrina"]Your signature is true,VN defeated US army in 1973,USwas so disgraced,Hochiminh is the greatest man in the world.why was US so beligerent?[/quote]

Please don't discuss this kind of things here!!

this question has been seen already:
http://www.artofproblemsolving.com/Forum/post-356581.html#p356581

and just saying Ramsey theory doesn't help-that is not a magic word!


Here are the facts

-there is a village, at least 10 people
-everyone in the village has one cat
-1 cat = 1 person
-at least one cat is sick
-if a cat is sick you have to kill it, but you cant look at your own cat
-people do not communicate, only meet in the center of the village each day and look at others cats
-cant kill anothers cat that you think is sick
-you have to kill your cat the moment you think its sick
-on the tenth night, people start killing their cats
-at least 1 cat is sick

the question is, "how many cats were sick"?

the [url=http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/]blue-eyed islanders problem[/url] sounds so much better when you write it in sentences.

sorry for the lack of sentances but I am struggling with this problem, can anyone offer any help or a solution I can work with?

click on the link in my post.  alternately, think inductively.

John123,

use more clearer language.  Forget about your punctuation.

For instance, "everyone in the village has one cat" COULD mean
"All the villagers are sharing a total of one cat (that presumably
the one cat wanders around the village)," but you don't want to
allow that meaning in this case.

And instead of this equality "1 cat = 1 person," you meant the
equivalent of "There's 1 cat per person and 1 person per cat,"
I take it.

sorry

everyone in the village has one cat, so if there are 10 people, there are 10 cats. Also, every person has at least and no more than 1 cat.

is it 9?

because at least 1 cat is sick , thats where i got it from

i shall solve a similar problem as a hint/solution

at a party of logicians, each has a star taped to his back. Each logician knows that he has a star, and he knows that it is either gold or blue, and everyone can see everyone elses back(the star). logicians are not allowed to communicate the color of any star. Each person knows that there is at least gold star, and the host of the party announces that anyone who can prove that they have a gold star(assuming all the logicians use perfect logic, and have not drunk to much...) will win a prize. he announces "anyone who has a gold star shall win a prize!" no one claims it. he announces this repeatedly, and on the 8th time, several logicians claim their prize. how many logicians had a gold star?

well, first, if i were a logician and did not observe a gold star on someone else's back, i could conclude i have it since there is at least one. furthermore, if there were exactly one gold star in the crowd, the person with it would be able to prove that they have it. since no one claims the prize the first time the host makes an announcement, there are at least 2 gold stars. the host makes an announcement the second time. if there were two gold stars, one of the people with them would see only 1, and knowing there are at least 2 would claim his prize, but since no one claims the prize we know there are at least 3 gold stars, and so on. if it takes n announcements, there are n gold stars.


vector
linear algebra
matrix
superior algebra
superior algebra unsolved

The problem is to prove linear independency on a group of n, n-vectors.
Is it sufficient to take those n-vectors and place them as the rows of a matrix,
forming a matrix A, nxn. Now in solving the homogenous system Ax=0, one need only find one solution aside from the trivial to prove dependence.

I stemmed an argument that you only need to take the determinant of A to determine if there is any non-trivial solution. If det(A) = 0, there is a non-trivial solution, if det(A)!=0 there is only the trivial solution.

If the above is in fact true, I was then asked the question:

so if the vector space is n-dimensional, and n goes to infinity, then how do you find the determinant?

Is such a thing even defined?
And is my argument for a finite-dimensional proof for linear independence still valid?

well this criterion is well-known, sure it's true.

but what do you mean that for a n-dim. vector space n tends to infinity? if n is not fixed, what should be the dimension of your vector space? if you look for determinants for infinite matrices resp. endomorphisms of infinite-dimensional vector spaces, you could type "infinite determinants" into google ;). hitherto, no purely algebraic approach seems to be known.


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

[quote="Stiger"]Let $ G$ be a finite group.
Let $ H$ be a subgroup of $ G$ & $ K$ a normal subgroup of $ G$.
If $ (G: K)$ and $ |H|$ are relatively prime, then $ H$ is a subgroup of $ K$[/quote]

$ K$ is normal and then $ HK< G$ and $ |HK| \equal{} |H| \plus{} |K| \minus{} |H \cap K|$

Let's consider the projection of $ HK$ in $ G/K$. Then $ G/K > \pi(HK)$ and $ |\pi(HK)|\mid [G: K]$. But $ |\pi(HK)| \equal{} |H| \minus{} |H \cap K|$ and so $ \pi(HK)\mid |H|$.

Then $ |\pi(HK)|\mid ([G: K], |H|) \equal{} 1$, $ \pi(HK)\equal{}\{1\}$ and $ |H| \equal{} |H \cap K|$. And so $ H<K$.

It suffices to check that $ H \subset K.$

Let $ x$ be an element of $ H$ and let $ k$ denote the order of $ Kx$ in $ G/K.$ Since  $ k$ is a divisor of $ |\langle x\rangle|$ and $ |\langle x\rangle|$ divides any integer that annhilates $ x$, we obtain that $ k$ is a divisor of $ |H|$. On the other hand, Lagrange's theorem implies at once that $ k$ divides the order of $ G/K$, i. e., $ [G: K]$. Therefore, $ k$ is a common divisor of $ |H|$ and $ [G: K]$. The coprimality assumption allows us to conclude that $ k \equal{} 1$. Thus, $ K \equal{} (Kx)^{1} \equal{} Kx$ which implies in turn that $ x \in K$. [b]Done.[/b]

[quote="Riquelme"][quote="vict85"]

$ |HK| \equal{} |H| \plus{} |K| \minus{} |H \cap K|$

[/quote]

Is this part proven by exhibiting an explicit bijection between $ HK$ and $ H \cup K$?

Is the fact that $ |HK| \equal{} \frac {|H| \cdot |K|}{|H \cap K|}$ of any use to us here?[/quote]

You're right... I've confused the sum with the multiplication...  :blush:

And $ \pi(HK) \equal{} \frac {|H|}{|H \cap K|}$ because $ K < HK$.


We're given a function such that f(x) = x(x+1). What does this equal to?

\[\frac {f(x-1)f(x+1)}{f(x)}\]

A) f(x)
B) f(x)-1
C) f(2x)-3f(x)
D) 9*f({x-1}/3)
E) :sqrt: {f(x-2)f(x+2)}

[hide]first simplify it

{[(x-1) 2 +(x-1)][(x+1) 2 +(x+1)]}/(x 2 +1)

[(x 2 -x)(x 2 +3x+2)]/[x(x+1)]

[x(x-1)(x+2)(x+1)]/[x(x+1)]
then u can cancel stuff and get
(x-1)(x+2)
and then u just try out the answers and u will find (D) works[/hide]


Eh, it seems to be a trend to make language posts.  And I really like french.  So it seems sensible to post a french forum.  I personally am only a third year in high school, but I watch movies in french all the time to learn new things (like "M*rde!"  (if i can say that) which I learned from "Spaceballs") and reading books in french for what I can get out of them.

Je vais commencer la conversation.

Qu'aimez-vous faire pendant votre temps libre?

J'espère que ma grammaire a raison.

I speak French.  It's far from perfect but I'm able to read and write some stuff (officially it should be my second language  :whistling:

J'aimerais améliorer mon français. Je lis quelques bandes dessinées du Japon en français comme Dragonball et Rurouni Kenshin, parce qu'on ne veut pas faire une traduction néerlandaise et je ne peux pas acheter de traduction anglaise non plus  :roll:. Chaque jour, je regarde un peu la télévision française ou wallonne, mais pour le reste : la plupart des articles ou livres que je lis et la plupart des films ou programmes que je regarde sont en anglais, avec soutitrage en néerlandais.

[quote="funcia"]Heheheh...i am bad at french...how about Je suis maladie? *cries* I still don't get if french is an inflected language or not.[/quote]
"la maladie" is a noun , meaning "the disease, the sickness"
"malade" is an adjective, meaning "sick"

so "Je suis malade."="I am sick."

I don't know when exactly a language is to be called inflective, but I'd say French is inflective to be blunt, and definitely a lot more than English.

Thanks! I now know malade..and maladie alot better.
Inflected=latin with verb endings and all..
not inflected=english

i guess people can see how i am trying to teach myself..

ooh, if THAT means inflected, I think French is very much inflected.

Le garçon est fort.(=The boy is strong.)
Les garçons sont forts.(=The boys are strong.)

La fille est forte.(=The girl is strong.)
Les filles sont fortes.(=The girls are strong.)

Those were adjectives whose ending depends on the gender and the number.

And you can't always hear it, but when written down, French has a different verb ending for about every person:
je parle
tu parles
il parle
nous parlons
vous parlez
ils parlent

Lol if you think [i]that's[/i] complicated, you should see when you get into the thirteen other tenses.  It's a nightmare!  I've had "501 French Verbs" for about 3 months now and still have trouble conjugating anything for a lot of those tenses (especially the conditionals).  I really only use three or four very often, those being present, passé composé, the imperfect, and future.

French idioms on the other hand are quite fun to learn!

For example, this morning, to practice my french, I was muttering to myself:

"Pourquoi est-ce que je me suis levé à six heures ce matin!?  J'ai trop de sommeil!!!" and then someone asked me "Why are you speaking french" to which I replied (without knowing whether or not it was grammatically correct or not) "Bon, il n'y a pas beacoup de personnes qui parlent français ici.  Alors, je dois parler seul."  I still have to check this with my french teacher, although if you know whether something in that was incorrect or not, tell me.

Point being that I learned J'ai sommeil (I'm sleepy) today.

[quote="fredbel6"][quote="futuremontrealian"]Lol if you think [i]that's[/i] complicated, you should see when you get into the thirteen other tenses.  It's a nightmare![/quote]
I am well aware of those tenses. :roll:   I don't really see how you can find as many as thirteen tenses ? :
le présent, le passé composé, l'imparfait, le plus-que-parfait, le futur simple, le futur antérieur, le mode conditionnel présent, me mode conditionnel passé, le mode subjonctif présent et passé, passé simple et passé antérieur?
Those are twelve.
I can make them of most nonexotic verbs, but I am still not certain when to use "le mode subjonctif" and I always conveniently dodge that "shady passé simple" :wink:

[quote]I've had "501 French Verbs" for about 3 months now and still have trouble conjugating anything for a lot of those tenses (especially the conditionals).  I really only use three or four very often, those being present, passé composé, the imperfect, and future.[/quote]
You do know that even for the irregular verbs there are a few rules that work most of times? For "le mode conditionnel" you just have to go "steal" from "le futur simple"

[quote]"Pourquoi est-ce que je me suis levé a six heures ce matin!?  J'ai trop sommeil!!!" and then someone asked me "Why are you speaking french" to which I replied (without knowing whether or not it was grammatically correct or not) "Bon, il n'y a pas beacoup des personnes qui parle français ici.  Alors, je dois parler seul."  I still have to check this with my french teacher, although if you know whether something in that was incorrect or not, tell me.[/quote]
First of all : are you a lady or gentleman? :wink:   I'll assume you are a gentleman.
It should be [b]"à six heures"[/b] and [b]"trop de sommeil"[/b].
I also think "pourqoui me suis-je levé" is correct, not sure...
[b]"Bon, il n'y a pas beacoup de personnes qui parlent français ici."[/b] (you have to use the plural, as personnes stands for "persons", and beaucoup when expressing quantity of nouns is followed by "de")
All in all it sounds pretty decent to me  :wink: .[/quote]

il y a quelques formes littéraires, je pense, comme plus-que-parfait de subjonctif (je fusse est la forme d'être ou quelque chose comme ça).

[quote="MysticTerminator"]
il y a quelques formes littéraires, je pense, comme plus-que-parfait de subjonctif (je fusse est la forme d'être ou quelque chose comme ça).[/quote]
Yes, "je fusse" and "j'eusse" are "subjonctif imparfait", and "subjonctif plus-que-parfait" is constructied with "subjonctif imparfait" :

"que j'eusse parlé", "que tu eusses parlé"...

But I have never worked with that in class, and in fact we barely used the "passé simple" :|

[quote="fredbel6"][quote="futuremontrealian"]Lol if you think [i]that's[/i] complicated, you should see when you get into the thirteen other tenses.  It's a nightmare![/quote]
I am well aware of those tenses. :roll:   I don't really see how you can find as many as thirteen tenses ? :
le présent, le passé composé, l'imparfait, le plus-que-parfait, le futur simple, le futur antérieur, le mode conditionnel présent, me mode conditionnel passé, le mode subjonctif présent et passé, passé simple et passé antérieur?
Those are twelve.
I can make them of most nonexotic verbs, but I am still not certain when to use "le mode subjonctif" and I always conveniently dodge that "shady passé simple" :wink:

[quote]I've had "501 French Verbs" for about 3 months now and still have trouble conjugating anything for a lot of those tenses (especially the conditionals).  I really only use three or four very often, those being present, passé composé, the imperfect, and future.[/quote]
You do know that even for the irregular verbs there are a few rules that work most of times? For "le mode conditionnel" you just have to go "steal" from "le futur simple"

[quote]"Pourquoi est-ce que je me suis levé a six heures ce matin!?  J'ai trop sommeil!!!" and then someone asked me "Why are you speaking french" to which I replied (without knowing whether or not it was grammatically correct or not) "Bon, il n'y a pas beacoup des personnes qui parle français ici.  Alors, je dois parler seul."  I still have to check this with my french teacher, although if you know whether something in that was incorrect or not, tell me.[/quote]
First of all : are you a lady or gentleman? :wink:   I'll assume you are a gentleman.
It should be [b]"à six heures"[/b] and [b]"trop de sommeil"[/b].
I also think "pourqoui me suis-je levé" is correct, not sure...
[b]"Bon, il n'y a pas beacoup de personnes qui parlent français ici."[/b] (you have to use the plural, as personnes stands for "persons", and beaucoup when expressing quantity of nouns is followed by "de")
All in all it sounds pretty decent to me  :wink: .[/quote]

Thanks for the tips.  I'll change it :)  (I probably typed it more incorrectly than I said it.  I never bothered looking it over.  Except once when I spotted a missing r at the end of something).

I understand many tenses are based off one another, however, I still find this confusing.  My class (in america, as you can probably tell) is not very intelligent (and everyone else has a horrible accent.  They can't even pronounce "r" in french).  Thus, I have to study on my own by watching movies in french and submerging myself in the language.

To name the fourteen tenses mentioned in 501 french verbs:
présent de l'indicatif * imparfait de l'indicatif * passé simple

futur * conditionnel * présent du subjonctif *

imparfait du subjonctif * passé composé * plus-que-parfait de l'indicatif

passsé antérieur * futur antérieur * conditonnel passé * passé du subjonctif

and last but not least plus-que-parfait du subjonctif.

The last 7 are compound and the first 7 are not.

My confusion with the less common ones (like plus-que-parfait du subjonctif) is in that I am still in the act of becoming skilled with the basic past and future tenses.

Here's a question (one i would like answered):  How does one say math in french?  I have seen two different ways, but for some reason, I always can't remember if either have an s at the end (I'm thinking of math(s) and mathematique(s)).  Mathematique could mean mathematic as in adjective but I don't know.

Oh and I was speaking to funcia when I said how the other 13 tenses were a nightmare.  Not you, fredbel. :)

I've had French for three more years but practically all non Flemish productions on TV are from the USA (and we use subtitles :!: ). The result is that my English is much better.
What I struggle with is the "liaisons", I think I am going to ask the French subcommunity for help one of these days.

As far as I know, "mathématique" is an adjective meaning "mathematical".  "les mathématiques" means "mathematics" and is a feminine noun in plural form.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110928[/url]
So I went to ask for some help with those "liaisons" in French on the French national forum.  I got help but the people there seem to be busy for a moment now.  Maybe some people here can help me out too?

Warning, this is a question about pronunciation!

[quote="Kurt Gödel"]Lol, I could help you, Frédéric, because I lived in Brussels for 10 years (and my father is French/Wallon). Although I'm pretty good compared to other Dutch people, my father keeps banging his head against the wall when I mix up Dutch and French into one sentence  :D He calls it amateurism, I prefer to say that I'm a real Brusseleir  :P[/quote]
I didn't know! :o   I thought you were 100% Fleming like me.  Well,  :D  just follow the link and I will gladly follow your advice for any of my (right now already more than 20) difficult questions.:surrender: (mathmanman hasn't responded to all yet, and mstoenescu seems to be in doubt too)

For your liason question, and after following the link, I can officially answer only one or two of those with reasoning.

Whenever you have an ils/elles + conjugated verb, the conjugated verb (assuming it ends in ent) will not carry over, since the ent is silent anyway.

Example:

Ils veulent aider
is said like
Eeel vull ayday, with heaviness on the ll in vull.  I only know that the t is silent in the first place, if this helps at all :)

I can readily admit pretty much that your french is superbly better than mine, but then again, you're european, so what can I expect except multilingualism?

Okay that is good too know, never do liaison with -ent (in 3rd person, plural)
 :wink:

With that phonetics thing, keep in mind that English isn't my native either.
I think in "aider" the "ai" doesn't sound like the "a" or "ay" in "way".
I'd say it sounds like "èder"

Without getting into details  :maybe:  I'd rather refrain from equating all of Europe with multilingualism.  I'd restrict myself to Flanders, the Netherlands, Sweden, Finland.... (Of course, it's just :whistling:  coincidence that all those regions use subtitles...)

[quote="fredbel6"][quote="futuremontrealian"]Whenever you have an ils/elles + conjugated verb, the conjugated verb (assuming it ends in ent) will not carry over, since the ent is silent anyway. [/quote]
futuremontrealian, sorry for bringing this up again... but what about inversion then :

De quoi parlen[b]t[/b]-ils?]
 :maybe:[/quote]

Ok, I have an answer, and this better not be a wasted effort.  It took me 15 minutes waiting for my french teacher to get off the phone and answer my question, which made me late for the next class and cost me my participation points for the day. :)

The t IS pronounced.  But only in inversion.

Parlen[b]t[/b]-ils is indeed said like parlteel.

As I thought, but seriously : participation points! :o   How stern!  You shouldn't take that risk!
But thanks!

lol, you forget that america is famous for its easy classes and lack of challenging education.  Participation usually makes up at least a fourth of one's grade here.  Which is probably why we americans have such a tendency for stupidity...

Oh well, I didn't regret the loss.  I have a 104% in the class anyway, ('twas trig) so I had no reason to care about a measly 15 points.

...sorry I'm distracted by this cool proof on the left...

well anyway...  does anyone know a good place to find a french pen pal (or french canadian)?  or really just anyone who speaks french better than I do to correspond with?


algorithm
number theory
Euclidean algorithm
relatively prime

Show that for any positive integers $ a$ and $ b$, the number
\[ (36a\plus{}b)(a\plus{}36b)\]
cannot be a power of two.

[hide="solution"]
Let $ (36a\plus{}b)(36b\plus{}a)\equal{}k$. We must prove $ k$ is not a power of 2 i.e. $ k$ has no prime factors greater than 2.

Assume $ k$ is a power of two, for later contradiction.

Case 1: $ a\equal{}b$
If $ a\equal{}b$, then $ 37|k$, and we have a contradiction.

Case 2: $ a\neq{b}$
WLOG, let $ b>a$. If $ k$ is a power of 2, then both $ 36a\plus{}b$ and $ 36b\plus{}a$ must powers of two. Let $ 36a\plus{}b\equal{}2^{i}$ and $ 36b\plus{}a\equal{}2^{j}$. Obviously, $ 36b\plus{}a>36a\plus{}b\Rightarrow2^{j}>2^{i}$. Thus we must have $ 2^{i}|2^{j}\Rightarrow36a\plus{}b|36b\plus{}a$.

We know that $ 36a\plus{}b|36b\plus{}1296a$, so by Euclidean Algorithm, $ 36a\plus{}b|(36b\plus{}1296a)\minus{}(36b\plus{}a)$.

This means $ 36a\plus{}b|1295a$. Since we assume $ 36a\plus{}b\equal{}2^{i}$ and 2 is relatively prime to 1295, we must have $ 36a\plus{}b|a$, which is clearly impossible and thus a contradiction.

We have covered all cases, so we are done.
[/hide]

[hide="Alternate"]Suppose both $ a$ or $ b$ are even. Let $ 2^{k}$ be the highest power of $ 2$ dividing both $ a$ and $ b$. If the expression is a power of $ 2$, it still will be after dividing out $ 2^{2}k$, which makes it so at least one of $ a$ or $ b$ is odd. Suppose WLOG that $ a$ is odd. Then the first factor is odd and greater than $ 1$. Contradiction.[/hide]


algebra
polynomial
vector
induction
algebra unsolved

Find all polynomials $P(x)$ such that
\[(x +1)P(x-1)+(x-1)P(x+1)=2xP(x)\]
for all real numbers $x$.

[quote="zaya_yc"]$ (x \plus{} 1)P(x \minus{} 1) \plus{} (x \minus{} 1)P(x \plus{} 1) \equal{} 2xP(x)$ .Find all polynamial $ P(x)$.[/quote]

Let $ P(x),Q(x)$ solutions and $ \lambda,\mu\in\mathbb R$ : $ \lambda P(x)\plus{}\mu Q(x)$ is a solution and so the set of solutions is a $ \mathbb R$-vector space.

Assume $ \exists P(x),Q(x),R(x)$ three independant solutions. It's possible (since independant) to find $ a\notin\mathbb Z$ and $ \alpha,\beta,\gamma\in\mathbb R$ not all zero such that :
$ H(a)\equal{}H(a\minus{}1)\equal{}0$ with $ H(x)\equal{}\alpha P(x)\plus{}\beta Q(x) \plus{}\gamma R(x)$

But then, since $ (x\minus{}1)H(x\plus{}1)\equal{}2xH(x)\minus{}(x\plus{}1)H(x\minus{}1)$, we get thru induction $ H(a\plus{}n)\equal{}0$ $ \forall n\in\mathbb N$ and so $ H(x)\equal{}0$ (since polynomial with infinitely many roots), which is impossible since $ P(x),Q(x),R(x)$ are three independant solution.

So the dimension of the vector space is at most $ 2$ and since $ 1$ and $ x^3\minus{}x$ are clearly two independant solutions, the dimension is $ 2$ and we got a basis.

Hence the general solution : $ \boxed{P(x)\equal{}ax^3\minus{}ax\plus{}b}$

[quote]and so the set of solutions is a $ \mathbb{R}$-vector space. [/quote]
sorry but what is a vector space?

[quote="aadil"] sorry but what is a vector space?[/quote]

See http://en.wikipedia.org/wiki/Vector_space

[quote=pco]
...Assume $ \exists P(x),Q(x),R(x)$ three independant solutions. It's possible (since independant) to find $ a\notin\mathbb Z$ and $ \alpha,\beta,\gamma\in\mathbb R$ not all zero such that :
$ H(a)=H(a-1)=0$ with $$ H(x)\equal{}\alpha P(x)\plus{}\beta Q(x) \plus{}\gamma R(x)$$
...
[/quote]

Why is this possible?

Basically, we have a system of two equations with three unknowns ($\alpha,\beta,\gamma$).

Rigorous proof :
$\exists$ at most finitely many $x$ such that $P(x)Q(x-1)=P(x-1)Q(x)$
(else $P(x)Q(x-1)=P(x-1)Q(x)$ $\forall x$ and so $\frac PQ$ is constant and $P,Q$ not independant)

So $\exists a\in\mathbb Z$ such that $P(a)Q(a-1)-P(a-1)Q(a)\ne 0$

So following system (where $\alpha,\beta$ are unknowns) has always at least a solution (since determinant is nonzero) :
$\alpha P(a)+\beta Q(a)=-R(a)$
$\alpha P(a-1)+\beta Q(a-1)=-R(a-1)$

And so $(\alpha,\beta,1)$ is a suitable triplet.

Let $Q(x) = P(x)-P(x-1)$, then we get $$(x-1)Q(x+1) = (x+1)Q(x);\;\;\;\;\;(*)$$ Since for $x=1$ we get $Q(1)=0$ and for $x=-1$, we get $Q(0)=0$ we can write $Q(x) = R(x)x(x-1)$. Plugging this in to $(*)$ we get $R(x+1)=R(x)$ for all $x$ so $R$ is constant polynomial. Now we have $$P(x+1)-P(x) = cx(x-1)$$
By comparing the degree we see that $\deg(P)=3$...


OK, so I'm given the function f(x,y) = x^3 + 3x^2 - 3xy^2 + 3y^2 - 4. In polar coordinates, it's f(x,y) = 3r^2 + r^3 cos(3 \theta) - 4. I'm asked to find three lines such that f(x,y) = 0. One of them is x = 1, as you can find by inspection. I can't find any others, and I don't know how to approach the problem.

I'm stumped, and I'd really appreciate some help.

NB: It's only a small problem worth a few marks, so there must be an elegant solution. I've thought of a few possible approaches, but they're all much too cumbersome to be the solution expected.

Edit: Blast it, I've gone and posted in the wrong forum. The problem is probably beyond high school level, but I'd still appreciate some help if anyone has any to offer.

It's clear that $ x \equal{} 1$ is the only line with equation $ x \equal{} c$ that works.  Every other line has an equation $ y \equal{} ax \plus{} b$ for some $ a, b \in \bf R$.  So, plug in and solve.


If a circle is divided with 1 line inside, we will get 2 areas
If a circle is divided with 2 lines inside, we will get 4 areas
If a circle is divided with 3 lines inside, we will get 7 areas
If a circle is divided with 4 lines inside, we will get 11 areas

See the pictures below

If the dividing continuous.....
a) How many areas (maximum) we will get if we use 100 lines?
b) How many areas (maximum) we will get if we use $ n$ lines?

ummm...where r the pics?
and btw....do the lines intersect [color=black]inside[/color] the circle at several points(hope u get my point) or do they intersect at the same point?

[hide]$ \frac{(1\plus{}n)n}{2}\plus{}1$

When there is no line, a new line produces one more region.

When there is already one line, a new line can intersects with the line, and the circle, creating two more regions.

When there are already two, a new line can intersects with those lines, and the circle, creating three more regions.

And so on.[/hide]


algebra
polynomial
Pascal\u0027s Triangle

That thing about Fibonacci numbers is pretty interesting. MAthworld can really explain some things well

[quote="chess64"]I told you, when you just  want to know what something is, look it up on MathWorld!

http://mathworld.wolfram.com/PascalsTriangle.html[/quote]

Thank you for telling me about this site. It is quite good.

Yeah, that site does have some interesting facts about Fibonacci Numbers.

[quote="anirudh"]What is Pascal's Triangle??? :?[/quote]
i learned about it in 6th grade.  you have a triangle and the top has the number 1.
then the second row has two ones becuase in Pascal's triangle when you have two numbers together you add them and the sum is the number right below the two numbers. Here is the example.

                                                            1
                                                           1 1
                                                          1 2 1
                                                         1 3 3 1
                                                        1 4 6 4 1

[b]you have to quote it to look at it[/b]

[quote="math92"][quote="JesusFreak197"]Pascal's Triangle gives the coefficients of a polynomial expansion, as well as the values for combinations.[/quote]
i don't understand...can you explain more? :?[/quote]

For example, $4\choose0$ is 1, $4\choose1$ is 4,. $4\choose2$ is 6, $4\choose3$ is 4, and $4\choose4$ is 1. That's 1 4 6 4 1. Does that look familiar?

Also, Let's expand $(x+y)^4$. It comes out to be $x^4+4x^3y+6x^2y^2+4xy^3+y^4$ Look at the coefficients. It is also 1 4  6 4 1, the fifth row of the triangle.

Yeah, and : ${{n}\choose{0}} = 1$  :gleam:

[quote="nat mc"][quote="math92"]thanks 236!
i get it now but what does $4\choose0$ mean?[/quote]
${{4}\choose{0}}$ means "four choose zero", it is combinations of 0 elements from a four element set. It is also written $_4C_0$[/quote]
thanks nat mc, i get it now. :lol:


Hey I have think homework questions that is worth a lot of points and it is just giving me a headache, I think Im over thinking about it, but I also think its wrong...

Ok the problem is,

Your mom and you go to a restaurant and you guys have a total of $20. You MUST tip 25% of the price of both meals. You guys also get the same meal that cost the same amount.

Write and equation for this problem,

My equation, c = cost of both meals
20+0.25c

Then second part of the problem is,

How much did both meals cost?

And I am having trouble with that, I dont know if my equation works, it just looks right to me but I dont know how I would use it.

Please help me solve this and tell me step by step procedure executing problems like these.

Thank you so MUCH

well, what type of equation is the problem asking for?

Also, the amount of meals that both meals cost can be anything!  It has nothing to do with the tip or the amount of money you have at first.

I'm guessing you left out some information.  Did the problem say you two spent EXACTLY 20 dollars on both meals?

maybe you should post this in high school basics.  should probably go there if it is just a typical high school problem.

anyways, if my assumption is correct (which seems like the only possible use of the fact of the $ \$$20 and the tip), here is the solution.

[hide]let c=cost of both meals.
c+(tip)=20  tip=0.25c
1.25c=20
c=16

So both meals would cost 16 dollars.[/hide]

Ill try to make it more understandable,

So you and your mom go out to eat and you guys have ONLY $20 (as a total, not each). We both get a meal that cost the same amount. The question is what is the most amount of money you two can spend while leaving a tip that must be 25% of the total meal cost.
Example
Both meals cost
$10, a total of $20 spent. We have to give 25% more of
$20. So the total spent and given a tip is $25.
But for this problem, it cannot be $25 because you guys only have $20.

Let the amount you each pay be $ m$.  To find the maximum value of $ m$ such that $ 2m \plus{} .5m$ is less than or equal to $ \$20$, we set them equal.  Therefore, we have that $ 2m \plus{} .5m \equal{} \frac {5m}{2} \equal{} 20$.  Thus, $ \boxed{m \equal{} \$8}$.

Note: Your sum is $ 2m \plus{} (.25)(2m) \equal{} 2m \plus{} .5m$.

it is asking for cost of BOTH meals, though.  so it would be 8*2=16


Can anyone help me on this?

Solve for x:  $|\frac{x^{2}+2x}{x-2}|<|x+2|<|x-3|$

thanks
part of answer:[hide]they did
if $-2\le{x}<2$

$\frac{x^{2}+2}{-x+2}<x+2<-x+3$

$x^{2}+2<-x^{4}<x^{2}-5x+6$

but aren't they suppose to switch the signs cuz they're multipling by -x+2 which is negative.

shouldn't it be :
$x^{2}+2>-x^{4}>x^{2}-5x+6$     ?
also how come they didn't do
if $x<3$ ?  [/hide]  I'm not sure if this should be in the high school basic forum or in the intermediate forum.

[hide]Lets start with $|x+2|<|x-3|$ as $|a-b|$ is the distance between $a$ and $b$ the inequality translates to the question: find all numbers that are closer to (-2) than to 3. As the midpoint of $(-2,3)$ is $\frac12$ the answer is $x<\frac12$.
As for ${\frac{|x^{2}+2x|}{x-2}}<|x+2|$ rewrite it as $\frac{|x\parallel x+2|}{|x-2|}<|x+2|$. Obviously $x=-2$ is not a solution and when $x \ne-2$ we can divide both sides with $|x+2|$ and multiply by $|x-2|$ to obtain $|x|<|x-2|$. This I hope u understood how to "solve" :) . Finaly we get $x<\frac12$ and $x \ne-2$ are the solutions. [/hide]

hmm your way makes sense, but the answer is[hide]$\frac{1}{2}<x<1$.

 i get $x^{2}+2<-x^{2}+4<x^{2}-5X+6$ by multiplying by $-x+2$  (see top).

I'm getting  $0<2x^{2}-5x+2$ when i move all the terms to the other side   for $-x^{2}+4<x^{2}-5X+6$, but they have $0[color=orange: 8a6277a4da]>[/color: 8a6277a4da]2x^{2}-5x+2$  why did the sign change?  [/hide]

all the time u reffer to somthing "devine" you call [color=red]"they" [/color]. O.K. "they" as anybody human make mistakes just check with $x=0$ we get $0<2<3$  and this is a solution so obviously [color=red]"they"[/color] made a mistake. :D


calculus
integration
trigonometry
derivative
calculus computations

Find the maximum and minimum value of $ F(x) \equal{} \frac {1}{2}x \plus{} \int_0^x (t \minus{} x)\sin t\ dt$ for $ 0\leq x\leq \pi$.

$ F(x)\equal{}\frac{1}{2}\plus{}\int_{0}^{x}(t\minus{}x)\sin t\ dt\equal{}\frac{1}{2}\plus{}\int_{0}^{x}(x\minus{}t)(\minus{}\sin t)\ dt$

$ F(x)\equal{}\frac{1}{2}\plus{}\bigg[(x\minus{}t)\cos t\bigg]_{0}^{x}\plus{}\int_{0}^{x}\cos t\ dt\equal{}\frac{1}{2}\minus{}x\plus{}\sin x$

$ F'(x)\equal{}\cos x \minus{}1\rightarrow F'(0)\equal{}0$ and $ \forall x\in (0,\pi],F'(x)<0$

$ \therefore F_{max}\equal{}F(0)\equal{}\frac{1}{2},F_{min}\equal{}F(\pi)\equal{}\frac{1}{2}\minus{}\pi$ :)

Sorry I made a typo, please edit your solutuion. :oops:

Thanks in advance.

kunny

hello kunny, we have $ F(x)\equal{}\minus{}\frac{x}{2}\plus{}\sin(x)$ and the first derivative is $ F'(x)\equal{}\minus{}\frac{1}{2}\plus{}\cos(x)$, we find $ x_{max}\equal{}\frac{\pi}{3}$, so we get the maximum value as $ F\left(\frac{\pi}{3}\right)\equal{}\frac{\sqrt{3}}{2}\minus{}\frac{\pi}{6}$ and the minimum value as $ F(\pi)\equal{}\minus{}\frac{\pi}{2}$.
Sonnhard.

My solution for the edited problem is exactly the same with Dr Sonnhard' solution. :)

I wonder whether we could differentiate $ F(x)$ directly or not.  :maybe:

You can use $ \frac{d}{dx}\int_a^x f(t)\ dt\equal{}f(x)$ for some constat $ a$ and the differentiation of product.


According to Irodov Problems in General physics (problem 3.39):

We can calculate the field of a dipole by taking the gradient of the voltage of a dipole.
Here is my solution to that problem (the problem just asks to calculate the field strength of a dipole as a function of r and $ \theta$, where $ \theta$ is the polar angle where $ \theta \equal{} 0$ marks the pole parallel to $ \vec p$.  V is the potential relative to infinity.

Here is my work:
\[ V (r, \theta) \equal{} \frac {k p \cos {\theta}}{r^2}\]

\[ E_r \equal{} \minus{} \frac {\partial V}{\partial r}\]

\[ E_r \equal{} \minus{} \frac {2kp \cos {\theta}}{r^3}\]
Similarly:
\[ E_{\theta} \equal{} \minus{} \frac {1}{r} \frac {\partial V}{\partial \theta}\]

\[ E_{\theta} \equal{} \frac {kp \sin {\theta}}{r^3}\]

\[ \vec E (r, \theta) \equal{} \minus{} \frac {2kp \cos {\theta}}{r^3} \hat {r} \plus{} \frac {kp \sin {\theta}}{r^3} \hat {\theta}\]
And
\[ E (r, \theta) \equal{} \sqrt {{E_r}^2 \plus{} {E_{\theta}}^2} \equal{} \frac {kp}{r^3} \sqrt {1 \plus{} 3{\cos}^2 {\theta}}\]
I went into hyperphysics and their formula is much simpler:

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dipole.html

At wikipedia, I read a much more complicated formula:
\[ \vec E \equal{} k( \frac {(3 \vec p \cdot \hat {r}) \hat {r} \minus{} \vec p}{r^3} \minus{} \frac {1}{3 {\epsilon}_0} \vec p \delta (\vec r)\]
http://en.wikipedia.org/wiki/Dipole
Can somebody explain the discrepancy?

hey!
the formuale that u derived and the formulae in wiki are both correct..wiki gives a coordinate free expression while u give it in polar coords.
u can derive one formulae from the other.
on the other hand hyperphysics has a simpler expression because it dusnt give u the formulae for arbit $ \theta$ but only for $ \theta\equal{}0,\frac{\pi}{2}$ which u can get from ur expression as well


I believe you may be thinking of the following area formula (although I do not know what it is called.)  The points (X i, Y i) are coordinates in the x-y plane.

Area = ½ Summation {i=1 to n} (X i •Y i-1 − X i-1•Y i) where the points (X i, Y i) are the vertices of an n-sided polygon.  Note: You must let (Xo, Yo) = (Xn, Yn)

Please use the following formula in my previous post (just above this).

Area = $\sum_{i=1}^{n}\;(x_i * y_{i-1})-(x_{i-1} * y_i)$,  with $\ (x_0, y_0)=(x_n, y_n)$

Split out of the Mathcounts forum; was in reply to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73496 .


The midpoints of the three sides of an equilateral
triangle are connected to form a second triangle. A
third triangle is formed by connecting the midpoints
of the second triangle. This process is repeated until
a tenth triangle is formed. What is the ratio of
the perimeter of the tenth triangle to the perimeter
of the third triangle? Express your answer as a
common fraction.

1. I haven't started studying for Mathcounts yet this school year.
2. Experiment with smaller cases. For example, find the perimeter of the first triangle, then the second, etc. You should see a pattern.
3. If you don't know how to find the perimeter, you should note that each side of the next triangle (the one most recently drawn in) is half the length of the opposite side of the previous triangle. Infact, you can just use that fact to find the perimeter of the tenth.

I know the answer, i just wanted to see how many people would get it. Btw, please hide your answers. :wink:

Me has started earlier than August. :D  :D  :D

EEEEEEEEEEEEEEEEEEEEEYEAH!!!!!

I started in August, but accidentally clicked September...


function
real analysis
real analysis unsolved

Let $ f$ be a simple function on $ [a,b]$,then for every $ \epsilon > 0$,there is a step function $ g$ on $ [a,b]$ such that $ f(x) \equal{} g(x)$ except on a set of measure less than $ \epsilon$

Approximate the simple function as a linear combination of characteristic functions over open sets. The structure of open sets is simple enough that you can do an explicit construction. It's messy and boring though.

Do you mean use the theorem for every finite measurable set $ E$ and $ \epsilon>0$ there exist a set $ U$ which is the union of finite open interval such that $ m(E\cup U \minus{} E\cap U) < \epsilon$,Yes,it is messy and boring.

[quote="Kalle"]I was just thinking of that every measurable set has an open set containing it with only $ \epsilon$ more measure.[/quote]
But in this case,how can you assert the function you constructed is a step function?

Isn't the sum of two step functions a step function?

But a step function mean(In my book):
$ f$ is a step function defined on $ [a,b]$ which means there is a partition $ a \equal{} a_1 < a_2 < \cdots < a_n \equal{} b$ such that $ f$ is constant on $ (a_i,a_{i \plus{} 1})$ for $ i \equal{} 1,2,\cdots,n \minus{} 1$
Then how to finish your construction?


If $ x$ is $ 125\%$ of $ y$, then what percent of $ x$ is $ 2y$?

x is 125% of y, so 80% of x is y, and 160% of x is 2y.