inequalities
geometry
circumcircle
cyclic quadrilateral
triangle inequality
geometry proposed

In a triangle $ ABC$, the angle bisectors of angles $ CAB$, $ ABC$, $ ACB$ intersect the circumcircle  at $ P, Q, R$ respectively.
 Prove that $ AP\plus{}BQ\plus{}CR > AB\plus{}BC\plus{}AC$.

Let $ PB \equal{} PC \equal{} L.$ By Ptolemy’s theorem for cyclic quadrilateral $ ABPC$ we have

$ AB \cdot L \plus{} AC \cdot L \equal{} AP \cdot BC \Longrightarrow AP \equal{} \frac {L(AB \plus{} AC)}{BC}$

On the other hand, by triangle inequality we get $ PB \plus{} PC > BC$ $ \Longrightarrow 2L > BC$

Therefore, $ AP > \frac {_1}{^2}(AB \plus{} AC).$  The cyclic sum yields:

$ AP \plus{} BQ \plus{} CR > \frac {_1}{^2}(2AB \plus{} 2BC \plus{} 2CA) > AB \plus{} BC \plus{} CA.$


[quote="TCAtect"]The problem and solution:

http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_3

The solution claims that the second solution for y gives non-real roots, though I calculated it to be otherwise (final answer: 720?). I'm not sure if I have a misunderstanding of non-real roots. Can anyone correct me where I'm wrong or clarify the solution?[/quote]

Yes non-real roots in the [u]orignial[/u] equation, not the $ x^2\plus{}18x\plus{}30\equal{}\minus{}6$ equation.

Plugging the two roots you get from that equation back into the main equation you get the RHS to be sqrt of a negative number


97 Vestavia Comprehensive

#2. Two poles are 50 feet apart on level ground. One pole is 10 feet taller than the other. A cable hangs from the top of the poles in the shape of a parabola. It's lowest point is 10 feet above a point on the ground 20 feet from the shorter pole. Find the height of the taller pole.

Thanks in advance!

Let the taller pole have height $ k$, so then the shorter pole has height $ k\minus{}10$.

Place the diagram on the coordinate plane with the ground lying on the x-axis, point directly under the lowest point of the parabola at the origin, and the shorter pole to the left of the taller pole.

Let the parabola have equation $ y\equal{}ax^2\plus{}10$.

$ a(\minus{}20)^2\plus{}10\equal{}k\minus{}10$

$ a(30)^2\plus{}10\equal{}k$

$ 400a\plus{}20\equal{}900a\plus{}10$

$ 500a\equal{}10$

$ a\equal{}\frac{1}{50}$

therefore, $ k\equal{}\frac{900}{50}\plus{}10\equal{}28$


I read that with the change of base formula, you can write something like $ \log_ab$ as $ \frac{\log b}{\log a}$ if your calculator only handles logs with base 10. Unfortunately, my book didn't explain why it works - so can someone please explain why to me? Thanks in advance!

take base c logarithm in both side at $ a^x\equal{}b$

it becomes $ \log_c{a^x}\equal{}\log_c{b}$

which is $ x\log_c{a}\equal{}\log_c{b} \implies x\equal{}\frac{\log_c{b}}{\log_c{a}}$

and when $ a^x\equal{}b \implies x\equal{}\log_a{b}$

so $ \log_a{b}\equal{}\frac{\log_c{b}}{\log_c{a}}$

This is also helpful in other circumstances, not just for calculators. Maybe a problem will become much easier if, I don't know, you write it in base 5 instead of base 3. That's when the Change of Base formula comes in.


Hi, I tried searching for this on the forum but couldn't get anything.

How can I get text to appear to the side of a picture, instead of below it?

Thanks!

You could use minipages as in:
[code]\begin{minipage}{3in}
\includegraphics[width=2in]{picture.jpg}
\end{minipage}
\begin{minipage}{3in}
text goes here. New lines using \\
moves the text up
\end{minipage}[/code]

We use the [b]picins[/b] package when typesetting our textbooks.

[url]http://texcatalogue.sarovar.org/entries/picins.html[/url]

Just mentioning another package.

wrapfig with English explanation You can see the results in
my book at http://www.potto.org

Genick


How many pairs (m,n) satisfying 10<=m<n and m+n<=99 have the property that m+n and m-n have the same digits in reverse order?

[hide="Here is a bit of useful information:"]
We've got $n > m \geq 10 \ \wedge n+m \leq 99$. So,
\[n+m=10a+b \ \bigwedge \ n-m = 10b+a \Longrightarrow 2n=11(a+b) \ \bigwedge \ 2m = 9(a-b) \]
This means that $11 | n$ and $9 | m$ and $\ a \equiv b \bmod 2$. Someone can take over from here; I'm tired.  :)
[/hide]

[quote="cincodemayo5590"][quote="matt276eagles"]I think the problem statement is incorrect.  If $m<n$ then $m-n<0$.  Also are $m$ and $n$ necessarily integers?[/quote]

I thought that the problem statement was that the two numbers were $n-m$ and $n+m$, since they would be both positive; lol, that shows you how much I pay attention to exactly what I'm reading.  :P  My guess also is that they are positive integers.[/quote]

yes, they are.


If $ a,b,c>0$, then

$ \frac{a^2\plus{}a\minus{}c^2}{b\plus{}c}\plus{}\frac{b^2\plus{}4b\minus{}a^2}{c\plus{}a}\plus{}\frac{c^2\plus{}9c\minus{}b^2}{a\plus{}b}\geq4$

[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a^2 \plus{} a \minus{} c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 4b \minus{} a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 9c \minus{} b^2}{a \plus{} b}\geq4$[/quote]
Because $ \frac{a}{b\plus{}c}\plus{}\frac{4b}{a\plus{}c}\plus{}\frac{9c}{a\plus{}b}\geq\frac{(a\plus{}2b\plus{}3c)^2}{2(ab\plus{}ac\plus{}bc)}\geq4$
and $ \sum_{cyc}\frac{a^2\minus{}c^2}{b\plus{}c}\geq0\Leftrightarrow\sum_{cyc}(a^4\minus{}a^2b^2)\geq0.$


vector
trigonometry
calculus
calculus computations

Let $ \hat{a}$ and $ \hat{b}$ be unit vectors in the xy plane making angles $ \theta$ and $ \phi$ with the x axis, respectively. Show that $ \hat{a} \equal{} \cos \theta \hat{i} \plus{} \sin \theta \hat{j}$, $ \hat{b} \equal{} \cos \phi \hat{i} \plus{} \sin \phi \hat{j}$, and using vector algebra prove that $ \cos(\theta \minus{} \phi) \equal{} \cos \theta \cos \phi \plus{} \sin \theta \sin \phi$.

The first part intuitively makes sense. I am not sure about the second part.

R.H.S. can be regarded as scalar product of some two vectors.

So that implies that $ \hat{a} \cdot \hat{b} \equal{} \cos(\theta \minus{} \phi)$?


geometry
circumcircle
trigonometry
geometry proposed

Let $ ABCD$ be a convex quadrilateral. Points $ P,Q$ and $ R$ are the feets of the perpendiculars from point $ D$ to lines $ BC, CA$ and $ AB$, respectively. Prove that $ PQ\equal{}QR$ if and only if the bisectors of the angles $ ABC$ and $ ADC$ meet on segment $ AC$.

[quote="lasha"]Let $ ABCD$ be a convex quadrilateral. Points $ P,Q$ and $ R$ are the feets of the perpendiculars from point $ D$ to lines $ BC, CA$ and $ AB$, respectively. Prove that $ PQ \equal{} QR$ if and only if the bisectors of the angles $ ABC$ and $ ADC$ meet on segment $ AC$.[/quote]

ABCD should be [b]cyclic[/b] quadrilateral
it is the 4th problem from IMO 2003

My solution:
Using the Simson's line theorem, we get that the points P, Q and R are collinear.
By Menelaus Theorem (for triangle ABC which is intersected by R-Q-P) and similarity of triangles ARD and CPD we get that PQ=QR iff AB:BC=AD:DC, that is, PQ=QR iff the angle bisectors of ABD and ADC meet on AC. Q.E.D.

ABCD can be not cyclic and the statement remains valid.

1).The bisectors of the angles $ \angle{ABC}$ and  $ \angle{ADC}$ meet on segment in segment $ [AC]\Leftrightarrow |AB|.|CD| = |AD|.|BC|$
2). If $ R\mbox{ - is circumradius of} \triangle {ABC}$, then:
$  \begin{array}{c}|QR| = |AD|.\sin{A} = |AD|.\frac {|BC|}{R} \\
|QP| = |DA|.\sin{C} = |AD|.\frac {|AB|}{R} \\
|AB|.|CD| = |AD|.|BC|\end{array}\}\Rightarrow |QP| = |QR|.$

using the Simson's line we get that  $P,Q,R$ are collinear, so  quadrilateral $AQDP$ is cyclic so $\angle PDQ= \angle BAC= \angle BDC$ because $ABCD$ is cyclic, so if bisectors meet on $AC$ it means $\frac {AB}{BC}= \frac {AD}{CD}$ so it means $ABCD$ is harmonic so $BD$  is symmedian $\triangle ACD$ is simillar to $\triangle PRD$ and if $BD$ is symmedian $DQ$ is median because they they are isogonals and angles are same


my TI-83 says $i^{i}$ is real. can someone tell me how to compute $i^{i}$?

[b]Definition:[/b]  $a^{b}= e^{a \ln b}$

Now, $\ln i = \ln e^{i \frac{\pi}{2}}= i \frac{\pi}{2}$ (we are restricting the range of $\ln$ so that it is a valid function).  Hence

$i \ln i =-\frac{\pi}{2}$
$i^{i}= e^{i \ln i}= e^{-\frac{\pi}{2}}$

Write $i$ as $\cos (\pi/2)+i \sin (\pi/2) = e^{i\((\pi/2)}$
so $i^{i}$ becomes $e^{i*i*\pi/2}= e^{-\pi/2}$
(Or that is, at least,the principle value)

(You can use to find the power of any complex number, just write it down as $re^{i\theta}$

[quote="t0rajir0u"][b]Definition:[/b]  $a^{b}= e^{a \ln b}$ [/quote]
Wait a second... you inverted a and b!
$a^{b}=e^{\ln a^{b}}=e^{b\ln a}$!  :wink:
Anyway, it doesn't affect what you've written in your answer!

Sorry, my mistake.  I had it right the first time and rewrote it differently  :oops:

$i^{i}$ has multiple answers... (remember fractional exponents like 1^{1/5} give multiple answers, so it is reasonable for there to be multiple answers for a complex exponent

Does anyone know how to solve this without using $e$ or $ln$?

Exponentiation for complex exponents (and actually, even for real exponents) isn't well-defined without reference to $\ln$.


factorial
modular arithmetic
induction
number theory
greatest common divisor

An integer $k=n(n+1)(n+2)\cdots (n+7)+7!$ $\forall n\in\mathbb{N}$ . Prove that $k$ cannot be represent as sum of square of two numbers .

Let $k=x^2+y^2$. $k=n(n+1)\cdots(n+7)+7! \Rightarrow k \equiv 0 \pmod7$. So $x^2+y^2 \equiv 0 \pmod7$ so $x \equiv y \equiv 0 \pmod7$ which can be proven looking at cases of $x=7a, \ 7a+1, \ 7a+2, \ \cdots$ and adding it to $y=7b, \ 7b+1, \ 7b+2, \ \cdots$, in which none are $\equiv 0 \pmod7$ unless $x=7a$ and $y=7b$. So then we have $k=49a^2+49b^2=49(a^2+b^2)=n(n+1)\cdots(n+7)+7!$ so $a^2+b^2=\frac{n(n+1)\cdots (n+7)}{49}+\frac{6!}{7}$ which means that $n \not\equiv 0 \pmod7$.
Can't go any further right now. Maybe look at $n(n+1)\cdots (n+7) \pmod7$ for $n \equiv 1,\ 2,\ 3 \ \pmod7$ since it's symmetric (the case for $n \equiv 1 \pmod7$ is the same as $n \equiv 6 \pmod7$ and so forth).

by logic I am guessing the solution requires the use of this identity:
[quote="shyong"]not sure whether this helps or not , but i think we can use

$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ or $=(ac-bd)^2+(ad+bc)^2$

and write $10^{15}$ as $(1^2+1^2)^{15}(2^2+1^2)^{15}$[/quote]
Becuase that is what Shyong recently posted in answering a similar problem, right before posting this problem with the source of "own"... :lol:

You can notice that $128|n(n+1)(n+2)\cdots (n+7)$

 :)

[quote="Ahduhduh"]7!=n(n+1)...(n+7) for n=1, but 128|7!?[/quote]

Actually, for $n = 1$, $n(n+1)\cdots(n+7) = 8!$.

[hide]
We claim that $128|n(n+1)\cdots(n+7)$. Since there are $8$ consecutive terms, at least $4$ must be divisible by $2$, $2$ by $4$, and $1$ by $8$. That gives us $4+2+1 = 7$ powers of $2$, so $128|n(n+1)\cdots(n+7)$. Let $n(n+1)\cdots(n+7) = 128x$.

We have $7! = 2^4 \cdot 3^2 \cdot 5 \cdot 7 = 16y$ where $y \equiv 3 \pmod{8}$ (by calculation).

Then $128x+16y = 16(8x+y)$ so $8x+y$ must be the sum of two squares. But $8x+y \equiv y \equiv 3 \pmod{8}$ and squares are only $0,1,4 \pmod{8}$ so it cannot be the sum of two squares, as desired.[/hide]

nice, does anyone have any way to explain how people get these insights into wihch bases to use??

In a problem that involves $d^{th}$ powers, take $\bmod m$ such that $d | \phi(m)$.

http://mathworld.wolfram.com/TotientFunction.html
http://mathworld.wolfram.com/EulersTotientTheorem.html

You mean prove Euler's Totient Theorem?   :huuh:

This is not a trivial thing to do...

no. Is it stated in his theorem that when solving a problem like this to take certain mods?
if so, I guess I will have to try to research that myself. nevermind.

Short explanation, then:

Euler's Totient Theorem gives, for $(a, m) = 1, a^{ \phi(m) } \equiv 1 \bmod m$.  If you have a problem that involves $\phi(m)$ powers, then obviously taking it $\bmod m$ forces a definite value $\bmod m$ on one side of the equation, and from there it is usually easy to prove no solutions.

Now, if instead you have a problem that involves $d | \phi(m)$ powers, they are restricted by the fact that $a^{ \frac{ \phi(m) }{d} d} \equiv 1 \bmod m$, and so they themselves can only take on certain values (although not just $1$.)

Example, because the above is really unclear:

Suppose I wanted to prove that $a^2 + b^2 = 3(c^2 + d^2)$ has no integer solutions.  Now, obviously $(a, b, c, d) = 1$, or else we could divide through by the $gcd$.

Observing that $\phi(4) = 2$, I take the equation $\bmod 4$.  Clearly, $a^2 \equiv 0, 1 \bmod 4$.  This means that the LHS $\equiv 0, 1, 2 \bmod 4$ but that the RHS $\equiv 0, 3, 2 \bmod 4$, and observing that if both sides shared a factor of a power of $2$ we could merely divide through, we get that the LHS $\equiv 1 \mod 4$ but the RHS $\equiv 3 \bmod 4$ - Contradiction.

Example #2:

$\phi(25) = 20$
$x^{20} \equiv 1 \bmod 25$ (Euler's Totient Theorem)
$x^{10} \equiv 1, -1 \bmod 25$ (because, squared, it must be $\equiv 1$)
$x^5 \equiv 0, 1, -1, 7, -7 \bmod 25$ (because, taken to the fourth power, it must be $\equiv 1$)


[quote="SnowStorm"]Source: AHSME 1973 #10

If $n$ is a real number, then the simultaneous system below has no solution if and only if $n$ is equal to
\begin{eqnarray} nx + y &=& 1 \nonumber \\ ny + z &=& 1 \nonumber \\ x + nz &=& 1 \nonumber \end{eqnarray}

$\text{(A)} \ -1 \qquad \text{(B)} \ 0 \qquad \text{(C)} \ 1 \qquad \text{(D)} \ 0\mbox{ or }1 \qquad \text{(E)} \ 1/2$[/quote]

[hide]$\text{(A)} -1$
adding the three equations $=x+y+z+n(x+y+z)=3, (n+1)(x+y+z)=3$. $n\neq-1$. [/hide]


A fair coin is flipped 7 times. What is the probability that at least 5 consecutive flips come up heads?

5 consecutive heads = 3 ways
6 consecutive heads = 2 ways
7 consecutive heads = 1 ways

therefore its 6/2^7=3/64

answer : 3/64

i think ur solution is not based on 'consecutively'.

...although you must put 7C5, 7C6, 7C7 in front of them...

it says 'consecutively'.

there are 3 possible ways where you can flip 5 heads consecutively, 2 ways for 6 heads, and 1 way for 7 heads.

We have $3$ cases where 5 consecutive heads can come up:
HHHHH--
-HHHHH-
--HHHHH

We are going to use the formula $\frac{Success's}{Possibilities}$ to find the probability that the first case, the second case, and the third case work.

[b]HHHHH-- case:[/b]
The total possibilities is $2^7$.
$\frac{Success's}{2^7}\implies \frac{Success's}{128}$

The success's have to have the first 5 be heads, and the last two be anything.  There is 2 ways for the first - to be fulfilled and 2 ways for the second - to be fulfilled.  $2*2=4\text{ success's}$
$\frac{4}{128}$ is the probability for the case HHHHH--.
We note that -HHHHH- will have the same probability ($\frac{4}{128}$) as HHHHH--, because it is just a derivation off off the first case.
Therefore we will have $\frac{4}{128}+\frac{4}{128}+\frac{4}{128}\implies 3\frac{4}{128}$ being our probability.
\[3*\frac{4}{128}\]
\[3*\frac{\frac{4}{4}}{\frac{128}{4}}\]
\[3*\frac{1}{32}\]
\[\frac{3}{32}\]
is how I got my solution.

[quote="JayZz428xX"][quote="shadow123blast"]can u help me get it the answer on AOPS says it is 1/16
but i thought the answer was $\frac{17}{128}$
something like $(3)\frac{1}{2}^5+(2)\frac{1}{2}^6+\frac{1}{2}^7$
which simplifies to $\frac{3}{32}+\frac{1}{32}+\frac{1}{128} = \frac{17}{128}$[/quote]
Are you forgetting to simplify $\frac{17}{128}$?[/quote]
well seeing as 17 is a prime number $\frac{17}{128}$ can't be simplified any further furthermore anyone knows that 17 is odd and 128 can't be divisible by an odd number
so i don't see ur point on this problem?
If you mean something else tell me but u aren't making any sense asking me to simplify 17/128.
But that is besides the point i actually figured out where we are all getting the problem wrong it says consecutive so the others may or may not be heads if they aren't consecutive
the answer i finally figured out
first you have HHHHHH $\frac{1}{128} chance$
then BHHHHHB $\frac{1}{128}$
then HHHHHBB and BBHHHHH $\frac{2}{128}$
then HHHHHHB and BHHHHHH $\frac{2}{128}$
finally HHHHHBH and HBHHHHH $\frac{1}{128}$
thats it now we simplify $\frac{1}{128}+\frac{1}{128}+\frac{1}{128}+\frac{2}{128}+\frac{2}{128} = \frac{8}{128}$
which we simplify to $\frac{1}{16}$
thus now there are no other possible ways to create another situation where there are 5 consecutive heads or more the answer remains at $\frac{1}{16}$

[quote="JayZz428xX"][quote="shadow123blast"]can u help me get it the answer on AOPS says it is 1/16
but i thought the answer was $\frac{17}{128}$
something like $(3)\frac{1}{2}^5+(2)\frac{1}{2}^6+\frac{1}{2}^7$
which simplifies to $\frac{3}{32}+\frac{1}{32}+\frac{1}{128} = \frac{17}{128}$[/quote]
Are you forgetting to simplify $\frac{17}{128}$?[/quote]
$\frac{17}{128}$ is already is simplest form.

The problem with the other solutions is that we overcount possibilities like HHHHHHT. Shadow123blast's last solution is correct, by finding the cases:

HHHHHHH ($\frac{1}{128}$)
HHHHHHT, THHHHHH ($\frac{2}{128}$)
TTHHHHH, HHHHHTT, THHHHHT ($\frac{3}{128}$)
HTHHHHH, HHHHHTH ($\frac{2}{128}$)

Which sums to $\frac{8}{128}=\frac{1}{16}$.


I'm stuck on these questions and have no idea how to approach them. Some hints are appreciated.

Ten students have been nominated for positions of secretary, treasurer, social convenor, and fundraising chair. In how many ways can these positions be filled if the Norman twins are running and plan to switch positions on occasion for fun since no one can tell them apart?

[hide][/hide]

How many different ways are there of arranging seven green and eight brown bottles in a row, so that exactly one pair of green bottles is side-by-side?

[hide][/hide]

In how many ways could a class of 18 students divide into groups of 3 students each?

[hide][/hide]

What is the highest power of 2 that divides evenly into 100!?

[hide][/hide]

do ya want hints or solutions?
for hints i have a big one for all of them [hide="hint"]$ \dbinom{n}{r}$[/hide]
shouldn't this be in the MC forum?

i solved all of them except for question 3... i'm not getting the same answer as the answer key... is it wrong?

3.In how many ways could a class of 18 students divide into groups of 3 students each?
i did $ \dbinom{18}{3} \cdot \dbinom{15}{3} \cdot \dbinom{12}{3} \cdot ... \cdot \dbinom{3}{3}$ which is not $ 190590400$

for question 4, i did it a really lame way... listing all the even numbers from 1 to 100 and seeing how many ways $ 2^x$ can divide evenly into each of those even numbers.... then add all $ x$s up... is there a faster approach?

[quote="Poincare"]
shouldn't this be in the MC forum?[/quote]

'Twas a rather pleasant shock for me as well when I realized the high school basics forum (and even some of the problems on hs intermediate!) was of comparable difficulty to the MC forum... especially over the summer when some rather difficult problems were presented in the marathon over there...

[quote="instantnoodles"]
3.In how many ways could a class of 18 students divide into groups of 3 students each?
i did $ \dbinom{18}{3} \cdot \dbinom{15}{3} \cdot \dbinom{12}{3} \cdot ... \cdot \dbinom{3}{3}$ which is not $ 190590400$
[/quote]

After that you have to divide by $ 6! \equal{} 720$ to account for the fact that the groups could be picked in different orders.

[quote]
for question 4, i did it a really lame way... listing all the even numbers from 1 to 100 and seeing how many ways $ 2^x$ can divide evenly into each of those even numbers.... then add all $ x$s up... is there a faster approach?[/quote]

There is essentially no better way to do it, afaik. With some tricks that are gained by experience, you'll be faster... for example, a faster way would be to compute
$ \left\lfloor \frac {100}{2^1} \right\rfloor \plus{} \left\lfloor \frac {100}{2^2} \right\rfloor \plus{} \left\lfloor \frac {100}{2^3} \right\rfloor \plus{} \left\lfloor \frac {100}{2^4} \right\rfloor \plus{} \left\lfloor \frac {100}{2^5} \right\rfloor \plus{} \left\lfloor \frac {100}{2^6} \right\rfloor$

Here are my solutions (in case anybody needs them)
1. $ \frac {10! \minus{} 9!2!}{6!}$
$ 10!$ are the number of ways you can arrange the $ 10$ students, $ 9!2!$ are the number of ways the Norman Twins can switch, divide by $ 6!$ because there's only $ 4$ slots available.. There are other solutions to this problem.. feel free to post them.

2. Arrange the brown bottles in a row.
let $ B$ represent the brown bottles
$ \minus{} B \minus{} B \minus{} B \minus{} B \minus{} B \minus{} B \minus{} B \minus{} B \minus{}$
There are $ 9$ available slots.
Now put the pair of green bottles into the available slots
There are $ 5$ green bottles left and $ 8$ more slots available.
The answer is
$ 9 \cdot \dbinom{8}{5}$

3 & 4. refer to leoxnlin's post

[quote="ThetaPi"]

[quote]$ \left\lfloor \frac {100}{2^1} \right\rfloor \plus{} \left\lfloor \frac {100}{2^2} \right\rfloor \plus{} \left\lfloor \frac {100}{2^3} \right\rfloor \plus{} \left\lfloor \frac {100}{2^4} \right\rfloor \plus{} \left\lfloor \frac {100}{2^5} \right\rfloor \plus{} \left\lfloor \frac {100}{2^6} \right\rfloor$[/quote]

Please explain how you got this. I am inexperienced.[/quote]

Note that $ \left\lfloor \frac{a}{b} \right\rfloor$ gives the number of multiples of $ b$ less than or equal to $ a$.

Each even number has a factor of 2: $ \left\lfloor \frac {100}{2^1} \right\rfloor$

So far we've given each even number one "2" in it. But multiples of four have (at least) 2 "2"s in them... so we add one to each count for the multiples of four: $ \left\lfloor \frac {100}{2^2} \right\rfloor$

... and so on for multiples of 8, 16, etc.

[quote="leoxnlin"][hide][quote="ThetaPi"]

[quote]$ \left\lfloor \frac {100}{2^1} \right\rfloor \plus{} \left\lfloor \frac {100}{2^2} \right\rfloor \plus{} \left\lfloor \frac {100}{2^3} \right\rfloor \plus{} \left\lfloor \frac {100}{2^4} \right\rfloor \plus{} \left\lfloor \frac {100}{2^5} \right\rfloor \plus{} \left\lfloor \frac {100}{2^6} \right\rfloor$[/quote]

Please explain how you got this. I am inexperienced.[/quote]

Note that $ \left\lfloor \frac {a}{b} \right\rfloor$ gives the number of multiples of $ b$ less than or equal to $ a$.

Each even number has a factor of 2: $ \left\lfloor \frac {100}{2^1} \right\rfloor$

So far we've given each even number one "2" in it. But multiples of four have (at least) 2 "2"s in them... so we add one to each count for the multiples of four: $ \left\lfloor \frac {100}{2^2} \right\rfloor$

... and so on for multiples of 8, 16, etc.[/hide][/quote]

I actually forgot that $ 2^7>100$, thus the confusion.

[quote="ThetaPi"][quote="Wickedestjr"]1. I don't really understand what the Norman twins have to do with this.
2. Basically, you subtract the compliment. The total number of ways to arrange the fifteen bottles is 15 factorial, which is 1,307,674,368,000. The total number of ways that the bottles can be arranged with no two green bottles touching is 8 factorial times 7 factorial, because there are eight factorial ways to arrange the brown bottles and 7 factorial ways to arrange the green bottles. Which is 203,212,800. Subtract 203,212,800 from the larger number to get the answer; [color=orange]1,307,471,155,200[/color]
3. Well the answer is (18 choose 3) x (15 choose 3) x (12 choose 3) x (9 choose 3) x (6 choose 3), which I don't feel like calculating right now, but anybody that didn't know the answer before can calculate it now.
4. The solution is really quite hard to explain, but the answer is [color=orange]2^97[/color].[/quote]

For question three, you will need to divide $ {18 \choose 3} \times {15 \choose 3} \times {12 \choose 3} \times {9 \choose 3} \times {6 \choose 3} \times {3 \choose 3}$ by $ 6! \equal{} 720$ as the order in which the groups of 3 are chosen doesn't matter.

[quote]$ \left\lfloor \frac {100}{2^1} \right\rfloor \plus{} \left\lfloor \frac {100}{2^2} \right\rfloor \plus{} \left\lfloor \frac {100}{2^3} \right\rfloor \plus{} \left\lfloor \frac {100}{2^4} \right\rfloor \plus{} \left\lfloor \frac {100}{2^5} \right\rfloor \plus{} \left\lfloor \frac {100}{2^6} \right\rfloor$[/quote]

Please explain how you got this. I am inexperienced.

Will someone post an alternative and more detailed solution to the first problem on the Norman twins? I don't understand it.[/quote]

Thanks for catching that mistake. I knew that there was something wrong about that answer, but I thought that it was dividing by 6, instead of 720, but I wasn't quite sure that that was correct either, so I just left it as it was.


Prove that for: $ a,b,x,y,z>0,$ the following double inequality holds:
[size=75]$ (a\plus{}b)\cdot\frac{(xy\plus{}yz\plus{}zx)}{x\plus{}y\plus{}z}\cdot \left[\frac{x}{(ay\plus{}bz)^2}\plus{}\frac{y}{(az\plus{}bx)^2}\plus{}\frac{z}{(ax\plus{}by)^2}\right]\geq$
$ \geq\frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by}\geq \frac{3}{a\plus{}b}$[/size]

The second inequality is an old well known generalization of the Nesbitt's inequality (Romania TST 1986). It's easy to prove it.

$ LHS\equal{}\frac{x^2}{x(ay\plus{}bz)}\plus{}\frac{y^2}{y(az\plus{}bx)}\plus{}\frac{z^2}{z(ax\plus{}by)}\stackrel{Titu's\ lemma}{\geq}$
$ \geq\frac{(x\plus{}y\plus{}z)^2}{(a\plus{}b)(xy\plus{}yz\plus{}zx)}\geq\frac{3(xy\plus{}yz\plus{}zx)}{(a\plus{}b)(xy\plus{}yz\plus{}zx)}\equal{}RHS$

I'm interested in seeing your solutions to the first inequality. This short proof could be a hint! I do prefer to give you the whole expression of the inequality, without additional conditions (which could make it nicer, but much more difficult). For example:

Let $ a,b,x,y,z>0$ be such that $ (a\plus{}b)\cdot (xy\plus{}yz\plus{}zx)\equal{}x\plus{}y\plus{}z$.
Prove the following inequality:

$ \frac{x}{(ay\plus{}bz)^{2}}\plus{}\frac{y}{(az\plus{}bx)^{2}}\plus{}\frac{z}{(ax\plus{}by)^{2}}\geq\frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by}$

[quote="marin.bancos"] For example:

Let $ a,b,x,y,z > 0$ be such that $ (a \plus{} b)\cdot (xy \plus{} yz \plus{} zx) \equal{} x \plus{} y \plus{} z$.
Prove the following inequality:

$ \frac {x}{(ay \plus{} bz)^{2}} \plus{} \frac {y}{(az \plus{} bx)^{2}} \plus{} \frac {z}{(ax \plus{} by)^{2}}\geq\frac {x}{ay \plus{} bz} \plus{} \frac {y}{az \plus{} bx} \plus{} \frac {z}{ax \plus{} by}$[/quote]

It's easy By caushy shwarz   :)

[quote="marin.bancos"]Do it, please![/quote]

$ \left(\frac{x}{(ay\plus{}bz)^{2}}\plus{}\frac{y}{(az\plus{}bx)^{2}}\plus{}\frac{z}{(ax\plus{}by)^{2}}\right)(x\plus{}y\plus{}z)$$ \ge\left(\frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by}\right)^2$

and hence it suffices to show that

$ \frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by} \ge x\plus{}y\plus{}z$

we have the identity :

$ \sum_{cyc} x(ay\plus{}bz)\equal{}\sum_{cyc}(axy\plus{}bxz)\equal{} \sum_{cyc} (axy\plus{}bxy)\equal{}(a\plus{}b)(xy\plus{}yz\plus{}xz)\equal{}x\plus{}y\plus{}z$

Thus by caushy shwarz

$ \left(\frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by}\right) \sum_{cyc} x(ay\plus{}bz) \ge (x\plus{}y\plus{}z)^2$

which is equivalent to
$ \frac{x}{ay\plus{}bz}\plus{}\frac{y}{az\plus{}bx}\plus{}\frac{z}{ax\plus{}by} \ge x\plus{}y\plus{}z$

which ends our proof  :)

Thank you for your proof, [b]Abdek[/b]!
My solution to the last problem:
$ LHS \equal{} \frac {x^3}{[x(ay \plus{} bz]^2} \plus{} \frac {y^3}{[y(az \plus{} bx)]^2} \plus{} \frac {z^3}{[z(ax \plus{} by)]^2}\stackrel{My\ lemma}{\geq}$
$ \geq \underbrace{\frac {x \plus{} y \plus{} z}{x(ay \plus{} bz) \plus{} y(az \plus{} bx) \plus{} z(ax \plus{} by)}}_{ \equal{} 1}\cdot\left[\frac {x^2}{x(ay \plus{} bz)} \plus{} \frac {y^2}{y(az \plus{} bx)} \plus{} \frac {z^2}{z(ax \plus{} by)}\right] \equal{}$
$ \equal{} RHS$
You can see [b]my lemma[/b] here: [url]http://www.mathlinks.ro/viewtopic.php?t=310721[/url]


How many integers are greater than 87 and less than 321?

This is 88 thru 320, inclusive, or $ 320 \minus{} 88 \plus{} 1 \equal{} \boxed{233}$ numbers.


Let a,b,c,d be complex numbers satisfying

a+b+c+d=a^3+b^3+c^3+d^3=0. Prove that a pair of the a,b,c,d must add up to 0.

Eliminate $ d$, and you get $ a^{3}+b^{3}+c^{3}-(a+b+c)^{3}=0$.

Use the factorization from to the other topic to realize that is just $ 3(a+b)(b+c)(c+a)=0$.

[quote="t0rajir0u"]So you're only solving for the special case $ d = d^{3}$?

[hide="Hint"] As a polynomial in $ d$, we know that $ d+(a+b+c) | d^{3}+(a^{3}+b^{3}+c^{3})$. [/hide]

Why is this in HSB, anyway?[/quote]

huh?

Maybe I wasn't clear enough

$ 0=a+b+c+d\implies d=-a-b-c$

$ a^{3}+b^{3}+c^{3}+d^{3}=0\implies a^{3}+b^{3}+c^{3}+(-a-b-c)^{3}=0$

which factors into

$ (-3)(a+b)(b+c)(c+a)$
[this was another question that he posted]

which is equal to $ 0$

hence the conclusion.

Sorry, it didn't occur to me you meant "eliminate" in the context of solving systems.  Anyway, yes, that's more or less the first solution I came up with, but the polynomial method is very general so I want to see how well it pans out here.


function
logarithms
calculus
calculus computations

let $ f$ be continuous function on $ [a;b]$ and there exist $ f'$ on $ (a;b)$; $ f(x) \ne 0$, $ \forall x \in (a;b)$.
Prove that: $ \exists c \in (a;b)$ such that :
$ \frac {f'(c)}{f(c)} \equal{} \frac {1}{a \minus{} c} \plus{} \frac {1}{b \minus{} c}$

$ f$ must keep the sign, assume wlog it's positive.
Define $ g(x) \equal{} \log [f(x)(x\minus{}a)(b\minus{}x)]$, $ x\in(a,b)$. $ g(a) \equal{} g(b) \equal{} \minus{}\infty$, so $ \exists c\in (a,b)$ st $ g'(c) \equal{} 0$.

[quote="zhoraster"]$ f$ must keep the sign, assume wlog it's positive.
Define $ g(x) \equal{} \log [f(x)(x \minus{} a)(b \minus{} x)]$, $ x\in(a,b)$. $ g(a) \equal{} g(b) \equal{} \minus{} \infty$, so $ \exists c\in (a,b)$ st $ g'(c) \equal{} 0$.[/quote]

it' s not clearly ìf you say : " $ g(a) \equal{} g(b) \equal{} \minus{} \infty$, so $ \exists c\in (a,b)$ st $ g'(c) \equal{} 0$."

i think , if we consider $ g(x) \equal{} f(x).(x \minus{} a)(x \minus{} b)$ then $ f(a) \equal{} f(b) \equal{} 0$ , use Rolle theroem we have .  :oops:


algebra
polynomial
number theory unsolved
number theory

find P(x)=ax^4+bx^3+cx^2+dx+e such that a,b,c,d,e are integers and
P(3)=5,P(20)=105

If $P$ is a polynomial with integers coefficients, then
$a-b|P(a)-P(b)$, thus $20-3=17|P(20)-P(3)=100$ which is impossible.

bye


Find all natural n such that $n!$ is not divisible by $n^{2}$.

[quote="mathnerd314"][hide]This means that n only has one proper factor. So n has to be prime.
...Is that right?[/hide][/quote]
 i had some trouble with this question myself. I [b]think[/b] n must be prime, but you should get a second opinion. :P

[quote="me@home"][hide=" :D$_{Click}$"]
The question rephrases to, $\text{Find all natural n such that }n\not | (n-1)!$.
All divisors $d$ of $n$ besides itself satisfy $d\leq n \implies d \in \{1,2,...,n-1\}$. Assume that $n$ isn't prime, i. e. has divisors between $1$ and $n$. Consider any of these divisors $d$. $d\in \{2,...,n-1\}, \frac{d}{n}\in \{2,...,n-1\}$. We make the further assumption that there exists some $d$ such that $\frac{n}{d}\not =d$. This implies that, since two numbers exist in $(n-1)!$ which multiply to $n$, $n | (n-1)!$.
Notice that we have [i]proved[/i] that $n|(n-1)!$ - except not for two "strategically" unchecked cases:
[b]All divisors $d$ between $1, n$ equal $\frac{n}{d}$[/b]
$d=\frac{n}{d}\implies d^{2}=n$. So there exists at most one such $d$; therefore it is the only divisor between $1, n$; moreover, it is the squareroot of $n$. This case tells us that if $\boxed{n=p^{2}}$ (the square of a prime) then $n\not | (n-1)! \implies n^{2}\not|n!$
[b]$n$ is prime, i. e. there exist no $d\not \in \{1, n\}$[/b]
$\boxed{n=p}\implies n\not | (n-1)! \implies n^{2}\not|n!$
[/hide][/quote]
[hide=" :D$_{Click}$ "]thanks[/hide]

[hide="Is it too late to retract a two-year-old flaw in my "solution"?"]
Clearly primes work. Suppose n=ab (1<a,b<n) is composite.
Then ab | (n-1)! if a,b are distinct. If we are forced a=b, the same conclusion holds as long as 2a<n=a^2.
Thus the one non-prime that works is n=4 (because the above inequality fails).
[/hide]
I wouldn't normally say this, but for the sake of correctness thanks for reviving a post you shouldn't have revived.


Let $ a$ , $ b$ be two positive numbers. Prove easily that $ (a \plus{} b)^4\le \left(a^2 \plus{} 3b^2\right)\left(b^2 \plus{} 3a^2\right)$ .

Prove its "[u]strengthening[/u]" : $ (a \plus{} b)^4\le\boxed {(a \plus{} b)^3\sqrt {2\left(a^2 \plus{} b^2\right)}\le \left(a^2 \plus{} 3b^2\right)\left(b^2 \plus{} 3a^2\right)}$ .

[quote="Virgil Nicula"]$ \{a,b\}\subset\left(0,\infty \right)\ \Longrightarrow\ (a \plus{} b)^4\le\boxed {(a \plus{} b)^3\sqrt {2\left(a^2 \plus{} b^2\right)}\le \left(a^2 \plus{} 3b^2\right)\left(b^2 \plus{} 3a^2\right)}$ .[/quote]
The left follows from the Cauchy Schwarz Inequality. For the right, by AM-GM Inequality, we obtain
\[ 2(a \plus{} b)^3 \sqrt {2(a^2 \plus{} b^2)} \le (a \plus{} b)^2[(a \plus{} b)^2 \plus{} 2(a^2 \plus{} b^2)] \equal{} (a \plus{} b)^2(3a^2 \plus{} 2ab \plus{} 3b^2)
\]
It suffices to show that
\[ 2(a^2 \plus{} 3b^2)(3a^2 \plus{} b^2) \ge (a \plus{} b)^2(3a^2 \plus{} 2ab \plus{} 3b^2)
\]

\[ \Leftrightarrow 6a^4 \plus{} 20a^2b^2 \plus{} 6b^4 \ge 3a^4 \plus{} 8a^3b \plus{} 10a^2b^2 \plus{} 8ab^3 \plus{} 3b^4
\]

\[ \Leftrightarrow 3a^4 \minus{} 8a^3b \plus{} 10a^2b^2 \minus{} 8ab^3 \plus{} 3b^4 \ge 0
\]

\[ \Leftrightarrow (a \minus{} b)^2(3a^2 \minus{} 2ab \plus{} 3b^2) \ge 0
\]
which is true. :)

[quote="Virgil Nicula"][color=darkblue][b][u]Method II (shortest !).[/u][/b] Let $ a$ , $ b$ , $ c$ be three positive numbers. Apply [b]C.B.S.[/b] :

$ \left\|\begin{array}{c} (a\cdot c \plus{} c\cdot b)^2\le\left(a^2 \plus{} c^2\right)\left(c^2 \plus{} b^2\right) \\
 \\
(a\cdot b \plus{} c\cdot c)^2\le\left(a^2 \plus{} c^2\right)\left(b^2 \plus{} c^2\right)\end{array}\right\|\ \bigodot\ \Longrightarrow$

$ \boxed {\ c(a \plus{} b)\left(ab \plus{} c^2\right)\le\left(a^2 \plus{} c^2\right)\left(b^2 \plus{} c^2\right)\ }$ .

For $ c: \equal{} \sqrt {\frac {a^2 \plus{} b^2}{2}}$ obtain our  inequality ![/color] :lol:[/quote]
Very nice for sure, Mr. Nicula! :)


let $ (O)$ a circle with radius $ R$. find all quadrilaterals $ ABCD$ inscriptibles on $ (O)$ such that $ AB.BC.CD.DA \geq 4R^4$

Are you sure it is $ 4R^2$ ?  $ 4R^4$ seems better?

suppose arc AB is the longest arc that does not contain other points among the four arcs AB,BC,CD and DA. if the center of circle is outside the quadilateral, then AB is less than or equal to 2R. Obviously, the product of other three segment is not more than $ R^3$.

If the center of circle is in ABCD, then obviously we get t maximum when four segments are equal, so also true. :)

$ X \equiv AC \cap BD$ is the diagonal intersection and $ \psi \equal{} \widehat{(AC, BD)}$ the angle between the diagonals.

$ AB \cdot BC \cdot CD \cdot DA \equal{}$

$ \equal{} \left(AX \cdot \frac{\sin \psi}{\sin \widehat{ABX}}\right) \cdot \left(XC \cdot \frac{\sin \psi}{\sin \widehat{XBC}}\right) \cdot (2R \sin \widehat{CBD}) \cdot (2R \sin \widehat {DBA}) \equal{}$

$ \equal{} 4R^2 \cdot AX \cdot XC \cdot \sin^2 \psi \equal{} 4R^2 (R \plus{} OX) (R \minus{} OX) \sin^2 \psi \equal{}$

$ \equal{} 4R^2 (R^2 \minus{} OX^2) \sin^2 \psi \le 4R^4$

with equality only for $ OX \equal{} 0$ and $ \psi \equal{} 90^\circ,$ when $ ABCD$ is a square.


$x\in\mathbf.{R}$
$solve$the$equation$
$x=6+\sqrt{6+\sqrt{x}}$ :rotfl:

Make a substitutioin $y=x-6$ to get the equation:
$y^{4}-12y^{2}-y+30=0$, and we note that $y>0$ since $y=\sqrt{6+\sqrt{y+6}}>0$
it is easy to verify that $y=3$ and $y=-2$ are solutions, $y>0$ so only $y=3$ works. Then using Vieta's theorem if the other two solutions are $y_{1}, y_{2}$ we have $y_{1}*y_{2}=-5$, $y_{1}+y_{2}=-1$ so because $y_{1}, y_{2}> 0$ the only other solution is $y=\frac{-1+\sqrt{21}}{2}$
so the solutions for $x$ are $9$ and $\frac{11+\sqrt{21}}{2}$

I think you get the same polynomial when $y=\sqrt{x}$. :maybe:

[quote="rem"]so the solutions for $x$ are $9$ and $\frac{11+\sqrt{21}}{2}$[/quote]in fact, the latter is no solution... ;)

I will translate arkhammedos's post into readable text:
I present the solution in this form: The equation is $x=6+\sqrt{6+\sqrt{x}}$
Let $y=6+\sqrt{x}$ then the equation will be $x=6+\sqrt{y}$ we have $6=x-\sqrt{y}$ and $6=y-\sqrt{x}$. Therefore $x-y=\sqrt{y}-\sqrt{x}$, therefore:
$(\sqrt{x}-\sqrt{y}) (\sqrt{x}+\sqrt{y})=x-y$
$(\sqrt{x}-\sqrt{y}) (\sqrt{x}+\sqrt{y}+1)=0$
We cannot have $(\sqrt{x}+\sqrt{y}+1)=0$ since left side is always greater than $0$. Therefore:
$\sqrt{x}-\sqrt{y}=0$
$\sqrt{x}-(x-6)=0$
$x-\sqrt{x}-6=0$
$\sqrt{x}=-2,3$ but $\sqrt{x}$ is positive hence $\sqrt{x}=3$ and so the only solution is $x=9$.

That's a really nice solution, arkhammedos! :thumbup:

[quote="boxedexe"]$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y$...typo

Masoud Zargar[/quote]
thank you. The typo is now edited. :)


geometry
parallelogram
geometry proposed

In plane, let be given the fixed segment $AD$. Parallelogram $ABCD$ moved, always satisfies consition $\frac{AC}{AD}=\frac{BD}{BA}$. Find the locus of points $B$ and $C$.

i am not sure but if i understand your problem,it is so easy.and the locus is two circles with centers $A$ and $D$.
here is my solution:
$\frac{AC}{AD}=\frac{BC}{BA}$ $\rightarrow$ $\frac{sinCDA}{sinACD}=\frac{sinBAD}{sinACB}$ $\rightarrow$ $sinACD=sinACB$ $\rightarrow$ $\angle ACD=\angle ACB$ $\rightarrow$ $CB=CD=AB=AD$(?) $\rightarrow$ $AB$ and $CD$ are fixed.


number theory
Diophantine equation
algebra proposed
algebra

for each positive integer define
                            $T_{n}$ = $\frac{n(n+1)}{2}$
detemine all ordered pairs $(m,n)$ where $m > n$ such that
                    $T_{m}$ - $T_{n}$ =$2^{k}$
for some positive integer k.

For $m>n$, the equation reduces to $(m-n)(m+n+1)=2^{k+1}$.
Note that the two factors in the lhs have opposite parities and that $m+n+1>1$, so that $m-n =1$ and $m+n+1 = 2^{k+1}$.
This leads to the solutions $(2^{k},2^{k}-1).$

Pierre.

If you observe closely, you'll see that the term $T_{n}$ is strictly related to the sum of the first $n$ integers (you might know that since this is a very known recurrence). So, we can rename this cute thing ;-)

$T_{n}= \sum_{i=1}^{n}{i}$

Knowing that $m>n$, we can kill all the $n$ terms that compose the right side of the equation, and it remais like:

$\sum_{i=n+1}^{m}{i}=(m+n-1)(m-n)=2^{k+1}$

and we have only to solve this diophantine equation, like the other guy told above =] =]


the heavier object in an atwood machine has a mass which is twice the mass of lighter block.
the tension in the string is $ 16N$ when the system is relased from rest.Calculate the decrease in gravitational potential energy after $ 1$second the system is released from rest

Let $ m_1$ be the mass of the lighter object and $ m_2$ be the mass of the heavier object. We are given that $ m_2 \equal{} 2 m_1$. The equations of motion are,

$ m_1 a \equal{} T \minus{} m_1 g$

and

$ m_2 a \equal{} m_2 g\minus{} T$

where $ a$ is the positive acceleration of each object, $ T \equal{} 16$ is the tension, $ g$ is gravity, and the positive direction is up. You can use these equations to solve for $ a$:

$ a \equal{} \frac{(m_2 \minus{} m_1)g}{m_1\plus{}m_2} \equal{} \frac{g}{3}$

Plug this back into either equation to find that $ m_1 \equal{} \frac{12}{g}$. The distance each object travels after $ t$ seconds is  $ \frac{a t^2}{2}$ so after one second they travel $ \frac{a}{2}$ meters. We assume that initially the gravitational potential energy is $ 0$. So, after $ 1$ second the gravitational potential energy is $ m_1 \frac{a}{2} g \minus{} m_2  \frac{a}{2} g$. You can plug in for those variables to get something that only depends on $ g$.

Yeah I had a similar solution.
Infact we get that the decrease in gravitational potential energy is $ 2g$.

forgive me if i am dumb . FibonacciFan can you tell me for what you have assumed gravitational potential energy as zero ie: for the entire system of two objects (also earth if you want) or for any one object. i think we cannot assume the gravitational potential energy for the system to be zero becoz both the object will not be of the same level and one separation between the objects will be some non zero height.so one object will have more potential energy than other .....so we cant assume the g potential of the system to be zero right? :(

Agr_94_math you told that you had a "similar" post ...not the same one as FibonacciFan .so can you post your solution. i hope that you have not assumed  gravitational potential energy of the system to be zero.(forgive me :?:   if we can assume so and pls reason out your statement)

Potential energy is always relative to some arbitrary base height. The formula for gravitational potential energy is $ mgh$ and $ h$ depends on where the base height is. Only differences in potential energy are physically meaningful, which is why you can choose the base height to be whatever is most convenient. And for any system, you can always find a base height such that the system has $ 0$ gravitational potential energy.

sorry but i still didnt get u
ok suppose the base height is the height where that biggermass is located and let the lighter block be at height $ h$ from the base height...........if u take the initial potential energy of the bigger block which is at the base height to be zero then the potential energy of the lighter block is$ mgh$..so the initial potential energy of the system is $ mgh \plus{} 0$.then how can you take the initial potential energy of the system to be zero .

If the two objects start out at different heights, then you would take a base height in between them to get 0 gravitational potential energy.

Conservation of energy give us that the required decrease in potential energy equals the increase on the kinetic energy

\[ {1\over2}3mv^2 \equal{} {{18} \over g}{{g^2}\over 9} \equal{} 2g\]

where $ m$ is the lighter block's mass and $ v \equal{} at \equal{} {g\over3}.1$ is the scalar velocity of the sistem after 1 second.


Determine all pairs (a,b) of postive integers with x^2+1 is divisible by y  and    y^3+1 is divisible  by x


algebra
polynomial
inequalities
algebra unsolved

We have $ a_{n-1}=\sum_{i=1}^n r_i$ and $ a_{n-2}=\sum_{1\leq i<j\leq n}r_i r_j$. The inequality is equivalent to
\[{ \frac{\sum_{i=1}^n r_i}{n}}\geq\sqrt{\frac{\sum_{1\leq i<j\leq n}r_i r_j}{\binom{n}{2}}}\mbox{,}\]
which is the well-known Mac-Laurin-Inequality.  :)

$ (n\minus{}1)(\sum_{i\equal{}1}^{n} r_i)^2 \ge 2n (\sum_{1 \le i < j \le n} r_ir_j) \Longleftrightarrow \sum_{1 \le i < j \le n} (r_i\minus{}r_j)^2 \ge 0$

[quote="trejos"]Suppose the polynomial $ x^{n} \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_{1} \plus{} a_{0}$ can be factorized as  $ (x \plus{} r_{1})(x \plus{} r_{2})...(x \plus{} r_{n})$, with $ r_{1}, r_{2}, ..., r_{n}$ real numbers.
Show that $ (n \minus{} 1)a_{n \minus{} 1}^{2}\geq\ 2na_{n \minus{} 2}$[/quote]
Let $ f(x) \equal{} x^{n} \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_{1} \plus{} a_{0}$. Since $ f(x)$ has $ n$ real roots, $ f'(x)$ must have $ n \minus{} 1$, $ f''(x)$ must have $ n \minus{} 2$, ..., and $ f^{(n \minus{} 2)}(x)$ must have $ 2$ .
Since $ f^{(n \minus{} 2)}(x) \equal{} \frac {n!}{2}x^2 \plus{} (n \minus{} 1)! a_{n \minus{} 1}x \plus{} (n \minus{} 2)!a_{n \minus{} 2}$ has two real roots: $ ((n \minus{} 1)!a_{n \minus{} 1})^2 \ge 4 \left ( \frac {n!}{2} \cdot (n \minus{} 2)!a_{n \minus{} 2} \right ) \iff$
$ (n \minus{} 1)(a_{n \minus{} 1})^2 \ge 2n a_{n \minus{} 2}$ and we are done :)

Just take the derivative n-2 times.
FACT:if a polinomial has n real roots, its derivative has at least n-1.
Then the 2nd degree polinomial that results from taking n-2 derivatives has 2 real roots.
Then just use Bhaskhara's formula and the fact that the determinant has to be positive


This one is a direct result of an inequality discussed at least twice. Guess what!

I think that I have the resut
Problem<=>  \sum x 2 + \sum x
>= ( \sum x) 2

 give f(x_1,x_2,...x_n)=\sum x 2 + \sum x
>= ( \sum x) 2
give x_n>=x_n-1>=....>=x1
we can prove that
f(x_1,x_2,...x_n)>=f(x_1,x_2,...,x_n-1+x_n/2,x_n-1+x_n/2)


Let $ c_{1},c_{2},\dots c_{2006}$ be the cards,that lie in the box,such that card $ c_{i\plus{}1}$ lies under $ c_{i}$,for all $ i\in[1,2005]$.There are $ k$ empty box,and you can take one card $ c_{i}$ from any box,and put it in empty box,or put it on another card $ c_{i\plus{}1}$.Find the min. value of empty box,such that there is a way to put all cards from one box to another one.

Well with $ k \equal{} 3$, this is just the Towers of Hanoi problem, if I understood correctly...

You must find the minimum value of k.What is the Towers of Hanoi problem?What does it mean? :blush:

[quote="JBL"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97754

The difference between this problem and the Towers of Hanoi problem is that a given card can be put only on a card which is exactly one larger, i.e. $ c_{1}$ can be placed on an empty spot or on $ c_{2}$, but not on $ c_{4}$ (which is necessary for the conventional solution of Towers of Hanoi).

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

Oh, I see...

Hmm...yeah I see that makes it complicated


function
calculus
integration
linear algebra
matrix
algebra
rational function

$ \frac {1}{\prod\limits_{i \equal{} 1}^{m}(x \plus{} \alpha_{i})} \equal{} \sum\limits_{i \equal{} 1}^{m}\frac {A_{i}}{x \plus{} \alpha_{i}}$
where $ \alpha_{i} \neq \alpha_{j}$ $ \forall i \neq j$
find   $ A_{i}$ by $ \alpha_{1}, \alpha_{2},..., \alpha_{m}$

It's simple:

$ A_i \equal{} \prod_{j \in \{1,...,m\} \setminus i}\frac {1}{\alpha_j \minus{} \alpha_i}$

This can be viewed as:

$ \frac {1}{(x \plus{} \alpha_1)(x \plus{} \alpha_2)...(x \plus{} \alpha_n)...(x \plus{} \alpha_m)} \equal{} \frac {C_1}{x \plus{} \alpha_1} \plus{} ... \plus{} \frac {C_n}{x \plus{} \alpha_n} \plus{} ... \plus{} \frac {C_m}{x \plus{} \alpha_m}$

$ \frac {1}{(x \plus{} \alpha_1)(x \plus{} \alpha_2)...(x \plus{} \alpha_m)} \equal{} \frac {C_1(x \plus{} \alpha_n)}{x \plus{} \alpha_1} \plus{} ... \plus{} C_n \plus{} ... \plus{} \frac {C_m(x \plus{} \alpha_n)}{x \plus{} \alpha_m}$

Set $ x \rightarrow \minus{} \alpha_n$ and now:

$ \frac {1}{(\alpha_1 \minus{} \alpha_n)(\alpha_2 \minus{} \alpha_n)...(\alpha_m \minus{} \alpha_n)} \equal{} C_n$

[quote="milin"]It's simple[/quote]
yes, therefore I wrote simple problem . :)
 another question (also simple)
$ \big{(} \frac {1}{\prod_{i \equal{} 1}^{m}x \plus{} \alpha_{i}}\big{)}'_{\alpha_{1}} \plus{} \big{ (}\frac {1}{\prod_{i \equal{} 1}^{m}x \plus{} \alpha_{i}}\big{)}'_{\alpha_{2}} \plus{} ... \plus{} \big{(}\frac {1}{\prod_{i \equal{} 1}^{m}x \plus{} \alpha_{i}}\big{)}'_{\alpha_{m}} \equal{} \sum_{i \equal{} 1}^{m}\frac {D_{i}}{(x \plus{} \alpha_{i})^{2}}$
find $ D_{i}$ by $ \alpha_{1},...,\alpha_{m}$

It's also simple!  :wink:

Just multiply with $ (x\plus{}\alpha_i)^2$ and set $ x \rightarrow \minus{}\alpha_i$. This gives the same result, only negative. In other words:

$ D_i\equal{}\minus{}A_i$

[quote="Timestopper_STG"]I can't catch up :maybe: Can you explain the first one in more detail :blush:[/quote]

Sure. First, denote: $ S\equal{}\{1,2,...,n\}$. Now, the left hand side is a rational function of the form:

$ LHS\equal{}\frac{1}{f(x)}\equal{}\frac{1}{(x\plus{}\alpha_1)(x\plus{}\alpha_2)...(x\plus{}\alpha_n)}$

We know that:

$ f(\minus{}\alpha_i)\equal{}0\qquad \forall i \in S$

Now, the right hand side is a sum of rational functions, and each of the functions is of the form:

$ \frac{C_i}{x\plus{}\alpha_i}$

If we multiply the equation with

$ (x\plus{}\alpha_j) \qquad j \in S$

the LHS will becomes $ \frac{x\plus{}\alpha_j}{f(x)}$. We know that the $ f(x)$ has a part in it looking like $ (x\plus{}\alpha_j)$ so they cross each other out. Denote the new function $ f_j(x)$. Now, for the RHS. The rational function of the form:

$ \frac{C_j}{x\plus{}\alpha_j}$

becomes

$ \frac{C_j}{x\plus{}\alpha_j}(x\plus{}\alpha_j)\equal{}C_j$

and every other becomes:

$ \frac{C_i(x\plus{}\alpha_j)}{x\plus{}\alpha_i}$

Now, if we set $ x\equal{}\minus{}\alpha_j$ the LHS becomes $ f_j(\minus{}\alpha_j)$. At the same time each rational function on the RHS containing $ x\plus{}\alpha_j$ becomes $ 0$, and only $ C_j$ "lives". So, we have:

$ C_j\equal{}f_j(\minus{}\alpha_j)\equal{}\prod_{k \in S \setminus j}\frac{1}{\alpha_k\minus{}\alpha_j}$

Hope you got it.. If not, try writing the procedure for $ n\equal{}3$ (or some other) and see the practical part of the solution.

good milin.
your way also used when we study integration of some functions like something this $ \frac {1}{(...) (...)(..)(...)}$
there also exist another way more difficult  (so called : Method of uncertain factors ) about linear equations where all coefficient can be expressd as fraction of determinant some matrix's, noted that given matrix is symmetric  and determinant is polylinear cososymetric function therefor determinant can be dividid  $ (x_{j} \minus{} x_{i})$ $ i \neq j$ and e.t.c.
let's write another problem (not so simple) :
consider :
$ (n\geq 4)$ and $ \beta_{i,j}\neq \beta_{p,j}$ for all $ i\neq j$

$ \prod \limits _{i \equal{} 1}^{n}\frac {1}{(x \plus{} \alpha_{i})^{\beta_{i,j}}}$=$ \sum_{i \equal{} 1}^{n}\sum _{k \equal{} 1}^{\beta_{i,j}}\frac {A_{j,i,k}}{(x \plus{} \alpha_{i})^{k}}$
for all $ j$
i note that $ \beta_{i,j}$ $ \in N$
consider  $ (j) \equal{} (\beta_{1,j},\beta_{2,j},...,\beta_{n,j})$
and by $ (j \plus{} 1)$ denoted $ (\beta_{1,j \plus{} 1},\beta_{2,j \plus{} 1},...,\beta_{n,j \plus{} 1})$ where
as set $ \{\beta_{1,j \plus{} 1},\beta_{2,j \plus{} 1},...,\beta_{n,j \plus{} 1}\} \equal{} \{\beta_{1,j},\beta_{2,j},...,\beta_{n,j}\}$
for example $ (j) \equal{} (1,2,3,4,5)$ then  $ (j \plus{} 1) \equal{} (5,1,2,3,4)$ or $ (j \plus{} 1) \equal{} (2,3,1,5,4)$or something like this . and e.t.c
we consider all different changing of $ (j)$ it's equal to $ n!$ and for all $ (j) \equal{} (\beta_{1,j},\beta_{2,j},...,\beta_{n,j})$ we consider
$ \prod \limits _{i \equal{} 1}^{n}\frac {1}{(x \plus{} \alpha_{i})^{\beta_{i,j}}} \equal{} \sum_{i \equal{} 1}^{n}\sum _{k \equal{} 1}^{\beta_{i,j}}\frac {A_{j,i,k}}{(x \plus{} \alpha_{i})^{k}}$
what can you say about
$ \sum_{i,j} A_{j,i,k}$ ?can we say that it's equal to $ 0$ for all $ k \equal{} 1,...max \{\beta_{i,j}\}_{i \equal{} 1}^{n}$?
for example :
consider
$ \prod _{1}^{n}\frac {1}{x \plus{} \alpha_{i}} \equal{} \sum _{1}^{n}\frac {A_{i}}{x \plus{} \alpha_{i}}$
then easly seen that $ \sum_{i} A_{i} \equal{} 0$


AMC
USA(J)MO
USAMO
AIME
geometry
MATHCOUNTS
AMC 10

Meh, predictions?

I had arithmetic error on 1, arithmetic error on 6, misbubble on 8, which left me with a score of 6.

So I have a 198 index...with the 10A.

Does anyone think I have a decent chance at making USAMO?

Do you think it's only going to be floor score, or index and floor score for <10th graders? Because I got 4 retarded mistakes which leaves me with 8. and then 204.5 with the 10b. Does it matter whether u took the a or the b?

EASY AIME??? ARE YOU KIDDING ME?????
I thought the AIME was really evil this year, especially #4 and #5, which took me like 20 minutes each to solve.
I think I got a 9.

[quote="Yongyi781"]I [i]thought[/i] I got a 12, but I really ended up with a 9...[/quote]

I [i]thought[/i] I got an 11, but I really ended up with a 6...

[quote="E^(pi*i)=-1"]I doubt it.  This was an easy AIME.  I'm going to predict an index of 207-210 and a floor of 7.[/quote]

I'm going with floor 6 index 205 since it was harder than last year but still somnewhat easy

[quote="Yongyi781"]I [i]thought[/i] I got a 12, but I really ended up with a 9...[/quote]

ditto

asdf

[quote="Yongyi781"]... which means your other 2 errors were "smart" errors?[/quote]

No, just bad counting.

ugh all u guys saying 205 r scaring me w/ my 204.5 .......

[quote="xpmath"]I thought this was a fairly easy AIME, although it was harder than last year's. Unfortunately, I had the flu and couldn't take it officially. I'm betting the floor will be 7 and the index will be ~205.[/quote]

Hmm how could the floor not be six? Couldn't someone have gotten a 150/6?

Dangit I should have gotten a 150 AMC so I could pull that off

Well, it's the AMC 12 that decides the floor, I think. Plus, I personally think that someone with a 150 on the 12 should be able to get at least a 7...we'll just have to see.

I got 224, can I breathe easy and consider myself in?
(I'm a senior, first time if I get in :D)

While I agree that the AIME II was much harder than the AIME I, I don't think you can necessarily use the words "overwhelming majority," perfect628.
Yeah, there's a poll about it and a few dozen posts saying that it was harder, but that may not be the overwhelming majority or even the majority.
Of course, I'm hoping that it is.

I got an 8 on AIME II (officially) while getting a 126 on AMC 10A.  You might notice that I sucked on my AMC score; this is because I'm just naturally better at AIME than AMC, and because I've been taking WOOT (so I've been focused on Olympiad math and above, not even AIME math) so I was totally out of shape for the much lower levels of the AMC this year (then I did some massive studying for the AIME on the day before so that I could get back in shape; I know that cramming isn't generally recommended but I think it doubled my score).
So my index sits at the ambiguous number of 206, as a 9th grader.  Do you guys think I will make it to USAMO?  The past indices for 10th grade and below have been lower than 206, but maybe this year (because of the apparently "easy" AMCs) the index will rise.  Hopefully they won't even use an index, because my 8 should let me pass :).

216.5 (AMC12A - 136.5, AIME II - 8) (assuming no bubbly errors)

EDIT: 2000th POST.

They better not make the index 204.5. I got a 204~
Just to tell you guys, the poll as of the last time I checked (two minutes ago) said that about 80-some % said AIME II was harder. I guess it depends on your definition of overwhelming, but that's a pretty large majority at least.
I'm biased - I just want the AIME II index to be lower because that's the one I took  :lol:

I got 120 AMC 10 B, 6 AIME. I could have gotten 26 more points if I had remembered to double or halve during three different problems. In total stupid errors cost me 42 points.

how could it be a floor of 6? has anyone confirmed that there is a 150/6? i say floor 7, ~204 index.

[quote="Quickster94"]I believe that the USAMO teacher handbook says invites go out 4/10[/quote]
Invitations via e-mail go out on 4/10, and invitations via mail go out little bit later.

@MathAndKnowledge
- I'm in the exact same situation as you, except I'm in 8th grade and got 8 on AIME I. :D

I'm thinking the highest possible numbers will be 210 index and 7 floor.

[quote="zman"][quote="Zeph"]We had a confirmed 144/6, I think, so an index of 204 would mean that the floor would be 6.[/quote]

Not necessarily true, if my understanding is correct: The frist cutoff is, say 210 with lowest AIME = 7; that is, floor = 7, then they only consider all the left overs with an AIME score of 7 or above, conseqently the second cutoff could be lower than 204, say 195 = 125 + 7*10. And the 144/6 with index 204 will miss the cut.[/quote]

Well, obviously if the index cutoff is 210, the 144/6 will miss the first cut.  Zeph is just saying that [b]if[/b] 204 was the index cutoff for AMC 12, then the floor would most likely be 6.

Yes, I forgot to say "cutoff" after index. My post should've said:

We had a confirmed 144/6, I think, so an index cutoff of 204 would mean that the floor would be 6.

I know one who has 144(12)/5 and 150(10)/4, they all qualified USAMO last year but certainly not this year. :maybe:

personally i think AMC 10 was MUCH MUCH easier than last year. i think the top 1% on A and B were (just a guess, not based on any stats). 123 and 125 respectively.

AIME I was much easier than both AIMEs last year and AIME II was harder (but not much harder) than both AIMEs. so overall i think the cutoff (for 10) will be around 208-210.

for 12, i don't know, maybe around 204-206.


I can't figure this out
If f: X>Y is a bijective function, prove that its inverse is unique.

It's fairly trivial.  Suppose there exist two inverses $g, h$ and suppose that $g(y) = x$.  Then $y = f(x)$.  But because $f(h(y)) = y$ and $f$ is bijective it follows that $h(y) = x$, so $g, h$ are identical.


ratio limit geometry rectangle trigonometry probability Gauss

what is the golden ratio? what is the purpose of it? and how do you derive it ?

Let [tex]F_k[/tex] denote the kth Fibionacci (probably not spelled right) number.  The golden ratio phi is defined as

[tex]
\begin{eqnarray*}
\displaystyle\phi & = & \lim_{k \to \infty} \frac{F_k}{F_{k-1}}\\
 \displaystyle      & = & \lim_{k \to \infty} \frac{F_{k-2} + F_{k-1}}{F_{k-1}}\\
  \displaystyle     & = & 1 + \lim_{k \to \infty} \frac{F_{k-2}}{F_{k-1}}\\
   \displaystyle    & = & 1 + \frac{1}{\phi}
\end{eqnarray*}
[/tex]

Solving for [tex]\phi[/tex] gives [tex]\phi = \frac{1+\sqrt{5}}{2}[/tex]  Note that we want the positive solution since the limit is a ration of two positive numbers.  As for the use, it is used in a formula for the kth Fibionacci number, and the Greeks thought that the best looking rectangles had a side ratio 1:[tex]\phi[/tex]

Only use \displaystyle at the beginning of your equation (not at every line in the eqnarray), and then it will come out correctly.

The golden ratio, what a great number.  To put what nr said in words: it can be defined as the number you get when you take the ratio of two consecutive fibonacci numbers(bigger on top).  Then keep doing this without bound.  So basically the bigger the fibonacci numbers you have then the closer it is to phi(the golden ratio).  This is kind of just a property of the golden ratio though.  It is believed that the first definition was given by Euclid.  He said that a line is in "mean and extreme ratio" if when divided into two sections the ratio of the larger section to the smaller is equal to the ratio of the whole line segment to larger piece.

There are many more interesting facts about it and I could go on forever.  If you really want to delve deep into the spectacular characteristics of it I recommend reading the Golden Ratio.  I can't remember who wrote it but it is good.  If you don't want to do that use google or ask us more questions.

My favorite definition is that it is the solution to x=1-1/x.

Just a brief historical comment. That definition you mentioned joml was known before Euclid. In fact it was very famous to the early Greeks. There are a lot of things commonly attributed to Euclid that he did not discover, because he wrote of them in [u]The Elements[/u]. But [u]The Elements[/u] was really more like an almanac/textbook of all of the mathematics known to the Greeks up until that point. Some of its results were original by Euclid, but many were known before him.

For an interesting historical discussion of Greek mathematics and the Golden Ratio I recommend the the few sections about it in [u]A History of Mathematics[/u] by Carl Boyer. Its actually a very interesting read.

For the Greeks,  the Golden Ratio (which got its name much much later) was a geometric concept, not algebraic.

Here is the original motivation, written as a challenge:

Given an AxB rectangle with A<B.  Remove an AxA square.  Prove that if the original rectangle and remaining rectangle are similar, then B/A equals the Golden Ratio.

A second challenge:  Prove that the ratio of the diagonal of a regular pentagon to its side-length is the Golden Ratio. (Trig not required)

"The Golden Section" by Hans Walser, available from the Math. Association of America  ( http://www.maa.org/Welcome.html ) describes its application to fractals, number theory, probability and 2D and 3D geometry - all math. in it is precalculus - good read!

Mathworld has a nice page on the Golden Ratio: [url]http://mathworld.wolfram.com/GoldenRatio.html[/url]

There was a program on the Discovery Science Channel the other day on the Golden Ratio. As Gauss and others said, the Greeks were fascinated by the Golden Ratio, and it is used in many classical buildings. It also recurs in nature (they showed a picture of a fish and many other animals to demonstrate this). [/list]

Don't they use the golden ratio to make the best looking note card?

Hmmm... 5/3 (standard small notecard) is approximately the golden ratio.

And it has also been said that the distance between human features (like the eyes) can follow the Golden Ratio!


Let x,y,z be distinct non-negative real numbers. Prove that:
$ \frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(y\minus{}z)^2}\plus{}\frac{1}{(z\minus{}x)^2}\geq\frac {4}{xy\plus{}yz\plus{}zx}$and when does the equality hold？


A machine for making pizza puts

salami on every 18th pizza
pinapple on every 21st pizza
peppers on every 20th pizza

how many pizzas have toppings on them?
how many pizzas have pepper and salami but not pinapple on them?

Could you please show me how you arrived at the answer?

This has already been posted by you here: [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=41145]http://artofproblemsolving.com/Forum/viewtopic.php?t=41145[/url]. Please try to not post the same topic twice, especially in the same forum. Refer back to the old topic if you have further questions.


Google Inc. has been in the news recently due to its "slower" fourth-quarter revenue growth. During this year, the sequential revenue growth rate for the company was 7.5 percent in the second quarter and 15.1 percent in the third quarter. In order to continue an arithmetic sequence of growth for the quarters, what would the growth for the fourth quarter have to be?

This is a well-constructed problem. Good Job!

thanks, 4everwise, do you want to try and solve it?

It's already been solved and there are no other ways to approach it, so I'm guessing he didn't feel the need to solve it again.


no and this is what i mean about random junky posts... i bet you just geussed

Posted by MrMcGeek

The answer is [hide]pi / 4[/hide]. I don't know if there's an easy proof of this, but I do know that it's obtained from the series of the arctan function:

[hide]arctan x = :Sigma:i = 1:inf: (-1)i + 1 * x2i - 1 / (2i - 1).

Plugging x = 1 gives

arctan 1 = 1 - 1 / 3 + 1 / 5 - 1 / 7 + ...

Now, tan (pi / 4) = 1. The result follows.[/hide]

That was not particularly random or junky -- I mean, there wasn't any math to back it up, but it was at least math-related.  Anyhow, this isn't really intermediate level -- definitely the other series question isn't.  They don't have elementary solutions.  Especially not the other one.

Leeyuhang -- note that your answer is wrong right off the bat, since if we take the 1 at the beginning and then pair off terms after that, we see that we have 1 minus a bunch of positive numbers which can't add up to a number bigger than 1.

oh, i read it wrong. i don't really get this, what does pi have to do with these series?

That was the power series for arctan x when x was 1.  Look it up at MathWorld, I'm pretty sure it's there.  That is encountered later in calculus though so don't worry about it now.

hm, this question was in my geometry book....

of course its in the geometry book.  Its in a section where they talk about pi appearing in funny places.  And another exercise whats 1/(1^2)+1/(2^2)+1/(3^2)... and Leehuyang dont answer this if you look in the book

When I first found out about that series (1-1/3+1/5-1/7+...) when I was a little kid, I had my father make a computer program for it, and we let it run all day. We got about 10 digits of pi from it. It was rather disappointing.


This is an idea I got from the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url] message board.  Basically, you rate how hard your problem is on a scale of 1 to 10.  The material in the Intermediate forum should be from late AMC 10/middle AMC 12 to most of the AIME.

So, roughly speaking, the last 3 or 4 problems on the AIME would be 10's and the middle AMC problems should be in the lower third of the range.  I will post some examples of problems and their difficulty in here.

This system should only be used for challenge problems.  If you are asking for help you probably don't know how hard the problem is.  Only through the combined effort of everyone will this system be successful :)  It would good in the case of someone posing a problem that they are unsure of the difficutly for other members to offer suggestions to the difficulty.  That way we can always go back and edit it so that most problems will have a rating for them.

[size=150][b]Example Problem Ratings[/b][/size]

(AHSME 1987 #15)  [color=red]Difficulty Level: 3[/color]

If $(x,y)$ is a solution to the system

\[xy=6 \ \ \text{and} \ \ x^2y+xy^2+x+y=63,\]

find $x^2+y^2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44145]Click here to view the thread for the above problem.[/url]

(Mexican 1988 A3) [color=red]Difficulty Level: 6 or 7[/color]

Two circles of different radii $R$ and $r$ touch externally. The three common tangents form a triangle. Find the area of the triangle in terms of $R$ and $r$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44147]Click here to view the thread for the above problem.[/url]

(AIME 2001 #14)  [color=red]Difficulty Level: 10[/color]

Find the remainder when sum of all values of $\theta$, $0\leq \theta <360^\circ$, for which the complex number $z=\cos \theta +i\sin \theta$ satisfies $z^{28}-z^8-1=0$ is divided by 1000.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44146]Click here to view the thread for the above problem.[/url]

lol, your rating system do not have much effect on me. Since I know nothing about American's Contest. But I shall try to post the suitable problems.

I think after you see it in affect for a while, you will see about the range of what level the problem you post is.

In theory a rating system is a good idea, but I think in practice it would be very hard to have any uniformity to the ratings. Different people have vastly different opinions of how difficult a problem is, and a lot of the problems here don't come from AMC contests. We see arguments over whether a problem even belongs in a certain forum all the time. I imagine that arguments over how difficult a problem is [i]within[/i] a forum could become commonplace.

I suggest that we have enough division of the forums right now to just let the problems be put in the appropriate forum and leave it at that. The fact that there's now a Beginners Forum and a Pre-Olympiad forum helps to keep the problems posted here at a pretty uniform level.

Yeah, you're right.  After thinking about it, it seems like too much work and wouldn't be all that beneficial.


geometry
LaTeX
Asymptote
geometry theorems

which is the best 'easy to learn' geometry software to draw geometric diagrams for incorporating in geometry books. I have seen several ones but none produce the the drawings in a format suitable to be included with text. I am looking for something very simple and fast (preferably free), and not heavyweights like Mathematica or Matlab.

geogebra. you can convert it into latex code (i don't know how to do it, but i saw my friend did that).

Try Geometry Expressions 1.0
  It is simple and good software to draw figures.

Use asymptote, export to an .eps file, and import into the .tex document.

Although it might not be 'easy to learn,' though... but coding skill is valuable to have.

(1) Geogebra is not working stating it requires Java 1.4 or above. I have Java 1.6 installed.
(2) Geometry Expression is not free and not versatile enough.
(3) Asymptote is command-line program and not windows based.

Anyway thanks for suggestions.

Asymptote has windows ports. But yeah, it's CL-based. But so is LaTeX; that shouldn't be an obstruction.

Writing books directly in Latex is a lifetime job. We have to use some Latex converter to take care conversion to Latex.


Prove that centre of mass is a unique point

I started with O as origin.G as CM and G' as assumed 2nd CM in the same sysyem of particles.

OG=R=(1/M) sum(i) [m_i*r_i]

OG'=R'=(1/M) sum(k) [m_k*r'_k]

The problem is theere is no unique relation between r_i and r'_k

let $O$ be the origin
let  $P$ be the  centre of mass

$\vec OP$ $=$ $\frac{\int_{S}\vec r_{i}dm}{M}$

where $\vec r_{i}$ is position vector ,$S$ is the system  you are corncerned with amd $M$ is the total mass of the system

now when does the above expression hold true this clearly holds if
the following limit exists and is unique

$\vec OP$ $=$ $\lim_{max \Delta M_{i}\to\ 0}\lim_{n\to\ \infty}\frac{ \sum_{i=1}^{i=n}\Delta M_{i}\vec r_{i}}{M}$
now in your discrete form uniqueness is obvious because there can only be one $\vec r_{i}$ can only be one

I am not sure.Suppose a system is such that its particles are self -interacting.In that case,[b]r_i[/b] s will change to some other valye [b]r'_i[/b].Still R_cm should be unique.

i think u r confused with $\vec r_{cm}$ is 'unique' or it moves . :maybe:
anyways see by the definition of centre of mass it has to be unique. also since the particles are self intertacting no net force is exerted on the centre of mass so if it  were moving with  some  velocity initially it will continue to do so.

I want to prove that CM is unique.It may be a definition.But in a question paper it appeared.

what i am stressing is by definition centre of mass has to be unique.

well of course  u can give a proof as i have partially  done above but that's meaningless and like proving 1+1 is unique :)

OK.I see.A definition can not be questioned...

I hope to add something.
There is a formula for computing the center of mass, and this formula does not involve solving for any implicit variables. It is merely a function of the masses and positions of particles within the system. By the definition of a function, it cannot have two values when the same variables are used as input. Therefore, simply because G has a formula which is a function, the center of mass is unique.

Another check is contradiction can be found if you assume that there are exactly x (x => 1) centres of mass. It can be shown that, when you have x centres of mass (x => 1), you can show that there are at least x + 1 centres of mass. Therefore the original assumption of exactly x centres of mass cannot be true.


The real number x1,x2,...,x6 are chosen inside the interval [0,1].
Prove that

 (x1-x2)(x2-x3)(x3-x4)(x4-x5)(x5-x6)(x6-x1) \leq 1/16

Let y_i = x_i - x_(i+1)  (with x_7 = x_1), then :
y_1 + y_2  + ... + y_6 = 0  (1)
and -1  \leq y_i  \leq 1 for each i  (2).
We want to prove that y_1*y_2*...y_6  \leq 1/16. (3)

Now, we forgive the x_i's and prove (3) under (1) and (2).
note that (3) trivially holds if some y_i is 0 or if the number of y_i < 0 is odd. Thus, we may assume that the number of y_i < 0 is even. From (1), this number is 2 or 4. Since we may change each y_i into -y_i, we assume that there are exactly two of the y_i which are < 0, say y_1,y_2 < 0 (the problem is symetric in y_i).
Then : 0 < y_1*y_2  \leq 1 and 0 < -(y_1 + y_2) \leq 2.
It follows that :
P = y_1*...*y_6  \leq y_3*y_4*y_5*y_6  \leq [ (y_3 + y_4 + y_5 + y_6)/4 ] 4  (from AM/GM)
That leads to P  \leq ( -(y_1 + y_2)/4 ) 4  \leq (1/2) 4 = 1/16.

Pierre.


[b]Consider how you would answer the following question differently on this forum and how you would answer it on the SAT writing section. What examples would you use to support arguement? Would you respond differently on a test than you would on a forum? How so?[/b]

"How do you cherish good memories? Reminiscing about good times is depressing for me because I can not go back in time to relive happy times. I always imagine some day when I'm older I'll look back to the times when I did competitive math contests and swimming competitions in high school and be sad that I have moved on and put the past behind me... " -Mr. Elmo the Freaking Elmo

Do you think that it is better to just forget the past and focus on the present?
Plan and write an essay in which you develop your point of view on this issue. Support your position with reasoning and examples taken from your reading, studies, experience, or observations.

Well on a forum, I wouldn't write an entire essay and I wouldn't be as detailed.  On the test, I would definitely write more in-depth and write more.

Another thing is, people dont care as much on forums. On forums, they try to express their opinion. However, on the SAT their goal is not to express their opinion, it's to get the highest score. So on the SAT, people choose sides that are easier to support, not what they believe in. Also, they make up experiences that help support their sides (which is actually easy, as long as you don't call your friend Bob).

On the SAT, I would choose Elmo. On the forums, I would choose Err...mo?. I would choose Elmo on the SAT because it makes the most sense of all the other answer choices. I would choose Err...mo? on the forums because none of the answer choices really make much sense and because nobody really cares what you say on the forums (or not as much as on the SAT).

Outline: Answer Yes!!

I. Intro - Attention getter: Someone that drives a car and only focuses on what is in the rear mirrors is bound to crash story.

II. Eastern Mysticism
    1. The main teachings of Buddhism
    2. Lao Tzu and the Tao Te Ching
    3. Generalize to happiness as "in the moment" thinking

III. Personal Experience
    That time I got a bad grade on a test, and instead of moving on I dwelled on it. It caused me to suffer in my other classes. If I would just move on I would have continued to do well, and even brought that grade up.

IV. Literature
    1. Thematic content of The Bell Jar
    2. The symbolism of the the jar in The Bell Jar
    3. The actual outcome of Sylvia Plath

V. Conclusion: Rearrange the words of the intro!! You don't have time for something new.

NOTE TO SELF: Use big words and complex sentence structure to trick the grader into thinking I'm a good writer.

This is meant to be funny, but probably only is to me since I literally write something along those lines onto my scrap paper any time I write one of these things.

Actually, you probably should avoid complex sentence structures since they take more time to write and just shield from what you're trying to say, which will result in you losing points.

In real life I completely agree. Complex sentence structure is a pain to write well, and even more of a pain for the reader to decipher. It most often known as the "pompous style" or the "academic style." It is known as this for a reason. Most very well educated people fall prey to the pompous style and end up writing fluff with no content. But for something like the SAT where a grader will spend a max of 60 seconds reading and deciding upon a grade, the pompous style makes you sound like a great writer. I know. It's horrible. But that is the game of a standardized test.

[quote="Nerd_of_the_Ages"]So on the SAT, people choose sides that are easier to support, not what they believe in.[/quote]
hmm.. I always find that the side I believe in is the one that's easier to support. :)

I only use alternative words if it condenses the sentence. For example, in my previous sentence, I could've said "reduces the length of" in place of "condenses." :P Otherwise, I let the words flow naturally.

[quote="lingomaniac88"]I agree with the above post.  I prefer to call it pompous, because that's what it really is.  If this is what a teacher, professor, or standardized test grader expects, then I will write this way and die a little inside.  I prefer keeping my language simple yet powerful, and I believe this is the most effective style.  Using this style makes it easier for a reader to grasp a piece of writing.  This is especially necessary when you are arguing a position like in an SAT essay.

Has anyone ever read [i]Politics and the English Language[/i] by George Orwell?  He gives a good example of "simple is powerful" with Ecclesiastes 9:11.

[quote="George Orwell"]I am going to translate a passage of good English into modern English of the worst sort. Here is a well-known verse from Ecclesiastes:
"I returned and saw under the sun, that the race is not to the swift, nor the battle to the strong, neither yet bread to the wise, nor yet riches to men of understanding, nor yet favour to men of skill; but time and chance happeneth to them all."

Here it is in modem English:
"Objective considerations of contemporary phenomena compels the conclusion that success or failure in competitive activities exhibits no tendency to be commensurate with innate capacity, but that a considerable element of the unpredictable must invariably be taken into account."[/quote][/quote]

hah I understood the second one better
was that the way it's supposed to be

the first one uses way too many commas

it's like get to the point already

man this is probably why all my teachers are like 'USE SENTENCES THAT ARE LESS THAN THREE PAGES LONG' on my essays

Indeed.  In reference to above poster, yes sometimes the one you agree with is easier to support but it may not be.  What if you agree but you can't think of any references or proof, while for the other topic, you are neutral but have tons of references.

That, and you could always BS something, which is basically what they expect you to do.

I am convinced that 2 pages of miscellaneous quotations and large words will score a 12*.

I've heard of several people scoring 12's without a conclusion, and one case, having ended mid-sentence.

Not me, though; I made the stupid mistake of focusing on content, and wasted valuable energy I could've saved up for the 3rd CR section.

Now, the SAT essay approach that I [i]actually[/i] used is quite similar to what I would do on a forum -- pick a side, think of strong examples that aren't trivial or overused (if everyone's going to write about MLK, write about Gandhi), decide which order to present my arguments in, think of any applicable analogies/metaphors/sidenotes, and then write. And review. And review again.

[u]* Disclaimer:[/u] [i]If you use the above strategy and [b]don't[/b] get a 12, I'm not to blame. It's probably because you're not an Asian girl  :rotfl: (I have statistics to back this up)[/i]


Hi...
Will anyone help me whith this sum?
[img]http://img132.imageshack.us/img132/944/cot2nc4.jpg[/img]

$ \sum_{k \equal{} 1}^n\frac {\pi k}{2n \plus{} 1} \equal{} \frac\pi{2n \plus{} 1}\sum_{k \equal{} 1}^nk \equal{} \frac\pi{2n \plus{} 1}\left(\frac {n(n \plus{} 1)}2\right) \equal{} \boxed{\frac{\pi n(n\plus{}1)}{2(2n\plus{}1)}\equal{}\frac {\pi n^2 \plus{} \pi n}{4n \plus{} 2}}$


Prove that if the fraction

 a/b =1-1/2+1/3-1/4+......-1/1334+1/1335

is reduced to its lowest terms the a is a multiple of 2003.

This one can be used each year with a prime number ! Next is 2011  :)

S(n) = \sum 1/i, i = 1 .. n

1-1/2+1/3-1/4+......-1/1334+1/1335 = S(1335) - 2*S(667) = 1/668 + 1/669 + ... + 1/1135.

There is an even number of terms and 1/668+i + 1/1135-i = 2003 / (668+i)(1135-i).

Now 2003 is prime so we can't find factors in (668+i)(1135-i) to cancel it since (668+i),(1135-i) < 2003.

From which competition in 2003 does it come ?

I think it was the Pan African Olympiad in 2003.
It is based on an old IMO problem, I think 1979/1 or so.
And indeed, 1979 is prime :)

But Arne how is it possible that you seem to know every problems from every countries for the last 50+ years ??? Very impressive.
It's even the first time I hear about Pan African Olympiad. Is it an olympiad for the whole Africa with IMO level ?

It is much much easier than the IMO, and of course not all African countries participate :)
And actually, the problem is from the PAMO 2000, not 2003  :?

[url]http://www.saasta.ac.za/pamo/2002.html[/url]


[hide]
Say that Ott was given 1 dollars by each.
Thus Moe had 5 dollars.
Loki had 4 dollars
Nick had 3 dollars

Thus there are a total of 15 dollars, and Ott has 3 of them
$\frac 15$[/hide]

[hide]From what is given, we see that 1M/5=1L/4=1N/3, where M, L, and N stand for Moe, Loki, and Nick's money, respectively. Therefore, we can call Moe's money 5x, Loki's money 4x, and Nick's money 3x. After they each give to Ott (who has 3x) they each have 4x, 3x, and 2x. Put 3x over the sum of all the money and you get 3x/12x, or 1/4. [/hide]

Biffandoc, I don't think there's a total of [hide]15[/hide].

[b]biffanddoc[/b] added in otts money of $\$$3 in his example to the others original money, thus giving him $\$$15, when after subtracting he would have had $\$$12 like [b]easyas3.14159...[/b]  was saying

[quote="236factorial"]Loki, Moe, Nick, and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and NIck gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott have?

Source: AMC 8 2002, number 25[/quote]

[hide]he has 1/4 of the groups money. lets say, since each person gave him the same amount, they each gave him a dollar. he now has 3 dollars from each of his friends. this means Moe has 4 dollars, Loki now has 3 dollars, and nick now has 2 dollars. 4+3+2+3=12. 3/12 is 1/4.[/hide]

[quote]Loki, Moe, Nick, and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and NIck gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott have?[/quote]

[hide]Let us say that Loki, Moe, and Nick each gave Ott one cent. Either they are really bad friends, or they're pretty hard up.  :rotfl: Moe had five cents, Loki had four cents, and Nick had three cents. Totally, they had twelve cents (what a treasure). Ott had three cents from each of his friends, so he had one-fourth of all the group's money.[/hide]

Wow, this is easy for a number 25 on the AMC 8.


$ a,b,c>0$
$ \sum_{sym} a^5b \plus{} 3\sum_{sym}a^3b^2c \geq \sum_{sym} a^3b^3 \plus{} 2\sum_{sym}a^4bc \plus{} \sum_{sym}a^2b^2c^2$

[quote="Cai Ni"][quote="NextPeace"]$ a,b,c > 0$
$ \sum_{sym} a^5b \plus{} 3\sum_{sym}a^3b^2c \geq \sum_{sym} a^3b^3 \plus{} 2\sum_{sym}a^4bc \plus{} \sum_{sym}a^2b^2c^2$[/quote]

 :?: Dear NextPeace:Are you sure "sym".It seems wrong.[/quote]

It is true ineq. sym is symmetric.. I have two proofs.

[quote="NextPeace"]$ a,b,c > 0$
$ \sum_{sym} a^5b \plus{} 3\sum_{sym}a^3b^2c \geq \sum_{sym} a^3b^3 \plus{} 2\sum_{sym}a^4bc \plus{} \sum_{sym}a^2b^2c^2$[/quote]
$ \sum_{sym}( a^5b \plus{} 3a^3b^2c \minus{} a^3b^3 \minus{} 2a^4bc \minus{} a^2b^2c^2)\equal{}$
$ \equal{}(a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}\sum_{cyc}(a\minus{}b)^2ab(a^2\plus{}ab\plus{}b^2\minus{}ac\minus{}bc)\geq0,$ which is obvious.

good, planetarian_.

I made this problem by Jensen.
$ a\plus{}b\plus{}c\equal{}1, \ \sum_{cyc}a(a^2\plus{}bc)^2\geq(a^3\plus{}b^3\plus{}c^3\plus{}3abc)^2$(by Jensen.)
And just expand this expression.


inequalities
limit
linear algebra
matrix
calculus
integration
real analysis

Suppose for every m $ {\sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |} }$ converges and that $ \sum\limits_{m \equal{} 0}^N {\sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |} }$converges to some finite sum as n tends to infinity.

Then $ \sum\limits_{m \equal{} 0}^N {\sum\limits_{n \equal{} 0}^\infty  {a_{mn} } }$is a convergent series and $ \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {a_{mn} } }  \equal{} \sum\limits_{n \equal{} 0}^\infty  {\sum\limits_{m \equal{} 0}^\infty  {a_{mn} } }$, where the right hand side is also understood to be convergent.

In the proof, we shall assume that the theorem holds if $ \forall m\forall n,0 \le a_{mn}$.

[b]A Proof[/b]:

Let $ u_{mn}  \equal{} \frac{1}{2}(|a_{mn} | \plus{} a_{mn} )$ and $ v_{mn}  \equal{} \frac{1}{2}(|a_{mn} | \minus{} a_{mn} )$

$ \forall m,\sum\limits_{n \equal{} 0}^\infty  {u_{mn} }  \le \sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |}$
$ \forall m,\sum\limits_{n \equal{} 0}^\infty  {v_{mn} }  \le \sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |}$

By the comparison test the series on the left hand side of the inequality are understood to be convergent. From this, we see that

$ \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {u_{mn} } }  \le \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |} }$
$ \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {v_{mn} } }  \le \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {|a_{mn} |} }$

By the comparison test the series on the left hand side of the inequality are understood to be convergent. From this, we see that

$ \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {a_{mn} } }  \equal{} \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {(u_{mn}  \minus{} v_{mn} )} }  \equal{} \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {u_{mn} } }  \minus{} \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {v_{mn} } }$
$ \equal{} \sum\limits_{n \equal{} 0}^\infty  {\sum\limits_{m \equal{} 0}^\infty  {u_{mn} } }  \minus{} \sum\limits_{n \equal{} 0}^\infty  {\sum\limits_{m \equal{} 0}^\infty  {v_{mn} } }  \equal{} \sum\limits_{n \equal{} 0}^\infty  {\sum\limits_{m \equal{} 0}^\infty  {(u_{mn}  \minus{} v_{mn} )} }  \equal{} \sum\limits_{m \equal{} 0}^\infty  {\sum\limits_{n \equal{} 0}^\infty  {a_{mn} } }$

[b]Some questions[/b]
1) Is the proof correct?
2) Regardless of the answer to 1, are there simpler proofs of this fact?
3) What are the counter examples when the double series is not absolutely convergent?

I think that the proof is correct, although it could probably be tightened a streamlined a little. However, it's really just a corollary of the main theorem.

The main theorem is that for nonnegative $ a_{jk},$ we have

$ \sum_{j\equal{}0}^{\infty}\sum_{k\equal{}0}^{\infty}a_{jk}\equal{}
\sum_{k\equal{}0}^{\infty}\sum_{j\equal{}0}^{\infty}a_{jk}\equal{}
\lim_{N\to\infty}\sum_{j\equal{}0}^{N}\sum_{k\equal{}0}^{N}a_{jk}.$

This is the theorem that requires some effort to prove.

Another thought: you have clearly assumed that $ a_{jk}$ is real. How would you modify your proof for complex-valued (but still absolutely convergent) series?

In the class handout I wrote for all of this I also included the triangular partial sums as well as the square partial sums. That is,

$ \lim_{N\to\infty}\sum_{j\equal{}0}^N\sum_{k\equal{}0}^{N\minus{}j}a_{jk}.$

As for counterexamples, I think I used this: let $ a_{jk}$ be the following array:

$ \begin{matrix}1&\minus{}1&0&0&0&\cdots\\
0&1&\minus{}1&0&0&\cdots\\ 0&0&1&\minus{}1&0&\cdots\\ 0&0&0&1&\minus{}1&\cdots\\
0&0&0&0&1&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{matrix}$

For that, try adding up the row-first iterated sum, the column-first iterated sum, the limit of the square partial sums, and the limit of the triangular partial sums.

This pair of theorems (that interchanging the order in a double infinite sum is justified either if the terms are nonnegative or if the series converges absolutely) closely parallels the pair Monotone Convergence Theorem/Dominated Convergence Theorem (as applied to integrals of sums), which in turn parallels the the pair Tonelli's Theorem/Fubini's Theorem for dealing with double integrals.

For complex series, we can use the same proof by treating the real terms and the complex terms separately.

Is your class handout available online? I was actually most interested in the triangular partial sums but I have not yet attempted to modify my proof for that.

In your example, row-wise the sum is 0, column-wise the sum is 1. The triangular sum seems to be divergent since it apparently oscillates between 1 and -1. For the square partial sums it looks like we have 1. That is quite an rich example; I will show it to my TA.

EDIT: I think for the triangular sums, you can apply what has already been proved by extending the inner sum from N-j to N while interpreting those terms with indices higher than N-j as 0. Then you can perform the interchange using what has been proved and then have the inner sum range from 0 to N-k without changing the value.

The case I was most interested in is as follows. I hope the proof is also correct:

[b]Proof[/b]

If $ a_{kj} \ge 0,\forall j\forall k$ then let $ b_{kj} \equal{} 0,j < k$, $ b_{kj} \equal{} a_{kj} ,j \ge k$

According to the theorem for double series of positive terms stated in the OP,

$ \sum\limits_{k \equal{} 0}^\infty {\sum\limits_{j \equal{} k}^\infty {a_{kj} } } \equal{} \sum\limits_{k \equal{} 0}^\infty {\sum\limits_{j \equal{} 0}^\infty {b_{kj} } } \equal{} \sum\limits_{j \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^\infty {b_{kj} } } \equal{} \sum\limits_{j \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^j {a_{kj} } }$

The corresponding lemma for absolutely convergent double series may be modified to show that if $ \sum\limits_{k \equal{} 0}^\infty {\sum\limits_{j \equal{} k}^\infty {|p_{kj} |} }$or $ \sum\limits_{j \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^j {|p_{kj} |} }$
 converges then the other converges and $ \sum\limits_{k \equal{} 0}^\infty {\sum\limits_{j \equal{} k}^\infty {p_{kj} } } \equal{} \sum\limits_{j \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^j {p_{kj} } }$.

In particular we have the [b]change of base point theorem[/b], which is what I was trying to prove all along:

Suppose the power series $ \sum\limits_{k \equal{} 0}^\infty {a_n (z \minus{} a)^n }$has radius of convergence R>0 about a and some $ z_o$ satisfies $ |z \minus{} z_o | < r \equal{} R \minus{} |z_o \minus{} a|$. For every z satisfying $ |z \minus{} z_o | < r \equal{} R \minus{} |z_o \minus{} a|$ $ \sum\limits_{n \equal{} 0}^\infty {|a_n |(|z \minus{} z_o | \plus{} |z_o \minus{} a|)^n } \equal{} \sum\limits_{n \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^\infty {|a_n |} } \left( \begin{array}{l} n \\
k \\
\end{array} \right)|z \minus{} z_o |^k |z_o \minus{} a|^{n \minus{} k}$converges by definition of radius of convergence.

The above comments show that
$ \sum\limits_{n \equal{} 0}^\infty {\sum\limits_{k \equal{} 0}^n {a_n \left( \begin{array}{l} n \\
k \\
\end{array} \right)(z \minus{} z_o )^k (z_o \minus{} a)^{n \minus{} k} } } \equal{} \sum\limits_{k \equal{} 0}^\infty {\left( {\sum\limits_{n \equal{} k}^\infty {a_n \left( \begin{array}{l} n \\
k \\
\end{array} \right)(z_o \minus{} a)^{n \minus{} k} } } \right)} (z \minus{} z_o )^k$ converges.


Let $n \geq 2, n \in \mathbb{N},$ find the least positive real number $\lambda$ such that for arbitrary $a_i \in \mathbb{R}$ with $i = 1, 2, \ldots, n$ and $b_i \in \left[0, \frac{1}{2}\right]$ with $i = 1, 2, \ldots, n$, the following holds:

\[\sum^n_{i=1} a_i = \sum^n_{i=1} b_i = 1 \Rightarrow \prod^n_{i=1} a_i \leq \lambda \sum^n_{i=1} a_i b_i.\]


Given a circle of n lights, exactly one of which is initially on, it is permitted to change
the state of a bulb provided one also changes the state of every dth bulb after it (where d is
a divisor of n strictly less than n), provided that all n/d bulbs were originally in the same
state as one another. For what values of n is it possible to turn on all bulbs via a series of
moves of this kind?

The version of this problem without the condition "provided that all n/d bulbs were originally in the same
state as one another" seems also to work, but I forgot the solution in that case... can anyone think of it?


algebra
polynomial
Rational Root Theorem

Let f(x) be a polynomial with integer coefficients such that f(k)=k, where k is a positive integer, and f(0)=p, where p is a prime number and p!=k.  Find k.

[hide="I think this is right"]

I think k = 1, and p = 2.  f(x) = - x + 2.[/hide]

<shrugs> not so hard, once you think about it,

[quote="jen7"]Let f(x) be a polynomial with integer coefficients such that f(k)=k, where k is a positive integer, and f(0)=p, where p is a prime number greater than k.  Find k.[/quote]
is this for all p above k?

yes, this is true for all p greater than k

Here's a more rigorous approach:

[hide]Let p(x) = f(x) - k.  Then you know the constant term in p(x) is p-k.  p(x) has a root k, so by the rational root theorem, p-k must be a multiple of k.  Under normal circumstances, this is impossible, since GCF(p-k, k) = 1.  We must have k = 1.  p can be anything. [/hide]

Edit--actually, in general, shouldn't this still be true for p < k?

[quote="probability1.01"]actually, in general, shouldn't this still be true for p < k?[/quote]
but since k=1, there is no p<k

Yeah, but I mean you didn't have to say p > k in the problem.  Just p != k.


[quote="village"]$a,b,c\geq 0$ prove that $\frac{1}{a+b+c}$ $\geq$ $\frac{ab}{(1+a+ab)(ab+bc+ac)}$ $+\frac{ac}{(1+c+ac)(ab+bc+ac)}$
$+\frac{bc}{(1+b+bc)(ab+bc+ac)}$[/quote]
It's wrong: $a=\frac{1}{2},$ $b=4$ and $c=0.$ :wink:

[quote="village"]yeap that was wrong because i made mistake when i was writing. $a,b,c>0$[/quote]
It's still wrong: $a=\frac{1}{2},$ $b=4$ and $c\rightarrow0^{+}.$


algebra
polynomial
combinatorics unsolved
combinatorics

Help me to solve this problem.
Prove for every integer n:\[ \sum_{i=0}^n (-1)^i\binom{2n-i}{i}2^{2n-2i} = 2n+1 \]

Take the Vandermonde summation formula:

\[ \sum_{k=0}^{c}\binom ak \binom b{c-k} = \binom{a+b}{c} \]

Since both sides are polynomials in $a$ and $b$, we can plug in negative and rational values if we want.

Now specialize: $c=n$, $a=n-\frac 12$, $b=-n-1$ which gives your identity.


hint in spoiler:

[hide]try computing [tex]\displaystyle \large {(e^{i \theta})}^3 [/tex]two different ways.[/hide]

[hide]e^ix  = sinx  + cosx  wher x is any angle[/hide]

[tex]\displaystyle{\sin(Arg(z))=\frac{Im(z)}{|z|}=\frac{y}{\sqrt{x^2+y^2}}}[/tex], alternatively, if that's helpful.

I hope I'm answering the same question you're asking:

Because of the way angles are defined: y corresponds to sine, x corresponds to cosine; the imaginary part corresponds to sine of the argument, the real part corresponds to cosine.

Also, Rep, the first expression in your displayed equation is not correct.  It should be [tex]\cos(2 \theta) + i \sin(2 \theta)[/tex], not just [tex]\sin(2 \theta)[/tex].  Then in the final step you are just comparing coefficients of i.


Does anyone know where I can get .pdf copies of AMC12, AIME, and USAMO for the 2008 and 2009 competitions?  The ones that i downloaded off the site have weird problems when I download them: some of the problems involving Geogebra get corrupted.  Any help would be appreciated.

[url]http://www.math.ksu.edu/main/handbook/ProblemSets[/url]

for more go here:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286559[/url]

I only need strictly amc 12, aime, and usamo from 2008 and 2009.  also, anything in good quality before 2000.

The AoPSWiki doesn't have PDFs. You should try the Contests section. http://www.artofproblemsolving.com/Forum/resources.php?c=182


Find x of the equation:
$ x^4 \plus{} 4x^3 \plus{} 12x^2 \plus{} 18x \plus{} 24 \equal{} 0$.

$ LHS\equal{} (x^2\plus{}2x\plus{}3)^2 \plus{} 2(x \plus{}\frac{3}{2})^2 \plus{}\frac{21}{2} >0$
hence  equation hasn't root

Thank! I have a other solution of this problem. I will post it soon.

We have $ x^4 \plus{} 4x^3 \plus{} 12x^2 \plus{} 18x \plus{} 24 \equal{} 0$ $ \Longleftrightarrow$ $ (x^2\plus{}2x)^2\plus{}2(x\plus{}\frac{9}{4})^2\plus{}\frac{111}{8}\equal{}0$
hence equation hasn't root.


abstract algebra
vector
superior algebra
superior algebra unsolved

suppose R is an abelian ring.if the number of idempotents in R be finite.show the number of idempotents in R is the potent of 2?

this is because the idempotent elements form a boolean ring under the same multiplication and the new addition e & f = e + f - 2ef. observe that e & e = 0, i.e. we have a finite vector space over F2.