ratio
quadratics
geometric sequence
arithmetic series
algebra
quadratic formula
arithmetic sequence

just found this forum...looks pretty useful (i know I'll need the help)

1. In a given geometric progression all of whose terms are positive, any term is equal to the sum of the next two following terms.  Find the common ration

2. For a given arithmetic series, the sum of the first 50 terms is 200, and the sum of the next 50 terms is 2700.  What is the first term of the series.

3. Three numbers a, b, and c, none of them zero, form an arithmetic progression.  Increasing a by 1 or increasing c by 2 results in a geometric progression.  Find the value of b

I have worked all the problems out, and always get the wrong answers or dead-end.

umm for 2 just use the formula for the sum of an arithmetic series you get two equations and then you can solve for a

[hide="1."]
Let the geometric sequence be $a, ar, ar^{2},\ldots , ar^{n},\ldots$.

Then, $ar^{n}=ar^{n+1}+ar^{n+2}$. Solving for $r$, you get

$r^{2}+r-1=0$.

Find the roots of the equation, and use the fact that the terms of the sequence are positive to determine the common ratio.[/hide]

[hide="1"] Let's take three consecutive terms $a, ar, ar^{2}$ where $r$ is the ratio.  Then

$a = ar+ar^{2}\implies 1 = r+r^{2}\implies r^{2}+r-1 = 0$

Quadratic formula gives $r = \frac{-1 \pm \sqrt{5}}{2}$.  Because all the terms are positive, the ratio must also be positive, so $r = \boxed{ \frac{-1+\sqrt{5}}{2}}$. [/hide]
[hide="2"] Say that our first term is $a$ and our difference is $d$.  Then the first $50$ terms are

$a+(a+d)+(a+2d)+...+(a+49d) = 200$

And the next $50$ terms are

$(a+50d)+(a+51d)+...+(a+99d) = 2700$

Subtracting our equations,

$50d+50d+...+50d = 2500$

There are $50$ terms, so this equation becomes $2500d = 2500 \implies d = 1$.

Adding our equations (reversing one of the equations),

$(2a+99)+(2a+99)+...+(2a+99) = 2900$

Again there are $50$ terms so this equation becomes $100a = 2900-99 \cdot 50 =-2050$, so $a = \boxed{-20.5 }$. [/hide]
[hide="3"] We can write our arithmetic progression as $b-d, b, b+d$, where $d$ is the difference.

The problem tells us that $b-d+1, b, b+d$ is a geometric progression, so

$\frac{b+d}{b}= \frac{b}{b-d+1}\implies b^{2}= b^{2}-d^{2}+b+d \implies d^{2}= b+d$

The problem also tells us that $b-d, b, b+d+2$ is a geometric progression, so

$\frac{b+d+2}{b}= \frac{b}{b-d}\implies b^{2}= b^{2}-d^{2}+2b-2d \implies d^{2}= 2b-2d$

So $b+d = 2b-2d$.  This gives $b = 3d$.  Plugging in to the first equation, $d^{2}= 3d+d = 4d \implies d = 0, 4$.

Since $d = 0 \implies b = 0$ but the problem tells us none of $a, b, c$ are zero, we conclude that $d = 4, b = \boxed{12}$.  Our original progression was

$8, 12, 16$

Which becomes the geometric progression

$9, 12, 16$

(With common ratio $\frac{4}{3}$) when we add $1$ to $a$, and becomes the geometric progression

$8, 12, 18$

(With common ratio $\frac{3}{2}$) when we add $2$ to $c$. [/hide]

[hide="3."]
Assume $a<b<c$ and the sequence is thus $a,b,c$.

Then, you have the equations

$\frac{a+c}{2}=b$,

$\sqrt{(a+1)c}=b$,

$\sqrt{a(c+2)}=b$.

Solve to get $a=8$, $b=12$, $c=16$.[/hide]

so for two i believe the correct answer is:
[hide]
$a=-\frac{41}{2}$[/hide]

EDIT: i didnt know [b]t0rajir0u[/b] posted the full solution to 2

wow...u guys are awesome.  I wish i could think conceptually like this!!!

Thank you so much, I've found a great source of information

how about this one?

the first term of an arithmetic series of consecutive integers is k^2+1.  Find the sum of 2k+1 terms of this series in terms of k

okay
[hide]
using the formula for the sum of an arithmetic sequence
$S=\left(2k+1\right)\left(\frac{k^{2}+1+k^{2}+1+2k}{2}\right)$
$S=\left(2k+1\right)\left(k^{2}+2k+1\right)$
$S=\left(2k+1\right)\left(k+1\right)^{2}$

[/hide]

[quote="SM4RT"]okay
[hide]
using the formula for the sum of an arithmetic sequence
$k=\left(2k+1\right)\left(\frac{k^{2}+1+k^{2}+1+2k}{2}\right)$
$k=\left(2k+1\right)\left(k^{2}+2k+1\right)$

solve for k
[/hide][/quote]

I'm not sure why you have a $k$ on your LHS.

ya my bad..i thought the sum of the series was k

[quote="SM4RT"]okay

using the formula for the sum of an arithmetic sequence
$k=\left(2k+1\right)\left(\frac{k^{2}+1+k^{2}+1+2k}{2}\right)$
$k=\left(2k+1\right)\left(k^{2}+2k+1\right)$

solve for k
[/quote]
I think you mean $S=\left(2k+1\right)\left(k^{2}+2k+1\right)$.

hey thanks a lot guys...again

and one more problem, I think I have most of it:

If:
$x, 2x+2, 3x+3$ forms a geometric sequence, what are the numerical values of the first six terms?

I've found that
$a=x$, $ar=2x+2$, $ar^{2}=3x+3$ and from there I'm lost on what to do

[hide="Method"]
Solve for $x$ in $\sqrt{x(3x+3)}=2x+2$.

By the way, if you don't understand where this is coming from, it's due to the fact that in a geometric sequence a term is the geometric mean of its two neighbors.[/hide]
[hide="Answer"]
You get $(x+1)(x+4)=0$ and this yields $x=-1$ or $x=-4$.

If $x=-1$, the sequence is

$-1, 0, 0,0,0,0\ldots$.

If $x=-4$, the sequence is

$-4,-6,-9,-13.5,-20.25,-30.375,\ldots$.
[/hide]

[quote="SM4RT"]i get this..
[hide]
$\frac{2x+2}{x}=\frac{3x+3}{2x+2}$, since its geometric
$4(x+1)^{2}=3x(x+1)$ EDIT:2x+2=2(x+1) 3x+3 = 3(x+1)..i'm sure u see this but just in case
$(x+1)(4x+4-3x)=0$
$(x+1)(x+4)=0$
$x=\boxed{-4}$
x=-1 doesnot work..
EDIT: well it does but ur terms are -1,0,0,0,0...

P.S. this is how i learned latex..i just quoted the other person and saw how they did it..it worked for me
[/hide][/quote]

haha, same, I think I've got most of it down pat.  The solution is right on line with what I thought (the -1 answer), I just didnt realize it was extraneous

This is weird...

I get how you guys got $x=-4$

which would yield the answer $-4,-6,-9,-13.5,-20.25,-30.375$

however, taking the next term, which I suppose would be $4x+4$, the answer would be $-12$, not $-13.5$ and similarly for $5x+5$ the answer would be $-15$ not $-20.25$

either I'm going insane, or something's not right here

The next term is not necessarily $4x+4$.

You can only assume what you're given.

thats true because if it is then its not geometric..


At http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61209
Enescu proposed the following problem:

Find a function $f: \mathbb R\rightarrow\mathbb R$ which satisfies
$f(f(x))=-x, \forall x \in \mathbb R$ and is also continuous at x=0.

Here is a generalization I found to this problem:

Find a function $f: \mathbb R\rightarrow\mathbb R$ which satisfies
$(f\circ f\circ ... \circ f)(x)=-x, \forall x \in \mathbb R$ and is also continuous at x=0 (the "$\circ$" operation has been applied $n$ times, where $n \in \mathbb N, n\geq 2$).

Also, another thing to think about: if $f: \mathbb{R}\to \mathbb{R}$ satisfies $f(f(x))=-x,\quad \forall x\in\mathbb{R},$ then there are infinitely many points where $f$ is discontinuous?


calculus
derivative
calculus computations

If using derivative to find max ...for example maximum profit and the answer comes down to 1000/3 and I have to use decimal .... what should I do? the rounding process?

If it is dollars, round to two decimal places, i.e. to the nearest cent.


I'm just interested to know how many of you will even possibly experience a White Christmas

[quote="ZetaX"]Read http://www.mathlinks.ro/Forum/viewtopic.php?t=116340 and think if this poll has any use ;)[/quote]

Why are you singling out this particular user when tons of other users have created polls recently?

I don't know how you can predict if you'll have a White Christmas or not this early but I'd say probably not.

Because most of the others have at least any meaning...

And even if it's not the only one, e.g. if 100 murderes have commited murder, you are still not allowed to do, and the judge will probably laugh at you if you say the others where sent free, too.

Considering that it has already snowed here, I will almost definetly have a white Hannukah :wink:

This should have probably been made in G&FF

[quote="goldendomer"]There's a 50-50 chance that I will have a white christmas. I live in NC in the snow doesn't come till January. If it does snow, it wont be hard. :([/quote]

How is it 50-50 then?

I'll also say I'd definitely have a white Hanukkah but one specific day is less likely.

Right now I'm living in Melbourne, Australia, and it doesn't snow here ever.  Ever.  Most kids have never seen snow before - it is a big deal when they "go to snow".  The other day somone said that "it feels so much like Christmas."  It was 37 C/97 F and sunny...  It was so wrong to me, since in Wyoming, USA we always have white Christmases (sp?).  White winters for that matter...

Australia's winters are during Wyoming's summers.

We got our first snow on October 24th, in MI  :D

Yes certainly.

[quote="i_like_pie"]San Francisco = 0% chance of snow. :(

[quote="renatzu"]I can only remember it happening once.  I live near New York City, the snow doesn't really start coming until late January.[/quote]
I used to live near NYC. It snowed in June once. :| Seriously. A few years ago.[/quote]

have you ever seen that disney movie about these girls stealing a snow maker and i think they might live in san fransisco?
-jorian


number theory
greatest common divisor
function
geometry
geometric transformation
linear algebra
matrix

Prove that if $f,g$ are entire functions with no common zero, then we can find entire functions $a,b$ such that $af+bg=1$.

Excuse me,Moubinool, why the functions $a,b$ are entire?

Because $f^2+g^2$ is an entire function which does not have any zeros. If $t$ is such a function, then $\frac 1t$ is a function defined everywhere, and everywhere differentiable, so it's entire. What's wrong with that? :?

Why does $f^2 + g^2$ have no zeros?  What if $f(r) = i$ and $g(r) = 1$?

Oh, sorry, for some reason, I was thinking only of real values :). I don't know what came over me.. :blush:

First of all, let's prove the following: consider a sequence
$ z_{1},z_{2}, \ldots \in \mathbb{C} $, such that $ lim z_{j} =
\infty $, and a sequence $ k_{1},k_{2}, \ldots \in \mathbb{N} $
and a for every $ j \in \mathbb{N} $ a collection of complex
numbers $ w_{j0},w_{j1},w_{j2}, \ldots , w_{j k_{j} } $. Then
there exists an entire function $ f(z) $, such that for every $ j
\in \mathbb{N} $ we have $ f^{(l)}(z_{j}) = w_{jl} $ for $ 0
\leqslant l \leqslant k_{j} $. First of all, applying translation,
we can assume that all $ z_{j} \neq 0 $. Now first we will
construct a collection of entire functions $ g_{jl} $, $ j=1,2,
\ldots $, $ 0 \leqslant l \leqslant k_{j} $, such that $
g_{jl}^{(l')}(z_{j'}) = 1 $ if $ l=l',j=j' $ and $ 0 $ otherwise.
To do it, take some $ j $. Then consider a sequence of complex
numbers which consists of all $ z_{j'} $, when $ j' \neq j $, such
that $ z_{j'} $ appears exactly $ k_{j'} $ times. Then consider
the Weierstrass product for this sequence and denote it by $
g_{j}(z) $. Then if we consider the functions $ g_{j}(z),
(z-z_{j})g_{j}(z), \ldots , (z-z_{j})^{k_{j}}g_{j}(z) $, then it
is easy to see that we can obtain  $ g_{jl}(z) $, $ 0 \leqslant l
\leqslant k_{j} $ as their linear combinations. Now let us prove
the following: suppose that we have some $ \varepsilon > 0 $, some
$ a \in \mathbb{C} $,$ R > 0 $ and $ k \in \mathbb{N} $, such that
$ |a| < R $. Then there exists an entire function $ F(z) $, such
that $ |F(z)| \leqslant \varepsilon $ for $ |z| \leqslant R $ and
$ F(a) = 1 $, $ F'(a) = F''(a) = \ldots = F^{(k)}(a) = 0 $. First
of all, we can consider only the case $ a = 1 $. Take now some
natural $ n $, and consider $ \frac{1}{(1-w)^{n+1}} = 1 +
\begin{pmatrix}
  n+1 \\
  1
\end{pmatrix}
w +
\begin{pmatrix}
  n+2 \\
  2
\end{pmatrix}
w^{2} + \ldots +
\begin{pmatrix}
  n+k \\
  k
\end{pmatrix}
w^{k} +
\begin{pmatrix}
  n+k+1 \\
  k+1
\end{pmatrix}
w^{k+1} + \ldots $. Then if we take $ F(z) = z^{n+1} ( 1 +
\begin{pmatrix}
  n+1 \\
  1
\end{pmatrix}
(1-z) +
\begin{pmatrix}
  n+2 \\
  2
\end{pmatrix}
(1-z)^{2} + \ldots +
\begin{pmatrix}
  n+k \\
  k
\end{pmatrix}
(1-z)^{k} +
\begin{pmatrix}
  n+k+1 \\
  k+1
\end{pmatrix}
(1-z)^{k+1} )$, then we have that $ F(1) = 1 $, $ F'(1) = F''(1) =
\ldots = F^{(k)}(1) = 0 $. Also we have for $ |z| \leqslant 1 $
that $ |F(z)| \leqslant C_{k} |z|^{n} n^{k+1} $ for some constant
$ C_{k} $. So we see that for $ n $ big enough we will have $
|F(z)| \leqslant \varepsilon $ for every $ |z| \leqslant R $.

Consider now some $ j \in \mathbb{N} $. By what we have shown now,
there exists an entire function $ F_{j}(z) $, such that $ |w_{jl}
F_{j}(z)g_{jl}(z)| \leqslant \frac{1}{2^{j+l}} $, for $ |z|
\leqslant \frac{|z_{j}|}{2} $, for every $ 0 \leqslant l \leqslant
k_{j} $, and such that $ F_{j}(z_{j}) = 1 $, $ F_{j}'(z_{j}) =
F_{j}''(z_{j}) = \ldots = F_{j}^{ (k_{j} + 1 ) }(z_{j}) = 0 $.
Then we see that $ f(z) : = \sum_{ j \geqslant 1 , 0 \leqslant l
\leqslant k_{j} } w_{jl} F_{j}(z)g_{jl}(z) $ converges uniformly
on compact subsets of $ \mathbb{C} $ to an entire function, and
satisfies the desired properties.

 Now let's derive the solution from what we have proved. Suppose
we have two entire functions $ f,g $ with no common zero. Assume
that zeroes of $ g(z) $ are $ z_{1},z_{2}, \ldots $ with
multiplicities $ k_{1},k_{2}, \ldots $ respectively. Let us find
an entire function $ a(z) $, such that the function $ G(z):= 1 -
a(z)f(z) $ has zeroes $ z_{1},z_{2}, \ldots $ with multiplicities
at least $ k_{1},k_{2}, \ldots $ respectively. This means that for
each $ j $, we must have $ a(z_{j})f(z_{j}) = 1 $ and $
a'(z_{j})f(z_{j}) + a(z_{j})f'(z_{j}) = 0 $, $ a''(z_{j})f(z_{j})
+ 2a'(z_{j})f'(z_{j}) + a(z_{j})f''(z_{j}) = 0 , \ldots ,
a^{(k_{j})}(z_{j})f(z_{j}) + k_{j} a^{(k_{j}-1)}(z_{j})f'(z_{j}) +
\ldots + k_{j} a'(z_{j})f^{(k_{j}-1)}(z_{j}) +
a(z_{j})f^{(k_{j})}(z_{j}) = 0 $. Using the fact that $ f(z_{j})
\neq 0 $, we can inductively find what $ a(z_{j}), a'(z_{j}),
\ldots , a^{(k_{j})} $ must be equal to. So we obtain the problem
which we have already solved, for the function $ a(z) $, so we can
find such a function. Therefore we have found an entire function $
a(z) $, such that the function $ G(z):= 1 - a(z)f(z) $ has zeroes
$ z_{1},z_{2}, \ldots $ with multiplicities at least $
k_{1},k_{2}, \ldots $ respectively. Therefore there exists an
entire function $ b(z) $, such that $ G(z) = 1- a(z)f(z) =
b(z)g(z) $, so $ a(z)f(z) + b(z)g(z) = 1 $.

I hope it is correct.

By the Mittag-Leffler theorem, you can find a meromorphic function M whose principal parts occur only at the zeros of g, and whose principal part a a zero z' of g is the same as the principal part of 1/(fg) at z'.  In this way, h = M - 1/(fg) is holomorphic at the zeros of g, so A = Mg and B = -hf are both entire.  Then Af + Bg = Mfg + 1 - Mfg = 1.


geometry
circumcircle
rectangle
incenter
geometry proposed

[quote="mathwizarddude"]Point H is arbitrary picked on the diameter AB of a circle with center O. Another circle is inscribed as the picture shows. Prove that EC bisects HEB.

http://img175.imageshack.us/img175/9498/geomxa3.png[/quote]
[hide="Solution"]
Let the circle with diamter $ AB$ be $ \mathcal{C}$ and let the small circle tangent $ \mathcal{C}$ be $ \mathcal{S}$

Invert around $ E$ with radius $ EH$.  Let the inversion map point $ P$ to $ P'$ and any figure $ \mathcal{F}$ to $ \mathcal{F'}$.  Now, $ \mathcal{C'}$ becomes the line $ A'B'$.  $ A'$ and $ B'$ become points on $ AE$ and $ EB$ so that $ HA'\perp EA$ and $ HB'\perp EB$ respectively since $ EH\perp AB$.  Also, the segment $ AB$ is mapped to the circumcircle of $ \triangle HA'B'$, call it $ \mathcal{H}$.  Since \[ \angle HA'E \plus{} \angle HB'E \equal{} 90 \plus{} 90 \equal{} 180\] we have that $ EA'HB'$ is cyclic, so $ E$ lies on $ \mathcal{H}$.  Furthermore, observe that\[ \angle A'EB' \equal{} 90 \equal{} \angle EA'H \equal{} \angle EB'H\]so $ EA'HB'$ is a rectangle.  Let the intersection between the diagonals of this rectangle be $ Q$, which means that $ Q$ is the center of $ \mathcal{H}$.  Finally, $ \mathcal{S'}$ becomes a circle tangent to $ \mathcal{H}$, $ B'Q$, and $ HQ$.  Let the center of $ \mathcal{S'}$ be $ K$.  Since $ \mathcal{S'}$ is tangent to $ B'Q$ and $ HQ$, we have that $ KQ$ bisects angle $ B'QH$.  The tangency point between $ \mathcal{S'}$ and the circumcircle of the rectangle$ \mathcal{H}$ is $ C'$ since $ C$ with the tangency point between line $ AB$ and $ \mathcal{S}$.  Since $ \mathcal{H}$ is tangent to the $ \mathcal{S'}$, we have that $ Q$, $ K$, and $ C'$ are collinear, so $ QC'$ bisects $ \angle A'QH$.  Thus, $ C'$ is the midpoint of minor arc $ A'H$, so $ EC'$ bisects $ \angle HEB'$, so $ EC$ bisects $ \angle HEB$, as desired. [/hide]

[quote="plane geometry"]1.A.P.Q are collinear
2.AE^2=AC^2=AP*AQ=AH*AB
3.Angle AEC=Angle ACE
DONE![/quote]

How do you prove 2 (i.e., your claim on $ AE^2\equal{}AC^2\equal{}AP*AQ\equal{}AH*AB$)?

Take a look at http://www.mathlinks.ro/viewtopic.php?t=146656

[quote="mathwizarddude"][quote="plane geometry"]1.A.P.Q are collinear
2.AE^2=AC^2=AP*AQ=AH*AB
3.Angle AEC=Angle ACE
DONE![/quote]

How do you prove 2 (i.e., your claim on $ AE^2 \equal{} AC^2 \equal{} AP*AQ \equal{} AH*AB$)?[/quote]
you can prove A,P,Q are collinear(notice O1P∥OB)  then PKBQ concircle   so AE^2=AC^2=AP*AQ=AH*AB

Inversion in circle (A) with center A and radius AE takes the line AB to itself, the circle (O) to the line EH and the line EH to the circle (O). The circle (O1) tangent to all 3 goes to itself, so that it is perpendicular to the inversion circle (A) and (A) goes through the tangency point C. EA is tangent to the circumcircle of the right triangle EHB, (A) is the E-Apollonius circle of this triangle, EC bisects the angle <HEB.

Dear Mathlinkers,
let 0 the circumcircle of AEB, D the second intersection of EH with 0.
Accordling to the Sawayama lemma (See Http://perso.orange.fr/jl.ayme  vol.1, Sawayama and Thebault's theorem), PC go through the incenter of BED.
BED is B-isosceles ; then BH the B-altitude of BED is also the B-bissector of BED and C is the incenter of BED.
We are done.
Sincerely
Jean-Louis


modular arithmetic
floor function
logarithms
number theory proposed
number theory

Great problem: prove that there exists infinitely many n such that n, n+1 and n+2 are squarefree.

Can anybody tell me what does 'squarefree' means?

Sorry for my poor English.

It is a number that is not divisible by the square of any prime number;

I even haven't any idea to prove the problem about infinitively existence of $two$ consecutive squarefree numbers. :(
 [b]harazi[/b]could you possibly help me with some hint?

Isn't the desired result an immediate consequence of the fact that the 'probability' that a given positive integer is squarefree is $\frac 6 {\pi ^2} $, which is more than $\frac 1 2$  :?

Pierre.

That works if we want to show that there are infinitely many pairs of consecutive squarefree numbers, right? :? I mean, it doesn't work anymore for triples. The problem with pairs was posted before and it was solved using the exact same idea. Moreover, in that old thread, I believe the problem we have right here was also mentioned (but with no proof).

But how can you have a finite number of consecutive non-squarefree triples if the density of the non-squarefree  is more than $0.5$?

Pierre.

Am I totally misunderstanding this? :?

First of all, we want triples of squarefree numbers, not non-squarefree numbers, right? Second of all, think about the set of naturals giving residues $\pm 1\pmod 3$. This set has density $\frac 23>\frac 12$, and yet it contains no triples of consecutive numbers. I'm sorry, Pierre, if I am misunderstanding you :(. That's been happenning to me a lot lately :).

Are you both kidding?  :?  Of course this problem is an immediate corollary of the fact that the square-free integers have density $\frac{6}{\pi^2}>\frac{1}{2}$. But instead into [i]triples[/i], break the positive integers into [i]quadruples[/i] of consecutive integers $\{4k+1,4k+2,4k+3,4k+4\}$. Each such quadruple certainly contains the non-square-free number $4k+4 \equiv 0 \pmod{2^2}$, so at most three numbers in a quadruple are square-free. Because the square-free integers have density $>1/2$, it cannot be that all quadruples from some point onwards contain at most two square-free elements. Hence, there are infinitely many $k$ for which $4k+1,4k+2,4k+3$ are all square-free. That's all!

--Vesselin

Wow! :) I didn't see it, unfortunately. Perhaps this is what Pierre meant. Thanks, vess! :)

[quote="vess"]break the positive integers into [i]quadruples[/i] of consecutive integers $\{4k+1,4k+2,4k+3,4k+4\}$. Each such quadruple certainly contains the non-square-free number $4k+4 \equiv 0 \pmod{2^2}$, so at most three numbers in a quadruple are square-free. Because the square-free integers have density $>1/2$, it cannot be that all quadruples from some point onwards contain at most two square-free elements.[/quote]

Yes, that's exactly what I meant (even if I wrote non-squarefree instead of squarefree  :blush:  ).

Pierre.

Then I must apologize yet again... :( Sorry I didn't see that  :blush:  :blush:

I don't know why you will have to apologize Grobber. Anyone may confuse, and my 'explanations' were a bit concise.

Now, Harazi, is that the solution you were expected? Or do you have a more constructive one?

Pierre.

No, in fact this is the solution I have found long time ago. Thanks for remembering it to me. I simply forgot it, though it is so easy.  :?

Well done, Vess-Pierre-Harazi !!!!!!!!! :lol:

What about quadruples of consecutive square-free numbers?  Are there an infinite number of them?

:D Cute joke. Among any four consecutive numbers, there's one divisible by $4$.

However, a similar (only (very!) slightly trickier) density argument shows that there are infinitely many quadruples of consecutive integers, each of which is of the form $2^{\varepsilon}q$ with $q$ a square-free odd integer and $\varepsilon \in \{0,1,2\}$. [Can you see this?] More generally, I managed to generalize the density argument to show the following:

[b]Generalization. [/b] Let $n$ be a fixed positive integers. Then, there are infinitely many $n$-tuples $(k,k+1,\ldots,k+n-1)$ of consecutive integers, none of which is divisible by a prime power of the form $p^{\max \big ( 2, 1 + \lfloor \log_p n \rfloor \big)}$.

Yet more generally, I think (but I haven't done it, hence the labeling 'Conjecture') that the proof can be extended to show this:

[b]Conjecture. [/b] Let $n; a_1, a_2 \ldots,a_n$ be fixed positive integers. Then, there are infinitely many $n$-tuples of the form $(k+a_1,\ldots,k+a_n)$ such that none of their components are divisible by a prime power of the form $p^{\max ( 2, 1 + A(p) )}$, where $A(p) := \min_{x \in \mathbb{Z}} \max_{1 \leq i \leq n} \mathrm{ord}_p \, (x + a_j)$.

Can anyone prove the conjecture?

--Vesselin

I haven't thought about the generalizations. As for the first one, the one with $2^{\varepsilon}$, we can look at $9$-uples of the form $9k+1,9k+2,\ldots,9k+9$. This set contains $9k+9$, which is inelligible, and also a multiple of $8$, which is, again, inelligible. It's easy to see now that if we are only allowed to have consecutive triples of our numbers for large enough $k$, we would cover at most $6$ numbers in such a $9$-uple (again, for sufficiently large $k$'s). However, the density of the set we're looking at is $\frac 7{\pi^2}>\frac 23=\frac 69$, so we get a contradiction.

This (working with the $9$-tuples) is exactly what I did, too!  :)

i have no solution for this pro, but i think new pro interesting
[b]NEW PRO[/b]: show that there exist infinite squares number $a,b,c$ such that $a+b,b+c,c+a$ are both squares numbers


AMC
AMC 10
AMC 10 B
AMC 12
ARML
AIME
email

See http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-1-1012-dir,pic.shtml

The state listings of results are up. :)

I just looked at some of the stats, and noticed that the table of what answers were given appears to have the wrong denominator. At least in CA, the sum of each row was less than 10%; this probably means that the denominator was the number of takers nationwide.

First in KS on 12 with a 132 on A test  :) .

Our team score is one question off from the #1 teams in both California and Florida  :D

Edit:  Wow, a Chinese school has a 450 team score  :|

#1 in Massachusetts on AMC 10 B  :D

Nice work on getting those statistics up for us, CAMC!

2nd place in FL on AMC10B, who got a 144?

And thanks for all the statistics. :D

I got a 144 on AMC10B in Florida...

But I'm from Michigan.

First in Minnesota for the AMC 12 with a 144 on 12A.   :D
Oh, I'm happy.  Very happy.

[EDIT]

I also seem to be tied for first in Region 5.  Yum.

[/EDIT]

ya first in fl on 10a

well tied for first

144 was karlanna on the 10b

Not so stellar, but out of all 9th graders in Cali, I'm #8, so that's not too shabby. 12A, by the way.

OMG!! second in Indiana in 10a (with a 135). the only person who beat me was undefined117 (with a 139.5). i was the top scoring girl in both  the a and b, though :wow:. i am totally amazed.

p.s. to break a tie for first in the AMC 8 using the AMC 10 scores, will it matter that some ppl took the a and others took the b? cuz i think Fanatic's score on the b was better than my score on the a (by 3 points)...

i was like 15th in Indiana, on 10a, but i was second among 7th graders on 10a. :play_ball:  :wow:

1st in whole entire state on 10B :lol: . Another kid from my national team was 2nd :P

[quote="Pineappleperson"]YAY!

1st in michigan for amc 10

if that's anything to brag about..........[/quote]

Tied with 5 other people.

[quote="mwpl11"][quote="undefined117"]Second in IN (look at sig) after a 10th grader..

I SOOO should've taken 12B, except I found out about it a month later :mad:

goal for next year: perfect score! mwahaha.[/quote]

WHO LET U TAKE THE 10 AND 12??? That is soo not fair. Did Mr. A let you take both? Mrs. H didn't.[/quote]

it was practice, but yes, he had a spare copy he lent to me.

The reason that 8th graders have the highest scores in Indiana for AMC 10 is because most good 9th and 10th graders in IN take the 12 (like me, I'm in 10th grade and I got a 144 on the AMC 12A)

I only know that 5 or so 9th and 10th graders in Indiana took the 12s..

I know about ten at one school (not mine)

What-Terre Haute South?

What I meant is that only 5 or so could get 130s by taking the 10.  Most of the people that took the 12 as 9th or 10th graders probably cant get as high as the 8th graders on the 10.  For exapmle, out of 7th, 8th, 9th, 10th graders I got the 4th highest amc 10A score but I also got the 6th highest amc 12b score..

Highest 6th grade score in MI in 10A!
Nothing special :(

I sucked on the 12A. Ugh.

Tied for 55th with 14 other people.

[quote="nat mc"]yeah and our state winning 12 team was all <11 grade[/quote]

Lol our winning regional amc12 team was all >= 11. Things look grim for the future of our school, but hey we'll be back for one more year, and hopefully not lose to montgomery next year.

Yeah.  Two seniors and I (freshman) in my school qualified for the AIME, so I'm pretty sure our team score is pretty low.  We're the Math and Science Academy in Virginia Beach too  :roll:   We can't compete with TJ.

[quote="nat mc"]yeah we could win the region 4 years in a row (we won it last year and this year already)[/quote]

This is Rickards' 5th year in a row that we won regionals and we're definitely going to carry it to 6 before my year leaves :wink:. Although... we've never won nationals  :(. We were 2nd this year and 2nd in some other year I think.


A lonely wanderer, wounded with iron, I am smitten with war-blades, sated with strife,
Worn with the sword-edge; I have seen many battles, Much hazardous fighting, oft without hope
Of comforts or help in the carnage of war Ere I perish and fall in the fighting of men.
The leavings of hammers, the handiwork of smiths, Batter and bite me, hard-eged and sharp;
The brunt of the battle I am doomed to endure. In all the folk-stead no leech could I find
With wort or simple to heal my wounds; But day and night with the deadly blows
The marks of the war-blades double and deepen.

Who or what is the "I" referred to in the riddle?


Find all positive integers $ m$ and $ n$ such that $ 1 \plus{} 5 \times 2^m \equal{} n^2$

I think i have a solution but im not sure if im correct. Could someone please help me with this  =)

[hide="My Solution?"]$ 1 \plus{} 5 \times 2^m \equal{} n^2$

$ 5 \times 2^m \equal{} n^2 \minus{} 1$

$ 5 \times 2^m \equal{} (n \minus{} 1)(n \plus{} 1)$

Say $ n \minus{} 1$ is an odd number, then $ n \plus{} 1$ must also be an odd number

Two odd numbers multiplied together makes an odd number.

But $ 5$ is odd and $ 2^m$ is even, and so their product is even.

Therefore if $ n \minus{} 1$ is odd then there are no solutions

Therefore is there are any solutions, then $ n \minus{} 1$ must be even

Then i dont know where to go from there  :maybe: [/hide]

All help is very much appreciated  :lol:

[hide="How to go on- here it is"]

2 cases:

1) $ n\minus{}1\equal{}5\cdot 2^a,n\plus{}1\equal{}2^b$

$ 5\cdot 2^{a\minus{}1}\plus{}1\equal{}2^{b\minus{}1}$, so at least one of $ a,b$ must be one, as if they are bigger both than 1, we would get odd=even.

Now it's easy to see there are no solutions

2) $ n\minus{}1\equal{}2^a,n\plus{}1\equal{}5\cdot2^b$

$ 2^{a\minus{}1}\plus{}1\equal{}5\cdot2^{b\minus{}1}$, same reasoning as above, $ a\equal{}3,b\equal{}1,m\equal{}4,n\equal{}9$

[/hide]


solve: \[ x^{\frac{1}{6}}\plus{}x^{\frac{1}{3}}\minus{}2\equal{}0\]

i know that you can see that the answer is 1 just by looking at it, but how do you do it algebraically?

[hide="Solution"]Make the substitution $ u \equal{} x^{\frac {1}{6}}$. $ u^2 \plus{} u \minus{} 2 \equal{} 0\implies u \equal{} \frac { \minus{} 1\pm3}{2} \equal{} 1, \minus{} 2$.
Discard $ \minus{} 2$ for obvious reasons.
Then $ x \equal{} 1^3, \equal{} \boxed{1}$.[/hide]

Edit: Nice catch ;)

Edit: Right, Farenhajt.

Wait, no. First of all, $ ( \minus{} 2)^3 \equal{} \minus{} 8$, not 8.

And because of this, if you try to plug in $ x \equal{} \minus{} 8$, you can't get any defined value because $ ( \minus{} 8)^{\frac {1}{6}}$ is undefined.

So the only solution to this equation is $ x\equal{}\boxed{1}$.

First of all, if the final equation is $ u^2 + u - 2 =0$, then the substitution is $ u = x^{1/6}$, not $ u = x^{1/3}$

Hence the values for $ u$ yield $ x = u^6 = \{\begin{matrix}1 \\
64\end{matrix}$

But $ x = 64$ doesn't work, so the only solution is $ x = 1$.

[quote="21 7 15"][i]basically, let x^1/3=p
implying that p=1 and -2
x=1 and -8[quote]i am nobody since nobody is perfect i am perfect[/quote][/i][/quote]
Eh, spam basically? The problem has already been solved, and the errors have already been fixed.

Again, [b]-8 cannot be a solution since you can't take an even root of a negative number.[/b] That yields the reason as to why 1 is the only solution.

Hey, people, please re-read my post. All of you keep working with the wrong substitution.

Because he used my erroneous solution.
And... reposted it.
For some reason.

Anyway, I corrected it. Thanks for pointing that out (I don't know what happened that day :lol:)


We say that a 9x9 chessboard filled with empty spaces and numbers from 1 to 9 is sudokable iff we can fill the empty spaces with numbers from 1 to 9 so that the rules of the game sudoku are respected. Find the number of sudokable chessboards.

i created it  :lol: !! It came in my mind while i was solving a sudoku   :D !
I tried to solv, but i couldn't: it seems it is very hard!!!

Considering the fact that the number of valid ending position with all 81 numbers filled in is a fairly intractalble problem for humans (current answers are computer calculated), I wouldn't expect this to be doable at all.


What's wrong with putting in the polls about the teletubbies when all of a sudden the mods just lock them for no reason? It's not even causing flaming, lol.

The topic isn't serious. People find it hard to believe that people at your age would watch teletubbies. You'll see all the sarcasm in the teletubbies vs barney poll.

[quote="fedja"]Well, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214088]this one[/url] has hardly any more sense, of course, and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213927]this one[/url] resulted in hardly any more intelligent discussion, but, indeed, if the next thing you intend to ask is "What is your favorite webkin?", you may expect your poll to be locked or deleted again. I cannot state any written or unwritten forum rule that the teletubby poll would violate, but it is just does not belong here. If you really want to continue the teletubby thread, I can unlock it and move it to G&FF and you can have your fun (until some mod there gets tired of it). If you just wanted to make a parody on other polls in this subforum, your point is taken ;)

By the way, when my daughter was 9 months old, she seemed to perefer teletubbies over barney, but it is hard to be completely sure now. :)[/quote]
Maybe move those topics to G&FF thx :D

Well, you might be a little young on the forums to know this, but before, the Polls had a description, which said "[b]Administrative[/b] Polls" which implies a sense of seriousness in these topics.

I have unlocked your topic and moved it to G&FF, but I will lock it if it gets out of hand.


Given a,b >0 satisfied $ 27b\plus{}a\equal{}ab$ . Find the minimum of $ a^2\plus{}b^2$

$ 27b\plus{}a\equal{}ab$
$ a\minus{}ab\equal{}\minus{}27b$
$ a(1\minus{}b)\equal{}\minus{}27b$
$ a\equal{}\frac{27b}{b\minus{}1}$

From this we see that if $ a>0$, $ b>1$, and if $ b>1$, $ a>0$.

Let $ M\equal{}b^2\plus{}\left (\frac{27b}{b\minus{}1}\right )^2$
$ \frac{dM}{db}\equal{}2b\plus{}2\left (\frac{27b}{b\minus{}1}\right )\left (\frac{27(b\minus{}1)\minus{}27b}{(b\minus{}1)^2}\right )$
Setting $ \frac{dM}{db}\equal{}0$
$ 0\equal{}b\plus{}\left (\frac{27b}{b\minus{}1}\right )\left (\frac{27(b\minus{}1)\minus{}27b}{(b\minus{}1)^2}\right )$
$ 0\equal{}b^4\minus{}3b^3\plus{}3b^2\minus{}b\minus{}729b$
$ 0\equal{}b(b^3\minus{}3b^2\plus{}3b\minus{}730)$
$ 0\equal{}b(b\minus{}10)(b^2\plus{}7b\plus{}73)$
The only real roots are at $ b\equal{}0$ and $ b\equal{}10$
Ignore $ b\equal{}0$ since it does not satisfy the earlier criteria.
When $ b\equal{}10$, $ M\equal{}1000$.

The minimum of $ a^2\plus{}b^2$, satisfying all the criteria is $ 1000$, when $ a\equal{}30$ and $ b\equal{}10$.

Thanks you very much, your solve is very nice!


Fahrenheit 9/11 trailer is out!:

http://www.apple.com/trailers/lions_gate/fahrenheit_911/

I can't wait to see this movie.

Micheal Moore...uh oh.  Even though I am a democrat I don't really like his movies at all.  Though it is really scary when you think about the stuff the government does that you do not know about.

[quote]Anything to do with Fahrenheit 451?[/quote]

Yes, I think that the title "fahrenheit 9/11" is a based on that, if you think about the message of his movie and the plot of the book it makes sense.

[quote="JS1527"]Micheal Moore...uh oh.  Even though I am a democrat I don't really like his movies at all.  Though it is really scary when you think about the stuff the government does that you do not know about.[/quote]
It's also scary how many people don't check the facts and believe things that were made-up, simply because it was in a movie.


Peut on trouver deux entiers a,b et un nombre premier p aussi grand que l'on veut tel que
 (2^a) - (p^b)=1

Non, parce que j'imagine que $a$ et $b$ sont supposés plus grands que $1$, hein?
D'après la récente preuve de la conjecture de Catalan, la seule solution de $x^a-y^b=1$ où $x,y,a,b>1$ sont des entiers est $3^2-2^3=1.$

Bon, cela dit, j'ai pas cherché mais ton problème doit pouvoir se résoudre sans ce type de bombe atomique.

Pierre.

Comme par hasard hier apres avoir posté la question,je me suis rendu compte que le probleme n'a pas de solution mis a part 2^2-3^1 si je ne me trompe pas.En effet en ecrivant l'equation sous la forme
1+2+2^2  + 2^3 +....+ 2^(a-1) = p^b
Puis en ecrivant p an base 2 sous sa forme generale et en etudiant les puissances de p sous cette forme ,il est impossible de tomber sur le membre gauche de l'equation sous sa nouvelle forme.il y a toujours des trous dans la succession des puissances de 2.Cela dit ,je fais la remarque que je n'ai pas conduis les calculs à terme avec toute la rigueur qu'il faut.


How long do you guys do mathcounts and how long do you practice for it?

since 5th grade.
Georgetown Novice 5th place that year

Well I just started this year (8th grade) because that is when I heard about it.

I started at about December. I'm already getting 43s at chapter :)

and 36s at state :)

Um, since 6th grade, that is, ever since the program existed in my school. :D

Not as much as some people.
I'm a spelling bee geek.  :blush: I don't know if anyone on here is, either.  :P
Just started last year, been doin' some problems and such, gettin' pretty good scores.

~The_Scintillator

I don't exactly "practice" for MathCounts.  Really, this month, I've been practicing for HMMT & AMC & AIME.  I've only taken 2 MATHCOUNTS contests, 1 of which I got a 42 and the next (from yesterday) a 44.  But anyway, I think it is a wonderful competition, and I'm going to be competing, but it's not the main contest I'm readying for in February.

OVER 9000!
Hours.
Yeah.
Anyways, I started in umm, 6th grade.

I practiced a lot last year, but I don't really practice Mathcounts anymore (I haven't taken a practice test since October or something).

Alan and I have a lot of practices, around three every week...
two hour 1 and a half hour ones and one five hour one.
Total of 8 hours of MC practice a week.
-faints-

I don't practice for MathCounts, it practices for me.

Couldn't resist. Anyways, I practice for MathCounts for like 30 minutes a week, and I practice for real math competitions for about 2 hours every day.

err... 1.5 hrs a week. During school. =D

Oh, wait. Hmm..
We have an hour and a half two to three times a week during school. Oops.
And does reading my AoPS book count?

~*~The Scintillator~*~

I learned about it last year, and

I need to practice more!!!! T_T

1.5 hrs at school? What school?

Anyway, I dont practice much... take a few old Chapter tests, 38, 39, 37, 42

[quote="speedcuber"]i dont really practice that much....

most of the time i practice violin instead.[/quote]
...rofl
hmm some schools have mathcounts as an elective @lotsofmath

I feel so sad now...
I get no time in school to practice, except half an hour, once in a while
I practice like 3 hours a week
and my scores still stink
i get like 25s in chapter...
hopefully i'll do better than last year

Since fourth grade, i am in 5th grade now


[quote="xscapezaer"]If $ab = a-b$, find $\frac{a}{b}+\frac{b}{a}-ab$[/quote]

Is the answer 1?

[hide][quote="xscapezaer"]no its not one. Don't guess. Use algebra[/quote]

$ab=a-b$

$ab+b=a$

$b(a+1)=a$

$b=\frac{a}{a+1}$

$\frac{a}{b}+\frac{b}{a}-ab$

$\frac{a}{\frac{a}{a+1}}+\frac{\frac{a}{a+1}}{a}+{a}\cdot{\frac{a}{a+1}}$

$a\cdot{\frac{a+1}{a}+\frac{a}{a+1}\cdot\frac{1}{a}-{\frac{a^{2}}{a+1}}}$

$a+1+\frac{1}{a+1}-{\frac{a^{2}}{a+1}}$

${\frac{a^{2}+2a+1+1}{a+1}}-{\frac{a^{2}}{a+1}}$

$\frac{2a+2}{a+1}=2.$ [/hide]

Disregard what i said before, i was being idiotic

These are all good. I did it this way...

[hide]
$ab=a-b$
$a^{2}b^{2}= a^{2}-2ab+b^{2}$
$ab = \frac{a}{b}-2+\frac{b}{a}$
$\frac{a}{b}+\frac{b}{a}-ab=2$
[/hide]


That is sweet.

Since we are talking about PIE, I wonder if anyone can do this question.

In my high school of 100 students, 70 like strawberry, 75 like chocolate, and 80 like vanilla.  At least how many people like all three flavors?

Split from the "Spam" topic. Let's have only ONE problem per thread, perhaps?

I remember there was a question like this on the 90-91 nat sprint, which was the very first nat test that I did in preparation for the nats this year.

Except I had no clue how to do it and the numbers were lilke a billion x scarier

There was, I remember one from either '92,'93, and '94, that had question like that.  The numbers weren't that large though I thought.  So yeah.

[quote="pgpatel"][hide]25 is the minimum.
70 is the maximum.
[u]25[/u] is the answer[/hide][/quote]

How did you get 25? I got 15 for some reason.

SplashD, why don't you explain how you got 15?

[quote="pgpatel"][hide]25 is the minimum.
70 is the maximum.
[u]25[/u] is the answer[/hide][/quote]

Could you explain your solution to people who didn't get the answer and need help?  Thanks.

Oh yeah, I made a mistake in my calculations.

20 Chocolate/Strawberry
30 Chocolate/Vanilla
25 Vanilla/Strawberry
25 All 3

[quote="mathgeniuse^ln(x)"][quote="bpms"][hide]25 is the minimum.
70 is the maximum.
[u]25[/u] is the answer[/hide][/quote]

Could you explain your solution to people who didn't get the answer and need help?  Thanks.[/quote] It's simple. 70+75+80=225. So if each person had two things then there would be 225-200=25 stragglers left. So there would have to be 25 people who had done all three.

[quote="mathgeniuse^ln(x)"]Good Job!

Would you subtract 198 if there were 99 people only?[/quote]

Yeah, you would.

[quote="mathgeniuse^ln(x)"]Good Job!

Would you subtract 198 if there were 99 people only?[/quote] Yes because double of the number of people is what it is if everyone likes two of the things.

Okay Thanks

Do you think that there were any questions from this year that involved PIE?

[quote="ajai"]Is PIE in caps for a reason? Just wondering....[/quote]
In this case, "PIE" actually stands for Principle of Inclusion-Exclusion, and not the regular pie (which is what you'd probably think it was considering the posts in this topic  :P ). (more information here: http://www.artofproblemsolving.com/Wiki/index.php/PIE)

[quote="ajai"]Is PIE in caps for a reason? Just wondering....[/quote]

Yes.  It stands for Principle of Inclusion-Exclusion.  It is really useful in counting.  Nebula42 has a nice website.


Given two non-intersecting convex polygons, show that there is a line which separates them.

Er, do you mean two convex polygons? The problem statement in its current form is clearly false.

EDIT: If you mean convex, then it becomes something of a [hide="classic."]
Consider the convex hull of the polygons. Trivially, a set of adjacent vertices makes up each of the polygons. We can just choose the first and last of the vertices in one of the polygons, then construct a parallel arbitrarily close to it so that it doesn't intersect the other polygon and isn't contained in the first, and hence separates the two, as desired.[/hide]

Yes, I meant convex, sorry for not  mentioning that :oops: .
What do you mean by that arbitrary close parallel line,i believe that arbitrary close line is contained in  first polygon from which you have chosen first and last vertice, maybe do you mean line arbitrary close to another polygon?
If that case is true how do you know that this line isn't contained in first polygon, please explain that a bit more.

[quote="Temperal"]Well, the line through the first and last vertices clearly doesn't go through the other polygon (think about the definition of a convex hull). Hence, it's possible to construct a line parallel to it out of the first polygon but close enough such that it doesn't go through the second polygon.

Drawing a picture helps.  :)[/quote]

Still i don't see why it's possible to construct a line parallel to it out of the first polygon but close enough such that it doesn't go through the second polygon. Maybe i'm not seeing something obvious,but I would be grateful if anyone could clarify me  that :) .

I don't think it works - see attachment.

It must instead be this:
[quote="Ivob"]Given two non-intersecting convex polygons where one is not inside of the other, show that there is a line which separates them.[/quote]

Uh, if one is inside the other, then they clearly intersect. "Polygon" here refers to the area occupied by the shape, not the boundary.


acute
combinatorial geometry
Triangle
point set
counting
IMO
IMO 1970

Given $100$ coplanar points, no three collinear, prove that at most $70\%$ of the triangles formed by the points have all angles acute.

Does 'at most' imply giving an equality case as well? As I seem unable to find one...

No, equality is not achieved.  For example, in the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=14796]why 70%[/url], I showed that the fraction of acute triangles is at most $66.7\%$.

I'm a bit confused about this: Is this statement valid? :
"Since among the triangles formed by any 5 points at most 70% are acute, at most 70% of all the triangles are acute."
I can't seem to wrap my head around the statement, because it seems to be applying pigeonhole to a set of 3-element sets...

It doesn't work, because when you take two subsets of 5 points, they can both contain the same triangle.

It does work, cause each triangle is counted the same number.
showing the meaning one one person writing so easy is bad at this site, but I know it is difficult and made hundres of mistakes too

Is there any solution ?

[color=#408000]Moderator says: yes, follow the link given by Ravi B.[/color]


Let $P$ be a point that lies outside of circle $O$. A line passes through $P$ and meets the circle at $A$ and $B$, and another line passes through $P$ and meets the circle at $C$ and $D$. The point $A$ is between $P$ and $B$, $C$ is between $P$ and $D$. Let the intersection of segment $AD$ and $BC$ be $L$ and construct $E$ on ray $(PA$ so that $BL \cdot PE = DL \cdot PD$.

Show that $M$ is the midpoint of the segment $DE$, where $M$ is the intersection of lines $PL$ and $DE$.

[quote="lightrhee"]This is the fifth problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422487]the second round of 19th Korean Mathematics Olympiad[/url].
5.
P is a point that is outside of circle O. A line passes through P and meets the circle at A and B, and another line passes through P and meets the circle at C and D. A is between P and B, C is between P and D. Let the intersection of segment AD and BC be L and construct E on ray PA so that $BL \cdot PE = DL \cdot PD$. Show that M is the midpoint of the segment DE, where M is the intersection of line PL and DE.[/quote]

We have $\frac{BL}{DL}=\frac{PD}{PE} \Rightarrow$
Under the sin theorem we have $\frac{EM}{MD}=\frac{EP sin \angle BPL}{DP sin \angle DPL}=\frac{DL sin \angle BPL}{BL sin \angle DPL}$

$\frac{DL}{sin \angle DPL} * \frac{sin \angle BPL}{BL} = \frac{LP}{sin \angle PDL} * \frac{sin \angle PBL}{LP} = 1$  (as $\angle LDP=\angle PBL$)

Nice problem :)

Since $\frac{BL}{DL}=\frac{PD}{PE}=\frac{AL}{LC}$ $\Longleftrightarrow$ $S_{\triangle{PEL}}=S_{\triangle{PLD}}$ and the conclusion follows.


inequalities
real analysis
real analysis unsolved

Let $f(x)=\sum_{n=0}^{\infty} a_nx^n$ and $g(x)=\sum_{n=0}^{\infty} b_nx^n$ be 2 power series which verify:
1)$b_n$ are positive real numbers.
2)$\sum^{\infty} b_n=\infty$
3)$lim_{n\to \infty}\frac{a_n}{b_n}=1$(i.e $a_n$,$b_n$ are equivalent)
Show that $f(x)$ and $g(x)$ are equivalent near 1.

for any $e>0$ there exists $N$ such that for $n>N$, $|a_nx^n-b_nx^n|<b_nx^n$.
Then sum this inequalities : $|\sum_{N}^{+\infty}b_nx^n - \sum_{N}^{+\infty}b_nx^n|<e |\sum_{N}^{+\infty}b_nx^n|$. From here it is easy to conclude since $\sum b_n = \infty$.


calculus
integration
function
rotation
trigonometry
calculus computations

1. How do I calculate the integral of square root of (1+(9/4)*x^2) dx??
2. If had a question to find the volume of the following:
    Bounded by y = square root of x, x = 1, y = 0. Axis x = 1.

What does it mean for x = 1 and y = 0?? I know that axis x = 1 means that the function is rotated with respect to x = 1 right??

The curve $y=\sqrt{x}$ does not define a region on its own, so we add two lines and consider the region enclosed by all three curves. You are meant to rotate the region (not just the curve) around the line $x=1$.

$\int\sqrt{1+\frac94x^{2}}\,dx$

The trig substitution $x=\frac23\tan\theta$ will lead to a constant times

$\int\sec^{3}\theta\,d\theta,$

which is an annoying integral, but one that's usually well-covered in calculus textbooks.

I don't have the sec integral in my calculus book so is there any other way to do this

If you know some hyperbolic identities, the hyperbolic substitution $\frac32x=\sinh t$ can also be used. It's complicated either way.

For the $\sec^{3}$ integral, you integrate by parts.

$\int \sec^{3}{\theta}d\theta$

Use $1+\tan^{2}{\theta}= \sec^{2}{\theta}$ and rewrite the integral as $\int (\sec{\theta}+\sec{\theta}\tan^{2}{\theta})d\theta$.


Let $x,y,z$ be real positive numbers, such that $x+y+z=1$. Prove that:

$\frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\leq 1$

Moreover, we can replace $\leq$ with $<$ if $x,y,z$ are all stated to be positive.

[quote="jrshoch"]$x+yz=(1-(y+z))+yz=1-y-z+yz=(1-y)(1-z)$

Repeating similarly with every denominator:

$\frac{x^2}{(1-y)(1-z)}+\frac{y^2}{(1-x)(1-z)}+\frac{z^2}{(1-x)(1-y)}\leq 1$

Common denominator:

$\frac{x^2(1-x)+y^2(1-y)+z^2(1-z)}{(1-x)(1-y)(1-z)}\leq 1$

I'm going to bed but am I going in the right direction?[/quote]

Continue:

replace the $1-x$ by $y+z$, $1-y$ by $x+z$, and $1-z$ by $x+y$ in the numerator.  multiply out and we get something like $xy(x+y) + yz(y+z) + xz(x+z)$, replace them back by $1-x$, $1-y$, $1-z$ again we get $xy+yz+xz-3xyz$.  Multiply the denominator over and multiply out we have $xy+yz+xz-3xyz \le 1-x-y-z+xy+yz+xz-xyz$.  obviously the products of two variables cancel out.  Also $1 = x+y+z$ so those cancel out too.  divide by $-xyz$ we get $3 \ge 1$ which is obviously true, and we're done

are there any elegant way of doing this instead of algebra all the way??

[img]http://www.video-by.narod.ru/resh1.gif[/img]
x,y,z are real positive and x+y+z=1

Interesting, it seems that this inequality is independent of other famous inequalities?

Besides, if we relax the condition on x, y , z to be [b]non-negative[/b],
the equality sign of this inequality holds if and only if (x=0 or y = 0 or z =0).
This condition is [b]different from[/b] that of the other famous inequalities too.

Shall we give a [b]name [/b]to this inequality?!  :)


Please help me with something like this:

F:R\{0}-->R satisfies for each x apart from x=0 the following equation:

F(-x)/x + F(1/x) = x

Find F

 thanks


A permutation of the integers 1, 2, 3, ..., n is a listing of these integers in some order. For example, (3, 1, 2) and (2, 1, 3) are two different permutations of the integers 1, 2, 3. A permutation $(a_{1}, a_{2}, ..., a_{n})$ of the integers 1, 2, ..., n is said to be "fantastic" if $a_{1}+a_{2}+...+a_{k}$ is divisible by k for [i]each[/i] k from 1 to n.
For example (3, 1, 2) is a fantastic permutation of 1, 2, 3 because 3 is divisible by 1, 3 + 1 is divisible by 2, and 3 + 1 + 2 is divisible by 3. However, (2, 1, 3) is not fantastic because 2 + 1 is not divisible by 2.

a) Show that no fantastic permutation exists for n = 2000.
b) Does a fantastic permutation exist for n = 2001? Explain.

[hide=" a?"]

It suffices to show that $2000\nmid \frac{(2000)(2001)}{2}$. This is because regardless of the permutation, $a_{1}+a_{2}+...+a_{2000}$ will always sum to the same constant value.

$\frac{(2000)(2001)}{2}= 200100$
$2000\nmid 200100$

as desired.[/hide]

[hide="a"]
in order for a fantastic permutation to occur we must have atleast the entire sum $a_{1}+a_{2}+...+a_{n}$ be divisible by n.

For n = 2000
$S_{n}= \frac{n(n+1)}{2}= 1000 * 2001$

$\frac{1000 * 2001}{2000}= \frac{2001}{2}$

this leaves a remainder and thus n=2000 can not have a fantastic permutation.
[/hide]
[hide="b?"]
Taking n = 2001 instead of 2000 from a) we have that the sum is divisible by 2001..

Since it is a permutation there are 2001! combinations of the values so one permutation must yield a fantastic permutation if the entire sum is divisible by n..
I'm stuck here, I dont know how I should prove it.
[/hide]

[quote="max_tm"]Hmm I don't necessarily agree with your statement for b... You showed that every permutation of {1,2,...,2001} is a possible fantastic permutation, but I'm not sure if this guarantees that any are fantastic.[/quote]
Yeah thats where I am stuck.. I proved that there is a possibility that n=2001 could be a fantastic permutation but I dont know how to go about fully proving it.

Yeah I'm in the same boat  :P. I was going to go ahead and try some simpler cases (ie. show that n=5 has a fantastic permutation) but I must go do some AP English lit. practice testing, go on without me  :(.

not bad, not bad. nice solution.  hide it.   [hide=":|"]you so ugly, when you were born yo momma said 'what a treasure' and you daddy said 'yeah let's bury it'[/hide]


it's beens a mixed kinda day,atlast i found sum time to do math ,solveda nice one and cudnt solve sumthing that looked innocuous :D
here's the nice one
Consider the sequence y_2, y_3,… defined by the recurrence relation
(n+1)(n-2)*y_(n+1) = n(n 2  - n - 1)y_n - (n-1)^3*y_(n-1)
and initial conditions y2 = y3 = 1. Show that yn is an integer if and only if n is prime.
 and here's the innocuous one
find all primes p such that 2p+1 and 4p+1 are also primes
awaiting different solutions

For the first one I bet that y_n=(1+(n-1)!)/n. For the second one work modulo 3. You find that p=3.

[quote="galois"]reasonably good guess ,harazi and oops,i forgot the 2nd one was meant for the beginneres section
and i just looked up the source for the 1st one,it's herbert wilf,the man who wrote generatingfunctionolgy (an amazing work)
how many of u have it[/quote]

i have it, and as u said galois it really is a good book/document

I also have that bok, but I didn't read it. In fact, I had posted a link to it on the old version of the site, in the "Math Links" section.


function
inequalities proposed
inequalities

For $x_k$with $k=1,..,n$positive real numbers such that their sum is $1$ prove or confute the following ineq

$\displaystyle \sum{\frac{1}{x_k^2+1}} \le \frac{n^3}{n^2+1} $

With $x_k=\frac{1}{n}$ we have equality. Now let us suppose the maximum of the LHS does not occur there. Then we can take $x_i \neq x_j$ (let $x_i=a$ and $x_j=b$ for simplicity) and replace each by $\frac{a+b}{2}$. We will now show that this increases the LHS, which is inconsistent with it being at a maximum.

We need to show that $\frac{1}{a^2+1}+\frac{1}{b^2+1} \leq \frac{2}{(\frac{a+b}{2})^2 +1}$. Multiplying out, this is equivalent to
\[a^4+2a^3b+2a^2b^2+2ab^3+b^4+6a^2+4ab+6b^2+8 \leq 8a^2b^2+8a^2+8b^2+8\]
\[2(a-b)^2 \geq a^4+2a^3b-6a^2b^2+2ab^3+b^4\]
\[2(a-b)^2 \geq (a+b)^2(a-b)^2 + 2ab(a-b)^2\]
\[2 \geq (a+b)^2 + 2ab\].

But $2 \geq \frac{3}{2} \geq \frac{3}{2} (a+b)^2 \geq (a+b)^2 + 2ab$, as required.

Sorry Fiachra,But You proof $\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{1}{1+\frac({a+b}{2})^2}$
For all $a,b \le \frac{1}{2}$
But with function:$f(x)=\frac{1}{1+x^2}$
$f''(x)=-\frac{1+x^2-4x}{(1+x^2)^2}$
And we don't have $f''(x)\le 0$ for all $x \le \frac{1}{2}$
...maybe I'm wrong?
For this problem,Harazi andI have post two solutions.
I think better use form:
$\frac{1}{1+x^2}\le a+bx$
In general case.

Small typo in my post (minimum instead of maximium). hungkhtn, I effectively showed the function was concave, which I think was also what you showed.


Prove that it is possible to construct a unit cube with five non-overlapping tetrahedrons

Let's consider the unit square with vertexes ABCDEFGH, where ABCD and EFGH are parallel, similarly oriented squares.

Consider removing the tetrahedron with vertexes A, C, F, and H. The result is four other tetrahedrons, each formed around one of the remaining 4 vertexes. Their vertexes are BACF, DACH, EAFH, and GCHF.

This arrangement makes it easy to find the volume of any tetrahedron in which opposite edges have equal length.


To prove that (∂u/∂T)_P=c_P – Pβv where _P =>P constant;β=>co-eff. of vol exp.

I proved it for ideal gases.
Write d'Q=dh-vdP
Now expand d'Q with 1st law and du(in 1st law) in terms of dP and dT.Since du is a total differential.
Expand dh as total differential in dP and dT.

Now I used the property of ideal gases:(∂u/∂v)_T=(∂h/∂P)_T=0
The rest is a bit manipulation.

Can anyone say how to prove this in general?
The book does not mention specifically that "use ideal behaviour" or so.
There must be some way to get it.
Please help.

Again: [b]please[/b] use LaTeX. Maybe there is someone out there with enough patience to plough through ascii math symbols, but I don't think there's much people like that on this forum. Thank you.

I agree with [b]Djole[/b] . Please use $\text{\LaTeX}$ !!
Look for it on the site! It is not that difficult.
Now did you mean this:
$\hline$
To prove that $\left(\frac{\partial u}{\partial T}\right)_{P}=c_{P}-P\beta v$
$\beta$ is co-eff. of vol exp.

I proved it for ideal gases.
Write $d'Q=dh-vdP$
Now expand $d'Q$ with 1st law and $du$(in 1st law) in terms of $dP$ and $dT$.Since $du$ is a total differential.
Expand $dh$ as total differential in $dP$ and $dT$.

Now I used the property of ideal gases:$\left(\frac{\partial u}{\partial v}\right)_{T}=\left(\frac{\partial h}{\partial P}\right)_{T}=0$
The rest is a bit manipulation.

Can anyone say how to prove this in general?
The book does not mention specifically that "use ideal behavior" or so.
There must be some way to get it.
Please help.


calculus
integration
limit
calculus computations

I hope it is not well-known.

\[\int\limits_1^\infty\frac{\{x\}}{x^2}\ dx\]

$\{x\}$ means fractional part of $x$.

Enjoy!


geometry
rectangle
trigonometry
angle bisector
geometry proposed

Let  a rectangle $ ABCD$. In the interior of triangle $ ABD$ we take point $ K$ such that $ AK$ bisects angle $ A$ .Line $ BK$ meets line $ CD$ at $ E$ and line $ DK$ meets line $ CB$  at $ Z$. Prove that  $ DE\equal{}BZ$ .

  [color=blue][i]A synthetic solution in seeken !  [/i][/color]

 Babis

[quote="stergiu"]Let  a rectangle $ ABCD$. In the interior of triangle $ ABD$ we take point $ K$ such that $ AK$ bisects angle $ A$ .Line $ BK$ meets line $ CD$ at $ E$ and line $ DK$ meets line $ CB$  at $ Z$. Prove that  $ DE \equal{} BZ$ .

  [color=blue][i]A synthetic solution in seeken !  [/i][/color]

 Babis[/quote]
[hide] Say that DZ meets AB at N and BE hits CD at E.  Now, extend AK to hit BD at L.  By Ceva's Theorem, we have that
\[ (\frac {AM}{MD})(\frac {DL}{LB})(\frac {BN}{NA}) \equal{} 1
\]

\[ (\frac {AM}{MD})(\frac {BN}{NA}) \equal{} \frac {LB}{DL}
\]
Note that $ \frac {LB}{DL} \equal{} \frac {AB}{AD}$ by the Angle Bisector Theorem.  Thus,
\[ (\frac {AM}{MD})(\frac {BN}{NA}) \equal{} \frac {AB}{AD}
\]
Now, we have that by Menalus's Theorem (using triangle AND and line BKM) that
\[ (\frac {AM}{MD})(\frac {NB}{AB})(\frac {DK}{KN}) \equal{} 1
\]
Noting that triangles KDE and KNB are similar, we have that $ \frac {KD}{KN} \equal{} \frac {DE}{BN}$.  Using this, and multiplying both sides by $ \frac {AB}{AN}$, we obtain that
\[ (\frac {AM}{MD})(\frac {NB}{AN})(\frac {DE}{BN}) \equal{} \frac {AB}{AN}
\]

\[ (\frac {AB}{AD})(\frac {DE}{BN}) \equal{} \frac {AB}{AN}
\]

\[ \frac {AD}{AN} \equal{} \frac {DE}{BN}
\]
Triangles ADN and BZN, giving us that $ \frac{AD}{AN}\equal{}\frac{BZ}{BN}$.  Therefore,
\[ \frac {DE}{BN} \equal{} \frac {AD}{AN} \equal{} \frac {BZ}{BN}
\]

\[ \frac {DE}{BN} \equal{} \frac {BZ}{BN}
\]

\[ DE \equal{} BZ
\]
[/hide]

[quote="stergiu"][color=darkred]Let $ ABCD$ be a rectangle. Let $ K$ be an interior point of $ \triangle\ ABD$ such that

the ray $ [AK$ bisects $ \widehat {BAD}$. Denote $ \|\begin{array}{c} E\in BK\cap CD \\
\ Z\in DK\cap CB\end{array}$. Prove that  $ DE = BZ$ .[/color][/quote]
[color=darkblue]$ W1.\blacktriangleright$ [b]Why " Let $ ABCD$ be a [u]rectangle[/u] " ?
$ W2.\blacktriangleright$ Why " Let $ K$ be an [u]interior[/u] point of $ \triangle ABD$ " ?
$ W3.\blacktriangleright$ Why " the ray $ [AK$ [u]bisects[/u] $ \widehat {BAD}$ " ?[/b]

The [b]author[/b] of this problem was negligently, superficially in finishing off it !

Here is an another proof of this easy and nice problem ([b]more " tidy "[/b]). [/color]

[quote="Virgil Nicula"][color=darkred][b]An easy extension.[/b] Let $ ABCD$ be a parallelogram.

For a point $ M$ define $ \|\begin{array}{c} P\in BM\cap CD \\
\ R\in DM\cap CB\end{array}$. Prove that $ \frac {BR}{DP} = \frac {\sin\widehat {MAB}}{\sin\widehat {MAD}}\ .$
[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the intersections $ \|\begin{array}{c} E\in AM\cap BD \\
\ U\in AB\cap DM \\
\ V\in AD\cap BM\end{array}$. Apply the [b]Ceva's theorem[/b]

to the point $ M$ and $ \triangle\ BAD$ $ \implies$ $ \frac {ED}{EB}\cdot\frac {UB}{UA}\cdot\frac {VA}{VD} = 1\ \ (*)\ .$ Therefore,

$ \|\begin{array}{c} \frac {ED}{EB} = \frac {AD}{AB}\cdot\frac {\sin\widehat {MAD}}{\sin\widehat {MAB}} \\
 \\
AD\parallel BC\implies\frac {UB}{UA} = \frac {BR}{AD} \\
 \\
AB\parallel CD\implies\frac {VA}{VD} = \frac {AB}{DP}\end{array}\|\boxed {\times}$ $ \implies$ $ \boxed {\ \frac {BR}{DP} = \frac {\sin\widehat {MAB}}{\sin\widehat {MAD}}\ }\ .$[/color]

[hide="The relation (*)"][color=darkblue]
Otherwise, we can obtain the relation $ (*)$ if apply the [b]Menelaus' theorem[/b] :

$ \|\begin{array}{c} \overline {DMU}\mathrm {\ and\ }\triangle\ ABE\implies\frac {DE}{DB}\cdot\frac {UB}{UA}\cdot\frac {MA}{ME} = 1 \\
 \\
\overline {BMV}\mathrm {\ and\ }\triangle\ ADE\implies\frac {BE}{BD}\cdot\frac {VD}{VA}\cdot\frac {MA}{ME} = 1\end{array}\|\boxed {\div}$ $ \implies$ $ \frac {ED}{EB}\cdot\frac {UB}{UA}\cdot\frac {VA}{VD} = 1\ .$[/color][/hide]
[color=darkblue][b]Here are two nice and interesting consequencies.[/b][/color]

[color=darkblue][b]Consequence I.[/b][/color] [color=darkred]Let $ ABCD$ be a parallelogram. Let $ R\in (BC$, $ P\in (DC$ be two points for which denote $ M\in BP\cap DR$. Prove that $ BR=DP$ $ \Longleftrightarrow$ the point $ M$ belongs to the bisector of the angle $ \widehat {BAD}$.[/color]

[color=darkblue][b]Consequence II.[/b][/color] [color=darkred] Let $ ABCD$ be a parallelogram. Let $ R\in (BC$, $ P\in (DC$ be two mobile points so that $ BR=DP$. Ascertain the geometrical locus of the intersection $ M\in BP\cap DR$. [/color]

[quote="Virgil Nicula"][color=darkred][b]An easy extension.[/b] Let $ ABCD$ be a parallelogram.

For a point $ M$ define $ \|\begin{array}{c} P\in BM\cap CD \\
\ R\in DM\cap CB\end{array}$. Prove that $ \frac {BR}{DP} = \frac {\sin\widehat {MAB}}{\sin\widehat {MAD}}\ .$
[/color][/quote]
We denote as $ E,$ $ F,$ the orthogonal projections of $ M,$ on $ AB,$ $ AD$ respectively and as $ E',$ $ F'$ the ones, on $ CD,$ $ BC,$ respectively.

It is equivalent to prove that $ \frac {BR}{DP} = \frac {ME}{MF}$ $ (1)$ $ ($ see the [b][size=100]schema t=169109[/size][/b] $ ).$

Because of the triangles $ \bigtriangleup MBR,$ $ \bigtriangleup MDP,$ have $ \angle BMR = \angle DMP,$ we obtain that:

$ \frac {(MR)\cdot (MB)}{(MD)\cdot (MP)} = \frac {(MBR)}{(MDP)} = \frac {1/2\cdot (BR)\cdot (MF')}{1/2\cdot (DP)\cdot (ME')}$ $ ,(2)$

From $ BC\parallel AD$ $ \Longrightarrow$ $ \frac {MR}{MD} = \frac {MF'}{MF}$ $ ,(3)$ and from $ AB\parallel CD$ $ \Longrightarrow$  $ \frac {MB}{MP} = \frac {ME}{ME'}$ $ ,(4)$

From $ (2),$ $ (3),$ $ (4)$ $ \Longrightarrow$ $ (1)$ and the proof is completed.

Kostas vittas.


quadratics
number theory unsolved
number theory

Find the least positive integer $ m$ such that for some prime number $ p$, $ 7^{p}+3^{p}-m$ can be a perfect square.

[hide]I think of this when I read JBMO 2007-4, I misunderstand it as $ 7^{p}+3^{p}-4$, then I calculate that for $ m=0, 2, 3, 4, 5, 7, 8$, $ 7^{p}+3^{p}-m$ can not be a square for prime number $ p$. I had not prove that when $ m=1, 6$, $ 7^{p}+3^{p}-m$ can not be a square for prime number $ p$ yet.
The answer might be $ m=9$ for $ p=2$.[/hide]

[quote="zhaoli"]Find the least positive integer $ m$ such that for some prime number $ p$, $ 7^{p}+3^{p}-m$ can be a perfect square.[/quote]
Since $ 7^{2}+3^{2}-9 = 7^{2}$, it's enough to prove there aren't any positive integer $ m \le 8$ and any prime $ p \in \mathbb{N}$ such that $ 7^{p}+3^{p}-m$ is a perfect square. Indeed, suppose the contrary, so letting [E1]: $ 7^{p}+3^{p}= a^{2}+m$, where $ a \in \mathbb{N}$. Then obviously $ a$ and $ m$ have the same parity. If $ m \equiv 1 \bmod 2$, then $ (-1)^{p}+3^{p}\equiv m+1 \bmod 8$, viz $ m \equiv 1 \bmod 8$. Thus necessarily $ m = 1$ and $ 7^{p}+3^{p}= a^{2}+1$, which in fact makes $ a \equiv 0 \bmod 3$ and $ 7^{p}\equiv 1 \bmod 9$ (absurd as $ \mbox{ord}_{9}(7) = 6$).

So $ m \equiv a \equiv 0 \bmod 2$. Even $ 7^{n}+3^{n}\equiv 2 \bmod 4$, for any $ n \in \mathbb{N}$, and hence $ m \equiv 2 \bmod 4$. Now note that $ 7^{2}+3^{2}-m$ is not a square for both $ m = 2$ and $ m = 6$. Then in the following let $ p \ge 3$ as to equation [E1], and conclude that $ 7^{p}+3^{p}\equiv 0 \bmod 5$ and $ a^{2}\equiv-m \bmod 5$. If $ m = 2$, this is absurd, as $ -2$ isn't a quadratic residue $ \bmod\;\! 5$ (in fact, $ (-2)^{\frac{1}{2}(5-1)}\not\equiv 1 \bmod 5$). If finally $ m = 6$, argue from [E1] that $ a^{2}\equiv-1 \bmod 5$, which is absurd again for similar reasoning as above.

First, thank s.tringali for your interest on my problem.

I partly solve the problem as follows:
For $ p=2$, it is clear that $ m=9$ is the least positive integer.
So let $ p \geq 3$, then $ p$ is odd. Let $ A(m)=7^{p}+3^{p}-m$.
For $ m=4, 8$,  $ A(m) \equiv 2 (\bmod 4)$.
For $ m=3, 7$,  $ A(m) \equiv 3 (\bmod 4)$.
For $ m=2, 5$,  $ A(m) \equiv 2 (\bmod 3)$.
Also for $ m=2, 3$,  $ A(m) \equiv 3, 2 (\bmod 5)$.
For $ m=1$, $ A(1)$ is divisible by $ 3$, as it is a square, it should be divisible by $ 9$. so $ 7^{p}\equiv 1 (\bmod 9)$, so $ p$ is divisible by $ 3$, so $ p=3$. But $ 7^{3}+3^{3}-1=369$ is not a square.

As for $ m=6$, I have not think of an idea.

By the way, $ a^{2}\equiv-1 (\bmod 5)$ can be ture for $ a \equiv 2 or 3 (\bmod 5)$. Would you please explain your last statement more clearly?


[hide="Number 2"]
Let these be x. We have that
$(4n)^{x}=4^{x}n^{x}=40\sqrt{3}$ and $(3n)^{x}=3^{x}n^{x}=45$.  Dividing, we get that $\frac{45}{40\sqrt{3}}=\frac{3\sqrt{3}}{8}=(\frac{3}{4})^{x}$  It is easy to see that x is simply 1.5. Then we have that $((4n)^{\frac{3}{2}})=40\sqrt{3}\rightarrow n^{\frac{3}{2}}=5\sqrt{3}\rightarrow n^{3}=75$[/hide]

[quote="samath"][hide="Number 2"]
Let these be x. We have that
$(4n)^{x}=4^{x}n^{x}=40\sqrt{3}$ and $(3n)^{x}=3^{x}n^{x}=45$.  Dividing, we get that $\frac{45}{40\sqrt{3}}=\frac{3\sqrt{3}}{8}=(\frac{3}{4})^{x}$  It is easy to see that x is simply 1.5. Then we have that $(n^{\frac{3}{2}})=\frac{40}{3}\rightarrow n=(\frac{40}{3})^{\frac{2}{3}}=\sqrt[3]{\frac{1600}{9}}\rightarrow n^{3}=\frac{1600}{9}$[/hide][/quote]

Check your work. You made a small mistake.  :wink:


probability
geometry
3D geometry
tetrahedron

Here are some easy problems from the MMPC

Two integers are chosen randomly between 0 and 80 inclusive. The sum of these integers is divided by 9. What is the probability that the remainder is 3?

A geometric figure has four vertices, A,B,C,D. The distances between the verticies are AB=1, AC=1,AD=11/20,BC=1,BD=11/20. This figure is a?

A - triangle
B - square
C - quadrilateral other than a square
D - tetrahedr0n
E - does not exists.

[hide]I'm guessing D, tetrahedr0n, because It can't be a quadrilateral since AB=1, AC=1, and BC=1, plus there's four points so it's not a triangle. Judging from the fact that AD=11/20 and BD=11/20 (also from your name  ), I say it's a tetrahedr0n. Can anyone confute me?  [/hide]

Did I use it right now?

Oh, and I added the "Easy" in there cause it's not much of a challenge  , no offence. By the way, what does "MMPC" stand for?

[hide]

There are 81 numbers between 0 and 80 inclusive. We are asked to find those that are 3 mod 9. Because they are consecutive numbers, and 81 is a multiple of 9,  it's safe to assume that exactly 1/9 of the numbers are of residue class 3, so the answer is 1/9.

Assuming the points are coplanar, it is a quadrilateral other than a square.

[/hide]

[color=cyan]The second one definitely is "not enough information."[/color]

:?: It could be C and D, I guess, but ABCD would be concave quadrilateral, but that's probably OK. So maybe E means what Joel said...any advise?

But Southpaw, the problem says that you must consider the "sum of these" two chosen integers, not the integers between 0 and 80 themselves.

Good point; by combinatrics...er, combinations, the only length left undefined is CD...so if it's less than 9/20th or something, then it'll be impossible...

MMPC is Michigan Mathematics Prize Competition. These problems are from part 1, where there are 40 such problems. you are right tare, they are pretty easy. the real challenge is part II, where they can get quite a bit harder.

Ever wonder why Michigan is so good at math? (beside from the point that it has a big population)

Anyways, this could be turned into a question like this: There is an equilateral triangle ABC. There is point P within the triangle. If AP=BP=x, then what is CP?

[color=cyan]It need not be concave -- it can be a convex quadrilateral or a concave quadrilateral or a variety of tetrahedra.  If the other length were given, that would do it.  I also thought 11/20 might have been meant to be replaced with something smaller than 1/2.[/color]

[quote="TheSouthpaw"]You're right. I didn't read the question carefully. However, I have a hunch that I'm right anyways, lemme do it out:

There are 9*9 = 81 ways to sum two numbers and get  3 mod 9. This is because for every integer a, there is another integer b for its "3-inverse." There are 81*81 ways of summing two numbers from the set 0 - 80. Thus the ansewr is 81/81:^2: = 1/9[/quote]

Maybe it's just me, but isn't 81/81:^2: equal to 1/81?

lol.. how is it easy if you got it wrong?  :lol:

yeah yeah, I [b]thought[/b] it was easy...*sigh* changing back title...now.

looking back, i included all the problem data. strange, huh?

[quote="rrusczyk"]For that second question - maybe you left out length CD by mistake?[/quote]
The second one's still unsettled. We get two answers currently, C and D.

these answers are directly from the official solutions:
(1). Among the integers from 0 to 80, the 9 possible remainders each occur 9 times. Thus all 9 remainders are equally likely when the sum of two numbers is divided by 9.

(2) ANS : E  - Triangle ABC is equilateral . Its altitude has length  3^(1/2)/2. So the distance from a vertex to the center is 2/3*3^(1/2)/2=11/(3^(1/2)) which is greater than 11/20. Thus D cannot be 11/20 from A, B, C.

oh shoot - i DID copy the problem down wrong. well thats the last time anyone should listen to me. ok for the second problem, we also have CD=11/20. (Like Mr. Rusczyk mentioned.) wow that was so stupid. at least we had an good debate over the problem.

9 has 9 rem.
in 4 ways the rem.s add upto 3. from 9 possible choices of rem.s 2 can be chosen in 36 ways.
thus probability=4/36=1/9.

[quote="tetrahedr0n"]the answer to the first one is right. the second one is strange.... i typed the problem exactly as it appeared. the solutions packet says it cannot exist.[/quote]

but if it is a tetrahedron it can exist.
as in 2d  plane it cannot exist.


complex analysis
complex analysis unsolved

Prove that
\[ \sum_{m\equal{}\minus{}\infty}^{\infty}\sum_{n\equal{}\minus{}\infty}^{\infty}\frac{1}{(m^2\plus{}a^2)(n^2\plus{}b^2)}\equal{}\frac{\pi^2}{ab}\coth (\pi a)\coth (\pi b)\]

[hide]Use [url=http://www.mathlinks.ro/viewtopic.php?t=254323]this[/url] theorem.[/hide]

That thing obviously splits into a product of two times $ \sum_{m \in \mathbb Z} \frac{1}{m^2\plus{}a^2} \equal{} \frac{\pi}{a}\coth(\pi a)$. Well, calculating that kind of sum is just using the residue theorem and integrating about a circle whichs radius tends to $ \infty$, as you say in that post.
But why did they square it¿

I don't know, maybe they wanted to tangle who will solve it, i just wrote how it is in the original. It is from [url=http://rapidshare.com/files/196322311/Whittaker_K.T.__Watson_G.N._A_course_of_modern_analysis__CUP__1935__T__609s__MCet_.djvu.html]Whittaker K.T., Watson G.N. A course of modern analysis (CUP, 1935)(T)(609s)[/url] page 136.


Hi everyone,

Could someone please help with this problem?

It says:

You are going to play a game. First you roll a standard six-sided die. Then using the number on the die, you toss that number of coins. For example, if you roll a 4 then you will toss 4 coins.

a. Determine all the possible outcomes for this game

I found 6 possible outcomes 1,1 2,2 3,3 etc. because you can only toss the number of coins you rolled on the die.

b. suppose you win 1 dollar for each head that you toss. Determine the probability distribution for the amount of money won.

I got 10.50 or less because half of 21 is 10.50 from adding up all the possible heads.

d. Suppose you are setting up a booth at a school carnival and you want to rais 2.00

I would really appreciate any help that you could give me

Thank you very much

And we all would really appreciate if you used the preview option before pushing the "submit" button. ;) Please, edit your post to make it readable and then we'll think of your question :)


geometry
3D geometry
sphere
calculus
derivative
function
Pythagorean Theorem

1.The surface area of a sphere  is decreasing at the constant rate of 3pi cm^2/s At what rate is the volume of the sphere decreasing at the instant its radius is 2cm?
Side note: I am aware that the equation is (4/3)(pi)r^3 and just apply 2 from the start seems too good to be true. The answer is -3(pi) cm^3/s

2. A ladder 13 ft. long rest against a vertical wall and is sliding down the wall at the rate of 3ft/s at the instant the foot of the ladder is 5 ft. from the base of the wall. At this instant how fast is the foot( or base) of the ladder moving away from the wall? Side note this problem looks like a pythag. theoremx^+y^2=r^2. the answer comes out to 7.2 ft./s. on this one. though I know what formula to use for the certain situation, I really do not know the steps in solving these problems.

You just want to carefully write down everything you know and then plug in so

$ S(r)=4\pi r^{2}$ which means $ \frac{dS}{dt}=8\pi r\frac{dr}{dt}=-3\pi$ so the rate at which the radius is changing is $ \frac{dr}{dt}=-\frac{3}{8r}$. Now use this in your volume formula:

$ V(r)=\frac{4}{3}\pi r^{3}$ and thus $ \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=\frac{-3\pi}{4}r$. Just evaluate this at $ r=2$ to get $ \frac{dV}{dt}\big|_{r=2}=-3\pi$, with your units of course.

The general principle of related rates problems:

(1) Write down an equation which is [i]always[i] true; the sphere volume formula or the Pythagorean theorem, in these cases.
(2) Differentiate.
(3) Use the information which is true at this specific moment (such as that radius of 2, or the foot of the ladder 5 feet from the wall) to solve for the rate of change you want.

The order is important: if we differentiate something which is only true at one moment, we don't get a true equation.

I see how -3pi is the answer, for the second, I used the pythag. theorem, and tried to find x and y(which did not work). And I do not know of any other equation for that problem. please help.

The Pythagorean theorem does work. Show us what you tried.

ok I tried to find y  and x in this case,x^2+y^2=r^2. I know the 13 ft. has to be r, So for y, i got y=(-2x-2rdr/dt)/2 and for x, x=(-2y-2rdr/dt)/2. I do not think this makes any sense. Also. I isolated dr/dt for it to be -2x-2y-2r=dr/dt. I am not sure if I am getting the right idea though.

$ x$ and $ y$ are changing. $ r$ is not; the ladder is always exactly 13 feet long. You derivatives don't make sense, and you're not trying to solve for $ x$ or $ y$.

ok, so what would be the first step then if I am using just one equation instead of 2 like the last one?

$ x$ is a function changing with time and $ y$ is a function changing with time, but $ r$ is not so differentiating the Pythagorean Thm leads to $ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$


Just to keep stuff recent, what are your current favorite songs?

 :)

when youre gone-avril lavigne

face down-red jumpsuit apparatus

the great escape-boys like girls

Guns and Roses
Sweet Child of Mine
November Rain
Patience

All American Rejects
Move Along
It Ends Tonight


How come there is no American flag on my posts?


linear algebra
matrix
search
linear algebra unsolved

let A be $ n\times n$ matrix such that each row has exactly 2 entries distinct 0: one entry on the main diagon is 2006, and the other is 1. prove that A is invertible

For a more general result see http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1488690471&t=32062.
A similar proof than Gershgorin Circle Theorem applies here as well.


analytic geometry
trigonometry
integration
Euler
function
calculus
calculus computations

find the volume of solid which is bounded by $\left(\frac{x}{a}\right)^\frac{p}{q}+\left(\frac{y}{b}\right)^\frac{p}{q}+\left(\frac{z}{c}\right)^\frac{p}{q}=1$, where $p,q$ are coprime integers.

Since the points of solid M are  symmetrical to the  coordinate planes, the volume will be:
\[V=8\iiint_{\substack{M^'}} dxdydz\]
where M' is the eighth part of the solid placed in the first octant.
Using the generalized sferical coordinates:
\[x=a\rho\sin^\alpha\vartheta\cos^\beta\varphi\qquad y=b\rho\sin^\alpha\vartheta\sin^\beta\varphi\qquad z=c\rho\cos^\alpha\vartheta\\
\qquad d\\
0\le\rho\le 1,\qquad 0\le\vartheta\le\pi/2\qquad 0\le\varphi\le\pi/2\]
with $\alpha=\beta=2q/p$
Then the Jacobian is:
\[
\frac{D(x,y,z)}{D(\rho,\vartheta,\varphi)}=\frac{4q^2}{p^2}
\cos^{\frac{2q-p}p}\vartheta\sin^{\frac{4q-p}p}\vartheta
\sin^{\frac{2q-p}p}\varphi\cos^{\frac{2q-p}p}\varphi\]
and the volume:

\[
\begin{array}{cc}
V=\frac{32q^2}{p^2}abc\int_0^{\pi/2}\cos^{\frac{2q-p}p}\vartheta
\sin^{\frac{4q-p}p}\vartheta\,d\vartheta\int_0^{\pi/2}
\sin^{\frac{2q-p}p}\varphi\cos^{\frac{2q-p}p}\varphi\,d\varphi
\int_0^1\rho^2\,d\rho&=&\\
\dfrac{8abcq^2}{3p^2}B\bigg(\frac{2q}p,\frac qp\bigg)
B\bigg(\frac qp,\frac qp\bigg)&=&\\
\dfrac{8abcq^2}{3p^2}
\dfrac{\Gamma^3(q/p)}{\Gamma(3q/p)}\end {array}\]
where $\Gamma \;\text{and}\; B$  are the Euler functions tied by known relation:
\[\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=B(x,y)\]


For what value of $ x$ does $ 2^{12} \equal{} (\frac{1}{8})^x$?


Is there any n>3 such that $\prod_{k=1}^{n}(1+k^2)$ is a square?

Notice that for n=3, the product is 100.

[quote="harazi"]I didn't solve it, but I showed that there are finitely many n or which that expression could be a square. Since I don't find this trivial and since it was just an observation and not a proof of mine, I will post it:
   The idea is that for large n the largest prime factor of that expression is at least 2n+1. This follows from a theorem of Cebasev which states that  if p(n) is the largest prime factor of the product then $\frac{p(n)}{n} $ tends to infinity as n grows.  I won't prove that  :D . But I will use it. There is M such that for all n>M we have $ p(n)>2n+1 $. Then it is easy to see that $ p(n) $ must appear with exponent 1 in the product. Indeed, it cannot appear with exponent more than 1 in any factor of the product and it cannot be factor of $ i^2+1 $ and $ j^2+1$ for different i and j. So, part of the open question is proved.
   I know some of you might say this post is useless, but I'm too tired now to deicde wheter it's useless or not.[/quote]

Harazi, could you post the paper for Cebasev's theorem? Thanks.


Suppose that $ (x_n)$ is a sequence in $ \mathbb{R}$. Define a sequence $ (y_n)$ by $ y_n\equal{}\frac{x_1\plus{}x_2\plus{}\cdots\plus{}x_n}{n}$, $ \forall n \in \mathbb{N}$. Prove that if $ (x_n)$ converges to $ x \in \mathbb{R}$ then $ (y_n)$ converges to $ x$.

This is pretty much an exact replay of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167852]this thread[/url].

As for moldovan's comment, see this comment from that other thread:

[quote="jmerry"][quote="TTsphn"]Use Stolz lemma[/quote]
That's circular- this [i]is[/i] the theorem (Cesaro-Stolz is how I usually see it here).[/quote]


An organic compound A does not undergo aldol condensation and is used with some other components as a corrosion inhibitor for steel.A is prepared from B and C by treating them with OH-(base) and heating them.C answers iodofrom test.B on treating with Ph3P=CH2 gives the monomer D of a widely used polymer.  Using D's polymer,petri dishes are made. D reacts with E to form another Polymer F which is used in the manufacture of gaskets.  E reacts with G to from another polymer H.E on ozozolysis gives you glyoxal and 2 moles of methanal.

find A,B,C,D,E,F,G,H :?:  :wink:

 :D :)  :D  :P

B-benzaldehyde
D-styrene
E-butadiene
F-BunaS
G and H- there could be so many
A-benzylideneacetophenone(guessing)
C-acetophenone(guessing)

Sorry VARUNARASU

include this data

polymer H is used in footwear, adhesives, sealants.


projective geometry
geometry solved
geometry

Quadrilateral $ABCD$ is circumscribed around a circle with center $O$. The foot of the
perpendicular drawn from $O$ to $AC$ is $M$. Prove that the distances from $O$ to $BM$ and $DM$ is equal.

I change the notations:

[color=blue][b]Problem.[/b] Let ABCD be a circumscribed quadrilateral and I its incenter. Denote by M the orthogonal projection of the point I on the line BD. Prove that the point I has equal distances from the lines AM and CM.[/color]

[i]Solution.[/i] Let the diagonals AC and BD of the quadrilateral ABCD meet each other at E. Let U be the harmonic conjugate of the point E with respect to the segment AC. Then, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=135302#p135302]http://www.mathlinks.ro/Forum/viewtopic.php?t=20538 post #2[/url], we have $IU\perp BD$. Thus, the point U lies on the perpendicular to the line BD from the point I. The foot of this perpendicular is, of course, the orthogonal projection M of the point I on the line BD.

Now, as the point U is the harmonic conjugate of the point E with respect to the segment AC, the lines ME and MU are harmonically conjugate lines with respect to the angle AMC. On the other hand, these two lines ME and MU are perpendicular to each other (this follows from $IU\perp BD$). If two lines are harmonically conjugate to each other with respect to an angle and perpendicular to each other, then they are the two angle bisectors (the internal one and the external one) of this angle. Hence, the two lines ME and MU are the two angle bisectors of the angle AMC. Particularly, this yields that the point I, lying on the line MU, lies on one of the two angle bisectors of the angles AMC. Hence, this point I has equal distances from the edges AM and CM of this angle. The problem is solved.

 Darij


http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html

what areas of this do you think and neccessary to learn?

[quote="Mr. Diaz"]Chances are the person ranked above you knows it better than you do.[/quote]
I really have to disagree with the philosophy behind memorizing a sheet of formulas; I highly advise you to read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]Mr. Rusczyk's article[/url] on the matter.  It goes against the entire philosophy of this website.

At the same time, most of this stuff is legitimately worth knowing, but I would advise you to understand why, for example, the diagonals formula works.

okay. sorry, i read it, i completely disagree. i am learning these tricks and im also learning why they work. that question is all i need answered. which is the most useful in MC?

All MathCounts problems have some little trick that, if you know, it won't take you long to do the problem. A lot of these tricks involve formulas, and trying to derive formulas during the test gives you a big disadvantage. So neither is more important; you need both.

If I remember correctly, all I've ever actually [i]intentionally[/i] memorized were the squares and cubes. The rest all came from practice.

see my goal is just to finish the test. ive never finished an MC test in my life. help...PLEASE!!! i only have til monday.

Here are a few tips.

If you're a cautious person like me, and you always want to check your answer before moving on to the next problem, don't check your answer unless the problem is something like counting triangles in a figure, where it's really easy to miscount and it really is likely that you'll get it right if you double check.

If you read a problem, and it seems really hard, reread it more carefully this time. That way, you won't be spending time solving the wrong problem.

Go as fast as you can on the test. (unless you are the kind of person who makes lots of mistakes if you go too fast.) Because sometimes, I'll be taking a practice sprint round and then   realize that I wasn't trying to go fast. It needs to be a sprint. that's why they call it the sprint round. (I sometimes forget this  :P )

I hope this helps!

On the target though, do make sure to check it. I make tons of mistakes on the target rounds because of that (though maybe it's just because it's a practice round and I'm too lazy to time it, so the instant i finish, I go to the next set)


It is given a triangle $ ABC$.
$ M$-inner for ABC.
$ MN \bot AC (N \in AC); MP \bot BC (P \in BC)$.
$ \angle MAC\equal{}\angle MBC$
$ AQ\equal{}QB (Q\in AB)$
To proove that $ NQ\equal{}PQ$

Let X and Y be the mp. of AM and BM respectively.
QY=XM=XN. PY=YM=QX
also in rt. $ \triangle XQN$ and $ \triangle QYP$.
angle(PYM=BMP=AMN=MXN). and angle(QYM=QXM). so angle(QXN=QYP).
thus $ \triangle QYP$ congruent $ \triangle QXN$ by S.A.S.
thus QP=QN


[quote="Carcul"]Show how 3-ethylbenzonitrile can be prepared from ethylbenzene.

(Source: created by Carcul)

Carcul[/quote]

[b][i]1.NITRATION (PARA PRODUCT ISOLATED)
2. KMnO4 (p-nitro benzoic acid)
3. SOCl2 AND NH3 (p-nitro benzamide)
4. P2O5 (p-nitro benzonitrile)
5. von Richter Synthesis (to get 3 cyano benzoic acid)
6. SOCl2 , (CH3)2Cd
7. Mozingo reduction[/i][/b]

Wouldn't the cyano also be reduced? You could also have some problems in some of those steps. Anyway, following the hint I gave only two steps are enough...

How about:

1. Ethylbenzene + HNO3 + H2SO4 --> p-ethylnitrobenzene
2. p-ethylnitrobenzene + NaNO2 + HCl --> C2H5(C6H5)N2+ Cl-
3. Diazonium salt + CuCN --> p-ethylbenzonitrile

That's correct, except for the fact that I asked for the synthesis of [b]3[/b]-ethylbenzonitrile, not [b]4[/b]-ethylbenzonitrile.  :juggle:

Oh oops!  Sorry, here's a revised synthesis.

1. ethylbenzene + CrO3 ---> acetophenone
2. acetophenone + HNO3 + H2SO4 --> m-nitroacetophenone
3. m-nitroacetophenone + Zn(Hg) + HCl --> m-ethylaniline
4. m-ethylaniline + NaNO2 + HCl --> CH3CH2 (C6H4) N2+ Cl-
5. diazonium salt + CuCN --> m-ethylbenzonitrile

I will use very simple reaction (finished in 1 minute)but may not be practical
1) Br2/hv to dibromide
2) H2O/H+ to acetophenone
3) CO/HCL/AlCl3 (Gatterman-Koch)
4) glycol/H+
5) Zn(Hg)/HCl
6) hydrolyzed
6) Tollen
7) PCl5
8) NH3
9) P2O5

Infinitesimal yield

Is this a plausible method ? I don't know how to reduce both together , clemmenson reduction ..followed by Hs/SnCl4 is one method I know . can that be done ?

That looks perfectly ok, but observe that the keto group should be reduced first, otherwise imine formation can be considerable.


Do there exist a finite set $A\,(|A|\geq 2)$ of distinct points in the plane $\mathbb{R}^{2}$ such that for every $x,y\in A,x\ne y$ there exists $z\in A$ such that $z$ is in the normal passing through the midpoint of the line segment $xy$?

The three vertices of an equilateral triangle?

Pierre.

Also a regular pentagon -- perhaps the question should be to ask for all possible sizes of such a set?


Compute the sum:

{m/n}+{2m/n}+...+{(n-1)m/n}, where m,n are from N*, and (m,n)=1.

this one is kinda trivial!  :?

Solution:

Lets denote S={m/n}+{2m/n}+...+{(n-1)m/n} [b](1)[/b]
and we write it again like: S={(n-1)m/n}+...+{2m/n}+{m/n} [b](2)[/b]

because (m,n)=1 we have that the numbers k*m/n and (n-k)m/n with a<=k<=n-1 are not integers and their sum k*m/n+(n-k)m/n=m which is natural,

adding (1) and (2) we get that: 2S=n-1 and S=(n-1)/2

cheers! :D :D :D :D


calculus
derivative
analytic geometry
graphing lines
slope
function
limit

Find the Derivative of these functions.....

1. 4 sqaure root x
2. 3 cos x
3. 4 sin x
4. -3e^x
5. 4lnx      l= one of those funny loop things
6. 3/x^3

7. 5x^4 - 4x^-5 + 6x^1/3 - 6x^1/3
8. 3sin x- 4cos x + 3e^x- 5lnx     Again, 1= funny loop thing.
9. x^4 sin x
10.( x^3+x^2+x+1)^1/2
11. 3sin(x^2) - 2cos(3x)
12. x^4 / x^2+2x+3
13. e^2x ln(x^2+1)
14. 5000(1-e^ -0.03t)
15. (x+1)^3 / x^2
16. 4x^3(3x-4)^2

17.Let y= f(x)+ x^4 +8x^3 + 16x^2

a) Find dy/dx = y' and factorize it
b) Find d^2y/ dx^2 =y"
c) Find the rate of change of y with respect to x at x= 1 and at x= -1
d) What is the slope of the tangent to y=f(x) when x+-3
e) Find y when x= -3
f) Find the equation of the tangent to y=f(x) at x= -3. You should use the results from d) and e)
g) At which values of x is the tangent horizontal?
h) At what points does the tangent to y= f(x) have slope 12?
(Hint: From (d), x= -3 is a solution of f'(x) =12.)

18. Find the values of x for which f' (x)= 0. ( The functions are said to be "stationary" at these values of x)

a) f(x) = 2x^2 +3x+4
b) f(x) = 2x^3 - 3x^2 +1
c) f(x) = 4x^3(3x-4)^2
d) f(x) = (x+1)^3 /x^2
e) f(x) =  3x+4/5x-7

First of all, this should be in the Calculus Forum.

Now, show us what you've gotten so far and we'll tell you which ones are right, and point out what you're doing wrong on the ones that aren't right.

[hide="Hint"]plug them into $ f'(x)\equal{}\lim_{x\to 0} \dfrac{f(x\plus{}\triangle x) \minus{} f(x)}{\triangle x}$.[/hide]

That's bad advice. She should be developing and learning rules (which are based on that limit) so that she doesn't have to evaluate it every time she wants to take a derivative.

When you start plugging into that limit, then you will see how to calculate it faster with trig variables etc.

There is something to be said for proving some basic derivatives yourself, but most students are not equipped with the problem-solving capability to, for example, independently derive the product or chain rules, nor should they be expected to.

Two things:

- Do your own homework.
- If you still can't figure it out, ask about concepts, not answers.

As Mahatma Gandhiji used to say "Apna kaam khud karo" (meaning-do your work yourself) i also advice you do your own homework. :D  :D


Let $ABCD$ be a square and $M,N$ points on sides $AB, BC$ respectively such that $\angle MDN = 45^{\circ}$. If $R$ is the midpoint of $MN$ show that $RP =RQ$ where $P,Q$ are points of intersection of $AC$ with the lines $MD, ND$.

Maybe...

$\angle PDN = 45^{o}= \angle PCN$. Then $PDCN$ is cyclic and $\angle PCD = \angle PND = 45^{o}$.
So $PN \bot MD$. Since $MR = RN$ and $\angle MPN = 90^{o}$ we have that $MN$ is the diameter of the circumscribed circle of $MPQN$. Then $RM = RN = RP = RQ = r$.

how does $\angle PDC = 45^{o}$ it's said $\angle PDN =45^{o}$?

prove $\text{AMQD}$ and $\text{DPNC}$ are cyclic. angle bash some more. and then prove $\text{MNQP}$ is cyclic, upon noticing $\angle \text{MPN}= 90$
$\implies \text{MN}$ is diameter. $\text{R}$ is centre hence $\text{RP}=\text{RQ}$ oops sorry. i thought ur soln had an error. but it was actually a typo. sorry  :blush:

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20I.pdf p. 16-18.

Sincerely
Jean-Louis

It's already clear that, $DCNP$ is cyclic, which implies that, $\angle MPN=\angle MCN=90^{\circ} \implies BMPN \text{ is cyclic}$ Also, $MR=RN \implies R \text{ is the center of circle passing through } BMPN$, Now since, $$\angle DPN=2\angle DBN \implies PD=BP \implies \angle PBC =\angle MDB +45^{\circ}=\angle NDC+45^{\circ}=180^{\circ}-\angle AQN$$ $ \implies Q \text{ also passes through } BMPN \implies RP=RQ=RN=RB=RM$

[quote=jayme]Dear Mathlinkers,

also at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=500164

Sincerely
Jean-Louis[/quote]


function
calculus
integration
analytic geometry
geometry
3D geometry
sphere

Hi all,

I'm having some difficulty demonstrating the following:

Show that neither the inclusion $ L^2(\mathbb{R}^d) \subset L^1(\mathbb{R}^d)$ nor the inclusion $ L^1(\mathbb{R}^d) \subset L^2(\mathbb{R}^d)$ is valid.

I've tried considering the function $ f(x)\equal{}|x|^{\minus{}\alpha}$ for $ |x| \leq 1$ and $ |x|>1$, when $ f(x)$ is in $ L^1(\mathbb{R}^d)$ for $ |x| \leq 1$, but I'm not really getting anywhere.  I feel like this is the right direction to be going, but I'm stuck.  Please help!

This is the right direction - you just have to pick the values of $ \alpha$ that make the argument work. Can you show us your specific computations?

That's sort of my problem - making it a rigorous argument.  I have the inkling that $ f(x)\equal{}|x|^{\minus{}\alpha}$ is integrable whenever $ \alpha < d$.  I can't perform the calculations, though.  I think the real problem is that I don't quite understand what is going on here.  The text that I'm working out of is quite terse, and it's difficult to fully grasp a concept when it's presented in that way.  I've tried to approximate $ f$ by simple functions; then I just get an ambiguous expression like $ \sum_{k\equal{}1}^{\infty} a_k m(E_k)$ for the value of the integral, for some $ a_k$ constants and $ E_k$ measurable sets.

I'm not looking for a quick answer.  Instead, I would like to really understand how to show these things.

convert to polar coordinates and integrate the same way you would in a calculus course.

The polar coordinates give you that $ dx \equal{} dx_1\,dx_2\dots dx_d \equal{} r^{d \minus{} 1}\,d\omega\,dr$

If we integrate $ d\omega$ over all possible angles, we get the $ d \minus{} 1$ dimensional measure of the unit sphere in $ \mathbb{R}^d.$ While an exact value can be given for that, it's would be beside the point here - just call it $ a_d.$

For instance, let $ f(x) \equal{} |x|^{ \minus{} \alpha}$ for $ |x| < 1$ and $ 0$ otherwise.

$ \|f\|_1 \equal{} a_d\int_0^1r^{d \minus{} 1}r^{ \minus{} \alpha}\,dr \equal{} \begin{cases}\frac {a_d}{d \minus{} \alpha} & \alpha < d \\
\infty & \alpha\ge d\end{cases}.$

If you want to be even a little more rigorous, you can integrate over $ B(0,1)\setminus B(0,\varepsilon)$ and see what happens when $ \varepsilon\downarrow 0$.


1 ) Find all $ x,y,z \in \mathbb{N}^* / \{1\}$ such that $ (x\minus{}1)(y\minus{}1)(z\minus{}1) \mid xyz\minus{}1$ .
2 ) Find all $ x,y,z \in \mathbb{N}^*/ \{1\}$ such that $ (x\minus{}y)(y\minus{}z)(z\minus{}x) \mid xyz\minus{}1$ .

Problem number 1 is exactly the first problem of IMO 1992..


Solve in terms of $ a$ and $ b$, assuming $ ab\neq \minus{} 1$

$ \frac {x \plus{} y \minus{} 1}{x \minus{} y \plus{} 1} \equal{} a$

and

$ \frac {y \minus{} x \plus{} 1}{x \minus{} y \plus{} 1} \equal{} ab$

[hide="Answer"]$ x \equal{} \frac {a \plus{} 1}{ab \plus{} 1}$ and $ y \equal{} \frac {a(b \plus{} 1)}{ab \plus{} 1}$[/hide]

[hide="Solution"]

First, we must assume $ x\neq y\minus{}1$

From the first equation: $ x\plus{}y\minus{}1\equal{}ax\minus{}ay\plus{}a$
$ x\equal{}(y\minus{}1)\frac{a\plus{}1}{a\minus{}1}$

From the second equation: $ y\minus{}x\plus{}1\equal{}abx\minus{}aby\plus{}ab$
$ x\equal{}\frac{y\plus{}1\plus{}aby\minus{}ab}{ab\plus{}1}$

Now compare 2 equations for x, eliminating it, and solve for y. Return result for y into one of them, and you got x. [/hide]

[hide="Simpler Solution"]
Note that you can simplify the LHS in the second equation like so:

$ \frac {y \minus{} x \plus{} 1}{x \minus{} y \plus{} 1} \equal{} \frac { \minus{} (x \minus{} y) \minus{} 1 \plus{} 2}{x \minus{} y \plus{} 1} \equal{} \minus{} 1 \plus{} \frac {2}{x \minus{} y \plus{} 1}.$

Applying that, we have

$ \frac {2}{x \minus{} y \plus{} 1} \equal{} ab \plus{} 1\implies x \minus{} y \plus{} 1 \equal{} \frac {2}{ab \plus{} 1}\qquad \text{(1)}.$

Applying (1) on the first equation, we have

$ x \plus{} y \minus{} 1 \equal{} a(x \minus{} y \plus{} 1) \equal{} \frac {2a}{ab \plus{} 1} \qquad \text{(2)}.$

Adding (1) + (2), we get

$ 2x \equal{} \frac {2(a \plus{} 1)}{ab \plus{} 1}\implies x \equal{} \frac {a \plus{} 1}{ab \plus{} 1}.$

Subtracting (2) from (1), we get

$ 2y \minus{} 2 \equal{} \frac {2(a \minus{} 1)}{ab \plus{} 1}\implies y \equal{} \frac {a \minus{} 1}{ab \plus{} 1} \plus{} 1\implies y \equal{} \frac {a(b \plus{} 1)}{ab \plus{} 1}.$

Therefore, $ (x,y) \equal{} \left(\frac {a \plus{} 1}{ab \plus{} 1},\frac {a(b \plus{} 1)}{ab \plus{} 1}\right).$
[/hide]

[hide]we have $ a\plus{}1\equal{}\frac{2x}{x\minus{}y\plus{}1}$
$ ab\plus{}1\equal{}\frac{2}{x\minus{}y\plus{}1}$
$ a\minus{}1\equal{}\frac{2y\minus{}2}{x\minus{}y\plus{}1}$

$ x\equal{}\frac{a\plus{}1}{ab\plus{}1}$

$ \frac{a\minus{}1}{ab\plus{}1}\equal{}y\minus{}1$
$ y\equal{}\frac{a\minus{}1}{ab\plus{}1}\plus{}1$
$ y\equal{}\frac{ab\plus{}a}{ab\plus{}1}$[/hide]


Mr. Smith and Mr. Jones are two maths teachers, who meet up one day. Mr. Smith lives in a house with a number between $13$ and $1300$. He informs Mr. Jones of this fact, and challenges Mr. Jones to work out the number by asking closed questions.

Mr. Jones asks if the number is bigger than $500$. Mr. Smith answers, but he lies.

Mr. Jones asks if the number is a perfect square. Mr. Smith answers, but he lies again. (What a dishonest man...)

Mr. Jones asks if the number is a perfect cube. Mr. Smith answers and (feeling a little guilty) tells the truth for once.

Mr. Jones says he knows that the number is one of two possibilities, and if Mr. Smith just tells him whether the second digit is $1$, then he'll know the answer. Mr. Smith tells him and Mr. Jones says what he thinks the number is. He is, of course, wrong.

What is the number of Mr. Smith's house?

does it matter if 13 to 1300 inclusive or exclusive?

between means exclusive.  I have the solution!
[hide]So, first we list the cubes and squares between 13 and 1300, not too hard if you have them memorized (like me  :lol: ).  Let us assume that Mr. Smith says his number is bigger than 500, and that it is both a perfect square and a perfect cube.  Then the only possible answer is 729, which can not be right because it says he has two solutions.  So instead, we try and make it not a perfect square, but a perfect cube, and we are left with the solutions 512 and 1000.  Since either one of these must be wrong, the solution must be less than 500, a perfect cube, and a perfect square.  The only solution is 64.[/hide]

I figured that it doesn't.
Here is my solution
[hide]First, I made a truth table type thing, where the three conditions were answered true or false.  x>500, x=k^2, x=k^3.
After looking at possibilities, I came to the conclusion that x>500, x><k^2(doesn't equal), x=k^3 were the three answers.  Switching the answers to x<500, x=k^2, x=k^3, I got that x = k^6.  This yields with the 13<k^6<1300 restriction: k = 2 or 3.  Using the fact that k^6<500, you get that 3^6 is 729, which is obviously greater than 500.  The solution is that the house number is 2^6, or 32.[/hide]

Reply #3 is correct, Reply #4 has a minor miscalculation.