Here's something I found:

Outline a synthesis of [i]o[/i]-tolylmethyl alcohol from benzyl chloride.

What is o-tolylmethyl alcohol? is it this , if it is then I'll try the synthesis.

attempt #1:
1. H2SO4
2. Gatterman-Koch
3. Wolff-Kishner
4. Acidify
5. Thionyl Chloride!

Yes Lordvoldemort, that's the correct compound.

Jdlkem, if you were my student that answer would get a red cross on it. You have to explain your steps, not just throwing in a set of reagents.

[quote="jdlkem"]:lol:

1. H2SO4 - sulfate protecting group at the para position of the benzyl chloride
2. Gatterman-Koch - adding an aldehyde group to make a benzaldehyde.
3. Wolff-Kishner - reduce the carbonyl (basic, so protecting group remains unaffected)
4. Acidify - neutralize any base + kick off the sulfate
5. Thionyl Chloride! - change the hydroxyl group to a chloride group (with retention of stereochemistry)[/quote]

Um, I think he wanted the reverse lol (i.e., benzyl chloride -> o-tolylmethyl alcohol)...or maybe I don't get the last step :(

:huuh:

[quote="Jdlkem"]1. H2SO4 - sulfate protecting group at the para position of the benzyl chloride[/quote]

Sulfate is $ \ce{SO4^{2\minus{}}}$ while the group you are thinking is $ \ce{SO_{3}H}$. And you know that the ortho position also reacts, right?

[quote="Jdlkem"]2. Gatterman-Koch - adding an aldehyde group to make a benzaldehyde.[/quote]

What are the reagents?

[quote="Jdlkem"](basic, so protecting group remains unaffected)[/quote]

What about the chloromethyl group?

[quote="Jdlkem"]Thionyl Chloride! - change the hydroxyl group to a chloride group (with retention of stereochemistry)[/quote]

What does that have to do with the product I asked? And what stereochemistry?

[hide="Method?"]
1. Sulphonate to get SO3H at the para position
2. Gatterman's formylation to get CHO group at the ortho position
3. Add acid to eliminate the SO3H group
4. Wolff Kishner to get the required compound.  :maybe:
[/hide]

[hide="Comment"]:police:

"Sulphonate to get SO3H at the para position" should be "Sulphonate to get a mixture of SO3H at the ortho and para positions and then separate".

"Wolff Kishner to get the required compound."

Isn't the formyl group reduced to methylene? And what about the chloromethyl group?[/hide]

[hide="Continuing from 3rd step of the method described by madness"]
4. React with some zinc in presence of HCl and separate the compund from ZnCl2.
5. React with formaldehyde in presence of conc. NaOH and separate sodium formate form the mixture to get our product.
Cannizaro reaction.[/hide][/hide]

I am not following you. What is the product from your step 4?

compound after 4 is same as that of 3 but here, there is no chlorine atom in the molecule (replaced by H)

:whistling:  Do you really think that will work? You have a formyl group on the molecule, acidic conditions, a nucleophile...

By the way my synthesis uses only two steps. :juggle:

Or why dont we perform step 5 first and then step 4?

Or why don't we try to kill to rabits with a single blow? Why don't you use $ \ce{LiAlH4}$?

Here's my synthesis:

[hide]1) Benzyl chloride + Mg/ether -----> A

2) A + formaldehyde -----> o-tolylmethyl alcohol
[/hide]


Let $k=\prod_{i=1}^{n}\prod_{j=1}^{n} i^j$. Find the smallest $n$ such that $2007^{2007}$ divides $k$.

[hide]
the whole thing simplifies to $(n!)^{\frac{n(n+1)}{2}}$.  Since $2007^{2007}$ has 2007 223's n must be at least 223 because it is prime.  We have enough 3's to handle the $3^{4014}$ and $\frac{224*223}{2}$ is definitely greater than 2007.  So $n = 223$.
[/hide]


Can anyone think of an intuitive way of proving that the product of four consecutive integers is not the square of an integer?

x(x+1)(x+2)(x+3)=(x^2+3x+1)^2 - 1, and one less than a perfect square cannot be a perfect square.

I think that's pretty intuitive. ;)

Four consecutive positive integers, that is.


Hi everyone,

After a bit of Googling, I hope I've found the right forum to answer this question.  I apologize if something similar has been asked before or if this isn't exactly the right place to ask it.

My strong suit has never been math; I'm much more of a liberal arts type of guy.  However, I need to pass a basic aptitude test that contains a math section in which you are not allowed to use a calculator or scratch paper.  All the problems are uniform motion problems (d=rt) and all the problems have four multiple choice answers.  There is a time limit.

Here's an example: A plane has flown for 3 hours with a ground speed of
210 knots. How far did the plane travel?

Obviously I know how to solve this, but I'm interested in tips to decrease the time it would take me to solve these problems in my head, without the aid of a calculator or scratch paper.  Basically, if math gods could explain a good thought process to use to solve the problem.  On the test, the numbers get less nice as you progress.

[quote="DeltaNui"]Hi everyone,

After a bit of Googling, I hope I've found the right forum to answer this question.[/quote]
So you're basically concerned with your arithmetic skills

- "The Arithmetic Game" at [url]http://zetamac.com/arithmetic/[/url] has been posted before I think
- Memorize $1^{2}$ through $19^{2}$ at least
- Multiplication of $11$: $(10a+b)(11) = 100a+10(a+b)+b$
- Recognize multiples of $2$, $3$, $4$, $5$, $8$, $9$, $11$ at least (for multiple choice)
- Try some kind of approximating activity


Determine all triples of positive integers $(a,b,c)$ such that $a \leq b \leq c$ and $a +b + c + ab+ bc +ca = abc +1$.

it is easy to check that $a>1$ , or else when $a=1$ it gives us $b+c=0$ which is impossible

now setting $2\leq a\leq b\leq c$
regroup we have

$a=\frac{1-b-c-bc}{1+b+c-bc}$ $\implies a+1=\frac{2(1-bc)}{1+b+c-bc}=\frac{2(bc-1)}{bc-b-c-1}$

since $a+1\geq 3$ $\implies \frac{2(bc-1)}{bc-b-c-1}\geq 3$........(*)

(i) if $(b-1)(c-1)=1$ or when $b=c=2$ , $a=-7$ impossible
(ii) if $(b-1)(c-1)=2$ or when $b=2,c=3$ it is also impossible.
(iii) if $(b-1)(c-1)>2$ then from (*) we have

$\frac{2(1-bc)}{1+b+c-bc}=\frac{2(bc-1)}{bc-b-c-1}\geq 3$ $\implies (b-3)(c-3)\leq 10$

Case work :
a) when $(b-3)(c-3)=10$ so $a=2,b=5,c=8$ or $a=3,b=4,c=13$
b) when $0\leq (b-3)(c-3)\leq 9$ by case work there is also no solution .

So out answer is only $(a,b,c)=(2,5,8)$ and $(3,4,13)$

* is there a faster method ? I believe there is other than case work i done ....

AS per the official solution, there are three triplets and as far as your answer is concerned --- the one containing 13 should be [hide] $(\ 2 , 4, 13 )\$ [/hide]

Since $2\leq a\leq b\leq c,$ we have $\frac{a}{b}\leq 1,\ \frac{a}{c}\leq 1$ and $\frac{2}{c}\leq 1.$

Thus $a+b+c+ab+bc+ca=abc+1\ \cdots [*]\Longleftrightarrow abc=a+b+c+ab+bc+ca-1$

$\Longleftrightarrow a=\frac{a}{b}\cdot \frac{1}{c}+\frac{1}{c}+\frac{1}{b}+\frac{a}{c}+1+\frac{a}{b}\leq \frac{1}{b}\cdot 1+\frac{1}{c}+\frac{1}{b}+1+1+1$

$=\frac{2}{b}+\frac{1}{c}+3\leq 1+\frac{1}{2}+3=4.5$, yielding $a=2,3,4.$

Case 1: $a=2$

From $[*],$ we have $bc=3b+3c+1\Longleftrightarrow (b-3)(c-3)=10.$ Since $-1\leq b-3\leq c-3,$ we have $(b-3,c-3)=(1,10),\ (2,5),$ yielding $(b,c)=(4,13),\ (5,8).$

Similarly,

Case 2: $a=3\ ;\ \ 2bc-4b-4c=2\Longleftrightarrow (b-2)(c-2)=5,\ 0\leq b-2\leq c-2,$ yielding $(b,c)=(3,7).$

Case 3: $a=4\ ;\ \ 3bc-5b-5c=3\Longleftrightarrow (3b-5)(3c-5)=34,\ 1\leq b-2\leq c-2,$ yielding $(b,c)=(2,13),$
which isn't suitable for $a\leq b\leq c.$

Therefore desired answer is $(a,b,c)=(2,4,13),\ (2,5,8 ),\ (3,3,7).$

I just notice that when we change  the equation into

$(a+1)(b+1)(c+1)=2(abc+1)$

the question can be solved using same method as here

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75213[/url]

[quote=aayush-srivastava]Hey(sorry for reviving this thread)
But we can also solve this by substituting $a-1=p,b-1=q,c-1=r$.
Only 2 cases arise and we are done immediately..[/quote]
I know,you are just copying the official solution,or whatsoever :cool: .


Let $ a,b,c >0$ and $ a\plus{}b\plus{}c\equal{}3.$Prove that
$ \frac{a^2 \plus{}1\plus{}2b^2}{a\plus{}1}\plus{}\frac{b^2\plus{}1\plus{}2c^2}{b\plus{}1}\plus{}\frac{c^2\plus{}1\plus{}2a^2}{c\plus{}1} \geq 2(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}).$

[hide="original inequality"] $ \frac{a^2 \plus{}1\plus{}2b^2}{a\plus{}1}\plus{}\frac{b^2\plus{}1\plus{}2c^2}{b\plus{}1}\plus{}\frac{c^2\plus{}1\plus{}2a^2}{c\plus{}1} \geq \sqrt{2}(\sqrt{a^2\plus{}b^2}\plus{}\sqrt{b^2\plus{}c^2}\plus{}\sqrt{c^2\plus{}a^2})$[/hide]

[quote="4865550150"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3.$Prove that
$ \frac {a^2 \plus{} 1 \plus{} 2b^2}{a \plus{} 1} \plus{} \frac {b^2 \plus{} 1 \plus{} 2c^2}{b \plus{} 1} \plus{} \frac {c^2 \plus{} 1 \plus{} 2a^2}{c \plus{} 1} \geq 2(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}).$

[hide="original inequality"] $ \frac {a^2 \plus{} 1 \plus{} 2b^2}{a \plus{} 1} \plus{} \frac {b^2 \plus{} 1 \plus{} 2c^2}{b \plus{} 1} \plus{} \frac {c^2 \plus{} 1 \plus{} 2a^2}{c \plus{} 1} \geq \sqrt {2}(\sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2})$[/hide][/quote]
Actualy your inequality is weaker than the original problem and we dont need the condition too :wink:
[hide="solution"]$ a^2\plus{}b^2 \ge 2ab$,and $ b^2\plus{}1 \ge 2b$,so $ a^2\plus{}1\plus{}2b^2 \ge 2ab\plus{}2b\equal{}2b(a\plus{}1)$
So th $ LHS \ge 2(a\plus{}b\plus{}c)$
So we have to prove that:
$ a\plus{}b\plus{}c \ge \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$ ,which is trivial[/hide]


Aha - I see you chose [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php#gettingstarted]the second option[/url] for Getting Started.  Had you chose the first option, you would have [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints]read this[/url].


How can someone, who don't have any idea  how Rubic cube is solved , learn to solve it?

practice. it took me a total of about 10 hours to learn it. i just took a sheet of paper with algs on it, did the same alg, undid it, did it again, until i could do it without the paper. then i moved on to the next alg

As ive been saying, watch Dan Brown's video tutorial on Youtube. its da bomb.

....

First buy a cube. Then read the booklet that comes with it.
Method 2:
Get a cube. Then go search it on the internet for a beginner's method.
Youtube videos suck.


Define the operation $ \star$ by $ a \star b \equal{} (a \plus{} b)b$. What is
$ (3\star5) \minus{} (5\star3)$?


geometry
trigonometry
angle bisector
trig identities
Law of Cosines

[quote="xenon_nightmare152"]Let ABC be the right triangle with vertices of A(0, 2), B(1, 0), and C(0, 0). If D is the point on AB such that the segment CD bisects angle C, what is the length of CD?[/quote]

[hide]AB lies on line y=-2x+2
CD lies on line x=y
The two lines intersect at point (2/3,2/3)

The distance from the origin to tat point is $\frac{2\sqrt{2}}{3}$[/hide]

I understand jli's solution, but how would you find the equation of an angle bisector in general?

The most useful Theorem to use would be the angle bisecting formula, which states that for a triangle $\triangle ABC$, if $\angle B$ is bisected, call the point where the angle bisector crosses $\overline{AC}$ $D$.  Then, you have $\frac{AD}{DC}=\frac{AB}{BC}$.  This should do the trick for a general triangle.

[quote="xenon_nightmare152"]Let ABC be the right triangle with vertices of A(0, 2), B(1, 0), and C(0, 0). If D is the point on AB such that the segment CD bisects angle C, what is the length of CD?[/quote]

[hide]We can do this, by using the angle bisector theorem and law of cosines.

By Angle Bisector we get $AD=\frac{2\sqrt{5}}{3}$.

NOw by law of cosines we get

$CD^{2}=\frac{20}{9}+4-2*\frac{2\sqrt{5}}{3}*2*\frac{2}{\sqrt{5}}=\frac{8}{9}$.

Hence,

$CD=\frac{2\sqrt{2}}{3}$[/hide]

I agree with you, b-flat.
[hide="Using the theorem..."]From the Triangle-Angle-Bisector Theorem,
$\frac{AC}{AD}=\frac{BC}{BD}$

$\frac{2}{AD}= \frac{1}{BD}$

$2BD = AD$

Since $AB=AD+DB=\sqrt{5}$,

It's easy to see that $AD=\frac{2\sqrt{5}}{3}$, and $DB = \frac{\sqrt{5}}{3}$.

From Stewart's Theorem,
$(DC)^{2}(AB)+(AD)(DB)(AB) = (AC)^{2}(DB)+(CB)^{2}(AD)$

Applying this to our triangle, and substituting,

$(DC)^{2}= \frac{8}{5}$

$(DC)^{2}= \frac{40}{25}$

$(DC)^{2}= \frac{2\sqrt{10}}{5}$[/hide]

[quote="jli"][quote="vishalarul"]I agree with you, b-flat.
[hide="Using the theorem..."]From the Triangle-Angle-Bisector Theorem,
$\frac{AD}{DC}=\frac{AB}{BC}$

$AB=\sqrt{1^{2}+2^{2}}= \sqrt{5}$
$BC=1$
$AC=AD+DC=2$

Putting this all in our equation,
$\frac{AD}{DC}=\sqrt{5}$
$AD = DC\sqrt{5}$
Substituting,
$DC\sqrt{5}+DC = 2$
$DC = \frac{2}{\sqrt{5}+1}= \sqrt{5}-1$.
That's our answer. [/hide][/quote]

i think D is on segment AB[/quote]

Yes, it is on segment AB, aka, the hypotenuse.

Oh, I see my mistake.  :blush: I'll edit it now.


calculus
derivative
function
analytic geometry
graphing lines
slope
integration

I want to know, how I can show following statement:

If f : R^n -> R is a sobolev function such that its weak derivative
DR has continuous representative, then DR is classical derivative
of f in R^n.

Thank you very much.

This belongs in advanced topics, and you should start a new thread for a new question.

DR being the weak derivative means $\langle DR, \varphi\rangle = \langle f, \varphi'\rangle$ for all $\varphi \in C_0^\infty$, right?

Construct a sequence $\varphi^E_n(x)$ that is the characteristic function of $E$ and then just continued down linearly to 0 with slope $-n$. Then you can use continuity of DR and two sets $E_1, E_2$ that differ on an appropriately shaped set to show that $f$ is absolutely continuous, and then you just have to take the derivative of $f$ and show that it is equal to DR up to a set of measure 0.

I hope that's the correct approach. Also, what is the definition of Sobolev function? Perhaps I should have used that somehow?

[quote="Xevarion"]Also, what is the definition of Sobolev function?[/quote]
The original poster didn't supply enough detail. Most typically, a Sobolev space is a set of functions such that the original function and a certain number of its (weak, of course) derivatives belong to $L^p$ (usually $L^2$) of a certain domain. The Soblolev norm is obtained by adding the norms of the function and of this certain number of derivatives (or perhaps adding the squares of the norms and taking the square root.) A Sobolev space is a Banach (usually Hilbert) space.

There are Sobelev embedding theorems, which say that for a certain number of derivatives in $L^p$ in a certain dimension, the function can be represented as a continuous function. Such theorems are important in the study of PDE's, especially elliptic PDE's.

That's really interesting! Thanks, Professor Merryfield.

I realized that the test function you need for this problem is a little more complicated than teh one I suggested. Well I mean, the one I suggested isn't continuously differentiable, but you just approximate it by good functions (easier to approximate its derivative and show that the integrals of those functions approximate the original function). So then you can show that it is legitimate to use that function.

Then you have your test functions that in the limit basically give you the Fundamental Theorem of Calculus (you just have to shape the set $E$ appropriately I guess).


Let $AB$ be a chord of circle $O$. $C,D$ are points on the circle on the same side of the chord, and $AC=CB$. Is it always true that $AC+CB>AD+DB$?

Use sine Law like $\frac{CB}{sin{CAB}=2R}$ etc so it simplifies to finding if it's possible to have:
$sin(x)+sin(y)\geq 2sin(z)$ when $x+y=2z$
So let wolog $x\geq y$ so $x=z+p$, $y=z-p$ with $p>0$ then:
$LS=sin(z+p)+sin(z-p)$
$=2sin(z)cos(p)\leq 2sinz$ so only case when $sin(x)+sin(y)\geq 2sin(z)$ is when we got $cos(p)=1$ so $p\geq 90^{o}$ but $z\leq 90^{o}$ so have $y\leq 0$ impossible.

So result is true.

Let $BD<AD$ take $P$ on AD such that $AP=BD$. Triangles $APC$ and $BDC$ are congruent so $CP=CD$. Let $M$ be the midpoint of $PD$. $AM=AP+\frac{AD-AP}{2}=\frac{AD+BD}{2}$.
And obviosly $AC>AM$.

[color=darkblue][u][b]Very nice the Delta's proof ![/b][/u] It is the [size=117][b][u]top-proof[/u][/b][/size]. Here is another proofs, mine.[/color]
[quote="Radio"] [color=darkred]Given is an $A$ - isosceles triangle $ABC$ inscribed in the circle $w=C(O,R)$. Ascertain the position of the point $M\in w$ for which the line $BC$ doesn't separate the points $A$, $M$ and the sum $(MB+MC)$ is [b]maximum[/b].[/color] [/quote]
[color=darkblue][u][b]The first proof.[/b][/u] Denote $BC=a$ ([u]constant[/u]) and the [u]fixed[/u] point $D\in w\ ,\ O\in AD$.
For the [u]mobile[/u] point $M$ denote $m(\widehat{BMC})=2\phi$ ([u]constant[/u]) and for
the [u]variable[/u] triangle $BMC$ denote the its semiperimeter $p$ and the its incircle $C(I,r)$.
From the well-known relations $\{\begin{array}{c}I\in C(D,k)\mathrm{\--\ a\ fixed\ circle}\\\ OI^{2}=R^{2}-2Rr\\\ r=(p-a)\tan \phi\end{array}$ we get :
$(MB+MC)$- [b]maximum[/b] $\Longleftrightarrow$ $r$- [b]maximum[/b] $\Longleftrightarrow$ $OI$- [b]minimum[/b] $\Longleftrightarrow$
the distance from the point $O$ to the circle $C(D,k)$ is [b]minimum[/b] $\Longleftrightarrow$
$I\in OD$ $\Longleftrightarrow$ $M: =A$ $\Longleftrightarrow$ $MB=MC$.

[u][b]The second proof[/b][/u] ([u]very short ![/u]). I"ll use the same notations as in the first proof.
From the [u]Ptolemeu's theorem[/u] applied to the cyclical convex quadrilateral $MBDC$
we get $MD=\frac{k}{a}\cdot (MB+MC)$. Thus, $(MB+MC)$ is [b]maximum[/b] $\Longleftrightarrow$
$MD$ is [b]maximum[/b] $\Longleftrightarrow$ $M: =A$ $\Longleftrightarrow$ $MB=MC$.[/color]


1) Prove that $1+3+5+7+\cdots+(2n-1)=n^{2}$.

2) Show that $n$ lines in a plane can divide it into at most $\frac{n^{2}+n+2}{2}$ regions.

3) Prove that for all positive integers $n$, $n^{3}+(n+1)^{3}+(n+2)^{3}$ is divisible by $9$.

4) Suppose that $x+\frac{1}{x}$ is an integer.  Prove that $x^{n}+\frac{1}{x^{n}}$ is also an integer for all natural numbers $n$.

5) A bank has an unlimited supply of $3$-peso and $5$-peso bills.  Prove that it can pay its customers any integer number of pesos greater than $7$.

6) Prove that a square can be dissected into $n$ squares for $n\geq 6$.

[quote="E^(pi*i)=-1"]4) Suppose that $x+\frac{1}{x}$ is an integer.  Prove that $x^{n}+\frac{1}{x^{n}}$ is also an integer for all natural numbers $n$.[/quote]
[hide]
Let $P_{n}$ be the claim that $f(x;n) = x^{n}+x^{-n}$ is an integer.  Then given that $P_{1}$ is true,

$f(x;1)^{2}= x^{2}+2x(x^{-1}) x^{-2}= f(x;2)+2$,

from which it follows that $P_{2}$ is true.  Therefore, there exists an integer $m \ge 2$ such that $P_{1}, \ldots, P_{m}$ is true.  Now consider

$f(x;m)f(x;1) = (x^{m}+x^{-m})(x+x^{-1}) = f(x;m+1)+f(x;m-1)$.

Rearranging, we find

$f(x;m+1) = f(x;m)f(x;1)-f(x;m-1)$.

Since each $f$ on the right-hand side is an integer, then the left-hand side must also be an integer.  Hence the truth of $P_{1}, \ldots, P_{m}$ implies the truth of $P_{m+1}$.  By the induction hypothesis, it follows that $P_{n}$ is true for all positive integers $n$.

Furthermore, it is obvious that $f(x;0) = 2$ (we cannot have $x = 0$), so $P_{0}$ is true, and since $f(x;-n) = f(x;n)$, it follows that if $f(x;1)$ is an integer, then $f(x;n)$ is true not only for positive integers $n$, but for all integers $n$.[/hide]

[quote="E^(pi*i)=-1"]Yay!  Someone posted in the Maryland forum!  Nice solution, too.  :)[/quote]

Too bad they're from California  :wink: .

1
For n=1 this is obviously true.if this is true for any n=k then,
1+3+5+...+2k-1=k^2.
Adding (2k+1) on both sides of this expression we get,
1+3+5+...+(2k-1)+(2k+1)=k^2+2k+1=(k+1)^2.
This proves the claim.

2
the claim is obviously true for n=1 as 1 line divides a plane into 2 regions.
Now, if k lines divides the plane into at most (k^2+k+2)/2 regions the (k+1)th line can divide all k lines atmost creating (k+1) new regions as for intersecting a line it creates a new region and it creates 1 more region before intersecting any lines.
So,(k+1) lines will divide the plane into
(k.k+k+2)/2+(k+1)=(k.k+k+2+2k+2)/2=((k.k+2k+1)+(k+1)+2))/2
=((k+1)^2+(k+1)+2))/2 regions.
This completes the proof.

4
The base case n=8 is true since it can be paid with a 3 and a 5 pesso coin.
Now if k pesso is payable,for any k>7 .If k is not a multiple of 3,then k+1 pesso can be paid by taking away a 5 pesso coin and adding two 3 pesso coin.If k pesso is paid with only 3 that is k=3m then (k+1) pesso is payable by taking away 3 3pesso coins and inputting 2 more 5 pesso coins this is possible since there are 3 3pesso coins in the least multiple of 3 greater than 7.
This completes the proof.


geometry
function
integration
calculus
calculus computations

Find the area of the region that is bounded by two functions $ y\equal{}x^2$ and $ y\equal{}x^3$ and by a line $ y\equal{}1$.

I got $ \int^1_0 |x^3\minus{}x^2| dx \equal{} \frac{1}{3}\minus{}\frac{1}{4}\equal{}\frac{1}{12}$

is it right?

The answer is right, but for $ 0\leq x\leq 1$, we have $ x^3\leq x^2\leq 1$ so you don't need to use the absolute value.


Every time I click on the "view posts since last visit" link, I find exactly 300 new posts. Since I don't believe that there's a secret conspiracy on MathLinks for making the number of new posts a multiple of 100, it's quite obvious that there's a limit to the automatic search of new posts (and that MathLinks is a very frequented forum  :D ). How can I see [b]all[/b] new posts? Can I select the forums from which I will read all new posts (e.g., I'm more interested in Calculus than in Inequalities and I want to read the more interesting topics as first).

I didn't find any option about this in "Preferences" so I wonder if it's possible.

Thanks
GM

Given the large number of users, forums and posts per day, we cannot "remember" more than 300 new posts per person. So it's nothing we can help with right now, sorry.

[quote="Xevarion"]Could you filter it so that for each user, you only "remember" a new post if it passes the filter? (The filter is user-defined)[/quote]It would mean changing the way the current system works, and I don't have the time for that right now. I will do it at a given time in the future (ie add it on the wish list).

[quote="GioMott"]Ok thank you very much. Another question: if in the menu Profile-->Preferences-->Reading topics I choose "Yes" for the option "Keep messages unread" the messages I haven't read will still be marked unread at each successive visit?

Bye!
GM[/quote]Yes, but they are subject to the same 300 posts limit. Considering that right now on average we have 600 new posts every day, if you come back in more than 12 hours, those will be gone.


function
real analysis
real analysis unsolved

Prove that every continuous function on a compact interval is a uniform limit of step functions*.

*We define a step function to be a function that is piecewise constant, $f(x) = \sum_{j=1}^{n}c_{j}\chi_{[a_{j},b_{j})}$ where $[a_{j},b_{j})$ are disjoint intervals. ($\chi$ denotes the characteristic function of the interval)

Just follow exactly the usual proof that a continuous function is Riemann integrable


[color=darkred]Given a quadrilateral [b]ABCD[/b].[b]AD[/b] intersects [b]BC[/b] in point [b]W.X,Y[/b] are the midpoints of diameters[b] AC,BD[/b].
Let show area of triangle [b]XYZ[/b] with [b]S1[/b] and show area of quadrilateral[b] ABCD [/b]with [b]S2[/b].
Find [b][u]S1:S2[/u][/b].[/color] :wow:

Let $ABCD$ be a quadrilateral with $AD$ not paralell to $BC$ and $W$ a exterior point of the quadrilateral that belong to the intersection of extension of $AD$ and $BC$, forming the triangle $ABW$ that do not depends of how $AB$ and $DC$ are oblique.

Once we deduce that $A',C',D',X',Y'$ are the ortogonal projection  point(altitude) of the points $A,C,D,X,Y$ in that order,and $A_\Delta...$, $A_\square...$ the area of a respective triangle and a respective quadrilateral,we first have that:
$\displaystyle\frac{AA'}{AW}=\frac{DD'}{DW}\ -> DD'=\frac{AA'.DW}{AW}$ and:

$\displaystyle A_\Delta_{ABW}=A_\Delta_{ABC} + A_\Delta_{ACD} + A_\Delta_{DCW}$ thus:

$\frac{BC.AA'}{2} + \frac{CW.AA'.DW}{2.AW} + \frac{AD.CC'}{2}= \frac{(BC + CW).AA}{2}'$ $-> CC'=\frac{CW.AA'}{AW}$

$A_\square_{ABCD}= A_\Delta_{ABW} - A_\Delta_{DCW}=\frac{BW.AA'}{2} - \frac{CW.DW.AA'}{2.AW}=\frac{AA'}{2}(\frac{AW.BW - CW.DW}{AW}$

$A_\Delta_{XYW}=A_\Delta_{ABW} - A_\Delta_{YBW} - A_\Delta_{AXW} - A_\square_{ABYX}$

Since we have by similar triangle  that $XX'=\frac{CC'}{2}$ and $YY=\frac{DD'}{2}$,we got:

$A_\Delta_{YBW}=\frac{BW.YY'}{2}=\frac{BW.DW.AA'}{4.AW}$

$A_\Delta_{AXW}=\frac{AW.XX'}{2}=\frac{AW.CW.AA'}{4.AW}$

$A_\square_{ABYX}=A_\Delta_{ABX} + A_\Delta_{BYX}$

->$A_\Delta_{ABX}=\frac{A_\Delta_{ABC}}{2}=\frac{BC.AA'}{2}$

->$A_\Delta_{BYX}=\frac{A_\Delta_{BYD}}{2}=\frac{A_\Delta_{ABW} - A_\Delta_{DBW} - A_\Delta_{AXD} - A_\Delta_{ABX}}{2}$ in consequence:

$A_\square_{ABYX}=\frac{AA'}{8}.(\frac{BW.AW + CW.AW -2.BW.DW - AD.CW + 2.AW.BC}{AW})$ calculing:

$A_\Delta_{XYW}=\frac{AA'}{8}(\frac{3.AW.BW - 3.AW.CW + AD.CW - 2.AW.BC}{AW})$

$=\frac{AA'}{8}.(\frac{AW.BW - CW.DW}{AW})$ thus:

S1:S2=$\frac{\frac{AA'}{8}.(\frac{AW.BW - CW.DW}{AW})}{\frac{AA'}{2}.(\frac{AW.BW - CW.DW}{AW})}=\frac{1}{4}$

Thank you so much for solve that.
and you (amir2) how did you know that S1:s2=1:4?

That problem was in Yugoslavia (Serbia and Montenegro) national olympiad many years ago, i think 1982, so i wonder where did you read it?

I haave read it long time ago in one of my geometry books but i think it is not important. ;)

Well maybe it's not, i asked that because notation is totaly same (ok quadriliteral is $ABCD$ everywhere but letters like $X$, $Y$ and $Z$ was not so popular in 1982 in my country, that is only problem in that decade with that letters, if you understand what i am talking,) and that's why i am asking  :)  :)  :)


vector
analytic geometry
combinatorics unsolved
combinatorics

Let $ M$ be a set with $ m$ elements (where $ m > 1$). Let $ A_1,A_2,...,A_n$ be subsets of $ M$ such that the numbers of common elements of any two of them equal $ r \geq1$. Prove that $ m \geq n$.

(This is the non-uniform version of Fisher's inequality. The original, uniform version assumes that all the sets have the same size. The proofs are actually pretty similar).

Represent each set $ A_i$ by a vector $ v_i \in \mathbb{R}^m$ by having the $ k$th coordinate of $ v_i$ be 1 if $ k \in A_i$ and 0 otherwise. For instance if $ M \equal{} \{1,2,3,4,5,6,7\}$ and $ A_1 \equal{} \{2,3,6,7\}$ then $ v_1 \equal{} (0,1,1,0,0,1,1)$. We will show that the vectors $ v_1,\ldots,v_n$ are linearly independent which implies $ n \leq m$.

Let $ |A_i| \equal{} N_i$ be the number of elements in $ A_i$. The inner product $ \langle v_i, v_j \rangle$ can be calculated as follows:

$ \langle v_i, v_j \rangle \equal{} \begin{cases} N_i & \mbox{if}\,\, i \equal{} j \\
r & \mbox{if}\,\, i \neq j \end{cases}$

Suppose $ a_1, \ldots, a_n$ are real numbers such that

$ a_1 v_1 \plus{} a_2 v_2 \plus{} \cdots \plus{} a_n v_n \equal{} 0$.

Then we have:

$ 0 \equal{} \langle \sum_{i \equal{} 0}^n a_i v_i , \sum_{j \equal{} 0}^n a_j v_j \rangle \equal{} \sum_{i \equal{} 0}^n a_i^2 N_i \plus{} \sum_{i \neq j} r a_i a_j \equal{}$

$ \equal{} r \left(\sum_{i \equal{} 0}^n a_i\right)^2 \plus{} \sum_{i \equal{} 0}^n a_i^2 (N_i \minus{} r) \equal{} A \plus{} B$.

Notice that we must have $ N_i \geq r$ for all $ i$, so $ B$ is nonnegative, and $ A$ is obviously nonnegative as well. Thus the only way $ A \plus{} B \equal{} 0$ is if $ A \equal{} 0$ which means $ a_i \equal{} 0$ for all $ i \in \{1, \ldots, n\}$.

This proves that the vectors $ v_1, \ldots, v_n$ are linearly independent so their number cannot exceed the dimension $ m$.


trigonometry
function
calculus
derivative
calculus computations

I want to solve the problem the following equation

F(x) = x*(1-sin(x/x))^5/3

Can you help me?

F(x) = x*(1-sin(x/x))^5/3
1, Are you sure that is right? x/x=1 and should be replaced directly by 1.
2, It's not even an equation. It's a FUNCTION.

Yes you have right. It's a function and I did a mistake : the correct one is the following:

F(x) = (x*(1-sin(x)/x)^5/3) - A

So what are we to do? Solve the equation $x\left(1-\frac{\sin x}{x}\right)^{5/3}-A=0$? It is unsolvable analytically and fairly easy to solve numerically by the bisection or Newton method (I would prefer bisection because the derivative of this particular function may vanish and, unless you are sufficiently close to the root from the beginning, the Newton method may easily go astray). You have finitely many roots for all $A$ (probably, at most 3, but I'am not completely sure about it for small values of $A$). You can get some crude estimates for the roots if you want but they are fairly simple too. Not much else can be said.

Or have I misunderstood you completely and we are to do something else? Then, please, tell exactly what.  :?


I;m currently a high school senior and I took the SAT for the first time today. I'm a little disheartened because I know I was capable of doing much better. I made silly mistakes on math, and with a more-than-likely harsh curve, I'll probably end up with a 750 if I'm lucky.

My writing and math sections are typically very strong and only get thrown off due to silly mistakes. CR is where I flop. I get anywhere from 680-750, but I'm certain I can do a lot better.

The CR section is incredibly difficult for me. I can usually get the vocab and easy/medium passages with no problem, but the section with the harder vocab and ridiculous reading passage is where I flop. Sometimes I can narrow the hard ones down to 2/3 choices but more often than not, I THINK I've narrowed it down to two choices when in fact I've eliminated the correct answer as an option.

I've done like EVERY official collegeboard practice test available, including the free official practice ones they release online every year. Are there any other ways I can practice the passage-based reading questions?

You're pissed about a 750?  You really need to relax.

As far as critical reading, the best things you can do are to keep doing exams and keep reading in general.  Reading will expose you to some vocab that you may not ordinarily see.  The more exams you do, the more used you should become to College Board's style of questions, and in particular, how they try to trick you.  You should go back and look at the tests you have taken and figure out why you are getting questions wrong.  Chances are, they are mistakes that you can improve on, if not fix.

[quote="JRav"]You're pissed about a 750?  You really need to relax.[/quote]

honestly if you're applying to one of the top schools, being pissed about losing fifty easy points is a legitimate response

Contrary to possible opinion, we live in a very different time and age than, I have to say, a few years ago, when it comes to college applications.

Popular belief is that a few GPA tweaks here and few SAT points and a few more APs and some lab work and the list goes on and on...will guarantee entrance into selective schools.  The number of schools that people are applying to (of which they have no legit reason to apply) is pretty ridiculous.

Sometimes, getting good at CR is just a matter of maturity, or maybe a different mindset.  Imo, it's like ETS and the Collegeboard want to brainwash you, even though they know you can probably perfectly read any article published anywhere(especially with your current score), with exception of seriously technical, specialized articles.

I know people with scores lower than 750 on all 3 sections who gained entrance into top schools, Ivy and non-Ivy.  The admissions committees really do know how to look beyond the numbers.

Yes.

But for those with obsessive parents who want to squeeze every last ten points out of their child...

They need to stop being helicopter parents.  It doesn't help.

To do well on CR you need to read in areas that you are not used to reading, and insure that you understand-remember it is easy to do something and not benefit, especially if you do it in a quick, undirected way. At the lowest level you should look up unfamiliar words. The other basic thing is that you should read short pieces and try to summarize each paragraph and eventually the whole article.

Since you are  a senior you probably have better things to do with your time. But better late than never.

Los is correct.  I'll admit that I did not do as well as I hoped the first time I took it.  However, the second time I took it, I got my score up 70 points just by doing practice exams.  As I said before, it really does help.

well, personal experience.

although i have to admit that statement may not be applicable everywhere, it is definitely at least a trend for those whose parents push them to work really hard.

Some comments:

# of colleges to which applicants are applying are up
the number of APs taken has also risen.

i may be biased though, when it comes to this.

[quote="zapi2007"]well, personal experience.[/quote]

so you applied to colleges several years ago as well?

[quote="blahblahblah"][quote="zapi2007"]Contrary to possible opinion, we live in a very different time and age than, I have to say, a few years ago, when it comes to college applications.[/quote]

And how, exactly, do you know this?[/quote]

[quote="blahblahblah"] so you applied to colleges several years ago as well?[/quote]

Probably, it would have been much better had you asked these questions via a PM. It unnecessarily clutters the thread!

[quote="Blkmage"] The CR section is incredibly difficult for me. I can usually get the vocab and easy/medium passages with no problem, but the section with the harder vocab and ridiculous reading passage is where I flop. Sometimes I can narrow the hard ones down to 2/3 choices but more often than not, I THINK I've narrowed it down to two choices when in fact I've eliminated the correct answer as an option.[/quote]

I don't know whether you're a native or a non-native speaker of English, but if you happen to be a native speaker of English-- I would be quite surprised, because the CR section is, at most times, relatively more difficult to ace for non-native speakers of English. Personally, I observe that the Critical Reading (CR) section of the SAT is somewhat analogous to the verbal section of the general GRE. In order to excel at the sentence completion and reading comprehension components of the (CR) section you must develop or rather inculcate a habit of reading a lot.

Particularly, you must inure yourself to reading "dull, uninteresting and banal" articles. By doing this, you would be able to read those "banal, uninteresting" reading comprehension passages with better comprehension. Also, whenever you attempt an RC question in some mock test, just check as to where you went wrong and why a particular option is correct.

For acing the "harder" or tougher sentence completion questions you should have a pretty decent vocabulary. You need to memorize some word lists, but you can develop an extensive vocabulary only through reading. For answering tough sentence completion questions, you need to have knowledge of finer shades of meaning of common place words and you need to know more words. By answering the sentence completion questions quickly you can save some time for those dreadful RCs.

I myself am a non-native speaker of English, but by doing all those which I had suggested to you-- I had managed to score in the 99% percentile range on the GRE verbal section. I hope my advice or suggestion is helpful.

At the end of the day, SAT scores don't matter much and they shouldn't be accorded extremely high importance. But, getting a perfect SAT or GRE score is really nice-- it boosts your confidence a lot!

[quote="Arachnotron"]If it makes you feel any better, I got a 2350 on my SAT and didn't get into any ivies, whereas my friends with 1900-2150 did...[/quote]

To the [b] OP [/b], the above instance clearly drives home the point that SAT scores shouldn't be accorded very high importance or weight age.

Well i think for somethings they are given a lot of weight.-since the other measures of "success" in high school are pretty arbitrary, and there are to my knowledge very few things in the way of national competitions that you could win..say, if you wanted to study history at yale. Mostly, you just need a decent gpa, good standing as a student, and good sat scores. maybe play the tuba.

For math, bio, cs,physics, etc etc etc, our system is more refined-more tests, competitions, etc.

So if the [b]OP[/b] was applying to a math school, and had high sat scores, and was from a state like MA,CT,NY,CA where there is always a large pool of similar candidates, you probally aren't strong enough to get into a good math school. If you are from a traditionally weak math state (south or maybe midwest), you'd surely get in.

Los

I believe a CR score above 650 is almost same as that above 700---So, don't be nervous---
No matter which college you enter, it is just a beginning of your career---Hard work is ahead of you---
And that much is true, for our  non-native English speakers, CR is much more difficult---
But I like CR Section best---the passages are interesting and challenging---


calculus
integration
LaTeX
logarithms
calculus computations

I haven't integrated in years and need some help

int(1 - t/(100-x-t))/ (1-t/(100-x)) from t=0..100-x

thanks

A pretty version:
$ \int_0^{100-x}\frac{1-\frac{t}{100-x-t}}{1-\frac{t}{100-x}}\,dt$
(Quote my post to see how the $ \text{\LaTeX}$ works)

I'd start with some algebraic simplification:$ \frac{1-\frac{t}{100-x-t}}{1-\frac{t}{100-x}}= \frac{100-x}{100-x-t}\cdot\frac{100-x-2t}{100-x-t}= \frac{(100-x)(100-x-t)}{(100-x-t)^2}$
Let $ L=100-x$ for convenience- it's just a constant.

Substituting $ s=100-x-t=L-t$,
$ \int_0^{L-c}\frac{L(L-2t)}{(L-t)^2}\,dt= -L\int_L^c\frac{2s-L}{s^2}\,ds =L\int_c^L\frac 2s-\frac{L}{s^2}\,ds =L\left[2\ln s+\frac 2s\right]_c^L$
$ =2L\ln L+2-2L\ln c-\frac{2L}{c}$
As $ c\to 0$, this diverges to $ -\infty$. As atomicwedgie said, there's a singularity at the upper limit which makes the integral diverge.

[quote="atomicwedgie"]Is this a life contingencies integral?  If so, it's not set up correctly.  There is a singularity at t = 100-x.  If indeed it is correct as written, this integral is divergent.[/quote]

I met with my teacher, and it was set up wrong.  Thanks for the help.


I just looked up the Fibonacci Sequence on the internet, and it gave me this complicated formula, can someone explain it?

http://www.geocities.com/cyd_conner/binetsFormula.gif

and

http://www.geocities.com/cyd_conner/successorFormula.gif

thanx :D

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[color=darkred][size=75][Note:  This thread was split by moderator rcv from parent thread due to advanced content.][/size][/color]
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[quote="Matthew"]

By the way, the second formula you gave is flawed (try x=1 for example).[/quote]

sorry, didn't really get the formula, so couldn't see whether or not it's flawed.... but do you really have to memorize the whole formula? isn't it kinda long?

I think finding about the formula is too advanced..

 :?

[quote="LuCy4EvA"]I just want to ask a question: what's the formula for the nth number in the Fibonacchi Sequence.

and middle school people that's in math team and math count should know basic information about Fibonacchi, but we never learned how to find the nth number[/quote]

Lucy - Here is the basic idea and the method  -  how the formula  for this,  comes around.  It is not hard, yet a very poweful  method which  can be applied to many such situations.

Basically  for the time being, let us start with  something like:

  Let us assume  a series whose nth term   is given by $x^n$  that is  $x_n = x^n$

and we further assume that it just so happens that: (Suppose we are lucky enough to select a $x$ such that) :

$x^2 = x + 1$    ( Eq 1 )

which when multipllied by $x^n $ will give:

$x^{n+2} = x^{n+1} +x^{n} $
It will  satisfy the propertiy like $x_{n+2} = x_{n+1} + x_n $   (eq 2)

In fact, any x which satisfies eq 1 will satisfy  (eq 2). In fact if you take any constant, say  $A$  and have a term
$T_n =  A*x_n $  that will work too.

So all we have to do is to solve the eq 1.
The roots of eq 1 are  $\frac  {1+\sqrt 5}{4} $ and  $\frac {1 - \sqrt 5} {4}  $

Let us call them $k_1$  and   $k_2$

(Remember  $k_1$ and $k_2$ are just two real numbers whose value we know )

So any  term $T_n$ given by  $A k_1^n  + B k_2^n $  will satisfy eq 2  (If you don't believe me , just try it out.. This is the most important part to understand.. since  we have $k_1^2 = k_1 + 1$ and $k_2^2 = k_2+1$ we have  $T_{n+2} = T_{n+1} + T_n$

We are almost done now.. all we have to do is to check that $T_1$   is  $1$    and   $T_2$ is $1$
(with next terms o- eg  2, 3, 5 etc will  be automatically generated)

So we write  when n=1

$ 1 = A k_1^1 + B k_2^1$

and when n=2

$ 1 = a k_1^2 + B k_2^2 $

Remember you know  $k_1$ and $k_2$  .. so just solve for A and B and just substitue and you will get the   FAMOUS formula  from the internet ..which you posted ..

You will find that you will get something like   $A = -B =  1/(2k_1) $    where  $k_1 = \frac {1+\sqrt 5} {4} $
and $k_2 = \frac {1-\sqrt 5} {4} $ and the nt term is given by:

$T_n =  A ( k_1^n - k_2^n) $ which is same as you have seen...

Hope this is help ful and understandable .... (let me know it it was..)...

P.S - I know it may be a little advanced for this forum .. but then you can always skip it (or save it and read it later) .. i but still hope that some may find it helpful enough ...

Mods - If you find the message not fit for this form please feel free to move/edit/delete/ this message.. thanks.


prove that:
                   $\sqrt{\frac{x}{y+1}} +\sqrt{\frac{y}{z+1}} +\sqrt{\frac{z}{x+1}}$ $\geq {3}/ \sqrt{2}$
 where $x,y,z >0$ and $xyz=1$
                        iwant to have a solution

I think u post it already. See http://www.mathlinks.ro/Forum/topic-60340.html . :D


Find all pairs (x; y) of real numbers satisfying the system of equations

$x^3 + 1 - xy^2 - y^2 = 0$;
$y^3 - 1 - x^2y + x^2 = 0$.

  Darij

Equations equivalent to:

(x+1)(x^2 - x + 1 - y^2) = 0  (1)
(y-1)(y^2 + y + 1 - x^2) = 0  (2)

---
If $x = -1$, then $y = -1, 0, 1$; if $y = 1$, then $x = -1, 0, 1$ can easily be checked to be the only solutions.
Now suppose $x \neq -1$, $y \neq 1$.

$x^2 - x + 1 = y^2$ and $y^2 + y + 1 = x^2$ imply $x - y = 2$.  Substitute $x = y+2$.  Then $y = -1$ is implied
in both equations.  This implies $x = 1$.

In summary, the solutions are $(-1,-1), (-1,0), (-1, 1), (0,1), (1,1), (1,-1)$.

hope thats right, i blew through it quite fast.. but the general idea on how to knock down the cases is the same:  get rid of x=-1 and y = 1, then use the big brackets.

Adding both equations, this factors as
	\[(x-y)(x^2-y^2+x+y)=0.\]
	So either $x=y$ or $(x+y)(x-y)=-(x+y)$. Therefore, either $x=-y$ or $x-y=-1$. All of them lead to equations that are easily solvable. The solutions are thus solutions are $(-1,-1), (-1,0), (-1, 1), (0,1), (1,1), (1,-1)$.


The area of an equilateral triangle is $ 24\sqrt3$ square centimeters. What is the number of centimeters in the perimeter of the triangle? Express your answer in simplest radical form.

$ \frac{\sqrt{3}s^2}{4}\equal{}24\sqrt{3}$
$ s^2 \equal{}96$
$ s\equal{}4\sqrt{6}$

The perimeter is $ 12\sqrt{6}$

[quote="kz190"]The formula for finding the area of an equilateral triangle given the side length is $\frac{s^2\sqrt3}{4}$. The given area is $24\sqrt3$. You can simplify it to $s^2\equal{} 96$.

$\sqrt96\equal{} {4\sqrt6}$

$4\sqrt6\cdot 3 \equal{} \boxed{12\sqrt3}$[/quote]
How is $4\sqrt6\cdot 3$ equal to $12\sqrt3$? The correct answer is $12\sqrt6$, which dragon96 already stated in 2008. Please don't revive threads and then post an incorrect answer.


analytic geometry
algebra unsolved
algebra

Let S be the point of intersection of the two lines $ l_1 : 7x \minus{} 5y \plus{} 8 \equal{} 0$ and $ l_2 : 3x \plus{} 4y \minus{} 13 \equal{} 0.$ Let $ P \equal{} (3, 7), Q \equal{} (11, 13),$ and let $ A$ and $ B$ be points on the line $ PQ$ such that $ P$ is between $ A$ and $ Q,$ and $ B$ is between $ P$ and $ Q,$ and such that \[ \frac{PA}{AQ} \equal{} \frac{PB}{BQ} \equal{} \frac{2}{3}.\] Without finding the coordinates of $ B$ find the equations of the lines $ SA$ and $ SB.$


A more general form of this inequality was on one of the USA TSTs not so long ago. It's certainly true.

$\left(ac+bd-1\right)^{2}\geq 0$

so because of your inequality, the rest follows I think (since you would get a negative number otherwise).

[quote="amcavoy"] (since you would get a negative number otherwise).[/quote]

And if $a^2 + b^2 - 1 < 0$ and $c^2 + d^2 - 1 < 0$?

[quote="silouan"]It is a special case of ACZEL ineq here http://www.mathlinks.ro/Forum/viewtopic.php?t=68963[/quote]

Good link!


Consider the list of statements that are created. Let $ P(n)$ mean the statement that all elements of a $ n$ element set are the same. The inductive argument states the following:

$ P(1) \implies P(2)$
$ P(2) \implies P(3)$
$ P(3) \implies P(4)$
$ \ldots$

Let's go through this step by step and see where the error is. We find it very quickly! $ P(1) \implies P(2)$ is clearly false. The other steps are all correct.

Going back and examining the argument, we notice that the argument $ a_1 \equal{} \ldots a_n$ and $ a_2 \equal{} \ldots a_{n \plus{} 1}$ $ \implies a_1 \equal{} \ldots a_{n \plus{} 1}$ doesn't work here, because $ n \equal{} 1$, and the two statements $ a_1 \equal{} a_1$ and $ a_2 \equal{} a_2$ do not imply anything about their relationship.


[quote="snorida47x"]What is the range of g(x)= |x+2|?

Could you please explain in detail how exactly how you found the range of this function.

I have also graphed this function on a program, here is the link to the graph: http://www.yousendit.com/transfer.php?action=download&ufid=78C61B5102C94F4A[/quote]
Of course this function can reach only nonnegative values, because $|x+2|\geqslant 0$ by definition. Conversely, any nonnegative number can be value of this function: we have $f(y-2)= |y-2+2|= |y|=y$ since $y$ is nonnegative. Therefore the range of $g$ is $\langle 0,\infty)$.
I think the problem is too easy for this forum and should be moved somewhere.


majmooe bala ro tajziye mikonim:
$\sum_{i=1}^{10}|x+y-10i|\sum_{j=1}^{10}|3x-6y-36j|\sum_{k=1}^{10}|k(19x+95y-95k)|$
hala age deghat konid ke tabee:
$f(x)=|x-a_{1}|+|x-a_{2}|+...+|x-a_{2n}|, a_{1}<a_{2}<...<a_{2n}$
be ezaaye har $x$ dar $[a_{n},a_{n+1}]$ be dast miyad.hala tooye haselzarbe balayi majmooe avval baraye$50\leqq+y\leq60$ va majmooe dovvom baraye $180\leq3x-6y<216$ kamtarin meghdaar momken ast.
baraye be dast avordane kamtarin meghdare sevvom dobare ye tabei barhasbe $t=19x+95y$ dar nazar migirim va shib ha ro hesaab mikonim.(khodetoon inasho hesaab konid).age hesaab konid shin ta noghteye $t=95$ barabare ba $-1-2-...-10$ va dar in noghte $-1$ mishe $+1$.sepas dar $t=2*95$, $-2$ be $+2$ tabdil mishe va hami tor ta akhar.(age bekham tozihe kaamel bedam bayad 2 barabar benevisam :( ).jayi ke shib az manfi be mosbat tabdil mishe $t=7*95$ ast.pish az in noghte shib barabare $1+2+3+4+5+6-7-8-9-10=-13$ ast va bad az in noghte shib barabare ba $1+2+3+4+6+7-8-9-10=+1$.
(khodayish masaleye mosibatie na? :P  :rotfl: )
hala age dastgah na moadele haye $50\leqq+y\leq60$ va $180\leq3x-6y\leq216$ va moadeleye $19x+95y=7*95$ ro hal konim.hala halleshoon ba khodetoon vali javabe moadele va namoadeleye avval ine:$53.75\leqq\leq66.25$.az moadele va namoadele dovvom ham be dast miyad:$52.857\leqq\leq61.428$.chon ina javabe moshtarak daran pas kamtarin meghdare ebarate morede nazar barabar ast ba haselzarbe kamtarin meghdar haye $3$ majmooe morede nazar.kamtarin meghadre inmajmoo ha barabare ba(be tartib) $250,900,,10640$ ke age zarbeshoon konid javabe akhar mishe:
$2394000000$
ajab minimumi!!!be darde hichi nemikhore :wink: manzooram ine ke kheili zeshte


Write the polar equation in rectangular form $ r\equal{}\minus{}12sin(\theta)$
how do i do this?

[hide="Hints"]Using the formulas $ x^2 \plus{} y^2 \equal{} r^2$, $ x \equal{} r\cos \theta$, and $ y \equal{} r\sin \theta$ are three common formulas that help in converting many polar equations to rectangular form.[/hide]

[hide="Solution"]$ r \equal{} \minus{}12 \sin \theta$ implies $ r^2 \equal{} \minus{}12r \sin \theta$. Using the above formulas, this yields $ x^2 \plus{} y^2 \equal{} \minus{}12y$, so $ x^2 \plus{} y^2 \plus{} 12y \plus{} 36 \equal{} 36$, so $ x^2 \plus{} (y\plus{}6)^2 \equal{} 36$.[/hide]


[quote="Summation"]A)  In the plane with 4 circles, where they CAN vary in size relative to each other, what is the maximum number of intersection points all together possible?

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

B)  In the plane with 4 ellipses, where they CAN vary in size and/or shape relative to each other, what is the maximum number of intersection points all together possible?

>

***  Attempting A) or B), but not both, is enough
of a mathematical workout in my subjective opinion for the
"Puzzles & Brainteasers" section.  ***[/quote]

A) two circles can intersect in at most 2 points, so the answer is $2\binom{4}{2}=\boxed{12}$

B) two ellipses can intersect in at most 4 points, so the answer is $4\binom{4}{2}=\boxed{24}$

I agree with your two respective values.


Euler
number theory solved
number theory

prove:

for every natural number N (N!=0) and decinal digit x , x!=0 , there exists a natuaral number K (K!=0) , such that (NK) is a number whose digits are x and 0 only.

thank you.

It should be moved to number theory...
Write $N=2^k 5^l m$ with $gcd(10,m)=1$. Then $K=x \cdot 2^l 5^k \frac{10^{\phi(9m)}-1}{9}$ will do by the theorem of Euler-Fermat (since $m$ divides $\frac{10^{\phi(9m)}-1}{9} = 1+10+10^2+...+10^{\phi(9m)}$).

Edit:
simpler solution:
the numbers $x,x+10x,x+10x+100x,...$ can't have all different residues $\mod N$, so just take two with same residue and subtract.


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]

Let $ ABCD$ be a quadrilateral inscribed in a circle. The diagonal $ BD$ bisects $ AC$. If $ AB \equal{} 10$, $ AD \equal{} 12$ and $ DC \equal{} 11$, find $ BC$.

let, AC int. BD at X. AX=CX=m.
since t.AXB similar t.DXC implies m/DX=10/11
and. t.AXD similar t.BXC implies m/DX=BC/AD=BC//12
thus, 10/11=BC/12
BC=120/11


inequalities
trigonometry
function
calculus
calculus computations

Prove that
\[ \left|\dfrac{\pi}{4} \minus{} \arcsin x\right|\leq\dfrac{\pi}{4}\sqrt {1 \minus{} 2x\sqrt {1 \minus{} x^2}}\,\forall x\in [0,1]
\]
and
\[ \arcsin x\leq\dfrac{x}{\dfrac{2}{\pi} \plus{} \dfrac{\pi}{12}(1 \minus{} x^2)}\,\forall x\in [0,1].
\]

I solve the a) the problem b is samilar.
Let $ x \equal{} \sin{\alpha}$ where $ \alpha\in [0,\frac {\pi}{2}]$
The inequalities equivalent :
$ |\frac {\pi}{4} \minus{} \alpha|\leq \frac {\pi}{4}|\sin( {\alpha} \minus{} \cos {\alpha}|$
$ \Longleftrightarrow |\frac {\pi}{4} \minus{} \alpha|\leq \sqrt {2}\frac {\pi}{4}|\sin( {\alpha} ) \minus{} \frac {\pi}{4})|$
Consider the function :
$ f(y)\equal{}\frac{y}{\sin {y}}$


there is no base 1, string of zeroes is a contradiction because string of zeroes is zero! so no, not every number is equal to zero. lowest base is 2. that clear?

That's what I thought, and I don't think there's a base zero either  :lol:

:P  :P Are you sure? I thought there was a base 0. :P  :P

Just kidding


number theory
greatest common divisor
least common multiple
induction
number theory unsolved

Let $ x_1,x_2,...,x_k$ be positive integers such that $ \gcd (x_1,x_2,...,x_k)=1$. For all positive integers $ k$, let $ s_k=x_1^k+x_2^k+...+x_n^k$. Prove that

$ \gcd (s_1,s_2,...,s_n)$ divides $ \text{lcm} (1,2,...,n)$

assume that the case is not so. than there exists such a prime number $ p$ that the largest exponent $ a$ of $ p$ that divides $ gcd(s_i),1\leq i\leq n$ is grater than that $ b$ of $ lcm(1,2,...n)$. denote $ f_l \equal{} \sum_{sim} x_1x_2x_3.....x_l$ obviously $ p^a$ devides $ s_i$ for every $ 1\leq i\leq n$

by newton-girard formulae we have
$ s_1 \equal{} f_1$,, $ s_2 \equal{} s_1f_1 \minus{} 2f_2$,, $ s_3 \equal{} s_2f_1 \minus{} s_1f_2 \plus{} 3f_3$... etc
which yields us $ p^a$ devides $ if_i$ for every $ 1\leq i\leq n$

because $ p^a|nf_n$ we know (due to the definition of $ p$) $ p|f_n \equal{} x_1x_2...x_n$ meaning $ p$ devides one of the $ x$es. wlog $ x_1$
$ p^a|(n \minus{} 1)f_{n \minus{} 1}$ also implies $ p|f_{n \minus{} 1}$ which implies (due to the upper row) $ p|x_2x_3x_4...x_n$ meaning (wlog) $ p|x_2$
$ p^a|(n \minus{} 2)f_{n \minus{} 2}$ again implies $ p|f_{n \minus{} 2}$ and again due to the upper row $ p|x_3x_4x_5...x_n$ meaning wlog $ p|x_3$
we complete this by induction $ p|x_1,x_2...x_k$ implies $ p|x_1,x_2....x_k,x_{k \plus{} 1}$ leading to $ p|x_i$ for every $ 1\leq i\leq n$ which is in contradiction with $ gcd(x_1,x_2....x_n) \equal{} 1$


The determinant is divisible by the Vandermonde determinant, and the ratio is a symmetric function of degree 1, and therefore a multiple of $\Sigma a_i$.  You can determine the multiple by evaluation at some specific set of $a_i$, or by introducing some ordering on monomials and considering the "highest" terms.

In this book there was a hint: Use the Viet's formulas and open the determinant by the last column.
 How it understand?
 :whistling:


calculus
integration
geometry
search
calculus computations

Consider points $ P$ which are inside the square with side length $ a$ such that the distance from $ P$ to the center of the square equals to the least distance from $ P$ to each side of the square.Find the area of the figure formed by the whole points $ P$.

It's the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1823532652&t=302873 no?

Hi [b]Kunny![/b]
[hide][u][b]My Hint:[/b][/u]
[img]http://img38.picoodle.com/i576/duytuan119/id_89e_u9n46.jpg[/img]
$S=S_{GHIJKLM}$[/hide]

Thank you for the attached figure.
This would help someone solve the problem.

it is obvious that this shape would be symmetric wrt x and y axis

so we can focus on the first quadrant and multiply our answer by 4

to make the origin closer to $(x,y)$ than the line $y=a$

you need $\sqrt{x^2+y^2} \le a-y$
which is $y\le  -\frac{1}{2a}x^2+\frac{a}{2}$

and to be closer to the origin than to $x=a$ we symmetrically get
$x\le -\frac{1}{2a}y^2+\frac{a}{2}$

since the equations are symmetric, we know they will intersect on the line $y=x$

so we solve $x=-\frac{1}{2a}x^2+\frac{a}{2}$

$-2ax=x^2-a^2 $ or $x^2+2ax-a^2=0$

$x=\frac{-2a\pm \sqrt{4a^2+4a^2}}{2} = (\sqrt{2}-1)a=\alpha$

so we calculate $\int_0^\alpha -\frac{1}{2a}x^2+\frac{a}{2} dx + \int_\alpha^{\frac{a}{2}} \sqrt{a^2-2ax}dx$

$= -\frac{1}{6a}\alpha^3+\frac{a}{2}\alpha - \frac{1}{3a}(a^2-2a\alpha)^{\frac{3}{2}}$

$=a^2\left[ -\frac{(\sqrt 2 -1)^3}{6} + \frac{\sqrt 2 -1}{2} - \frac{1}{3}(1-2(\sqrt 2 -1))^{\frac{3}{2}}\right]$

$=\frac{a^2}{6}\left[ -(2\sqrt 2 - 6+3\sqrt 2-1)+ 3(\sqrt 2 -1)-2(3-2\sqrt 2 )^{\frac{3}{2}}\right]$

$=\frac{a^2}{3} \left[ 2 -\sqrt 2 - (3-2\sqrt 2)^{\frac{3}{2}}\right]$

so the total area should be $\frac{4a^2}{3}\left[ 2 -\sqrt 2 - (3-2\sqrt 2)^{\frac{3}{2}}\right]\approx .2287\cdot a^2$


calculus
derivative
calculus computations

Let $ f(x,y) \equal{} Ax^{2}\plus{}2Bxy\plus{}Cy^{2}\plus{}2Dx\plus{}2Ey\plus{}F$, where $ A > 0$, $ B^{2}< AC$.  Prove that there exists $ (X,Y)$ which minimizes $ f$ over $ \mathbb R^{2}$, and that
$ f(X,Y) \equal{} DX\plus{}EY\plus{}F$.

edit: typos fixed, sorry

I would understand the problem better if you had $ 2Bxy$ instead of $ 2Bx$ and $ B^{2}<AC$ instead of $ B^{2}<C$.

A minimizer exists because $ f\to\infty$ away from the origin -- indeed, $ Ax^{2}\plus{}2Bxy\plus{}Cy^{2}\equal{}A(x\plus{}(B/A)y)^{2}\plus{}(C\minus{}B^{2}/A)y^{2}$. Find the critical point by solving a linear system of equations.

straightforward differentiation yields a positive definite Hessian and X = (BE-CD)/(AC-B^2), Y = (BD-AE)/(AC-B^2), substitution proves the second part of the claim[/code]


Suppose that $n\geq 3$ is the odd number and $t,k$ are the least of numbers such that $nt$ and $nk+1$ are the perfect squares. Prove that $t,n\geq \frac n4$ if and only if $n$ be the prime number. :lol:  :)

I believe he means $t, k \geq \frac{n}{4}$.

What he means is that $t$ is the smallest positive integer such that $tn$ is a perfect square, and $k$ is the smallest positiv integer such that $kn+1$ is a perfect square.

For example, if $n=19$, then $t=19$, which is definitely $\geq \frac{19}{4}$ and $k=5$, which is also $\geq \frac{19}{4}$. If $n=63$, then $t=7$, and $k=1$. So both can only be satisfied iff $n$ is a prime.


Hey folks,

For all of you that are interested in how the Georgia team did this year, well, we could have done better, but nothing to be ashamed of. The team placed 43rd overall. Sitan was our top placer at 99th. When you do some of the analysis of the results, it rapidly becomes apparent that this was both a difficult and closely contested event this year. Just a couple of mistakes here and there and we might have moved up 10-15 places.

For those of you that are both ARML and MATHCOUNTS veterans, I have a question. Why does Georgia do so well at ARML and typically poorly at MATHCOUNTS? I have some of my own opinions, but would be interested in hearing yours.

See you tomorrow,
Fulton

Partly we don't have any really good coaches [no offense to you, but consider say Mr. Fischer or Mr. Boyd], and partly we don't have experienced people. I'm not sure, but I think the last time we had a two-time Mathcaunter was when Harrison was on the team, and I'm not sure of that. And there's really no substitute for going once, being defeated, and coming back with a real drive to succeed at nats.

Gosh, gee, thanks a lot. Consider the fact that I didn't hardly know what MATHCOUNTS was until mid year!

So what do we do to change all of this? How much impact do you think the "state" coach can have? It seems to me that unlike ARML where we have you guys for 3 or 4 years, the MATHCOUNTS groups tend to be a year at a time. Can we make an impact with the local coaches? What's the best way to prepare for MATHCOUNTS? It seems very different from what I would consider "normal" math team. I think that speed is emphasized rather than depth of knowledge. I know that I spent the first part of the year working with my middle schools talking about Geometry and modular arithmetic. Very much NOT emphasizing speed and memorization.

ARML is a more of a team event, both in the team size and the nature of the competitive events.  Thus the ARML team is more of a community than MATHCOUNTS provides.  And yes, MathCounts is much more shallow and speed-oriented (apart from the Masters' Round).  Also, in Georgia, my impression is that we have a much better high school competition math environment than we have a middle school environment.  Many middle schoolers go to few tournaments except MathCounts competitions, whereas in high school there are many excellent tournaments year-round for people to attend.  Also, the AMCs are much better publicized at the high school level than at the middle school level, so most middle school students will see few questions approaching easy AIME difficulty.

If you look [url=http://www.artofproblemsolving.com/Wiki/index.php/Georgia_mathematics_competitions]here[/url], you can see a list of most of the math competitions in Georgia.  The vast majority of the tournaments cater to high school students.  Most of the tournaments for middle schoolers are primarily high school tournaments which either regularly include middle schools in the JV division or have a separate division for middle schoolers.  Most of their business, however, comes from high schools.  The only exceptions, in fact, are the Berry College tournament, MathCounts, and the middle school state tournament.  Why is this the case?  Most tournaments are either run by high schoolers or college students and faculty.  Naturally, both are more interested in high school tournaments, where the curriculum is less restricted, they can ask harder problems, and business is already better anyhow.  Why do middle schoolers not run tournaments for each other?  First, the middle schools are much more restrictive than the high schools.  Second, middle schoolers would have trouble getting the manpower necessary to run a tournament of any size, due to the fact that middle schoolers usually don't have math honor societies that require service hours.  There is also the problem that more dedicated, motivated teachers are bound to be more interested in teaching high school than middle school, and to be honest, I think that in general (not everybody :wink: ), we in high schools don't do as good a job helping and encouraging middle schoolers as we should.

So in short, I think it is part of the underlying structure of the mathematical community here.

And to give MathCounts credit, I really don't think you have to memorize for it.  Most of their problems do emphasize picking a creative approach (albeit in a very brief time period) rather than having a list of memorized facts, if I remember aright.


ratio
geometry
trigonometry
AMC
AIME
MATHCOUNTS
angle bisector

I think 1337 posted something like this a while ago...

(Solution courtesy of The Quatto Master 6000)
Draw a line through the 75 angle that divides it into a 60 degree angle and a 15 degree angle. Now you have two triangles - one 30-60-90 triangle and one isosceles 15-150-15 triangle. I'm too lazy to finish it off, but it's trivial now.

[quote="13375P34K43V312"][hide="oooh i know"]

man too much geometry for one day

we know it's 15-75-90 duh
So yeah extend the short side to make a 30-60-90. Then let the short leg of that be 1, because it's a ratio and it doesn't matter. then the short leg of the 15-75-90 is 1-x, the rest of it is x, and the long leg is $\sqrt{3}$ by the 30-60-90, and the hypotenuse is 2. then BAM angle bisector theorem

$\frac{1}{2}=\frac{1-x}{\sqrt{3}}$
$\sqrt{3}=2-2x$
$2x=2-\sqrt{3}$
$x=1-\frac{\sqrt{3}}{2}$

so then it's easy now $1-x=\frac{\sqrt{3}}{2}$ so yeah divide that by the hypotenuse and use pythagoras to get the hypotenuse is $\frac{\sqrt{13}}{2}$ so the ratio is $\frac{\sqrt{3}}{\sqrt{13}}=\frac{\sqrt{39}}{13}$ yeah it looks like I did something wrong.[/hide]

dude that's such an algebra-heavy/long solution...just use the law of sines[/quote]
it's not law of sines when it's a right triangle :roll:
[quote="Ubemaya"]I think 1337 posted something like this a while ago... [/quote]
I don't know if l33t posted it, but I posted it here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=trig&t=124768

[quote="13375P34K43V312"]He might mean reflecting over the hypotenuse. Now I see the other thread, it seems that I made a mistake in my solution. Did anyone catch it?

for the law of sines thing, i mean you could use it, but i guess you could just use ratios as it's a right triangle...[/quote]
Like, you've been doing too much AIME trig.
Trig originated in the right triangle, so the answer is just $\sin 15$

Please do not post any trigonometry in this Forum.  No matter how easy you think it is, it doesn't belong in the MATHCOUNTS Forum.  Thank you.

[quote="Ubemaya"](Solution courtesy of The Quatto Master 6000)
Draw a line through the 75 angle that divides it into a 60 degree angle and a 15 degree angle. Now you have two triangles - one 30-60-90 triangle and one isosceles 15-150-15 triangle.[/quote]
Now, there's a little bit of algebra to simplify the answer.  But that's OK, because algebra is allowed in the MATHCOUNTS Forum.  Trigonometry is not allowed in the MATHCOUNTS Forum.

[hide="hint"]
$\frac{\text{short leg}}{\text{long leg}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

Now, you can solve for what the problem actually asked for.  :)
[/hide]

Ok, using Pyth. Thm. I got hyp (longest side) ^2=12+x
so the longest side is sqrt(12+x)
The ratio of the shortest side to the longest is 1/sqrt(12+x)
multiplying to get rid of the radical in the denominator, do I leave it as sqrt(13)/13 or sqrt(12+x)/12+x?  What if they want s instead of x as the variable?  What is the answer?

[quote="LawOfSigns"]Ok, using Pyth. Thm. I got $(\text{hyp (longest side) })^{2}=12+x$
so the longest side is $\sqrt{12+x}$
The ratio of the shortest side to the longest is $\frac{1}{\sqrt{12+x}}$
multiplying to get rid of the radical in the denominator, do I leave it as $\frac{\sqrt{13}}{13}$ or $\frac{\sqrt{12+x}}{12+x}$?  What if they want s instead of $x$ as the variable?  What is the answer?[/quote]

I have no idea what your $x$ is...

[hide="Answer"]
From rcv's diagram:
hyp=$\sqrt{1+7+4\sqrt{3}}=\sqrt{8+4\sqrt3}=\sqrt6+\sqrt2$ (actually there is a small algebra involved in figuring this simplification out)
shortest leg=$1$

Therefore the ratio is $\frac{1}{\sqrt{6}+\sqrt2}=\frac{\sqrt6-\sqrt2}{4}$[/hide]

...too...many...triangles

Whatever, so after u turn the 75-15-90 into a 30-60-90, make sure u turn the 15-150-15 into 2 15-75-90s  wait a minute...

AIME discussion belongs [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]here[/url].

[quote="Ubemaya"][quote="perfect628"]That's one way...there's, imo, an easier way...[/quote]

What would this way happen to be?[/quote]
Yes, I'm wondering too. Has your way been used, or do you have another great method?

Construct 30-60-90 and use Angle Bisector Theorem.

[quote=beastgert]And bpms quit aops in 2007 wow these people are veterans[/quote]

back when people hid their solutions, did not double post, and did not revive extremely old threads. -_-

As one said...
[quote=a87321dc0]/request lock
Please don't revive threads that are past 6 months old :what?: .
@above techguy2 asked if you could factor four out of $\sqrt{8+4 \sqrt{3}}$, not about 15-75-90 triangles[/quote]

[quote=equinox145111]There's nothing wrong with reviving a post. The user had a legitimate response.[/quote]

If you need something, please post a new thread

what is the ratio of the sides in a 15-75-90 triangle?

[quote=mathwhiz16]Am I the only one that memmed the ratio

$\frac{\sqrt{6}-\sqrt{2}}{4}$, $\frac{\sqrt{6}+\sqrt{2}}{4}$, 1?[/quote]

[quote=asp211]$\frac{\sqrt{6}-\sqrt{2}}{4}$, $\frac{\sqrt{6}+\sqrt{2}}{4}$, $1$

This I find the easiest to memorize because I find it easy for the hypotenuse to be a rational number[/quote]

agreed.

The ratio of the sides?

It isn't.  I trust you've seen the problems that have 15-75-90 triangles, with nastier side ratios, to complicate them?

[quote=Kwow]Hey, um... This seems to be an Alcumus problem, so please post this on the Alcumus forum... after you've solved it...[/quote]

the OP was made in 2007 ;)

[quote=perfect628]A right triangle's two acute angles are in the ratio $1: 5$. What is the ratio of the shortest side length to the longest?[/quote]

memming sine and cosine of 15 helps ALOT

[quote=GameMaster402]memming all the multiples of 15 for sine and cosine, and knowing how to use the unit circle efficiently is really helpful.[/quote]

and if your good with a pentagon and golden ratio mulltiples of 18 as well

here's a nicer ratio for 15-75-90..

1 : $2 + \sqrt{3}$ : $\sqrt{2} + \sqrt{6}$

edit: oops sorry, didnt see that someone already mentioned it


inequalities
trigonometry
inequalities unsolved

Let $ABC$ be an acute triangle. Prove that: \[\left(\frac{\cos A}{\cos B}\right)^{2}+\left(\frac{\cos B}{\cos C}\right)^{2}+\left(\frac{\cos C}{\cos A}\right)^{2}+8\cos A\cos B\cos C\ge 4\]

[quote="arkhammedos"]sorry but whats $y.x. z$ are.[/quote]
They are positive reals.


Hi, I have another question coming from one of my chemistry books  that I'm not sure on how to solve.  I tried the problem and I was hoping if anyone can please offer me a hand with this.  That would be great.  Thanks again.

Question: A piece of chalk (mainly calcium carbonate) is placed in $250mL$ of $0.433M$ $HCl$.  All the $CaCO_{3}$ reacts, releasing carbon dioxide gas, and leaving a clear solution.  $50.00mL$ of the solution is pipetted into another flask.  $39.1mL$ of $0.0549M$ $NaOH$ is required to titrate the $HCl$ remaining in this $50.00mL$ portion.  What was the original mass of $CaCO_{3}$ in the piece of chalk?

Well I began by writing down the balanced chemical equation: $CaCO_{3}+2HCl\rightarrow H_{2}O+CO_{2}+CaCl_{2}$
Initially moles of acid is $n_{HCl}=CV=(0.433M)(0.25L)$
After the reaction, there were some moles of acid remaining as it it was the excess reactant.  The number of moles which remained can be calculated by finding the number of moles required by the base $NaOH$ to titrate it.
Thus, the residual moles of acid from the first equation = $C_{NaOH}V_{NaOH}=(0.0391L)(0.059M)$
The amount of $HCl$ that reacted in the first equation = (Initial amount of moles) - (moles needed to titrate it)
$\Rightarrow$ Amount Consumed$=(0.433M)(0.25L)-(0.0391L)(0.059M)$
This amount consumed also equals to the moles of the limiting reactant.  Hence, $2\times n_{CaCO_{3}}=(0.433M)(0.25L)-(0.0391L)(0.059M)$
Solving for $n_{CaCO_{3}}=\frac{(0.433M)(0.25L)-(0.0391L)(0.059M)}{2}$
Finally, the mass of $CaCO_{3}=n_{CaCO_{3}}M_{CaCO_{3}}$

I tried this approach and got an answer of about 5.3-g.  However, my answer did not concur with the back of my book.  I'm hopping to hear some advice from you guys, thanks.

You have titrated 50 mL (1/5) of the initial solution.


If $ \sqrt{x \minus{} 1} \minus{} \sqrt{x \plus{} 1} \plus{} 1 \equal{} 0$, then $ 4x$ equals:

$ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 4\sqrt{\minus{}1}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ 1\frac{1}{4}\qquad
\textbf{(E)}\ \text{no real value}$

[hide="Solution"]

$ \sqrt {x \minus{} 1}$ $ \plus{} 1$ = $ \sqrt {x \plus{} 1}$

Square both sides $ \rightarrow$ $ x \minus{} 1$ + $ \sqrt {x \minus{} 1}$ + $ \sqrt {x \minus{} 1}$ $ \plus{} 1 \equal{} x \plus{} 1$.
Which is $ x \plus{} 2$$ \sqrt {x \minus{} 1}$ $ \equal{} x \plus{} 1$
$ \rightarrow$ $ \sqrt {x \minus{} 1}$ $ \equal{}$ $ \frac {1}{2}$
$ \rightarrow$ $ x \minus{} 1$ $ \equal{}$ $ \frac {1}{4}$
$ \rightarrow$ $ x$ $ \equal{}$ $ \frac {5}{4}$
So $ 4x \equal{} 5$ $ \rightarrow$ $ A$

[/hide]


Compute the number of integer values $ n$, $ 1 < \equal{} n < \equal{} 2004$ such that for each value of $ n$ the folling inequality has exactly two integer solutions

$ |\frac {nx}{2004} \minus{} 1| < \frac {n}{2004}$

so you want $ f(x) = \frac {nx}{2004} - 1$ to have a small absolute value right

that is, $ - \frac {n}{2004}\< f(x)\< \frac {n}{2004}$
$ \implies 2004 - n < nx < 2004 + n$

so $ \frac {2004}{n} - 1 < x < \frac {2004}{n} + 1$

so there are at most 2 solutions, because the "range" is 2, and there is 1 solution iff $ \frac {2004}{n}\in \mathbb{Z}$. (there's always a solution though because the "range" is >1)

so $ 2004 = 2^2\times3\times167$, so there are $ 3(2)(2) = 12$ ways that are bad

so... $ 1992$ is the answer.


Ai co cai ebook extremal combinatorics pmmmm.

Ban nay o dau ra ma cai ten nghe kho hieu qua day. Day la dien dan toan nen phai giu tu ngu dung muc, khong nen ham ho, bay ba


function
Euler
abstract algebra
number theory
totient function
relatively prime
number theory open

For those who are unfamiliar with the $ \phi$-function (Euler's totient function), $ \phi(n)$ is defined as the number of positive integers that are relatively prime to $ n$ where $ \phi(1)$ is defined to be $ 1$. Carmichael's Conjecture states that $ \phi$ is never one-to-one; that is, for any $ n \in \mathbb{N}$ there exists $ m \in \mathbb{N}$  such that $ \phi(n)\equal{}\phi(m)$ and $ n \not\equal{} m$. It has been proven for all $ n$ up to $ 10^{10000000}$. A full proof remains unknown.

Have any of you ever worked on Carmichael's Conjecture before? If so, what were your results?

The most famous paper regarding Carmichael's conjecture is due to Kevin Ford (a professor in the University of Illinois). The famous paper can be read here http://www.math.uiuc.edu/~ford/wwwpapers/sierp.pdf

In the paper he proves that for any $ k \geq 2$ and any $ m$ there exist at least $ k$ values of $ x$ such that $ \phi(x) \equal{} m$. His techniques weren't enough to determine whether Carmichael's conjecture is true or not, but he did prove that if it were false, there would be a positive proportion of numbers that would be counterexamples (that means a lot of counterexamples).

There has been recent work on a generalization of Carmichael's conjecture to unit groups. A thesis coming out this year works on that.

[quote="Quique"]The most famous paper regarding Carmichael's conjecture is due to Kevin Ford (a professor in the University of Illinois). The famous paper can be read here http://www.math.uiuc.edu/~ford/wwwpapers/sierp.pdf[/quote]
Interesting. Thanks for the post! Until now I was only familiar with Carl Pomerance's work on the conjecture. In his [url=http://www.math.dartmouth.edu/~carlp/PDF/carmichaelconjecture.pdf]paper[/url] he proves that if there exists $ n \in \mathbb{N}$ such that $ (p\minus{}1)|\phi(x)$ implies $ p^2|n$ for every prime $ p$, then $ \phi(x)\equal{}n$ has only one solution for $ x$. Hence if such an $ n$ exists, it is a counterexample to Carmichael's Conjecture. Indeed, it is likely that no such $ n$ exist.

[quote="Quique"]In the paper he proves that for any $ k \geq 2$ and any $ m$ there exist at least $ k$ values of $ x$ such that $ \phi(x) \equal{} m$.[/quote]
I think what you meant to say here is that for any $ k \geq 2$, there exists an $ m$ such that $ \phi(x)\equal{}m$ has at least $ k$ solutions for $ x$.

[quote="Quique"]There has been recent work on a generalization of Carmichael's conjecture to unit groups. A thesis coming out this year works on that.[/quote]
Do you have any further information on this paper?

[quote]Do you have any further information on this paper?[/quote]

The thesis is by Jonathan Bayless working under Carl Pomerance at Dartmouth College. You could probably find it by requesting your library to get it lent from the Dartmouth library.

Here's the abstract of the talk Bayless gave at the Maine/Quebec Number Theory Conference:
"Jonathan Bayless, Dartmouth College (grad): "An analogue of Carmichael's conjecture."
Carmichael's conjecture that there is no unique preimage for the Euler function is well-known and widely believed. We show that the analogue for the function U, which maps an integer n to the class of Abelian groups isomorphic to the unit group mod n, is false and give a lower bound on the number of counterexamples up to x. "


2. Let $a_1, a_2, \cdots$ be a convex sequence of real numbers. which is to say $a_{k-1}a_{k+1} \geq 2a_k$ for all $k \geq 2$. prove that for all $n \geq 1$
$\frac{a_1+a_3+\cdots +a_{2n+1}}{n+1} \geq \frac{a_2+a_4+\cdots +a_{2n}}{n}$

3. Let $a_0, a_1, \cdots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1} \leq a_i^2$ for i=1,2,... Prove that for each $n\geq 1$
$\frac{a_0+a_1+\cdots +a_{n}}{n+1}\frac{a_1+a_2+\cdots +a_{n-1}}{n-1} \geq \frac{a_0+a_1+\cdots +a_{n-1}}{n}\frac{a_1+a_2+\cdots +a_{n}}{n}$

4. Given a sequence ${{x_n}_{n=0} }^{\infty}$ with xn > 0 for all n>=0 such that the sequence ${{a^n x_n}_{n=0} }^{\infty}$ is convex for all a>0. Show that the sequence ${{log x_n}_{n=0} }^{\infty}$ is also convex.

5. How should the numbers 1,...,n be arranged in a circle to make the sum of the products of pairs of adjacent numbers is as large as possible? As small as possible?

I ould be glad if some one give solns!

2. By putting $a_{2i}=\frac{a_{2i-1}+a_{2i+1}}{2}$ we make right side as big as it can be for constant $a_1,a_2,\ldots,a_{2n+1}$ so also for constant left side. Now our ineq comes to:
$\frac{a_1+a_3+\ldots+a_{2n+1}}{n+1}\geq\frac{\frac{a_1+a_3}{2}+\frac{a_3+a_5}{2}+\cdots+\frac{a_{2n-1}+a_{2n+1}}{2}}{n}$
and after multiplying it by $n(n+1)$ we have:
$n(a_1+a_3+\cdots+a_{2n+1})\geq (n+1)(\frac{a_1}{2}+a_3+a_5+\ldots+a_{2n-1}+\frac{a_{2n+1}}{2})$
which is equivalent to:
$(n-1)(a_1+a_{2n+1})\geq 2(a_3+a_5+\ldots+a_{2n-1})$
We will prove this but you can see that this is true cause of convexity...

Now let's have positive integers $i<j$. We know that for each $i\leq k\leq j$ we have:
$\frac{a_{k-1}+a_{k+1}}{2}\geq a_k$
Adding these ineqs together we get:
$\frac{a_{i-1}+a_{i+1}}{2}+\frac{a_{i}+a_{i+2}}{2}\cdots+\frac{a_{j-1}+a_{j+1}}{2}\geq a_i+a_{i+1}
+\ldots+a_{j}$
which is same as:
$\frac{a_{i-1}}{2}+\frac{a_{i}}{2}+a_{i+1}+\cdots+a_{j-1}+\frac{a_{j}}{2}+\frac{a_{j+1}}{2}\geq a_i+a_{i+1}+\ldots+a_{j}$
by deleting many terms at both sides we get:
$\frac{a_{i-1}+a_{j+1}}{2}\geq \frac{a_i+a_j}{2}$
Using this more times we can get:
$\frac{a_{i-1}+a_{j+1}}{2}\geq \frac{a_i+a_j}{2}\geq \frac{a_{i+1}+a_{j-1}}{2}\geq \ldots$
But that $\ldots$ does not mean: to infty!

Applying this we have ineualitis:
$\frac{a_1+a_{2n+1}}{2}\geq \frac{a_3+a_{2n-1}}{2}\\
\frac{a_1+a_{2n+1}}{2}\geq \frac{a_5+a_{2n-3}}{2}\\
\ldots\\
\frac{a_1+a_{2n+1}}{2}\geq \frac{a_{2n-1}+a_3}{2}$
Summing all these ineqs and multiplying by two we get:
$(n-1)(a_1+a_{2n+1})\geq 2(a_3+a_5+\ldots+a_{2n-1})$
so the problem is solved.

3. is equivalent to $(a_0 + \cdots +a_n)(a_1 + \cdots + a_{n-1}) \ge (n^2-1)a_na_0$, this can be proved by amgm and $a_ia_{j+k} \ge a_ja_{i+k}$ which can be concluded from the given condition

I was just writing a solution to 3. but you've written in 2 lines what I was about to write in much more lines :D  :P

4. Given condition gives $x_{k+1}a^2-2x_k a+x_{k-1} \ge 0$ true for all a, so the discriminant must be smaller than 0, which is equivalent to the log convexity.


15 "i" symbols (where i = sqrt(-1)), 1 "^" for exponentiation, unlimited pairs of parentheses for grouping and multiplication, and unlimited plus signs for addition, as long as an allowable arithmetic expression is formed.   No other symbols/characters may be used.

Rules:

No "ii" form or longer allowed.  Parentheses must be used.
Edit:  "(i)(i)" is definitely allowable.   However, the forms of
"i(i)" and/or "(i)i" MAY be allowable, depending on their relative
placement.

Express the answer in the form of A^B, A,B being positive integers.

Answers with fewer pairs of parentheses are simpler.

At any stage in your work, only allow any exponents to be real values.

let's see....
$((i+i)(i)) =-2$ and we're left with 12 to maximize the exponent. $(i+i+i)(i+i+i)(i+i+i)(i+i+i) = 81$, however, (-2)^81 is a very SMALL number. This can easily be remedied by chaning the exponent to 3*3*4*2 instead of 3*3*3*3 so we have (-2)^(72) = 2^72.

This was formed as:

((i+i)(i))^((i+i+i)(i+i+i)(i+i+i+i)(i+i)) = 2^72

Is there a larger number?

Oh yeah, I can also take one out of the 4 in the exponent and put it in the base to get

((i+i)(i+i))^((i+i+i)(i+i+i)(i+i+i)(i+i)) = 4^54 = 2^108

Which is larger than 2^72.

Anyone which to verify? My current answer is 4^54 = 2^108

Impressive, thought-out work of the prior poster.

In the rules, I recall to myself that there is not a restriction on the base, A, in A^B, being held to a real number.  With that in mind,
I offer:
[hide]I'm going to allow a base with "i" in it, but then the real
exponent (so far as I know) will have to be a multiple of 4 to get
rid of the "i" when the base is raised to it.
(i+i+i)^((i+i+i)(i+i+i)(i+i+i+i)(i+i)) = (3i)^((3i)(3i)(4i)(2i))=
(3i)^(72i^4) = (3i)^(72*1) = (3i)^72 = (3^72)(i^72) =
(3^72)*1 = 3^72 ~ 2^114.1173, which is greater than 2^108,
of the prior attempt.[/hide]


can you post your solution because I want to know how the inequality works  in this situation?

let $ f(x,y,z)\equal{}(x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (z \plus{} x)^2$. Then

$ f(x,y,z) \leq f(1, xy,z) \leq f(1, 1, xyz)$.

$ f(x,y,z) \leq f(1, xy,z)$  becouse

1. $ 1\plus{}x^2y^2 \geq x^2\plus{}y^2 \Leftrightarrow (1\minus{}x^2)(1\minus{}y^2) \geq 0$.

2. $ 1\plus{}xy \geq x\plus{}y \Leftrightarrow (1\minus{}x)(1\minus{}y) \geq 0$.

In fact there is more general result:
If $ a_1 \geq a_2 \geq \ldots \geq a_{n} >0$ and $ b_1 \geq b_2 \geq \ldots \geq b_{n} >0$ and

$ \prod_{i\equal{}1}^{j}a_{i} \geq \prod_{i\equal{}1}^{j}b_{i}$ for $ 1 \leq j \leq  n$, then

$ a_1\plus{}a_2\plus{} \ldots \plus{}a_{n} \geq b_1\plus{}b_2\plus{} \ldots \plus{}b_{n}$ and

$ \prod_{i\equal{}1}^{n}(a_{i}\plus{}c) \geq \prod_{i\equal{}1}^{n}(b_{i}\plus{}c)$ if $ c > 0$ and

$ \prod_{i\equal{}1}^{n}(a_{i}\minus{}c) \leq \prod_{i\equal{}1}^{n}(b_{i}\minus{}c)$ if $ c > 0, a_1a_2 \cdot \ldots \cdot a_{n}\equal{}b_1b_2 \cdot \ldots  \cdot b_{n}, a_1 \geq a_2 \geq \ldots \geq a_{n} >c$.

I think there is a topic with these inequalities.

[quote="dmitin"][b]Nak[/b]

[quote]$ f(x,y,z) \leq f(1, xy,z)$[/quote]

Do you apply the statement
[quote]
If $ a_1 \geq a_2 \geq \ldots \geq a_{n} > 0$ and $ b_1 \geq b_2 \geq \ldots \geq b_{n} > 0$ and

$ \prod_{i \equal{} 1}^{j}a_{i} \geq \prod_{i \equal{} 1}^{j}b_{i}$ for $ 1 \leq j \leq n$, then

$ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_{n} \geq b_1 \plus{} b_2 \plus{} \ldots \plus{} b_{n}$[/quote]
to numbers $ a_1,a_2,a_3\quad \equal{} \quad (1 \plus{} xy)^2,(xy \plus{} z)^2,(z \plus{} 1)^2$ ([b][color=blue]reordered[/color][/b] so that $ a_1 \geq a_2 \geq a_3$)
and numbers $ b_1,b_2,b_3\quad \equal{} \quad (x \plus{} y)^2,(y \plus{} z)^2,(z \plus{} x)^2$ ([b][color=blue]reordered[/color][/b] so that $ b_1 \geq b_2 \geq b_3$)?

If so why are conditions
[quote]1. $ 1 \plus{} x^2y^2 \geq x^2 \plus{} y^2 \Leftrightarrow (1 \minus{} x^2)(1 \minus{} y^2) \geq 0$.

2. $ 1 \plus{} xy \geq x \plus{} y \Leftrightarrow (1 \minus{} x)(1 \minus{} y) \geq 0$.[/quote]
enough?[/quote]

No, i don`t. I mean

$ f(1,xy,z) \minus{} f(x,y,z) \equal{} 2(x^2y^2 \plus{} 1 \minus{} x^2 \minus{} y^2) \plus{} 2z(xy \plus{} 1 \minus{} x \minus{} y)$


$1983$cities are served by ten airlines. All services are both ways. There is a direct service between any two cities.

Show that at least one of the airlines can offer a round trip with an odd number of landings.


- identifies x with -x for each x in the equator of S2.
Thank you all!

Alternative: $ H_m(X) \equal{} 0$ for all $ m > 2$. Now, this space is the quotient of the plane by a totally disconnection action of the cyclic group of order two (projective space). This means that $ \pi_1(X)\cong\mathbb{Z}/2\mathbb{Z}$. We know $ H_1(X)$ is its this groups abelianization (but the group is abelian anyways so they're isomorphic). Now, $ H_2(X)$ is given by orientability, it is free of rank 1 if the surface is orientable and trivial otherwise. Is the projective plane orientable?

This isn't $ \mathbb{R}P^2$. This is $ S^2$ with an action on its equator $ S^1$; basically it's $ S^1$ with two $ 2$-cells attached each with a degree 2 attaching map. You get homology groups of $ \mathbb{Z}, \mathbb{Z}/2,$ and $ \mathbb{Z}$, in contrast with the projective plane which doesn't have a $ \mathbb{Z}$ in dimension 2.

WOW i totally skipped that part of the problem, I saw identify $ x$ with $ \minus{}x$ and then posted. SORRY.

Can we reason by analogy that $ S^n$, with the antipodal points of $ S^{n\minus{}1}$ identified, is realized by one $ 0$-cell, one $ n\minus{}1$-cell wrapped on the $ 0$-cell, and two $ n$-cells wrapped twice around the $ n\minus{}1$ cell?

Then $ H_n \equal{} \mathbb Z$, $ H_{n\minus{}1} \equal{} \mathbb Z/2 \mathbb Z$, $ H_0 \equal{} \mathbb Z$ and $ H_i \equal{} 0$ for all other $ i$.

Not quite; I have used above (implicitly) that $ S^1$ with antipodal points identified is homeomorphic to $ S^1$. In general $ S^{n\minus{}1}$ with antipodal pts identified is $ \mathbb{R}P^{n\minus{}1}$; the construction (which I'll call $ X$) on $ S^n$ you have described has $ \mathbb{R}P^{n\minus{}1}$ as a subcomplex; computing the homology of $ X$ involves computing the homology of $ \mathbb{R}P^{n\minus{}1}$, which isn't pretty with $ \mathbb{Z}$-coefficients. In particular, for small $ m$, $ H_m(X) \equal{} H_m(\mathbb{R}P^{n\minus{}1})$.

Over $ \mathbb{Z}/2$, however, the homology of $ \mathbb{R}P^{n\minus{}1}$ is nicely behaved. Perhaps the calculation isn't so bad for $ H_n(X;\mathbb{Z}/2)$.

So how would the $ n$-cell glue onto the $ n\minus{}1$ cells?

There would be two $ n$-cells like in the $ S^2$ case.

To attach an $ n$-cell one must give a map $ S^{n\minus{}1} \to \mathbb{R}P^{n\minus{}1}$; in this case it is the covering map where antipodal points are identified.

So if we dealt with $ S^3$, we'd have one $ 0$-cell $ v$, one $ 1$-cell $ a$, one $ 2$-cell $ A$, and two $ 3$-cells $ U$ and $ L$?

And since $ d(A) \equal{} 2a$ is injective, that means that $ d(U) \equal{} d(L) \equal{} 0$.

So for this new space $ X$, we'd get $ H_0(X) \equal{} \mathbb Z$, $ H_1(X) \equal{} \mathbb Z/(2)$, $ H_2(X) \equal{} 0$, and $ H_3(X) \equal{} \mathbb Z^2$.

Is this right?  And also, would this be a weird proof that this space $ X$ is not a manifold? (since manifolds have top dimensional homology either $ \mathbb Z$ or $ 0$.)

Really sorry for forget logging! Thank you all!


Five distinct points are identified on a circle.  How many distinct lines are determined by these points?

Each line contains $ 2$ points, so there are $ \binom{5}{2}\equal{}\frac{5\cdot4}{2}\equal{}\boxed{10}$ distinct lines that are determined by these points.


Hello,

I copied and pasted the exact lines from the pdf file that documents the new version of the geometry package:
[code]
[asy]
import geometry; size(150); defaultpen(linewidth(0.8));
point A=(0,0), B=(2*Cos(40),2*Sin(40)); line l1=line(A,B);
draw(l1, linewidth(bp));
line l2=rotate(100,A)*l1;
draw(l2, linewidth(bp));
[/asy]
[/code]

[asy]import geometry; size(150);
point A=(0,0), B=(2*Cos(40),2*Sin(40)); line l1=line(A,B);
draw(l1, linewidth(bp));
line l2=rotate(100,A)*l1;
draw(l2, linewidth(bp));[/asy]
It does something very basic, but it does not work. Can someone explain why?

Thanks, because this is pretty irritating. I've been trying everything at a time and it does not work.

Hi,
[quote="darkprince"]
I copied and pasted the exact lines from the pdf file that documents the new version of the geometry package:
[code]
[asy]
import geometry; size(150); defaultpen(linewidth(0.8));
point A=(0,0), B=(2*Cos(40),2*Sin(40)); line l1=line(A,B);
draw(l1, linewidth(bp));
line l2=rotate(100,A)*l1;
draw(l2, linewidth(bp));
[/asy]
[/code]
[/quote]
I don't see these lines in the last version of geometry_en.pdf. Can you give me the page number please ?
[quote="darkprince"]
It does something very basic, but it does not work. Can someone explain why?
[/quote]

It does what you ask it : draw lines on an 0x0 picture size.
Adding an object with real dimensions, for example a circle((0,0),1), and all becomes luminous :-)
[quote="darkprince"]Thanks, because this is pretty irritating. I've been trying everything at a time and it does not work.[/quote]

Try this one (does not work on APS but works on my computer):
[code]import geometry; size(150); defaultpen(linewidth(0.8));
point A=(0,0), B=(2*Cos(40),2*Sin(40)); line l1=line(A,B);
draw(l1, linewidth(bp));
line l2=rotate(100,A)*l1;
draw(l2, linewidth(bp));
draw(circle((0,0),1),invisible);[/code]

Hello,

I tried it on Asymptote in my computer and it works, but it does not work with AoPS forums, unless you say import geometry_dev, which probably means the forum is still running on an older version of Asymptote.

Thank you very much for responding! I appreciate it.


In every square inch, there are 240 1-inch threads. This is equal to 20 feet. With 20 yards of thread, 3 square inches of fabric can be woven.

[hide=Solution] 20 yards is $20 \cdot 36$ inches. Now, each square inch uses up $24 \cdot 10$ so dividing gives $2 \cdot \frac{9}{6}=3.$


I think it would be a good idea if there was some system like what there is at another website. If you have new private messages, when you visit the website a popup window will appear saying you have new private messages. Because I keep forgetting to check my PM place thingy.

Actually.. now that I think about it.. didn't this site have it already? I'm sure it had it already. Wheres it gone? Is it just me?

its just you. when i login, theres a window that pops up saying that i have a new message. i believe that this may have been an option and you might have turned it off by accident.

bleu is right - check out your profile; the option should be there.

actually, it didnt work for me and i had the option turned on. but then i realized that my popup blocker was on, so just turn that off for this site at least

Hmm...some poppers can't be turned off for one site though. Like mine for example.

[quote=TethysTide]Is that feature really that old?[/quote]

dont bump posts especially ones from 2003 just leave it alone
someone lock this pls

[quote=Rabbit47][quote=physicskiddo]They did this to prove someone wrong on an MSM forum haha[/quote]

Honestly, feel like people should leave these forums alone. Don't bump posts from over 10 years ago that are basically completely irrelevant to the present.[/quote]

I completely agree, I just wanted to give context


calculus
integration
trigonometry
real analysis
real analysis unsolved

I am having problems solving this integral. Actually i am not in touch with calculus since some time so i forgot the basics too.  :(

Can anyone please solve this one with methods. I need it very urgently  :maybe:

[url=http://imageshack.us][img]http://img169.imageshack.us/img169/626/untitledbv3.jpg[/img][/url]

Cheers,
Dipen

The integral in Problem #10, $\int_{0}^{\infty}\frac{e^{-x}\sin(x)}{x}dx$, located here is similar.  I'm sure one can apply similar methods to the integral you gave.

http://web.mit.edu/hmmt/www/datafiles/solutions/2007/scalc07.pdf

It's not going to come from finding an antiderivative.

Here's what I'd do: Note that $\frac{\sin ax}{x}=\int_{0}^{1}\cos(axy)\,dy$. The integral becomes
$\int_{0}^{\infty}\int_{0}^{1}e^{-3x}\cos(axy)\,dy\,dx$
Switch the order (justified by absolute convergence):
$\int_{0}^{1}\int_{0}^{\infty}e^{-3x}\cos(axy)\,dx\,dy= \int_{0}^{1}\frac{3}{9+a^{2}y^{2}}\,dy=\frac1a\arctan\left(\frac{a}{3}\right)$


if $ \ T: {\mathbb{R}}^{n_1} \times ... \times {\mathbb{R}}^ {n_k} \to {\mathbb{R}}^m$ be multilinear(k_linear) then for every $ \   (a_1,...,a_k) \in {\mathbb{R}}^{n_1} \times ... \times {\mathbb{R}}^{n_k}$ we have :
$ DT_a(h_1,...,h_k) \equal{}T(h_1,a_2,...,a_k)\plus{}T(a_1,h_2,a_3,...,a_K) \plus{}...\plus{} T(a_1,...,a_{k\minus{}1},h_k)$

Try to show that if two $ a_i$ are replaced by $ h_i$, say $ T(h_1,h_2,a_3,..,.a_k)$, then $ |T(h_1,h_2,a_3,..,.a_k)| \leq C|(h,k)|$ for some constant $ C$. Then you can work from definition to show the differential.


logarithms
number theory
least common multiple
number theory unsolved

prove that, for n>2,  the least common multiple of 1,...,n is greater than 2^{n-1}.

Sorry for my ignorance, but what does the $o$ symbol mean? ie what is $o(1)$?


As  a begginer, i  have a  question (again). Can anybody please  tell me  some  details  about  the command \includegraphics*[viewport=100 200 500 650]{my image}?

I would like to know  what  are  the  numbers  after  viewport, and  why  when i  change them  the figure  dissapears, or appears  the half of it, or appears in  different page.
What are  the  ''best''  numbers  to have  the figure in the center (under the problem)  , or beside  a problem ?

Thanks  again.

Don't use viewport. Rather use [height=1cm, width=1cm] or just one of them (the other length will be proportional).

you can also give the dimensions according to your text

[width=0.5\textwidth]

I always use this method


At the start of the game there are $ r$ red and $ g$ green pieces/stones on the table. Hojoo and Kestutis make moves in turn. Hojoo starts. The person due to make a move, chooses a colour and removes $ k$ pieces of this colour. The number $ k$ has to be a divisor of the current number of stones of the other colour. The person removing the last piece wins. Who can force the victory?

[hide]Let $r=2^ki$ and $g=2^lj$, where $i,j$ are odd. Let's call $2^k$ and $2^l$ [i]even parts[/i] of $r$ and $g$. If the even parts are equal, the second player can force the victory. Otherwise, the first can win. Of course, if $r=0$ or $g=0$ (not both), the first wins. The strategy follows:

If w.l.o.g $k>l$, the player in this turn can change $(2^ki,2^lj)$ for $(2^ki-2^l,2^lj)$, delivering a position with equal even parts.

Now assume $k=l$: then, for any $d=2^mt$ dividing $g$ (clearly $m \le k$), we make $(2^ki-2^mt,2^lj)$, and these two numbers necessarily have different even parts: if $m=k$, the first coordinate is $2^k(i-t)$, a multiple of $2^{k+1}$ (as $i$ and $t$ are both odd), and if $m < k$, we write it as $2^m(2^{k-m}i-t)$, which has even part less than $2^k$.

The number of stones decreases at every turn, and the winning player (who always delivers equal even parts) never delivers zero. So, eventually he will receive a zero, remove the remaining stones and win.[/hide]


pe bune ca mi`a placut la nebunie problema asta :) chiar merita incercata, este superba
$ABCD$ este un tetraedru regulat, $M$ si $N$ sunt puncte diferite in planele $(ABC)$ si $(ADC)$. sa se arate ca segmentele $MN$, $BN$ si $MD$ sunt laturile unui triunghi.

cine a propus problema asta ? stie cineva ? pe bune ca o problema asa de frumoasa nu am intalnit de mult (acum poate sunt eu subiectiv - dar asta e parerea mea)

eu am facut-o cu proiectii, dar ,sincer, nu prea imi place solutia asta. are altcineva o solutie interesanta?

eu cred ca am gresit la ea, ca mi`a iesit destul de scurt
sunt curios cum ai facut cu proiectii, scrie rezolvarea :)

exista o solutie absolut remarcabila la problema asta :D

mai m-am gandit la o solutie:
deosebim doua cazuri:
1.cand fie BN sau DM este latura minima(aici am demonstrat).
2.MN este latura minima (aici mai am de lucru).
totusi nu sunt sigur ca voi ajunge undeva cu idee asta.

la solutia cu proiectii nu postez inca rezolvarea. vreau mai intai sa vad o solutie cu adevarat frumoasa si apoi va voi arata solutia mea.

[quote="maky"]eu cred ca am gresit la ea, ca mi`a iesit destul de scurt
sunt curios cum ai facut cu proiectii, scrie rezolvarea :)

[b][color=blue]exista o solutie absolut remarcabila la problema asta [/color][/b]:D[/quote]

eu am o carte IMO Compendium, pe care majoritatea cred ca o stiti. te referi cumva la solutia aceea sau ai gasit alta mai frumoasa?

la aia ma refer :)

eu am zis asa
[hide="solutia mea probabil gresita"]
pentru inceput e clar ca $MN<BN+DM$ ca alea sunt mai mari decat inaltimea si $MN$ e mai mica decat latura.
acuma hai sa presupun ca $MN+BN\leq DM$. atunci am ca $BN\leq DM-MN\leq DN \ \ (\star)$, cu egalitate doar daca $M\in[DN]$.
fie acum $X$ mijlocul lui $[BD]$. $(ACX)$ este planul mediator al segmentului $[BD]$, si fie acum $P$ intersectia lui $BN$ cu planul $(ACX)$. am ca $BN=BP+PN=DP+PN$.
insa $BN\leq DN$, deci $DP+PN\leq DN$, insa inegalitatea e invers in mod normal, deci $DP+PN=DN$, adica $P\in[DN]$, si am egalitate in $(\star)$.
de aici se deduce usor ca $M\equiv N$, contradictie.
deci $MN+BN>DM$, si analog $MN+DM>BN$, deci sunt laturile unui triunghi.
[/hide]

[hide="ideea mea"][b]Cazul I[/b]:
$MN$ este cea mai mare dintre laturi.
[u]Concluzia[/u]:$MN<BM+DN$
[u]Demonstratie[/u]: Proiectam punctele $N$ si $M$ pe $(BCD)$ in $N'$ si $M'$. Exprimam lungimile segmentelor $BM$, $DN$ si $MN$ in functie de proiectiile segmentelor pe $(BCD)$, precum si de $MM'$ si $NN'$. Concluzia rezulta imediat.

[b]Cazul II[/b]:
$DN$ este cea mai mare dintre laturi ( cazul cu $BM$ se trateaza la fel)
[u]Concluzia[/u]:$DN<BM+MN$
[u]Demonstratia[/u]: $BM\geq h$ (1), unde $h$ este inaltimea tetraedrului. Proiectam $D$ pe $(ABC)$ in $N'$. Valoarea maxima a lui $NN'$ este $a^{2}/3$.
Exprimam $DN$ in functie de $DN'$ si de $NN'$. Dupa calcule si folosindu-ne de (1) se ajunge la
$a<MN+(a\sqrt{6})/2$, propozitie evident adevarata.[/hide]

Hai sa mai punem inca o dimensiune  :D. Fie $E$ astfel incat $ABCDE$ este simplexul din ${\mathbf R}^{4}$ (adica toate distantele dintre puncte sunt egale - cum e triunghiul echilateral in plan si tetraedrul regulat in spatiu). Atunci, din simetrie, $BN=NE$ si $MD=ME$, deci segmentele respective sunt lungimile laturilor triunghiului $MNE$.


geometry
integration
calculus
calculus computations

The old City of London can be approximated by a circle of radius 1 mile with its center at St. Paul's cathedral. Moving away from the center along a radius, the population density can by approximated by $ p(x) = \ 5000*{e}^{\-0.25x}$ people per square mile. Find the exact population of the city.

Consider a small segment of the city of width $ dx$ at a distance $ x$ from the center of the city.Thus,the population of this segment will be
$ dP=p(x)$*Area of the segment

$ dP=\ 5000*{e}^{\-0.25x}*2{\pi}xdx$

Thus the total population P will be sum of the small populations $ dP$

$ P=\int_{0}^{1}5000*{e}^{\-0.25x}*2{\pi}xdx$

Solve for P.


$ x_1 ,x_2,...,x_n \in R^{\plus{}}$ dar rabete $ \sum \frac{1}{1\plus{}x_i} \equal{}1$ sedgh mikonand . sabet konid
$ \sum \sqrt{x_i} \geq (n\minus{}1) \sum \frac{1}{\sqrt x_i}$

 :)

This is [color=blue][b]"Vojtech Jarnik", 2002[/b][/color] inequality. You can found it:

$ \bullet$ [url=http://www.mathlinks.ro/viewtopic.php?p=1401950#1401950/]Vojtech Jarnik 2002 problem 3[/url]