geometry
inequalities
ARML
similar triangles
algebra
difference of squares
special factorizations

[color=cyan]1.  a/b, a*b, a+b, and a-b are all real numbers.  Find both ordered pairs, (a, b), such that three of those four numbers are equal.

2.  What is the largest integer which divides
(n^2)*(n^2 - 1)*(n^2 - 4) for all integers n?

3.  Two of the altitudes of a right triangle have lengths 4 and 5.  What is the largest possible length of the third altitude?[/color]

JBL wrote:1.  a/b, a*b, a+b, and a-b are all real numbers.  Find both ordered pairs, (a, b), such that three of those four numbers are equal.

2.  What is the largest integer which divides
(n^2)*(n^2 - 1)*(n^2 - 4) for all integers n?

3.  Two of the altitudes of a right triangle have lengths 4 and 5.  What is the largest possible length of the third altitude?

[hide]#1. Either the first pair of expressions are equal or the last pair of expressions are equal. The latter case gives b=0. Therefore, a=a=a*0 or a/0. The only a that will work is a=0, which gives (0,0). This makes the first expression undefined, but the last three are all equal. If the first pair of expressions are equal, then b=:pm:1. If b=1, then we have a=a=a:pm:1, which is false. If b=-1, when we have -a=-a=a:pm:1, or 2a=:pm:1, or a=:pm:1/2. This gives the two ordered pairs (1/2,-1) and (-1/2, -1). So...I find three ordered pairs. Unless you don't want me to count (0,0) since that would make one of the expressions undefined.

#2. This can be factored with difference of squares as n*(n-2)*(n-1)*n*(n+1)*(n+2). Clearly, 5 divides this (we have 5 consecutive factors). Also, 8 must divide it (we have to have one multiple of 4, and one other multiple of 2). 9 must divide it (either there is a 3 in each of two factors that are three apart, or there is a 3 in n, which is squared).  Multiplying these altogether gives that 360 must divide it. This is the largest integer that can divide it since if n=3, the expression is 360.

#3. In a right triangle, two of the altitudes are the length of the sides, while the last altitude is the altitude to the hypotenuse. This last altitude will have length ab/c, where a and b are the lengths of the legs of the triangle, and c is the length of the hypotenuse. This can be proven by similar triangles, or finding the area of the right triangle with ab/2=cx/2, where x is the length of the altitude. So, either:

1) 4 and 5 are the two legs, making the hypotenuse :sqrt:41, and the last altitude 4*5/:sqrt:41=20:sqrt:41/41.

2) 4 is one leg and 5 is the altitude, making c=4b/5. Obviously the hypotenuse cannot be shorter than one of the legs.

3) 5 is one leg and 4 is the altitude, making c=5b/4. Solving with Pythagoras yields b=20/3 as the length of the last altitude. This is longer than the previous length we found, and is thus the answer.[/hide]

[color=cyan]There are several problems with your number 1.  Try it again.[/color]

[hide] the altitude drawn to the hypotenuse in a triangle with legs a and b and hyp c is (ab/c). not to hard to derive. if the given altitudes are one side (lets say the side is 5) and the hyp alti, then 4=5a/c. a :^2: +25 = c :^2:. 4c=5a by the first thing. multiplying the second eq by 16 yields 16a :^2: +400= 25a :^2:, or 9a :^2: = 400, so a=20/3. this is one possibility for the last alti. if a is 5, then b also = 20/3. if the alti to the hypotenuse is 5 and a is 4, then we have 5=4b/c. b :^2: + 16 = c :^2:. 5c=4b. multiplying by 16, we get 25b :^2: +256=16c :^2:. negative answer, so no dope there! actually dont think the altitude to  the hyp can be greater than a side, because then hyp would be = to a side. anyways, 20/3. [/hide]

i know there is an inequality with [hide]altitude inverses[/hide], but i felt like taking the long route.

Correct answer -- you missed one possiblity, though -- [hide]the altitudes of length 4 and 5 could both be legs of the triangle.  Your assertation that the altitude to the hypotenuse cannot be greater than one of the legs is correct, because then the other leg would necessarily be greater in length than the hypotenuse, which we know to be false.[/hide]

i think i covered that joel. but maybe i didnt. but im sure i thought of that first.

[color=cyan]You may have thought of it, but you didn't write it up.[/color]

OK, vale (OK in Spanish), doesn't matter. It sounds too edgy...
I mean, it doesn't matter who thought up of the idea first. The important thing is that you were able to come up with that idea, not when.

[color=cyan]Does that mean anything?  I have no idea what you're talking about.[/color]

Joel, do you have the older ARML book (pre 1988-1994).  If so, could you post some of those questions sometime?

[color=cyan]I'm afraid I don't -- nearly all my high school math contest books were on loan from my high school and I had to give them back when I graduated.  I might at some time have an opportunity to lay my hands on one, or at least pieces of some of the contests, but not any time soon, unfortunately.[/color]

max: i dont think its a big deal really...

If anyone else has the ARML book i was talking about, could you post some sometime?  That would be great, thanks.

[color=cyan]well, I was sort of hoping other people would respond.  Oh, well.  Number 1 is correct, not including the solution (0, 0) since 0/0 is clearly not a real number.  Number 2 is very nicely done.  Number 3 is also correct.[/color]


Prove that:
X^2*y+y^2*z+z^2*x+x*y*z<=4/27
where x; y; z are nonnegative real numbers satisfying x+y+z=1

let me rephrase it for easy reference
prove that
$\sum_{cyclic}x^{2}y$ $+$ $xyz$ $\leq$ $\frac{4}{27}$
if $x$ $+$ $y$ $+$ $z$ $=$ $1$ and $x,y,z$ are non negatives

first homogenize it and then apply muirhead inequality to prove

$\sum_{cyclic}x^{2}y$ $+$ $xyz$ $\leq $ $\frac{4}{27}(x+y+z)^{3}$

Well it's cyclic so I don't know how you'd use Muirhead.

Looks like there are 2 equality cases, namely (1/3, 1/3 1,3) and (0, 1/3, 2/3) so it doesn't seem trivial. Stronger version of Canada '99 I see... but anyways using the same argument:

Let $f(x,y,z) = x^{2}y+y^{2}z+z^{2}x+xyz$ and WLOG $x$ is the largest of the three. Since $f(x,y,z)-f(x,z,y) = (x-y)(y-z)(x-z)$ and we want to maximize $f$, we might as well assume $y \geq z$. Then, we have $f(x+z, y, 0)-f(x,y,z) = z(x-y)(y-z) \geq 0$ conveniently so we may assume $z = 0$ and it's not hard to see that $x^{2}y \leq \frac{4}{27}$.

Putting this together, we get the desired inequality and the two equality cases (up to permutation) above...


because it's obviusly:
look:
$ \sum \equal{} \sum \frac {1}{2^{2n}(2n)!} \plus{} \frac {1}{2^{2n \plus{} 1}(2n \plus{} 1)!} \equal{} \sum \frac {1}{2^{n}n!} \equal{} e^{\frac {1}{2}}$(because we summation odd and even )
I use fact that :
$ e^{x} \equal{} \sum\frac{x^{n}}{n!}$ and instead $ x$ I put $ \frac {1}{2}$


function
floor function
trigonometry
algebra
polynomial
algebra solved
Functional Equations

First time posting, hope its right...  :blush:
First, we show f(0) = 0
Since f(f(0)) = f(0) + f(0)*f(0) and f(f(0)) = f(0) + f(x)*f(-x) + x^2
f(0)*f(0) = f(x)*f(-x) + x^2 for all real x   Let f(0) = c, so f(c) = c + c^2
c^2 = f(x)*f(-x) + x^2   let x = c, so c^2 = f(c)*f(-c) + c^2, so either f(c) = 0 or f(-c) = 0.
If f(-c) = 0, then f(f(-c)) = f(-c) + f(0)*f(-c)   so f(0) = 0
If f(c) = 0, c^2 +c = 0 so c = 0 or c = -1.
If c = -1, then f(f(x)) = f(x) + f(x)*f(0) = 0  So f(x) = 0 for all x in f's range
But then 0 must be f's range, since f(some element) = 0, so f(0) = 0, and we have a contradiction.
Since f(0) = 0
f(f(x)) = f(x)  but f(x) = x for all reals only in f's range
x+y = f(x+y) + f(x)*f(y) -xy  so if y = 0, f(x) = x, for all real x...

[quote="matamata"]x+y = f(x+y) + f(x)*f(y) -xy  so if y = 0, f(x) = x, for all real x...[/quote]

I'm sorry, but I don't understand how you get this line.

Since $f(f(x)) = f(x) \forall x$ we have $f(x)f(y) = xy$.  Now the range of $f$ cannot be $\{0\}$ only, hence take any $x \neq 0$ in the range of $f$.  Then we have $xy = f(x)f(y) = xf(y)$ and hence $f(x) = x \forall x$.

ack! I knew I'd do something stupid like that...
What you did is how I solved it on paper, but while typing it up, I thought that "f(f(x)) = f(x) implies f(f(x+y)) = x+y, it's even simpler!", sorry  :blush:

[quote="dudeldai"]functional equations are a new topic for me ... so i have a question

it's quite easy to proove that f has to be injective.
now we let f(x+y)=-f(x)*f(y)
and plugging in the initiial equation we get f(-f(x)*f(y))=-xy.
since f is injective -->f(x)*f(y)=x*y
and finally for x=y , f(x)^2=x^2 , f(x)=x.

is it possible ? please help me i'm tryn to get better in this stuff .
thx[/quote]
1) Show us your proof that $f(x)$ is injective. It's true but it does not seem so simple (I had to go up to the solution $f(x)=x$ to prove injectivity)

2) what is the meaning of "let $f(x+y)=-f(x)*f(y)$" ?
If you mean that $f(x+y)=-f(x)f(y)$ $\forall x,y$, then it's false
If you mean that you choose some specific values $x,y$ such that  $f(x+y)=-f(x)f(y)$, then :
2.1. You must we aware that you did not prove that such $x,y$ exist
2.2. You must be aware that your conclusions will only be available for these specific values of $x,y$ and will not be general conclusions

3) How can you conclude from $f(-f(x)f(y))=-xy$ that $f(x)f(y)=xy$ ????? This is absolutely not a conclusion of the [not-yet-proved] injectivity.
Injection allow you to conclude from $f(a)=f(b)$ that $a=b$. That's all.

4) If you succeed to prove that $f^2(x)=x^2$, you can not immediately conclude $f(x)=x$. The only thing you could then conclude is that :
$\forall x$, either $f(x)=x$, either $f(x)=-x$
For example, the function $f(x)=(-1)^{\lfloor 32\sin x\rfloor}x$ is such that $f^2(x)=x^2$ $\forall x$

thx for helping me! i still have a question.
So in point 2) you want to say that if i want to substitute a value for f
for example in my case f(x+y)=f(x)f(y)(i see now that it is totaly wrong) i have to show that it is true for
every x,y  to make general conlcusions and finally find all possible functions or can i even find some specific
x,y  to find general conclusions?I mean when we try to find out more about a function most of the time we first try to make specific conclusions like f(1) and f(0) ecc.. but with these we can still make genral conclusions in the end..so my question : if i find an equation like f(x+y)=f(x)f(y) and i prove it for  some specific x,y .. plugging it in the initial  equation can i make general conclusions ???

i made some strange conclusions yesterday i was really tired .. sorry

If you choose $x,y$ with a given constraint, then your conclusion is valid only for those $x,y$.

[quote="hxtung"]Find all function $f:R \rightarrow R$ such that:

$f(f(x+y))=f(x+y)+f(x)f(y)-xy;\forall x,y \in R$.[/quote]
Here is my solution: Replacing $y=0$ we get $f(f(x))=f(x)[1+f(0)]$ So for all $t \in Range$ we have $f(t)=ct$ where $c=1+f(0).$
The original equation forces c to be 1. So for all $t \in Range$ we have $f(t)=t.$
Now the original equation becomes
$f(x+y)=f(x+y)+f(x)f(y)-xy \Longrightarrow f(x)f(y)=xy$ fixing y for a value s.t $f(y) \neq 0$ (we have such a point since $f \equiv 0$ is not a solution) we get $f(x)=ax$ and original equation gives $a=1.$
So $f(x)=x$ for all $x \in \mathbb{R}.$

[quote="ozgurkircak"]Here is my solution: Replacing $y=0$ we get $f(f(x))=f(x)[1+f(0)]$ So for all $t \in Range$ we have $f(t)=ct$ where $c=1+f(0).$
The original equation forces c to be 1. [/quote]
Why this last sentence ?

Pluging $f(x)=cx$ $\forall x\in f(\mathbb R)$ in original equation, we get $(c-1)f(x+y)=f(x)f(y)-xy$ $\forall x,y\in\mathbb R$ and this does not allow to conclude $c=1$

Replacing $x,y$ by $f(x),f(y)$ in the above equation, we get $(c-1)f(f(x)+f(y))=(c^2-1)f(x)f(y)$ but this still does not allow to conclude $c=1$

thank you very much for your remark.
I now see that I made a mistake by taking granted if $a, b \in Range$ then $a+b \in Range$ which I didn't prove. And by considering the equation like a polynomial equation in $f(x)$ and $f(y).$
To complete my solution I need to prove two things:
$a,b \in Range \Longrightarrow a+b \in Range$ also there are an infinite number of elements in the Range.

A different approach:

Let $P(x,y)$ be the assertion.
$P(0,0) \implies f(f(0))=f(0)+f(0)^2$
$P(f(0),-f(0)) \implies f(f(0))=f(0)+f(f(0))f(-f(0))+f(0)^2$
Subtracting the first equation from the second gives:
$f(f(0))f(-f(0))=0$

Case $1$: $f(-f(0))=0$:
$P(0,-f(0)) \implies f(0)=0$

Case $2$: $f(f(0))=0$:
$P(0,f(0)) \implies f(0)=0$

Now as we have in general that $f(0)=0$, we take:
$P(x,0) \implies f(f(x))=f(x) \forall x$, which makes the original equation:
$f(x)f(y)=xy$

Let $y=x$, to get:
$f(x)^2=x^2$, so $\forall x$, either $f(x)=x$ or $f(x)=-x$.

Let $f(c)=-c$ for some $c$, and take:
$P(c,0) \implies c=-c$, so $c=0$

Therefore, $f(x)=x$

I know i'm reviving an old  thread but this is my first post , please proceed to look at this

plugging x,-x we get f(x)f(-x)=f(0)^2 -x^2
f(x)f(-x)=(f(0)-x)(f(0)+x)
so one of the solutions if f(x)=f(0)-x and f(-x)=f(0)+x , clearly f(x)=x satisfies this so it is a solution

is this considered one of the "solutions" and if this is an exercise worth 5 points how much points would i get with this trick?
thanks in advance

\[P(x;o) \Rightarrow f(f(x))=f(x)(c+1)  \text{here c=f(o)assume that} c\neq o then from original equation we have: f(x+y)c=f(x)f(y)-xy (1)  in(1) P(c;-c) \Rightarrow f(a)=o here a={c,-c} in (1) P(x;a) \Rightarrow f(x+a)=-frac{a}{c}x 1) case a=c \Rightarrow f(x)=-x-a here x\toa -2a=f(a)=o contradiction to our assume 2)case a=-c \Rightarrow f(x)=x-a checking it in original equation we get a=o this is also contradiction to our assume.So we have c=o from(1) we get f(x)f(y)=xy y\tob that f(b)%Error. "neqo" is a bad command.
\Rightarrow  f(x)=kx while checking we get k=1 So the solution is f(x)=x .\]


geometry
circumcircle
incenter
ratio
symmetry
geometric transformation

Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.

Let $ M_b$ be midpoint of $ AC$.
We have $ \angle FAM_b = \angle FBC$, $ \angle AM_bF\equiv \angle DM_bF = \angle DEF = \angle BEF = \angle BCF$
$ \Rightarrow \triangle FAM_b\sim \triangle FBC$, $ (i)$. Thus, through the [b]spiral similarity[/b] with center $ F$, denote this $ f$ then:
$ f: \triangle FAM_b\mapsto \triangle FBC$
$ \Rightarrow f: A\mapsto B$, $ M_b\mapsto C$
$ \Rightarrow f: A\mapsto M_b$, $ B\mapsto C$
$ \Rightarrow f: \triangle FAB\mapsto \triangle FM_bC$, $ (1)$
But it is easy to notice that: $ \triangle BAF\sim \triangle BM_bC$ (a.a), $ (2)$
$ (1),(2)\Rightarrow \triangle CM_bB\sim \triangle FM_bC$
$ \Rightarrow \frac {CB}{CF} = \frac {CM_b}{FM_b} = \frac {M_bB}{M_bC}$, $ (*)$.
With the same argument, we get $ \triangle BAM_b\sim \triangle BFC$
$ \Rightarrow \triangle BAM_b\sim \triangle AFM_b$ (thanks to $ (i)$).
$ \Rightarrow \frac {BA}{AF} = \frac {M_bB}{M_bA}$, $ (**)$
But $ M_bA = M_bC \Rightarrow \frac {BA}{BC} = \frac {FA}{FC}$ (Combine $ (*)\&(**)$)
$ \Longrightarrow ABCF$ is a harmonic quadrilateral. Therefore, $ BF$ passes through the Pole of $ AC$ wrt $ (O)$, where $ O$ is the circumcenter of $ \triangle ABC$. Then we will get the result of the problem.
Our proof is completed then :lol:

Let $ M$ the midpoint of $ AC$ , $ O$ the circumcenter of $ \Omega$ and let the tangent to $ \Omega$ at $ B$ meets $ AC$ at $ R$ .
Then $ BOMR$ is cyclic (let $ Q$ its circumcenter) . Then $ R,Q,O$ are collinear . So if the circumcircle of $ BOMR$ meets $ \Omega$ at $ F'$ , then $ BF'\bot RO$ . So by polars $ RF'$ is tangent to $ \Omega$ . By well known theorem we have that $ BF'$ is the $ B$ symmedian. Finally if $ BE$ meets the circumcircle of $ BOMR$ at $ S$ , then $ SM\equal{}SF$ . But also because $ RF'\equal{}RB$ we have that $ D$ is the incenter of $ F'BM$ so $ SD\equal{}SM\equal{}SF$ and because $ DM\bot ME$ we also have $ SD\equal{}SM\equal{}SF\equal{}SE$ . This shows that $ MDF'E$ is cyclic and so $ F\equiv F'$ and we are done .

External bisector of $ \angle B$ cuts $ CA$ at $ Y.$ $ I, I_b$ are incenter and B-excenter of $ \triangle ABC.$ $ (E)$ is a circle with center $ E$ and radius $ EA \equal{} EC \equal{} EI \equal{} EI_b.$ The cross ratio $ (B, D, I, I_b)$ is harmonic and $ E$ the midpoint of $ II_b$ $ \Longrightarrow$ $ \overline{BD} \cdot \overline{BE} \equal{} \overline{BI}\cdot \overline{BI_b}$ $ \Longrightarrow$ $ BY \perp BDE$ is radical axis of $ \omega, (E).$ Pairwise radical axes $ EF, CA, BY$ of $ \Omega, \omega, (E)$ meet at their radical center $ Y.$ Since $ DF \perp EFY$ and $ BY \perp BDE,$ $ BDFY$ is cyclic. Its circumcircle is the B-Apollonius circle of $ \triangle ABC$ $ \Longrightarrow$ $ BF$ is the B-symmedian.

[color=darkred][b][u]PP[/u].[/b] Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$ . The line $ BD$ intersects the circumcircle $ \Omega$ of
$\triangle ABC$ at $ B$ and $ E$ . Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$ .[/color]

[color=darkred][b][u]Lemma[/u].[/b] Let $ ABC$ be a triangle with the circumcircle $ w$ . Denote the following points : the middlepoint $ M$ of the side $ [BC]$ ;

the point $ D\in (BC)$ for which $ \angle DAB\equiv\angle DAC$ ; the point $ E$ for which $ \{A,E\} \equal{} AD\cap w$ ;  the $ A$-symmedian $ AL$ ,

where $ L\in (BC)$ ; the point $ S$ for which $ \{A,S\} \equal{} AL\cap w$ . Then $ SD\perp SE$ , i.e. the quadrilateral $ MDSE$ is cyclically.[/color]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ XX$ - the tangent in $ X\in w$ to $ w$ .  Denote $ T\in BB\cap CC$ . Observe that $ \overline {MET}\perp BC$ and $ \angle ASB\equiv\angle ACB$ .

From the [u]first well-known property[/u], $ T\in \overline {ALS}$ and the line  $ \overline{ALST}$ is the $ A$-s[u]ymmedian[/u] in $ \triangle ABC$ $ \Longrightarrow$ $ \left\|\begin{array}{c} \angle BAS\equiv\angle MAC \\
\ \angle SAE\equiv\angle DAM\end{array}\right\|$ .

From the [u]second well-known property[/u], the division $ \{A,S;L,T\}$ is [u]harmonically[/u]. Since $ ML\perp MT$ results $ \angle DMA\equiv\angle DMS$ .

Show easily that $ \triangle BAS\sim\triangle MAC$ (a.a.) . Thus, $ \frac {AB}{AS} \equal{} \frac {AM}{AC}$ , i.e. $ AS\cdot AM \equal{} AB\cdot AC$ .

From the [u]third well-known property[/u] $ AD\cdot AE \equal{} AB\cdot AC$, obtain $ \boxed {AS\cdot AM \equal{} AD\cdot AE \equal{} AB\cdot AC}$ , i.e. $ \frac {AS}{AE} \equal{} \frac {AD}{AM}$ .

Since $ \angle SAE\equiv\angle DAM$ obtain $ \triangle SAE\sim\triangle DAM$ and in consequence, $ \angle DMA\equiv\angle SEA$ . Since $ \angle DMA\equiv\angle DMS$

and $ \angle SEA\equiv\angle SED$ obtain  $ \angle DMS\equiv\angle SED$ , i.e. the quadrilateral $ MDSE$ is cyclically. In conclusion, $ SD\perp SE$ .

[b][u]Remark[/u].[/b] Here are another interesting metrical relations : $ \left\|\begin{array}{ccc} ABS\sim CMS & \implies & \boxed {SB\cdot SC \equal{} SA\cdot SM} \\
 \\
AMC\sim CMS & \implies & \boxed {MA\cdot MS \equal{} MB^2}\end{array}\right\|$ .

The [u]proposed problem[/u] is an immediate consequence of the above lemma.[/color] [u]See [b]PP13[/b][/u] from [url=http://www.artofproblemsolving.com/blog/101045][color=red][b]here[/b][/color][/url]

Very nice application of Reim's Theorem jayme.Though a very obvious result,  it has great application in many synthetic geometry results.
Thanks.

[quote="jayme"]Dear Mathlinkers,
1. Let A' be the midpoint of BC, X, Y the second meetpoint of AA', EA' with the circumcircle of ABC and Te the tangent to this circumcircle at E.
2. Te // DA'
3. The circle with diameter ED passes through A'
4. By a converse of Reim's theorem applied to this circle and the circumcircle, FD goes through Y;
5. By a converse of Pascal's theorem applied to EAXFYTe, FX //DA'  (BC)
and we are done...
Sincerely
Jean-Louis[/quote]
Do you have any document about [b]Reim's theorem[/b], [b]Jayme[/b]? I am so excited about this. :blush:

See for example my website
http://perso.orange.fr/jl.ayme   then: A propos
You will see all the possibility.
Sincerely
Jean-Louis

Dear friends.
I have a short solution for the nice problem,here :) .
[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled1-4.jpg[/img]

Denote by $ A'$ the midpoint of segment $ AC$.

$ EO$ meets $ (O)$ again at $ P$

By ${ \angle {DFE} = 90^0}$ ,we have $ F,D,P$ are collinear.

From $ BPEF$ is cyclic and $ BDA'P$ is cylic ,we have :

$ \angle {EBF} = \angle {EPF} = \angle{DBA'}$.

Therefor,$ BF$is the symmedian line of the triangle $ BAC$.

All are very nice solutions indeed, I must  admit.

   T.Y.
   M.T.

After the [b]Ma29[/b]'s [u]shortest proof[/u], this problem seems very simple ...
In the another proofs appeared many nice properties of this configuration.
[b]Thanks to all[/b] for their interest and proofs.

Hello my friends.
I think all proof above are very nice. :)
Here is another solution to this problem:
[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-108.jpg[/img]

Denote $ V \equal{} EF \cap AC$.

We have :$ \bar {VD} .\bar{VA'} \equal{} \bar{VF}.\bar{VE} \equal{} \bar{VA}.\bar{VC}$  (1)

But $ A'$ is the midpoint of segment $ AC$.(2)

From (1) ,(2) and [b]Maclaurin's theorem [/b], we obtain the division $ {VDAC}$ is harmonically,so $ V,B,P$ are collinear.

Denote $ Q \equal{} BC\cap {PA}$ and denote by $ H$ the intersection of two tangent to $ (O)$ at $ A,C$ ,respectively.

We have $ Q,D,H$ lie on the polar of $ V$ wrt $ (O)$,i.e $ Q,D,H$ are collinear.

By [b]Pascal's theorem[/b] applied to six point $ A,A,P,C,B,F$ ,we obtain $ B,F,H$ are collinear.

Therefore ,$ BF$ is the symedian line of the triangle  $ ABC$.

It can also be viewed as an easy application of : http://www.mathlinks.ro/viewtopic.php?p=1082899#1082899

Indeed, from the above problem it follows that the line $ FM$ (where $ M$ is the midpoint of $ AC$) meets the circumcircle of $ \triangle{ABC}$ at a point $ T$ such that $ BT\parallel{}AC$. By symmetry,
\[ \angle{MBC} \equal{} \angle{MTA} \equal{} \angle{ABF}.
\]

We can use my found property at the proposed problem from [url=http://www.mathlinks.ro/viewtopic.php?t=275869][color=red][b]here[/b][/color][/url].

We will solve this problem in the $A$-indexing. Let $K$ be the symmedian point, $I$ be the incenter, and $M$ be the midpoint of $BC$. It suffices to show that $A,K,F$ are colinear. We will do so with phantom points.

First, note that $M$ lies on $\omega$ since $\measuredangle DME=90^\circ$ as $M,E$ lie on the perpendicular bisector of $BC$. Define $F'=AK\cap\omega$. We will show that $F'DME$ is cyclic, implying $F'=F$ which finishes the problem.

We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Since $F'$ is on the $A$-symmedian, it can be parameterized by $F'=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (F')=0$. Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$, so plugging in $F'$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$. So $$F'=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$$

Of course, $M=(0:1:1)$. By the angle bisector theorem, $D=(0:b:c),I=(a:b:c)=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right),I_a=(-a:b:c)=\left(\frac{-a}{-a+b+c},\frac{b}{-a+b+c},\frac{c}{-a+b+c}\right)$. By Fact 5, $E$ is the midpoint of $II_a$ so $E=\frac{I+I_a}2=\frac12\cdot\frac{1}{(a+b+c)(-a+b+c)}[(-a^2+ab+ac,-ab+b^2+bc,-ac+bc+c^2)+(-a^2-ab-ac,ab+b^2+bc,ac+bc+c^2)]=(-a^2:b^2+bc:c^2+bc)$. Plugging in $D,M,E$ into the general circle equation, $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)$, gives

[list]
[*]$M:-a^2+2(v+w)=0$
[*]$D:-a^2bc+(b+c)(bv+cw)=0$
[*]$E:-a^2bc(b+c)^2+a^2b^2(c^2+bc)+a^2c^2(b^2+bc)+(-a^2u+(b^2+bc)v+(c^2+bc)w)(-a^2+(b+c)^2)=0$
[/list]

We can solve the first two equations with substitution (or Cramer's Rule) to get $v=\frac{a^2c}{2(b+c)}$ and $w=\frac{a^2b}{2(b+c)}$. Substituting these values into the third equation, we get

\begin{align*}-a^2bc(b+c)^2+a^2bc(b+c)^2+(-a^2u+a^2bc)(-a^2+(b+c)^2)&=0\\
a^2(-u+bc)(-a^2+(b+c)^2)&=0\\
u&=bc.\end{align*}

Hence, $\omega$ has equation $$-a^2yz-b^2zx-c^2xy+[(bc)x+\left(\frac{a^2c}{2(b+c)}\right)y+\left(\frac{a^2b}{2(b+c)}\right)z](x+y+z)=0.$$ To see if $F'\in\omega$, we must check that $\text{Pow}_\omega (F')=0$ but $$-a^2(2b^2)(2c^2)-b^2(2c^2)(-a^2)-c^2(-a^2)(2b^2)+[(bc)(-a^2)+\left(\frac{a^2c}{2(b+c)}\right)(2b^2)+\left(\frac{a^2b}{2(b+c)}\right)(2c^2)](-a^2+2b^2+2c^2)=0\Rightarrow 0+[-a^2bc+a^2bc](-a^2+2b^2+2c^2)=0$$ which is clearly true. So $F'=F$, finishing.

Let $T$ be midpoint of arc $AC$ witch contains $B$.
Claim: $EM$ and $FD$ meet at $T$.
Proof : It's obvious that $E,M,T$ are collinear. $\angle DFE = \angle 90 = \angle TFE$ so $F,D,T$ are collinear as well.

Claim: $TBDM$ is cyclic.
Proof : $\angle TMD = 90 = \angle TBE = \angle TBD$.

Claim: $ABM$ and $FBC$ are similar.
Proof : $\angle BAM = \angle BFC$ and $\angle BCF = \angle BTF = \angle BTD = \angle BMA$.

Now we have $\angle ABM = \angle CBF$ so $BF$ is symmedian.

[hide=Headsolved while i was having a bath] [quote=Reworded problem]Let be given a triangle $ ABC$ and its internal angle bisector $ AD$ $ (D\in BC)$. The line $ AD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ A$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ AF$ is the symmedian line of triangle $ ABC$.[/quote]

Let $M$,$M_A$ denote the midpoint of $BC,\widehat{BAC}$ respectively.
Clearly $M_AADM$ is cyclic and $M \in \odot(DE)$.By radical axis theorem on $\odot(M_AADM),\odot(DE)$ and $\odot(\triangle ABC)$ we get that $M_AA,BC,EF$ are concurrent.
Finally,
$$-1=(E,M_A;B,C)\stackrel{A}=(D,AM_A\cap BC;B,C)\stackrel{E}=(A,F;B,C)\text{ }\blacksquare$$[/hide]
[hide=Headsolved while reading Phy][quote=Reworded problem]Let be given a triangle $ ABC$ and its internal angle bisector $ AD$ $ (D\in BC)$. The line $ AD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ A$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ AF$ is the symmedian line of triangle $ ABC$.[/quote]
Say $M$ be the midpoint of $BC$,so $M \in \odot(DE)$.Say $MF \cap \odot(\triangle ABC)=A'$.By Reim we have $AA' || BC$.Finally,
$$-1=(P_{\infty},M;B,C)\stackrel{A'}=(A,F;B,C)\text{ }\blacksquare$$[/hide]

Denote the circle with diameter $DE$ as $\omega$. Perform a $\sqrt{bc}$ inversion around $A$ and a reflection across the A-bisector. For any point $S$, let $S^{\ast}$ denote its image under these transformations. Then, $\Omega$ becomes $B^{\ast}C^{\ast}$ which is also $BC$. Line $AD$ remains unchanged. $D$ was originally the intersection of $AD$ and $BC$, so now it is the intersection of $AD$ and $\Omega$, which is $E$, so $D^{\ast}=E$. Similarly, $E^{\ast}=D$.\\

Claim: $\omega^{\ast}=\omega$. Note that $\omega$ passed through $D$ and $E$. Since $D$ and $E$ inverted to each other, $\omega^{\ast}$ also passes through $D$ and $E$. However, since the centers of $\omega$ and $\omega^{\ast}$ are collinear with $A$, this forces $\omega^{\ast}$ to be the same circle.\\

Now, consider $F$. Originally, it was the intersection of $\Omega$ and $\omega$ other than $E$, so $F^{\ast}$ is the intersection of $\Omega^{\ast}$ (which is $B^{\ast}C^{\ast}$) and $\omega^{\ast}$ (which is still $\omega$) other than $E^{\ast}$. Now, let $M$ be the midpoint of $BC$. Then, $\angle DME=90$, so $M$ lies on $\omega$. Thus, $F^{\ast}=M$ since it is the second intersection, so $AF^{\ast}$ is a median and thus $AF$ is a symmedian, as desired.

switch the center to $A$
[hide=solution]
We invert around $A$ with radius $\sqrt{AB\cdot AC}$.
Let $M$ be the intersection of $\omega$ with $BC$. Because $DE$ is the diameter of $\omega$, then $EM\perp BC$, so $M$ is the midpoint of $BC$.
[list]
[*] $(\triangle ABC)$ and $BC$ invert to each other.
[*] $D$ and $E$ invert to each other.
[*] $\omega$ inverts to itself (by PoP).
[/list]
Hence $M$ inverts to $F$, so $AF$ inverts to $AM$. A symmedian and $AD$ are inverses, so $AF$ is a symmedian.

Let $\omega \cap \Omega=T,\omega\cap AC=K,DT\cap \Omega = R$
Note that $RT\bot AC,RB\bot BT\Rightarrow \angle DRU=\angle DBU=\angle ERT$
Consider $AC\bot KE\Rightarrow K$is the mid-point of $AC$
Hence $KD=UD\Rightarrow BF$ is the symmedian line of $\triangle ABC$

B indexed geo wtf. Switch to A indexed geo.

Let $M$ be the midpoint of $\overline{BC}$, so $DEFM$ is cyclic. Since a $\sqrt{bc}$ inversion swaps $\overline{BC}$ and $(ABC)$, it also swaps $D$ and $E$, since both lie on the $A$-angle bisector. Therefore, since $\overline{DE}$ is a diameter of $\omega$, $\omega$ is fixed under inversion. Furthermore, $M$ gets swapped with the point $F' \in (ABC)$ such that $\overline{AF}$ is a symmedian, since lines $\overline{AM}$ and $\overline{AF}$ are preserved under inversion and are reflections of each other about the $A$-angle bisector. On the other hand, since $\omega=(DEM)$ is fixed, we have $(DEM)=(DEF')$, implying $DEF'M$ cyclic. Since $F'$ and $F$ are both distinct from $E$ and lie on both $(ABC)$ and $(DEM)$, it follows that $F=F'$, i.e. $\overline{AF}$ is a symmedian. $\blacksquare$

Note that $F$ maps to the midpoint of $AC$ under $\sqrt{ac}$ inversion at $B$ and reflection over the $B-$ angle bisector.

$\sqrt{bc}$-inversion swaps $D, E$, thus $(DE)$ is orthogonal to the circle of inversion. It follows that the midpoint $M$ of $\overline{BC}$ and $F$ are swapped, yielding the desired conclusion.

Take a $\sqrt{bc}$ inversion. Since this swaps $(ABC)$ and $BC,$ we have $D,E$ are swapped, so the circle with diameter $DE$ is fixed, and $F$ gets sent to the foot from $E$ to $BC,$ which is the midpoint of $BC,$ so $AF$ is a symmedian.

We consider an inversion around $A$ with radius $\sqrt{bc}$ followed by a reflection around the line $AE$.

We can see that athis sends $B$ and $C$ themselves and same with $D$ and $E$. The circumcircle now becomes the line $BC$ and the circle with diameter $DE$ does not move. Thus, it remains to prove that the second intersection of the circle with diameter $DE$ intersects $BC$ again at the midpoint of $BC$. Thus is obvious by considering the line from the circumcenter, $O$, to $E$ must have that $OE \perp DC$ and obviously is the midpoint. $\blacksquare$

Let $M$ be the midpoint of $BC$ and $N$ be the top point of $(ABC)$. Then
[list]
[*] $\angle EFD = \angle EFN = 90$, so $NDF$ collinear.
[*] $\angle NAE = \angle DMN$, so $NADM$ is cyclic.
[/list]
Thus $AF$ is a symmedian, as
\[\measuredangle MAD = \measuredangle MND = \measuredangle DAF. \quad \blacksquare\]

We solved the $A$-centered version of this problem.

Perform a $\sqrt{bc}$-inversion. Then note that $D\leftrightarrow E$ which means that $\odot(DE)$ is fixed under this inversion. Now $BC \leftrightarrow \odot(ABC)$ and thus, $F\leftrightarrow \odot(DE)\cap BC$. But note that $\odot(DE)\cap BC$ is simply the midpoint of $BC$. Thus $AF$ is isogonal to the $A$-median and so, $AF$ is the $A$-symmedian. :yoda:

Perform a force-overlaid inversion (complete with the reflection about the angle bisector) about $B$.  Let the resultant image of any object $X$ be $X'$. Note that $F', E'$ must lie on line $A'C'$, and also that $90^\circ = \angle DFE = \angle BFE - \angle BFD - \angle BE'F' - \angle BD'F = \angle E'F'D'$; and thus $D'F' \perp A'C'$; however, as $D'$ is the midpoint of arc $A'C'$, $F'$ is nothing but the midpoint of $A'C'$. Thus $BF$ is the reflection of the median about the angle bisector, i.e. the symmedian. $\square$

Let $E'$ be the antipode of $E$ in $(ABC)$. Since $\angle EBE' = 90 ^{\circ}$, we have $(A,C;D,BE' \cap AC)=-1$.
Also, $F$ lies on $DE'$, so $(A,C;F,B) \stackrel{E'}{=} (A,C;D,BE' \cap AC) = -1$, which implies that $BF$ is a symmedian of $\Delta ABC$.


AMC
AIME
USA(J)MO
USAMO
AMC 10
email
percent

Just wondering how many people qualified for the AIME.

sure

i was surprised at the cutoff for amc 12b

Barely on both the 10a and the 12b

[hide="10a"]
126
[/hide]

[hide="12b"]
105
[/hide]

If you look at the polls for scores, half of the people say "I got perfect".  I think that a lot of them are telling the truth, but there are plenty of people, like me, that click view results and think "I am stupid" and don't vote.

Now if most people say that they got a really high score, what about just qualifying...It is going to be like 5% don't qualify instead of 5% do qualify.

Yes, I did qualify.  Last year, I missed the cutoff by .5.  This year, I missed the A cutoff by 11.5 (due to really careless errors).  I got 109.5 on the B.

[quote="indianamath"]what the heck!!! i checked my answers online and i got 138, but my teacher said i got 132. :wallbash_red:  :wallbash_red:  :wallbash_red: DId i BUBBLE something WRONG?[/quote]

They are both good scores....it won't really make a difference for USAMO.  :D

Yeah, I made, 12a-138 and 12b-141.

[quote]what the heck!!! i checked my answers online and i got 138, but my teacher said i got 132.    DId i BUBBLE something WRONG?[/quote][/quote]

Your teacher still Mrs. Arnold?
It doesnt make a difference for USAMO, but that means he (a 7th grader) didnt beat everyone on the amc10 at the high school, which wouldnve been sad, considering we are actually a good high school (20 AIME qualifiers or so..this year) :D

[quote]Did anyone else's school have a sharp decrease in AIME qualifiers? Our school went from maybe 50 students to barely 20, two of whom qualified only on the B. I know that the scoring was designed to achieve this; however, I didn't expect the difference to be so large.[/quote]

well, our school had 30+ qualifiers last year, and this year 20 or so qualified through the A (dont know about the B)

Yup, I made it by a few points.

[quote="davidyko"]Yes sir, I made it; 112.5 on the 12A.
And we only have two qualifiers including myself at my high school. (I'm school winner!  :ninja: ) Yes, my school sucks at math.[/quote]

Except for the fact that I took the 12B instead of the 12A, I'm in [i]exactly[/i] the same situation.

i got 121.5 on the 10a :(
112.5 on the 12b :)
EDIT:  1000th post :D

for people not quite at that level some who are elite are who made it this year, it is of great prestige to them. All I have to say.

[quote="MysticTerminator"]hmm I too made AIME

yay?[/quote]

what'd you get?

I voted in this poll a while ago, but since AIME is in two days, might as well actually post. I made AIME quite easily, even as a freshman:

12A - 138
12B - 144 (tied for school winner at TJ w00t [better yet, it was another freshman :P])

No perfects from TJ this year, on either test :(

Now to (hopefully) get double digits on AIME, although an 8 would be safe considering my index would be a 224.

[quote="PenguinIntegral"][quote="indianamath"][quote="MysticTerminator"]hmm I too made AIME

yay?[/quote]

duh :|[/quote]
I congratulate you

making AIME is hard

don't diss him[/quote]

dude he won the USAMO in the entire country!!!(I think, or something like that) :P

[quote="indianamath"][/quote][quote="PenguinIntegral"][quote="indianamath"][quote="MysticTerminator"]hmm I too made AIME

yay?[/quote]

duh :|[/quote]
I congratulate you

making AIME is hard

don't diss him[/quote][quote="indianamath"]

dude he won the USAMO in the entire country!!!(I think, or something like that) :P[/quote]
No, that was Brian Lawrence.

[quote="indianamath"][/quote][quote="PenguinIntegral"][quote="indianamath"][quote="MysticTerminator"]hmm I too made AIME

yay?[/quote]

duh :|[/quote]
I congratulate you

making AIME is hard

don't diss him[/quote][quote="indianamath"]

dude he won the USAMO in the entire country!!!(I think, or something like that) :P[/quote]

Dude what that doesn't mean he can't be happy about making AIME too

[quote="probability1.01"][/quote][quote="indianamath"][/quote][quote="PenguinIntegral"][quote="indianamath"][quote="MysticTerminator"]hmm I too made AIME

yay?[/quote]

duh :|[/quote]
I congratulate you

making AIME is hard

don't diss him[/quote][quote="indianamath"]

dude he won the USAMO in the entire country!!!(I think, or something like that) :P[/quote][quote="probability1.01"]

Dude what that doesn't mean he can't be happy about making AIME too[/quote]
Well it does mean that you can't say "Making AIME is hard" ... without , you know, something exploding.

I couldn't make it at AMC 10A, I got 114, and the school winner got 115.5.
But I made it in AMC12B, I got 108.

I am the only one who qualified, but I am the youngest one who took the test (7th grade)

[quote="Pineappleperson"]i qualified for aime with a 144. (9th grade)

really thought i coulda made USAMO this year........

too bad i got like a 3 or 4 on the AIME....... :(  :o[/quote]

you might make USAMO! some really good people who can get over 7 got like 4-5 this year.

i made...
150 amc10a
112.5 amc12b

but i failed aime

Ok. I did much wose than I expected on the AIME. I was hopin' for at least a five, but I got a 3. I made 6 really stupid mistakes... :wallbash_red:


inequalities
geometry
inequalities unsolved

A line intersects the sides AB and BC of △ABC at points M and K.
If the area of the triangle MBK is equal to the area of the quadrilateral
AMKC, prove that
\[ \frac{|MB|\plus{}|BK|}{|AM| \plus{} |CA| \plus{} |KC|}\geq \frac{1}{3}.\]

[quote="mkrnic"]A line intersects the sides AB and BC of △ABC at points M and K.
If the area of the triangle MBK is equal to the area of the quadrilateral
AMKC, prove that
\[ \frac {|MB| \plus{} |BK|}{|AM| \plus{} |CA| \plus{} |KC|}\geq \frac {1}{3}.\]
[/quote]

$ \frac{BM}{AB}\equal{}\gamma$ , $ \frac{BK}{BC}\equal{}\alpha$ , $ \alpha\gamma \equal{} \frac{1}{2}$ , $ (\frac{1}{2} \le \alpha,\gamma \le 1)$

$ CA < AB\plus{}BC$

Let : $ \frac{1}{2} \le \alpha \le \gamma \le 1$

$ LHS > \frac{\alpha \plus{}\gamma \frac{AB}{BC}}{(2\minus{}\alpha) \plus{} (2\minus{}\gamma)\frac{AB}{BC}} \ge  \frac{\alpha}{2\minus{}\alpha} \ge \frac{1}{3}$

Equality holdes for : $ \alpha \equal{} \frac{1}{2}$ and $ \frac{AB}{BC} \rightarrow 0$


abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that every abelian group G is a module over the ring of integers, Z.

To show this, the following must be satisfied:
1. x(c+d) = xc+xd
2. x(yc) = (xy)c
3. (x+y)c = xy+yc

for all x,y in Z and c,d in G.

I don't know how to go about this. To prove the first one I started writing
x(c+d), let c+d=g in G
....then..?

A good rule of thumb is to understand why a result is true before you try to prove it.  What does a module over $ \mathbb{Z}$ look like?  Is there a natural way to interpret the product of an integer and a group element in a group?  (It will help to write the group in additive notation.)


[b]XLII OLIMPIADA MATEMATICA ESPANOLA - Fase Local 2006[/b]

[size=150][u][b]Primera sesion[/b][/u][/size]

[b][u]Problema 1[/u][/b]

Se da un triangulo isoceles $ABC$, con el angulo recto en $C$, y los catetos de longitud $2$. Un arco de circulo $l$ centro en $A$ divide al triangulo en dos partes de la misma area, mientras que el arco de circulo $m$ con centro en $B$ es tangente al circulo $l$ en un punto de la hipotenusa $AB$.
Hallar el area de la porcion del triangulo no cubierta por los sectores circulares correspondientes a los dos arcos.

[b][u]Problema 2[/u][/b]
Se suponen conocias las raices reales de las $n$ ecuaciones de segundo grado que se indican :

           $Ecuacion$               $\to$                   $Raices$
$x^2 + a_1x + b_1 = 0$      $\to$     $x_0,x_1$
$x^2 + a_2x + b_2 = 0$      $\to$     $x_0,x_2$
                                          $...$
$x^2 + a_nx + b_n = 0$      $\to$     $x_0,x_n$

Encontrar, razonadamente, las raices de la siguiente ecuacion

$x^2 + \frac{a_1 + a_2 + ... + a_n} { n } x + \frac{b_1 + b_2 + ... + b_n} { n } = 0$

[b][u]Problema 3[/u][/b]

En el triangulo $ABC$ se traza la bisectriz interior $CD$ se sabe que el centro del circulo inscrito en el triangulo $BCD$ coincide con el centro del circulo circunscrito del triangulo $ABC$.
Calcular los angulos del triangulo $ABC$.

[size=150][u][b]Segunda sesion[/b][/u][/size]

[b][u]Problema 4[/u][/b]

Encontrar, razonadamente, dos numeros enteros positivos $a$ y $b$ tales que:

$b^2$ sea multiplo de $a$,
$a^3$ sea multiplo de $b^2$,
$b^4$ sea multiplo de $a^3$,
$a^5$ sea multiplo de $b^4$,
pero $b^6$ no sea multiplo de $a^5$

[b][u]Problema 5[/u][/b]

Un numero $x$ positivo verfica la ecuacion:

$x^2 + \frac{ 1 } { x^2 } = 7$

Demostrar que

$x^5 + \frac{ 1 } { x^5 }$

es entero y calcular su valor.

[b][u]Problema 6[/u][/b]

Se considera la ecuacion

$| x - 1 | > ax$

donde $a$ es un parametro real.

a) Discutir la ecuacion segun lo valores de $a$.
b) Caracterizar los valores de $a$ para los cuales la inecuacion tiene exactamente DOS soluciones enteras.

Aqui estan los enlaces para los hojas escaneadas, por si alguien las quiere :)
[code]
http://img9.imageshack.us/img9/9918/hoja10el.gif
http://img299.imageshack.us/img299/6347/hoja28ur.gif [/code]

Issam, no podria escibir los problemas ud??
es que sale muy chiquito y no se puede leer...  :D

Gracias  issam!!

Carlos Bravo ;)
Lima Peru

donde podemos poner las soluciones a los problemas?? en este topico??


function
integration
calculus
derivative
vector
real analysis
real analysis theorems

Stokes Theorem and Change of Variable Theorem are the two basic results in Calculus. Could we possibly take Stokes Theorem as a particular case of Change of Variable Thoerem or the other way round, i.e. could we proof one theorem from another?  :?:

The right way to see Stokes' Theorem is that it is the Fundamental Theorem of Calculus. [I don't see it as a change of variables.]

Consider this result: for a sufficiently nice function $f,$

$\int_{a}^{b}f'(x)\,dx=f(b)-f(a).$

We have a set - in this case, the one-dimensional set that is the interval $[a,b]$ - and its zero-dimensional oriented boundary - the points $a$ and $b,$ distinguishing which point lies on which end. On the left, we have a one-dimensional integral of the derivative of $f$ over the interior of the set in question; on the right we have the zero-dimensional integral of $f$ itself over the boundary.

Another version of the FTC in higher dimensions is the Fundamental Theorem of Line Integrals: If $C$ is a curve from the point $\vec{p}_{0}$ to the point $\vec{p}_{1}$ and $f$ is a sufficently smooth function, then

$\int_{C}\vec{\nabla}f(\vec{r})\cdot d\vec{r}=f(\vec{p}_{1})-f(\vec{p}_{0}).$

Let's look at the theorems available in a typical "Calculus III" course:

One dimensional integral in $\mathbb{R}^{1}:$ Fundamental Theorem of Calculus.

One dimensional integral in $\mathbb{R}^{2}:$ Fundamental Theorem of Line Integrals.

Two dimensional integral in $\mathbb{R}^{2}:$ Green's Theorem.

One dimensional integral in $\mathbb{R}^{3}:$ Fundamental Theorem of Line Integrals.

Two dimensional integral in $\mathbb{R}^{3}:$ Stokes' Theorem.

Three dimensional integral in $\mathbb{R}^{3}:$ Divergence Theorem (Gauss).

Upon seeing that, it would be fair to ask: can we find four such theorems in $\mathbb{R}^{4},$ and five such theorems in $\mathbb{R}^{5}?$ Indeed we can, and we can find one theorem that generalizes them all. The generalization also usually goes by the name of Stokes.

There's a skinny little book - it used to be published by Dover, although I haven't looked for it lately - by Spivak entitled [i]Calculus on Manifolds.[/i] Essentially the entire book is devoted to the generalized Stokes' Theorem, and the language needed to state and prove it.

[quote="Kent Merryfield"]The right way to see Stokes' Theorem is that it is the Fundamental Theorem of Calculus. [I don't see it as a change of variables.].......[/quote]

Well of course, that is the well-known interpretation which can be found in any standard textbook including the book by Spivak. What I am wondering is that whther we can make a prove of Stokes Theorem, somehow by change of variable theorem

Look, there is such a question in "Jerrold E. Marsden & Anthony J. Tromba: Vector Calculus3Rd Edition, Freeman & Co, ISBN 0-7167-1856-1", P504:

*30.  Use Green's Theorem to prove the change of variables formula in the following special case:

${\int_{D}dxdy =\int_{D*}\vert \frac{\partial (x,y)}{\partial (u,v)}\vert dudv }$

for a transformation $(u,v)\mapsto (x(u,v),y(u,v))$ [the same notation used as in the book]


Did ayone else see it? Well, I guess it is still techinically going on, but yeah. It's really cool. The moon has turned a shadowy orangish colour. Anuone know when the next total lunar eclipse is?

On average there are two lunar (and solar) eclipses every year.  They may not be total. Next one  is a penumbral eclipse on  April 2005. Next total lunar eclipse in in Mar 2007. .. you can  easily google and find out  the  tables:
say at:
http://sunearth.gsfc.nasa.gov/eclipse/lunar.html

Ahhh..

I was ata math competition, so I didn't see it. How ironic.

i saw it and the moon looked really cool

i think my science teacher said it was on the 26, but that day was all cloudy and i couldnt see it....

this morning i saw the moon and it seemed REAL big and REAL bright!!! (it looked real cool & CLEAR)

on friday, the moon was all yellow and HUGE and i still didn't know why at that time


[hide="hint"]Put $y=(\sqrt{x^{2}+1}+x)^{5}$ we have $y+\frac{1}{y}=123$. That is all  :wink: [/hide]

Bad title. Locked.

http://www.mathlinks.ro/Forum/viewtopic.php?t=144404


hello, or do you mean
$ \frac{(abc\plus{}1)^2}{(a\plus{}1)(b\plus{}1)(c\plus{}1)}\geq\frac{(abc\minus{}1)^2}{(ab\plus{}a\plus{}1)(bc\plus{}b\plus{}1)(ca\plus{}a\plus{}1)}$?
Sonnhard.

no...i mean exactly d same as given in the question

Let us consider two cases:

The first case is when $ abc \le 3.$ In this case, we have
$ (ab\plus{}a\plus{}1)(bc\plus{}b\plus{}1)(ca\plus{}c\plus{}1) \ge (a\plus{}1)(b\plus{}1)(c\plus{}1),$
and
$ abc\plus{}1\minus{}(abc\minus{}1)^2 \equal{}abc(3\minus{}abc) \ge 0,$
so the inequality holds.

For the second case, we have $ abc>3.$ Then, by using the estimation
$ (ab\plus{}a\plus{}1)(bc\plus{}b\plus{}1)(ca\plus{}c\plus{}1) \ge (ab\plus{}a)(bc\plus{}b)(ca\plus{}c)\equal{}abc(a\plus{}1)(b\plus{}1)(c\plus{}1),$
it suffices to prove that
$ abc(abc\plus{}1) \ge (abc\minus{}1)^2,$
which is true because
$ abc(abc\plus{}1)\minus{}(abc\minus{}1)^2\equal{}3abc\minus{}1>0.$
The proof is completed.


The number 360 in base 7 is equal to 2k0 in base 9. What is k, where k is a digit?

I have no idea what the problem is about....Can somebody explain me what number in base means ...... thank you..

I got how to solve the problem but I still dont get the definition of base...what is it for...?


function
geometry
3D geometry
sphere
complex analysis
complex analysis unsolved

If $ g$ is a non-constant entire function such that $ g(0)\equal{}1$, then there exists a number $ r>0$ such that $ g(z) \ne 1$ for all $ z$ satisfying $ 0<|z|<r$.

Thank you  :roll: .

[quote="leshik"]By contradiction... then apply Uniqueness theorem :wink:[/quote]

You mean that by letting $ h(z) \equal{} g(z) \minus{}1$ then by contradiction, we can construct a sequence $ \{z_n\}_{n \geq 1}$ which are the zeros of $ h$ tending to $ 0$. Then $ h$ must be a constant by the Identity Theorem?

You know this result for when $ g(0) \equal{} 0$. Well there is nothing special about the value zero! The plane (or the riemann sphere) look the same at every point, so whether the value at a point is 0, 1, any complex number (or even infinity) does not matter for the analysis.


algebra polynomial calculus integration quadratics search

Canadian Math Olympiad #1, 1999:

Solve in real numbers:

4x2 - 40[x] + 51 = 0 where [x] is the greatest integer less than or equal to x.

Though first appeared on the 1985 AHSME problem 30 (very surprising that Canada would use the exact same problem that appeared on an AHSME only 14 years ago).

I dunno, I can't say either way about that.  There is of course a higher standard of proof for the CMO than the AHSME.

[hide] let x=a+e (a is integral, e is fractional)

4(a+e)(a+e) - 40a + 51 = 0

4a :^2: + 8ae + 4e :^2: = 40a+51

a :^2: + 2ae + e :^2: = 10a+12.75

a+2e+(e :^2: / a) = 10 + 12. 75

[a+2e+(e :^2: / a)]=[10+12.75/a]...the (e :^2: / a) is near zero, so cases

if a>12.75, a=10 --> e=0.618, x=10.618..[/hide]

are there multiple solutions? i found more than one... did i do that right? if its right so far, i'll finish it off.

Well, you started off okay.  But at the point where you divided by a, I feel like you went off track.  First of all, e2/a doesn't really have to be near zero -- what if a were 1?  What if e was near 1 and a was equal to 1?

Also, you have a line which says "if a > 12.75, a = 10."  I think we can both agree that this is not true no matter what.

There are multiple solutions.  Your other solutions might certainly be correct, but what you've got isn't horribly rigorous (due to the approximation).

hint: [hide] you may want to approach the problem by trying to put an upper and lower bound on x, and then considering discrete intervals [m, m+1), and solve for x in each interval.[/hide]

yeah, i should have written all my thoughts out. the 12.75/a thing was a mistake, and i knew that e :^2: /a didnt have to be near zero.. i'll do the rest of the cases here

[hide] if the 2e...expression is >1, and a>12.75, then a+1=10-->a=9, e=-0.21, x=8.78, doesnt work

if the 2e...expression is >2, and a>12.75, then a+2=10-->a=8, e=0.2, x=8.2, which works!

if the 2e...expression is <1, and a<12.75, then a=10+something,

guess and check time. a=11 satisfies a=[10+12.75/a], but doesnt work otherwise. nothing less than 10 would satisfy that, so we must let the 2e...expression be <2. so then a+1=[10+12.75/a]. 10 and only 10 satisfies this, x=9.34is not another solution

finally, if the 2e...expression is <3 and a<12.75, then a+2=[10+12.75/a] a=9 satisfies that, and only that, and x=8.78, doesnt work. so solution is x=8.2[/hide]

sorry, this isnt terribly neat or legible, and probably not entirely rigorous. or right. and a could be negative.

s0mp -- you still have the same problem in many cases of having a = 9 when a > 12.75.  Also, your answer of 8.2 doesn't quite work -- check it in the original equation.  It's off by 0.04, which works out to the e^2 term -- you were still setting it equal to zero.

Also, remember that 0 :le: e < 1, so that means that 0 :le: 2e < 2 necessarily.  Your beginning idea (splitting x up) can definitely be turned into a solution (as can zscool's hint), but you're definitely still having some problems with it.

This is really going to make me look like an idiot:

[hide]

By any chance is the answer 28? ( dont even ask how i got that )[/hide]

One of the nice things about this problem is that you can check your answer in the original equation.  Is (4*28 :^2: - 40*[28] + 51) equal to zero?  (No, actually.  But if you write down your method, we might be able to figure out where you ment wrong.)

There are multiple answers, and it turns out none of them are rational, although there isn't any particularly good reason that they shouldn't be.  You can prove there aren't any rationals by noting that there were rational roots, we would have 4*rational^2 + integer = integer -> 4*rational^2 = integer -> rational is integer or integer over 2, and then you can deal with those cases easily.

Try starting with s0mp's method, but finish it differently after giving a name to x's integer and fractional parts.

You used Google, of course. Don't. Solve the problems on your own, or don't post.

[quote="Scrambled"]the answer for [x] is 8 and the answer for x =  :sqrt: 67.25

dont ask how i got it[/quote]

Ok, I won't, and you are wrong (sure, that value may work, but you haven't solved the problem..)

Scrambled -- you really should post solutions, not answers.  Having a correct answer doesn't really say anything, whereas any solution (correct or not) can lead to suggestions and an eventual correct solution and understanding of how to get it.

ok fine.  I have a really weird way of solving this problem, and here it is:

[hide]First, assume x and [x] are the same thing and plug this into the quadratic equation.  you 8.5 and 1.5.  The answer for [x] will either be 8 or 1.  1 cant work because any positive number with forty subtracted cannot equal 40.  So [x] is 8.  lets say y is x :^2: , because x cant equal an integer or anything .5, or 4x :^2: -40x is even.  4y=320 - 51.  So 4y = 279.  so y=67.25.  So x :^2:  is 67.25.  so x is root 67.25.  Yes this is not the best way, and this probably isnt the right way either.[/hide]

By the way, i must say sumthing to confuted.  If i did search on google(which i didnt) i would come up with more than one answer, because JBL says there is more than one answer.  Besides, im almost certain that if you searched for the canadian math olympiad thingie on google, then it would tell you the solution, so i would have posted that solution on this thread.  Not only the answer.  Obviously, confuted thinks my brain is scrambled...

Hmm.. I don't think you can just say x=[x] and then solve for [x] out of that.. you are missing lots of solutions.

There, see, now I can give you suggestions.

It's an interesting thought, and you're close, but there's a little problem -- your approximation of x = [x] is sufficiently rough that it leaves open solutions that have a value of [x] close to, but not equal to, those that you found.  I mean, think about what happens if you have a number 7 and something.  The x2 part will be smaller, but -40[x] will increase (by 40, in fact), allowing the possibility of a smaller root than those you imagined.  Similarly for roots near 1.  There's too much wiggle-room.  So although the root you found is correct, it's not the only one.

Ok, I'm going to post a couple of hints following on from what scrambled tried.. if you don't want to, please don't read them, and if you do, read one at a time and then go and try the problem again..

1: [hide]When you let x=[x], you would have proved that there cannot be an integer solutions for x. How can we relate the formulas 4x^2 - 40[x] + 51 and 4x^2 - 40x + 51? This will give you bounds for x.[/hide]

2: [hide]Unfortunately, since x can be any real number, there are still infinite possibilities for x. Which part of the original equation aren't there infinite possibilites for? Try each of them, and you are done.[/hide]

after actually trying the problem on paper   , i found my own method.

[hide] 4x :^2: -40[x]+51. if x is negative, then 4x :^2: is positive, as is 40[x] and 51, so it can never be zero. now we are left to x being positive. let x=a+e, where a is integral and e is fractional. so:

4(a+e) :^2: -40[a+e]+51=4a :^2: +8ae+4e :^2: -40a+51=0

8ae+4e :^2: =40a-51-4a :^2:

you can complete the square of the rhs to show that its roots are a=1.5,8.5. we want it to be positive, which it is when 1.5<a<8.5

it is also easily shown that 0 :le: lhs :ge: 8a+3.

taking the gif's of both sides, we get

[8ae+4e :^2: ]=40a-51-4a :^2: , and we can replace lhs by:

8a+k=40a-51-4a :^2: , a quad equation which must have integer roots. solving for a...

4a :^2: -32a+51+k=0, and by the quad formula, after simplification:

a=(8- :sqrt: {13-k})/2, or (8- :sqrt: {13-k})/2

now just trial and error, but remembering that k is at most -3...

k=-3, a=6,2

k=-12, a=impossible (since a must be an integer)

k=-23, a=7 (remembering that 8a+k :ge: 0)

k=-36, a=impossible

k=-51, a=8, and beyond that, the 8a+k :ge: 0 will be violated.

so a=2,6,7,8 and x=29/4, 189/4, 229/4, 269/4, square roots of, of course. [/hide]

did i make any mistakes?

Good job.  You did make mistakes, but they were quite minor.

"0 :le: lhs :ge: 8a+3."  should be "0 :le: lhs < 8a + 4"

"k is at most -3" should be "k is less than 4"

"the 8a+k :ge: 0 will be violated" should be "the 1.5<a<8.5 will be violated"

Everything else is fine, though.  Good job!

Yep that solution is correct. Heres the way that I did it, basically the same way but without needing to split up x into a+e.

[hide]

0 = 4x^2  - 40[x] + 51 >= 4x^2 - 40x + 51 = (2x-3)(2x-17). So this instantly gives 1.5 < x < 8.5, and thus 1 <= [x] <= 8. Plug each of these values into the original equation, solve for x and check whether [x] does indeed equal that value, and you get those four solutions.

[/hide]

(blah, edited a typo - type - thing)

you said you cant just say [x] = x.

But in your solution, you did...


Not long time ago,someone,I don't remember who, posted an inequality:

   Let $a,b,c>0$ with the product $1$.Then, prove that: $a^2 +b^2+c^2\geq a+b+c $. Well this inequality is very easy.I even gave a proof on that topic just using Cauchy-Schwarz's inequality and AM-GM. Here is a stronger one:

 $*$ Let $a,b,c>0$ such that $ a+b+c\geq3 $.Prove the inequality: i) $a^2 +b^2+c^2\geq a+b+c $. ;) ;)  ;)  ;)

                                                                                ii) $a^3+b^3+c^3\geq a^2+b^2+c^2 $  ;) ;)  ;)  ;)

i don't know whether it's that what you mean but i think it's trivial :?
i) by cauchy
$ (x^2 + y^2 + z^2)( 1 + 1 + 1) \geq (x + y + z)^2$
which is equivalent to
$ (x^2 + y^2 + z^2) \geq \frac {(x + y + z)^2}{3} \geq \frac {(x + y + z)^2}{(x+y+z)} \geq x+y+z$

ii) we can assume that $ x \geq y \geq z $ then $ x^n \geq y^n \geq z^n $ with n is integer
by chebyshev we have
$ x^3 + y^3 + z^3 \geq \frac 13 ( x + y + z)( x^2 +y^2+z^2) \geq x^2 +y^2+z^2 $

Yes,you're right.both of them are trivial.I've posted them just to see the difference between them and Kedlaya's inequality.  ;)


Recently, the following poll was conducted in another thread on this forum with results as of 12 Jul 2006:

AMC math contest problems provide the greatest motivation for further study of which of the following fields?
Algebra	  17% 	[ 3 ]
Number Theory	 23% 	[ 4 ]
Geometry	  29% 	[ 5 ]
Probability / Counting	 29% 	[ 5 ]
Statistics / Data Analysis 0% 	[ 0 ]
Total Votes : 17

As a follow-up, what characteristics of a problem can motivate the study of a particular field?

What have problem writers done to motivate the fields ranked highly in this poll?

What problem characteristics are lacking in fields on this poll that are not ranked highly?

What might be done with extra-curricular activities such as math contests to better motivate the future study of fields not ranked highly?

Well probability is part of statistics and there are ALOT of probability in the AMC contests.

[quote="Chocoboom"]I dont think i've ever sen a statistics questions mainly because its impossible to do those without a calculator. Like z-scores and standard deviations.[/quote]

Those are possible to calculate, just really tedious. It took me 3 minutes to calculate the standard deviation and z-scores for the set $S=\{2,3,6,7\}$ And that has only 4 elements, which is a fairly small set.

[quote="Altheman"]well, i am starting study of statistics for my job (also with analytic geo and discrete math), and believe it or not, my obscure problem solving background has helped me, i am able to understand the concepts clearly, and i am accustomed to reading math notation, also i can connect ideas in the book to problem solving ideas, namely the study of frequency graphs relates to expected value...also look here for my post that tells about how i became interested in math

[url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=95726[/url]

just a side note, it takes a certain type of person who enjoys challenges to do problem solving, and also the child has to make the decision on their own to pursue problems, in my opinion, you should just present the material to students (possibly through teachers)[/quote]

All math are connected  :lol:


number theory
prime numbers
number theory unsolved

(Based on http://www.mathlinks.ro/Forum/viewtopic.php?p=1438386#1438386)

Let $ a_n$ be the minimum $ k \in \mathbb{N}$ such as $ \sqrt [i]{\frac {k}{i}} \in \mathbb{N}$  ,   $ \forall{i}\in \{2,\cdots,n \plus{} 1\}$ and primes.

Clearly $ a_1 \equal{} 2$ and from that other problem $ a_2 \equal{} 648$

Find $ a_3$ and if possible an explicit formula for $ a_n$

[quote="ZetaX"]You can't have that $ \frac n2$ and $ \frac n4$ are both perfect squares.[/quote]

Sorry for the mistake, I have edited for primes $ i$

Let the prime numbers in that set are $ p_1,p_2,...,p_k$ ($ p_1 \equal{} 2, p_2 \equal{} 3...$). $ p_1a_1^{p_1} \equal{} p_2a_2^{p_2} \equal{} ... \equal{} p_ka_k^{p_k}$. Obviously $ a_1 \equal{} p_1^{s_{11}}p_2^{s_{12}}p_3^{s_{13}}...p_k^{s_{1k}}$,  $ a_2 \equal{} p_1^{s_{21}}p_2^{s_{22}}p_3^{s_{23}}...p_k^{s_{2k}}$, ..., $ a_1 \equal{} p_1^{s_{k1}}p_2^{s_{k2}}p_3^{s_{k3}}...p_k^{s_{kk}}$. Then $ k \equal{} p_1^{b_1}p_2^{b_2}p_3^{b_3}...p_k^{b_k}$. To find $ min$ k we have to find the minimum $ b_i$'s. For example $ b_1\equiv \minus{} 1(mod p_1),$ $ b_1\equiv 0(mod p_2)$, $ b_1\equiv 0(mod p_3)$,..., $ b_1\equiv 0(mod p_k)$ which is [b]Chinese remainder theorem[/b]. Analogically for the others.
Remark: we can do the same think if the $ i$'s are relatively prime.


In the expansion of $ (xy\minus{}2y^{\minus{}3})^{16}$, find the coefficient of the term that does not contain $ y$.

[hide]I think that the term that doesn't contain a $ y$ is the term where the $ y$'s in the numerator and denominator cancel out. - But how do I find this term?[/hide]

Let the number of $ xy$'s and $ \minus{}2y^{\minus{}3}$'s be $ a$ and $ b$. Then you want $ a\plus{}b\equal{}16$ and $ a\minus{}3b\equal{}0$...

We can throw out the x since we're just finding the coefficient.
Let $ a$ be the number we raise $ y$ to and let $ b$ be the number we raise $ \frac{\minus{}2}{y^3}$ to. Thus:
$ a\plus{}b\equal{}16$
$ a\equal{}3b$
Thus $ a\equal{}12$, $ b\equal{}4$. The coefficient of this is $ (\minus{}2)^{4}(\binom{16}{4}) \equal{} 29120$

Does $ a\equal{}3b$ because of the $ \minus{}3$ in $ 2y^{\minus{}3}$?

[quote="math154"]Let the number of $ xy$'s and $ \minus{} 2y^{ \minus{} 3}$'s be $ a$ and $ b$. Then you want $ a \plus{} b \equal{} 16$ and $ a \minus{} 3b \equal{} 0$...[/quote]

why $ a \minus{} 3b \equal{} 0$... ?

[quote="Octahedr0n"]Does $ a \equal{} 3b$ because of the $ \minus{} 3$ in $ 2y^{ \minus{} 3}$?[/quote]
Yes, a must be 3 times as much b. a=3b also leads to a-3b=0.


Let $ n\geq 3$ be an integer, and suppose that the sequence $ a_1, a_2, ...,a_n$ of positive real numbers  satisfies $ a_{i \minus{} 1} \plus{} a_{i \plus{} 1} \equal{} k_ia_i$ for some positive integer $ k_i$ (Here $ a_0\equal{}a_n,a_{n\plus{}1}\equal{}a_1$). Prove that $ 2n\leq \sum_{i \equal{} 1}^{n}k_i\leq 3n$

I don't undestant
[quote="tdl"]Let $ n\geq 3$ $ 2n\leq \sum_{i \equal{} 1}^{n}k_i\leq 3\to n\le 1$[/quote]

It is not true. Let $ k_i\equal{}k>3$, then $ a_i\equal{}x_1^i\plus{}x_2^i\in Z$ is solution, were $ x_{1,2}\equal{}\frac{k\pm \sqrt{k^2\minus{}4}}{2}$.


MATHCOUNTS
trigonometry
ratio
LaTeX
Pythagorean Theorem
geometry

Right $ \Delta POQ$ is similar to $ \Delta XYZ$. Given that
$ \overline{XY} \equal{} \sqrt{189}$ units, how many units are in the length of $ \overline{ZY}$?

[asy]pair P,O,Q,X,Y,Z; unitsize(.5cm);
O = (0,0); P = (8,0); Q = (0,4);  Y = (13,0); X = (23,0); Z = (13,4);
draw(P--O--Q--P); draw(X--Y--Z--X);
label("P",P,E);label("X",X,E);label("Y",Y,W);
label("O",O,W);label("Q",Q,N);label("Z",Z,N);[/asy]

I don't think enough info is given, shoudln't there be a side length of QOP or another one of XYZ?  Because there's no way to solve this using mathcounts level stuff, only trig :D

I don't think you can solve this with trig, either.  :roll:

Well you have one angle, and one length...

I think you need at least three pieces of information to solve for a triangle, unless you have the ambiguous SSA case.

assume integers, and you know it's 6. forget other triangle

Oh here is the other info: length QP is 10 and QO is 4

[hide]If QP=10 and QO=4 then OP= sqrt84 from Pythagorean theorem

These two triangles are similar so we find the common ratio between the sides XY/OP = sqrt189/sqrt84 = 1.5

1.5*QO=6= ZY

ANSWER: 6

(i need to learn latex)

[/hide]


ratio
geometry
rectangle
quadratics
algebra
quadratic formula

The ratio of the length to the width in a rectangle is the same as the ratio of the diagonal to the length. What is the length of the diagonal if the width of the triangle is 2?

I would approach it like this:

[hide="Answer"]$2^2+x^2=y^2$, $\frac{2}{x}=\frac{x}{y}\Rightarrow 2y=x^2$, so we have $2^2+2y=y^2\Rightarrow y^2-2y-4=0$, $\boxed{y=1+\sqrt{5}}$, which is equivalent to $\sqrt{6+2\sqrt{5}}$.[/hide]

[quote="solafidefarms"][hide]Let the length be $l$ and the width $w$. The diagonal is then equal to $\sqrt{l^2+w^2}$.

We are given that $\frac{l}{w}=\frac{\sqrt{l^2+w^2}}{l}$. We can multiply this equation by $lw$ to yeild the equivalent equation $l^2=w\sqrt{l^2+w^2}$. Squaring both sides gives $l^4=w^2l^2+w^4\rightarrow l^4-w^2l^2-w^4=0$ We may substitute 2 for $w$ to yeild $l^4-4l^2-16=0$, and we can apply the quadratic formula to learn $l^2=\frac{4\pm 4\sqrt{5}}{2}=2\pm 2\sqrt{5}$. The square of a number must be positive, so $l^2=2+2\sqrt{5}$. We are looking for the diagonal, $\sqrt{l^2+w^2}$, so we do not need to find $l$ if we know $l^2$.

We already know $w$ and thus $w^2$ so the diagonal is $\sqrt{6+2\sqrt{5}}$, which seems reducable but I've forgotten what it comes out to.[/hide]

Billy[/quote]
As JesusFreak197 pointed out, $\sqrt{6+2\sqrt{5}}$ is indeed reducable.
Generally, $\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}$
(For proof, simply square both sides.)
In this case, $a=1$ and $b=5$, so that's why $\sqrt{6+2\sqrt{5}}=1+\sqrt{5}$.


what are d8 and d5 species......???

Square planar geometry

This should be similar to the octahedral case.  In fact, one can imagine slowly removing the point charges along the z axis until they are completely gone.  The figure below shows the correlation diagram:

Based on the energies of the d orbitals for a square planar complex, one would predict that this arrangement would be quite popular for d8 species.

reference....-  http://bouman.chem.georgetown.edu/S02/lect32/lect32.htm

no. of d electrons = 8 and 5 respectively for d8 and d5


state if your birthday is coming up soon, or if it was in the past week, or if its today so we can pm/email you birthday "presents"...

My birthday is within the next 12 months!!!1

My aunt is going to have a baby any time now (and I will be gone for over a week so dont get mad at me for not posting on AoPS) does that count? My birthday isn't coming for 4 months (its around Halloween and my older sister tried to return me because she thought I was a vampire.... gotta love those older siblings. I wish so much that I was a vampire.... *sigh*) I cant wait! I think I'm getting a DS....

My birthday is next year.

I am giving info until someone guesses my birthday correctly.

GUESS MY BIRTHDAY

It's sometime next week.

[hide="I think it is..."]
sometime next week[/hide]

9th?

I said my birthday in another thread that I will not reveal. MUAHAHAHAHAHAHHH!!! :diablo:

8th!

The message is too small. Please make the message longer before submitting.

YESSSSSSSSSSSSSSSSSSSSS lotsofmath is right!!

Guess my birthday?  Here's a clue:

It's after December 31st.

YES.  I was born on New Years.   :)

It's "great" how I can never have a party with any of my friends on my birthday...So much fun.  \endsarcasm.

maybe someone should start a "what month were u born in?" poll
or maybe a guess the birthday game, like guess the number game
i dunno lol

My birthday is in October. I have the same birthday as Niels Bohr.

The following conversation actually occurred, though the words were probably slightly different:

Person 1: It's your birthday today?
Person 2: Yes.
Person 1: Really? I thought that was like a year ago!


Βάζω μια άσκηση γεωμετρίας σχολικού επιπέδου γιατί μου άρεσε πολύ ο τρόπος λύσης της :)

Εστω παραλληλόγραμμο $ ABCD$. Ένας κύκλος περνάει από το  $ A$ και τέμνει τα τμήματα $ AB,AC,AD$ στα σημεία $ B_1, C_1, D_1$ αντιστοίχως. Να δείξετε ότι:
$ AB*AB_1 \plus{} AD*AD_1 \equal{} AC*AC_1$

Η αλήθεια είναι ότι είναι εύκολη :oops:  αλλά είναι πολύ κομψή :wink:

 [color=red]EDITED (mod)[/color]

Δε νομίζω ότι η άσκηση είναι σωστή. Γιατί αν $ r \equal{} AB \equal{} AC \equal{} AD$ είναι:
$ \frac{AB*AB_1}{r} \plus{} \frac{AD*AD_1}{r} \equal{} \frac{AC*AC_1}{r} < \equal{} > AB \plus{} AD \equal{} AC$, που δεν ισχύει λόγω της τριγωνικής ανισότητας.

Μια εναλλάκτική διατύπωση είναι η εξής:
Εστω παραλληλόγραμμο $ ABCD$. Απ' το σημείο $ A$ φέρουμε κύκλο που τέμνει τις πλευρές $ AB,AC,AD$ στα σημεία $ B_1, C_1, D_1$ αντιστοίχως. Να δείξετε ότι:
$ \vec{AB}*AB_1 \plus{} \vec{AD}*AD_1 \equal{} \vec{AC}*AC_1 (I)$.

Λύση:
$ (I) < \equal{} > \vec{AB}*r \plus{} \vec{AD}*r \equal{} \vec{AC}*r < \equal{} > \vec{AB} \plus{} \vec{AD} \equal{} \vec{AC}$, που ισχύει γιατί έχουμε παραλληλόγραμμο

[quote="maniopas"]Δε νομίζω ότι η άσκηση είναι σωστή. Γιατί αν $ r \equal{} AB \equal{} AC \equal{} AD$ είναι:
$ \frac {AB*AB_1}{r} \plus{} \frac {AD*AD_1}{r} \equal{} \frac {AC*AC_1}{r} < \equal{} > AB \plus{} AD \equal{} AC$, που δεν ισχύει λόγω της τριγωνικής ανισότητας.
[/quote]

Ισως δεν εξήγησα πολύ την άσκηση...
Κατ' αρχάς φαντάζομαι εννοείς αν $ r \equal{} AB_1 \equal{} AC_1 \equal{} AD_1$, το οποίο όμως δεν γίνεται γιατί τα σημεία $ A,B_1,C_1,D_1$ είναι ομοκυκλικά. Το $ A$ δεν είναι το κέντρο του κύκλου!!!

Οπότε νομίζω η διατύπωση της άσκησης μπορεί να μέινει όπως έχει αρκεί να δειυκρινιστεί ότι το $ A$ είναι σημείο του κύκλου.

Αφού λες από το Α φερνουμε κύκλο εννοείς οτι το Α ειναι κέντρο του κύκλου άρα εννοεις που να περνάει απο το Α

Πράγματι είναι εύκολη αλλά δεν είναι κομψή, τουλάχιστον αυτό που έκανα εγώ είναι να πάρω το αντιδιαμετρικό του Α και μετά διανύσματα, προβολή και εσωτερικό και βγαίνει σε μια γραμμή. :)

Έχεις υπ' όψιν σου κομψή γεωμετρική λύση? Η αλήθεια είναι ότι δεν το 'ψαξα γεωμετρικά καθόλου... :(

Κι εγώ διανυσματική λύση είχα στο νου... είναι και στο βιβλίο της κατεύθυνσης αυτή η άσκηση. Μου άρεσε γιατί δείχνει πώς ένα καθαρά γεωμετρικό πρόβλημα λύνεται με διανύσματα :)

Μια καθαρά γεωμετρικη λύση ίσως περπατησεί με μετρικές σχέσεις αλλά δεν έχω προλάβει να τη δώ . Με διανύσματα γινεται πολύ ευκολη.

Όπως ακριβώς βγαίνει με διανύσματα, βγαίνει και με τριγωνομετρία. Αλλά σε μια καθαρά γεωμετρική λύση, η διαδικασία που θα ακολουθήσουμε θα έκανε μέσα (έστω και έμμεσα) της απόδειξη της τριγωνομετρίας που που θα χρησιμοποιούσαμε για να τη λύσουμε.

[quote="bilot4"]...Οπότε νομίζω η διατύπωση της άσκησης μπορεί να μέινει όπως έχει, αρκεί να δειυκρινιστεί ότι το $ A$ είναι σημείο του κύκλου.[/quote]
Νομίζω η πρόταση είναι σωστή έτσι και αποδεικνύεται συνθετικά όχι δύσκολα.

[hide]Αν $ C'_{1}$ και $ C''_{1},$ είναι τα σημεία τομής της διαγώνιας $ AC,$ από τους περιγεγραμμένους κύκλους $ (O_{1}),$ $ (O_{2}),$ περί τα τρίγωνα $ \bigtriangleup BB_{1}C_{1},$ $ \bigtriangleup DD_{1}C_{1}$ αντιστοίχως, αρκεί να αποδειχθεί ότι $ AC'_{1} \equal{} CC''_{1},$ από $ (AB_{1})\cdot (AB) \equal{} (AC'_{1})\cdot (AC_{1})$ και $ (AD_{1})\cdot (AD) \equal{} (AC_{1})\cdot (AC''_{1}).$

Αυτό σημαίνει ότι πρέπει να αποδείξουμε ότι $ BC'_{1}\parallel DC''_{1}$...και για αυτό το τελευταίο, είναι απαραίτητο να αποδείξουμε ότι τα σημεία $ B_{1}$ και $ C_{1}$ και $ B'_{1}$ είναι συνευθειακά, όπου $ B'_{1}\equiv CD\cap (O_{2}),$ ή ότι είναι συνευθειακά τα σημεία $ D_{1}$ και $ C_{1}$ και $ D'_{1},$ όπου $ D'_{1}\equiv BC\cap (O_{1}).$[/hide]

Κώστας Βήττας.

[quote="bilot4"]Εστω παραλληλόγραμμο $ ABCD$. Απ' το σημείο $ A$ διέρχεται κύκλος που τέμνει τις πλευρές $ AB,AC,AD$ στα σημεία $ B_1, C_1, D_1$ αντιστοίχως. Να δείξετε ότι:
$ AB*AB_1 \plus{} AD*AD_1 \equal{} AC*AC_1$[/quote]

Από το εγγεγραμμένο τετράπλευρο $ AB_1C_1D_1$ προκύπτει $ \angle B_1AC_1 \equal{} \angle B_1D_1C_1$ και $ \angle B_1C_1D_1 \equal{} 180^0 \minus{} \angle B_1AD_1$

Επομένως τα τρίγωνα $ ABC, D_1C_1B_1$ είναι όμοια, αφού έχουν $ A \equal{} D_1$ και $ B \equal{} C_1$

Τότε θα είναι $ m\cdot AB \equal{} D_1C_1,\ \ m\cdot BC \equal{} C_1B_1, \ \ m\cdot AC \equal{} D_1B_1$, όπου $ m$ ο λόγος ομοιότητας

Από Πτολεμαίο παίρνουμε $ AB_1\cdot C_1D_1 \plus{} AD_1\cdot B_1C_1 \equal{} AC_1\cdot B_1 D_1 \equal{} >$

$ AB_1\cdot m\cdot AB \plus{} AD_1\cdot m\cdot BC \equal{} AC_1\cdot m\cdot AC \equal{} >$

$ AB_1\cdot AB \plus{} AD_1\cdot BC \equal{} AC_1\cdot AC \equal{} >$

$ AB_1\cdot AB \plus{} AD_1\cdot AD \equal{} AC_1\cdot AC$

Λουκάς

Όμορφη και δυνατή λύση, διδακτική για τον τρόπο χρήσης του λόγου της ομοιότητας.

Απολαμβάνεις το πως ένας ευέλικτος νους πετυχαίνει τη βέλτιστη λύση, που θα μπορούσε ίσως εύκολα να τη σκεφτούμε, αλλά όμως εξίσου εύκολα μπορεί να μας ξεφεύγει.

Λουκά σ' ευχαριστώ πολύ, Κώστας Βήττας.


abstract algebra
superior algebra
superior algebra unsolved

1. In $R$, define $a$ $\circ$ $b$ $=a+b-ab$. Show that this binary operation
is associative, and that $(R,o)$ is a monoid with zero as the identity
element.
2. An element $a \in R$ is called left (resp., right) $quasi-regular$ if $a$ has
a left (resp., right) inverse in the monoid $(R,o)$ with identity. If $\circ$
is both left and right quasi-regular, we say that $a$ is quasi-regular.
(1) Show that if $ab$ is left quasi-regular, then so is $ba$. (2) Show that
any nilpotent element is quasi-regular. (3) Show that, if $R$ has an
identity $1$, the map $(R,\circ)\to (R,x)$ sending $a$ to $1-a$ is a monoid
isomorphism. In this case, an element $a$ is left (right) quasi-regular
iff $1-a$ has a left (right) inverse with respect to ring multiplication.
3. A $I\subseteq R$ is called quasi-regular (resp., left or right quasi-regular) if
every element of $I$ is quasi-regular (resp., left or right quasi-regular).
Show that if a left ideal $I\subseteq R$ left quasi-regular, then it is quasi-
regular.
4. Define the Jacobson radical of $R$ by $rad R =$ {$a \in R: Ra$ is left quasi-
regular}. Show that $rad R$ is a quasi-regular ideal which contains
every quasi-regular left (resp., right) ideal of $R$. (In particular, $rad R$
contains every nil left or right ideal of $R$.) Show that, if $R$ has an
identity, the definition of rad R here agrees with the one given in the
usual definition.
5. A left ideal $I\subseteq R$ is said to be regular (or modular) if there exists
$e \in R$ which serves as a "right identity modulo $I$."; i.e., $re = r$
$(mod I)$ for any $r \in R$. Show that if $I \subset R$ is a regular left ideal,
then $I$ can be embedded in a regular maximal left ideal of $R$. Finally,
show that, if $rad R\neq R$, then rad $R$ is the intersection of all regular
maximal left (resp., right) ideals of $R$. (Hint. If $e \in R$ is not left
quasi-regular, then $I =$ {$re-r : r \in R$} is a regular left ideal not
containing $e$.)
6. A left $R$-module $M$ is said to be simple (or irreducible) if $R * M \neq 0$
and $M$ has no R-submodules other than $(0)$ and $M$. Show that $_{R}M$
is simple iff $M \cong R/m$ (as left $R$-modules) for a suitable regular
maximal left ideal $m$. Show that $rad R$ is the intersection of the
annihilators of all simple left $R$-modules.
7. Show that $rad(R/rad R) = (0)$, and that, if $I$ is an ideal in $R$, then,
viewing $I$ as a ring, $rad I = I \cap rad R$. This shows, in particular,
that a ring $R$ may be equal to its Jacobson radical: if this is the case,
$R$ is said to be a radical ring. Show that Я is a radical ring iff it has
no simple left (resp., right) modules.

The first problem is simpl. what's about other?


geometry
3D geometry
tetrahedron
ratio
geometric transformation
homothety
geometry solved

PERU is a tetrahedron. X is a point inside.
 The centroids of XERU, PXRU, PEXU,PERX are P', E', R', U' respectively.

  Show that the volume of P'E'R'U' is 1/64 the volume of PERU.

I'm not sure, but it shouldn't be too hard if you employ the determinant formula of the volume of a tetrahedron.

Choosing some arbitrary origin $O$, we have that
\[\vec{OP'} = \frac{\vec{OX} + \vec{OE} + \vec{OR} + \vec{OU}}{4}\]
and
\[\vec{OE'} = \frac{\vec{OP} + \vec{OX} + \vec{OR} + \vec{OU}}{4},\]
so
\[\vec{P'E'} = \vec{OE'} - \vec{OP'} = \frac{\vec{OP} - \vec{OE}}{4} = -\frac{1}{4} \vec{PE}.\]

Similarly,
\[\vec{P'R'} = -\frac{1}{4} \vec{PR}, \quad \vec{P'U'} = -\frac{1}{4} \vec{PU},\]
so tetrahedron $P'E'R'U'$ is tetrahedron $PERU$ scaled by a factor of $-1/4$.  Therefore, the volume of tetrahedron $P'E'R'U'$ is $1/4^3 = 1/64$ that of $PERU$.

Grobber, the problem is much simpler than you think...

Let G be the centroid of the tetrahedron PERU, and let P", E", R", U" be the centroids of the triangles ERU, PRU, PEU, PER. Then, we know that the centroid G of the tetrahedron PERU lies on the line EE" and divides the segment EE" in the ratio EG : GE" = 3. Hence, GE" : EG = 1/3, so GE" : GE = -1/3 (with directed segments). In other words, the point E" is the image of the point E under the homothety with center G and factor -1/3. Similarly, the points R", U", P" are the images of the points R, U, P under this homothety. Thus, the tetrahedron P"E"R"U" is the image of the tetrahedron PERU under this homothety, and since the factor of the homothety was -1/3, the volume of the tetrahedron P"E"R"U" must equal $\displaystyle \left( -\frac13\right) ^3=-\frac{1}{27}$ of the volume of the tetrahedron PERU (hereby, I use directed volumes, so that the minus sign in $\displaystyle -\frac{1}{27}$ stands for the fact that the orientation of the tetrahedron P"E"R"U" is opposite to that of PERU).

Similarly, the centroid E' of the tetrahedron PXRU lies on the line XE" and divides the segment XE" in the ratio XE' : E'E" = 3. This ratio yields (after some algebraic transformations) XE' : XE" = 3/4. Hence, the point E' is the image of the point E" under the homothety with center X and factor 3/4. Similarly, the points R', U', P' are the images of the points R", U", P" under this homothety. Therefore, the tetrahedron P'E'R'U' is the image of the tetrahedron P"E"R"U" under this homothety, and since this homothety had the factor 3/4, the volume of the tetrahedron P'E'R'U' is $\displaystyle \left( \frac34\right) ^{3}=\frac{27}{64}$ times the volume of the tetrahedron P"E"R"U"; but since the latter volume is $\displaystyle -\frac{1}{27}$ of the volume of the tetrahedron PERU, we see that the volume of the tetrahedron P'E'R'U' equals $\displaystyle \frac{27}{64}\cdot \left( -\frac{1}{27}\right) =-\frac{1}{64}$ times the volume of the tetrahedron PERU.

  Darij


is my question at
http://www.mathlinks.ro/Forum/viewtopic.php?t=127805

not complicated enough?.  I'll give the full solution if there are any requests. If not then I suppose you all want somethin more challenging?

The question you posted isn't hard and you did get several answers. And you can post more challenging problems if you want.


The greatest value which the function

f(x)=(3sin{(10x + 11)^2 -7)^2

takes, as $ x$ varies over all real values equals?

$ f(x)=(3\sin{(10x + 11)^2 -7)^2}$

Same thing as the previous thread you made, except we are now trying to find the maximum of $ |3\sin (10x + 11)^2 - 7|$.

We do not want to combat the $ -7$ constant, but rather we want to add to its "negativeness". However, this is not possible since the $ \sin$ function is ultimately squared, so we try to make that 0. Since it can be reached, the maximum is just $ (-7)^2 = \boxed {49}$.

[quote="Yongyi781"]

We do not want to combat the $ \minus{} 7$ constant, but rather we want to add to its "negativeness". However, this is not possible since the $ \sin$ function is ultimately squared, so we try to make that 0. Since it can be reached, the maximum is just $ ( \minus{} 7)^2 \equal{} \boxed {49}$.[/quote]

Yes, sorry; I did it then realized it was wrong and changed it thus posting it twice accidentally. The second post was my actual question, however I am afraid I don't quite understand what you mean about making the sin function 0?

No wait; do you mean that if we take the max of sin, which is 1, then this will leave us with a smaller value (namely 36), than if we took sin to be 0?

Yes, since the $ \sin$ function is squared, we cannot possibly have a value of $ 3\sin (10x \plus{} 11)^2$ to be negative, so the minimum it can have (thus maximizing the negativeness of the overall function) is 0.


logarithms
inequalities
induction
integration
inequalities proposed

Prove the equations

\begin{eqnarray*}
&&\text{\textbf{a)}\ \ \ \ \ \ \ \ \ \ }\lg \left( n+1\right) >\frac{3}{10n}%
+\lg n; \\
&&\text{\textbf{b)}}\;\;\;\;\;\;\;\;\;\;\lg \left( n!\right) >\frac{3n}{10}%
\left( \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-1\right)
\end{eqnarray*}
for every natural number $n.$ (Hereby, $\lg $ means the logarithm with base $%
10.$)

I hope I didn't miscalculate, but it is quite boring :

a) Let $f(x) = lg(x+1) - lg(x) - \frac 3 {10x}$. Then $f'(x) = \frac 1 {10 \cdot Ln(10) \cdot x^2(x+1)} [(3\cdot Ln(10) - 10) x  + 3\cdot Ln(10)] > 0$ since $3\cdot Ln(10) - 10$. Thus $f$ is increasing on $(0; +\infty)$. Moreover $f(1) = lg(2) - 0,3 > 0$ thus $f(x) > 0$ for all $x \geq 1$.
Thus $f(n) > 0 $ for all integers $n > 0$, and we are done.

b) Using a) for $k=2,3, \cdots , n-1$ and summing, we deduce that $lg(n) > \frac 3 {10} (1 + \rac 1 2 + \cdots + \frac 1 {n-1})$. (1).

In another hand, the desired inequality clarly holds for $n =2$. Now assume it holds for some $n$.
Then, from the induction hypothesis and a) :
$lg((n+1)!) = lg(n!) + lg(n+1) > \frac {3n} {10} (\frac 1 2 + \cdots + \frac 1 n - 1) + \frac 3 {10n} +  ln(n)  = \frac 3 {10} (\frac n 2 + \frac n 3 + \cdots \frac n n - n + \frac 1 n) + lg(n) =  \frac 3 {10} (\frac {n+1} 2 + \frac {n+1} 3 + \cdots \frac {n+1} n + \frac { n+1} {n+1} -( n+1) + \frac 1 n + 1 - \frac 1 2 - \frac 1 3 - \cdots  -  \frac 1 {n+1}) + lg(n)  = \frac {3(n+1)} {10} (\frac 1 2 + \frac 1 3 + \cdots \frac 1 n + \frac 1 {n+1} - 1) + \frac 3 {10} (\frac 1 n + 1 - \frac 1 2 - \frac 1 3 - \cdots \frac 1 {n+1} ) + lg(n)$.

Thus, it suffices to prove that $\frac 3 {10} (\frac 1 n + 1 - \frac 1 2 - \frac 1 3 - \cdots - \frac 1 {n+1} ) + lg(n) \geq 0$.
But, from (1), it suffices to prove that $\frac 3 {10} ( 2 - \frac 1 {n+1}) > 0$, which is clearly true.
This ends the induction step and the proof.

Pierre.

a) For $n=1, 2$ this is true. Assume $n\geq3$.
Using $f(x)-f(y)=(x-y)f'(z)$ for some $z\in(x,y)$ get $\lg(n+1)-\lg n=\frac{\lg e}z>\frac{\lg e}{n+1}>\frac3{10n}$, $z\in(n,n+1)$.

b)for any $x\in[i,i+1]$, $\frac1x>\frac1{i+1}$, therefore $\frac1{i+1}=\int_i^{i+1}\frac1{i+1}dx<\int_i^{i+1}\frac1x dx=\ln(i+1)-\ln i$. Summing get $\sum_{i=2}^n\frac1i<\ln n$.
In the same way $\ln(i+1)>\int_i^{i+1}\ln x dx=(i+1)\ln(i+1)-i\ln i-1$ and $\lg(n!)=\lg e\sum_{i=2}^n \ln i>\lg e(n\ln n-n)$ the initial inequality follows from $\lg e>\frac{10}3$.


[quote="nutz_for2.718281828"]What are the users that are called "bots"?[/quote]

They're not users, they're bots (or robots)

Judging from their names, it seems they're bots from search engines so they can crawl through forums and index.

[quote="akalra1"]Alexa?

dont think so[/quote]Alexa is a search site.

"bots" is short for "robots", which are things that scan through websites to evaluate the content in them and what they find, they report back to their server.  This allows search engines like google to know what is on each web page so your searches are more accurate.  They were here before, but now, you can tell when they are online and where they are in the forum (or blogs).


calculus
integration
logarithms
algebra
partial fractions
special factorizations
calculus computations

Dear all,
how to find integration of this

(3x - 7) / (x^2 - 6x + 13) ?

thanks alot

[hide]$\int\frac{3x-7}{x^2-6x+13}\ dx=\int\frac{3x-9}{x^2-6x+13}\ dx + \int\frac{2}{x^2-6x+13}\ dx$

Just use $u=x^2-6x+13$ on the first and partial fractions ([b][u]ugh![/u][/b]  No integer solutions) on the second one.[/hide]

I'm always in favor of making "little" substitutions - substitution that do not in and of themselves solve the problem, but make it easier to see what to do next.

The first step is to realize that that denominator is (in real terms) an irreducible quadratic. Completing the square yields

$x^2-6x+13=x^2-6x+9+4=(x-3)^2+4.$

The first "little" substitution to try is $u=x-3.$ Then $x=u+3$ and $dx=du.$

$\int\frac{3x-7}{x^2-6x+9}dx=\int\frac{3(u+3)-7}{u^2+4}du= \int\frac{3u+2}{u^2+4}du.$

Splitting this into two integrals by letting $\frac{3u+2}{u^2+4}=\frac{3u}{u^2+4}+\frac{2}{u^2+4}$ accomplishes the same thing as what inteluser said - the only difference is the visibility of the motive.

The finish of this should involve both a logarithm and an arctangent.

Ooh, that's [i]much[/i] prettier.  I'll have to remember that.

It has nothing to do with that.  That equation doesn't even have real roots, much less integral roots.


conics
parabola
calculus
algebra
system of equations

Hello, first post here, nice to meet you…I sincerely hope this is in the right place, or it’s not too much trouble to move it.

Anyway, I’m posting because I’m trying to design an animation system in Basic (very novice programmer) and one of the things I’m trying to do requires coding that will let the program determine a parabola based on three points.

I know this is possible, because my graphing calculator can do it and I’ve learned how myself (though admittedly I messed up last time I tried…), but I’m having trouble solving for a, b, and c in general. There are just too many steps that take too long to write out, and I’m sure that I’ll mess up somewhere, what with all the parentheses and quotients and negative signs (Those messed me up in calculus, believe it or not).

So, the base equation is, in this-variant-of-Basic parlance:

y = a*x*x+b*x+c
or
y = ax^2+bx+c

Let us call my three points:

x(1),y(1)
x(2),y(2)
x(3),y(3)

as this is how I refer to them in the program, being unable to make use of that most wonderful device that is ‘subscript’. You see how already the parentheses appear? Anyway:

a = (y-b*x(1)-c)/(x(1)*x(1))
b = (y-a*x(2)*x(2)-c)/x(2)
c = y-a*x(3)*x(3)-b*x(3)

I’m fairly certain this is the correct approach, and hope that I have these relatively simple operations correct. My problem, then, is this:

When I attempt to solve these equations through substitution, I get bogged down in writing so many operations and parentheses; the equation takes up multiple lines with multiple quotients and I’m sure I’ll mess something up.

Can any of you offer advice on how to solve this system of equations without working myself to a frustrated, tearful mess? I have the software Maple 10, if that can be of help; a TI-83 Plus; this forum that I don’t really know how to use; some programming skill; some algebra ability; and you.

To make notation clearer, let the points be $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$. Then all of them satisfy the equation of parabola: for any $i$ we have $ax_{i}^{2}+bx_{i}+c=y_{i}$. So, writing it for $i\in\{1,2,3\}$, we have: \[ax_{1}^{2}+bx_{1}+c=y_{1},\]  \[ax_{2}^{2}+bx_{2}+c=y_{2},\]  \[ax_{3}^{2}+bx_{3}+c=y_{3}.\] This is simple system of linear equations for $a,b,c$ and I hope you can solve it.  :D

Simple.

Ok, there must be something I'm missing.

No, for some reason it balloons in complexity when I try to solve it.

[quote="Farenhajt"]You must express one variable from one equation and then substitute that into the other two. For example, from the first equation you get $c=y_{1}-ax_{1}^{2}-bx_{1}$, then plug that into the second and third and regroup to get the system in $a,b$ only. Then go on.[/quote]
Or use determinants if you are patient.
edit: brutal method (useful for curves of arbitrary degree, not necessairly parabolas) is decribed in [url]http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html[/url].

The 'brutal method' looks more exactly what I want--being able to give curves of arbitrary degree could be quite useful for animation.

Unfortunately, when I simply copy the one for second degree equations into my program, it gives the wrong values. The curves don't come close to my points, though they are parabolas.


i want to make a latex file.. i have installed TexnicCentre and Mictex and i have created one file but i have a lot of doubts..-
1) After i installed the above two all my pdf files automatically changed format.. so does that mean i cant use  a pdf editor for those files?
2) I wanna copy some things (info) from a texworks file that i have and not written by me.. how can i do it?? i just do know how to copy the text!
3) i plan on writing a proj kinda on inequalities.. which is better and easier? writing a pdf file or a Texworks file? and what about Word and Djvu ??
someone please help..

I think you misunderstand what is happening. The files used by TexnicCenter and Texworks are just text files that can be read in any editor and have the filename filename[i].tex[/i]. When you press the compile button you get filename[i].pdf[/i] which is purely for reading - it is very difficult to edit pdf files but they have the advantage that they will look the same on every computer.
If you need to change the pdf file, you alter the tex file and then compile again.

LaTeX is designed to be very mathematically friendly but has a steep learning curve. If you are willing to learn it then you will get much better results than anything else; the output is of typeset quality so is comparable to a printed book.

hello i do know latex! thts ob seeing i have some 250  sthng posts..
ok i see..
3) ?? can i or not copy text from a pdf file? or do i have to rewrite the content if say .. i am doing myself a similar file?

You can copy the text from a pdf, but not mathematics. In fact you can save the pdf as a text file but again you will only get the text not the mathematics. You can include pdf pages as images in your tex document (see [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfpages]pdfpages package[/url]) but you won't be able to edit those images.


In the argand diagram shown below, $ \vec{OP}$ represents the complex number z.

Label $ \vec{PA}$ with the complex number it represents

[url=http://img72.imageshack.us/my.php?image=28024931hc2.jpg][img]http://img72.imageshack.us/img72/8416/28024931hc2.th.jpg[/img][/url]

THey have it as $ 1-z$.. why? I thought $ z-1$

my diagram is right or wrong?

[url=http://img72.imageshack.us/my.php?image=81470414gh6.jpg][img]http://img72.imageshack.us/img72/6232/81470414gh6.th.jpg[/img][/url]

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

I'm not sure, maybe think of it as 1-z = 1+(-z)?

How do u do subtraction of complex no. using vectors?

Well, isn't (-z) just (z) in the opposite direction?

archimedes1 , yes, a "negative" vector is just a whose direction has been changed by $ \pm \pi$.
bos1234 , do not really know what your question is though , your diagram shows $ 1 = \vec{z}+\vec{PA}$ or rearranged $ \vec{PA}= 1-\vec{z}$, vectors need to be "head-to-tail" to be added (including adding a negative vector)
$ \vec{PA}= 1-\vec{z}= 1+\left(-\vec{z}\right)$ that arrangement implies that the $ \vec{z}$ must be negated with "head-to-tail" being perserved.

Oh ok. I understand it now

so in this qn (same qn)

If I draw the diagram like this (head to tail).. the resultant is -z-3?
[url=http://img176.imageshack.us/my.php?image=55854445wa5.jpg][img]http://img176.imageshack.us/img176/1438/55854445wa5.th.jpg[/img][/url]

but if i draw the vectors head to tail like this it becomes z+3??

[url=http://img154.imageshack.us/my.php?image=40611874ph4.jpg][img]http://img154.imageshack.us/img154/3682/40611874ph4.th.jpg[/img][/url]

The first diagram should have the resulting vector in the opposite direction, what you have now is $ -\vec{z}=\vec{OP}-3$
The second diagram looks correct.

If $ z_{1}=3+2i$, $ z_{2}=-4+7i$ show $ z_{1}z_{2}$ on an Argand diagram.

[url=http://img407.imageshack.us/my.php?image=21680792dr7.jpg][img]http://img407.imageshack.us/img407/2991/21680792dr7.th.jpg[/img][/url]

How does the book mark the pt $ z_{1}z_{2}$ so quickly like this?

What is the diagram trying to show?  :huh:
$ \vec{z}_{1}\vec{z}_{2}= \left(-3 \cdot 4-2 \cdot 7 \right)+\left( 3 \cdot 7+^{-}2 \cdot 4 \right) i =-26+13i$.


I'm going to try to get into AMC 9.  My question is: do you need to know Alg 2, or is it stuff like Alg 1/ Geometry?

Last time I checked, there was no AMC-9... do you mean AMC-8 or AMC-10?

Yes, I don't think there is an AMC 9. The only ones I know are AMC 8, AMC 10, and AMC 12.

The AMC 8, 10, and 12 are, [u]by no means[/u], based on school mathematics. The AMC's are about problem solving. However, if you must try to compare, the AMC 8 assumes concepts up to Pre-Algebra, the AMC 10 up to Algebra I and Geometry, the AMC 12 up to Precalculus.

However, there are many, many, many topics that would never appear in a school textbook that are tested on these exams, such as number theory and advanced counting methods (more toward the AMC 10 and 12 than the 8, but sure).

oh, yeah I meant AMC 10 or 12...heh heh.


Prove that: $ \boxed{\frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right ) \equal{} \frac {n \plus{} 1}{n \plus{} 2}}$

Locked. See the "Zero tolerance" thread!


Simplify the fraction $ \displaystyle\frac{\dfrac{3x}{4x\minus{}4}}{\dfrac{9x^2}{x\minus{}1}}$.

$ \displaystyle\frac{\dfrac{3x}{4x\minus{}4}}{\dfrac{9x^2}{x\minus{}1}}$ is $ \displaystyle\frac{\dfrac{3x}{4x\minus{}4}}{\dfrac{36x^2}{4x\minus{}41}}$, so it becomes $ \frac{3x}{36x^2}$, which is $ \frac{1}{12x}$

answer : $ \frac{1}{12x}$


analytic geometry
combinatorics unsolved
combinatorics

[quote="outback"]Does there exist an infinite set $ M$ of straight lines on the coordinate plane such that
(i) no two lines are parallel, and
(ii) for any integer point there is a line from $ M$ containing it?[/quote]
I think there is a missing requirement, otherwise, this problem is trivial.

Here's the original problem:

[quote]Does there exist an infinite set M of straight lines on the coordinate plane
such that
(i) no two lines are parallel, and
(ii) for any integer point there is a line from M containing it?
Does the answer change if we add one more condition:
(iii) any line from M passes through at least 2 integer points?[/quote]

I still don't understand, it is still quite easy if I understand correctly.

Let the lines through $ 2$ axes.
For positive $ a,b$, let a line through $ (a,b)$ and $ (a\plus{}1,b\plus{}2^a3^b)$
For $ a$ positive, $ b$ negative let a line through $ (a,b)$ and $ (a\plus{}1,b\plus{}5^a7^{\minus{}b})$
For $ b$ positive, $ a$ negative let a line through $ (a,b)$ and $ (a\plus{}1,b\plus{}11^{\minus{}a}13^b)$
For both of $ a,b$ negative let a line through $ (a,b)$ and $ (a\plus{}1,b\plus{}13^{\minus{}a}17^{\minus{}b})$

[quote="Akashnil"]I still don't understand, it is still quite easy if I understand correctly.

Let the lines through $ 2$ axes.
For positive $ a,b$, let a line through $ (a,b)$ and $ (a \plus{} 1,b \plus{} 2^a3^b)$
For $ a$ positive, $ b$ negative let a line through $ (a,b)$ and $ (a \plus{} 1,b \plus{} 5^a7^{ \minus{} b})$
For $ b$ positive, $ a$ negative let a line through $ (a,b)$ and $ (a \plus{} 1,b \plus{} 11^{ \minus{} a}13^b)$
For both of $ a,b$ negative let a line through $ (a,b)$ and $ (a \plus{} 1,b \plus{} 13^{ \minus{} a}17^{ \minus{} b})$[/quote]

That seems to work...  :)

For example
\[ M \equal{} \{d_{ij}: i, j \in Z, \text{ line} \  d_{ij} \text{ \ passes through } \ (0,0) \ \text{ and \ } (i,j)\}
\]
Is this correct?!


And what grade are you in? I figure if anybody gets perfect scores, they are likely to be an AoPSer, right??

[quote="guyinPA"]re: comment about scores by native english speakers - you think this applies even to the Math side??  Based on the people I've met at various math contests, I don't think I'd agree with that[/quote]
but reading and expecially  writing are easier for native english speakers.

The SAT math does test English comprehension to a degree. If you don't understand what the question means, you won't get it right- and SAT questions are tricky.
Of course, anyone with problems there is going to have a low verbal score.

I got a perfect on the Physics Subject Test, if that counts.

I took the SAT Reasoning today . . . I probably got a perfect unless I misbubbled something or they don't like my essay.

The writing section gives a fair amount of leeway in getting an 800--from what I understand, a perfect essay will buy a few errors on the MC, and a clean MC will allow for an off-base essay. A friend dislikes writing/English significantly more than I do, and we both got 800 on the writing section (although not on the 01/2008 exam.)

Has anyone here gotten 800 on the Latin SAT II?

I did pretty crappily (cumulative of 1850, but I've in 7th grade, so it doesn't really matter), but my brother got perfect on both math and CR

wow...beasts.
i'm in 7th grade and i got
800 math
660 writing
580 CR

hey gj lawerence! =D
join the perfect math/phail at reading club!
i hear bshen also got perfect math?

I think I just revived an old topic...
but:
I had 800 writing, w/ perfect multiple choice and 10 essay
720 math (Don't laugh). Guess that was a sign...
and 740 verbal

I took it in 7th grade.

hmm this academicsrule guy is crazy.

i phailed. majorly.

i got 690 on critical reading
590 on writing
690 on math (i know, i suck.)
im not entirely sure what happened to my math and writing.
oh well.

I got an 800 in the Math section (big werp) as an eighth grader, and got an 2120 overall.

Yo I had a pretty bad CR score. That's what I'm really worried about

lol I'm a girl.
Maybe that's why I suck at math? j/k. *cough* I never knew a girl has never won mc nats... :whistling:

[quote="tokenadult"]Perfect scores are rare. I'd expect them mostly among native speakers of English.

[url=http://www.collegeboard.com/prod_downloads/highered/ra/sat/composite_CR_M_W_percentile_ranks.pdf]SAT scores in class of 2007[/url][/quote]

Wow, the 600 (combined score) is horrible!
127 failed

@DonkeyKingKong
Anyone who pretends to be "Black" or "Ghetto" are either obsessed with rap or they have issues.

200 is the minimum score for any section of the SAT right?

i got 800M 780CR 660W in 7th grade and havent taken it since. :(
i have a feeling im a lot worse now though

My friend matt got a perfect in math. 800 710 w 640 cr

Wow I'm going into 7th and I got the worst scores ever: (Not in Jan though)
[hide="Crappy scores"]
SAT
M: 690
R: 500
W: 500
[/hide]

At least that beats 1060 that someone I know got (he's a cool person though).