geometry
symmetry
angle bisector
geometry unsolved

[quote="AndrewTom"]Two circles touch internally at $ M$. A straight line touches the inner circle at $ P$ and cuts the outer circle at $ Q$ and $ R$. Prove that angle $ QMP \equal{}$ angle $ RMP$.[/quote]

Let $ l$ be a line That touch the Out and the incircle at $ M$ and Let $ L$ be a point in $ l$  and Let $ I$ be the intersection of $ (MP)$ with The out circle , Then $ \angle LMP \equal{} \angle QPM$ and $ \angle LMP \equal{} \angle LMI \equal{} \angle MRI$ and Thus $ \angle QPM \equal{} \angle MRI \equal{} \angle RPI$ in $ \triangle RPI$ and in $ \triangle MIR$  we have $ \angle RPI \plus{} \angle PIR \plus{} \angle PRI \equal{} \angle PMR \plus{} \angle MRI \plus{} \angle MIR$ Which implies that $ \angle PMR \equal{} \angle PRI$ and since $ M,Q,R$ and $ I$ are cocyclique we get $ \angle QRI \equal{} \angle PRI \equal{} \angle QMP \Rightarrow \angle QMP \equal{} \angle PMR$

[i][b]Figure:[/b][/i]:

[img]http://img62.imageshack.us/img62/327/swe2r.png[/img]

You can see this lemma http://www.mathlinks.ro/Forum/viewtopic.php?t=323691

Or invert about a circle centered at M, so then we reduce it to a very simple diagram whose solution is by symmetry only.

Cheers,

Rofler

why cant we use tangent secant property out here ? can get the answer in 2 steps

1) It happens the two circles are homothetic at $ M$, hence the tangent to the external circle at the point of intersection of $ MP$ with it will be parallel to $ RQ$, done.
2) It has been posted before (few times?)
3) If the subject circles are externally tangent to each other, $ MP$ is the external angle bisector of $ \angle QMR$ then.

Best regards,
sunken rock


I am told that there is an olympiad taking place in Sudbury on May 15th.  Does anyone know what it is called, and where I can find more information on it? (previous exams and scores, anything like that)

Maybe you are thinking of the Ontario Math Olympiad for Grade 7/8 students??? As far as I know (not much) this competition is for teams of 4 that won their regional competitions... and I don't think there is any info about it online... Not too sure...

Seems likely that's it, thanks.  I'm supposed to be in contact with the organizers, but they've been very slow about sending me information.

Looks like there's some background and sample questions on the O.A.M.E. website.

Isn't it in perterburouh (miss spelling) ths year? Thats what the people at the contest told our team.

Actually, I've found the contest I was looking for.  It's very local (just one school) and not an olympiad.  Thanks for the help, though.


linear algebra
matrix
quadratics
inequalities

let A,B are square matric  satify : A*B + 3*A+2008B=0;
 PROVE A*B=B*A; and det(3*A*A+3*A*B+2008B*B)>0

a) $ A\equal{}\begin{pmatrix}
a_{11}&\cdots&a_{1n}\\
\vdots&\ddots&\vdots\\
a_{n1}&\cdots&a_{nn}
\end{pmatrix}$, $ B\equal{}\begin{pmatrix}
b_{11}&\cdots&b_{1n}\\
\vdots&\ddots&\vdots\\
b_{n1}&\cdots&b_{nn}
\end{pmatrix}$

so $ AB\equal{}\begin{pmatrix}
a_{11}b_{11}\plus{}\cdots\plus{}a_{n1}b_{n1}&\cdots&a_{11}b_{1n}\plus{}\cdots\plus{}a_{1n}b_{nn}\\
\vdots&\ddots&\vdots\\
a_{n1}b_{11}\plus{}\cdots\plus{}a_{nn}b_{n1}&\cdots&a_{n1}b_{1n}\plus{}\cdots\plus{}a_{nn}b_{nn}
\end{pmatrix}$

and $ BA\equal{}\begin{pmatrix}
b_{11}a_{11}\plus{}\cdots\plus{}b_{1n}a_{n1}&\cdots&b_{11}a_{1n}\plus{}\cdots\plus{}b_{1n}a_{nn}\\
\vdots&\ddots&\vdots\\
b_{n1}a_{11}\plus{}\cdots\plus{}b_{nn}a_{n1}&\cdots&b_{n1}a_{1n}\plus{}\cdots\plus{}b_{nn}a_{nn}
\end{pmatrix}$

JRav: Was that supposed to have anything to do with this problem? If you just want to practice, that's what the Test Forum is for.

----

We have:

$ AB\plus{}3A\plus{}2008B\equal{}0$

$ AB\plus{}3A\plus{}2008B\plus{}6024I\equal{}6024I$

$ (A\plus{}2008I)(B\plus{}3I)\equal{}6024I.$

Since $ A\plus{}2008I$ and $ B\plus{}3I$ are scalar multiples of inverses of each other, they commute. It follows that $ A$ and $ B$ commute.

Since $ A$ and $ B$ commute, they are simultaneously upper triangularizable over $ \mathbb{C}.$ If we knew that the eigenvalues were real, we would have the second part because $ 3x^2\plus{}3xy\plus{}2008y^2$ is a positive definite quadratic form. However, I see nothing to guarantee the eigenvalues are real.

At the moment, I'm not sure I believe the second part, although I haven't constructed a counterexample yet.

For the second part, if $ z$ is a root of $ 3x^2\plus{}3x\plus{}2008$, then because $ z$ is nonreal and $ A,B$ commute, $ 3A^2\plus{}3AB\plus{}2008B^2\equal{}3(A\minus{}zB)(A\minus{}\overline{z}B)$.

If $ A,B$ were specifically real matrices, then $ A\minus{}\overline{z}B\equal{}\overline{A\minus{}zB}$ and determinants preserve complex conjugation, so we would get $ \text{det}(3A^2\plus{}3AB\plus{}2008B^2)\equal{}3|\text{det}(A\minus{}zB)|^2\geq 0$. (The strict inequality doesn't hold without extra conditions, since there's a stupid choice of $ A,B$ that would make the determinant 0.)

However, if $ A,B$ can be any complex matrices that satisfy the equation, this method doesn't work, but I don't think $ 3A^2\plus{}3AB\plus{}2008B^2$ would necessarily even have a real determinant.


Let $ABC$ a triangle such that $|AB| \neq |AC|$, and let $D$ the intersection between the tangent of the circumcircle of $ABC$ in $A$ and the line $BC$. Let $E$ (resp. $F$) the point on the perpendicular bissector of $[AB]$ (resp. $[AC]$) such that $BE$ (resp. $CF$) is perpendicular to $BC$. Show that $D,E,F$ are colinear.

What is "mediator"?
Sorry again for my poor knowledge and horizon.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=3659 .

  darij

[b]Nice problem ![/b] It is wellknown that $\frac{DB}{DC}=\left(\frac cb\right)^2.$

$\triangle BEO\sim ABC\sim COF\Longrightarrow \frac{BE}{R}=\frac cb\ \wedge\ \frac{CF}{R}=\frac bc\Longrightarrow\frac{BE}{CF}=\left(\frac cb\right)^2.$

Therefore $\frac{DB}{DC}=\frac{BE}{CF}$ and from the Thales' theorem results $D\in EF.$

[u]Remark.[/u] The point $D$ belongs to the tangent in the point $O$ at the circumcircle of the triangle $OEF$ !. See http://www.mathlinks.ro/Forum/viewtopic.php?t=67243

\[ \underline {\overline {\left| \ Merry\ Christmas\ !\ \right| }}\ \wedge \ \underline {\overline {\left| \ A\ Happy\ New\ Year\ !\ \right| }}  \]

It's exactly the same proof that I have found. :)

See[i] Labiliau[/i], and the mentioned extension.


calculus
integration
limit
logarithms
absolute value
real analysis
real analysis solved

$f:[0;1]\rightarrow ]0;+\infty[$ continous

$u_n=(\int_{0}^{1}f^\frac{1}{n}(t)dt)^n$

It is known that $\lim u_n = exp(\int_{0}^{1}\ln f)$

Prove that

$\lim \;(\frac{u_n}{e^{\int_{0}^{1}\ln f}})^n=exp(1/2\int_{0}^{1}(\ln f)^2)-1/2(\int_{0}^{1}\ln f)^2)\geq 1$

Thank's for the correction Kent.

—————————

Another question in 2020

Find an equivalent of

$y(n)=(\frac{u_n}{e^{\int_{0}^{1}\ln f}})^n - exp(1/2\int_{0}^{1}(\ln f)^2)-1/2(\int_{0}^{1}\ln f)^2) $

Correction, Moubinool:

$\lim\,\left(\frac{u_n}{e^{\int_0^1\ln f}}\right)^n= \exp\left[\frac12\int_0^1(\ln f)^2-\frac12\left(\int_0^1\ln f\right)^2\right]$

Some preliminaries:

$f$ continous and $f:[0,1]\mapsto(0,\infty)$ implies that the range of $f$ is bounded away from zero and that $\ln f$ is a bounded function.

I'm going to be using Taylor polynomials. It doesn't matter that nothing was said about the differentiability of $f$; what matters it that the exponential and logarithm are smooth.

The "big-O" notation will be uniform: that is, a statement that a certain quantity is, for instance, $O(n^{-3})$ would mean that there is a constant $M$ such that for all values of $x$ and for all $n$, the absolute value of that quantity is $\le M/n^3.$

Notation: let $g=\ln f$ and let $v_n=\ln u_n.$ Then

$v_n=n\ln\left(\int_0^1e^{g/n}\right).$

$\ln\left[\left(\frac{u_n}{e^{\int_0^1\ln f}}\right)^n\right]= n\left(v_n-\int_0^1g\right).$

Now, the argument:

$v_n=n\ln\left(\int_0^1 e^{g/n}\right).$

$e^{g/n}=1+\frac gn+\frac12\,\frac{g^2}{n^2}+O(n^{-3}).$

$\int_0^1 e^{\g/n}=1+\frac1n\int_0^1 g+\frac1{2n^2}\int_0^1 g^2+O(n^{-3})$

$\ln(1+x)=x-\frac{x^2}2+O(x^3)$, so

$\ln\left(\int_0^1 e^{g/n}\right)=\frac1n\int_0^1 g- \frac1{2n^2}\left(\int_0^1 g\right)^2+ \frac1{2n^2}\int_0^1 g^2+O(n^{-3})$

Multiply by $n$:

$v_n=\int_0^1g+\frac1{2n}\left(\int_0^1g^2- \left(\int_0^1g\right)^2\right)+O(n^{-2})$

Finally, $n\left(v_n-\int_0^1g\right) = \frac12\left(\int_0^1g^2-\left(\int_0^1g\right)^2\right)+O(n^{-1})$

Taking limits as $n\to\infty$ finishes our result.

Expressing the same result (with the same proof) in slightly different language:

Suppose $X$ is a random variable whose range is contained in $[a,b]$ with $0<a<b<\infty.$ Let $u_n = \left(E(X^{1/n})\right)^n.$

Then $\lim\,\left(\frac{u_n}{e^{E(\ln X)}}\right)^n= e^{\frac12\mathrm{Var}(\ln(X))}.$

Since the variance of any random variable is nonnegative, it is clear that this result is $\ge 1.$


Hello, this is  my problem : I  create  a  figure ussing a  greek  program and i put  greek letters on  points . But when i  use  go2pdf to  turn  it into a  pfd  file (to insert it in a  tex  document)  the  letters  appeared  ''Chinese''....(with all  respect to  Chinese  people, but i  want to make myself clear). What  can i  do  about it?
Thank  you.

When you go to print using go2pdf, hopefully in the print dialog box there is an option for "Printer Preferences" or something like that.  Click that, then (hopefully) there is an option which is something like "Embed all fonts in PDF", which should be checked.  (Or perhaps the converse, "Do not embed fonts in PDF", which should be unchecked.)  But that's basically what you probably need to do -- make sure that your original document's fonts are carried over to the pdf document -- and there must be a switch somewhere which makes it happen.

(I don't know if these options actually exists, since I don't use go2pdf.)


calculus
integration
logarithms
trigonometry
calculus computations

\[ \int\limits_0^{\frac{\pi }
{4}} {\log (1 \plus{} \tan \theta )} d\theta
\]


[quote="symmetry"]T = period of pendulum
L = length of pendulum [in feet]

Solve for L if T = 2 seconds.

2 = 2PI(sqrt{L/32})[/quote]

[hide]
$2 = 2\pi\sqrt{\frac{L}{32}}$
To get rid of the radical, square both sides
$4 = \pi^2\frac{L}{8}$
Divide by pi^2, multiply 8
$\frac{32}{\pi^2} = L$

$\frac{32}{\pi^2}$?
[/hide]

[quote="ccy"]
$\frac{32}{\pi^2}$?
[/quote]
yes, that is the answer

[quote="symmetry"]T = period of pendulum
L = length of pendulum [in feet]

Solve for L if T = 2 seconds.

2 = 2PI(sqrt{L/32})[/quote]

[hide]
$2=2*\pi(\sqrt{\frac{L}{32}})\\\frac{1}{\pi}=\sqrt{\frac{L}{32}}\\\frac{1}{\pi^2}=\frac{L}{32}\\L=\frac{32}{\pi^2}$
[/hide]

Good stuff here.  Thank you all.  This questions was really tricky.


What is the maximum number of intersection points in a diagram with two circles of unequal diameter and a straight line?

The two circles can together have up to 2 intersection points.

When the straight line is drawn, it can have up to 2 intersection points with each circle.

Total maximum number of intersection point is $ 2\plus{}2\plus{}2\equal{}\boxed{6}$ intersection points.


modular arithmetic
number theory unsolved
number theory

what is the remainder when
2^1990 divided by 1990?


induction
function
algebra
domain
algebra unsolved

Let $ f:\mathbb{Q}\to\mathbb{Q}$ satisfying $ f(2x^{2}\plus{}y^{2})\equal{}2f(x^{2})\plus{}f(y^{2})\minus{}100$ and $ f(100)\equal{}200$. I've proved that $ f(nx^{2})\equal{}nf(x^{2})\minus{}50(n\minus{}1)$, by induction, and then tried to generalize to $ f(\frac{n}{m}x^{2})\equal{} nf(\frac{x^{2}}{m})\minus{}50(n\minus{}1)$ but I failed to prove this too by induction. Then it would trivially follow that $ f(x)\equal{}\frac{3}{2}x\plus{}50$. Is there a faster way to do this?

Yes we will solve eqution function on $ x\geq 0$
If $ x<0$ then it take arbitrary value.

[quote="MellowMelon"]First of all, you can't even get all rationals from $ x^2$, $ y^2$, or $ 2x^2 \plus{} y^2$, particularly the negatives (but also some positives like $ 2/3$). The values you can't get can map to anything on $ f$. So this function has loads of solutions.

Also, making the substitution $ g(x) \equal{} f(x) \minus{} 50$ cleans things up. The equation becomes $ g(2x^2 \plus{} y^2) \equal{} 2g(x^2) \plus{} g(y^2)$ with $ g(100) \equal{} 150$ and $ g(0) \equal{} 0$.[/quote]

Allowing $ x,y \in \mathbb{C}$ I think you can obtain all rationals, with some limitations, of course. Or am I wrong?

Anyway, how can be solved that induction? By fixing $ m \in \mathbb N$ it's complicated.. Isn't there some "double" induction?

Well, if $ x,y \in \mathbb{C}$, of course you could obtain all rationals, even all complex numbers. But allowing $ x,y \in \mathbb{C}$, there is still the restriction that $ x^2$ and $ y^2$ are both in $ \mathbb{Q}$, since $ f$ takes a rational number. So we may as well just change the problem to $ f(2x \plus{} y) \equal{} 2f(x) \plus{} f(y) \minus{} 100$ for rationals $ x,y$. This of course looks a lot easier to solve.

Can you clarify exactly what $ x,y$ are?

I know nothing about them, except that $ x^2,y^2 \in \mathbb Q$, as you said..

If I proved that $ f(nx^2)\equal{}nf(x^2)\minus{}50(n\minus{}1)$ by induction on $ n$, is it correct to apply induction on $ m$ in $ f(\frac{n}{m}x^2)\equal{}nf(\frac{x^2}{m})\minus{}50(n\minus{}1)$?

[quote="Stewie"]Let $ f: \mathbb{Q}\to\mathbb{Q}$ satisfying $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ and $ f(100) = 200$. I've proved that $ f(nx^{2}) = nf(x^{2}) - 50(n - 1)$, by induction, and then tried to generalize to $ f(\frac {n}{m}x^{2}) = nf(\frac {x^{2}}{m}) - 50(n - 1)$ but I failed to prove this too by induction. Then it would trivially follow that $ f(x) = \frac {3}{2}x + 50$. Is there a faster way to do this?[/quote]

Hello Stewie,

I think you've not well understood what Mellowmellon tried to explain you :

1) the domain of validity of the equation $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ is very important to know :
If this equation is valid $ \forall x,y\in \mathbb{Q}$, it's very different of the case where this equation is valid $ \forall x,y\in\mathbb{R}$ such that $ x^{2}$ and $ y^{2}\in\mathbb{Q}$ or the case where this equation is valid $ \forall x,y\in\mathbb{C}$ such that $ x^{2}$ and $ y^{2}\in\mathbb{Q}$

2) without any precision, we generally must consider that we have $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ $ \forall x,y\in\mathbb{Q}$
Then, there are, for example, no couple or rationals numbers such that $ 2x^{2} + y^{2}=5$. So, $ f(5)$ can't be involved in any equation and so $ f(5)$ may have any value.
As a consequence, your induction for $ f(nx^{2}) = nf(x^{2}) - 50(n - 1)$ is likely wrong since it would imply $ f(1)=\frac{103}{2}$ and $ f(5)=\frac{115}{2}$
As another consequence, it's useless to try to find the other induction you're looking for, since the result is wrong.

So, I suggest you :

a) Give us the fundamental precision about the domain of validity of the functional equetion :
a1) : $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ $ \forall x,y\in\mathbb{Q}$
a2) : $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ $ \forall x,y\in\mathbb{R}$ such that $ x^{2}$ and $ y^{2}\in\mathbb{Q}$
a3) : $ f(2x^{2} + y^{2}) = 2f(x^{2}) + f(y^{2}) - 100$ $ \forall x,y\in\mathbb{C}$ such that $ x^{2}$ and $ y^{2}\in\mathbb{Q}$
a4) : Other

b) If the answer to first question is a1), then post your induction demo for $ f(nx^{2}) = nf(x^{2}) - 50(n - 1)$ and we'll check it because I think it's wrong for $ f(5x^{2})$

And I apologize if I made a mistake in this message.


Here's a problem that once came to my mind, and I asked to my math teacher.

Can the numbers 1, ..., 2n, for $n\in\mathbb{N}$, be paired in such a way that the sum of the members of each couple is a perfect square?
Or, better formulated: for which values of n is this possible?

Example: it is trivially true for n=4: the pairs are (1,8), (2,7), (3,6), (4,5), and their sum is in all cases 9.

Now, THIS is a beginner's problem. :D  :D  :D

Well, I don't know how easy or hard could this problem be. As I said earlier, it just came to my mind. If it was divine inspiration, God forgot to tell me the difficulty level :D :D :D.
No, really, I don't know, but it looks to me simple enough (when stated).


calculus
derivative
function
Putnam
real analysis
real analysis unsolved

Let $ f: R-->R $ be a function satisfying :
 $ f' , f'' , f''' > 0 $ for all x and $ f(x) \geq f'''(x) $ for all x
 Prove that : $ f'(x) < 2f(x) $

Two suggestions for romano: learn to source properly, or learn to research for sources. Either way, must learn to give credit to competition problems by sourcing them. unknown = i don't know and will not look for it.


Solve for x:

((log base 3 of x) ^ 2) + (log base 3 of (x^2)) + 1 = 0

This one has me stumped.


Find the distance from the point (1,1,1) to the plane containing (1,0,0) , (0,2,0), and (0,0,4).


[size=150]You should give examples ，Given concrete expression。
BQ[/size]


δινεται η εξισωση $ ax\plus{}by\equal{}c$
$ a>0$,$ (a,b)\equal{}1$.
ν.δ.ο. η εξισωση εχει μια τουλαχιστον μια ακεραια λυση $ (x_0,y_0)$
me $ y_0$  ανηκει στο συνολο{(0,1,2,3,...,α-1)}

Η εξισωση ειναι γραμμικη διοφαντικη εξισωση και οι συντελεστες των μεταβλητων της ειναι πρωτοι μεταξυ τους, οποτε οι ακεραιες λυσεις της εξισωσης δινονται απο τους τυπους: $ y \equal{} f(t) \equal{} y_0 \minus{} at$ και $ x \equal{} x_0 \plus{} bt$ οπου ο $ t$ ειναι μεταβλητος ακεραιος και $ (x_0, y_0)$ ειναι μια ακεραια λυση. Το ευρος του διαστηματος $ [\frac {y_0 \minus{} a}{a}, \frac {y_0}{a}]$ ειναι ισο με $ \frac {y_0}{a} \minus{} \frac {y_0 \minus{} a }{a} \equal{} \frac {a}{a} \equal{} 1$, αρα στο διαστημα αυτο υπαρχει ενας τουλάχιστον ακεραιος αριθμος $ u$

Διακρινουμε 2 περιπτωσεις:

1) Αν ο ακεραιος $ u$ ειναι το ενα ακρο του διαστηματος τοτε και το αλλο του ακρο θα ειναι επισεις ακεραιος. Οποτε για $ t \equal{} \frac {y_0}{a}$ στους τυπους των λυσεων τις διοφαντικης εξισωσης προκειπτει η ακεραια λυση της εξισωσης $ (x, y) \equal{} (m, k)$, με $ k \equal{} y_0 \minus{} a\frac {y_0}{a} \equal{} 0$ που ανηκει στο συνολο {0, ... α-1}.

2) Αν ο ακεραιος $ u$ δεν ειναι ακρο του διαστηματος τοτε θα ανηκει στο εσωτερικο του διαστηματος αυτου. Τοτε για $ t \equal{} u$ προκειπτει η ακεραια λυση της εξισωσης $ (x, y) \equal{} (m, k)$ με $ k \equal{} f(u) \equal{} y_0 \minus{} au$ και $ m \equal{} x_0 \plus{} bu$, ακομα η σηναρτηση $ f(t)$ απο την οποια προκειπτουν τα $ y$ που επαληθευουν τη διοφαντικη εξισωση για τις διαφορες τιμες του $ t$ ειναι γνησιως φθινουσα με $ f'(t) \equal{} \minus{} a < 0$, αρα αφου $ \frac {y_0 \minus{} a}{a} < u < \frac {y_0}{a}$ προκειπτει οτι $ f(\frac {y_0}{a}) < f(u) < f(\frac {y_0 \minus{} a}{a}) \leftrightarrow 0 < k < a$. Οποτε, αφου ο $ k$ ειναι ακεραιος, εχουμε $ 0 < k \leq a \minus{} 1$ και η αποδειξη ολοκληρωθηκε.

$ Qed$

Ειχα βαλει μια λανθασμενη λυση αλλα τη διορθωσα και νομιζω πως ειναι σωστη τωρα. Check it :P

σωστη μου φαινεται nick!nice :)
στο μυαλο μου εχω μια λυση με χρηση ευκλειδιας διαιρεσης(take it as a hint... :wink: )


Find a closed form for $ \sum_{n \equal{} 1}^{\infty}3^{n \minus{} 1}\sin^3\left(\frac {a}{3^{n}}\right)$.

[hide="Incorrect solution"]
Seeing cubed sines, we think of the identity $ \sin 3\theta \equal{} \minus{} 4\sin^3\theta \plus{} 3\sin\theta\implies \sin^3\theta \equal{} \frac {3\sin\theta \minus{} \sin 3\theta}{4}$.

Applying this, the sum becomes
\[ \frac {1}{4}\left[\sum_{n \equal{} 1}^{\infty}3^n\sin\left(\frac {a}{3^n}\right) \minus{} \sum_{n \equal{} 1}^{\infty}3^{n \minus{} 1}\sin\left(\frac {a}{3^{n \minus{} 1}}\right)\right]\]
That difference of sums telescopes, leaving only $ \boxed{ \minus{} \frac {1}{4}\sin a}$.
[/hide]

[hide="Answer"]
The answer is supposed to be $ \frac {a \minus{} \sin a}{4}$.
[/hide]

Edit:
Oops, turns out series like these that alternate and whatever can't be manipulated so harshly (http://en.wikipedia.org/wiki/Riemann_series_theorem and so on)
If you take the sum to a number $ k$ and then take the limit as $ k$ grows boundlessly, a $ \plus{}a/4$ appears.

You are very close. Just use your telescoping idea to calculate the partial sum of the series  $ \sum_{n = 1}^{m}3^{n - 1}\sin^3(\frac {a}{3^{n}}) =  \frac {1} {4}  ( 3^m\sin \frac {a} {3^m} - \sin a  )$.
Now, $ \lim_{m \to \infty} 3^m\sin \frac {a} {3^m} = a$, and you get your answer.

The reason is the series is not absolutely convergent (for which any manipulations and permutations are allowed).

Yeah, I realized the same thing.
Thanks!


For training, a cyclist rides the same distance 3 days a week,
twice the distance 2 days a week and three times the distance
1 day a week. What portion of the weekly total distance did the
cyclist ride the day she rode the farthest? Express your answer
as a common fraction.

For sake of ease, let's say that the distance the cyclist rides 3 days a week is 1 mile. That means two days a week, she rode 2 miles, and 1 day a week, 3 miles. That makes a total of 3(1) + 2(2) + 1(3) = 10 total miles. The farthest distance she rides in a week is 3 miles, so our answer is $ \boxed{\frac{3}{10}}$.


calculus
analytic geometry
integration
calculus computations

suppose that F= < (x)*(e^y)*(z),  (y^2)*(e^x)*(y)*(z), (x^2)+sin(xy+z^2) + 42>.  consider the ellipsoid (x^2)/(a^2)+ (y^2)/(b^2) +(z^2)/(c^2)=1.  Calculate $\iint_{E}\mbox{(\nabla\times\bold{F})}\cdot\bold{dS}$

$\iint_{E}(\nabla\times\bold{F}) \cdot \bold{dS}= \underset{\text{inside }E}{\iiint}\nabla \cdot (\nabla\times\bold{F}) \, dV = 0$

because $\nabla \cdot (\nabla\times\bold{F}) = 0$.

sos440 already answered it using the divergence theorem, but here's a second argument using Stokes' Theorem.

Take some closed curve $C$ on the ellipsoid $E.$ (It doesn't really matter what closed curve. If you want to visualize it, take the intersection with some coordinate plane.) That curve splits the the ellipsoid into two surfaces. Both of those surfaces have $C$ as the boundary curve, but in order to keep a consistent orientation of $C,$ one of those surfaces (call it $E_{1}$) is oriented outward and the other (call it $E_{2}$) is oriented inward.

Then by Stokes $\int_{C}\bold{F}\cdot\bold{dr}=\iint_{E_{1}}(\nabla\times\bold{F})\cdot\bold{dS}=\iint_{E_{2}}(\nabla\times\bold{F})\cdot\bold{dS}$

But $\iint_{E}(\nabla\times\bold{F})\cdot\bold{dS}= \iint_{E_{1}}(\nabla\times\bold{F})\cdot\bold{dS}-\iint_{E_{2}}(\nabla\times\bold{F})\cdot\bold{dS}=0.$


What PNT means ?

I think that max is 27/8.
The problem reduces to show that
$(x^2+y^2+z^2)^3\geq8(x^3y^3+y^3z^3+z^3x^3)$.

PNT=probably no true solution! Seriously speaking, this inequality is damn hard and I cannot find anything without tough computations. Vasc, did you prove it?

Imposible ! I think both darkmaster and PNT haven't any solution !

And I think that you are not friend of them, right?  ;)  :D

Seriously people, I'm not an inequality fan, so I shouldn't be postong here (it's somewhat like spamming :)), but will someone pleeeease (:D) tell me who (or what :maybe:) PNT is?


prove that from every set of 200 integers we can choose a subset of 100 with the total sum divisible by 100.

I'm going to bump this because I've seen this problem in several different places, thought about it for a long time, but haven't been able to solve it.  I can't find a solution anywhere, either.  I can solve the problem if we switch 100 and 200 with $ p$ and $ 2p$, but this doesn't seem to help because my method doesn't generalize to non-primes.

Can anyone solve this?

This is a rather hard problem, in my mind. It appears in Peter Winkler's book under "toughies", and Winkler mentions it is actually a published theorem by $ \mbox{Erd\"{o}s}$, Ginzburg and Ziv. It also appeared in the 1970 All Soviet Union Mathematical Competition, which I find quite surprising given the complexity of the solution. Perhaps there's a clever trick we're all missing.

I will explain how to reduce the problem to the prime case, which you already solved. If needed I'll add the solution for the prime case.

Following Winkler, I'll call a set of integers "$ n$-flat" if its sum is divisible by $ n$. Consider the following two statements:

$ A(n)$. Every set of $ 2n$ integers contains an $ n$-flat subset of size $ n$.
$ B(n)$. Every $ n$-flat set of $ 2n$ integers can be split into two disjoint $ n$-flat subsets of size $ n$.

The original problem is to prove $ A(100)$. It is obvious that $ A(n)$ implies $ B(n)$. But $ B(n)$ also implies $ A(n)$: given any set of size $ 2n$, replace one element with some other integer so that the set becomes $ n$-flat. The modified set now contains two disjoint $ n$-flat subsets, by $ B(n)$; one of them does not contain the modified integer, so it is contained in the original unmodified set so $ A(n)$ is true.

The strategy of the proof is now to assume $ A(a)$ and $ B(b)$ and deduce $ B(ab)$.

Let $ S$ be a set of $ 2n \equal{} 2ab$ integers which is $ n$-flat. Note that $ S$ is also $ a$-flat. Pick any subset of $ 2a$ integers in $ S$. Using $ A(a)$ we know that this subset contains an $ a$-flat set of size $ a$. Call it $ S_1$. Remove $ S_1$ from $ S$ and repeat: pick a subset of size $ 2a$ and use $ A(a)$ to find another subset of size $ a$ which is $ a$-flat. This can be continued to generate disjoint sets $ S_1, S_2, \ldots, S_{2b \minus{} 1}$ all of which have size $ a$ and are $ a$-flat. But the remaining set, call it $ S_{2b}$, is also $ a$-flat since $ S$ itself is $ a$-flat.

Since each $ S_i$ is $ a$-flat, the sum of its elements is $ ak_i$ for some integers $ k_i$. The total sum of those $ ak_i$ is the total sum of $ S$, which is divisible by $ ab$. Therefore the sum of the $ k_i$ is divisible by $ b$. So the numbers $ k_i$ constitute a set of $ 2b$ integers that is $ b$-flat. Use $ B(b)$ to split the $ k_i$ into two subsets, each of which is $ b$-flat.

Let $ P$ be the union of the sets $ S_i$ which correspond to the $ k_i$'s in one of those subsets, and let $ Q$ be the similar union for the other subset. Both $ P$ and $ Q$ have size $ ab$, and each of them has sum divisible by $ ab$. Thus $ P$ and $ Q$ provide the required splitting of $ S$ into two $ n$-flat subsets of size $ n$.

We've now shown that $ A(a)$ and $ A(b)$ implies $ A(ab)$. To probe $ A(n)$ in general it remains to prove $ A(p)$ for $ p$ prime. I'll save that for a separate post, unless matt276eagles steps in first to explain his approach.

Thanks randomgraph, that's a really nice proof.

Here's my proof for primes:

[hide]Suppose we have a set of $ 2p$ integers.  If $ p$ of these integers are in the same residue class modulo $ p$, then these integers will have sum divisible by $ p$.  Therefore assume that no $ p$ of the $ 2p$ integers are in the same residue class modulo $ p$.

Then we can partition the $ 2p$ integers into $ p$ pairs such that each pair consists of two distinct integers modulo $ p$.  To prove this we use induction to prove a stronger statement, that given a set of $ 2n$ integers such that no $ n \plus{} 1$ are in the same residue class modulo $ p$.  The base case $ n \equal{} 1$ is obvious.  For the inductive step, if any residue class appears $ n$ times, then we use one of these elements to make a pair.  There can only be two such residue classes, so we can make a pair and still preserve the property that no more than half of the remaining set is in the same residue class.

Now we prove that given $ n$ pairs ($ n\le p$) with distinct integers modulo $ p$, there are at least $ n$ distinct possible sums modulo $ p$ when we take one element from each pair.  Again we use induction.  The base case $ n \equal{} 1$ is obvious.  If $ n$ pairs give us a set $ S$ of $ n$ distinct sums modulo $ p$, then take another pair $ (a,b)$ and consider the sets $ A \equal{} \{x: x \minus{} a\in S\}$ and $ B \equal{} \{x: x \minus{} b\in S\}$.  The set $ A\cup B$ has elements that are all possible sums when taking one element from each of $ n \plus{} 1$ pairs.  Note that $ A$ and $ B$ are cyclic shifts of each other, so for some $ 0<k<p$, if $ x$ is in $ A$ then $ x\plus{}k$ is in $ B$.

Suppose $ A$ and $ B$ are the same sets.  If $ x$ is in $ A$, then $ x\plus{}k$ is in $ B$ and therefore in $ A$ also, and continuing this process we find that $ x\plus{}mk$ is in $ A$ for all $ m$.  This cycles through all residues modulo $ p$, so $ A$ and $ B$ are both the complete set of residues modulo $ p$, and we're done.  If $ A$ and $ B$ are not the same sets, then $ A\cup B$ has at least $ n\plus{}1$ elements as desired.  Our induction is complete, so in particular given $ p$ pairs of distinct integers modulo $ p$, there are at least $ p$ distinct possible sums modulo $ p$ when we take one integer from each pair, which is what we wanted.[/hide]

EDIT: I explained the last part of the proof a little more; I'm pretty sure this works, but it may not be the best way.

[quote="randomgraph"]$ \mbox{Erd"{o}s}$[/quote] Use \H to get the correct accent marks: $ \mbox{Erd\H{o}s}$

Thanks - you're right, I was using the wrong accent.

At first I thought you were showing me how to do this without entering LaTeX and using mbox, but I guess that's not possible...

Well, Wikipedia articles tend to have correct accent marks, so you can go to the article and copy and paste.  There might be some way to do it directly as text, but if so, I don't know it.


Contact (via PM or email) one of the instructors of the class.

I ddint' think so. It costs 500 dollars.

[quote="rrusczyk"][quote="chess64"]Once we've enrolled in IS, how do we get the free course without paying $\$120?[/quote]

If you don't contact us as joml mentioned, we'll contact you.  We'll send out an email near the beginning of the IS asking for everyone's selection of course for the IS.[/quote]

MCrawford has already contacted me. :)


For many of us, myself included, the only thing that we realli need a boost on as far as verbal goes is vocab.

I was wondering...what vocab lists do/did u use?...and what did u think of them?

[quote="MithsApprentice"]For many of us, myself included, the only thing that we realli need a boost on as far as verbal goes is vocab.[/quote]

I think the Russians have a proverb "repetition is the mother of learning" (actually, now that I think of it, that's probably a Latin proverb originally), so I'll repeat myself. There may not be a royal road to mathematics, but there is DEFINITELY a royal road to high SAT I verbal scores: READ, READ, READ, READ, and READ. Jettison TV time for more reading time. Read the books that interest you, and read other books that you are curious about. Jettison video game time for more reading time. Read a few good magazines about grown-up topics. Jettison calculator game time for more reading time. Read "classic literature" that is supposed to be good for you, choosing the titles YOU like. To sum up, READ, READ, READ, READ, and READ.

By the way, did I mention how useful it is to read to prepare for the SAT I verbal test?

I would like to back up what tokenadult said -- a more effective strategy than studying lists is to try to read a variety of things to accumulate vocabulary and grammar.  I might recommend in addition that you read with a dictionary on hand, so every time you come across a word you don't know, you can look it up.  That would be very useful.  This of course works better over a longer period of time, but not necessarily well at all on a short-term basis.  Knowing a little Latin is handy, too, but that's probably overdoing it.

As a long-term strategy, reading is the best way to go.  But for the short term (less than a year before the test), I would recommend a specific focus on vocabulary.  One method is to take each of the practice tests in the book "10 Real SATs".  For each word there that you don't know (particularly on the question stems and correct answers), write down its definition on a note card and try to remember it.  The SAT often reuses words from previous tests.

I dunno if this makes any sense...but I think it would help to memorize a few hundred to a thousand or so words before we begin that reading campaign...

Personally, if i already know what the word means, then seeing it in the context of a passage would just help me to remember it b/c I rmr that I've seen it before. However, just reading with a dictionary and looking up words that I dont know doesnt help too much b/c I'll know that word that I just looked up...yet after a few weeks, it leaves the memory bank since I never spent a lot of time focusing on trying to remember it.

Instead, you might read and when you come across a word you don't know write it down on a list and then study that list.  If you really do read a lot, however, eventually it will all get in there.

[quote="cats..."]does it help that much to learn the vocabulary?  I think it helps somewhat, but i think even more important is being able to think your way out of the answer, without fully knowing the definitions.[/quote]

I must sadly disagree with this theory on the verbal section of the SAT, especially on analogies... If you don't know what the words mean, no reasoning in the world is going to get you the right answer. Even knowing half of the words will not get you the right answer. I spent many months, including all of last summer studying nothing but vocabulary in attempts to bring my 590 in verbal up. I only got it to a 610, o well :\, I never read as a child or until 11th grade as a matter of fact; that was my problem. Reading is something that must be started early or it doesnt help much, but it's definitely the best way to go.

As for vocab list... I have found that the book "30 Days to a more powerful vocabulary" works wonders... It didn't increase my SAT score much but i'm just stupid and there is no hope for me. My vocab however has improved emensly.

In addition, I spent last summer coding a vocabulary quiz system based off of the Caplain "Word Power" second edition. I managed to type out about 450 of the words and definitions into the database and the system works great. Unfortunately my webhost shut down on me and I do not have one currently. If I get one in the near future, I will put my vocab online quiz system back up for anyone to use.

a lot of times i don't know a lot of the words, but i usually do well in analogies / sentence completions.  I really do think you can reason your way through w/out knowing all of the words (obviously you need to know SOME of the words in the answer choices) and then you can make a bridge and get it down to 1 or 2 possibilities.

[quote="paladin8"]well, next year, you can say bye to those friendly analogies... i think this will greatly decrease the need for vocab words, except for some in sentence completion[/quote]

Paladin is referring to the [url=http://www.collegeboard.com/student/testing/newsat/about.html]planned changes in the SAT[/url], which begin next school year. The only way to know FOR SURE what effect the new format of the SAT will have is to do a detailed statistical study of before-and-after performance (something I'm sure ETS will do), but it seems to me that the royal road advice is STILL to READ, READ, READ, and READ. It's quite interesting to contrast the girls and the boys in the gifted homeschooling support group I am a member of here in Minnesota. The girls are all avid readers and generally mix up the genres of what they read quite well, and they CRUSH on the SAT I verbal section from an early age. Some of the boys are avid readers too, but some waste altogether too much time playing computer games (in my opinion as a member of the older generation) and don't take advantage of the lavish public library system here to learn broadly through reading. Even some of the boys who read a lot read altogether too much fantasy genre fiction and not enough genuine literature.

I'll have to see how my son fares on the SAT I (the scores should be out in a week or so) after three years of living in a non-English-speaking country and a few years of reading impairment because of strabismus. Fortunately, he now reads better and better day by day, and reads a lot. I thought it was unseemly (and inexpedient) to have him memorize word lists, so I just made sure we had a varied supply of library books in the house and had him read at liberty whenever he was caught up with (or trying to avoid [grin]) his math homework. Based on the practice tests in the book 10 Real SATs, that regimen alone, and increased familiarity with the test, was enough to raise his score substantially at age eleven. We'll see how he did on the real thing, but I remember from my days before the era of test prep that the verbal section of the SAT was EASY at age sixteen, after a childhood spent reading hundreds of library books.

Good luck to everyone still taking the "old" SAT I or the "new" SAT I.

[quote="paladin8"]well, next year, you can say bye to those friendly analogies... i think this will greatly decrease the need for vocab words, except for some in sentence completion, but thats the ez part... as for the new writing, ugh, bad news... stupid graders dont like giving out so many 1600s eh? well blah.[/quote]
Yeah, that's not so good. It was actually possible to reason your way through those without knowing what all the words meant, provided that you were somewhat literate in the language. Even without knowing what a word means, it's usually easy to tell if it's a verb, noun, or adjective/adverb - if the first half of the analogy compares two nouns, you better believe that the second half of it will as well. Reasoning like this, and knowing a little bit of vocabulary, was enough to get me a perfect score on the analogies.

can anybody summarize for me all the changes that are going to happen to the SAT
the only thing i heard was that they are going to add a writing section.


geometry
circumcircle
Euler
incenter
geometry solved

Hint: Prove that I, O and T are in the same line (where O is the circuncenter of ABC).

In my figure IT dose not perpendicular with EF . Something wrong here ???
 what does "center of gravity" mean ? Is it the same meaning with " centroid " ?

[quote="jarod"] what does "center of gravity" mean ? Is it the same meaning with " centroid " ?[/quote]
Of course.

Denote the middlepoints $M$, $N$ of the segments $[BC]$, $[EF]$ respectively. Then, from a well-known property, $AI\parallel MN$ a.s.o.

Can you post your solution more detail,I don't know your well-known property

[quote="bui_haiha2000"]Let ABC is a acute triangle.AI,BI,CI are bisector start from A,B,C. The line IX is perpendicular to the BC (X in the [BC]) The line IY is perpendicular to the AC( Y in the [CA]),the line IZ is perpendicular to the AB( Z in the [AB].
Let E in [AB] and F in [AC] satisfy that : BE=BC=CF.Let T is the center of gravity of triangle XYZ.
             Prove that IT is perpendicular to EF.[/quote]

The problem is almost completely solved on the forum:

The fact that $IO\perp EF$, where O is the circumcenter of triangle ABC, was proven at http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 ; see also http://www.mathlinks.ro/Forum/viewtopic.php?t=52596 , http://www.mathlinks.ro/Forum/viewtopic.php?t=30955 , http://www.mathlinks.ro/Forum/viewtopic.php?t=52563 , http://www.mathlinks.ro/Forum/viewtopic.php?t=20406 , http://www.mathlinks.ro/Forum/viewtopic.php?t=19133 , http://www.mathlinks.ro/Forum/viewtopic.php?t=44258 for different proofs and other properties of the configuration.

The fact that the line IO coincides with the line IT can be seen as follows:

First, since the points X, Y, Z lie on the incircle of triangle ABC, the center I of this incircle is the circumcenter of triangle XYZ. Also, the point T is the centroid of triangle XYZ. Thus, the line IT is the line joining the circumcenter and the centroid of triangle XYZ, i. e. the Euler line of triangle XYZ.

On the other hand, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 , the orthocenter of triangle XYZ, the incenter I of triangle ABC and the circumcenter O of triangle ABC are collinear; thus, the line IO passes through the circumcenter I and the orthocenter of triangle XYZ, so it must coincide with the Euler line IT of triangle XYZ.

  Darij


That's actually very useful, since all the past AMCs typically have something about that year (such as 2001, which had something about the largest possible sum of three different factors of it)

It might also be useful to know $2005=5\cdot 401$ and $2007=3^2\cdot 223$, as you may well be using those (although most people probably knew the $2005$ one from last year :P).

for the AMC its not really useful since you can TI-89 F2 Factor
but for AIME it helps

Since this topic is prime factors of 2006!!, I might as well point out that the prime factors of 2006!! are all the prime numbers less than or equal to 2006!. I will give a rough estimate by the prime number theorem that there are about $10^{3300}$ of them. I would also like to emphasize that you do not need to know all of them for the AMC.  ;)

[quote="eryaman"][quote="qweretyq"]for the AMC its not really useful since you can TI-89 F2 Factor[/quote]Unless you don't have a TI-89...[/quote]

In which case you would just download a simple factoring program...

Haha 2006!! !

That was what I was thinking when I read this thread.

[quote="deej21"]You can use an 89 on the AMC?  I thought the limit was 83/84...[/quote]

You certainly can (it's SAT permitted so it's AMC permitted as well).

Anything without a QWERTY keyboard is permitted for the AMC, I believe.


Let ABC be triangle .The bisector BF cuts the bisector CE at I .The line passing through I,perpendicular to EF ,cuts EF,BC respectively at P,Q .Find Angle BAC. Known that IQ=2IP
I have already sloved this pro,but my sol is very long and not good.
Do you have nice idea! :)
---
result : <BAC=60 o

This problem was posted here before not too long ago. Try looking for it (amfulger answered, but he didn't solve it completely; however, he showed that it works for A=60 deg).

oh,I known that! I sloved it. But I'm looking nicer new sol :)


What is the probability that a randomly drawn positive factor of $ 60$ is less than $ 7$?

$ \textbf{(A)}\ \frac{1}{10} \qquad
\textbf{(B)}\ \frac{1}{6} \qquad
\textbf{(C)}\ \frac{1}{4} \qquad
\textbf{(D)}\ \frac{1}{3} \qquad
\textbf{(E)}\ \frac{1}{2}$

60=2^2*3*5
There are 3*2*2=12 factors of 60
1,2,3,4,5,6 are all less then 7
6/12=1/2 therefore answer is E

Or...
$ 7<\sqrt{60}<8$, and neither $ 7$ nor $ 8$ is a factor of 60, so the probability is $ \dfrac 12$.

All of $1, 2, 3, 4, 5, 6$ are factors of $60$, and since the prime factorization of $60$ is $2^2 \cdot 3 \cdot 5$, it has $3 \cdot 2 \cdot 2 \rightarrow 12$ factors, so $\frac{6}{12} \rightarrow \boxed{1}{2}$.


Prove the inequality $ \sum_{k\equal{}1}^{n\minus{}1} \frac{n}{n\minus{}k}\cdot \frac{1}{2^{k\minus{}1}}<4\ (n\geq 2)$.

[hide="(somewhat ugly) solution"]
Proof by induction.
Let $ S_n \equal{} \sum_{k \equal{} 1}^{n \minus{} 1} \frac {n}{n \minus{} k} \cdot \frac {1}{2^{k \minus{} 1}}$.
Note that $ S_2 \equal{} 2 < 4$.
Define for all $ n \in \mathbb{N}_{\geq 2}$: $ R_n \equal{} \sum_{k \equal{} 0}^n \frac {1}{n \minus{} k} \cdot \frac {1}{2^k}$.
For $ n \in \mathbb{N}_{\geq 2}$ we have $ S_{n \plus{} 1} \equal{} \sum_{k \equal{} 1}^n \frac {n \plus{} 1}{n \plus{} 1 \minus{} k} \cdot \frac {1}{2^{k \minus{} 1}} \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} \frac {n \plus{} 1}{n \minus{} k} \cdot \frac {1}{2^k} \equal{} 1 \plus{} \frac {S_n}{2} \plus{} R_n$.
By induction we have $ S_n < 4$, but $ S_n < 4 \implies \frac {S_n}{2} \plus{} 1 < 3 \implies S_{n \plus{} 1} < 3 \plus{} R_n$.

Hence if we can prove that $ R_n \leq 1$ we are done.
Note that $ R_2 \equal{} 1 \leq 1$.
For $ R_n$ we have the recursion formula (if $ n \in \mathbb{N}_{\geq 2}$): $ R_{n \plus{} 1} \equal{} \sum_{k \equal{} 0}^n \frac {1}{n \plus{} 1 \minus{} k} \cdot \frac {1}{2^k} \equal{} \frac {1}{n \plus{} 1} \plus{} \frac {R_n}{2}$.
Hence (obviously) $ R_n \leq 1 \implies R_{n \plus{} 1} \leq 1$ if $ n \in \mathbb{N}_{\geq 2}$, so we are done.
[/hide]


inequalities
logarithms
inequalities unsolved

Setting:                     $ \sqrt {a \plus{} b \minus{} c} \equal{} x$    or $ x^2 \equal{} a \plus{} b \minus{} c$

                                  $ \sqrt {b \plus{} c \minus{} a} \equal{} y$    or  $ y^2 \equal{} b \plus{} c \minus{} a$

                                  $ \sqrt {c \plus{} a \minus{} b} \equal{} z$    or  $ z^2 \equal{} c \plus{} a \minus{} b$  where $ x,y,z > 0$ of course we get that:

      $ a \equal{} \frac {x^2 \plus{} z^2}{2}, b \equal{} \frac {x^2 \plus{} y^2}{2}, c \equal{} \frac {y^2 \plus{} z^2}{2}$ and $ x^2 \plus{} y^2 \plus{} z^2 \leq 3$.

Therefore our desired inequality becomes:

                        $ \frac {x^2 \plus{} 3}{x} \plus{} \frac {y^2 \plus{} 3}{y} \plus{} \frac {z^2 \plus{} 3}{z} \geq 12$.

 The equality case is possible when $ x \equal{} y \equal{} z \equal{} 1$.Can anyone prove my last inequality coz i got stuck now :( Thanks.
It seems easy but i can't figure it now.I will try again soon.

~Kostas

hi new_member ,you can use am-gm to prove your last inequality :

$ \sum(x\plus{}\frac{1}{x})\plus{}2\sum\frac{1}{x}\geq 6\plus{}2\frac{9}{\sqrt{3(x^2\plus{}y^2\plus{}z^2)}}\equal{}12$

:)

But $ x^2\plus{}y^2\plus{}z^2 \le 3$.  :wink:

Omg I can't believe i couldn't see this proof with AM-GM.Anyways Mehdi thanks a lot for your proof.
Felix,you're absolutely right.I edited it now.Sorry for my mistake.

Small proof of how we get $ x^2 \plus{} y^2 \plus{} z^2 \leq 3$

"From Cauchy-Schwarz we have that $ 3(a^2 \plus{} b^2 \plus{} c^2) \geq (a \plus{} b \plus{} c)^2$ or $ a \plus{} b \plus{} c \leq 3$ which after our subsitution  becomes the inequality that felix wrote."

~Kostas

But then the desired inequality doesn't become
\[ \sum \frac{x^2\plus{}3}{x} \ge 12.\]

Haha,Great.you're right.I guess i need some sleep :(
I will look at it again to see how i can fix things.Thanks for your corrections.

By simply substituting the values for a,b,c to the equation $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ we get:

     $ (x^2 \plus{} y^2)^2 \plus{} (y^2 \plus{} z^2)^2 \plus{} (z^2 \plus{} x^2)^2 \equal{} 12$.

Doing a few calculations we see that the above is equivalent to:

    $ x^4 \plus{} y^4 \plus{} z^4 \plus{} (x^2 \plus{} y^2 \plus{} z^2)^2 \equal{} 12$

But $ x^2 \plus{} y^2 \plus{} z^2 \leq 3$ therefore we get : $ x^4 \plus{} y^4 \plus{} z^4 \geq 3$

Hence our desired inequality to prove is :

   $ \frac {2x^2 \plus{} y^2 \plus{} z^2}{x} \plus{} \frac {2y^2 \plus{} x^2 \plus{} z^2}{y} \plus{} \frac {2z^2 \plus{} x^2 \plus{} y^2}{z} \geq 12$

By multiplying by $ x^2,y^2,z^2$ on both of the numerator and denominator of the fractions above respectively we get the following:

      $ \frac {2x^4 \plus{} (xy)^2 \plus{} (xz)^2}{x^3} \plus{} \frac {2y^4 \plus{} (xy)^2 \plus{} (yz)^2}{y^3} \plus{} \frac {2z^4 \plus{} (yz)^2 \plus{} (xz)^2}{z^3} \geq 12$

Therefore it suffices to prove now the inequality;

    $ (\frac {(xy)^2}{x^3} \plus{} \frac {(xy)^2}{y^3}) \plus{} (\frac {(yz)^2}{y^3} \plus{} \frac {(yz)^2}{z^3}) \plus{} (\frac {(zx)^2}{x^3} \plus{} \frac {(zx)^2}{z^3}) \geq 6$.

Hence at this point i am spending my time trying to figure this inequality for $ x,y > 0$ :

        $ \frac {(xy)^2}{x^3} \plus{} \frac {(xy)^2}{y^3} \geq 2$ which is equivalent to : $ x^2y^5 \plus{} x^5y^2 \geq 2x^3y^3$.

Once again I am stuck,so if anyone could see anything or any mistakes until this point please let me know.

~Kostas

Could someone plz post a solution to this problem because it actually seems that I can't finish it by myself.Thanks for your help.

~Kostas

[quote="alan43"]Given a triangle with sides $ a$,$ b$,$ c$ satisfying  $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Show that
$ \sum \frac {b \plus{} c}{\sqrt {b \plus{} c \minus{} a}}\geq 6$[/quote]
Let $ a \plus{} b \minus{} c \equal{} z^2,$ $ a \plus{} c \minus{} b \equal{} y^2$ and $ b \plus{} c \minus{} a \equal{} x^2,$ where $ x,$ $ y$ and $ z$ are positive numbers,
$ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2$ and $ xyz \equal{} w^3.$
Hence, $ \sum_{cyc}(x^4 \plus{} x^2y^2) \equal{} 6$ and $ \sum_{cyc} \frac {b \plus{} c}{\sqrt {b \plus{} c \minus{} a}}\geq 6\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {2x^2 \plus{} y^2 \plus{} z^2}{x}\geq12\sqrt [4]{\frac {\sum(x^4 \plus{} x^2y^2)}{6}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y \plus{} x^3z \plus{} 2x^2yz)\geq12xyz\sqrt [4]{\frac {\sum(x^4 \plus{} x^2y^2)}{6}}\Leftrightarrow$
$ \Leftrightarrow9u^2v^2 \minus{} 6v^4\geq\left(4\sqrt [4]{\frac {27u^4 \minus{} 36u^2v^2 \plus{} 9v^4 \plus{} 2uw^3}{2}} \minus{} u\right)w^3.$
Let $ f\left(w^3\right) \equal{} \left(4\sqrt [4]{\frac {27u^4 \minus{} 36u^2v^2 \plus{} 9v^4 \plus{} 2uw^3}{2}} \minus{} u\right)w^3.$
Hence, $ f$ is an increasing function.
Thus, $ f$ gets a maximal value when $ w^3$ gets a maximal value, which happens when two numbers from $ \{x,y,z\}$ are equal.
Id est, for the proof of $ \sum_{cyc}(x^3y \plus{} x^3z \plus{} 2x^2yz)\geq12xyz\sqrt [4]{\frac {\sum(x^4 \plus{} x^2y^2)}{6}}$ enough to check only one case:
$ y \equal{} z \equal{} 1,$ which gives $ g(x)\geq0,$ where
$ g(x) \equal{} \ln(x^3 \plus{} x^2 \plus{} 3x \plus{} 1) \minus{} \ln x \minus{} \frac {1}{4}\ln(x^4 \plus{} 2x^2 \plus{} 3) \minus{} \frac {3}{4}\ln6.$
$ g'(x) \equal{} \frac {(x^2 \plus{} x \plus{} 1)(x^4 \plus{} 3)(x \minus{} 1)}{(x^3 \plus{} x^2 \plus{} 3x \plus{} 1)(x^4 \plus{} 2x^2 \plus{} 3)x},$ which gives $ x_{min} \equal{} 1.$ Shown!  :)

Thank you so much.It seems that uvw-method is really powerful I didn't know it before.


function
AMC
AIME
binomial coefficients
algebra
binomial theorem

How useful are combinatorical identities and generating functions on the AIME? There isn't much time left so I want to prioritize what I study.

I would not try to cram in gf's.  Combinatorial identities on the other hand are very useful.  Of the top of my head you should know and understand how to compute binomial coefficients, interpreting problems in terms of binomial coefficient (e.g. block walking and committee forming), Pascal's Identity, the binomial theorem, and $\binom{n}{k}=\binom{n}{n-k}$.

Well, I just did that off the top of my head so it is by no means all that you should focus on.  I would go through AIME tests between now and the AIME, see where you are weak, and focus on that topic.  Don't hesitate to post questions on the boards.

Good luck on the AIME!

See chapter 17 of AoPS Volume 2 for useful combinatorical identities.


There are 8 identical looking coins. Two of which are defected. They could be heavier or lighter than the normal coins. Using a standard two-arm balance, what is the minimum number of scalings you need to identify these two defected coins?

Have A Great Day!

Are you allowing for one coin to be lighter and at the
same time the other coin to be heavier than the main coins?

Yes, it can be. Let's say "a" to be the weight of regular coin; x and y to be weights of two defected coins. The following can happen:

x<y<a ; x<a<y; a<x<y;  x=y<a; a<x=y

The question is how one can design a scalingscheme to identify these points weighting x, and y respectively.

Have A Good Day!

It seems a bit quiet here :). Does anyone find out the solution?

Have A Great Day!


1)A box contains 20 yellow discs,9 red discs,and 6 blue discs.If discs are selected at random,then the smallest number of discs that need to be selected to be assured of selecting at least two discs of each color is?
2)Find the value of A+B that satisfies
  (6^30+6^(-30))*(6^30-6^(-30))=(3^A*8^B)-(3^(-A)*8^(-B))

[quote]
1)A box contains 20 yellow discs,9 red discs,and 6 blue discs.If discs are selected at random,then the smallest number of discs that need to be selected to be assured of selecting at least two discs of each color is?
[/quote]
[hide="1"]
Take 20 yellow and 9 red discs. Then take two blue discs.
20 + 9 + 2 = [b]31[/b].
[/hide]

[hide="the solution to 1"]the worst case is that you pick all 20 yellows.

then you pick all 9 reds

then you can only pick 2 more blues

[hide="the answer is 31"]31[/hide][/hide]

The source is
[url=http://www.ouc.bc.ca/instructors/clee/bcchsmc/2004%5CJuniorPrelim_Feb06_2004.pdf]here[/url]

for question 1:

assume worst case. you need to take all 20 yellow ones, 9 red ones, and then 2 blue discs to make sure u have 2 of each. therefore total is 31.

for question 2:

expand.
LHS=6^60-6^-60
RHS=3^A*2^3B-3^-A*2^-3B

equating the two, it's obvious that A=3B=60 (since u need to combine the bases 3 and 2 of LHS to make the 6 of RHS)

therefore, A=60, B=20

Question 2 is silly -- for almost any choice of A (maybe every choice -- didn't think too hard about it), I should be able to find an appropriate choice of B that makes the expression work.  Each such combination will give me a different value of A + B.

Here's my guess at the "solution" to 2

[hide="Solution"]Both sides multiplied out are
6^60 - 6^(-60) = (3^A)(2^(3B)) - (3^(-A))(2^(-3B))
so the left side is also (2^60)(3^60) - (2^(-60))(3^(-60))
so a = 60 and b = 20
A+B = 80[/hide]

Where does everyone get nice fonts, I want a nice font but I am probably to dumb to use it.
[/hide]


I'll add my thoughts after 8 AM Thursday about the actual test and my score but...

What is the cutoff for USAMO for people who take AMC 10?

(As in you need score X on AMC10, and score Y on AIME to get a combined score Z which is necessary to make USAMO)

That doesn't get announced until after the AIME.

Generally, one could make it either by having an index score (AMC + 10*AIME) of some number (has been in the 220 range in the past) or higher,

OR

having an AIME score above some number, determined by the lowest AIME score of all those AMC12 participants who are invited to USAMO.

Yeah, I've been in this long enough to know that; I was asking for an estimate - as the cutoff for 10 is slightly skewed because they take an uneven number.

I don't really know what you mean by that, but it greatly depends on the difficulty of the AIME.  I think that for 9th and 10th graders, the cutoff the past few years has been close to 190 or 200, but I'm not positive of that.

Effectively, the only thing that matters for AMC qualification for those in 10th grade and below is the AIME score.  Last year and the year before, an 8 and a 6 respectively on the AIME were good for USAMO.

*sigh* i dun think i did that good... :(


geometry
circumcircle
inradius
incenter
geometry solved

Let $ A_{1},B_{1},C_{1}$ be points on the sides $ BC,CA,AB$ of a triangle $ ABC$. Lines $ AA_{1},BB_{1},CC_{1}$ intersect again the circumcircle of triangle $ ABC$ at $ A_{2},B_{2},C_{2}$, respectively. Prove that:
\[ \frac{AA_{1}}{A_{1}A_{2}}+\frac{BB_{1}}{B_{1}B_{2}}+\frac{CC_{1}}{C_{1}C_{2}}\ge \frac{3s^{2}}{r(4R+r)}\]
where $ s,r,R$ are the semiperimeter, inradius, and circumradius of triangle $ ABC$, respectively.

That's very nice proof Virgil, I only note that for all $ A_{1}\in BC$ and $ A_{2}=AA_{1}\cap (O)$ then
$ \frac{AA_{2}}{A_{1}A_{2}}\ge \frac{AA_{2}'}{A_{1}'A_{2}'}$ with $ A_{1}'=IA\cap BC,A_{2}'=IA\cap (O)$ with incenter $ I$ and now we can see the minimum of sum
$ \frac{AA_{1}}{A_{1}A_{2}}+\frac{BB_{1}}{B_{1}B_{2}}+\frac{CC_{1}}{C_{1}C_{2}}$ is
$ \sum\frac{AA_{2}'}{A_{1}'A_{2}'}$ sum and get min iff $ AA_{1},BB_{1},CC_{1}$ are concurrent at incenter.

I have just put up my [url=http://www.cip.ifi.lmu.de/~grinberg/O49.pdf]proposed solution[/url] online (should have done this a longer time ago).

  darij


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

[quote="apollo"]Let $ a,b,c \geq 0$ and $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} \equal{} 2$
Prove that $ a\sqrt {\frac {ab}{a^2 \plus{} 1}} \plus{} b\sqrt {\frac {bc}{b^2 \plus{} 1}} \plus{} a\sqrt {\frac {ca}{c^2 \plus{} 1}} \leq \sqrt {\frac {3}{2}}$[/quote]
Let $ a^2 \equal{} \frac {x}{y \plus{} z},$ $ b^2 \equal{} \frac {y}{x \plus{} z}$ and $ c^2 \equal{} \frac {z}{x \plus{} y},$ where $ x,$ $ y$ and $ z$ are non-negative. We obtain:
$ \sum_{cyc}a\sqrt {\frac {ab}{a^2 \plus{} 1}} \equal{} \sum_{cyc}\sqrt {\frac {x\sqrt {\frac {xy}{(x \plus{} z)(y \plus{} z)}}}{x \plus{} y \plus{} z}} \equal{} \sum_{cyc}\left(\frac {x}{x \plus{} y \plus{} z}\cdot\sqrt {\frac {(x \plus{} y \plus{} z)\sqrt {\frac {xy}{(x \plus{} z)(y \plus{} z)}}}{x}}\right)\leq$
$ \leq\sqrt {\sum_{cyc}\sqrt {\frac {xy}{(x \plus{} z)(y \plus{} z)}}}\leq\sqrt [4]{3\sum_{cyc}\frac {xy}{(x \plus{} z)(y \plus{} z)}}\leq\sqrt {\frac {3}{2}}.$
I have found mistake in my proof. I'll edit it soon.
The last inequality is wrong.
But $ \sqrt {\sum_{cyc}\sqrt {\frac {xy}{(x \plus{} z)(y \plus{} z)}}}\leq\sqrt {\frac {3}{2}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{xy(x\plus{}y)}\leq\frac{3}{2}\sqrt{(x\plus{}y)(x\plus{}z)(y\plus{}z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^2y\plus{}x^2z)\plus{}2\sum_{cyc}x\sqrt{yz(x\plus{}y)(x\plus{}z)}\leq\frac{9}{4}(x\plus{}y)(x\plus{}z)(y\plus{}z).$
But $ 2\sum_{cyc}x\sqrt{yz(x\plus{}y)(x\plus{}z)}\leq\sum_{cyc}(xy(x\plus{}z)\plus{}xz(x\plus{}y)).$
Thus, it remains to prove that: $ \sum_{cyc}(x^2y\plus{}x^2z)\plus{}\sum_{cyc}(xy(x\plus{}z)\plus{}xz(x\plus{}y))\leq\frac{9}{4}(x\plus{}y)(x\plus{}z)(y\plus{}z),$ which obviously true. :)

My solution: $ \because \frac{1}{1\plus{}a^2}\plus{}\frac{1}{1\plus{}b^2}\plus{}\frac{1}{1\plus{}c^2}\equal{}2$
$ \therefore \frac{a^2}{1\plus{}a^2}\plus{}\frac{b^2}{1\plus{}b^2} \plus{}\frac{c^2}{1\plus{}c^2}\equal{}1$
by Cauchy-Schwarz; $ L.H.S. \leq \sqrt{ab\plus{}bc\plus{}ca}\sqrt{\frac{a^2}{a^2\plus{}1}\plus{}\frac{b^2}{b^2\plus{}1}\plus{}\frac{c^2}{c^2\plus{}1}}$ $ \equal{}\sqrt{ab\plus{}bc\plus{}ca}$
It's remain to prove $ ab\plus{}bc\plus{}ca \leq \frac{3}{2}$
Cauchy-Schwarz; $ \frac{a^2}{1\plus{}a^2}\plus{}\frac{b^2}{1\plus{}b^2} \plus{}\frac{c^2}{1\plus{}c^2} \geq \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}3}$ $ \rightarrow ab\plus{}bc\plus{}ca \leq \frac{3}{2}$


Let $ ABC$ be a triangle,$ AA'$ the projection from $ A$ to the side $ BC$,$ BB'$ the projection from $ B$ to side $ AC$,
$ CC'$ the projection from $ C$ to $ AC$.It is known that the middles of these three projections are colinear.What is the type of the triangle?

[quote="Cronus"]It should'nt be of any special kind, would it(I ain't sure)?The only forget-me-not property of the triangle is AA' bisects the angle at  B'A'C' and so on.[/quote]
It must be of a special kind:[hide] it must be a triangle with a 90 degrees angle <prove it>[/hide]

Well, I misunderstood the triangle as the orthic but I still don't see why it should be of any special kind.

It should be a special kind....I don't agree with the fact that in any  triangle the middles of the projections from the vertices to the sides are collinear

Well, the diagrams I drew suggest that they are collinear. Could you please explain how you got the right-angled triangle?

The middles $ H_1,H_2,H_3$ of the projections are on the median triangle for $ \Delta ABC$.It isn't possible that 2 middles be the mase.The only posibility for the 3 points to be collinear is that two of them to be the vertices of the median triangle and the last one to be on the segment determined by the 2 points


what happens if someone gives a "simple" challenge question as humor or sarcasm like what is 2+2

The topic gets moved to the Games & Fun Factory forum the first time. The next time the topic gets deleted :)


[quote="NoSoupForYou"]$\sqrt{3}$.[/quote]Err...do you have a solution?


Let be given a sequence {x_n} defined by

          x_0 = 0, x_n = x_{n-1}/2004 + (-1)^n, for all n >= 1

Find  lim (x_n) 2 .
       n->oo

[quote="Namdung"]Let be given a sequence $ \{x_n\}$ defined by

          $ x_0 \equal{} 0, x_n \equal{} \frac {1}{2004}x_{n \minus{} 1} \plus{} ( \minus{} 1)^n$, for all $ n \geq 1$

Find  $ \lim_{n\to\infty} x_n^2.$[/quote]

$ x_n \equal{} \frac {1}{2004}x_{n \minus{} 1} \plus{} ( \minus{} 1)^n\Longleftrightarrow x_n \minus{} \frac {2004}{2005}( \minus{} 1)^n \equal{} \frac {1}{2004}\left\{x_{n \minus{} 1} \minus{} \frac {2004}{2005}( \minus{} 1)^{n \minus{} 1}\right\}$

$ \therefore x_n \minus{} \frac {2004}{2005}( \minus{} 1)^n \equal{} \left(\frac {1}{2004}\right)^n \left\{x_0 \minus{} \frac {2004}{2005}( \minus{} 1)^0\right\}$

$ \Longleftrightarrow x_n \equal{} \frac {2004}{2005}\left\{( \minus{} 1)^n \minus{} \left(\frac {1}{2004}\right)^n\right\}$

$ \therefore \lim_{n\to\infty} x_n^2 \equal{} \lim_{n\to\infty} \left(\frac {2004}{2005}\right)^2\left\{1 \plus{} \left(\frac {1}{2004^2}\right)^n \minus{} 2\left( \minus{} \frac {1}{2004}\right)^n\right\} \equal{} \boxed{\left(\frac {2004}{2005}\right)^2}.$


Need your help guys. Real glad to be here. here is my question.

Two birds - one 30 ft tall and one 40 ft tall - are 50 ft apart. A bird sits on the top of each tower. At the same instant, each bird flies at the same rate to a fountain that is between the towers; the birds arrive at the fountain at the same time. How far from each tower is the fountain?

[quote="jmccray"]Need your help guys. Real glad to be here. here is my question.

Two[b] towers [/b]- one 30 ft tall and one 40 ft tall - are 50 ft apart. A bird sits on the top of each tower. At the same instant, each bird flies at the same rate to a fountain that is between the towers; the birds arrive at the fountain at the same time. How far from each tower is the fountain?[/quote]

It should be towers right? A bird 40 ft tall would be like  :what?:

[hide="One way"]Draw a diagram first ;) It should look like
[code]       |
       |
|      |
|      |
|______|
A    X B[/code]

Now label a point A, X, and B on the base as shown.  Let $AX=50-x$ and $BX=x$

Use the pythagorean theorem to get $30^2+(50-x)^2=x^2+40^2$

Solving gives $x=18$.

So the two distances are $18, 32$  [/hide]

EDIT: changed my code

Sincw the birds got to the fountain at the same time, at a constant rate, it implies that the distances must be congruent... knowing that apply pythagorean theorem on both right triangles and set them equal to each other by their hypothenuses and solve ....

$30^2+x^2=40^2+(50-x)^2$
$900+x^2=1600+2500-100x+x^2$
$900=4100-100x$
$100x=3200$
$x=32$

So the distances are 32 and 18


[b]1.[/b] $(1+\frac{a}{b})^n+(1+\frac{b}{a})^n\geq 2^{n+1}$

[b]2.[/b] [i]Generalization:[/i] $\forall k \in\mathbb{N}$ $(k+\frac{a}{b})^n+(k+\frac{b}{a})^n\geq 2(1+k)^n$

[b]3.[/b] Could anyone find a more generalized form?

Equivalent form: $\frac{(1+\frac{a}{b})^{n}+(1+\frac{b}{a})^{n}}{2} \geq 2^{n}$. AM-GM yields: $LHS \geq sqrt{((1+\frac{a}{b})(1+\frac{b}{a}))^{n}} \Longleftrightarrow sqrt{(2+\frac{a}{b}+\frac{b}{a})^{n}}$ and $\frac{a}{b}+\frac{b}{a} \geq 2$, we are done.

[quote="Ezbakhe Yassin"]

[b]3.[/b] Could anyone find a more generalized form?[/quote]
true for $k \geq 0$  :)


Find all positive integer $ n$ such that
we can rearrange $ 1,2,...,n$ as $ a_1,...,a_n$ so that $ |a_i \minus{} a_{i \plus{} 1}| \equal{} 3,4or5$
$ i \equal{} 1,2,...,n$ and $ a_{n \plus{} 1} \equal{} a_1$

My answer is $ 7,8,9$ and $ \ge13$
Am I right :?:


In how many ways may the letters from A to H be arranged so that $A$ precedes $B$, $C$ comes before $D$, $E$ precedes $F$, and $G$ comes before $H$.

-mandelbrot 1994 meet 1 #6.

I'm obviously unsure if my solution is right, but here's a stab at the problem.

[hide]We regard each pair as an individual unit. Because each unit takes up 2 slots out of 8 and there is exactly 1 arrangement for each pair so that it does not violate our criteria, the number of ways is ${8\choose 2}{6\choose 2}{4\choose 2}{2\choose 2}(1)^4 = \boxed{2520}$ ways.[/hide]

Though now that I think about it, the answer I get seems a little too big....

That is correct.

Here's another way to do it [hide]- there's a bijection between instances when A is greater than B and instances when B is greater than A, and similarly for the other pairs. Thus, we can simply count with out the restrictions and then divide by 2 for each pair - so you get $8!/2^4 = 2520$.[/hide]


calculus
integration
logarithms
real analysis
real analysis unsolved

The first integral I wish to know if there is another method to solve it besides my own.

$ \int_0^{1}\ln(x)\ln(1 + x)\,dx$

$ = \int_0^{1}\int_0^{x}\frac {\ln(x)}{1 + y}\,dy\,dx$

$ = \int_0^{1}\int_{y}^{1}\frac {\ln(x)}{1 + y}\,dx\,dy$

$ = \int_0^{1}\bigg[\frac {x\ln(x)}{1 + y} - \frac {x}{1 + y}\bigg]\bigg|_{y}^{1}\,dy$

$ = \int_0^{1}\frac {\ln(y)}{1 + y}\,dy + \int_0^{1}\bigg[ - \ln(y) - \frac {2}{y + 1} + 1\bigg]\,dy$

$ = \int_0^{1}\frac {\ln(y)}{1 + y}\,dy + 2 - 2\ln(2)$

Now

$ \int_0^{1}\frac {\ln(y)}{1 + y}\,dy$

$ = \int_0^{1}\ln(y)\sum_{n = 0}^{\infty}( - 1)^yy^n\,dy$

=$ \sum_{n = 0}^{\infty}( - 1)^n\int_0^{1}\ln(y)\cdot{y^n}\,dy$

$ = \sum_{n = 0}^{\infty}( - 1)^n\bigg[\frac {x\ln(x)}{1 + y} - \frac {x}{(1 + y)^2}\bigg]\bigg|_0^{1}\,$

$ = - \sum_{n = 0}^{\infty}\frac {( - 1)^n}{(n + 1)^2}$

$ = - \sum_{n = 1}^{\infty}\frac {( - 1)^{n - 1}}{n^2}$

$ = - \left(1 - 2^{1 - 2}\right)\cdot\zeta(2)$

=$ - \frac {\pi^2}{12}$

Now putting the two together we get

$ \boxed{\int_0^{1}\ln(x)\ln(1 + x)\,dx = 2 - 2\ln(2) - \frac {\pi^2}{12}}$

Any other solutions would be awesome.

And could someone point out my mistake in this one?

$ \int_0^{\pi}\ln\left(a + b\cos(x)\right)\,dx$

$ = \int_0^{\pi}\ln(a) + \int_0^{\pi}\ln\left(1 + \frac {b}{a}\cos(x)\right)\,dx$

Now let

$ J(\theta) = \int_0^{\pi}\ln\left(1 + \frac {b}{a}\theta\cos(x)\right)\,dx$

$ \Rightarrow{J'(\theta) = \frac {d}{d\theta}\int_0^{\pi}\ln\left(1 + \frac {b}{a}\theta\cos(x)\right)\,dx}$

$ = \int_0^{\pi}\frac {\partial}{\partial{\theta}}\bigg[\ln\left(1 + \frac {b}{a}\theta\cos(x)\right)\bigg]\,dx$

$ = \int_0^{\pi}\frac {b\cos(x)}{\theta{b}\cos(x) + a}\,dx$

$ = \frac {1}{\theta}\bigg[\int_0^{\pi}\frac {\theta{b}\cos(x) + a}{\theta{b}\cos(x) + a}\,dx - \int_0^{\pi}\frac {a}{\theta{b}\cos(x) + a}\,dx\bigg]$

Now omitting some calculations I get

$ = \frac {1}{\theta}\bigg[\pi - \frac {a\pi}{\sqrt {a^2 - (b\theta)^2}}\bigg]$

So then seeing that $ J(0) = 0$ and $ J(1)$ is our desired integral we see that

$ \int_0^{\pi}\ln\left(1 + \frac {b}{a}\cos(x)\right)\,dx =\pi\ln(a)+ \int_0^{1}\bigg[\pi - \frac {a\pi}{\sqrt {a^2 - (b\theta)^2}}\bigg]\,d\theta$

But it doesn't. What am I doing wrong?

[quote="Mathstud28"]

Now omitting some calculations I get

$ \equal{} {\color{red}\frac {1}{\theta}}\bigg[\pi \minus{} \frac {a\pi}{\sqrt {a^2 \minus{} (b\theta)^2}}\bigg]$

So then seeing that $ J(0) \equal{} 0$ and $ J(1)$ is our desired integral we see that
[/quote]
$ \int_0^{\pi}\ln\left(1 \plus{} \frac {b}{a}\cos(x)\right)\,dx \equal{} \pi\ln(a) \plus{} \int_0^{1}{\color{red}\frac 1\theta }\bigg[\pi \minus{} \frac {a\pi}{\sqrt {a^2 \minus{} (b\theta)^2}}\bigg]\,d\theta$

[quote]
$ \int_0^{\pi}\ln\left(1 \plus{} \frac {b}{a}\cos(x)\right)\,dx \equal{} \pi\ln(a) \plus{} \int_0^{1}{\color{red}\frac 1\theta }\bigg[\pi \minus{} \frac {a\pi}{\sqrt {a^2 \minus{} (b\theta)^2}}\bigg]\,d\theta$[/quote]
Ahh thanks!

But alas, it still isn't right, actually, I think it is more wrong now  :D .

Oh, yes, either the left hand side should be the original integral, or there should be no $ \pi\ln(a)$ on the right. I completely missed that :blush:.

So there is something wrong with my work? Do you know specifically where?

Didn't I just tell you? :? $ \pi\ln(a)$ is the difference between the original integral and $ J(\theta)$, so if you add it, it means that you return to the original integral, and if you put $ J(\theta)$ on the left, it should not be added. Sorry if I wasn't clear.


In $\triangle ABC$, $D$, $E$ and $F$ are points on $AB$, $BC$ and $CA$, respectively, such that $AD: DB=BE: EC=4: 1$ and $CF: FA=3: 2$. If $P$, $Q$ and $R$ are the midpoints of $AE$, $BF$ and $CD$ respectively, find the area ratio of $\triangle PQR$ and $\triangle ABC$.

come on chess..what have you been asking about recently  ;)

[hide="hint"]mass points[/hide]

Here is a step that you have to be able to do in order to solve frt's problem.

[hide="..."]Let there be a $\triangle ABC$ with $D$, $E$, and $F$ existing on $AB$, $BC$, and $CA$, respectively such that $\frac{BD}{DA}=d$, $\frac{CE}{EB}=e$, and $\frac{AF}{FC}=f$. Let $AE\cap BF=X$, $BF\cap CD =Y$, and $CD \cap AE=Z$. Find the ratio of $\frac{[XYZ]}{[ABC]}$.[/hide]


trigonometry
inequalities
function
calculus
analytic geometry
geometry
perimeter

Prove: $\sin{A} < A <\tan{A}$ where $A\in(0,\frac{\pi}{2})$

The proof that sin A < tan A is fairly simple, I think.  Just say that ( 1 / cos A ) > 1 for all A that fit 0 < A < pi / 2.  However, I am not sure how to prove the middle part.  I am sure this is where the real challenge of the proof lies but I am new at this and just wanted to see if what I already said makes any sense or not.

but then how do you prove $\displaystyle \frac{1}{\cos x} > 0$?

There is some vicious circle in devious_'s proof. He uses $(\sin x)'=\cos x$. I just wonder how can he prove it?  ;) Actually, $\sin x\leq x\leq \tan x$ is a fundamental inequality and it has a nice geometrical proof, shich doesn't refer to calculus.

I'm just curious, but why can't we prove it by graphing the function in an xy system?

[quote="devious_"][hide]
Let:

$f(x) = \sin x - x$

Then:

$f'(x) = \cos x - 1$

From AM-GM we get:

$f'(x) + 1 \geq 2\sqrt{\cos x}$

By considering $\cos x$ in $(0, \frac{\pi}{2})$, we get that $f(x)$ is decreasing in that interval.

Now, since $f(0) = 0$:

$\sin x - x < 0$

$\sin x < x$
[/hide][/quote]

Well, this is a nice solution, but i don't think that calculus is allowed in this forum, and besides, not every high school students know derivative....

I know a very easy solution, that is, draw a $circle$ in the Cartesian coordinate system, then u can get it with observation.

I just wonder if there are any other proof?? This is a basic thing and i think there should be more than two solutions

[quote="xxreddevilzxx"]but then how do you prove \displaystyle \frac{1}{\cos x} > 0?[/quote]

For all $x$ in $(0,\frac{\pi}{2})$ the quantity $\cos (x)>0$. Clearly, over the same interval $\frac{1}{\cos (x)}>0$

[quote="Gyan"]Or simply use geometic proof  As suggested above ...: O is center, A and B are two  points in the  perimeter,  Say  OBX is rightangle triangle, where X is on the line OA (and OX<OA)  and  OBY is also a right triangle (Point B is  middle of line OY and OY>OB)

Now area of OYA> area of arc OAB > area of OBX
(As long as angle >0)
If the circel has radius 1 than this is same as saying sin(x) < x < tan (x)  (if x>0)
QED![/quote]

That's how i solved it...

Is there any others ways of doing so?? I wanna see some different proof..


The sum of five positive real numbers is 100. Prove that there are two numbers among them whose difference is at most 10.


trigonometry
geometry
3D geometry
algebra
factorization
sum of cubes

Here is a nice trig identity that you are to prove that I found in my textbook... Its actually really easy... accodring to my teacher... but I did it, and my solution seems to be too long... so could you please post more elegant solutions... if possible?

Prove that:

$1-\frac{\sin^{2}x}{1+\cot x}-\frac{\cos^{2}x}{1+\tan x}= \sin x \cos x$
[hide]
$LS = 1-\frac{\sin^{2}x}{1+\cot x}-\frac{\cos^{2}x}{1+\tan x}$
$= 1-(\frac{\sin^{2}x (1+\frac{\sin x}{\cos x})+\cos^{2}x (1+\frac{\cos x}{\sin x})}{(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})})$
$= 1-(\sin^{2}x+\frac{\sin^{3}x}{\cos x}+\cos^{2}x \frac{\cos^{3}x}{\sin x})(\frac{\sin x\cos x}{2\sin x \cos x+1})$
$= 1-( \frac{\sin x \cos x+\sin^{4}x+\cos^{4}x}{\sin x \cos x})(\frac{\sin x\cos x}{2\sin x \cos x+1})$
$= \frac{2\sin x \cos x+1-\sin x\cos x-\sin^{4}x-\cos^{4}x}{2\sin x\cos x}$
$= \frac{\sin x \cos x+(1-\sin^{2}x)(1+\sin^{2}x)-\cos^{4}x}{2\sin x\cos x+1}$
$= \frac{\sin x \cos x+\cos^{2}x(1+\sin^{2}x)-\cos^{4}x}{2\sin x\cos x+1}$
$= \frac{\sin x \cos x+\cos^{2}x+\cos^{2}x \sin^{2}x-\cos^{4}x}{2\sin x\cos x+1}$
$= \frac{\sin x \cos x+\cos^{2}x \sin^{2}x+\cos^{2}x(1-\cos^{2}x)}{2\sin x\cos x+1}$
$= \frac{\sin x \cos x+\cos^{2}x \sin^{2}x+\cos^{2}x\sin^{2}x}{2\sin x\cos x+1}$
$= \frac{\sin x \cos x+2 \cos^{2}x \sin^{2}x }{2\sin x\cos x+1}$
$= \frac{\sin x \cos x(1+2 \sin x\cos x)}{2\sin x\cos x+1}$
$= \sin x \cos x$
$= RS$ as required.[/hide]

[quote="Nawahd Werdna"]Here is a nice trig identity that you are to prove that I found in my textbook... Its actually really easy... accodring to my teacher... but I did it, and I found that the Left Side was actually equal to $\sin x \cos x+1$

Prove that:

$1-\frac{\sin^{2}x}{1+\cot x}-\frac{\cos^{2}x}{1+\tan x}= \sin x \cos x$[/quote]

The identity is correct as written.  The simplest way to prove it is to note

$\sin x (1+\cot x) = \sin x+\cos x = \cos x (1+\tan x)$,

and so the left-hand side is

$1-\frac{\sin^{3}x+\cos^{3}x}{\sin x+\cos x}$.

If we factor the numerator as a sum of cubes, we note the factor $\sin x+\cos x$ cancels the denominator, giving

$1-(\sin^{2}x-\sin x \cos x+\cos^{2}x)$,

at which point the identity becomes obvious.


Let a_1<a_2<...a_43<a_44 be positive integers not exceeding 125. Prove that among the 43 differences      d_i=a_i+1-a_i (i=1,2,...43) some value must occur at least 10 times.

Uhhh ATTC that each difference occurs at most nine times. Then $ a_{1}\geq1$, $ a_{10}\geq10$, $ a_{19}\geq28$, $ a_{28}\geq55$, $ a_{37}\geq91$, and finally $ a_{44}\geq126$, so we have a contradiction.


limit
real analysis
real analysis unsolved

Does there exsist such nonsmooth curve $ \gamma$  defined on $ [a,b]$ that the following relation holds:

$ \lim_{t_1\to t_2}\frac {s(t_1, t_2)}{|t_1 \minus{} t_2|} \equal{} \plus{} \infty$.
Where $ s(t_1,t_2)$ is length of curve(piece of $ \gamma$) in$ [t_1,t_2]\in[a,b]$.

There are curves that have no non-trivial rectifiable subarcs (so $ s(t_1,t_2)\equal{}\plus{}\infty$ for all $ t_1<t_2$). Is that what you are asking about?

take (almost) any path of a Brownian motion. Or you can think of fractal things like the Koch snowflake.


For natural numbers $m,n$ such that $GCD(m,n)=1,$ find $GCD(5^m+7^m,\ 5^n+7^n).$
Note that $GCD(m,n)$ denotes the Greatest Common devisor of $m$ and $n.$

[quote="{x}"]Let $a,b,m,n$ be four positive integers with $(a,b)=1$.
Then
\[ (a^m+b^m,a^n+b^n)=a^{(m,n)}+b^{(m,n)}  \]
Knowing this we solve the problem.

The theorem in question has been proved at last a dozen times on mathlinks.[/quote]
You are not right: $a,b$ must be odd.

The theorem {x} speaks of is with $a^m-b^m , a^n-b^n$ I think. But the proof is evry similar.
And see http://www.mathlinks.ro/Forum/viewtopic.php?t=67642 please.

Yes Rust, you are correct.
ZetaX, indeed I confused them.
I'm tired :(

Probably, because I can't see Rust posting in this topic :).

Oh ! Indeed, I begin to confuse nicknames :o
I hope I didn't offend anyone, and in particular Pavel Kozlov  :blush:

Can anyone finish  this problem? What level do you think about this problem?


Aight here is the problem. It was from a contest that was finished today, so im not cheating or anything.

summation of x=1 to 2003 of sin(x * theta)
divided by
summation of x = 1 to 2003 of cos(x*theta)

I tried this using e^itheta and taking imaginary and real parts, but this didnt work (at least i couldnt get it to a simplified enuff version that i was happy with it).

[quote]I tried this using e^itheta and taking imaginary and real parts, but this didnt work (at least i couldnt get it to a simplified enuff version that i was happy with it).[/quote]

Actually, I think that that is indeed how to do it. I remember asking folk here about a similar problem months ago on the AIME. It was the same problem except with 44 instead of 2003. I know 44 is a nicer number, but still...

the real part of the summatin when using e^itheta(k) is easy to evaluate, but the imaginary part poses some problems. First, it must be rationalized, but then there is no imaginary part in the denominator, so isnt it therefore going to be a zero in the denom? Can you evaluate it without rationalizing it?

Anyone with any ideas please respond.

Be careful, Jelyman -- Re(x + y) = Re(x) + Re(y), but the same is NOT true for multiplication or division.  You have to first add it up using e^(i*theta), and then switch back to sines and cosines in order to multiply the denominator by its conjugate before you can take the real and imaginary parts.

i converted it back to cos and isin first before conjugating, but thats the problem, u have to conjugate which gets rid of all imaginary terms in the denominator. So, um, can someone please help me out here?

Here's a hint, jelyman.  The denominator turns out to equal cos(2004 theta / 2) sin (2003 theta / 2) / sin(theta / 2).  You can prove that either by looking at exponentials, as you suggested, or by using trig identities.  There is a similar formula for the numerator.

Jelyman -- you have to take the real and imaginary parts [i]before[/i] you divide the two expressions together.

Maybe this helps:

If z=cosx+isinx then cosx=(z^2+1)/2z and sinx=(z^2-1)/2iz.

yes i know u have to take the imag and real parts before dividing, but thats not what has been my problem. The main problem I have with this is when i use the sin summation. We're taking the imaginary part of a fraction here, and to do this we must rationalize the denominator. Here is an example. We have
(a + bi)/(c + di).... We cant just take b/d as our imaginary part, can we? We have to rationalize teh denom, but then when we take our imag part, we dont have any imaginary part in the denominator. So isnt this the same as dividing by zero? Eh? Eh? NUdge? Nudge? elbow, elbow?  :roll: ....

This is my only problem, I dont know how to get rid of this bump.

I'm not sure I understand.  Let's take your example of finding the imaginary part of (a + bi) / (c + di).  As you say, we can rationalize the denominator by multiplying top and bottom by the conjugate c - d i.  That gives us (a + bi) (c - di) / (c2 + d2).  To find the imaginary part of that, we can take the imaginary part of (a + bi) (c - di) and divide by c2 + d2.  We are not dividing by the imaginary part of c2 + d2 (which is 0 as you point out); we are dividing by c2 + d2 itself (a real number).

Right -- you realized you don't do Im(w/z) = Im(w)/Im(z), but then after divided you tried to do exactly that.  the fact that there aren't any imaginary things in the denominator is good -- you just take the (imaginary part of the numerator) over the denominator, as Ravi B. said.

thats wat i thought it was, but this calculus kid in my school (me and him are at the top, if not the best, in our school) said that i cant do that and i am dividing by zero. That bum just cost us a point. Thanx for the help.


Let a, b, c, x, y, z be six nonnegative reals such that $a\geq b\geq c$, and such that either $x\geq y$ or $z\geq y$. Let k be a real such that $k\geq 1$. Prove that

$x\left(a^2+kbc\right)\left(a-b\right)\left(a-c\right)+y\left(b^2+kca\right)\left(b-c\right)\left(b-a\right)$
$+z\left(c^2+kab\right)\left(c-a\right)\left(c-b\right)\geq 0$.

  Darij

Darij, arqady's Lemma kill that! ;)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67096

Arqady, can you please post a proof of your lemma? :)
(I think you already did it, but I can't find it  :blush: )

[quote="mathmanman"]Arqady, can you please post a proof of your lemma? :)
[/quote]
 The arqady's Lemma ( :lol: ):
Assume $a\geq b\geq c>0,x>0,y>0,z>0.$
If $(a-c)x\geq(b-c)y$ or $(a-c)z\geq(a-b)y$ then
$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)\geq0.$
Proof:
Let $(a-c)x\geq(b-c)y.$
Hence, $z(c-a)(c-b)\geq0$ and $x(a-b)(a-c)+y(b-a)(b-c)=$
$=x(a-b)(a-c)-y(a-b)(b-c)=(a-b)((a-c)x-(b-c)y)\geq0.$
Id est, $x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)\geq0.$
Let $(a-c)z\geq(a-b)y$
Hence, $x(a-b)(a-c)\geq0$ and $y(b-a)(b-c)+z(c-a)(c-b)=$
$=z(a-c)(b-c)-y(a-b)(b-c)=(b-c)((a-c)z-(a-b)y)\geq0.$
Id est, $x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)\geq0.$
Hence the Lemma is proved. :D

Ah, thanks!
This lemma is indeed nice and seems to be useful ;)


Consider the dark square in an array of unit squares, part of which is shown. The ﬁrst ring of squares around this center square contains $ 8$ unit squares. The second ring contains $ 16$ unit squares. If we continue this process, the number of unit squares in the $ 100^\text{th}$ ring is

$ \textbf{(A)}\ 396 \qquad
\textbf{(B)}\ 404 \qquad
\textbf{(C)}\ 800 \qquad
\textbf{(D)}\ 10,\!000 \qquad
\textbf{(E)}\ 10,\!404$

[asy]unitsize(3mm);

defaultpen(linewidth(1pt));

fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray);
fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray);
fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black);

for(real i=0; i<=9; ++i)
{
draw((i,0)--(i,9));
draw((0,i)--(9,i));
}[/asy]

[hide]The number of squares in the n(th) ring is $ (2n\plus{}1)^2 \minus{} (2n\minus{}1)^2 \equal{} 8n.$[/hide]

[hide="Solution"]
We can partition the $ n^\text{th}$ ring into $ 4$ rectangles: two containing $ 2n\plus{}1$ unit squares and two containing $ 2n\minus{}1$ unit squares.  So there are $ 2(2n\plus{}1)\plus{}2(2n\minus{}1)\equal{}4n\plus{}2\plus{}4n\minus{}2\equal{}8n$ unit squares in the $ n^\text{th}$ ring.  Thus, the $ 100^\text{th}$ ring has $ 8 \times 100\equal{}800$ unit squares, which corresponds to answer choice $ \textbf{C}$.
[/hide]

[hide="Alternate Solution"]
We can make the $ n^\text{th}$ ring by removing a square of side length $ 2n\minus{}1$ from a square of side length $ 2n\plus{}1$.  So this ring contains $ (2n\plus{}1)^2\minus{}(2n\minus{}1)^2\equal{}(4n^2\plus{}4n\plus{}1)\minus{}(4n^2\minus{}4n\plus{}1)\equal{}8n$ unit squares.  Thus, the $ 100^\text{th}$ ring has $ 8 \times 100\equal{}800$ unit squares, which corresponds to answer choice $ \textbf{C}$.
[/hide]

Hope I helped!


ratio
Support
function
algorithm
MATHCOUNTS
number theory
prime factorization

I don't know where to put thhis so  a moderater can move it.

When I try to look for calculator programs on the web, it always (almost) makes it as a different file and my computor cant open it.  What should I do?

First of all, do you have a TI-83, TI-83 Plus, TI-89, or TI-92?

[quote="nat mc"]Calculator programs are awesome!! I have so many on mine.  Most of them I have written and some really good ones David wrote.[/quote]

cool...If you have like a looootttt of spare time, can you post them?
 :lol:
Thanx

[quote="NoSoupForYou"]Here is a good site for TI83 games and programs: http://www.geocities.com/tx3_tx5/home.html

PHOENIX: BEST TI83 GAME EVER:
http://www.angelfire.com/games/calc/phx.zip

Now the thing is, I have a TI-89 calculator, but I believe the procedure is the same for both calculators:

First, unzip the .zip files, if any.
Second, use the program that came with the link cable to transfer .83g files (using the Restore button)
Finally, transfer all remaining files (.83*) by right-clicking them in Windown Explorer, and choosing Send To... .[/quote]

yeah...I can't open the ones that actuall have all te transcripts and stuuff

That is a GREAT site!!!
There are also some cool apps on the [url=http://www.education.ti.com]TI Site[/url]

Also: If your calc crashes (as some do) and the flash apps on your calc go with it, usually all those programs should be accessable from either the CD or their site, but some programs (like cellsheet) you have to buy.  BUT, if you register your calculator at the TI site you can access a copy of the flash app without paying.

P.S.: for those who can't seem to download the programs onto your calc, you are supposed to drag and drop the file onto the TI Connect icon on yur desktop (with your calc plugged in, of course)

*looks at websites*

Interesting... it seems that most people write programs that are more of games, rather than things like prime factorization or Chinese remainder theorem or base conversion or stuff.

I prefer save my space for that kind of stuff, and of course, my list of the first 100 Fibonacci numbers. :)

Couldn't you just create a program that would calculate fibonacci numbers? I mean, if you're typing them in, and you accidentally hit clear on the 99th one...

Of course there's a program It goes like this

:Prompt X
:ClrList  LFIBON
:1$\rightarrow$LFIBON(1)
:1$\rightarrow$LFIBON(2)
:For(A,3,X)
:LFIBON(A-2)+LFIBON(A-1)$\rightarrow$LFIBON(A)
:End

It puts the first X fibonnaci numbers into X

I have a prime number list of 999. And a thingy that can detect prime nummbers up to infinity

Some other people had really good calculator's from last year, so...

[quote="Dnas"]You don't really need euclidean algorithm for mods... X-B*int(X/B) works. Probably don't really want a subroutine because in TI-BASIC for 83/84, subroutines (I hate that name... the correct term is function... but that's only in non-BASIC languages.... I hate BASIC... too wordy) can't take on arguments, so you have to reserve 3 variable or so for it and do the "order of operations" for the nested functions by hand.[/quote]

I meant a mod function for the Weekday program. You're right though, you don't need to use mods for the prime factorization program. I would bet that checking divisibility with the built in mod function on the TI-86 isn't much different time wise than checking X-B*int(X/B)=0.

And yep, that's what trial division means. Its checking 2 and then the odd numbers less than or equal to $[\sqrt{n}]$ for divisibility. I think this is probably the most efficient algorithm for numbers of the size the TI can handle (aside from storing a large list of primes and trial dividing by those).

[quote="sirorange"]Is anyone else just happy that they actually have a really good graphing calculator now? I know I am.[/quote]
Have you heard about all of TI's better calculators yet?

Hey, for those of us who don't spend that much money on calculators, it's nice to have a fairly good one.

I don't have that kind of money either--I got my 83 plus for $47

Wow, that's like 40% of the actual price - was it used?

A little bit, but no scratches.  It was on eBay.

[quote="bob123"]I already had a TI-89 Titanium (the best calculator :)), so I just wanted one that had a high monetary value so I could sell it :)[/quote]

You never use your TI-89....

[quote="Dnas"][quote="mathgeek2006"][quote="Dnas"]
BTW, did anyone find that the 84+ silvers they gave out this year are slower than last year's 83+ Silvers? My 83 beats the 84 by almost a second when doing 456123[/quote]
84 +'s have too much memory for their processor to handle.  Try comparing them in graphing.  That's about the only thing that the 84 + is better at.[/quote]
Hmmm... I haven't done too many tests, but the 83 seems to be better at graphing as well.[/quote]
Hmm...
That's strange.   I think that they're much better.  Maybe we just have different editions of the TI 83 plus (i know that there are a couple of editions of it.  If you compare mine to my brother's, mine has bold lettering and his doesn't.

[quote="bob123"]I already had a TI-89 Titanium (the best calculator :)), so I just wanted one that had a high monetary value so I could sell it :)[/quote]
To TI:  If you are reading this, I believe the silent majority of National MATHCOUNTS students are grateful for your generous gift.  At least half of this year's Colorado team brought borrowed graphing calculators to the competition.  I have little doubt they are grateful for your gift, and it seems unlikely they will be selling their keepsake any time soon.  On behalf of the many students who genuinely appreciate your gift, "thank you!"

Heh. Good point. I can see why this could lead to an incorrect assumption.

Thanks for the calculators! :)

I know some people not from CO that also borrowed graphing calculators. Thank you for the gift. TI-84+s are some of the best starting graphing calculators.


When starting a new unit, a teacher announced that each day after the first, the class must do twice the total number of problems that had been assigned in all previous days. At the end of six days, the students had completed one-third of the unit's problems. How many days did it take to complete the entire unit?

Suppose they did $ x$ problems on the first day. Then, in six days, they've done $ x \plus{} 2x \plus{} 4x \plus{} 8x \plus{} 16x \plus{} 32x \equal{} 63x$ problems, so there are $ 189x$ problems in total. The seventh day, they will do $ 64x$ problems, and then $ 128x$ on the eighth day, so it will take them $ 8$ days.

[hide]Actually, I don't believe your solution is correct. On the sixth day, they've done [b]63x[/b] problems. On the seventh day, they will have done [b]126x[/b] problems. That adds up to [b]189x[/b] problems, which is the total amount of problems so the answer is $\boxed{7}$ days. Or you could see that [b]1/3[/b] of the problems would be done on the sixth day so [b]2/3[/b] of the problems would be done on the seventh day and that adds up to the total amount of problems. [/hide]

Hope this helped,

-vdjejhehsw


Right now I am in high school, and going into Algebra II. I wanted some Algebra practice for next year/AMC 10, and I was recently looking at the AoPS books. I saw two algebra books, the Introduction and Intermediate series. I can do the post test on the Introduction series fine except problem # 9:

A rubber ball is dropped from a 100 ft tall building. Each time it bounces, it rises to threequarters
its previous height. So, after its first bounce it rises to 75 ft, and after its second
bounce it rises to 3/4 of 75 ft, and so on forever. What is the total distance the ball travels?

Answer: 700 (don't get why it is not infinite if the ball ounces forever)

Because of the post-test, I was thinking of choosing the Intermetidate series, but after seeing one of the problems from the Introduction to Algebra series posted and reading the excerpt on inverse functions, I realize there may be a lot of stuff in that book I don't know. I was wondering if any of you have an opinion on which book to buy? Thank you in advance for all your responses.

It is the infinate geometric series.

For example, 1+.5+.25+.125.........=2.

The formula is $ /frac {a}{1\minus{}r}$ where a is the first # and r is the common ratio between -1 and 1.  You should learn about it in algebra 2.  But you should be OK with the intermediate book.

Thanks for your advice on which book to choose. I have one question on the infinite series thing though. I looked deeper into this and I got another form of the formula:

a1/(1-r)=sum

For this problem, r could be 0.75 and the equation would S= 100/0.25, so why would the answer not be 400? If I did anything wrong please clarify me (I can tell you have a lot more experience in this than me).

after the first 100 foot fall each term in the sequence is repeated twice since the ball bounces up then falls down the same length. so the total distance is 2*400-100=700

1+1/2+1/4... is trivial without even knowing how to equate the sum of an infinite series.
S=1+1/2+1/4+1/8...
2S=2+1+1/2+1/4.....
2S-S=2[ANS.]


geometry
circumcircle
incenter
power of a point
radical axis
geometry unsolved

Let $O$ and $I$ be the circumcenter and incenter of triangle $\triangle ABC$ repectively. The incircle of $\triangle ABC$ touches sides $BC,CA,AB$ at $D,E,F$ respectively. Lines $FD$ and $CA$ intersect at $P$, $DE$ and $AB$ intersect at $Q$. $M,N$ are midpoints of line segments $PE$, and $QF$ respectively. Prove $OI\perp MN$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole paper[/url]

[quote="jackhui"]Let $O$ and $I$ be the circumcenter and incenter of triangle $ABC$ repectively. The incircle of $\triangle ABC$ touches sides $BC,\,CA,\,AB$ at $D,\,E,\,F$ respectively. Lines $FD$ and $CA$ intersect at $P,\,DE$ and $AB$ intersect at $Q.$ $M,\,N$ are midpoints of line segments $PE$, and $QF$ respectively. Prove $OI\perp MN$.
[/quote]
We have: $(PECA)=-1\Longrightarrow ME^{2}=MC\cdot MA\quad (1)$
Semilarly: $NF^{2}=NB\cdot NA\quad (2)$
Denote: $R$-circumradius of triangle $ABC$
We have:
\begin{eqnarray*}OI&\perp& MN\\ \Longleftrightarrow OM^{2}-ON^{2}&=&IM^{2}-IN^{2}\\ \Longleftrightarrow (OM^{2}-R^{2})-(ON^{2}-R^{2})&=&(IM^{2}-IE^{2})-(IN^{2}-IF^{2})\\ \Longleftrightarrow MC\cdot MA-NB\cdot NA&=&ME^{2}-NF^{2}\end{eqnarray*}
What is truly by $(1)$ and $(2)$

[quote="April"][quote="jackhui"]Let $O$ and $I$ be the circumcenter and incenter of triangle $ABC$ repectively. The incircle of $\triangle ABC$ touches sides $BC,\,CA,\,AB$ at $D,\,E,\,F$ respectively. Lines $FD$ and $CA$ intersect at $P,\,DE$ and $AB$ intersect at $Q.$ $M,\,N$ are midpoints of line segments $PE$, and $QF$ respectively. Prove $OI\perp MN$.
[/quote]
We have: $\begin{cases}(PECA)=-1\Longrightarrow ME^{2}=MC\cdot MA\\ (QFBA)=-1\Longrightarrow NF^{2}=NB\cdot NA \end{cases}$
$\Longrightarrow M,\,N$ lie on the radical axis of two circle $(O)$ and $(I)\Longrightarrow OI\perp MN$[/quote]

i got a solution using complex numbers...

i haven't been on geometry forum before so can you please tell me what does it mean $(PECA)=-1$ or at least where can i find the definitions? thanks...

It mean $P,\,E,\,C,\,A$ ia a harmonious division and you can find some problems and properties of its at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46146<]A use of harmonious division (four problems) (****)[/url]

Note that the circumferences with centres M,N and radii ME,NF meet at a point X that satisfies:
$\frac{XA}{p-a}= \frac{XB}{p-b}= \frac{XC}{p-c}$
and lies on the line IO (MN is the polar of X wrt the circumcircle and the incircle).

The question is:
Does this point have a name? (and maybe cool properties :) )

[quote="edriv"]MN is the polar of X wrt the circumcircle and the incircle[/quote]

Are you sure ?  :huh:

I'm sorry, obviously I was completely wrong!

All I can prove is that $X$ satisfies:
$\frac{XI}{XO}= \frac{r^{2}}{R^{2}}$.

I hope you find something more interesting...

I will prove that $I$ and $O$ both lie on the radical axis of the circles $(M)$ and $(N)$ with radii $ME$ and $NF$.
For $I$ we have $IE^{2}$ = $IF^{2}$.
For $O$, let $R$ be the intersection of $BC$ with $EF$ and $L$ the midpoint of $RD$. Then $P$, $Q$, $R$ lie on a line (Menelaus) and $LMN$ is the Newton-line of the complete quadrilateral formed by lines $DE$, $EF$, $FD$ and $PQR$. Using that $A$, $B$, $F$ and $Q$ are harmonic points and that $NA \times NB = NA^{2}$, one can prove that the power of O with respect to $(N)$ is $ON^{2}-NF^{2}= R^{2}$, where $R$ is the radius of $(O)$. In the same way, one can prove that $OL^{2}-LD^{2}= R^{2}$ (and that $OM^{2}-ME^{2}= R^{2}$), that is $O$ has the same power with respect to the three circles. So $O$ lies on the radical axis of the three circles, QED.
Moreover, since the power of $O$ to these three circles is $R{2}$, the three pairs of circles $((O) ; (L))$, $((O) ; (M))$ and $((O) ; (N))$ are orthogonal.

Benjamin

See how this problem is a consequence of this one:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=85987[/url]

The radical axis perpendicular to the line of the centers...

Clearly, the above solutions are the shortest. However, I will put in here some facts which probably the proposer used while creating this problem.

Here 's a fact:

[b]Fact[/b]Consider $P$ any point. Let $A_{1}B_{1}C_{1}$ be the cevian triangle of $P$ wrt $ABC$ and $A_{2}, B_{2}, C_{2}$ be the intersections of corresponding sides of $ABC$ and $A_{1}B_{1}C_{1}$. Then the midpoints of $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$ lie on a lint $l$ which is perpendicular to $OH_{P}$ where $O$ is the circumcenter of $ABC$ and $H_{P}$ is the orthocenter of $A_{1}B_{1}C_{1}$.

A straight consequence of this fact is the line $MN$ is perpendicular to $OH_{Ge}$ where $Ge$ is the Gergonne point of $ABC$. Also, a well-known lemma states that $O, I, H_{Ge}$ are collinear. Therefore $OI$ is perpendicular to $OI$.


Given reals $ a,b,c,x,y,z$ and $ (a\plus{}b\plus{}c)(x\plus{}y\plus{}z)\equal{}3$ $ (a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\equal{}4$
Proof $ ax\plus{}by\plus{}cz \ge 0$

[quote="hxy09"]Given reals $ a,b,c,x,y,z$ and $ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \equal{} 3$ and $ (a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 4$
Proof $ ax \plus{} by \plus{} cz \ge 0$[/quote]
It is Mathlinks contest 2005

Thank you very much,but can you give me a link? :)

[quote="hxy09"]Given reals $ a,b,c,x,y,z$ and $ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \equal{} 3$ $ (a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 4$
Proof $ ax \plus{} by \plus{} cz \ge 0$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70625


Prove that:

1) for all natural Nos k the sequence $ n^k$ diverges to infinity

2) the same for the sequence $ e^n$

Umm... what do you really mean for (1)?  For (2), it just follows from the fact that for each $ k\in\mathbb{N}$, you can always find some $ N\in\mathbb{N}$ such that $ e^{n}\geq k$ for all $ n\geq N$.

What is the definition of a sequence diverging to infinity?

A sequence $ (x_{n})$ diverges to $ \plus{}\infty$ if for all $ M\in\mathbb{Z}_{\plus{}}$, there exists some $ N$ such that $ x_{n}\geq M$ for all $ n\geq N$.


Solve the system of equations:
 $ x(x+y)^2= 9 $
 $ x(x^3-y^3) =7 $

It is easy to check that x=1, y=2 is a solution. Also (x+y)2 is non-negative, and its product with x is positive, so x must be positive. Hence y3-x3 is positive. So y > x and, in particular, y is also positive. Put k = y/x. Then we have x3(k+1)2 = 9, x4(k3-1) = 7. Hence 73(k+1)8 = 94(k3-1)3. Multiplying out, we get 6561k9 - 343k8 - 2744k7 - 29287k6 - 19208k5 - 24010k4 + 475k3 - 9604k2 - 2744k -6904 = 0. We know this must have a factor k-2, so we factorise as (k-2)(6561k8 + 12779k7 + 22814k6 + 16341k5 + 13474k4 + 2938k3 + 6351k2 + 3098k + 3452) = 0. All the coefficients in the second factor are positive, so it cannot have any roots k > 1. So the only (relevant) solution is k=2. That gives x3 = 1 and hence x = 1, y = 2.

if seems that $(x,y)=(1,2)$ is not a solution

for $x(x^3-y^3)=-7$ not $7$.

Sorry,boys .I think the second equation was like x(y3 - x3) = 7.That is my false.


how can you determine that a 65537gon is constructible?

It's because it is a Fermat number, that is, a number of the form $2^{2^{k}}+1$.  All $n$ gons of this form can be constructed using straight edge and compass.  This comes from the fact that you can only solve quadratics using SE&C.  Look up Constructible Polygon on wikipedia.