geometry
circumcircle
LaTeX
geometry proposed

These are problems related to Appollonius .
Here By construction ,i mean constructing by ruler and compass.
We denote:

$\mathcal L$=Tangent to a given line
$\mathcal P$=Crossing through a given point
$\mathcal C$=Tangent To a given circle

[color=darkblue]The Appollonuis problems:
Construct the following Circles:[/color]:
1.$\mathcal{P,P,P}$(Trivial)
2.$\mathcal{L,L,L}$(Trivial)
3.$\mathcal{P,L,L}$(Easy)
4.$\mathcal{P,P,L}$(Easy)
5.$\mathcal{P,L,C}$(Not easy)
6.$\mathcal{L,L,C}$(Not easy at all)
7.$\mathcal{P,C,C}$(Hard)
8.$\mathcal{L,C,C}$(Very Hard )
9.$\mathcal{C,C,C}$(Very hard and challenging indeed )

PS.For example:$\mathcal{P,P,P}$ means ,draw a circle which passes through three given points.

I have solved numbers $1,2,3,4,5,6$ by myself.
I didnt manage the $7,8,9$ cause i dont know how to deal with two circles :blush: .
Numbers $7,8,9$ are real challenges.
Who accept the challenges?(I dare say i dont, :D )

In my geometry book there are solutions for all Appollonius problems, but my English
is very bad and i can't write it down but the \textbf{MAIN} thing in all of these
solutions is this:

I don't know to write it in English, i think that it should look something like this
"INVERSE" so i will explain to you what i mean..

You have a circle with center $O$ and radius $r$. Given point $X$ somewhere in that
plane where is given circle, now you construct point $Y$ on line $OX$ such that
$r^2=OX \cdot OY$. In this way if u have line not passing through a center of given
circle and construct such point $Y$ for every point $X$ on that line you will obtain
a circle.

You should use that for solving your problems 7,8,9... (and that of course can be
used for solving all problems, for example if you use that you can solve problem 7
just like problems 3 and 5)

I hope so that you get it what i was trying to say.

Best wishes :)  :)  :)

CONTRACTION (the reduction of a some circle to one point):
$L.L.C.\rightarrow P.L.L.\rightarrow P.P.L.\rightarrow P.P.P.$
$L.C.C.\rightarrow \underline {\overline {\left| P.L.C.\right| }}\rightarrow P.P.L\rightarrow P.P.P.$
$C.C.C.\rightarrow P.C.C.\rightarrow \underline {\overline {\left| P.P.C.\right| }}\rightarrow P.P.P.$
The only problems are P.L.C. and P.C.C. (the nucleus of the Appollonius's problem). Afterwards I will offer the classical solutions for the two problems.
[b]Example.[/b]
$C_k=C(O_k,r_k),\ k\in \overline {1,3},\ r_1\le r_2\le r_3\ (C.C.C.)\Longrightarrow$
$O_1,\ C(O_2,r_2-r_1),\ C(O_3,\ r_3-r_1),\ i.e.\ P.C.C.$

[quote="levi"]CONTRACTION (the reduction of a some circle to one point):

[/quote]

well i am not sure that we think on the same thing but there is an easy way to find out...

in problem P,L,C do you first construct an arbitary circle with centar in given point A then two circles one image of a given circle (this one not passing through A) and an image of the given line (this one passing through A), then construct tangent to both of that circle and then from that tangent you construct the wanted circle

 :)  :)  :)  :)

Here are the solutions for P.L.C. and P.P.C (I suppose that $P\not\in intC)$.

$\underline {\overline {\left| P.d.\Gamma.\right| }}$. Let the point $P$, the line $d$, the circle $\Gamma =\Gamma (O,r)$ and the in great demand circle $\Phi$.
I note: $C\in d,\ OC\perp d,\ A,B\in \Gamma \cap OC,\ A\in (BC),$
$Y\in d\cap \Phi, X\in \Gamma \cap \Phi,\ R(second)\in PB \cap \Phi$.
$X\in BY,\ BR\cdot BP=BX\cdot BY=BA\cdot BC\Longrightarrow BR\cdot BP=BA\cdot BC$,
i.e the point R is the second intersection between the line BP with
 the circumcircle $\Theta$  of the triangle ACP.Therefore, $\Longrightarrow \underline {\overline {\left| P.R.\Gamma\right| }} .$

$\underline {\overline {\left| P.R.\Gamma.\right| }}$. Let the points $P,R$, the circle $\Gamma=\Gamma (O,r)$ and the in great demand circle $\Phi$.
I note $S\in \Gamma \cap\Phi$. A some circle which passes through $P$ and $R$ cuts the circle
 $\Gamma$ in the points $A,\ B$. But the lines $AB,PR$ and the tangent $XX$ in the point $X$
 at the circle $\Phi$ are concurrently. Therefore $\Longrightarrow \underline {\overline {\left| P.R.S.\right| }}$

[quote="levi"]CONTRACTION (the reduction of a some circle to one point):
$L.L.C.\rightarrow P.L.L.\rightarrow P.P.L.\rightarrow P.P.P.$
$L.C.C.\rightarrow \underline {\overline {\left| P.L.C.\right| }}\rightarrow P.P.L\rightarrow P.P.P.$
$C.C.C.\rightarrow P.C.C.\rightarrow \underline {\overline {\left| P.P.C.\right| }}\rightarrow P.P.P.$
[/quote]
I dont understand at all. Would you explain this a little? :)

well now i am sure that LEVI and I are not thinking on the same thing....
I wrote what i had to say so if you LUPIN understand then good for you if don't then never mind....

I have understand you M@re .and by your words we have the solutions.
But i was wondering what is the  meaning of levi ? :)

[quote="lomos_lupin"]I have understand you M@re .and buy your words we have the solutions.
But i was wondering what is the  meaning of levi ? :)[/quote]

well i am sure happy to hear that... and we all wonder what levi tried to say  :)  :)  :)

[i]Sir Arsene Lupin,[/i] may be I don't understand what say you or you did not understand mine. Levi is gamboling with Latex. Levi has not a pencil and a paper for to draw and to talk with you. Levi has a wretch computer. The Appollonius's construct problem is wellknown (J. Hadamard a.s.o.). I am sorry. I did not desire to annoy you. [i]With friendship, Levi.[/i]

[u]P.S.[/u] [b](Why?)[/b] The presumption of inocence and the extenuating circumstances
was invented by one mathematician. ([i]Gr. C. Moisil, a great Romanian mathematician[/i]) \[\bigcirc\Downarrow\Downarrow\infty\Downarrow\Downarrow\bigcirc\]

Wow ,what an answer, :) .
I really admire you levi cause of your latex abilities and i didnt want to offend you.
I apolosize if you take it in a bad manner but I really didnt understand this:
$\displaystyle L.L.C.\rightarrow P.L.L.\rightarrow P.P.L.\rightarrow P.P.P.
...$
Any way as you have said ,These problems are wellknow and i managed to find another nice solution so ,The problem is solved at all. :)

I think I understand What levi said.

When U want to solve $L.L.C$ first let circle   $\omega $   be tangent to 2 lines and circle.Enlarge radious of         $\omega$   till pass through center of $C$ now draw the parallel lines to $L_1,L_2$and tangent to    $\omega'$    the the problem becomes $L.L.C$ and the rest is like this,...

Thanks, Amir2 !. When we construct the tangents for two circles $C_1=C(O_1,r_1),\ C_2=C(O_2,r_2),\ r_1\le r_2$ we diminish the circle $C_1$ until it become the its centre $O_1$ afterwards we diminish the circle $C_2$ until it become the circle $C_3=C(O_2,\ r_2-r_1)$ Thus, the problem of the tangent of two circles becomes the problem of the tangent from the point $O_1$ to the circle $C_3$, i.e $C_1\ \Lambda\ C_2\Longrightarrow O_1\ \Lambda\ C_3$. Do you understand, Arsene Lupin ?
Unfortunately, now Arsene Lupin is sleeping or who know ?

Thanks, Amir2 !. When we construct the exterior tangents for two circles
 $C_1=C(O_1,r_1),\ C_2=C(O_2,r_2),\ r_1\le r_2$
we diminish (squeeze) the circle $C_1$ until it become the its centre $O_1$
afterwards we diminish the circle $C_2$ until it become the circle$C_3=C(O_2,\ r_2-r_1).$
Thus, the problem of the exterior tangents of two circles becomes the problem
of the tangents from the point $O_1$ to the circle $C_3$, i.e $C_1\ \Lambda\ C_2\Longrightarrow O_1\ \Lambda\ C_3\ \ (*)$.
 Do you understand, Arsene Lupin ?
If we speak about the interior tangents, then we fill the circle $C_2$ until it become
the circle $C_4=C(O_2,r_2+r_1)$ afterwards we diminish the circle $C_1$ until it
become its centre $O_1$. Thus, the problem of the interior tangents of the two circles
becomes the problem of the tangents from the point $O_1$ to the circle $C_4,\ i.e. \ \ (*)$.
Unfortunately, now Arsene Lupin is sleeping or who know ?
 In the Appollonius's problems appear and the lines.
These are translated on the one hand or the other hand with
the adequate length of the radius.
It is clearly, Arsene Lupin ? I hope that I didn't spoil
your the sleep (I am sorry !) Good luck to everybody !

Yup guys,You are defnitely clear now.
You see,by a little explanation its all clear. :) .
Thanks for the solutions.(although ,as i have said,I had found the solutions befor your explanations :D )

Ps.What is the meaning of $Arsene$ :?:  :?:  :D

My dear, where see you a [u]little[/u] explanation for be clearly? There was 17 rows in the last message. With regret, I must to say you that Arsene Lupin was a dreary character in a old feature film.

dreary, dreary , :!:  :!:
You know,I really dont like this word :flex:  :weightlift:  but anyway.
The problem Is solved so lets dont continue this topic.

I am sorry, I promise you that for the time being I will not use this word. Do you like the game with the words (pun) ? The [b]last in[/b] will be the [b]first out[/b] (L.I.F.O.)


preset 2n-1 natural numbers, demonsatrate can find n numbers have sum divisible n.

[quote="mathstarofvn"]preset 2n-1 natural numbers, demonsatrate can find n numbers have sum divisible by n.[/quote]Not High School Basics level. It's nice to invent problems but is it right? Let $ n \equal{} 3$. We have 5 natural numbers, say $ \{1,2,3,4,5\}$, then we can find 3 numbers, $ 1,2,4$, that the sum is, however, not divisible by 3...?  :what?:

I believe that the original problem would be to show that it is possible to find such a set of n numbers in the first set of 2n-1 numbers.

So a set of three numbers from the first five would be:
{1, 3, 5} for example, though others suffice since 3|9.

Now to prove that this is true for any such set, let us examine the set mod n.
Then it suffices to find an n element subset such that the sum of this subset is congruent to 0 mod n.

[WARNING.... NO RIGOROUS PROOF TO FOLLOW]
For n = 2, there are 4 different ways to select the original set (assuming order doesn't matter), from which it is always possible to select two elements who sum is 0 mod 2.

Now, it is also easy to see that this is true of larger numbers, though I have no way of proving it... :)

Sorry this is the first time i post topic so i have a mistake i have corrected my fault.(i am vietnamese so i am not good at English  :P )

This is a special case of Erdos-Ginzburg-Ziv Theorem. Definitely not High School Basics. Little kids may skip this problem :P


[quote="Scrambled"]It doesnt let me delete my own posts, even if no one has posted after me[/quote]It doesn't let you delete a new topic when you've made one. there were cases when people deleted (by mistake) the first post in a topic (that is the problem's statement) so in order to avoid that, the first post can only be deleted by an admin/mod in the forum.

Sometimes I can't delete a post at all, and it will be in the middle of a thread.  Why is that?

Ordinary users do not have the power to delete any post that has been replied to.

If your delete button vanishes for the [b]last[/b] post on a thread with multiple posts, then you have a real problem.


Point P is located on line AB.

a) Describe the locus of points that are:

1) 3 units from line AB

2) 5 units from point P

b) How many points satisfy BOTH conditions in part a above?

[hide="answer"]
a)
1) Two lines both parrallel to $\overline{AB}$ and 3 units away from it (one below and one above).

2) A circle of radius 5 with center $P$.

b) 4 points: each line creates a cord in the circle.
[/hide]

This is okay.  I now got the answers but have no clue how to find the answers on my own.

[size=150]How did you solve these questions?

What are the steps?[/size]

If you can't visualize it in your head, then draw a diagram.

[quote="Interval"]
[size=150]How did you solve these questions?

What are the steps?[/size][/quote]

3 units away from a line:

think about it. Draw a picture. Keep in mind that when they say distance to a line they always mean the distance to the line along a line perpendicular to that line, or the shortest possible distance to the line.

5 units away from a point:

think of 5 as a radius and the point as the center: it's have to be a circle

I made an artpad picture for you.
http://artpad.art.com/gallery/?jadolq1gma24

Set replay speed control to the fastest and watch it. It's not great, but it should explain the question.

The last word fell off, it is "points"


If $ a$, $ b$, $ c$, $ d$, and $ e$ are constants such that every $ x > 0$ satisfies
\[ \frac{5x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 4x \plus{} 7}{(x \plus{} 2)^4}
  \equal{} a \plus{} \frac{b}{x \plus{} 2} \plus{} \frac{c}{(x \plus{} 2)^2}
      \plus{} \frac{d}{(x \plus{} 2)^3} \plus{} \frac{e}{(x \plus{} 2)^4} \, ,\]
then what is the value of $ a \plus{} b \plus{} c \plus{} d \plus{} e$?

hello, we get
$ \frac {5x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 4x \plus{} 7}{(x \plus{} 2)^4} \equal{} 5 \minus{} \frac {48}{x \plus{} 2} \plus{} \frac {170}{(x \plus{} 2)^2} \minus{} \frac {260}{(x \plus{} 2)^3} \plus{} \frac {151}{(x \plus{} 2)^4}$
therefore we get
$ 5 \minus{} 48 \plus{} 170 \minus{} 260 \plus{} 151 \equal{} 18$
A nice problem!
Sonnhard.

[b]We don't need the partial fraction decomposition.[/b]

Let $ x\equal{}\minus{}1$.
So $ a \plus{} \frac{b}{\minus{}1\plus{}2} \plus{} \frac{c}{(\minus{}1\plus{}2)^2} \plus{} \frac{d}{(\minus{}1\plus{}2)^3} \plus{} \frac{e}{(\minus{}1\plus{}2)^4} \equal{} a\plus{}b\plus{}c\plus{}d\plus{}e$.
Thus $ a\plus{}b\plus{}c\plus{}d\plus{}e \equal{} \frac{5(\minus{}1)^4\minus{}8(\minus{}1)^3\plus{}2(\minus{}1)^2\plus{}4(\minus{}1)\plus{}7}{(\minus{}1\plus{}2)^4} \equal{} \boxed{18}$.

Right.  Maybe you should say a quick word about why you can substitute $ x \equal{} \minus{}1$, because the problem specified $ x > 0$.

Is there a better way to do it than just expanding out? I guess that doesn't take too long.

Kouichi's substitution method doesn't involve expanding out, and was my intended solution.  The only thing he should add is one sentence explaining why setting $ x \equal{} \minus{}1$ is legitimate.

Hmm, well because its a polynomial function already defined by an infinite number of points (since it is defined for any positive x), it should be similarly defined in the negative side. That is pretty nice  :P


Let $m,n$ be a positive integers.Find $(m,n)$ such that  $2^m-3^n=1.$

$(mod.3)$ we see that $m$ is even.
now, we have $3^n = (2^{m_1}+1)(2^{m_1}-1)$. (where $m = 2m_1$)
now, $(2^{m_1}+1)$ and $(2^{m_1}-1)$ are powers of three, with diference $2$.
so, $m_1 = 1$ and $n = 1 \Longrightarrow m = 2, n = 1$ are the only solution.


Functional Analysis
real analysis
real analysis unsolved

Let $ X$ and $ Y$ be two infinitely dimensional Banach space, $ T\in C(X\rightarrow Y)$, then prove
(1) Ran$ T\neq Y$;
(2) If $ M$ is a closed subspace of Ran$ T$, then dim$ M<\plus{}\infty$.

Is my proof right?

for (1):
  let $ \tilde{T}: X/\mathrm{Ker}T\rightarrow Y$, $ \tilde{T}\tilde{x}\equal{}Tx$, $ \forall x\in\tilde{x}\in X/\mathrm{Ker}T$. Then $ \tilde{T}$ is a compact operator. If $ \mathrm{Ran}T\equal{}Y$, then $ \tilde{T}^{\minus{}1}: Y\rightarrow X/\mathrm{Ker}T$ is bounded. Thus $ I_Y\equal{}\tilde{T}\tilde{T}^{\minus{}1}$ is compact, contradiction to $ \mathrm{dim}Y\equal{}\plus{}\infty$.

for (2):
  Since $ M$ is closed, $ \tilde{T}^{\minus{}1}(M)\subset X/\mathrm{Ker}T$ is closed. Thus $ I_M\equal{}SS^{\minus{}1}$ is compact, where $ S\equal{}\tilde{T}|_{\tilde{T}^{\minus{}1}(M)}$, which is in contradiction to $ \mathrm{dim}M\equal{}\plus{}\infty$.

Are they right? Thanks!


How many minutes are in 10$ \%$ of a 24-hour day?

24 hours are 1440 minutes.

so 10% of it is 1440/10, which is 144.

answer : 144


[b]  We take the triangle ABC with m(<BAC)=90, AM and CN medians.If we have m(<AOC)=90 and AC^2=2, calculate S(ABC), where {O}=AM intersectated with CN[/b]. :D

[quote="jin"]I think you want to know the area of ABC,that's $\sqrt{2}$.[/quote]
Why, explain me! :mad:

[quote="stancioiu sorin"][quote="jin"]It should not be in this forum.[/quote]
     It isn't right!You didn't read correctly!Please try again![/quote]
 :?
At least it looks like geometry... and the title looks like geometry, too... thus it should belong to the geometry forum...

[quote="jin"]$\angle MAC=\angle MCA=\angle CNA$
So $CNA$ and $BAC$ is similar to each other.
Suppose AN=x,AB=2x
$\frac{AN}{AC}=\frac{AC}{AB}$
Then x=1,$S_{ABC}=\frac{1}{2}AB\times AC=\sqrt{2}$.[/quote]
     You have some mistakes in writing your solution, but the answer is correct. :blush: I thought the area of ABC was 1, but I realised that my solution was wrong.Anywhay, it was an easy problem, but I wanted to solve it in many ways, but I think I can't. :roll:


Hey, on another post (which is now locked)  :(  a few of you wrote that people answer ctdown questions pretty quickly, sometimes even in about 8 seconds or so... my question is, does the timeclock start the second the question shows up on a screen (and is it one giant screen or mini- pc computer screens?) or does the Clock start the second AFTER the moderator finishes reading the question out loud?

Inquiring and anxious students want to know!!!!!!!!

The clock starts as soon as the question appears. At the countdowns I've been at, I found that most people were answering in under 10 seconds, but it seemed to me they were rushing, as there were a lot of wrong answers.

Wow, that seems superfast ... are the questions easy to SEE from wherever they are shown? Is it like a giant screen? And what happens if 2 people buzz in at the exact same time? Is it like Jeopardy? Whosever lights up first? Are there any "buzzer secrets"??? (you know, like how to buzz in quick -- or do you just raise your hands or something -- sorry, I've never done it before - does it differ between state and nats???

Well, I know some people who can answer a question in FIVE seconds, but the time starts when the question appears.

At my state, they have a projector from the ceiling projecting the questions onto a...hmmm...screen with about a 10 foot diagonal. The lowest part of the question is eye level while sitting or above, and you might be 10 feet from the screen.

Just estimates.

DON'T LISTEN TO THE GUY READING IT!!!!!! Scan the screen for important info-you'll beat someone not doing that 80% of the time. Iy is always projected so it's easy to read.

It's all practice.  I checked out my practice times from now to when I first did them - almost all of them increased significantly, even the ones under 3 or 4 seconds :D.  Skim the question, but make sure you don't leave out important facts (such as what they ask for).  If you are against a very good opponent, use the Daesun method.

Yeah, last year, I lost the second round at chapters. This year, I got 1st by a landslide. I've gotten much better. I didn't mean just look at the screen for half a second, then buzz in. I meant to quickly figure out what is being asked, and what info they give you. Don't waste time reading about Jane's grandmother or what kind of cake is being made with the flour.

A projector shows the question and peolpe in the audiance can solve while you solve too. The guy also says the question but in a sloow and monotone voice so usually "mathletes" just read ahead.

I disagree with not reading the extra information...sometimes they can pull a fast one on you and mix up answers so that you say the wrong answer (it [i]is[/i] correct but you did not answer the question)

[quote="perfect628"]Well, that method is, of course, imperfect. But, usually, you'll find all you need, and get the answer faster than our opponent. So, it cancels out, and you end up ahead. No matter how you approach CD, it's impossible to win w/o practice.[/quote]

of course at your MC practices you can have a copy of previous countdowns or the school round for 2007 as practice for countdown rounds...with the projector and everything

My chapter uses an official CD round (last year went 4th to 5th) and this year I got bumped to 3rd.  They have two comp monitors in front of us and a projector that's hard to see from the position a competitor is in.  Time seems to slow on the ones I do get, but if I don't get a problem then it seems the other person pressed the buzzer before I started work on the problem or just before I was going to buzz in.

Just curious, What is the Daesun method? (He was the national champ last year right?)

#1 thing to remember:
Don't raise your hand!
Hit the buzzer!

daesun/treething method

twitch and buzz

dude

i tried it at chapters this year

its such a PWNer

like seriously

always do it

there's no way you can lose with it unless the other person is using it or insane or you're just a plain n00b

skim problem, buzz, do it in your head while at the same time say a bunch of crap that makes absolutely no sense, when you get the answer, say it

its much easier at chapters because the problems are mad easy

but do it anyway

you might wanna practice before doing it though

i wouldn't advise doing it on your first time countdowning

especially in an official countdown

but its worth a try

if you know you can do it DO IT FTW

i'm done

what's FTW stand for?? And does the 3 seconds you're allowed to answer in start after you begin speaking or before?  In other words - you buzz - you say "I think the answer is 25!" -- if you say "[b]I[/b]" at second 3" will they allow you to say your answer?? Or do you need to be DONE answering by second 3?? What type of words did Daseun or others mutter before they gave answer?

well no one knows...but generally they give you the benefit of the doubt, unless it's really really obvious that you're trying to twitch

"my answer is..."
"like..."
"i think i got it this time..."
"uh...
"*coughcoughsneezecoughcough*..."

you get the point

ftw=for the win

[quote="Fanatic"]footnote about "ftw" [for the win] it's like semi facescious (sp?). Like the "lol" version of "to do best".

Anyhow, the "twitch" method can only work if you're really good. Some people just "have it in them" to do countdown. They're just that type of person. Some people that are extremely good at math [melanie wood comes to mind] can't do countdown.[/quote]

lol the twitch method works only for short-to-moderate problems...long word problems with/without complicated figures take a little more time...unless im not mistaken you have 3 seconds to give a full and proper response?

on and on another note:

do not drink a lot of caffiene (sp?) before CD!!!!!! umm i won't say now but my friend screwed up a simple problem because he was too hyper...lol

why don't they just move the table back?

Or using individual comp. screens would work, too. A new piece of paper for every Q? That seems impractical...

I think they mean that you get the print off similar to a sprint round. Thats only like a 1/2 page per round...

Oh nvm...they could look ahead...

DOH!

[quote="wxyz"]What problems can you do to practice countdown after you have done all the chapter and state countdown problems in the recent years (2002-2006)?  I kind of don't like to do the same thing over, but it is hard to find problems that are good for practicing countdowns.  Any ideas or suggestions? Thanks.[/quote]

probably not a good idea but even more previous years....

Excellent Question! I hope people will reply -- I'm at a loss too...

[quote="kyyuanmathcount"][quote="jhredsox"]If possible, you could do what Daesun Yim did. He set up a program in his calculator that at random would give him two two-digit numbers and he'd have to find the sum within ten seconds and they just kept on coming. If you would like to find more, go to search and search tripping in countdown or something, I remember I started a topic on it some time ago.

Wow. All in one sentence. Phew!

-jorian[/quote]

Sum would be trivial.  It was product - I have done that too, except my calculator programming skills don't allow me to have the 10 second limit.[/quote]

hehe i think i remember a 10 second limit....but i forgot as i haven't programmed BASIC in like a year...

[quote="jli"][quote="cat810930"]At my chapter, they actually print out the problems on a peice of paper for the two people that are answering because the screen is so close to the table that it's almost imposible to read the questions from up there. :)[/quote]

same here

did the format for countdown change this year or was it just my county?

woo hoo 1st![/quote]

LUCKY
I couldn't see anything...

And about the countdown round format, I think that each chapter chooses whether to use ladder or bracket format.  But if the countdown round is official, it has to be ladder format for obvious reasons.  My chapter (cd is unofficial) changed from ladder to bracket form this year.

I was the timekeeper at the Seattle chapter this year- I tried to start the clock when the question first appeared on screen, but that wasn't precise. We used a laptop running Powerpoint attached to a projector to show the questions.
The exact timing doesn't matter very much- the clock only ran down once or twice in all 9 matches. You're there to beat the other competitor, not the clock.
I called the three-second rule once. That didn't really require stopwatch-watching, since most competitors gave the answers within about 1 second.

[quote="jhredsox"][quote="kyyuanmathcount"][quote="jhredsox"]If possible, you could do what Daesun Yim did. He set up a program in his calculator that at random would give him two two-digit numbers and he'd have to find the sum within ten seconds and they just kept on coming. If you would like to find more, go to search and search tripping in countdown or something, I remember I started a topic on it some time ago.

Wow. All in one sentence. Phew!

-jorian[/quote]

Sum would be trivial.  It was product - I have done that too, except my calculator programming skills don't allow me to have the 10 second limit.[/quote]

i very much meant product. i was even thinking about accidently putting sum, but then i got side-tracked.

jorian[/quote]

Haha, no problem.  You don't know how many times I've done that on tests/contests.

Try not to use 3 seconds when practicing - under that pressure, many cannot finish solving in the 3 seconds.

[quote="jhredsox"]Remember not to get thrown off by the really easy ones that have happened at past competitions (no comment on the past chapter). But also don't buzz in even if it looks really simple if you're not sure it'll be real simple :D .

-jorian[/quote]
Really?
I always buzz in at county in under 5 secs regardless of whether I have any clue how to do the problem or not because usually I can get them in 8 secs.

[quote="jhredsox"]Remember not to get thrown off by the really easy ones that have happened at past competitions (no comment on the past chapter). But also don't buzz in even if it looks really simple if you're not sure it'll be real simple :D .

-jorian[/quote]

Buzz the moment that slide pulled up and then just say um...uh...um....[ANSWER GOES HERE]

We use a countdown-inspired round to finish off our local high school contest. There, the screen is mostly for the benefit of the rest of the audience; the competitors usually can't see it at all. We give them printed versions of the questions, one question per page. But we also tend not to use 10-second questions, and I'm willing to let a question go for at least two minutes, sometimes longer. We do have some kind of time limit after a first incorrect answer, but not before then.

[In 1996 - 11 years ago, wow! - jmerry never lost a countdown question, local or state.]

"Once I call on you, you will have three seconds to begin your answer."

Does "begin your answer" mean to say "uh"?  Does "begin your answer" mean to say "the answer is"?  Or does "begin your answer" mean to say the first syllable of the numerical answer?

I can't blame competitors for taking advantage of the rules your countdown moderator allows, but the safest bet is to be fastest and most accurate.

How do you get faster and more accurate?  Practice against your teammates.  Practice against your coach.  Practice against your older siblings.  Practice against your parents.  Practice against anybody you can find that's faster than you are.

[quote="rcv"] Practice against your older siblings.[/quote]
I personally found this very helpful.
In particular if this older sibling did mathcounts themself.

lol prolly memorizing squares (13-20) or something  :D

[quote="guyinPA"]Uh - while I think it's cool to impress friends (ok, geeky friends) why would being able to quickly multiply 2 digit numbers in your head help in  mathcounts? From the practice problems I saw, there weren't too many (if Any) problems that required multiplying large sums in one's head[/quote]
yes, but it sincerely helps with speed with numbers
jorian

I know rcv (rvc?) was quoting the below sort of tongue n cheek : "Once I call on you, you will have three seconds to begin your answer."

Does "begin your answer" mean to say "uh"? Does "begin your answer" mean to say "the answer is"? Or does "begin your answer" mean to say the first syllable of the numerical answer? "

but I think these are valid questions - I mean, if some contestants know how the rules work and others don't, that's not exactly fair, is it? Can anyone in the real "know" or someone from the Mathcounts organization, even, comment on the correct answers to the above questions?

(and yes, I get the point rcv (rvc) was making - practice lots, do your best, yada, yada, yada - yes, of course, that's obvious, but knowing the answers to the above questions ISN'T  :(  - I think that's why people ASK

I think knowing the multiplication table up to 20 or 30 WOULD help your speed.


Prove that the number below is a integer:

$ \frac{2006.4007.2003\plus{}1999}{(2005)^{2}}$

[hide]Let $ 2005 \equal{} x$

$ \frac {(x \plus{} 1)(2x \minus{} 3)(x \minus{} 2) \plus{} x \minus{} 6}{x^2}$

The constant term of $ (x \plus{} 1)(2x \minus{} 3)(x \minus{} 2) \plus{} x \minus{} 6$ is $ ( \minus{} 3)( \minus{} 2)(1) \minus{} 6 \equal{} 0$

The coefficient of the $ x$ term is $ ( \minus{} 2)( \minus{} 3) \plus{} (2)( \minus{} 2)(1) \plus{} ( \minus{} 3)(1) \plus{} 1 \equal{} 0$

$ \frac {(x \plus{} 1)(2x \minus{} 3)(x \minus{} 2) \plus{} x \minus{} 6}{x^2}$ is therefore, after simplification, a polynomial with integer coefficients.

so $ \frac {2006.4007.2003 \plus{} 1999}{(2005)^{2}}$ is an integer.[/hide]


Since they don't allow proving problems in Getting Started, I'll post this problem in Intermediate even though it really is for getting started:

Let $x=b^2+3j$ for some positive integers $b$ and $j$. Suppose that when $x$ is divided by 5, the remainder is 1. Show that $x=c^2+15k$ for some integers $c$ and $k$.

[hide]$x=b^2+3j and x=c^2+15k$

$x=5n+1$

Set c so that its square is 1 greater than a multiple of 5, because 15 is already a multiple of five. Thus we have c is 6 and n is 1. Therefore, you get $36 + 15 = x = 5(6) + 1 + 5(3) = 5(9) + 1$, matching the criteria for x.

So $c = 6$ and $k = 1$[/hide]


geometry
circumcircle
radical axis
power of a point
geometry proposed

If on $ \triangle ABC$, trinagles $ AEB$ and $ AFC$ are constructed externally such that $ \angle AEB\equal{}2 \alpha$, $ \angle AFB\equal{} 2 \beta$.
$ AE\equal{}EB$, $ AF\equal{}FC$.
COnstructed externally on $ BC$ is triangle $ BDC$ with $ \angle DBC\equal{} \beta$ , $ \angle BCD\equal{} \alpha$.
Prove that 1. $ DA$ is perpendicular to $ EF$.
                2. If $ T$ is the projection of $ D$ on $ BC$, then prove that $ \frac{DA}{EF}\equal{} 2 \frac{DT}{BC}$.

(1) $ (E)$ is a circle with center $ E$ and radius $ EA \equal{} EB.$ $ (F)$ is a circle with center $ F$ and radius $ FA \equal{} FC.$ $ DB$ cuts $ (E)$ again at $ X$ and $ DC$ cuts $ (F)$ again at $ Y.$ $ \angle AXD \equal{} \alpha, \angle AYD \equal{} \angle \beta, \angle XDY \equal{} \pi \minus{} (\alpha \plus{} \beta)$ $ \Longrightarrow$ $ X, A, Y$ are collinear. $ \angle YXB \equal{} \alpha \equal{} \angle DCB$ $ \Longrightarrow$ $ BCYX$ is cyclic with a circumcircle $ (P).$ Radical axes $ BX, CY$ of $ (E), (P)$ and $ (F), (P)$ intersect at the radical center $ D$ of $ (E), (F), (P)$ $ \Longrightarrow$ $ AD$ is radical axis of $ (E), (F)$ perpendicular to their center line $ EF.$

(2) If $ U, V, W \in XAY$ are feet of perpendiculars from $ D, E, F,$ then $ \frac {DA}{EF} \equal{} \frac {DU}{VW} \equal{} 2\frac {DU}{ XY} \equal{} 2\frac {DT}{BC}.$

Sorry for bumping this, but isn't $AFB=2$β, whereas you have used $AFC=2$β?

I think there must be some printing error as we constructed AFC so the angle which it is talking about must be angle AFC


This very nice inequality was communicated to me from Harazi.

For x,y,z,t positive real numbers and  t=min{x,y,z,t}  prove that

x^2+y^2+z^2-xy-yz-zx>=t(x+y+z-3(xyz)^1/3)

We can let $x \geq y \geq z \geq t$, and we must prove:
\[  \sum x^6- \sum x^3y^3 \geq z^3(x^3+y^3+z^3-3xyz) \]
\[ \sum (x-y) ^2  [(x ^2+xy+y ^2) ^2  -z^3(x+y+z)] \geq 0. \]

And now we can have $m=x-y $, $n=y-z$, and this one is not too difficult....
Am I right??? :?

[Admin Edit: Made Latex Compliant]

My solution is different. I believe you that your solution is good, but I won't check the computations!

I  dont  understand  Anh  Cuong's  computations....so  can  we  please have  your  hint  to  this  inequality  harazi?  Thanks

It's clear that it's enough to prove that [tex] x^2+y^2+z^2-(xy+yz+zx)\geq\sqrt[3] {xyz}\cdot (x+y+z-3\cdot\sqrt[3] {xyz}) [/tex]. Take [tex] xyz=1[/tex]. The inequality becomes [tex] x^2+y^2+z^2+3\geq x+y+z+xy+yz+zx [/tex] and follows from the lot discussed [tex]  a^2+b^2+c^2+abc+2\geq a+b+c+ab+bc+ca [/tex]

What do you say, Vali? Yes, I know , it can be better, but it's a start!

if you are talking about latex, it's great looking :) you might also try to edit the other posts above in order to make them more readable, as for the inequality itself, I will give it as try later ;)

and  how  you  prove a 2 + b 2 + c 2 + abc + 2  >= a+b+c+ab+bc+ca ?
thanks

I think I have put the solution to thi inequality three times so far. Consider f(a,b,c)=LHS-RHS. First you show you can choose one of the numbers, let it be a, smaller than 3 (otherwise is trivial). Then you show with cmputations that [tex] f(a,b,c)\geq f(a,\frac{b+c}{2},\frac{b+c}{2}) [/tex]. Then you show that [tex] f(a,t,t)\geq0 [/tex] by viewing it as a seond degree inequation in t, with negative discriminant.

I have tried to complete Anh Cuong's solution but I cannot.
Anh Cuong, can you please post your complete solution?

Thank you very much.

No, I mean a smaller than 3. It's enough.

Sorry  harazi....still  dont  get  the  point  ''otherwise is  trivial''  When  you  have time  give me  a  hint.

Now I have time. So, suppose that [tex] a,b,c\geq 3 [/tex]. Then [tex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 1 [/tex] so [tex] abc\geq a+b+c [/tex]. Since also [tex] a^2+b^2+c^2\geq ab+bc+ca [/tex] I think we are done.


Let a,b,c such that $ a,b,c\ge \minus{}4, a^2\plus{}b^2\plus{}2c^2\equal{}50$. Find the minximum of $ A\equal{}a\plus{}b\plus{}c$.

By Cauchy-Schwarz, $ A_{max}\equal{}5\sqrt{5}$ when $ (a,\ b,\ c)\equal{}(2\sqrt{5},\ 2\sqrt{5},\ \sqrt{5})$.

Sorry, I think it have minA=-11. Another solution, we have
$ (a\plus{}4)^2\plus{}(b\plus{}4)^2\plus{}2(c\plus{}4)^2\ge 0$
or $ 2a\plus{}2b\plus{}3c\ge \minus{}25$ (from $ a^2\plus{}b^2\plus{}2c^2\equal{}50$)
so, $ 3A\equal{}(2a\plus{}2b\plus{}3c)\plus{}(a\plus{}b)\ge \minus{}25\minus{}8\equal{}\minus{}33$ (from $ a,b\ge \minus{}4$)
$ A\ge \minus{}11$

[quote="Hong Quy"][quote="thegod277"]Let a,b,c such that $ a,b,c\ge \minus{} 4, a^2 \plus{} b^2 \plus{} 2c^2 \equal{} 50$. Find the minximum of $ A \equal{} a \plus{} b \plus{} c$.[/quote]
 let   $ x \equal{} a \plus{} 4 , y \equal{} b \plus{} 4$and $ z \equal{} c \plus{} 4$ (  with note that $ x,y,z\geq 0$)
equality becomes $ 8(x \plus{} y \plus{} 2z) \equal{} x^2 \plus{} y^2 \plus{} 2z^2 \plus{} 14$(**)
suppose that : $ x \plus{} y \plus{} z < 1$ (*).  From (*) we have  value all  $ x,y,z$ smaller $ 1$
$ x^2 \plus{} y^2 \plus{} 2z^2 \plus{} 14 < 8 \plus{} 8z$
  but  other hand : $ x^2 \plus{} y^2 \plus{} 2z^2 \plus{} 14 \geq 2(z^4 \plus{} 3) \plus{} 8 \geq 8z \plus{} 8$ ( by  AM-GM ineq )
 hence ; $ x \plus{} y \plus{} z \geq 1$
  Min $ (a \plus{} b \plus{} c) \equal{} \minus{} 10$[/quote]
 i have mistake
  hence ; $ x \plus{} y \plus{} z \geq 1$  $ \iff$  $ Min (a \plus{} b \plus{} c) \equal{} \minus{} 11$
 equality  occur  $ \iff a \equal{} \minus{} 4,b \equal{} \minus{} 4,c \equal{} \minus{} 3$


Let $ X_1, X_2,...$ be independently identically disributed random variables with mean $ E[X]$ and variance $ Var(X)$. Let $ N\equal{}\{0,1,2...\}$ be a random variable independent of all $ X_i's$. Let  $ S_n\equal{} \{ X_1 \plus{} X_2 \plus{}...\plus{} X_n \}$ if $ n\geq 1$ and $ 0$ otherwise. If $ E[S_N]\equal{}E[X]E[N]$
How do you show that $ Var(S_N)\equal{}E[N]Var(X)\plus{}(E[X])^2Var(N)$

We will use the fact that $ \text{Var}[Y] = \text{Var}[\text{E}[Y | X]] + \text{E}[\text{Var}[Y | X]]$.

In this case, $ Y = S$ and $ X = N$.

Now note:
$ \text{E}[S | N] = \text{E}[X_1 + X_2 + X_3 + \ldots + X_N | N] = \text{E}[X] \times N$
and
$ \text{Var}[S | N] = \text{Var}[X_1 + X_2 + X_3 + \ldots + X_N | N] = \text{Var}[X] \times N$
This is true since the $ X_i$'s are iid.

Then
\begin{eqnarray*} \text{Var}[S] & = & \text{Var}[\text{E}[S | N]] + \text{E}[\text{Var}[S | N]] \\
& = & \text{Var}[\text{E}[X] \times N] + \text{E}[\text{Var}[X] \times N] \\
& = & (\text{E}[X])^2\text{Var}[N] + \text{E}[N]\text{Var}[X] \end{eqnarray*}
since $ \text{E}[X]$ and $ \text{Var}[X]$ are constants.

N.B. You should be able to prove that $ \text{E}[S] =  \text{E}[X]\text{E}[N]$ using similar methods to above. These are standards results related to compound sums.


If $a$ and $b$ are positive integers,
 such that $(a^2 + b)/(b^2 - a)$ and $(b^2 + a)/(a^2 - b)$
 are both positive integers,
 find all a and b, if they exist.
 Otherwise, disprove their existence.

 P.S. : This problem has been doing the rounds here for some time,
 and no one knows the source. So if you've heard this before,
do let me know.

It's rather straighforward but there's some casework to do - I'll show main outline:

there are 3 cases:

1) $a^2-b=b^2+a$
2)$b^2-a=a^2+b$ (symmetrical to 1) )
3)none of the above holds then $a^2-b \leq \frac{b^2+a}{2}$ and $b^2-a\leq \frac{a^2+b}{2}$ from which we obtain $a^2+b^2 \leq 3(a+b)$ which also leads to some casework.


Anyone know when the results of the alternate SATs (June 18th) are going to be available?

I took alternate SAT II last year, and as far as I remember, it was posted online after a week or so from regular scores are up.

wasn't the test given the same week as ARML last year- like the wednesday after or something like that? This year, it was given alot later (2 weeks, isn't it), so i assume a similar delay would be there for the reporting of the grades...

On the College Board site, they said to check on July 2.


Hello!

I have a profit equation I need to figure out. It's based on a word problem about apple cider which goes like this: The math club decides to sell applce cider for a fundraiser.  They can get a good deal buying the bulk cider at 90 dollars  for 30 gallons.  Or They can buy for a nearby organic apple orchard.  The farmer has agreed to lower his price for the club to 5 dollars per gallon.  You get paper cups only in batches of 100 at 2 dollars each.

Then it asks you to find out essentially how many cups (8 oz)  the bulk cider will be.
[hide="Reveal Answer"](480 8oz cups).[/hide]

Then it asks you the total cost of the cider and cups for the bulk cider.
[hide="Reveal Answer"](90+10 dollars for 500cups = 100.00 total)[/hide]

Then it asks the same thing about the organic
[hide="Reveal Answer"](30 gal. * 5 dollars per gallon = 150.00+ 10 dollars for 500 cups = 160.00 total)[/hide]

Now's where the question gets a bit muddled for me.
Next it tells you that the club wants to sell the bulk cider for 1.75 per cup.  But if they go with organic cider they could charge 2.50 because it's higher quality.  They ask to write a Profit equation (Profit = Sales - Cost) for each type of cider in terms of nubmer fo cups sold.  c=number of cups sold and p=profit.
Here are my two equations:
Bulk:  P=1.75c-100.00
Organic: P=2.5c-160.00

Firstly, does this equation sound right?

Next it asks (this is where I have the trouble) to use the equations from the last question to find out how many cups the club needs to sell to make a profit on the bulk or the organic.  Here's what I did, and please help me figure out what I did wrong (cause it just doesn't look right)
Bulk: P=1.75c-100
        100+P = 1.75c-100+100
      100/1.75+P/1.75= 1.75c/1.75
       57.14     + P/1.75  = c

Where did I go wrong?  I know I did something terribly wrong.  Help.  Cracking Up

 :wallbash:

[quote="easyas3.14159..."]Your two equations sound right...I think all you have to do is find the value of c for both equations so that $p>0$. For the bulk, the minimum value of c is 58, and for the organic it's 65.[/quote]

How did you get to your conclusions?  I just have this huge mental block.  I feel like I aught to be posting to an "elementary school" forum instead.  I feel like such a dunce.

Anyway...can you spell it out for me?

I've re-written my equations to be p>1.75c-100  and p>2.5c-160

But I still just don't know if I'm going in the right direction, because after that, I've got no idea what to do. :?

[quote="easyas3.14159..."]Sorry if i didn't make it quite clear (me->algebra dunce).This is how I thought of the problem:
1)I took your original two equations with P representing the profit, whether it's positive or negative; and c as the number of cups sold.
2)Since we want P to be positive so the club can make money, we plug in values of c to find the least positive value that'll make P positive.
3)You can try numbers from that point, and you'll probably get 58 and 65.[/quote]

Thank you !  I am never really sure what to do with algebraic equations.  Sometimes they can be solved, other times (like this one) you have to guess at numbers.  I'm never too sure about guessing since it seems so "unscientific".  HA!

Thanks!  I've got it now!
Your are awesome....I just wish I could take you with me to my tests!
:)

[quote="heide.rumble"]Sometimes they can be solved, other times (like this one) you have to guess at numbers.  [/quote]

This problem requires no guessing, just rounding.  The equations:

$P_{bulk} = 1.75c - 100$
$P_{organic} = 2.5c - 160$

The last step of your work on $P_{bulk}$ is this:  $57.14 + \frac{P}{1.75} = c$.  This is not actually wrong!  You just forgot to interpret this result correctly.  The question asks how much needs to be sold so that a profit is made - that is, $P > 0$.  Putting $P = 0$ into this equation, we get $c = 57.14$.  Now, obviously we can't sell a fractional number of cups, so the smallest value of $c$ that makes $P$ positive is $\boxed{58}$ (just round up!)

Similarly, for $P_{organic}$, we find

$64 + \frac{P}{2.5} = c$.  Now, we want $P > 0$, and selling $64$ cups means $P = 0$, so we need to sell one more, or $\boxed{65}$.

So see, you can solve this just like any other algebra problem, you just need to interpret your result at the end.  :)


(B.Frenkin) Consider the triangles such that all their vertices are vertices of a given regular 2008-gon. What triangles are more numerous among them: acute-angled or obtuse-angled?

I'm too bad at Combinatorics so please !

Label the vertices of the polygon $A_0,A_1 \cdots A_{2007}$

We will explicitly find the number of obtuse angle triangles

Let the triangle chosen be $A_iA_jA_k$ with $0 \le i < j < k  \le 2007$

Then it is easy to see that the triangle is obtuse iff either of the three angles are obtuse
Obviously the three cases so formed are disjoint

[b]CASE 1[/b] $ \angle A_iA_kA_j$ is obtuse

$\implies j-i>1004$

No. of such triplets $(i,j,k)$ are
$\sum_{m=1}^{1002} \frac{m(m+1)}{2}$

[b]CASE 2[/b] $ \angle A_jA_iA_k$ is obtuse

$\implies k-j>1004$

Again,
No. of such triplets $(i,j,k)$ are
$\sum_{m=1}^{1002} \frac{m(m+1)}{2}$

[b]CASE 3[/b] $ \angle A_iA_jA_k$ is obtuse

$\implies k-i<1004$

No. of such triplets is
$1004 \times \binom{1003}{3}$

After that I think it should be easy

Hope I have made no mistakes  :P


Let $k$ be a positive integer and $M_k$ the set of all the integers that are between  $2 \cdot k^2 + k$ and $2 \cdot k^2 + 3 \cdot k,$ both included. Is it possible to partition $M_k$ into 2 subsets $A$ and $B$ such that

\[ \sum_{x \in A} x^2 = \sum_{x \in B} x^2.  \]


Find solutions or proof that no solutions exist to the equation
n(p-1)p - m(q-1)q = 2
where p,q are two odd primes, n and m positive integers

It depends. The equation writes $ n\left ( \frac {p\minus{}1}{2}p \right ) \minus{} m\left ( \frac {q\minus{}1}{2}q \right ) \equal{} 1$. If $ \frac {p\minus{}1}{2}p$ and $ \frac {q\minus{}1}{2}q$ are co-prime, then yes, there exist infinitely many solutions (e.g. $ p\equal{}5$, $ q\equal{}3$).
If not, clearly there is no solution. This may happen when $ p \mid q\minus{}1$ (e.g. $ p\equal{}3$, $ q\equal{}7$) or similarly $ q \mid p\minus{}1$, or when $ \gcd(p\minus{}1, q\minus{}1) > 1$ (e.g. $ p\equal{}11$, $ q\equal{}31$), or a combination of the two (e.g. $ p\equal{}7$, $ q\equal{}43$).


Prove the following without calculus... Please... :noo:
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...<3$

Nice method. Are there other ways to approximate this sum? For example, can we prove that
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...<2.9$?

Define $S_{n}= \frac{1}{0!}+\ldots+\frac{1}{n!}$.  Then we can use exactly the same trick, carried out the the $n$th stage:

$S-S_{n}= \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots$
so
$(n+1)(S-S_{n}) = \frac{n+1}{(n+1)!}+\frac{n+1}{(n+2)!}+\ldots < \frac{n+1}{(n+1)!}+\frac{n+2}{(n+2)!}+\ldots = \frac{1}{n!}+\frac{1}{(n+1)!}+\ldots = S-S_{n-1}$
so that
$nS < (n+1)S_{n}-S_{n-1}$.  This will give you upper bounds "as good as you like," provided you're willing to calculate the $S_{i}$.  For $n = 2$ it gives $2S < 3S_{2}-S_{1}= 3(1+1+\frac12)-(1+1) = \frac{11}{2}$ so $S < \frac{11}4 = 2.75$.

This converges extremely quickly: the approximation for $n = 10$ gives $S < \frac{4697191}{1728000}\approx 2.71828182870\ldots$ which is good to 9 decimal places.

Are there better ways to approximate $e$ than with series? I think you can get a little better with the Taylor series evaluated at $x=2$.

Just a "cosmetic" appendix to [b]JBL[/b]'s solution:

\begin{eqnarray*}nS<(n+1)S_{n}-S_{n-1}&\iff& nS<nS_{n}+(S_{n}-S_{n-1})\\ &\iff& nS<nS_{n}+\frac{1}{n!}\\ &\iff& S<S_{n}+\frac{1}{n\cdot n!}\end{eqnarray*}

hence for practical upper bounds it's enough to calculate just $S_{n}$.

Ooh, that's much prettier.  And it tells you how good the approximations are, since $S_{n+1}= S_{n}+\frac{1}{(n+1)n!}< S$.

[quote="vishalarul"]Are there better ways to approximate $e$ than with series? [/quote]

As in, without series?  I would be interested in any closed forms you might have for $e$  :rotfl:


[quote="akalra1"]what are hidden users?

just curious[/quote]

You can choose to be hidden and you would appear to be offline

Go to Profile > Preferences > Privacy Choices.

Actually the new server doesn't say you're online or offline. It says you're hidden.

Why do we have the hidden feature anyway? It's so useless.

[quote="José"][quote="ZetaX"]:huh:
There are some users that simply don't want to be seen when online.
What's the problem in that¿[/quote]

Yes, but the question is: why do they doesn't want to be seen online?????[/quote]

Your relative, friend should not check back you are sitting in front of the PC all day. Somebody asked you for a topic reply and you are too lazy to do that etc.

is there a cure?

i better go to the doctor

The only cure is being observed by people.

Put what in the AoPSWiki?

Anyway, the hidden feature makes it so that the person does not show up on the list of people currently on.  This list can be found at the bottom of the index page.  However, if you are viewing a thread, it will say "Hidden" under a person's name if they are hidden and online.

No, you can see the status as hidden as joml88 said (not sure it was the same before the update).

[quote="SplashD"][quote="joml88"]Anyway, the hidden feature makes it so that the person does not show up on the list of people currently on.  This list can be found at the bottom of the index page.  However, if you are viewing a thread, it will say "Hidden" under a person's name if they are hidden and online.[/quote]

I thought that was only if yourself is hidden and you're reading one of your posts.[/quote]

strangely, in some topics, I could see that you were hidden, in some other topics it says you're offline...

Thats normal, see what joml88 wrote:
Online: The person is online and is shown in the list.
Hidden: The person is online but has decided not to be shown there.
Offline: Simply offline.

just a useless feauture that everyone gets excited about (like everyone was about AoPS :D JK)


My signature is [code][url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1566]A Mathematical Insight][/url][/code], but for some reason won't show up!

If you look at all my posts you can see that there is no signature even though I have submitted one?

Is there anything I'm doing wrong?[/code]

go in profile look in preferences and choose the "always attach signature button".

If not, just click the check box in the "attach signature" button on the bottom of the posting mode thing.


A very interesting observation  :D :

Consider a spaceship with mass of $m$, that needs to travel a distance of $d$ in less than $t$ seconds.
Given than there is not external force acting on spaceship.
Calculate the minimum work needed.

PS. Consider the force nonconservative.
If you consider it conservative, One can say that accerelate half of the way then deaccerelate the other half in the same manner so the total work is zero, So dont add them algebraically.

What do you mean when you say the spaceship has to travel a distance d
 :?:
if you mean the centre of mass has to travel a distance d then that's quite not possible [b](provided the spaceship does not have any initial velocity)[/b] because there is [b]no net external force [/b]acting on the [b]system(spaceship)[/b]

Since its average velocity is at least d/t, my guess is that you accelerate it to the speed d/t at the beginning.  Then by the work-energy theorem W=1/2 m (d/t)^2

Problem is clear.
An object  of mass $m$ needs to travel a distance $d$ under $t$ seconds.
Find the path with minimum work.

[b]"Given than there is not external force acting on spaceship. "[/b]

CAN YOU EXPLAIN THAT ?

[quote="lomos_lupin"][quote="pardesi"][b]"Given than there is not external force acting on spaceship. "[/b][/quote]

That means only the engine of spaceship acts on it.[/quote]

[b]well That's not at all clear from the way you stated [/b].further this is   quiet a vague problem. it doesn,t specify whether the spaceship is human operated[b](that's necessary because then you can accelerate and deccelarate  according to your own will,you may not even move straight)  [/b]


I'm looking for a solution for this problem:
Find all functions f:  (0;1)--->R :f(xyz)=xf(x)+yf(y)+zf(z) for all real numbers x,y,z in (0;1)
thanks :)

Let's evaluate $f(xy^4)$ in two different ways. It can expressed as $f(xy^4)=f(x\cdot y\cdot y^3)=xf(x)+yf(y)+y^3f(y\cdot y\cdot y)=xf(x)+yf(y)+3y^4f(y)$. On the other hand, $f(y\cdot y\cdot xy^2)=2yf(y)+xy^2f(xy^2)=2yf(y)+xy^2(xf(x)+2yf(y))$. So, $xf(x)+yf(y)+3y^4f(y)=2yf(y)+xy^2(xf(x)+2yf(y))$ or $xf(x)(1-xy^2)=yf(y)(1+2xy^2-3y^3)$. Denote $g(t)=1+2xt^2-3t^3$, therefore, $g(1)=-2(1-x)<0$ and $g(0)=1$. Thus, there exists $u \in (0;1)$ such that $g(u)=0$. Put $y=u$, then $xf(x)(1-xu^2)=0$ or $f(x)=0$. It is obvious that $f(x)=0$ fulfils the condition.


um...

anyone play America's Army???

if you do, how did you download it...

and tell me how......

Hmmm...I think I read an article about America's Army in a magazine or newspaper.  It similates being in the Army, right?

yeah :)

I played it before,

but after we formatted my computer

it was bye bye :(

I play on and off. It's pretty fun. Go to http://www.americasarmy.com for a list of mirrors.

Well, I've played it at a LAN party once. For a free game, it's alright.


Let $ x,y\in R$ such that $ x\plus{}y\equal{}3(\sqrt{x\minus{}2}\plus{}\sqrt{y\plus{}1}\minus{}1)$. Find the maximum of $ xy$.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=324570[/url]


it should be the same because we have nothing different(i mean laws of physics,yes ofcouse thsi can detect current the other can't) in a galvanometer than a resistor.
just apply kirchoff'[s laws

[quote="BanishedTraitor"][quote="drunner2007"]I'm having trouble determining the resistance of a certain configuration of resistors. It is almost like the wheatstone bridge, except that instead of having a galvonometer on the diagonal, there is another resister. How would you find the equivalent resistance of the configuration?

(just to clarify, you have a square with one of its diagonals. there is a resistor on each edge of the square, and the diagonal. two opposite corners of the square are connected to a circuit and there is a potential difference $ V$ between these opposite corners)[/quote]

Doesn't the resistance of the resistors matter as well? :maybe: because it can be a Wheatstone Bridge but it is also possible that there is a potential difference across the resistor in the diagonal section.[/quote]

Yeah they do. Assume that you have resistances $ R_{1}$, $ R_{2}$, etc.

even galvanometer has it's own ressistnace so it is one and the same provided u assumed previously the galvanometer's resistance

[quote="BanishedTraitor"]In that case, if it is not a Wheastone Bridge then you have to change the triangular circuit into a $ Y$-shaped circuit.[/quote]

can you explain in more detail? i've tried interpreting the circuit in many different ways

well is u r really patient enough then kirchoof's laws shold do
otherwise star-delta method might work out (ofcourse that's derived from kirchoff's law)

[quote="BanishedTraitor"]If I interpreted your question correctly (I hope I did) then you should use the $ Y-\Delta$ Transform. Here's a link:
[url]http://en.wikipedia.org/wiki/Y-%CE%94_transform[/url][/quote]

thank you that's the kind of thing i was looking for.


Euler
inequalities
inequalities unsolved

ABC  triangle , then $ \frac{b+c}{m_a}+\frac{c+a}{m_b}+\frac{a+b}{m_c} \leq\frac{2\sqrt{3}R}{r}  $
thanks!

Yes it is easy:

∑(b+c)/ma ≤ ∑(b+c)/ha = (ab+bc+ac)/S

To prove  (ab+bc+ac)/S ≤ 2√3 R/r

         or   (ab+bc+ac) ≤ √3 R(a+b+c)

we use Euler inequality:  R ≥ √(a2+b2+c2)/3 or R ≥ (a+b+c)/3√3

As a result :  √3 R(a+b+c) ≥(a+b+c)2/3 ≥ ab+bc+ac


calculus
integration
trigonometry
limit
function
complex analysis
algebra

Compute the following integral

$ \int_{\minus{}\infty}^{\infty} \frac{x\minus{}1}{x^3\minus{}1}dx$

The book says this is equal to $ \frac{2\pi}{\sqrt{3}}$. However, I am not getting this answer. I keep getting different answers. How do I evaluate this integral?

$ \frac{x\minus{}1}{x^3\minus{}1} \equal{} \frac{x\minus{}1}{(x\minus{}1)(x^2\plus{}x\plus{}1)} \equal{} \frac{1}{x^2\plus{}x\plus{}1} \equal{} \frac{1}{(x\plus{}1/2)^2 \plus{} (\sqrt{3}/2)^2}$.

Therefore

$ \int \frac{x\minus{}1}{x^3\minus{}1} \, dx \equal{} \int \frac{dx}{(x\plus{}1/2)^2 \plus{} (\sqrt{3}/2)^2} \equal{} \left(\frac{2}{\sqrt{3}}\right)^2 \int \frac{dx}{((2x\plus{}1)/\sqrt{3})^2\plus{}1}$,

and with the substitution $ u \equal{} (2x\plus{}1)/\sqrt{3}$, $ du \equal{} 2/\sqrt{3} \, dx$, we obtain

$ \frac{2}{\sqrt{3}} \int \frac{du}{u^2\plus{}1} \equal{} \frac{2}{\sqrt{3}} \tan^{\minus{}1} u \equal{} \frac{2}{\sqrt{3}} \tan^{\minus{}1} \frac{2x\plus{}1}{\sqrt{3}}$.

Then the definite integral is evaluated by taking the appropriate limits, giving the result $ \frac{2\pi}{\sqrt{3}}$.

As for integrating this via complex analysis, I will leave that to someone else to work out.

The complex approach is pretty simple. As atomicwedgie did, simplify $ f(x)\equal{}\frac{x\minus{}1}{x^3\minus{}1}\equal{}\frac1{x^2\plus{}x\plus{}1}\equal{}\frac1{(x\minus{}\omega)(x\minus{}\overline{\omega})}$.

Now, consider $ I(R)\equal{}\int_C f(z)\,dz$, where $ C$ is the real axis from $ \minus{}R$ to $ R$, together with the counterclockwise semicircle of radius $ R$ in the upper half-plane. On the semicircle $ |z|\equal{}R$, $ |f(z)|\approx\frac1{R^2}$, so that portion of the integral is at most about $ \frac{\pi}{R}$ and goes to zero as $ R\to\infty$. Therefore, $ \lim_{R\to\infty} I(R)\equal{}\lim_{R\to\infty}\int_{\minus{}R}^R f(x)\,dx\equal{}\int_{\minus{}\infty}^{\infty}f(x)\,dx$.
Now, we calculate $ I(R)$. $ f$ has two poles, at $ \omega$ and $ \overline{\omega}$, one of which is in the upper half-plane. For large $ R$, $ C$ encloses the upper pole $ \omega$, and therefore $ I(R)$ is $ 2\pi i$ times the residue of $ f$ at $ \omega$. This residue is $ \frac1{\omega\minus{}\overline{\omega}}\equal{}\frac1{i\sqrt{3}}$ (use the factorization or partial fractions), so $ I(R)\equal{}\frac{2\pi}{\sqrt{3}}$ for large $ R$ and $ \int_{\minus{}\infty}^{\infty}f(x)\,dx\equal{}\frac{2\pi}{\sqrt{3}}$.

With experience, most of what I just wrote is unnecessary. This is the elaborated version, providing some explanations for the things I'd just write down barely thinking in my own calculations.

Observe that you can use jmerry's method for all cases where you have a rational function and the degree of the numerator is at least 2 less than that of the denominator

Techniques of Contour Integration for evaluation of Real Integrals are really awesome!! :lol:


find all $ m,n,l \in N$ such that :
$ m |nl\minus{}1$ , $ n | ml\minus{}1$ , $ l | mn\minus{}1$

$ (m,n)\equal{}(n,l)\equal{}(m,l)\equal{}1$

$ mn \plus{} ml \plus{} nl \minus{} 1 \equal{} mnlk$

$ \frac {1}{l} \plus{} \frac {1}{m} \plus{} \frac {1}{n} \minus{} \frac {1}{mnl} \equal{} k$

$ \implies$ $ m \equal{} min\{m,n,l\}$  $ \implies$  $ m\leq 3$  the rest is easy


Give $ a,b,c \ge 0$.Prove that:
$ \frac{a}{\sqrt{a\plus{}b}}\plus{} \frac{b}{\sqrt{b\plus{}c}}\plus{} \frac{c}{\sqrt{c\plus{}a}} \le \sqrt{(1\minus{} \frac{2abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)})(a\plus{}b\plus{}c)}$

For $ a\equal{}b\equal{}c$ it doesn't hold


geometry
3D geometry
algebra
factorization
sum of cubes
number theory proposed
number theory

Find the least positive integers $m,k$ such that

a) There exist $2m+1$ consecutive natural numbers whose sum of cubes is also a cube.

b) There exist $2k+1$ consecutive natural numbers whose sum of squares is also a square.

The author is Vasile Suceveanu


parameterization
calculus
integration
function
real analysis
probability and stats

If X and Y  are independent exponential random variables with parameter  p. Prove  that  X+Y  distributes as a  gamma  with parameters 2 and p.

 I used  the following transromations U= X+Y and V=X-Y but when  i  marginalize my  integral  results to  be undefined. Im  probably  making a  very  stupid error.

The density function for a sum of independent random variables is the convolution of their two densities.

The direct approach: $ \rho_X(x)\equal{}\rho_Y(x)\equal{}pe^{\minus{}px}$, so
$ \rho_{X\plus{}Y}(x)\equal{}\int_t \rho_X(t)\rho_Y(x\minus{}t)\,dt\equal{}\int_0^x pe^{\minus{}pt}\cdot pe^{\minus{}px\plus{}pt}\,dt$
$ \equal{}\int_0^x p^2 e^{\minus{}px}\,dt\equal{}p^2xe^{\minus{}px}$


$ \left( y\frac{dy}{dx}\plus{}2x \right)^{2} \equal{}\left(y^2\plus{}2x^2 \right) \left(1\plus{}\frac{dy}{dx} \right) ^2$

$ g(x)\equal{}y(x)^2\plus{}2x^2, g'(x)\equal{}2yy'\plus{}4x$

$ (\frac{1}{2}g'(x))^2\equal{}g(x)(1\plus{}y'(x))^2$

$ \frac{\frac{1}{2}g'(x)}{\sqrt{g(x)}}\equal{}1\plus{}y'(x)$

integrating both sides dx we get

$ \sqrt{g(x)}\equal{}x\plus{}y(x)$

so $ y(x)^2\plus{}2x^2\equal{}x^2\plus{}2xy(x)\plus{}y(x)^2 \to x^2\equal{}2xy(x) \to y(x)\equal{}\frac{1}{2}x$

yay!

hello, what is with
$ y(x)\equal{}i\sqrt{2}x$
$ y(x)\equal{}\minus{}i\sqrt{2}x$
Sonnhard.

The general solution is

$ \sqrt{y^{2}\plus{}2x^{2}}\pm \left( x\plus{}y\right) \equal{}C$


geometry
trigonometry
similar triangles
cyclic quadrilateral
trig identities
Law of Sines
Law of Cosines

Given Quadrilaterial $ABCD$, where
$AB = 1$
$BC = \sqrt{2}$
$CD = \sqrt{2}$
$DA = \sqrt{3}$ and
$BD = 2$.

If $AC$ and $BD$ intersect at $E$, find measure of $DEC$ in degrees.

[hide]
tan(bdc)=(2^.5)/(2^.5)
tan(bda)=(1)/(3^.5)
bdc is 45 degrees
bda is 30 degrees

Let bca be x
bac is 75-x
acd is 90-x and
cad is x+15
bec is 135-x
bea is 45 +x

(3^.5)/sinx=1/sin(x+15)
sinx=(3^.5)sin(x+15)

[/hide]

[hide="Hint for a simple solution"]
What is special about quadrilateral $ABCD$?
[/hide]

Well I know DBA and BCD are right angles, and for some reason I just thought 'cyclic quadrilateral', but not sure where to go with this.

the answer's $75$, need a picture to explain, but drop perpendiculars

here it is, too lazy to fill out numbers but you can figure it out

note it's composed of a $30-60-90$ and $45-45-90$ triangle and just play around with similar triangles and angles and stuffs.....

I got $90$, but I think I just found a mistake in my work.  Anyway, I did figure out how to continue Windslicer's line of thinking.  After finding that it is a cyclic quadrilateral, you know that the triangles $BAE$ and $CDE$ are similar, and that $BEC$ and $AED$ are similar...  I originally did that a much more convoluted way, but once you see it's cyclic, you get that really easily because of power of a point...

i'm pretty sure the answer's $\boxed{75}$, especially from my diagram

Yeah, that seems more likely.  I assumed something I shouldn't have when I was working it out before (that mistake I mentioned), but I think I see how to go with it...  I just need to mess with it a little more before I actually get it.

Well, once we see that $ABCD$ is cyclic...
[hide]$\angle DBC=\angle DAC=45^\circ$, so $\angle DEC=\angle ADB+\angle DAE= 30^\circ +45^\circ =\boxed{75^\circ}$ ;)
[/hide]

yeah, i first tried this problem with trig but it really was getting my nowhere so then i thought angle chasing would be much better, and with that, i got it in like a minute.

or perhaps my geometry trig is pretty bad :blush:

[quote="Adunakhor"]Could someone post a trig solution?[/quote]

um, why would you want to, it's much more difficult that way

I think it's nice to know how to approach and solve a particular problem in more than one way  :D

well, i saw gauss202's solution to this problem. A simple way to do it is to use law of sines actually. I'll get a diagram in a little bit...hold on.....

okay, well, i didn't get a good diagram. But I hope you have superhuman abilities and follow through my work :D

We have $\frac{\sin{\theta}}{1}=\frac{\sin{60}}{x}$ and $\frac{\sin{(180-\theta)}}{\sqrt{3}}=\frac{\sin{30}}{x}$ and since ${\sin{\theta}=\sin{(180-\theta})}$, this simply comes down to solving for $x$. Once finding $x$ we can use law of cosines etc. and eventually come up with what $\theta$ equals which will be $75$.


The combustion of 54.9 grams of a sample of hydrocarbon in excess oxygen produces 38.0 grams of water. What is the percentage (by mass) os hydrogen in the hydrocarbon sample?

A. 2.11%
B. 4.22%
C. 7.69%
D. 69.2%

[quote="aidan"]Nice solution.. I was wondering how to factor in the possibility that it might be an alkane or alkene. So 2n or 2n + 2? But using x and y gets rid of the problem.  :)[/quote]

well , hydrocarbon doesnt necessary means it have to be alkane or alkene since the definition for hydrocarbon is a compound which consist of carbon and hydrogen only :P

True, but compounds other than alkanes and alkenes are not expected to come out in the Malaysian Chemistry Quiz.  :D  Unless they decide to be nasty, that is. It can only be slightly out of the syllabus, I think.

after you find the percent of hidrogen in the compoud u can determine the compound's brute formula....
and then after obtaining (CH)n you can prove that in a hydrocarbure n must be even...
and then you can try to figure out values for whitch the compound is a valid one...

for example: n=2;C2H2 acetylene
                    n=4;C4H4 cyclobutadiine
                    n=6 C6H6  benzene
                    n=8 C8H8 cyclooctotetratriine
for n>8 the compounds are quite unstable due the high number of nesaturation and the high tension in the cycle....
and i'm not quite sure that C4H4 is a stable compound...

Sefu, good observation about n being even.  For reference:

Cyclobutadiene is not the only compound with formula C4H4.  Any 4 carbon hydrocarbon with a total of 3 pi bonds and rings will have this formula... I.E. buta-3-en-1-yne (which should be stable, I believe)...  FYI, cyclobutadiene is antiaromatic, has a strained ring, and is hence exceedingly unstable... I believe it has only been isolated at very low temperature in clathrates, or molecular cages.

About multiple isomers of the same formula, this goes for the rest of the formulae.  For example, benzene is not the only (stable) compound with formula C6H6.  Any other 6 carbon hydrocarbon with a total of 4 rings and double bonds will have the same formula.  This argument can be extended to higher values of n, and indeed I believe there are plenty of stable compounds $(CH)_{n}$ for all even n.

Nice observations from sefu and gregx007.. though this is a bit hard for me since I'm still in high school. We haven't covered that much yet.  :)


If $z$ is a complex number, will the infinite geomitric series hold?
Namely, $1+z+z^{2}+...=\frac{1}{1-z}$.

Yes, I believe, as long as the series converges.

\begin{eqnarray*}S&=&1+z+z^{2}+\cdots\\-zS&=&-z-z^{2}-z^{3}-\cdots\\ S-zS&=&1\\ S(1-z)&=&1\\ S&=&\frac{1}{1-z}\end{eqnarray*}

[quote="Zellex"]I believe you meant $+$ in the denominator.

FYI, your proof there, has an assumption. If the assumption is false, it doesn't work.

For example, it is well known that an infinite geometric series converges to a number if $|r|<1$ ; however, in your proof, if $r=2$, the proof would still "work".[/quote]

No, I definitely meant $-$.

Yes, my proof has an assumption. That's why I said "as long as the series converges".

Oh, yeah, I think you mean $-$.

But having the series converge is vague. I'm just emphasizing a point that $r<1$.

[quote="JBL"][quote="junggi"]it works as long as $|z|\le1$ i think[/quote]

No, it doesn't converge for any value of $z$ with $|z| = 1$.[/quote]
whoa did i really type that?  :blush:

$\sum a_{n}$ converges implies $a_{n}\to 0$ (but not the converse!).  If $|z| = 1$ then $|z^{n}| = 1$ and so $z^{n}\not \to 0$, so the sum doesn't converge.

$\sum_{j=0}^{\infty}a^{j}$ is really the number that the sum $\sum_{j=0}^{n}a^{j}$ converges to as $n \rightarrow \infty$

$\sum_{j=0}^{n}a^{j}=\frac{1-a^{n+1}}{1-a}=\frac{1}{1-a}+\frac{1}{a-1}(a^{n+1})$

If $|a|<1$, then as $n \rightarrow \infty$, the second term will approach $0$, which leaves $\frac{1}{1-a}$. However, if $|a| \geq 1$, then the second term will not approach $0$. You can't randomly cancel it out if it doesn't approach $0$.

i think we need to bust out the triangle inequality


What is the largest of four consecutive multiples of 6 if their mean is 27?

the mean would also be the median in this case. since theyre consecutive, they must be the 2 closest multples of 6 near 27 are 24 and 30. so these are the 2 middle ones, and the last 2 are 24-6=18 and 30+6=36, so the largest is 36

We could also use Algebra.

$ \frac{x \plus{} x \plus{} 6 \plus{} x \plus{} 12 \plus{} x \plus{} 18}{4} \equal{} 27$

$ 4x \plus{} 36 \equal{} 108$

$ x \equal{} 18$

$ 18 \plus{} 18 \equal{} \boxed{36}$


you probably already know this, but an operation is either + - * or /. The following problem has 4 operations:

[img]http://img294.imageshack.us/img294/4522/untitled16wp.png[/img]

you have to subtract pressure from 980 (1), then divide that by 60 (2) and then multiply that by 80 (3) and then add 75 (4). that's 4 operations. i need to simplify it down to only 2 operations. i can only get it to three, which is:

wind speed = (4145 - 4pressure) / 3

any ideas maybe???? i need wind speed on one side by itself....


geometry
geometric transformation
reflection
trigonometry
ratio
incenter
circumcircle

Can somebody (perhaps darij) explain a little about Isogonal conjugates and Lines and all that stuff? I ll be thankful! :D  :D

Okay, let's try a do-it-yourself course on isogonal conjugates :-)

I am also writing this for the future (I will refer to this post to cite theorems).

Usually, the notion of isogonal conjugates is defined by the following theorem ([1], p. 55):

[b]Theorem 1.[/b] If ABC is a triangle, and P is a point in its plane, then the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively, concur at one point.

This point is called the [b]isogonal conjugate[/b] of the point P with respect to triangle ABC.

Note that the isogonal conjugate of the point P with respect to triangle ABC is not defined if P is one of the three vertices of triangle ABC.

If it is clear which triangle is being considered, one can just say "isogonal conjugate of the point P" instead of "isogonal conjugate of the point P with respect to triangle ABC".

Theorem 1 is almost immediate if you know the Ceva theorem in the trigonometric form. If you know the Ceva theorem in the "usual" form only, you can help yourself with proving the following (easy) fact:

[b]Theorem 2.[/b] If D is a point on the sideline BC of triangle ABC, and if the reflection of the line AD in the angle bisector of the angle CAB intersects the line BC at a point E, then $ \frac{BD}{DC}\cdot \frac{BE}{EC}= \frac{BA^{2}}{AC^{2}}$.

This is proven very easy using the sine law or using areas. By the way, Theorem 2 is a generalization of the fact that an angle bisector in a triangle divides the opposite side in the ratio of the adjacent sides.

[b]EDIT:[/b] Theorems 1 and 2 appear (with proofs) in http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 post #1 as Theorems 2 and 1, respectively.

Note that in Theorem 1, I spoke about the "the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively". Under "angle bisectors", we usually understand internal angle bisectors. However, if you replace one or more of these internal angle bisectors by an external angle bisector, nothing changes - the reflections are the same. Prove this for yourself (it's very easy).

Now a merely trivial fact ([1], p. 56):

[b]Theorem 3.[/b] If the isogonal conjugate of a point P with respect to triangle ABC is called Q, then the isogonal conjugate of this point Q with respect to triangle ABC is the point P again.

Hence, we can call two such points P and Q [b]isogonal conjugate points[/b] with respect to triangle ABC. Note that Theorem 3 does not make sense if the point P is a vertex of triangle ABC or lies on a sideline of triangle ABC; in fact, if the point P is a vertex of triangle ABC, then its isogonal conjugate Q is not defined, and if the point P lies on a sideline of triangle ABC - let's say P lies on the sideline BC -, then the isogonal conjugate Q of the point P is the vertex A (why?), and then, the isogonal conjugate of Q is not defined. In all other cases, i. e. when the point P is not a vertex of triangle ABC and doesn't lie on any sideline, the point Q and its isogonal conjugate are defined, and Theorem 3 holds.

Another very easy result:

[b]Theorem 4.[/b] A point P is its own isogonal conjugate with respect to triangle ABC if and only if P is the incenter or one of the three excenters of triangle ABC.

In other words, the mapping "point $ \rightarrow$ its isogonal conjugate with respect to triangle ABC" has only four fixed points: the incenter and the excenters of triangle ABC.

Another fact relating the notion of isogonal conjugates to remarkable points of the triangle is the following:

[b]Theorem 5.[/b] The orthocenter H and the circumcenter O of triangle ABC are isogonal conjugate points with respect to triangle ABC.

This is very easy to show by angle chase.

Now, it's time to admit that in the above, I have been not too careful. (You know, that's tactics :D .) In fact, when I said in Theorem 1 that "the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively, concur at one point", the words "concur at one point" meant "concur in the projective sense", i. e. "concur at one point or are parallel to each other". Actually, there are some positions of the point P such that these reflections are parallel to each other. What positions exactly? The answer is given by the following result:

[b]Theorem 6.[/b] The reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively, are parallel to each other if and only if the point P lies on the circumcircle of triangle ABC.

This theorem ([1], p. 56) is proven easily by angle chase. Now, instead of saying that the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA, respectively are parallel to each other, we can say that they concur at an infinite point. This infinite point is then called the isogonal conjugate of the point P with respect to triangle ABC.

On the other hand, we can apply Theorem 1 if the point P itself is an infinite point. The lines AP, BP, CP are, then, three parallel lines through the points A, B, C, respectively. And it turns out that the isogonal conjugate of an infinite point P with respect to triangle ABC lies on the circumcircle of triangle ABC.

Altogether we can conclude:

[b]Theorem 7.[/b] The mapping "point $ \rightarrow$ its isogonal conjugate with respect to triangle ABC" is an involutive mapping of the projective plane (without the vertices and the sidelines of triangle ABC) to itself. This mapping transforms the circumcircle of triangle ABC into the line at infinity, and conversely.

Okay, so we know what happens with the circumcircle and with the line at infinity under our mapping. But what happens with a line?

[b]Theorem 8.[/b] If a line g doesn't pass through a vertex of triangle ABC, then its image under the mapping "point $ \rightarrow$ its isogonal conjugate with respect to triangle ABC" is a conic passing through the vertices A, B, C. This conic is a ${ \{\begin{array}[c]{c}\text{ellipse}\\\text{parabola}\\\text{hyperbola}\end{array}}$ if and only if the line g ${ \{\begin{array}[c]{c}\text{doesn't intersect}\\\text{touches}\\\text{intersects at two points}\end{array}}$ the circumcircle of triangle ABC.

This theorem is not so easy to prove any more. A way is using trilinear or barycentric coordinates. I wouldn't call it elementary geometry; the reason I mentioned it was just that you asked me about "Isogonal conjugates and [b]Lines[/b]". However, I am not sure whether you really wanted to know what is the image of a line under isogonal conjugation. Probably you met the term "isogonal line" in another context. In fact, instead of saying "reflection of the line AP in the angle bisector of the angle CAB", many people just say "isogonal line of the line AP with respect to the angle CAB" (or just "isogonal line of the line AP" if it is clear which angle we are speaking about).

Anyway, the above were the fundamentals of the isogonal conjugates story. Another thing is the relationship between isogonal conjugates and reflections. It begins with the following fact (which I used at the IMO 2004, in the problem 5, by the way):

[b]Theorem 9.[/b] Let M, U, V be three non-collinear points, and let P be another point in the plane. Let U' and V' be the reflections of the point P in the lines MU and MV, respectively. Then, the perpendicular bisector of the segment U'V' is the reflection of the line MP in the angle bisector of the angle UMV.

Again, this is very easy to prove. In fact, it is almost trivial that PU' = PV', and all the rest is angle chasing.

Now, let P be a point in the plane of a triangle ABC, and let X, Y, Z be the reflections of the point P in the sidelines BC, CA, AB. Then, after Theorem 9, the perpendicular bisectors of the segments YZ, ZX, XY are the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA. Since the perpendicular bisectors of the segments YZ, ZX, XY concur at the circumcenter of triangle XYZ, it follows that the reflections of the lines AP, BP, CP in the angle bisectors of the angles CAB, ABC, BCA concur at the circumcenter of triangle XYZ. This not only gives a new proof of Theorem 1; this also yields that the isogonal conjugate of P with respect to triangle ABC is the circumcenter of triangle XYZ. We state this:

[b]Theorem 10.[/b] If X, Y, Z are the reflections of a point P in the sidelines BC, CA, AB of triangle ABC, then the isogonal conjugate of P with respect to triangle ABC is the circumcenter of triangle XYZ.

Just for fun, look at what happens when the point P lies on the circumcircle of triangle ABC. As we know from Theorem 7, the isogonal conjugate of P is then an infinite point. In other words, the circumcenter of triangle XYZ is an infinite point. This can only mean that the triangle XYZ is degenerate, i. e. the points X, Y, Z are collinear! So you have obtained:

[b]Theorem 11.[/b] If a point P lies on the circumcircle of triangle ABC, then the reflections of the point P in the sidelines BC, CA, AB are collinear.

Of course, from this you can conclude that the orthogonal projections of the point P on the sidelines BC, CA, AB are collinear, too. Thus you have got a new proof of the Simson line theorem.

Another nice property of isogonal conjugates is related not to reflections in sidelines, but to orthogonal projections on sidelines.

[b]Theorem 12.[/b] Let P and Q be two isogonal conjugate points with respect to triangle ABC. Let $ A_{P}$, $ B_{P}$, $ C_{P}$ be the orthogonal projections of the point P on the sidelines BC, CA, AB of triangle ABC, and let $ A_{Q}$, $ B_{Q}$, $ C_{Q}$ be the orthogonal projections of the point Q on the sidelines BC, CA, AB. Then, the six points $ A_{P}$, $ B_{P}$, $ C_{P}$, $ A_{Q}$, $ B_{Q}$, $ C_{Q}$ lie on one circle, and the center of this circle is the midpoint of the segment PQ.

This one is not trivial but still quite easy to prove ([2], or [1], p. 67-69). Now, remembering Theorem 5, we can let P be the orthocenter H of triangle ABC, and Q be the circumcenter O of triangle ABC. Then, the points $ A_{P}$, $ B_{P}$, $ C_{P}$ are the feet of the altitudes of triangle ABC, and the points $ A_{Q}$, $ B_{Q}$, $ C_{Q}$ are the midpoints of the sides of triangle ABC. And thus we conclude that the feet of the altitudes and the midpoints of the sides of triangle ABC lie on one circle - the nine-point circle!

Two more nice facts about isogonal conjugates and orthogonal projections are the following:

[b]Theorem 13.[/b] With the notations of Theorem 12, the lines AQ, BQ, CQ are perpendicular to the lines $ B_{P}C_{P}$, $ C_{P}A_{P}$, $ A_{P}B_{P}$, respectively.

[b]Theorem 14.[/b] With the notations of Theorem 12, we have $ PA_{P}\cdot QA_{Q}= PB_{P}\cdot QB_{Q}= PC_{P}\cdot QC_{Q}$.

Theorem 13 can be proven by a quick angle chase ([1], p. 64-65), but it can be also seen as a very easy corollary of Theorem 9 (using that $ B_{P}C_{P}\parallel YZ$, $ C_{P}A_{P}\parallel ZX$ and $ A_{P}B_{P}\parallel XY$, where X, Y, Z are the reflections of the point P in the sidelines BC, CA, AB). Theorem 14 is an easy consequence of similar triangles.

Finally, note that the [b]symmedian point[/b] of a triangle ABC is defined as the isogonal conjugate of the centroid. In other words, the symmedian point is the point of intersection of the reflections of the medians of the triangle in the corresponding angle bisectors. But the properties of the symmedian point are an own story, and I think the above was already quite much. Anyway, I'm sorry that it got long and boring, but I hope you will be able to find out what is important!

[b]References[/b]

[b][1][/b] Ross Honsbeger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], Washington 1995. Particularly chapter 7.

[b][2][/b] Alexander Bogomolny, [i]Isogonal image of the circumcircle: What is it?[/i], http://www.cut-the-knot.org/Curriculum/Geometry/Circumsogonal.shtml .

[b][3][/b] Alexander Bogomolny, [i]Ceva's Theorem[/i], http://www.cut-the-knot.org/Generalization/ceva.shtml , Remark to Corollary 6. [Warning: This site has some Java applets. If your browser has problems with Java, turn it off or use "Save Target As" rather than clicking on the link.]

[b][4][/b] Pat Ballew, [i]Isogons and Isogonic Symmetry[/i], http://www.pballew.net/isogon.html .

[b][5][/b] Eric W. Weisstein, [i]Isogonal Conjugate[/i], http://mathworld.wolfram.com/IsogonalConjugate.html .

[b][6][/b] Paul Yiu, [i]Notes on Euclidean Geometry[/i], http://www.math.fau.edu/yiu/Geometry.html , section 8.6.1.

  Darij

Wow! thank you very much Darij! :D  :D  :D  it was not long or boring, it was great!

The most well known izogonal points in a triangle are :
 1. O and H      (ortocenter and circumcircle)
 2. G and L      (gravity center and Lemoine(semidian) point)
 3. $O_9$ and K    (Euler nine-point circle and Kosnita point)
 4. $\Omega_1$ and $\Omega_2$  (Brocard points)

Right. And here are two other pairs of isogonal conjugate points:

 5. Ge and U, where Ge is the Gergonne point and U is the internal center of similitude of the circumcircle and the incircle of triangle ABC;
 6. Na and V, where Na is the Nagel point and V is the external center of similitude of the circumcircle and the incircle of triangle ABC.

  Darij

I have written a note on isogonal conjugation ("Isogonal conjugation with respect to a triangle"). It can be found both here and on my website [url]https://www.cip.ifi.lmu.de/~grinberg/geometry2.html[/url]. It contains proofs of Theorems 1, 3, 4, 5, 6, 9, 10, 12, 13, 14 from my post above. Have fun. I included an introduction into the notion of antiparallels because it shortens a lot of proofs I am writing for ML, and a fact (Theorem 19) which is actually another problem from ML.

Note that Isogonal.pdf is the actual note, while missing.pdf is an extra file containing some images which are incorrectly displayed in Isogonal.pdf (for some reason, some segments disappeared when the file was compiled into PDF).

  Darij

I have just scanned in some very basic hand-written notes on isogonal conjugates. Maybe some people might find them useful.


there is a sequence of numbers what is next three

7100,23,11,6,3,...

our math teacher said that if you are a 5th grader with good English you can do it.
however if you are a professor with bad eng you cannot.
as you see i'm turkish and cannot do it..
would you say it to me....

Is this supposed to be a trick problem?

Because the sequence is not arithmetic or geometric. In fact, it's not a variation either. Perhaps a polynomial equation...

I'm guessing it has something to do with the letters of the alphabet.  After all, he said only people good at English could solve it.

This is basically the variation that every number, when using this trick, will end up at four. (My friend calls it the "four is a cosmic number trick"):

Here are some examples:

$7100 = 23 \implies 23 = 11 \implies 11 = 6 \implies 6 = 3 \implies 3 = 5 \implies 5 = 4$
$18 = 8 \implies 8 = 5 \implies 5 = 4$
$1024 = 21 \implies 21 = 9 \implies 9 = 4$

Basically, using 7100 again, you can use the trick by saying:

7100 is 23, 23 is 11, 11 is 6, 6 is 3, 3 is 5, 5 is 4, and 4 is a cosmic number

Kinda corny, but fun :D

[hide="The answer"]Oh

I got it now, I knew it had to do with the letters, just kept on thinking syllables

But then I realized it was letters

Seven thousand one hundred
= 23 letters

twenty three
= 11 letters

eleven
= 6 letters

six
= 3 letters

Then it is 5, 4, 4

[/hide]

Great outside the box type thinking question

I'ma pose it to my math class tomorrow.


When a car's brakes are applied, it travels 7 feet less in each
second than the previous second until it comes to a complete stop.  A car goes 28 feet in the first second after the brakes are
applied.  How many feet does the car travel from the time the
brakes are applied to the time the car stops?

Continue the arithmetic sequence to $ 0$, and then add.

$ 28\plus{}21\plus{}14\plus{}7\equal{}7(4\plus{}3\plus{}2\plus{}1)\equal{}7(10)\equal{}\fbox{70}$.


maybe the solutions are complex (meaning a/b is complex)
lets see here
$ a^6\plus{}3a^4b^2\plus{}3a^2b^4\plus{}b^6\equal{}a^6\plus{}2a^3b^3\plus{}b^6 \implies 3a^4b^2\plus{}3a^2b^4\equal{}2a^3b^3 \implies 3b^2\plus{}3a^2\equal{}2ab \implies 3b/a\plus{}3a/b\equal{}2 \implies A\equal{}2/3$
you are correct
the fact that A is supposed to be $ \ge$ 2 for real numbers a and b shows you that the only solutions for a and b include complex numbers

Wait why does a/b or whatever have to be $ \geq{2}$

suppose a/b is real for some reason (which it is not!)
call this x
we have $ x\plus{}\frac{1}{x}\equal{}k \implies x^2\minus{}kx\plus{}1\equal{}0$
in order for there to be real solutions for x, the discriminant must be nonnegative
so $ k^2\minus{}4 \ge 0 \implies k \le \minus{}2 \text{or} k \ge 2$
but since $ x\plus{}\frac{1}{x}\equal{}\frac23$, then x must be complex

[quote="Nguyen Ngoc Linh"]I am confuse. please help.
Given: (a^2 + b^2)^3 = (a^3 + b^3)^2
Compute A=a/b + b/a.
I solved it and found out that A=2/3. but a/b + b/a >=2. Why?[/quote]

I think there isn't a number system $ (a,b)$ such that (a^2 + b^2)^3 = (a^3 + b^3)^2 (unless (0;0)), that's why  A=2/3. but a/b + b/a < 2.

Note: a/b + b/a >= 2 when a, b > 0.

Yeah, Euler. you're right. The two equations are equivalent. SO, if a,b are real numbers, the first equation is not true. That all I want you to tell me. Thank you very much Mathlinkers.


limit
integration
topology
function
trigonometry
real analysis
real analysis unsolved

If $ f \in L^1 (\mathbb{R})$ then there exists a measurable set $ A \subset \mathbb{R}$ with finite measure such that $ f$ vanishes at infinity outside this set, that is $ \lim_{|x| \to\infty,x\notin A}f\left(x\right) \equal{} 0$.

We can assume $ f\geq 0$. Let $ E_t \equal{} \{x \in R : |x| > t\}$. Choose $ t_1, t_2, ...$ such that  $ \int_{E_{t_k}} f dm < 1/2^k$. Set $ A_k \equal{} \{x\in E_{t_k} : f(x) > 1/k\}$. Then $ (1/k)m(A_k) < \int_{E_{t_k}} f dm$, so $ m(A_k) < k/2^k$. Now set $ A \equal{} \cup A_k$. Then $ m(A) < \infty$, and $ f < 1/k$ on $ E_{t_k} \minus{} A$, showing $ f$ vanishes at $ \infty$ off of $ A$.

Nice. I previously tried proving other statements of the form "If $ f$ is integrable then it "nearly" vanishes at infinity" with "nearly" being given other meanings, but apparently "outside a finite-measure set" is the only one that works. An easy exercise is to disprove the statement in the cases where "nearly" means:

1) Outside a countable set.
2) Outside a set of measure zero.
3) Outside a set of the first category (a countable union of nowhere-dense sets).

(One can construct a counterexample which nails all three at once.)

However, I'm curious to see if the same result holds in a more general setting. Suppose $ X$ is some nice topological space (say, locally compact and Hausdorff), equipped with its Borel algebra and some Radon measure. Do integrable functions on $ X$ vanish at infinity outside a finite-measure set? Recall that a real (or complex) valued function on a locally compact space vanishes at infinity if $ \forall \varepsilon > 0$ there exists a compact set $ K$ such that $ |f(x)| < \varepsilon$ for $ x \notin K$. [b]WWW[/b]'s proof works just as well if the space is $ \sigma$-compact and I suspect that $ \sigma$-compactness is necessary, to some degree.

Except for a set of measure $ 0$, the set where  $ f > 0$ is $ \sigma$ compact in this setting, so everything should work. Also the conclusion can be strengthened to "$ f \rightarrow 0$ at $ \infty$ off of sets of arbitrarily small positive measure".

I wonder if the result is true for improperly Riemann integrable functions as well, such as $ x \mapsto \frac{\sin (x)}{x}$ (though for this particular function the result obviously holds).

[quote="CommandoGuard"]I wonder if the result is true for improperly Riemann integrable functions as well, such as $ x \mapsto \frac {\sin (x)}{x}$ (though for this particular function the result obviously holds).[/quote]

No, $ f(x) \equal{} sin(e^x)\sqrt {x}$ is a counterexample on $ [0, \infty)$. If we allow noncontinuous integrands, we can even get  $ |f| \rightarrow \infty$ at $ \infty$: Let $ S_0 \equal{} 0$, $ S_n \equal{} \sum_{k \equal{} 1}^{n} \frac {1}{k}$. For $ x \in [S_{n \minus{} 1}, S_n)$, define $ f(x) \equal{} ( \minus{} 1)^n\sqrt {n}$.


logarithms
limit
inequalities
calculus
integration
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Determine whether the series converges or diverges.

$ \sum_{n\equal{}2}^{\infty}\frac{1}{(\ln n)^{\ln (\ln n)}}\equal{}\sum_{n\equal{}2}^{\infty}\frac{1}{e^{\ln (\ln n)\ln (\ln n)}}$

[b]3. The attempt at a solution[/b]

for $ u\equal{}\ln (\ln n)$

$ \sum_{u\equal{}\ln (\ln 3)}^{\infty}\frac{1}{e^{u^2}}$

from Root Test

$ \lim_{u\rightarrow\infty}\sqrt[u]{\frac{1}{e^{u^2}}}\equal{}\lim_{u\rightarrow\infty}\frac{1}{e^u}\equal{}0<1$
series converges

Your reasoning is wrong.

[hide="Look here"]As you have stated, $ (\ln n)^{\ln(\ln n)} \equal{} e^{[\ln(\ln n)]^2}$. Now, we know that $ \lim_{n \to \infty} \frac{n^2}{e^n} \equal{} 0$, and so for n big enough we have $ n^2 < e^n$. Now substitute on this inequality n by $ \ln(\ln n)$, and so

$ [\ln(\ln n)]^2 < e^{\ln(\ln n)} \equal{} \ln n$

$ \equal{}> [\ln(\ln n)]^2 < \ln n$

$ \equal{}> e^{[\ln(\ln n)]^2} < e^{\ln n}$

$ \equal{}> e^{[\ln(\ln n)]^2} < n$

$ \equal{}> \frac{1}{e^{[\ln(\ln n)]^2}} > \frac{1}{n}$.

Therefore, as the harmonic series diverges then our series, which is bonded bellow by the harmonic series, also diverges.[/hide]

Specifically, the flaw in your solution is that your substitution doesn't make any sense for a summation: increasing $ u$ by 1 has nothing to do with increasing $ n$ by 1.  Thus, the root test (which applies when we have a sequence with the index incrementing by 1, or more generally some constant) applied to a sequence indexed by $ u$ says nothing about our original sequence indexed by $ n$.  If this were an integral, what you did is equivalent to making a $ u$-substitution and forgetting that you also need to change the $ dx$.


We've had a discussion with our computer science teacher about these terms - he was the one asking the question - and we've decided that it's impossible to define either of them. Information and / or data could be almost anything, I think the best point of view is looking at them as undefined elementary terms

[quote="spx2"]all information is a bunch o bytes.
it is all the same.
depends on who is the reader tough...
the readers interpretation[/quote]

Information is defined as:

$I = -\lg{P}$.

[quote="whateverthanks"]
im just curios, i defined for myself what information is or data and i asked my computer teacher if he could define those terms and the definitions utterly escaped him, i was amazed because otherwise he was filled with all kinds of info and data about the computer and its related products, services and industries...can anyone here provide a credible definition of what information or data is..........
[/quote]

Well, there is certainly some differences between information and data. What I learned in the CS classes in my school is like [b]data are just unorganised identity, when you process it and make it useful in a way, then you can call it information[/b]

Like this is data -->

[quote]fahim, john, reid, 899, 777, mini, 999, 1000[/quote]

and this is information -->
[quote]
john 777
fahim 899
mini 999
reid 1000[/quote]

So there is some difference in meaning of data and information in term of CS. May be the term [b]informatics[/b] came from this through.

hi,

Information is a general term : it doesn't have any specific

definition for informatics .

[color=blue][size=200]Datas[/color][/size] are memory blocs where some informations that will be used by [color=red][size=200]instructions [/color][/size]are saved .

[color=blue][size=200]Instructions[/color][/size] are the operations which programs try to do.

 :lol: .

Hi toetoe,
May be you didnt notice i wrote [b]"MAY BE"[/b] before informatics.  :(

$data ----processing---\to information$

And think a while, isnt your definition for data is same (in a sense) as my definition ?? I took mine just from textbook, its even quite a regular question in the higher secondary level CS exams in my country! :D

May be you are right from your sense, but i just told what i know as a fact... i dont wish to argue dear!

i dont know why you make so much fuss about this...
i dont think this is either interesting or important as a question...
but then again it may be only my oppinion


In page 42 of the introduction to counting and probability, "objects" is spelled "objectts"
(problem 6.10)
But AOPS is still awesome!!

Also on page 42, problem 2.15, You first say that Brenda is a man then back to woman.

mihail says that there's a typo in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163088[/url]

i've discovered several typos in the aops vol 1&2. other than that, the books are flawless....   :D


Let $ (O_1)$ and $ (O_2)$ be two intersecting circles and $ A,$ $ B$ are denoted by their intersections. An arbitrary line passes through $ A$ intersects $ (O_1)$, $ (O_2)$ again at $ C$ and $ D$ respectively. Let $ M$ be a point on $ CD$. From $ M$, lines that respectively parallel with $ BC,$ $ BD$ intersect $ BD,$ $ BC$ at $ X,$ $ Y$. The lines through $ X$ and $ Y$ that perpendicular to $ BD,$ $ BC$ intersects $ (O_2),$ $ (O_1)$ at $ F$ and $ E$ respectively $  ($ $ F$ is on the different side of $ A$ wrt $ BD$ of $ (O_2)$, so as $ E$ $  )$. Prove that $ \angle EAF = 90^{\circ}$.

It was appeared in IMO shortlist 2002 :
http://www.mathlinks.ro/viewtopic.php?p=118681#118681


Can someone please help me find the pattern to this sequence of numbers?  I wrote it three years ago and now I can't solve my own problem :(.

0,0,0,0,1,2,3,4,6,8,0,2,4,6,9

[hide="pattern"]For 0, you add 0 three times, then you add 1 four times, then you add 2 5 times. You will add 3 6 times. So the next numbers in the pattern will be 12,15,18,21,24.[/hide]

but he's just taking the last digits, right?

Edit: Oh, and he adds 2 six times, so thats probably not the pattern.


I'm sitting with a simple problem, but our lecturer have confused me with the way he's writing the question and the assignment is due tomorrow.

I've found a tangent plane, and i'm now supposed use this one to do an approximation. However, there are three values given, therefore I don't know how to use them?! the point values are (2,2pi), but for the approximation i have (1.95) (15pi/8)^2 cos(15pi/8).

The 1.95 is for the x value (since the point value for x is 2), but I don't know if the full (15pi/8)^2 cos(15pi/8) is for the y value or if shall be splitted and the last part used as a z value!? The values are written in the same way as I did above, so no commas, paranthesis etc can hint about how to split them or wht goes where.

Sorry guys if this is a dummy question, I just don't know how to solve this before tomorrow morning...

/J

The right person to be asking is your professor ;)  Try to make a reasonable assumption about what it means, and explain what assumption you made.  (Note that probably the target point and the nearby approximate point should be close together; if your assumption doesn't have this property, try alternate interpretations.)

Or, at least, post the [b]full[/b] text of the problem exactly as you have it in your assignment (if you haven't been able to figure out what it means looking at the full thing, it is quite silly to assume that we'll be able to figure it out looking at the tiny portion of the text you provided ;)).


In a solution I read, one of the steps relied on

[tex]\displaystyle \sum_{n=1}^{\infty} \left( \sum_{j=1}^{n} \left(-\frac{1}{3}\right)^n \right) = \sum_{j=1}^{\infty} \left( \sum_{n=j}^{\infty} \left(-\frac{1}{3}\right)^n \right).[/tex]

What is the justification for the change of indices? Is there some theorem/property that I should know about this?

both cases are in effect

[tex]\displaystyle\sum_{n \ge j}\left(-\frac{1}{3}\right)^n[/tex]
so both are equal.

It's always good to break things into ordered pairs and count them two different ways, then equate your expressions (<i>particularly</i> effective in combinatorics)