I have quiet a few of problems of cryptarithm. So I decided to start a marathon on it.
Let's start with an easy, well known one.

[b]Problem 1[/b]
[color=yellow]_[/color]SEND
+MORE
______
MONEY

This is made by Henry Ernest Dudeney(famous puzzlist).

Good. Now for the next problem.

[b]Problem 2[/b]

[color=yellow]_[/color]EINS
[color=yellow]_[/color]EINS
[color=yellow]_[/color]EINS
+EINS
_____
[color=yellow]_[/color]VIER

Eins is one, vier is four in German. This problem is made by Alan Wayne.

[b]Problem 3 - someone put up 2 while i was writing this[/b]

SPIDER + SPINS + LINE + REELS + IN = INSECT, when I=5.

my friend Kevin asked me this, but I dont know where he got it from.

[hide="answer for the second problem"]1329 + 1329 + 1329 + 1329 = 5316[/hide]

[hide="Solution for problem 3"]405967+40524+1526+76624+52=524683[/hide]

[b]Problem 4[/b]
SPAM+SPAM+SPAM+SPAM+SPAM+SPAM=EMAIL

[b]Problem 5[/b]

Do you know the 'Four Squares Theorem'? It says that any non-negative integer can be expressed by sum of four square numbers.
SQUARE+SQUARE+SQUARE+SQUARE=NUMBER

[quote="Sunny"]Think this works

[hide="Problem 5"] 149082+149082+149082+149082=596328 [/hide][/quote]
It's correct.

[b]Problem 6[/b]
Erdos said that "A Mathematician is a machine for turning coffee into theorems".
COFFEE+COFFEE+COFFEE=THEOREM

831199+831199+831199=2493597

[color=white]_[/color]one
[u]+one[/u]
[color=white]_[/color]two

none of the letters are 1 or 2

[hide="7"]467+467=934[/hide]

Edit: I thought it was problem 6, but it was 7 i just realized.

[b]Problem 8[/b]
$\frac{EVE}{DID}=.TALKTALKTALK\cdots$
There are two solutions.

[b]Problem 9[/b]
FORTY
[color=yellow]__[/color]TEN
+[color=yellow]_[/color]TEN
______
SIXTY

I found out the source for the last one(#8). It was appeared in Mathematical Problems by Martin Gardner.

[quote="Shamik Banerjee"][code] 오천백삼
 천구백이
 천이백이
+  구백이
________
 구천백구

I don't understand Korean language :(  What follows is from base 10 number system :D

오 = 5; 천 = 1; 백 = 0; 삼 = 3; 구 = 9; and 이 = 2.
 5103
 1902
 1202
+ 902
-----
 9109

_________________

[/code]

Please, let me know the Korean surprise :?[/quote]

[code]
 오천백삼 means 5103
 천구백이 means 1902
 천이백이 means 1202
+  구백이 means 902
________
 구천백구  means 9109[/code]
Which happen to be same as the solution.

______________________________
EDITED: A typo found :oops:

[quote="lightrhee"]Problem 14

SIX+SIX+SIX=NINE+NINE[/quote]
I get
[hide="Problem 14"]
S=9
I=4
x=2
N=1
E=3
$942+942+942=2826=1413+1413$[/hide]

Here's a new problem.  Give 1 solution, I made this up so there may be more than one; I've only found 1 though.

[size=200]Problem 15[/size]

MORE + GOOD=GREAT

There is no solution. G=1, obviously, and M=(8,9). For either situation, R cannot be higher than 1, and it cannot be 1, so it must be 0. If M=9, then O+O<10, so O=(2,3,4). R=0, so E+D>10 to make R+O=A without O=A. For O=2, we have E=4, A=3, and 4+D=T, so D=(6,7,8), which means T=(0,1,2), all of which are impossible since both have already been used. For O=3, E=6 and A=4, and 6+D=T, so D=(5,7,8), which means T=(1,3,4), all of which have already been used. For O=4, E=8 and A=5, and 8+D=T, so D=(2,3,6,7), which means T=(0,1,4,5), all of which have been used.

If M=8, then O+O>10, so O=(6,7,9). For O=6, we have E=2, A=7, and D=(8,9), which gives T=(0,1), both impossible. For O=7, we have E=4, A=8, which is already impossible. For O=9, we have E=8, which is already impossible. Therefore, there are no answers.

Oops! My solution doesn't work after all  :(  :blush: I had M=A

Oh well.  Try to find a solution for it if M=A.

Well, if M=A, then I'll assume it was where M=8 and A=8, so we have M=8, G=1, R=0, O=7, A=8, E=4, D=9, T=3:

[code] 8704
+1779
_____
10483[/code]

Next problem:

[code] MERRY
 +XMAS
______
TURKEY[/code]

I haven't been able to find a way to do this, although my friend says it's possible. Maybe some of the letters have to be equal to each other, but he said that wasn't the case.

My first post there :lol:.
[quote]
  9ERRY
 +X9A0
______
10RREY[/quote]

I don't know what i schould do next.
I try for k=r.

I don't see how it can be true without letters equal to each other... I always get U = S = 0.

Yeah... if we go with that, is it possible?

[quote="Shamik Banerjee"][code] MERRY
 +XMAS
______
TURKEY[/code]

has got NO solutions in base 10 if the different alphabets represent different digits.[/quote]
Should we move on then?

[size=150]problem 16[/size]
FIVE+SEVEN+ELEVEN+TWELVE+FIFTEEN+TWENTY=SEVENTY

"Your right, Merry + Xmas = Turkey with base 10 is impossible, that's because the actual problem is

A + Merry + Xmas = Turkey

Have fun

P.S. I haven't yet solved it. :mad:

[quote]A + Merry + Xmas = Turkey [/quote]

[hide]97445+6928+2=104375[/hide]

So much guess and test for me  :(

New Problem:
[code]
       ESTMODUS
X       INREBUS
-----------------
      *********
     *********
     ********
   *********
  *********
 *********
*********
-----------------
********INREBUS
[/code]

Each of the ten digits corresponds to at least one letter, and there is precisely one case where two different letters are replaced by the same digit.

Found this
[img]http://www.tenproblems.com/wp-content/uploads/2006/10/inrebus.gif[/img]


Let  K subset of a Metric space (M,d)
Prove if K + {x}  Compact ==>K Compact

Sorry , It is a problem that my prof ask me to prove I assume it is true

If your $+$ means union then it is not true. Note that $+$ cannot be any kind of operation since you are in a general metric space. Such as, the problem statement is clearly incorrect.


Prove that :
(x²+y²+z²-yz-zx-xy)²=(x-y)²(x-z)²+(y-z)²(y-x)²+(z-x)²(z-y)²
with the easiest way possible.

:huh: what did you do to get this? factorize [(x-y)²+(y-z)²+(z-x)²]/2 ?


The digits 2, 3, 4, 7 and 8 will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by 11? Express your answer as a common fraction.

When a $ 5$-digit number $ \overline{abcde}$ (is that how you express it?) is divisible by $ 11$, $ a \plus{} c \plus{} e \equal{} b \plus{} d$.  This only happens when you have $ 2\plus{}3\plus{}7\equal{}4\plus{}8$.

Thus, you have $ 6$ ways to arrange $ 2$, $ 3$, and $ 7$ in the spots $ a$, $ c$, and $ e$.  You have $ 2$ ways to arrange $ 4$ and $ 8$ in the spots $ b$ and $ d$.  Thus, there are $ 12$ possible numbers.

There are $ 120$ integers that can be made without any restrictions, so our answer is $ \frac{12}{120}$, or $ \boxed{\frac1{10}}$.

Actually, $ a\plus{}c\plus{}e\minus{}(b\plus{}d)$ must be divisible by $ 11$, it doesn't necessarily have to be equal.  However, with these numbers, the only way to make it work is with a+c+e=b+d

The digits $2, 3, 4, 7$ and $8$ will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by $11$? Express your answer as a $common fraction.$
Uh, can someone like explain this in a bit more easier/simpler way please

The divisibility rule for $11$ is as follows: Take the sum of the first digit and then every other digit that follows and subtract the sum of the second digit and every other digit that follows. If the result is divisible by $11$ (which generally is very easy to tell once the result is in), the original number is divisible by $11$. In numbers, if we have a five digit number $abcde$ then we do $(a + c + e) - (b + d) = \alpha$. If $\alpha \mod 11 = 0$ then $abcde$ is divisible by $11$.


geometry
3D geometry
Pythagorean Theorem

A gecko is in a room that is 12 feet long, 10 feet wide and 8 feet tall.  The  gecko is currently on a side wall (10' x 8'), one foot from the ceiling and one foot from the back wall (12' x 8'). The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall.  What is the length of the shortest path the gecko can take to reach the fly assuming that it does not jump and can only walk across walls and the ceiling? express your answer in simplist radical form.

[hide]If you draw out the room and the locations of the gecko/fly, then open up the room as if it were a cardboard box, you can see that the two points are at the tips of a right triangle with leg lengths of 8 and 22. Using the Pythagorean Theorem, the distance is found to be $2\sqrt{137}$ feet.

EDIT: AHH nvm.. it's a right triangle of leg lengths 6 and 21.. distance is $3\sqrt{53}$ feet.[/hide]

neither of those are correct i believe
jorian

[hide]heh.. does $2\sqrt{130}$ feet sound okay?[/hide]

The links have solutions...
The correct answer is
[hide]$2\sqrt{113}$[/hide]

[hide]
ahh 14 and 16..

and sorry i didn't hide my answers before; i'll edit them out..
[/hide]

lemme check if that works
[hide]
14^2+16^2=452
sqrt452=2sqrt113

omg, wow, it works

[/hide]

jorian

[quote="philB"]ummm.... how are you guys getting 14 and 16?.... I got 6 and 20...[/quote]
That way works, but it isn't the shortest.
You aren't unfolding the cube right.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=233255&highlight=gecko#233255

I'm rather confused about which way is front, back, etc. I think I probably got them confused or something, because I got 2 sqrt(106). Would someone be kind enough to post a diagram of the  setup (not a net)?

If you click on the link in the post right above yours it will take you to a very detailed diagram better than any of us could probably make!!  Thanks to RCV for that picture 2 years ago!!

I guess Im not very good at visual thinking, because I couldn't really picture the box very well even with the diagram on the link.  :blush: Does anyone have a diagram of it before you unfold it into the net?

Just draw a box...and then label according to the directions


LaTeX probability email MATHCOUNTS \/closed

thats a nice idea.  It would be better if it was on the server though.

I would be nice! Maybe you can put a section with a kind of link where you can go and chat there.... but, what about the amount of people chating???

To chat with the people here ;)
There are some possibilities, like discussing math or just chat with other MathLinkers.
I don't think you can discuss here like you can in a chat.

Do you want LaTeX on it? Otherwise why wouldn't you just chat in AIM or MSN or sth like that?

Well, I personally can't (and wouldn't anyway) use AIM or MSN, so it might be nice.  The other thing is to get gmail, and you can chat with everyone else that has gmail inside your email.

At the moment, we (something between seven end eleven chatters till now) invite those that one of us has added in the Yahoo messenger.
This is definitely not the best system (and a bit hard to join when you don't know one of the already joined ones ;) ) and will not work for more than 23 people.
And $\LaTeX$ can just be given as the plain code, I think everyone can tex it for himself then.

The biggest problem really is to have some chatroom(s) that are free to join, without invite etc. and without client when possible.
The best in my oppinion is IRC, and not only one, but some channels (I can think of that mathematical discussions and simple chatting don't fit into one single room, so there should, at least when people get more, be two or three rooms).

[quote="ZetaX"]We (darij, mathmanman, me) think that MathLinks need some type of chat.
There are many possibilities, e.g.:
- here on the website, with own server etc.
- on some other server; e.g.: euirc ( http://html.chat.euirc.net/ , channel #MathLinks), where you don't need any registration nor a client
- on some messenger (but I think that ill not work for a lot of people)
What do you think about it¿[/quote]

There is no reason to have any chat here. No reason.

For personal information, use PM. Otherwise there is no point in having a chat here.

[quote="conceive"][quote="ZetaX"]Multi-person-PMs¿ And sending 5 per minute¿
A chat is totally different from talking via PMs and on the board.

But explain why there is no reason please...[/quote]

Why is there ML? To discuss math problems. To discuss them together in the threads. It is sufficient. We can post in every topics every second, we can quote every post of any other poster. There is not any reason why to have some real-time threads... like chat.

That's why.

conceive[/quote]
I'm sorry, but to do things like AoPS classes, real time communication is needed.

I second IRC. How does #aops in efnet sound?

[quote="PenguinIntegral"]
I'm sorry, but to do things like AoPS classes, real time communication is needed.[/quote]

Not neccessarily. You can post here in threads, this is sufficient.

You can post in every time, you can qoute every post of every topics (expect the locked ones). This is sufficient.

conceive

You would never discuss here like you do in a chat.
You can't mix up different things, can't chat about something else than what the topic is about (oh lets start a topic "How's the weather¿"), can't discuss in real-life-speed (thats about 100 times faster than here, or do you just sit there and hit the actualize-button¿), can't really discuss similar things, can't do stuff that is considered spam or SPAM here (and that's correct so, forums are not there for chat-like discussion), can't speak the same way (believe it or not), can't solve problems together (this is possible on the forum, but noone does it really), can't just find/meet friends there.

PS: we don't speak of "sufficient" conditions; with some time, everything is also possible by writting letters per hand, so whats the point on using this site¿

This may be easier for the unofficial Mathcounts Math Jam chats over AIM.

1. every one would need yahoo messenger.
2. to chat in YIM, you need invitations I think.
3. it's simply better to use protocols like IRC that were made for multi user chatting as they offer more possibilities.

[quote="ZetaX"]For those new to IRC there is some html-client:

- go to http://webchat.euirc.net/webchat/en/
- enter some nickname, leave the password and Username unchanged
- choose some random channel at the bottom, for example "#euirc - General information channel"
- click on "go"
- in the empty line you see type:
"[b]/join #mathlinks[/b]"
and press enter [and don't forget the single space between "\join" and "#mathlinks" ;) ].
- voila, there you are

Similarily, you could go to http://www.euirc.net/en/javachat.php and procede by more or less the same procedure.[/quote]Both not working here...

Then you will have to use some other client like mIRC, Miranda, ...

Datas:
- server: irc.euirc.net::6665-6669
- channel: #mathlinks

Surely I will not.
Either IRC, Jabber, or some "local" chat. Everything else is evil ;)

[quote="chess64"]let's revive this[/quote]

Hmm didn't you make that chatroom (the one that went dead)? How's it going?

MathLinks chats always are near to death as the normal user number is about 2.
(And what chatroom by Chess64 ¿)

I have just joined that #mathlinks channel. It is really dead :( Only 1 member and there is a silence mode enabled. Though in my POV it was a good idea to create such a channel.

Everytime someone opens a channel, it lasts about one week until most people don't visit it anymore and it just dies...

I've got the following error:
[18:26] *** Error: Closing Link: Myth[94.24.174.174] (You are not authorized to connect to this server)

 :huh:

euIRC has a couple of servers, I recommend to try another one. See http://www.euirc.net/en/server_admins.php .

Well, we may also change network, I just chose euIRC because I knew it had some working own webclient.


Laws in certain states require a $ 60\%$ approval rate from voters before issues are ``passed.'' If 1,140 people voted in favor of the law, and this represents $ 57\%$ of the total number of voters, how many voters were there in this election.

The first sentence in this question is not needed to solve this problem.

$ 57\%$ is $ 0.57$ in decimal form, so simply using the equation $ 0.57x \equal{} 1140$ we can solve for the answer.

$ 0.57x \equal{} 1140$

$ \boxed{x \equal{} 2000}$


Let $n$ be a positive integer. Show that there exist three distinct integers between $n^2$ and $n^2+n+3\sqrt{n}$, such that one of them divides the product of the other two.

First we will show that there's some integer $a$ satisfying $a(a+1) \leq n$ and $n-3\sqrt{n}+1 \leq a^2$.

Let $n=k^2+r$ with $0 \leq r \leq 2k$. If $r \geq k$, take $a=k$. Then $a(a+1)=k(k+1) \leq k^2+r$, and $k^2+r-3\sqrt{k^2+r}+1 \leq k^2+r-3k+1 \leq k^2-k+1 \leq k^2=a^2$. Otherwise, if $r<k$, take $a=k-1$. Then $a(a+1)=(k-1)k \leq k^2+r$, and $k^2+r-3\sqrt{k^2+r}+1 \leq k^2+k-3k=k^2-2k<(k-1)^2=a^2$.

Our claim is proven. Now, take $X=(n-a)(n+a+1)=n^2+n-a(a+1)$. Since $a(a+1) \leq n$, this is at least $n^2$. Now let $Y=X+n-a, Z=X+n+a+1$. Then obviously $X \mid YZ$. Besides, $Y \leq Z=n^2+2n-a^2+1 \leq n^2+n+3\sqrt{n}$. We're done.


geometry
circumcircle
trigonometry
trapezoid
geometry proposed

An isosceles  triangle $ABC (AB= AC)$   is inscribed in a circle .The tangents to the circle at points $A$  and $C$  intersect each other at point $D$. If  $\angle DBC = 30$ , prove that $\triangle ABC$  is equilateral .

  [u] Babis[/u]

 [i](I have done a solution to this problem , but I'm afraid my solution is a little long.Probably you will find an easier one.If nessesary , I'll post  my solution after your attemts)[/i]

hope i haven't done a mistake...

let $BD$ intersect the circumcircle at $E$. then the given condition implies that
 $\angle EAC=\angle ECD=30$.

let $\angle ABE=\alpha$, then
$\angle ACE=\angle EAD=\alpha$.

also, from $AB=AC$ it follows that $BC||AD$ from where we conclude that $\angle ADE=\angle ADB=\angle DBC=30$ and $\angle EDC=90-2\alpha$.

then, in $\triangle ADC$ by the trigonometric form of Ceva we have that

$\frac{\sin\alpha \sin\alpha\sin(90-2\alpha)}{\sin 30\sin 30 \sin30}=1$

or  $8\sin^2\alpha(1-2\sin^2\alpha)=1$ or $(4\sin^2\alpha-1)^2=0$,

which implies that $\alpha=30$, and we're done  :D

[color=darkred][b][u]A equivalent enunciation.[/u][/b] Let $ABCD$ be a trapezoid
$(AB\parallel CD)$ so that $AB=AD$ and $BC=BD$.
Prove that $m(\widehat {ACD})=30^{\circ}\Longleftrightarrow$ the triangle $BCD$ is equilateral.
[b][u]Remark.[/u][/b] Let $M\in (AB)$ and the intersection $N\in DM\cap AC$.
Then $MA=MB\Longleftrightarrow$ the quadrilateral $BCDN$ is cyclically.[/color]

[color=darkblue][b][u]The first (trigonometrical) Proof.[/u][/b]
Denote $m(\widehat {ABD})=m(\widehat {ADB})=m(\widehat {BCD})=m(\widehat {BDC})=u$.
Then $m(\widehat {CBD})=180^{\circ}-2u$, $m(\widehat {BAC})=m(\widehat {ACD})=30^{\circ}$,
$m(\widehat {ACB})=u-30^{\circ}$ and $m(\widehat {CAD})=150^{\circ}-2u$. Therefore,
${\sin \widehat {ADB}\cdot\sin \widehat {BAC}\cdot\sin CBD}\cdot\sin \widehat {DCA}=$ $\sin \widehat {BDC}\cdot\sin \widehat {CAD}\cdot\sin\widehat {DBA}\cdot\sin \widehat {ACB}\Longleftrightarrow$
$\sin u\cdot\sin 30^{\circ}\cdot\sin (180^{\circ}-2u)\cdot \sin 30^{\circ}=$ $\sin u\cdot\sin (150^{\circ}-2u)\cdot \sin u\cdot\sin (u-30^{\circ})\Longleftrightarrow$
$\sin 2u=4\sin (2u+30^{\circ})\sin u\sin (u-30^{\circ})\Longleftrightarrow$ $2\sin (2u+30^{\circ})\sin (u-30^{\circ})=\cos u\Longleftrightarrow$
$\cos (u+60^{\circ})-\cos 3u=\cos u\Longleftrightarrow\cos (u+60^{\circ})-\cos u=\cos 3u$ $\Longleftrightarrow$
$-\sin (u+30^{\circ})=\cos 3u\Longleftrightarrow\cos (u+120^{\circ})=\cos 3u\Longleftrightarrow$
$u+120^{\circ}=$ $3u\ \ \vee\ \ (u+120^{\circ})+3u=0^{\circ}\Longleftrightarrow \boxed {\ u=60^{\circ}\ }\ .$

[b][u]The second (metrical) proof.[/u][/b]
Denote $AB=AD=a$, $BC=BD=b$, $CD=c$, $AC=d\ .$
$\triangle ABD\sim\triangle BDC\Longrightarrow \frac ab=\frac bc\Longrightarrow \boxed {\ b^2=ac\ }\ .$
The Euler's relation $AC^2+BD^2=2\cdot AB\cdot CD+AD^2+BC^2\Longrightarrow$
$b^2+d^2=2ac+a^2+b^2\Longrightarrow \boxed {\ d^2=a^2+2ac\ }\ .$
$m(\widehat {ACD})=30^{\circ}\Longrightarrow AD^2=AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos \widehat {ACD}\Longrightarrow$
$a^2=c^2+d^2-cd\sqrt 3\Longrightarrow a^2=c^2+a^2+2ac-cd\sqrt 3\Longrightarrow$
$c^2+2ac=cd\sqrt 3\Longrightarrow c+2a=d\sqrt 3\Longrightarrow (c+2a)^2=3(a^2+2ac)\Longrightarrow$
$a^2+c^2=2ac\Longrightarrow \boxed {\ a=c\Longrightarrow a=b=c\ ,\ d=a\sqrt 3\ }\ .$
[b]Remark.[/b] $\boxed {\ m(\widehat {ACD})=\phi\Longleftrightarrow \cos 2\phi=\frac{2a^2+c^2}{2a(a+2c)}\ }\ \ (\phi =30^{\circ}\Longleftrightarrow a=c)\ .$

[b][u]The third (syntetical) proof.[/u][/b][/color] Soon.

[color=darkred][b][u]Another equivalent enunciation.[/u][/b] Let $ABCD$ be a isosceles trapezoid $(AB\parallel CD)$ for which $AB=2\cdot AD\ .$ Prove that $\boxed {\ m(\widehat {ACD})=30^{\circ}\Longleftrightarrow m(\widehat {BAD})=60^{\circ}\ }\ .$
If denote the middlepoint $M$ of the side $[AB]$ and cutout the triangle $BCM$ obtain the proposed problem.[/color]

[color=darkblue][b][u]Proof.[/u][/b] Denote $AB=2a$, $AD=BC=a$, $CD=b$, $AC=c$. From the Euler's relation $2\cdot AC^2=2\cdot AB\cdot CD+2\cdot AD^2$ results $c^2=2ab+a^2$. Therefore,
$m(\widehat {ACD})=30^{\circ}\Longleftrightarrow a^2=b^2+c^2-bc\sqrt 3\Longleftrightarrow a^2=b^2+2ab+a^2-bc\sqrt 3\Longleftrightarrow$
$c\sqrt 3=2a+b\Longleftrightarrow 3(2ab+a^2)=4a^2+4ab+b^2\Longleftrightarrow$
$a^2-2ab+b^2=0\Longleftrightarrow a=b=\frac{c\sqrt 3}{3}\Longleftrightarrow m(\widehat {BAD})=60^{\circ}\ .$[/color]

See above the second (metrical) method and a second equivalent enunciation. Soon there I will offer the third (syntetical) method for the first equivalent enunciation.


algebra
binomial theorem
number theory proposed
number theory

Find all $n$ for which $3^{n}-2$ is a perfect square.

Obviously n>0.
$3^{n}-2=a^{2}$,
$3^{n}=(a-\sqrt{-2})(a+\sqrt{-2})$
no of the brackets on RHS is divisible by 3, so
$(1+\sqrt{-2})^{n}=\pm a \pm \sqrt{-2}$.
The coefficient of $\sqrt{-2}$ on LHS is at least n (by binomial theorem) but on LHS it can be only $\pm 1$. So we have $n=1$ and the result follows:

$n=1, a=1$.

$n=3, a=5$.
There are negative numbers in the binomial sum.


let $f(f(x))=x$ how can i prove $f(x)=x$ $?$

Functions are not my strong side, but here is what I guess:
If the function is one-to-one, then $f(f^{-1}(x))=x$, which means that for your function $f(x)=f^{-1}(x)$, which is only true for $f(x)=x$

Am I right?

Well, we don't want to be mean to people, do we? :) What Harazi means is that there's no way you can draw that conclusion, because if $f(f(x))=x$ it does not follow that $f(x)=x$. Take $f(x)=-x$, for example. You can't even find the solutions to $f(f(x))=x$.


AMC AIME AoPSwiki search probability USA(J)MO USAMO

here are some Mock AIMEs that Aops users have written already: [url=http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME]AoPSWiki Mock AIME page[/url]
you can also search the forum for others.

[quote="1337h4x"]Now that AMC season is over, is anyone planning on writing a Mocking AIME?[/quote]

But the AMC season isn't over yet.  :D  Yesterday's exams were the A exams, and there's still a second round of them coming some time soon. For Mock AIMEs, look in the AoPS Wiki ([b]minsoens[/b] has provided the link) or in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513]this thread[/url] for a slew of mock AIMES.

[quote="randomdragoon"]A mocking AIME? Okay..

Question 1. You have a 1/300 probability of qualifying for the USAMO on any given AIME, because you are lazy and you don't study. The probability that you will qualify for the USAMO at least once if you, like the dirty cheater that you are, take both the AIME I and the AIME II is p/q, where p and q are integers and gcd(p,q)=1. Find p.

...[/quote]
:rotfl: Very clever  :thumbup:

[quote="tjhance"][quote="randomdragoon"]A mocking AIME? Okay..

Question 1. You have a 1/300 probability of qualifying for the USAMO on any given AIME, because you are lazy and you don't study. The probability that you will qualify for the USAMO at least once if you, like the dirty cheater that you are, take both the AIME I and the AIME II is p/q, where p and q are integers and gcd(p,q)=1. Find p.

...[/quote]

not an integer between 0-999  :rotfl:[/quote]

[hide="ummm..."]
Count the complement.  The probability that you won't qualify on a given AIME is $\frac{299}{300}$.  So the probability that you won't qualify on either is $\frac{299^{2}}{300^{2}}=\frac{89401}{90000}$.

The probability that you will qualify is 1 minus this, or $\frac{599}{90000}$.  599 and 90000 are relatively prime, so the answer is 599.
[/hide]

Make more mocking AIME problems, randomdragoon.  :lol:

if i have some free time, i might put together another test...i have a few problems that i have been thinking of, but i am not prepared to put together a whole test just yet...

when's AMC 10B??  what's the difference?

[quote="1337h4x"]Now that AMC season is over, is anyone planning on writing a Mocking AIME?[/quote]

 :D  Wow, I totally didn't mean to say "mocking." Maybe it was a Freudian slip.

If someone approaches me, I'm sure I can contribute a few problems.


The functions $ f(x)$ and $ g(x)$ are defined on the real axis so that they satisfy the following condition: for any real numbers $ x$, and $ y$, $ f(x\plus{}g(y))\equal{}2x\plus{}y\plus{}5$. Find an explicit expression for the function $ g(x\plus{}f(y))$

Let $ u\equal{}x\plus{}g(y)$. Clearly the range of $ u$ is all real numbers. So we can say for all reals $ u,y$,

$ f(u)\equal{}2(u\minus{}g(y))\plus{}y\plus{}5$

$ f(u)\minus{}2u\equal{}(y\minus{}2g(y))\plus{}5$

By fixing one variable and varying the other, it is clear that $ f(u)\equal{}2u\plus{}C_1$ and $ g(y)\equal{}\frac{1}{2}y\plus{}C_2$ where $ C_1$ and $ C_2$ are constants. Plug these back in to find the relation between $ C_1$ and $ C_2$. The answer should follow.


Consider the sequence defined by $a_k=\frac 1{k^2+k}$ for $k\ge 1.$  Given that $a_m+a_{m+1}+\cdots+a_{n-1}=1/29,$ for positive integers $m$ and $n$ with $m<n$, find $m+n.$

[hide="Answer"]$a_k=\frac{1}{k(k+1)}$. We have $a_k+a_{k+1}=\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}\left(\frac{1}{k}+\frac{1}{k+2}\right)\Rightarrow$
$\frac{1}{k+1}\left(\frac{2k+2}{k(k+2)}\right)=\frac{2}{k(k+2)}$
Again, $(a_k+a_{k+1})+(a_{k+2}+a_{k+3})=\frac{2}{k(k+2)}+\frac{2}{(k+2)(k+4)}=\frac{2}{k+2}\left(\frac{1}{k}+\frac{1}{k+4}\right)\Rightarrow$
$\frac{2}{k+2}\left(\frac{2k+4}{k(k+4)}\right)=\frac{4}{k(k+4)}$
Continuing this process, we see that the numerator is the number of terms in the sequence, while the denominator is the number of the first term multiplied by one more than the number of the last term. Therefore, we have $\frac{(n-1)-m+1}{m((n-1)+1)}=\frac{1}{m}-\frac{1}{n}=\frac{1}{29}$. Multiplying across to remove denominators, we have $mn+29m-29n=0\Rightarrow (m-29)(n+29)=-(29^2)$. Since $m<n$, we have $m-29=-1\Rightarrow m=28$ and $n+29=29^2\Rightarrow n=29\cdot 28$, so $m+n=28(29+1)=\boxed{840}$.[/hide]

[hide]
$\frac{1}{k^2+k}=\frac{1}{k}-\frac{1}{k+1}$ (partial fraction decomposition)
the sum telescopes and you are left with $\frac{1}{m}-\frac{1}{n}=\frac{1}{29}$.
$\frac{n-m}{mn}=\frac{1}{29}$, time to use simon's favorite factoring trick!
$mn = 29n - 29m$
$mn - 29n + 29m = 0$
$mn - 29n + 29m -841 = -841$
$(m-29)(n+29)=-841$
$(29-m)(n+29)=841$
since m<n, $n=841-29$ and $m=28$. thus our answer is $28+841-29=840$.
[/hide]

[quote="joml88"][hide="Solution"]
We begin with the identity $\frac 1{k^2+k}=\frac 1{k}-\frac 1{k+1}.$  Thus, we have

\begin{eqnarray*} a_m+a_{m+1}+\cdots+a_{n-1} &=&\left(\frac 1m-\frac 1{m+1}\right)+\left(\frac 1{m+1}-\frac 1{m+2}\right)+\cdots+\left(\frac 1{n-1}+\frac 1n\right)\\ &=& \frac 1m-\frac 1n\\ &=& 1/29\\ \Rightarrow mn+29m-29n &=& 0\\ \Rightarrow (m-29)(n+29)=-29^2 \end{eqnarray*}

We deduce that $m-29=-1$ and $n+29=29^2.$  Thus, $m+n=\boxed{840}.$[/hide][/quote]

On your second line, that should say $-\frac{1}{n}$ instead, right?

[hide="Yeah.."]

This is classic problem. It is easy to observe that the denominator is $k(k+1)$ so the whole fraction must be $\frac{1}{k^2+k} = \frac{1}{k} - \frac{1}{k+1}$

Then the sequence is $\frac{1}{m} - \frac{1}{m+1} + \frac{1}{m+1} - \cdots - \frac{1}{n-1} + \frac{1}{n-1} - \frac{1}{n} = \frac{1}{m} - \frac{1}{n} = \frac{1}{29}$

Now, this equation is same as $29n-29m = mn$ and moving $29n-29m$ to RHS and subtracting $-29^2$ from both sides, we get $-29^2 = (m-29)(n+29)$. Since $m < n$, we obviously needs to have $m-29 = -1$ and $n+29 = 29^2$. $m$ is easily 28. But doing $29^2 - 29$ is not fun. So, here is an easy way:

$29^2 - 29 = 29(29-1) = (30-1)(28) = 840-28$ and since $m$ was 28, $m+n = \fbox{840}$. [/hide]

p8: We have $a_k = \frac{1}{k} - \frac{1}{k+1}$, so our expression cancels out to be $\frac{1}{m} - \frac{1}{n}$. Then $\frac{n-m}{nm} = \frac{1}{29}$, or $29n-29m = mn \implies mn + 29m-29n$, or $(m-29)(n+29) = -841$. This means $m = 28, n = 812$, so $m+n=\boxed{840}$.


The base-10 numbers 217 and 45 are multiplied.  The product is then written in  base-6.  What is the units digit of the base-6 representation?

Look at it in mod 6:
217 = 1 (mod6)
45 = 3 (mod6)
so the answer is 1*3 = [b]3[/b].

That's almost exactly what I did, only I divided each number by 6

217/6=36R1

45/6=7R3

3*1=3

so the units digit in base 6 is 3

well that's the exact same thing...mod 6 and dividing by 6 are the same.

Not really...mod 6 represents the numbers that are a+6b...and dividing is just, u know, dividing. Findidng the remainder is the first step in modding a number out, though, if thats what you meant.


How many possible 9-digit zip codes are possible if the first digit
cannot be zero?

Dont we have to account for changing order so 9x1^8x9!??

Once youve picked all 9 numbers there are 9! ways to arrange them.

That would be if the numbers were distinct.  However, they are not...

what if we have 123456789? thats all distinct.

But that's not what my calculations are. There are $ 9$ choices for the first digit, and $ 10$ for the remaining $ 8$. All permutations of $ 123456789$ are all included in this.

Why do I think you were being stupid on purpose to get posts... :P

No ive always been bad at C & P. Thank God for Alcumus!

That's coming from someone that made nats... suspicious :ninja:

Did they have any question like this on state?


induction
algorithm
continued fraction
algebra unsolved
algebra

I did not try to prove it myself, but induction on the length of the continued fraction representation of the rational would probably work.

Yep.  In fact, this recursion basically converts the binary representation of $ n$ into a continued fraction.

[quote="dvtrung"]Let $ \{a_n\}$ be sequence satisfies $ a_1 \equal{} 1, a_{2n} \equal{} a_n \plus{} 1, a_{2n \plus{} 1} \equal{} \frac {1}{a_{2n}}$. Prove that all rational numbers are terms of this sequence[/quote]

hint:
binary representation + continued fraction + bijection(uniqueness)

1. All of positive integers occurs in {a_n} (n>=1), namely, a_(2^n)=n for n=1,2, ... .
2. All of positive rational numbers less than 1 occurs in {a_(2n-1)} (n>=1).
3. All of positive rational numbers larger than 1 occurs in {a_n} (n>=1).

proof)
1. Since a_2n = a_n + 1 and a_1 = 1, it is obvious by induction.

2. We are going to prove the statement by induction.
First, it is obvious when the denominator is equal or less than 2 since a_1 = 1 and a_3 = 1/2.
Suppose that the statement holds for all of the rational numbers whose denominator is less than a. (a>2)
Let b/a be a rational number(but not an integer) such that 0<b/a<1 and gcd(a,b)=1.
By division algorithm, there exist unique integers q and r such that a=qb + r with 0<r<b. (r cannot be 0 since b/a is not an integer.)
Then, a/b = q + r/b.
By the supposition, there exist odd number N such that a_N = r/b.
Thus b/a = 1/(a/b) = 1/(q + r/b) = 1/(q + a_N) = 1/a_(2^q*N) = a_(2^q*N + 1), and we are done.

3. Let x/y be an arbitrary rational number(but not an integer) larger than 1 with gcd(x,y) = 1.
By division algorithm, there exist unique integers u,v such that x = uy + v with 0<v<y.
Therefore x/y = u + v/y.
By 2., there exist odd integer m such that v/y = a_m.
Thus x/y = u + a_m = a_(2^u*m), and we are done.

1.,2., and 3. imply that every rational numbers occur in the sequence {a_n}.
Q.E.D.

[i]The sequence $(x_{n})_{n\geq1}$ is defined by $x_{1}=1$, $x_{2n}=1+x_{n}$ and $x_{2n+1}=\frac{1}{x_{2n}}$ for all $n\geq1$. Prove that for any positive rational number $r$ there exists a unique $n$ such that $r=x_{n}$.[/i]

 Note that all the terms of the sequence are positive numbers and that $x_{2n}>1$, $x_{2n+1}<1$ for all $n\geq1$. We prove by induction on $k\geq2$ the following statement: for all positive integers $a,b$ such that $\gcd(a,b)=1$ and $a+b\leq k$, there exists a term of the sequence equal to $\frac{a}{b}$. If $k=2$, then $a=b=1$ and $\frac{a}{b}=1=x_{1}$.
Suppose that the statement is true for some $k>2$, and let $a,b$ be coprime positive integers such that $a+b=k+1$. If $a>b$, then we apply the induction hypothesis to the numbers $a-b$ and $b$. Clearly, $\gcd(a-b,b)=1$ and $(a-b)+b=a\leq k;$ therefore, there exists $n$ such that $x_{n}=\frac{a-b}{b}$. But then
\[
x_{2n}=1+x_{n}=1+\frac{a-b}{b}=\frac{a}{b}.
\]
If $a<b$, then we apply the induction hypothesis to the numbers $b-a$ and $a$.
There exists $n$ such that $x_{n}=\frac{b-a}{a}$, and we obtain
\[
x_{2n+1}=\frac{1}{x_{2n}}=\frac{1}{1+x_{n}}=\frac{1}{1+\frac{b-a}{a}}=\frac
{a}{b}.
\]
We proved that for any positive rational number $r$ there exists $n$ such that $r=x_{n}$, but we still have to prove that $n$ is unique. This follows from the fact that the terms of the sequence $(x_{n})_{n\geq1}$ are pairwise distinct. Indeed, suppose not, and choose $n\neq m$ such that $x_{n}=x_{m}$, with minimal $n$. The observation at the beginning of the solution shows that $n$ and $m$ have the same parity. If $n=2n^{\prime}$ and $m=2m^{\prime}$, then we obtain $x_{n^{\prime}}=x_{m^{\prime}}$, contradicting the minimality of
$n$. The case $n=2n^{\prime}+1$ and $m=2m^{\prime}+1$ yields $x_{2n^{\prime} }=x_{2m^{\prime}}$, and we again reach a contradiction.

Edit: I've changed the notations because I've copy/pasted from my book :)


I heard about AMC since last year and i did took AMC10 by paying like 100bucks.

i want to take it again possibly both amc10 A and amc12 B(im 10th grade) but im worried about expenses

anyway my parents were ok w/ money problem so i talked to my math teacher and even though i took it last year she didnt seem interested. (She did not like it last year too)

i talked to her about it last month and all she said was she need information so i gave her the AMC website and then she said she dont have time so i printed out AMC10and 12 information FOR her.
But still i dont hear anything from her about taking AMC

what should i do??? i KNOW taking AMC10 or12 will look good on my college thing(if i get high scored) so i DID and AM studying.

wat i don understand is my school is not bad, it's good school.
(its Carroll High school in  southlake, tx)-preppy school lol

anyway i need help, what should i do?? do i just have to do samething i did last year? asking teacher and counsil to sign my registration and then paying for it????

Mod Edit: title

Maybe you could find another math teacher to talk to who seems more interested?  Last year, my math teacher didn't really seem excited about it, and I had to end up paying for it myself (I later found out she was an English major or something like that, so I think that had something to do with it).  But I found a teacher who seemed a lot more excited about these contests than last year, and hopefully I won't have to pay for any of it this year.

Or you could find other schools in your area that offer it, or talk to a local college about having them offer it.  There's a flyer you could give the college here - http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-AMC,CU+Bro.pdf

The name of the thread has a spelling error...
Here is a list of colleges offering the AMC...
EDIT: The list didn't show up...
http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml

Yes, to the original poster, correct the spelling of the thread title, and find a different school to take the tests at if need be.

Or the college. Looks like the easiest option ( a little drive looks like nothing compared with changing the school district's attitude towards math and math competitions).
They do allow students from other states (we have no college in our state but 3 colleges within 160 miles from us).
And with colleges looks that they pay the costs, so it is free to us.
(I guess they do it for recruiting purposes).

Sorry for reviving an old thread.

I'm in a similar situation, except my school already offered the AMC 12A for this year.  I want to take the AMC 12B at a neighboring high school.  Do I need to get approval from the AMC to take the 12B at a different high school if my school already took the 12A?  I know the AMC regards the B contest as a backup if you cannot take the A contest, so is the B contest restricted to students who are unable to take the A test?

I've also heard that you cannot go to a different school to compete in math if your native high school already has a math team.  Does this mean I cannot take the 12B contest at another school even though my high school was only registered for the A contest?

Thanks for the help


hello, a calculation shows that 19 is the right answer.
Sonnhard.

Yes, it's not very hard to see that 19 is the right answer.


Let $ x_1,x_2,x_3$ be the roots of: $ x^3\minus{}6x^2\plus{}ax\plus{}a\equal{}0$. Find all real numbers $ a$ for which $ (x_1\minus{}1)^3\plus{}(x_2\minus{}1)^3\plus{}(x_3\minus{}1)^3\equal{}0$. Also, for each such $ a$, determine the corresponding values of $ x_1,x_2,$ and $ x_3$.

[quote="moldovan"]Let $ x_1,x_2,x_3$ be the roots of: $ x^3 \minus{} 6x^2 \plus{} ax \plus{} a \equal{} 0$. Find all real numbers $ a$ for which $ (x_1 \minus{} 1)^3 \plus{} (x_2 \minus{} 1)^3 \plus{} (x_3 \minus{} 1)^3 \equal{} 0$. Also, for each such $ a$, determine the corresponding values of $ x_1,x_2,$ and $ x_3$.[/quote]

$ x^3\minus{}6x^2\plus{}ax\plus{}a\equal{}0$
$ (x\minus{}1)^3\plus{}(a\minus{}3)x\plus{}(a\plus{}1)\equal{}3x^2$
$ (a\minus{}3)(x_1\plus{}x_2\plus{}x_3)\plus{}(3a\plus{}3)\equal{}3(x_1^2\plus{}x_2^2\plus{}x_3^2)$

On the other hand:
$ x^3\minus{}6x^2\plus{}ax\plus{}a\equal{}(x\minus{}x_1)(x\minus{}x_2)(x\minus{}x_3)$
$ a\equal{}\minus{}x_1 x_2 x_3$
$ a\equal{}x_1 x_2 \plus{} x_2 x_3 \plus{} x_3 x_1$
$ 6\equal{}x_1\plus{}x_2\plus{}x_3$

$ x_1^2\plus{}x_2^2\plus{}x_3^2\equal{}(x_1\plus{}x_2\plus{}x_3)^2\minus{}2 (x_1 x_2 \plus{} x_2 x_3 \plus{} x_3 x_1)$

$ x_1^2\plus{}x_2^2\plus{}x_3^2\equal{}36\minus{}2a$

$ (a\minus{}3)6\plus{}3a\plus{}3\equal{}3 (36\minus{}2a)$
$ 9a\minus{}15\equal{}108\minus{}6a$
$ 15a\equal{}123$
$ a\equal{}\frac{41}{5}$

[quote=OlympusHero]Therefore, our expression is $6(36-3a) - a + (36-2a) + 3 \cdot 6 - 3 = 0$

[b]I am fairly sure the above user is correct. Could someone point out my mistake?[/b][/quote]

Equation, that you'll get is $6(36-3a)-3a-3(36-2a)+3\cdot 6 - 3 =0 \Leftrightarrow a = \frac{41}{5}$.

[quote=Vulch][quote=mszew]
$ (x\minus{}1)^3\plus{}(a\minus{}3)x\plus{}(a\plus{}1)\equal{}3x^2$
$ (a\minus{}3)(x_1\plus{}x_2\plus{}x_3)\plus{}(3a\plus{}3)\equal{}3(x_1^2\plus{}x_2^2\plus{}x_3^2)$[/quote]
How did you get these two equations?[/quote]

The first one is just equivalent to the $x^3-6x^2+ax+a=0$.
For the second one, in the equation $(x-1)^3+(a-3)x+(a+1)=3x^2$ substitute $x_1,x_2,x_3$ instead of $x$ and sum up the resulting three equalities (also, use that $(x_1-1)^3+(x_2-1)^3+(x_3-1)^3=0$).

after you get the value of a how do you solve for the roots? I have lots of equations
$x_1+x_2+x_3 = 6$
$(x_1)^2+(x_2)^2+(x_3)^2 = 36 - 82/5$
$(x_1)^3+(x_2)^3+(x_3)^3 = 93 - 246/5$
$x_1x_2x_3 = -41/5$
$x_1x_2 + x_1x_3 + x_2x_3 = 41/5$
but I cant find the actual numbers.

Assume the roots of the equation $x^3-6x^2+ax+a=0$ are $\alpha,\beta,\gamma.$ Given $\boxed{\sum_{cyc}{}(\alpha-1)^3=0.}$ We have to find the value of $a.$
By Vieta's Relation we get, $$\alpha+\beta+\gamma=6$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=a$$ $$\alpha\beta\gamma=-a$$
From the above relation we get, $$(\alpha-1)^3+(\beta-1)^3+(\gamma-1)^3=0$$ $$\left(\sum_{cyc}{}\alpha^3\right)-3\left(\sum_{cyc}{}\alpha^2\right)+3\left(\sum_{cyc}{}\alpha\right)-3=0.$$ $$216-21a-108+6a+15=0\implies\boxed{a=\frac{41}{5}}$$
Vieta's Relation!!!! :love:


couldn't you take a 2 out of 2x+2 to make the answer:
$2x(x+1)(x+3)$?

Duh!  :roll:  I knew you were smarter than me, math92  :blush:

[quote="math92"][quote="ifai"][quote="math92"]Factor this:
$2x^3+8x^2+6x$[/quote]
Factor is a noun, isn't it?[/quote]
um, i think its both a noun and a verb, its just the context you use it in.[/quote]
the verb form of factor is factorize.

[quote="ifai"][/quote][quote="math92"][quote="ifai"][quote="math92"]Factor this:
$2x^3+8x^2+6x$[/quote]
Factor is a noun, isn't it?[/quote]
um, i think its both a noun and a verb, its just the context you use it in.[/quote][quote="ifai"]
the verb form of factor is factorize.[/quote]
oh, well my teacher always said factor.  My teacher is a dum dum.


Solve for $ (a,b,c)$ :

$ 2^c\minus{}b\equal{}\overline{abc}$

If they are digits, one way is just listing possibilities c=0,...9  :wink:


limit
logarithms
calculus
calculus computations

Let $ k, p\geq 1$ be natural numbers. Find the limit:

$ \lim\limits_{n\rightarrow \infty}\frac{1}{n}\left(\frac{n\plus{}k\minus{}1}{1}\plus{}\frac{n\plus{}k\minus{}2}{2}\plus{}\cdots\plus{}\frac{k}{n}\minus{}\ln(n\plus{}p)!\right)$.

From Stoltz's lemma the above limit is equal to :

$ \lim_{n\to\infty} \left( 1\plus{}\frac{1}{2}\plus{}...\plus{}\frac{1}{n}\minus{}\ln(n)\plus{}\frac{k}{n\plus{}1}\plus{}\ln\left( \frac{n}{n\plus{}p\plus{}1}\right)\right)\equal{}\gamma$

That is the correct answer.

Now, I guess it is not hard to calculate the more general limit

$ \lim\limits_{n\rightarrow \infty}\frac{1}{n^{\alpha}}\left(\frac{n\plus{}k\minus{}1}{1}\plus{}\frac{n\plus{}k\minus{}2}{2}\plus{}\cdots\plus{}\frac{k}{n}\minus{}\ln(n\plus{}p)!\right)^{\beta}$,

where $ \alpha, \beta>0$.

By asymptotical methods $ H_{n} \approx \ln n + \gamma$ and using stirling formula I get that :
${ \lim ... \approx n^{\beta - \alpha}}\gamma^{\beta}$


[quote="jorgeston"]Prove that

$ e \equal{} 1\plus{}\sum_{n \equal{} 1}^{\infty}\prod_{i \equal{} 1}^{n}\frac{1}{i}$

 is very easy :lol:[/quote]
It is easy.

If you write it out you get,
$ 1\plus{}1\plus{}\frac{1}{2!}\plus{}\frac{1}{3!}\plus{}... \equal{} e$

EDIT: Fixed mistake mentioned by boxeded.

Sylow: $ 1\plus{}1\plus{}\frac{1}{2!}\plus{}\frac{1}{3!}\plus{}\hdots \equal{}e$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Rewrite the numerator of the integrand:
$ 2(x\cos(x) \minus{} \sin(x))^2 \equal{} 2x^2\cos^2(x) \minus{} 4x\cos(x)\sin(x) \plus{} 2\sin^2(x) \equal{}$
$ \equal{} x^2\cos(2x) \plus{} x^2 \minus{} 2x\sin(2x) \plus{} 1 \minus{} \cos(2x)$

Now after several simple partial integrations it boils down to the integral $ \int_0^{\infty}\frac {\sin(x)}{x}\ dx \equal{} \frac {\pi}{2}$


analytic geometry
graphing lines
slope
calculus
integration
Pythagorean Theorem
geometry

[hide]Imagine OB as the hypotenuse of one right triangle (with the y-axis being one side) and OA as the hypotenuse of another (with the x-axis being one side) We could use the pythagorean theorem to find out the distances of OA and OB, which have to be equal. Also, the distance of OA would have to be the square root of twice a square because the slope is 1. OB must have one leg be 7x as long as the other. By thinking of numbers that fit these criteria, we can see that point A can be (10,10) and point B can be (2,14). We can then find the slope of AB, which is -1/2. (E) [/hide]

[quote="236factorial"]Points A and B are in the first quadrant and O is the origin. If the slope of OA is 1, the slope of OB is 7, and the length of OA is equal to the length of OB, then the slope of AB is

a) -2/3
b) -3/4
c) -6/7
d) -1/6
e) -1/2[/quote]

[hide]Point slope has a form of $m=\frac{y_1-y_2}{x_1-x_2}$

Applied it here: $\frac{y_A}{x_A}=1$, so $y_A = x_A$

$\frac{y_A}{x_A}=7$ then $y_A= 7x_A$

We have $OA=OB$, then $\sqrt{(x_A)^2 + (y_A)^2} = \sqrt{(x_B)^2 + (y_B)^2}$

$\Rightarrow  \sqrt{(x_A)^2} = \sqrt{50(x_B)^2}$

$\Rightarrow  x_A = 5x_B$

Then $\frac{y_B - y_A}{x_B - x_A}=\frac{y_B - 5x_B}{x_B - 5x_B}==\frac{7x_B-5x_B}{-4x_B}=\frac{-1}{2}$

$\boxed{E}$[/hide]

[hide]Ugh, I stink at slope/coordinate plane questions...I found the shortest lengths for lines OA and OB which had integral coordinates: A(1,7) and B(5,5). The slope of AB is -1/2, answer [b](e).[/b][/hide]


function
algebra
domain
polynomial
algorithm
complex analysis
rational function

Suppose I have a rational (complex) function:

$f(z)=\frac{P(z)}{Q(z)}$

Where P(z) and Q(z) are polynomials.

In theory, this can be expanded via partial fractions.  What would  the general expression for the Laurent series in an annular  domain $D_{n}$ representing the annulus between $R_{i}$ and $R_{i+1}$ be?  Note each domain of definition or convergence takes the appearance of an annulus, the total of which looks like a target with concentric circles the radius of each representing a root of Q(z).  I think the general expression for the Laurent series $L_{D_{n}}$ can be written without knowing what the roots are.  That is, in terms of $f_{D_{n}}= L_{D_{n}}[w_{0},w_{1},w_{2},...w_{n},z]$ where the w's are the zeros of Q(z) and L represents the Laurent series.

I'm new here and not sure what constitutes an ok post.  I'm taking Complex Analysis this semester and this is not any assigment although I think it would have been a good one.  :)

Assume WLOG that $f$ is a proper rational function; the polynomial part can be handled by long division anyway.

Let's start by looking near the origin; if $z$ does not divide $q(z)$, what's the power series for $\frac{p(z)}{q(z)}$? The coefficients satisfy a linear recurrence; for example, $\frac{x}{1-x-x^{2}}=\sum_{n=0}^\infty F_{n}x^{n}$, where $F_{n}$ are the Fibonacci numbers. The recurrence is based on the coefficients of $q$, and the initial conditions come from $p$ and $q$ together.

Near $\infty$, we have a reflected version of the same idea: $\frac{x}{1-x-x^{2}}=-\frac{x^{-1}}{1+x^{-1}-x^{-2}}= \sum_{n=0}^{\infty}F_{-n}x^{-n}$.

In the middle (say, on an annulus including $|z|=r$), we need to separate the function so we can have two series. We need a factorization $q(z)=q_{1}(z)q_{2}(z)$, where $q_{1}(z)$ is never zero for $|z|\le r$ and $q_{2}(z)$ is never zero for $|z|\ge r$. We can apply partial fractions with these two terms: $\frac{p(z)}{q(z)}= \frac{p_{1}(z)}{q_{1}(z)}+\frac{p_{2}(z)}{q_{2}(z)}$. The first has a power series expansion around zero, and the second has an expansion around $\infty$.
Continuing my example, we have $\frac{x}{1-x-x^{2}}=\frac{s}{1+s^{2}}\left(\frac s{x-s}+\frac1{1+sx}\right)=\frac1{\sqrt{5}}\left(\frac{sx^{-1}}{1-sx^{-1}}+\frac1{1+sx}\right)$, where $s=\frac{\sqrt{5}-1}{2}$ is the positive zero of $1-x-x^{2}$. ($\frac{s}{1+s^{2}}=\frac{s}{2-s}= \frac{(\sqrt{5}-1)/2}{(5-\sqrt{5})/2}=\frac1{\sqrt{5}}$). The Laurent series between the two roots is $\frac1{\sqrt{5}}\sum_{n=1}^\infty s^{n}x^{-n}+\frac1{\sqrt{5}}\sum_{n=0}^\infty (-1)^{n}s^{n}x^{n}$.

The above shows that we need at least some knowledge of the roots to get all of the Laurent series- we need to be able to tell which roots lie inside or outside of any given circle, and factor the denominator accordingly. If we know the roots exactly, we also get the exact annulus of convergence for each series.
For theoretical purposes, you want the full partial fractions decomposition; this algorithm isn't worth the complications.

Thanks.  I'll spend time with your analysis. :)


superior algebra
superior algebra unsolved

What is the splitting field of x^2+3x+1?
and what is the splitting field of x^2+3ix+1?

Thanks in advance :)

Splitting fields over what? What are the base fields here?

Basically, take the base fields and adjoin the roots. Assuming you're being sensible and the coefficients are in the base field, adjoining one root automatically gets you the other.

opps, sorry, i should have mentioned

x^2+3x+1 over Q
and x^2+3ix+1 over Q(i)

The roots of x^2+3x+1 is 1/2(-3+sqrt5) and 1/2(-3-sqrt5)
But I don't know how to adjoin them together

Adjoin one, and you get the other automatically. You could simplify with rational operations, too; if we have $ \sqrt{5}$, we have those numbers as well.


geometry
symmetry
ratio
projective geometry
geometry theorems

I've been seeing a lot of that on the forum lately. I think I get what it is, but could someone tell me where and how it is generally used in Olympiad problems? And where did you guys learn it?

Also, is "polar" geometry is part of projective geometry or inversive geometry?

[quote="billzhao"]I've been seeing a lot of that on the forum lately. I think I get what it is, but could someone tell me where and how it is generally used in Olympiad problems? [/quote] well at least this year, knowing poles and polars would have been very usefull on the geo problems from the Romanian TSTs :)

We are given a circle $(O)$ and a point $P$. We draw a chord of the circle through $P$, cutting the circle at $A$ and $B$. The locus of point $X$ s.t. $(A,B;P,X)=-1$ is a line which is obviously (for symmetry reasons) perpendicular to $OP$ (O is the center of the circle $(O)$). This is called the circular polar of $P$ wrt $(O)$.

Let $\angle xOy$ be an angle and let $P$ be a point in the plane. Draw a line through $P$ which cuts $Ox$ and $Oy$ in $A$ and $B$ respectively. The locus of $X$ s.t. $(A,B;P,X)=-1$ is a line passing through $O$. This is called the angular polar of $P$ wrt $\angle xOy$.

A very useful fact: Let $A$ and $B$ be two arbitrary points (different from the center of the circle $(O),\ O$. Then $A$ is on the circular polar of $B$ iff $B$ is on the circular polar of $A$ (something similar can be done for angular polars, of course).

It's part of projective geometry. Here's a connection between the angular polar and the circular polar:

Let $ABCD$ be a quadrilateral inscribed in $(O)$. $E=AB\cap CD,\ F=AD\cap BC$. Then the angular polar of $E$ wrt $\angle AFB$= the circular polar of $E$ wrt $(O)$.

I don't quite understand your notation. What does $(A,B;P,X)=-1$ mean?

See also http://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtml for general facts regarding poles and polars.

[quote="grobber"]$(A,B;C,D)$ means cross-ratio and it's equal to $\frac{CA}{CD}$:$\frac{DA}{DB}$, where all the 2-letter combinations are oriented segments. The four points must be collinear, of course.

If we have four fixed lines and we cut them with a variable line, then the cross-ratio of the four intersection points doesn't change, so we can also speak about the cross-ratio of four lines.[/quote]

err...there is only one $B$ in that fraction...I believe that we have:

$(A,B;C,D)=\frac{AC}{BC}: \frac{AD}{BD}=\frac{CA}{CB}*\frac{DA}{DB}$ [directed lengths], etc

[color=darkblue][b]Poles and polars come in pairs. Poles are plane points; polars are straight lines in the same plane. Below, I define the pole/polar correspondence with respect to a circle.

So let there be a circle $ w=\mathcal R (O,R)$. Let $ A$ be an arbitrary point, except $ O$, whose inverse image in $ w$ is $ A'$, i.e. $ OA\cdot OA' = R^{2}$. The point $ A'\in OA$. Let $ a$ be the line through $ A'$ perpendicular to $ OA'$ (or $ OA$, which is the same.) Then $ a$ is called the polar of $ A$, while $ A$ is the pole of $ a$ w.r.t. to $ w$.

If $ A$ is outside $ w$, we may draw two tangents from $ A$ to $ w$. Then the polar $ a$ is the line through the two points of tangency. If $ A$ lies on $ w$, its polar $ a$ is just the tangent to $ w$ at $ A$.

 Buy this applet

The applet allows adding any number of pole/polar pairs. This can be done in two ways. You can add a pole which will appear as a small hollow circle that could be dragged. Its polar will be drawn automatically. It is also possible to add a polar - a straight (draggable) line defined by two draggable points. Its polar will then appear as a small filled circle.

Observe the main properties of poles and their polars:

The polar of a pole that lies inside the circle of reference lies in its entirety outsdie the circle.

The polar of a pole on the circle is tangent to the circle at the pole.

The polar of a pole that lies outside the circle of reference crosses the circle at two points. The lines joining the points to the pole are tangent to the circle.[/b][/color]

[color=darkred][b][u]If point $ A$ lies on the polar of point $ B$, then point $ B$ lies on the polar of point $ A$. (La Hire's Theorem.)[/u][/b][/color]

[color=darkblue][b]The pole of a line through two poles lies at the intersection of their polars.

The polar of a point which is the intersection of two polars, is the line through the corresponding poles.

The polars of three collinear points are concurrent.

The poles of three concurrent lines are collinear.

The latter properties are a direct consequence of La Hire's theorem. For example, assume points $ A,\ B,\ C$ with polars $ a,\ b,\ c$ respectively, lie on a straight line $ m$. Let $ M$ be the pole of $ m$. Since each of $ A,\ B,\ C$ lies on $ m$, each of $ a,\ b,\ c$ passes through $ M$. They are therefore concurrent. Conversely, let three lines $ a,\ b,\ c$ concur at point $ M$. Then each of $ A,\ B,\ C$ lies on the polar $ m$ of $ M$. They are therefore collinear.

Note that one needs only a straightedge to construct the polar of a point. That tangents to a circle from a point are straightedge constructible even in the absence of the center of the circle. (Compare that to [u]Steiner's theorem.[/u])

Let $ A'$ be the inverse image of point $ A$ lying outside the circle of inversion. $ A'$ then lies inside the circle and, by definition, $ OA\cdot OA' = R^{2}\ \ (1)$.

Find points $ S$ and $ T$ on the circle such that $ ST$ is perpendicular to $ OA$ at $ A'$. From $ (1)$ and the fact that they share an angle at $ O$, triangles $ OAT$ and $ OTA'$ are similar $ (SAS)$. Since angle $ OA'T$ is right, so is angle $ OTA$. In other words, $ AT$ is perpendicular to the radius $ OT$ and is therefore tangent to the circle at $ T$.[/b][/color]

[quote="msaeids"]does somebody have a collection of problems?[/quote]
 you may look a few of them here:


Let $a,b,c,d$ are positive real number such that  $a+b+c+d=1$ . Prove that:
$1\ge a^{2}+b^{2}+c^{2}+c^{2}+21\frac{1}{3}abcd \ge \frac{1}{3}$

Is it difficult? :D

[quote="The soul of rock"]It's mine :blush: . I have very nice solution of this, just use a classical way!

@zaizai-hoang: Do you know who am I?[/quote]
of course  :D I known you long-ago  :D
Can you post "very nice solution of this" in here ?  :wink:
PS: Why did you change nick ? :D


articles
Stanford
college
geometry
trigonometry
linear algebra
rich tag

The IQ numbers above are in Cattel scale (where the human average IQ is around 100).
Please, [b]DON'T TRUST ANY INTERNET IQ TEST OR UNOFFICIAL TEST[/b], just vote if your result comes from an official IQ test, for example Mensa test.

I want to see how IQ is related to the interest in mathematics, expecially in young people (mathematical interest, not mathematical talent, which is of course related to IQ).   :maybe:
So, I make the assumption that everyone here is interested in mathematics...   :D

[quote="Valentin Vornicu"]where do you find the "official" mensa test? :)[/quote]
Visit [url=http://ro.mensa.org/]Mensa Romania[/url] to know about Mensa in your country, and to learn how to take the test!  ;)
However, the test paper is secret, and cannot be found on the web... you have to physically go to Mensa and say that you want to take the test (you often have to pay a small sum for it, 8 Euros in Italy).

[quote="MindFlyer"][quote="SingTheSorrow"]Haha, I think it's hard to believe that anyone's IQ here is below 90...[/quote]
I think it's much harder to believe that a man who misses that ranking by just two points could become the [url=http://www.georgewalkerbush.net/presidentialiqs.htm]President of the USA[/url]![/quote] I hope that you know that IS a hoax right? :) anyway it may very well have the lowest IQ of all presidents (Bush that is) but it's certainly not 91.
At least that's what I think.

Hi,

personally I think that math, computer science, philosophy and nature sciences students are the most intelligent students. Just have a look at the people who take positions of leadership. What did they study ? Very often they studied math. But what the hell do I need differential equations or similar stuff in my later non-university career ? Just to solve problems. Life is problem-solving. In those studies you acquire the skills to think abstractly and solve challenging problems. Now I want to provide some short examples. Probably some of you know the software company SAP, the fourth largest Software company in the world. The current Chief Executive Officer and Member of the Executive Board at SAP AG is Prof. Dr. Henning Kagermann. He studied math and physics and got a professorship in theoretical physics later then before he joined SAP.

some information on Henning Kagermann:

http://www.tu-muenchen.de/menschen/kuratorium/kuratorium/henning.tuml
http://www.forbes.com/finance/mktguideapps/personinfo/FromPersonIdPersonTearsheet.jhtml?passedPersonId=207427
http://www.time.com/time/digital/cyberelite/38.html

The founder and supporter of the Hasso-Plattner-Institute (HPI), http://www.hpi.uni-potsdam.de/eng/ ,
institute where I study is Prof. Dr. hc Hasso Plattner. He is one of three founders of SAP. He studied communications engineering at the excellent German university in Karlsruhe (greetings to Darij Grinberg  :) ). The institute was created in the style of American top universities as Carnegie Mellon http://www.cmu.edu/ .

some information on Hasso Plattner:

http://www.businessweek.com/magazine/content/01_30/b3742087.htm
http://www.forbes.com/finance/mktguideapps/personinfo/FromMktGuideIdPersonTearsheet.jhtml?passedMktGuideId=137875
http://www.forbes.com/finance/mktguideapps/personinfo/FromMktGuideIdPersonTearsheet.jhtml?passedMktGuideId=137875

Many students who went to IMO got excellent mathematician. Some even received the most prestigious award in math: the fields medal, e.g. Laurent Lafforgue and W. Timothy Gowers.

some information on the FieldsMedal:

http://mathworld.wolfram.com/FieldsMedal.html

some information on Laurent Lafforgue:

http://www.ihes.fr/EVENTS/lafforgue/aboutLaf.html
http://www.iolfree.ie/~alexandros/articles/lafforgue.htm

some information on W. Timothy Gowers:

http://www.ams.org/new-in-math/cover/gowers.html
http://www.dpmms.cam.ac.uk/site2002/People/gowers_wt.html

The former German IMO first prize winner Reinhard Hppner participated in IMO in the late 60s. Later on he was elected as minister of one of the 16 German districts: Sachsen-Anhalt.

Recent German IMO participants also decided not to go into research but rushed into leading management postions and and top positions in computer science as project leader, boss in the research department of various companies etc. I feel many of them just get more attracted by these jobs as they can earn lots of money.

What do you think ?  :)

[quote="Valentin Vornicu"]
I hope that you know that IS a hoax right? :)
[/quote]
Sure I know, but maybe Mr. Bush doesn't! I mean, how can he know that 91<100? He doesn't have enough fingers to handle such huge numbers!  :D

Ok, ok, enough of that...  ;)

I agree that people that studied exact sciences learn to study anything else with ease, but I also think that this is not the most important. Take for example Bill Gates. I think he didn't finish any university, but he is somebody who we all know :D

[quote="DusT"]Take for example Bill Gates. I think he didn't finish any university, but he is somebody who we all know[/quote]

In fact, I beleive what people develop when they are at competences like IMO, is cleverness. And it's exactly cleverness what anubody needs for achieving anything (i.e: lots of money). For example, Bill Gates is very clever, maybe windows is not the best OS in the world, but he is one of the best sellers and dealers in the world. Otherwise he wouldn't be were he is  :cool:

IMHO,

IQ is very important for life and specially academics. However I read some extracts from a book about EQ, emotional quotient, which stated that this is the most important factor towards success. In fact it enunciated statistical results obtained throught experiments (by observing different people from their chilhood until adulthood).  EQ is basically the hability to deal with ones own emotions, to empathize with others and so on...
Obviously this is more influential in non-academic settings, yet it's role in academic careers isn't negligible at all.
As an example I read that Mr. Bush, although wasn't a very good student, was very popular and had a lot of friends when he studied at Yale.  The hability to organize people, charisma, character are just a few of the factors needed for a political leader.

There are a couple of things that i wonder about and wish somebody can give me some answers...

1) What's the IQ of the greatest men in mathematics, such as Erdos, Godel, Andrew Wiles, etc?
2) What do you guys think is the role of mathematicians in society?

psykyk

[quote="psykyk"]
1) What's the IQ of the greatest men in mathematics, such as Erdos, Godel, Andrew Wiles, etc?
[/quote]
Take a look [url=http://www.aceviper.net/aceviper_net/ace_intelligence/aceviper_famous_people_iq_list/aceviper_famous_people_iq_list.html]here[/url]! There are the extimated IQ's of many famous people, including a few mathematicians.

[quote="Valentin Vornicu"]it seems that some german president was a genius :P :)[/quote]
Sometimes genius borders upon madness.  :?

Are yoy talking about Hitler or Stalin (I mean smart, but mean)?
Hitler was a german leader :)

[quote=Ziqinorin]$\phantom{[/quote]}$[quote=LucBob][quote=Ziqinorin][quote=michaelchang1]so he doesn't care about low IQ things like this topic[/quote]

He does if he replied to this topic. I'd say that his claim that he has a 175 IQ is profoundly ludicrous. He can prove it if he wants to, but he still hasn't to this day.[/quote]

I don't see why you think it's so ludicrous[/quote]

Because he has yet to prove it.[/quote]

basically your saying anything that hasn't been proven is ludicrous

I don't even know, bro.

My flowery writing from a writing thing was like 80 and a crap research paper was "genius". Screw that.

IQ isn't really important unless you put it into focus, so yea, but I wouldn't know either way, I suppose in between somewhere, though I took a test and it told me I had at least 120-130???

[quote=Ziqinorin]He can prove it if he wants to. I don't care. I know that he's trolling.[/quote]

You have to provide evidence if you want to say that. You can't just blanty claim something.

Responding to @Ziqinorin:

This person was posting in College Math 13 years ago. Right now, they are probably an adult. But you, a first grader, are spreading rumors and insulting them???? Respect your elders, were you never taught that? Forget respecting elders, at least don't spread fake information about people. Like right now, I could say, "You aren't a first grader, you are an 8th grader with zero common sense who cheats on everything and trolls."

[quote=Ziqinorin][quote=derpy_chimp][quote=Ziqinorin]This person hasn't been online since last month. Disappointing.[/quote]

people have their life outside of aops too smh[/quote]

I was hoping for him/her to come back for just a single day to reply to my comment. It doesn't take much time. I do doubt that this person actually got a 175 IQ. He/she would probably have made the IMO with that IQ, but this person didn't.

C.Y., who was on the CMC USA team, has an IQ of 147. Just for comparison.[/quote]

I hope you understand that some people don't necessarily put that into effect and it's just a potential measure. So it really doesn't matter, as long as it may be decent and you work hard. Despite how good it is, it's not a clear indicator of true success. Sorry to disappoint, and yes, there's many possibilities but we'll never know so might as well deal with it instead of flamespeculating.

Lol what is this about this guy's just curious who can blame him, just answer the question + a bit of additional input and grace yourself out...

@asimov
He probably doesn't even care if people doubt his IQ. He probably forgot about this thread a long time ago.

@Ziqinorin
Someone might be very intelligent, but his talents might be somewhere other than math.

[quote=baker77][quote=Ziqinorin]He can prove it if he wants to. I don't care. I know that he's trolling.[/quote]

You keep accusing people of trolling but what about yourself? You do that ALL the time with no shame whatsoever. So I think your judgement on whether someone is a "troll" or not is moot.[/quote]

IKR???? This person said they are in 1st grade, then they said they are in 3rd grade, then they said that they achieved a really good score on the SAT R&W section, this person answers correctly instantly on puzzle threads in MSM, and never posts a solution, and for some reason they always have the highest IQ or score in everything but they post no proof. And when others get a higher score than them, they say that the other person is lying and trolling. Oh, and they don't talk like a first grader.

Wow, 38% of people have an IQ >200. I wish I could be like that...

[quote=asimov][quote=Ziqinorin]He can prove it if he wants to. I don't care. I know that he's trolling.[/quote]

Why don't you prove YOUR IQ?[/quote]

Oh I walked into the local preschool my bad

@bove those are probably trolls.

[quote=asimov]@2above

I don't think that's true, because there are people like Ziqinorin.

Anyways, IQ shouldn't matter, it's just a stupid thing that is subjective and makes no sense whatsoever.[/quote]

I [b]did[/b] walk into my local preschool. Oh dear.

Can you people actually calm down @Ziqinorin and @asimov?


Let $f$ be a function s.t. $ f(x-1) + f(x+1) = \sqrt{3} f(x) $ Prove that $f$ is periodic and find its fundamental period

it is quite simple to show $f(x)=f(x+12)$. and if $f(x)\neq 0$ for some $x$ it is also show by the same method, that there is no smaller period than $12$. it remains to somebody else to list out what i have done on a peace of paper :)

Your answer is correct and you are also right that the calculations are a bit tedious!!!

show me how??
i m a beginner
i dont know to show a function is periodic

[quote="i_like_pie"]Please refrain from reviving old topics in the future, especially anything [i]this[/i] old (2 years and 9 months).

As for your question, feel free to start a new topic. :)[/quote]I don't think there is any problem reviving an old topic where the details for solving the problem are not incorporated into the topic and his question to ask for the solution is perfectly legitimate. Also, as levans admin said before, there is no actual violation of forum netiquette or forum guidelines when an old topic is revived.

Back to topic, here is a little help: Try substituting different values: Let $ x\equal{}z\plus{}1$, let $ x\equal{}z\minus{}1$. Keep substituting more values if needed, and play around with the algebra, which is quite laborious in this problem, but it's doable.

Tedious?

[hide="Or..."] Let $ a \in [0, 1)$ and consider the sequence $ f_n \equal{} f(n \plus{} a)$.  Its characteristic equation is $ x^2 \minus{} \sqrt {3} x \plus{} 1 \equal{} 0$ which has roots $ \frac {\sqrt {3} \pm i}{2}$.  But this is just $ \omega, \omega^{ \minus{} 1}$ where $ \omega$ is the primitive $ 12^{th}$ root of unity $ e^{ i \frac {\pi}{6} }$.  Then

$ f_n \equal{} A \omega^n \plus{} B \omega^{ \minus{} n}$

for appropriate constants $ A, B$ that depend on $ a$.  The rest follows. [/hide]


Consider two different situations in which a guy Bob is harmed (suppose seriously injured or killed). Bob is some random person that you've never met before; you know nothing about him.

(A) You directly cause the harm to Bob. It's a free choice depending on no other circumstances, and it's no better or worse for you or anyone else (except Bob) for you to do it.

(B) There is some indirect action you can take (say, pushing a button) which will [i]prevent[/i] Bob from coming to this harm. You know exactly what will happen to Bob if you push the button and if you don't. As before, it's no better or worse for you or anyone else (except Bob) for you to do it.

Obviously, in either case, you should make the choice that saves Bob from harm. But is there any difference between the cases? Is it in some way better to protect someone from an impersonal source of harm than to refrain from harming him yourself? Is it in some way worse to harm someone intentionally rather than to just allow a preventable natural process to take its course and harm him?

The reason I'm asking this is that it seems like if we consider the hypothetical situations, they're pretty much the same. But in real life, most people make decisions based on the idea that protecting someone is good, while refraining from harming them is kinda neutral; and directly harming someone is really bad while allowing someone to be harmed is bad but not as much so.

What do you think is the reason for this disparity?

This question is similar to something called the trolly problem. It goes as follows:

Suppose that you are the controller of a train switch. Coming towards your switch is a fast moving train. Down the path that the train is currently set to travel, there are four men (whom you do not know) working on the tracks. They will not have time to get out of the way. Along the other path, there is a single man working on the tracks (whom you also do not know), and he will also not have time to get out of the way if the train comes down his path.

The question is in two parts. Is it morally [i]permissible[/i] to switch the direction of the tracks so that the train will avoid the four men in it's current path, and kill the man working on the other track? If the answer is yes to the first question, are you morally [i]obligated[/i] to switch?

This question becomes interesting when contrasted with the following scenario:

Suppose you work in a hospital where there are 4 sick patients, each in need of an immediate organ transplant in order to save their lives (different organs). There is one healthy person in the hospital who is a perfect match for each sick patient. Is it morally [i]permissible[/i] to kill the healthy person and give his organs to the sick patients? If the answer to the question is yes, are you morally [i]obligated[/i] to give his organs to the other patients?

It's a very good question and I think maybe because it can be extended to many situations not only involving serious harm. Actually in the case of serious harm like involving the life of other people like Nathan says, I speak for myself now, I would be very disappointed of my own behavior if I didn't judge me as bad for causing direct harm to Bob as for not pulling the switch that will save him. But there are plenty of other issues about which you are very sure you will or will not be taking direct actions, but you are actually presented with situations in which you could take [i]indirect[/i] action and you do not.

I guess it comes down to priorities, in Bob's case I think is equally wrong causing direct harm to Bob and pulling the switch. Actually, what is the limit between direct an undirect harm? If you are absolutely positive about Bob's harm if you pull the switch I don't see how you can make it more direct even if you hit him in the face. Assuming it is not that clear, then I could think that trying to save Bob from a [i]hypothetical[/i] danger, is not worth my time or my money or whatever I risk. Like, for instance, some people understand the world [i]might[/i] be at risk because of the green house effect but that possibility is not worth the nuisance of pulling the bikes from time to time. Sometimes people use this as a justification, since we're not sure whether the green house effect is actually a problem or not, it isn't [i]wrong[/i] to act as if it wasn't. I think that such actions are not right. If you have a dilemma that affects your actions, [i]you should resolve it[/i] if you can. What you should do is to make sure the green house is really not an issue and if you find out it is then act accordingly

[quote="gauss202"]
The question is in two parts. Is it morally permissible to switch the direction of the tracks so that the train will avoid the four men in it's current path, and kill the man working on the other track? If the answer is yes to the first question, are you morally obligated to switch?
[/quote]

Definitely in the first question, in the second I don't think you are obligated but I think it is what I would do. However this is a simpler question than the other. You will be charged (or whoever was responsible) for the killing of some people in either case, it comes down to decide whether it is four or one. And the thing that makes it different from the next case is that the action by which I decide four or one, namely pulling the switch, does not contradict any moral values of mine

In the other case, it is harder because the action that implies my decision is killing one healthy guy which is something in normal conditions I wouldn't do(as opposed to pulling switches which is my favourite passe-temps) In this case, almost sure I wouldn't do it. That is I don't think it is [i]permissible[/i] at least for some finite number of beneficiaries. The morality that impedes me also makes me think in other factors that could enforce my judgement as for example,  maybe the sick guys were hardened smokers or whatever. Maybe not but I rather not to take the chances

Ursula

For the original question, here is another way (not exactly the same but related) to approach the idea I'm trying to look at. In a philosophy paper I read recently this was referred to as "the Asymmetry." Basically, it is saying that preventing harm to someone is good, and helping someone out (benefiting them in some way) is also good. But you are obligated to prevent harm if you can, whereas you are not obligated to help someone out every time you can.

It seems like people generally accept the Asymmetry. But honestly I don't understand why. The main justification for not always helping people out when you can is uncertainty.
"If I expend the time/resources for this good deed, I might not be able to perform some other one later."
"There are too many possible good things I could do, and if I just tried to do every one that presented itself, I wouldn't be able to live my own life."
(Cosmopolitan problem) "Not everyone agrees on what is benefiting, and I don't want to risk doing things that I think would be helpful but the recipient actually doesn't want."

And so on. On the other hand, people seem to think, if we are ever presented with an opportunity to protect someone from harm, we are clearly obligated to do it. But I think this is untrue. I think all of the same reasons apply in the other direction as well.
"Maybe what I think is harmful this other guy actually wants." (* see hidden bit at end)
"There are so many people suffering harm in the world that if I tried to help all of them I wouldn't be able to live my own life."
And so on.

So, I believe that I am equally obligated to exercise good judgment about what situations I should act on when I see opportunities to benefit someone or to protect someone from harm. If I see a kid with a cookie and a bully comes to try to steal it from him, I should protect the kid and his cookie. But if I have a cookie and I don't want it and I see some kid who really wants a cookie, I should give it to him. And I think I am equally morally obligated in each case, and personally I am equally happy to do either.

To answer what [b]gauss202[/b] was saying about the trolley and the surgery: personally, in the trolley case, I think one is obligated to switch the car. But that's because I have a very rational and rigid view of things, which says if you have a clear and easy choice between two mutually exclusive options, and you can predict the (important part at least) of the outcome, you are obligated to choose the one that's better. So, if I think it is bad for someone to die young in an accident like this, it is worse for four people to experience this wrong than one. Therefore I must switch the trolley to the other track.

On the other hand, the surgery case is much less clear. I find this example very disturbing actually, because I have not yet been able to identify for myself according to my system of values any reasons that it's [i]unambiguously[/i] wrong to kill the healthy guy.
The strongest argument against it, in my opinion, is by looking at the situation from a global perspective. Four lives are worth more than one, all else being equal. But now all else is no longer equal -- the original guy is healthy, and the other four are seriously ill. That presents some justification for saying the healthy guy's life might actually contribute more to society etc than the four sick guys. Maybe the transplants will save them, but for how long? If you want to make it a bit more unrealistic, then you could say the four guys will all be pretty much normal guys after the surgery, but then it's much closer to the trolley case.
But it's still weird. Because I feel like even if I were absolutely sure that my organs could be harvested to save four guys' lives right now, and they all had about the same prospects in life that I do, I still wouldn't want to let a doctor kill me and harvest my organs. And this bothers me because either I haven't fully understood these two examples, or in the "me" case I am also influenced by selfish reasons (such as the instinct for self-preservation) and I am not sure that this is generally a good thing.

This also leads into my (*) though. Because even if I wouldn't want to do it, should we blame a doctor who does it when the healthy person consents? What if everyone (the sick patients, the doctor, the healthy patient, and the healthy patient's close family) all agree that it's morally right for the doctor to kill the healthy patient and save the sick ones, so the doctor does it. Should the doctor be tried for murder? When does one's rights over one's own person stop? There is another example which suggests that they [i]should[/i] stop eventually:
[hide="graphic so hidden"] You may have heard about that incident in Germany a few years ago, where two guys met in an Internet chat forum about cannibalism or something. They decided to get together in person so that one of the guys could kill and eat the other. They actually did this. As far as anyone could tell, the victim did actually do everything willingly.
(There are disputes about this. For example, the killer drugged the victim so that he wouldn't experience as much pain. One could argue that once he was heavily sedated, he was no longer capable of telling the killer that he was having second thoughts, once the blood started to flow. But still, presumably he consented to being sedated.)
But the killer was in fact tried and sentenced to quite some time in prison. I'm not sure if it was (the German equivalent of) first degree murder, though. It might have been a somewhat lighter one.
Anyway, most people's gut reaction is probably that the killer has in fact done something terribly wrong. But it's not immediately obvious why, since no one was "harmed" according to a definition of harm that excludes things that the "victim" wants. [/hide]

All right I've already written too much. I need to do homework! This is a great discussion though, thanks for answering. :)

This is the part 2 to the trolley question I've heard (partially copy and pasted from your post)

Suppose that you are the controller of a train switch. Coming towards your switch is a fast moving train. Down the path that the train is currently set to travel, there are four men (whom you do not know) working on the tracks. They will not have time to get out of the way. There is also a man standing at the side of the track, between the trolley and the four workers. If you push that man into the track, the collision between the trolley and that man will reduce the velocity of the trolley enough so that the four workers will have time to get out of the way. Is it morally permissable / obligated for you to do this?

These hypothetical questions are very sticky.

There is no fundamental difference between passive and active harm.  However, reality is fundamentally different from these idealised examples.  In reality, there are no passive or active harms, only harms.  Blame is a meaningless idea.

To add another hypothetical question (and an answer) to this discussion, I give the following (non-idealised, I hope) example:

Q>you are given the option to donate a life saving kidney to one of two people.  The kidney is not yours, but you are in complete control of who the recipient is.  You know neither of the potential recipients, and know only ther occupations.  You have no reason to suspect the people who have created this situation of any trickery.  One potenial recipient is a sandwich artist at Subway.  The other is a doctor.  Who should you give the kidney too?

A>the doctor.  Why?  Doesn't this value one person's life over another?  The correct way to make any moral choice to consider all courses of action, including what may be termed "inaction" but is actually relatively un-influential action, and the possible effects of all the possible outcomes of all the possible choices.  Implied in the first question are many other options.  There are ALWAYS an infinite number of choices available to an individual, however, we usually subconciously create a multpile choice format if we are going to chose an option consciously, as is the case here.

But, to answer the question:  the expected value of goodness brought about by the doctor's continued existence is greater than the expected value of the goodness brought about by the sandwich artists continued existence (assume this is true, if it is not, then it is a bad example, and the answer should be opposite)  The idea is that the doctor will probably contribute more and go on to save more people's lives.  Probably.  But that's all that matters, since there is never going to be an absolute certainty.

To rebute the objection I earlier posed "doesn't this value one person's life or interests over another's?"  NO.  It values every person equally, but considers all the people affected by the decision, which is, of course, all the people in the world.  You can not artifically isolate the situation.  When one considers ALL the people affected not just the most obvious one's the doctor (or the hypothetical "person with a higher expected value of goodness stemming from their contiued existence") is the correct choice.


How many terms does the following arithmetic sequence have? \[ 5.7,6.7,7.7,\ldots,21.7\]

Using the formula, $ \frac {21.7 \minus{} 5.7}{1} \plus{} 1 \equal{} \boxed{17}$

Or just subract 0.7 from each term to get 5-21, or 17 numbers.

EDIT: WOW PHAIL Thanks for catching that.

That should be $ 17$ terms from 5-21. The number of terms is $ 21\minus{}5\plus{}1\equal{}\boxed{17}$.

The formula is, in a sequence $ \{a_1, a_2, ..., a_n\}$:

$ \frac{a_n\minus{}a_1}{a_2\minus{}a_1} \plus{} 1$, where $ a_n$ is the last term, $ a_1$ is the first, and $ a_2$ is the second.  Of course, for the denominator for the fraction, you can just have the difference between any two terms.


An electric heater consumes $ 1.8MJ$ when connected to a $ 200V$ supply for 30 min. Find the power rating of the heater and the current taken from the supply.
where w=energy and p=power

$ w\equal{} \displaystyle\int^{1/2}_0 p.\,dt$

$ w\equal{}\frac{p}{2}$

how to proceed from there?

Current is easy to find once you know the power since power is defined to be
$ P\equal{}VI\equal{}I^{2}R\equal{}\frac{V^{2}}{R}\equal{}\frac{E}{t}$

Since we're using SI, we have to define $ t$ in terms of seconds, which is 1800.  Then
$ P\equal{}\frac{1800000\text{J}}{1800\text{s}}\equal{}1000\text{W}\equal{}(200\text{V})I\Rightarrow I\equal{}5\text{A}$


How many license plates consist of 2 letters followed by 3 digits?

i dont at all understand this problem. please reply. :huh:

We have $ 26$ possibilities for each of the $ 2$ letters, and $ 10$ possibilities for each of the $ 3$ digits. So,
          $ 26\times{26}\times{10}\times{10}\times{10}\equal{}\boxed{676000}$.

26 letters and 10 digits. They can use them as many times as they want for example-aa999 so-
26*26*10*10*10  :)

[quote=oadaegan][hide=easier way to do it]each of the 26 numbers can be used as many times as wanted so $26 squared$ and each of the 10 digits can be used as many times as wanted so $10 $ cubed$+26 $ squared$=676000$[/hide][/quote]

isn't that exactly how gaussintraining did it?

facepalm*

That is exactly how they did it; the essence of the solution is the same. Just because you used different notation (and bad latex) doesn't make it any different.

In fact, you actually messed up when you said there are "26 numbers," so you're solution is technically incorrect.\

Not to be super picky, but it's technically not correct to say you can use the numbers/digits(letters??) "as many times as wanted." You didn't even explain how that statement leads to $\ 26^2 $ or $\ 26squared $ as you call it, and you didn't explain where the 10 cubed comes from, if you can use the 10 digits "as many times as wanted."


How do you prove that there is a multiple of 3 in every list of 3-consecutive-odd-integers (like 7,9,11)?

Thanks!

[hide]
Case One
The first number is a multiple of three.
Numbers $ 0 \mod 3 \ , \ 2 \mod 3 \ , \ 1 \mod 3$.

Case Two
The middle number is a multiple of three.
Numbers $ 1 \mod 3 \ , \ 0 \mod 3 \ , \ 2 \mod 3$

Case Three
The last number is a multiple of three.
Numbers $ 2 \mod 3 \ , \ 1 \mod 3 \ , \ 0 \mod 3$

Regardless of the first odd integer picked, either it or the consecutive odd integers will be a multiple of three.
[/hide]

The three terms can be expressed as $ x$, $ x+2$, and $ x+4$. If $ x+4$ is divisible by three, then so is $ x+1$. Since $ x$, $ x+1$, and $ x+2$ are consecutive integers, at least one of them must be divisible by three.

[quote="jeez123"][quote="Treething"]The three terms can be expressed as $ x$, $ x+2$, and $ x+4$. If $ x+4$ is divisible by three, then so is $ x+1$. [/quote]

and.... ?

If $ x+4$ is divisible by three, then so is $ x+1$
If $ x$ is divisible by three, then so is ...?

Could you expand this a bit more?
Thanks![/quote]
you keep subtracting 3s..

[quote="jeez123"][quote="Treething"]The three terms can be expressed as $ x$, $ x+2$, and $ x+4$. If $ x+4$ is divisible by three, then so is $ x+1$. [/quote]

and.... ?

If $ x+4$ is divisible by three, then so is $ x+1$
If $ x$ is divisible by three, then so is ...?

Could you expand this a bit more?
Thanks![/quote]

$ x+4$ and $ (x+4)-3$ have the same remainder when you divide by 3.

We also know that one of $ x$, $ x+1$, and $ x+2$ is divisible by 3, which basically means that one of $ x$, $ x+4$, and $ x+2$ is divisible by 3.

Just put the odd integers into groups of three.  They look like this:

$ 6 \cdot 0+1, \boxed{6 \cdot 0+3}, 6 \cdot 0+5,$
$ 6 \cdot 1+1, \boxed{6 \cdot 1+3}, 6 \cdot 0+5,$
$ 6 \cdot 2+1, \boxed{6 \cdot 2+3}, 6 \cdot 0+5,$
$ 6 \cdot 3+1, \boxed{6 \cdot 3+3}, 6 \cdot 0+5,$

And so forth.  If it wasn't intuitively obvious to you now, it should be, and this is not a bad way to construct an elementary proof either.

Another (pretty simple) way:

Given three consecutive integers, exactly one of them is divisible by 3 (sure... think about their remainders when divided by 3) therefore (*) given 6 consecutive integers exactly two of them are divisible by 3.

Then... exactly 3 of the 6 numbers mentioned above are consecutive even integers and therefore may be written as $ 2(M)$ , $ 2(M+1)$ and $ 2(M+2)$ but... $ M, M+1, M+2$ are consecutive integers and that means that exactly one of them is divisible by 3, therefore exactly one of the 3 consecutive even integers is divisible by 3 and... the other integer mentioned in (*) is certainly  odd!   ;)


let $ABC$ be a triangle .for each point $P$in the area of it,prove:

$PB+PC<AB+AC$

[hide="Proof"]Extend $PB$ until it intersects $AC$ in $Q$. Then we have

\begin{eqnarray*}PC<PQ+QC &\Longrightarrow& PB+PC<PB+PQ+QC\\ &\Longrightarrow& PB+PC<BQ+QC<(BA+AQ)+QC\\&\Longrightarrow& PB+PC<AB+AC\end{eqnarray*}[/hide]


Hi, let's get the math-likers in Minnesota talking to one another about math in the Land of 10,000 Lakes.

Yea we need to revive this forum

I revived a thread that was inactive for 9 years!!!!!

The Minnesota forums seems to be way more active then it was a few months ago


geometry
3D geometry
tetrahedron
geometric transformation
reflection
similar triangles

I have a question about the first solution presented in the attached article. How to prove, that given two orthocentric and perspective triangles we can obtain from it an orthocentric tetrahedron and then prove Sondat's theorem (as in the article)? They present only the proof when the tetrahedron is given and I didn't know if it's enough (and if it is - why?).

Thanks in advance.

Dera limes,
your reference is not availaible.
Can you propose another way to see this article?
Thank in advance
Sincerely
Jean-Louis

OK I attached this article. I'm sorry I didn't know that the link was dead.

I found a synthetic solution for Sondat's theorem. Dear Jean Luis, is it similar to your proof ?

[b]Lemma:[/b] Given two similar triangles $ABC$ and $A_1B_1C_1$. Let $A_2, B_2, C_2$ be the midpoints of $AA_1, BB_1, CC_1$, respectively. Then $\Delta A_2B_2C_2\sim \Delta ABC \sim \Delta A_1B_1C_1$.
[b]Proof:[/b] Let $M, N$ be the midpoints of $AB_1$ and $AC_1$. It's easy to see that $\Delta A_2MN\sim \Delta A_1B_1C_1$.
Use direct angle we can show that $(AB, A_1B_1)=(AC, A_1C_1)$, which follows that $\angle A_2MB_2=\angle A_2NC_2$. So $\Delta A_2MB_2= \Delta A_2NC_2$. Therefore $\Delta A_2B_2C_2\sim \Delta A_2MN$ and we are done.

Back to [b]Sondat[/b]'s theorem, which is stated that:
Given a triangle $ABC$. An arbitrary line d intersects $BC, CA, AB$ at $A', B', C'$, respectively. Construct the paralogic triangle $A_1B_1C_1$ of $ABC$ wrt $d$. Then $d$ bisects the line joining the two orthocenters $H$ and $H_1$ of $\Delta ABC$ and $A_1B_1C_1$.
[b]Proof:[/b]
Let $X, Y, Z, H_2$ be the midpoints of $AA_1, BB_1, CC_1, HH_1$. It's easy to show that $\Delta A_1H_1B_1\sim \Delta AHB, \Delta B_1H_1C_1\sim \Delta BHC, \Delta C_1H_1A_1\sim \Delta CHA$. Applying the lemma above we get $\Delta XYZ\sim \Delta ABC\sim \Delta A_1B_1C_1$ and $H_2$ is the orthocenter of $\Delta XYZ$.
Let $L$ be the Miquel point of the completed quadrilateral $ABCA'B'C'$.
The result $X, Y, Z, O, L$ are concyclic is well-known so I leave it to the readers.
Note that $A', B', C'$ is the reflections of $L$ wrt $YZ, ZX, XY$, respectively then $d$ is the Steiner's line of $L$ wrt $\Delta XYZ$. This means d passes through the orthocenter of triangle $XYZ$ or $d$ passes through the midpoint of $HH_1$.
We are done.

Dear Lynh and Mathlinkers,
the last problem of Sondat is another problem than the first posted... which I have solved on my site...
Dear Lynh, your nice approach seems to me different and is a new way to solved this problem.
Sincerely
Jean-Louis


find all $p(x,y)$ such that: $2p(x,y)=p(x+y,x-y)$

[color=red]Edit by Megus: edit instead of reposting[/color]

The problem is wrong
$P$ ye motaghayere ast ya 2 motaghayer?

Tarhe khodete. :D


algebra
polynomial
Rational Root Theorem

[quote="vinny919"]I don't get the rational root theorem.  Does it say that any rational roots to a polynomial HAVE TO HAVE TO HAVE TO be numbers that divide into the last term divided by numbers that divide into the first term?

So does that mean if I have a polynomial like

x^83941239487 ....[a lot of other crap] +....+ 2[/tex]

then the rational roots can only be -2,-1,0,1, and 2?[/quote].

Yes, but only if all the other coefficients are also integers, or we could have:

([x terms]+1/2)(x-2), which clearly has a rational root of x=2, but has a trailing coefficient of -1.

Yup. But just note, 0 can't be a root, as 0 is not a factor of 2. You only can have -2, -1, 1, 2.

well I want to know the possible values for c so that $x^2 - cx^2 - 2 =0$ has a rational solution, so I just said that x = -2,-1,1, or 2, and solved for c in each case. is this correct?[/u]

Well, in the expression you gave, we can solve for x directly and get
\[x=\pm\sqrt{\frac{2}{1-c}}\]
so we just want the values of c that make that a rational number.  But, there are infinitely many such values of c... are you sure you don't mean $x^2-cx-2=0$ as your equation?

You're never supposed to use the cubic formula.  Ever.  Every problem involving a cubic is not meant to involve the cubic formula.  And, if you've ever seen the cubic formula, you'll realize that you're always better off not using it.

[quote="vinny919"]Whoa sorry, the equation was

$x^3 - cx^2 - 2 = 0$

So here, can x only be -2,-1,1, and 2? I don't think I'm supposed to use the cubic formula here.[/quote]

Keep in mind that there is a condition in the theorem you called 'rational root theorem', and that is to require the polynomial to have integer coefficients. Since your polynomial equation $x^3 - cx^2 - 2=0$ is monic ([i]monic[/i] means a polynomial with leading coefficient 1 ), [b]if[/b] your equation were to have a rational root, it must be an [b]integer root[/b] as well.

[quote="vinny919"]can x only be -2,-1,1, and 2?[/quote]

Yes and nothing else. Remember that after you plug in $x$, $c$ must be an integer or else rational root theorem is not bound to hold.

And there is no need to use cubic formula. It is of theoretical (and perhaps computational) interest than being practical because of the cumbersome calculations that you need to go through is not worth that effort. One reason it was made is to show that any cubic equation can be solved in terms of radicals.

So I worked it out for x = -2,...,2 and got that c = -5/2, -3, -1, 3/2.

I checked all these with a cubic equation calculator somewhere on the internet and indeed each has a rational root.  I don't get this problem!

Hmm. If the problem did indeed say find all values of c such that -that equation- has a rational root, then there are infinitely many values of c. Perhaps it meant integer values of c. (because as said before, you can only use the rational root theorem if c is an integer). If c isn't an integer, its still possible for it to have a rational root. Like, x = 3 is a root of $x^3 - \frac{25}{9}x^2 - 2 = 0$.

Then it might be that there is a pattern among the c's, such that they are all of the sam form?  Is that possible?

Just solve for c in terms of x. Then thats a solution for any rational x you want. You can't get any more general than that..

Let me finalize this topic because it is wasting people's time replying back and forth so much just to solve one easy problem.

Question: How do you use rational root theorem? (vinny919)

Answer: To use the theorem, you need to [b][u]satisfy[/u][/b] the conditions imposed by the theorem.

Conditions: The polynomial MUST have integer coefficients and MUST have a rational root. NOTHING ELSE.

Theorem: If a polynomial $p(x)=a_kx^k+a_{k-1}x^{k-1}+\cdots+a_0$ has integer coefficients and has a rational root for $x=\frac{m}{n}$, where $m$ and $n$ are nonzero integers, then $m$ must divide $a_0$ and $n$ must divide $a_k$.

What does this theorem imply? It means that if you have a polynomial with integer coefficients and [b]also have[/b] a rational root, then the numerator of this rational root must divide the last term and the denominator of this rational root must divide the first term. THAT'S IT, SIMPLE. NOTHING MORE.

BUT... No more but's. The theorem does not imply $p(x)$ have a rational root. That is one reason why we say [b]possible[/b] rational roots because this theorem does NOT guarantee a rational root.

How is it applied? Let's do your problem by finding all values of $c$ such that $x^3-cx^2-2=0$. Here is where I want to break down the problem into two subsections, according to whatever you are looking for.

If you are looking for [b]rational[/b] values of $c$, then all you got to do is solve for $c$: $c=\frac{x^3-2}{x^2}$. We know $x \neq 0$ because obviously $p(x)$ does not have zero as a root. If you plug in any nonzero rational number into $x$, $c$ would also be rational and therefore there are infinite number of values for $c$.

If you are looking for [b]integer[/b] values of $c$, then you can only hope that the possible rational roots are $x=\frac{\pm 1}{1},\frac{\pm 2}{1}$. We still solve for $c$ and get $c=\frac{x^3-2}{x^2}$; but after we plug in those possible rational roots, we have
$x=-1\Rightarrow c=-3$
$x=+1\Rightarrow c=-1$
$x=-2\Rightarrow c=-\frac{5}{2}$
$x=+2\Rightarrow c=\frac{3}{2}$
Therefore, as you guessed, $c=-3$ and $-1$ are the values that satisfy the rational root theorem (integer coefficients).

Hopefully, this clarifies a lot of your questions.