ARML AMC AIME HMMT algebra function domain

And while we're at it, how about questions like who the top team will be at each site, and who might win Division B?

In the West, it's been SFBA A for a while but we (Southern California A) came within 4 points last year, and Washington was right behind us.

[quote="Silverfalcon"]Lehigh Valley all the way!  :D[/quote]

I second that.

you guys are all under-estimating the hidden power of the WA team :D

we have had really well organized practices, and i think our team would do a lot better this year than last year (mainly becuase i joined it this year  :P ).

Lehigh 2006 Fire Team all the way!!!! I wouldn't quit that team last year if I knew it's going to win the national championship.

[quote="matt276eagles"][quote="Silverfalcon"]Lehigh Valley all the way!  :D[/quote]

I third that.  Go Lehigh Valley Fire and Ice! :lol:  I was on Ice last year, should be on Fire this year.

By the way, Silverfalcon, are you going to be on the Lehigh Valley team this year?  Were you at the Lehigh contest?[/quote]

I didn't know about the contest last year but this year, yes, I was on the contest this year and I got 5th in sophomores (I'm from Juniata).

I'm guessing you're the freshman from Radnor who got first right?  :D I was busy at the results day so I didn't get to talk to anyone and know anyone for real yet.

[quote="Silverfalcon"]I'm guessing you're the freshman from Radnor who got first right?  :D [/quote]

Good guess.  :lol:

[quote="darkquantum"]Now that TJ A's monopoly of ARML victories has been broken, the door is left open for another team to win, for Lehigh to repeat, or for TJ to reclaim its throne.

So..., any predictions?[/quote]

NYC A!!!  just for the ice cream :D

matt276eagles, how did the other guy (senior) from your school did on AIME? haven't being able to contact you guys.

I'm not sure, but I think he got a 10.  By the way, are you going to be on the Lehigh Valley ARML team this year, jiangwei?

Lots of lehigh people here.  I feel outnumbered, maybe I got to force the whole TJ math team to join this site, it would help a lot.  There are a lot of good contenders this year, lehigh is scary, exeter is really good, usual elite schools are all going to be good.  I can tell you that I feel that TJ is going to have a strong showing this year, especially if they finally decide to make the team and power rounds decently hard like they did at HMMT or something.  I'm so excited, ARML is just a lot of fun.  Good luck to all of you!

I am pretty sure I will join the Lehigh team, unless somehow i get kicked down to the ice team, which is highly possible. I was thinking about carpools with you guys, matt276eagles. I guess you and me can fill in the spots for the 3 seniors that graduated last year.

Maybe he meant a perfect scorer on [i]AIME[/i], not USAMO...
In fact, Chicago team has one person who got perfect on AIME.

Believe me, many people on our team (especially A team) are much smarter than me :P

Bah! MN Gold will pwn you all!  (with our grand total of... 2 USAMO takers on the team.  And at least one more with index >200).
Actually, we are looking pretty good so far...  At practices, we've had a flurry of perfect (and a few almost-perfect) team events, and we had a perfect relay last practice : all 13 or so teams got the correct answer at the three minute mark.  So, we'll have to see...

[quote="joml88"][quote="white_horse_king88"]And if I'm not totally lost, isn't Brian Lawrence from Maryland??? :huh:[/quote]

Yes, he is.

So, um, frt...? (edit: i meant karth :P )[/quote]

Hmm... that's interesting... I thought I saw a post earlier that said frt was the perfect USAMO scorer. Oh well, my mistake  :P

Go GA.

I don't expect us to win, but we might. We have made the top 15 for the past 15 years.

I'll probably end up on the B team, but that's ok for being a freshman.

[quote="alan"][quote="darkquantum"]where are all teh NYC kids? I think NYC doesn't like AoPS.  :([/quote]

maybe try posting on the nyc forum?  also, i think sho and kevin believe that AoPS is evil incarnate.  and most of the rest of them dont know or care that it exists.[/quote]

Hey I am from NYC!!!!! and I love AoPS!!!!

DE ALL THE WAY!!

If we actually get to send a team that is...

[quote="PTynan89"][quote="Karth"]Uh, ARML is over for this year.... North Carolina already won  :|[/quote]

I meant for next year, silly.[/quote]
What he means is that this thread was created for last year's ARML, and if you want to discuss 2007 ARML, you should create a thread of your own for it... silly. :wink:

This.... silliness..... will stop NOW! I demand it!

I'm going to go to the lehigh competition. i saw the top scores from last year. shouldn't be too hard to make at least ice.

!! WESTERN MASS FOR THE WIN !!

...

yeah...

...

of course...

...

...

 :(

There's another thread for ARML 2007... so please post there.  :)


What do you think about putting humor in math papers?

Go ahead and tell us how it turns out.
I personally wouldn't.

Lame jokes and silliness are probably not good. However, something cleverly witty surely can't hurt, unless it really disrupts the flow of your writing. Generally, what's important is a smooth and flowing style, and if jokes fit in well with how you write, it's okay. People need to be able to understand what you write and not get confused by random junk mixed in. If the jokes don't seem like random junk mixed in for confusion, they should be okay.

Personally, I think people need to have better senses of humor about serious topics. My analysis professor today gave an example of a Borel measure that assigns infinite measure to balls of finite radius, and he denoted the measure function by $\mu_{\text{bad}}$. It was brilliant. There's nothing wrong with mixing in some fun along with work.

i doubt the teachers like it. i put the upper class made the lower class push the wagon on a history class paper and the teacher didn;t lik eit

It's always amusing to put humor in math papers :).

I SAY YES! As long as what you are saying isn't total crap..scratch that...as long as it makes sense.

What about jokes in published math papers? Katz and Tao once proved that two seemingly unrelated conjectures are equivalent to the Erdos Ring problem: does there exist a subring of $\mathbb R$ which is a Borel set of Hausdorff dimension strictly between $0$ and $1$? They stated this result as follows:
[quote]1.15. The main result.
[i]One Ring to rule them all,
One Ring to find them,
One Ring to bring them all,
and in the darkness bind them.[/i][/quote]
By the way, the ring problem was since solved by Edgar and Miller: any proper Borel subring of $\mathbb R$ has Hausdorff dimension zero.

This is from MathReviews (2001):
[quote]Spike Milligan wrote of a certain poet that he tortured the English language, yet had still not managed to get it to reveal its meaning. Trying to fathom the paper under review is similarly frustrating. The reviewer has read through the paper several times, and on each occasion has become more and more confused.[/quote]


search
function
calculus
algebra
polynomial
geometry
geometric transformation

"it's final! wearing a hat that covers over 20% of your face will increase your gansterness by 70% while reducing your uglyness by 60%"

lol.. said by my friend

dont tell me on one is into humor here..

Why did the chicken cross the road?

Because -----!   Ahahahahahah hah ha h... no?

No.

How many Neanderthals does it take to light a fire?

In the words of Robin Williams: [hide]None, they don't have it yet...[/hide]

1-800 Casino

[quote]If you have a gambling problem, please call 1-800-GAMBLE

[/quote]

Question Posed: I have an ace and a 6, the dealer shows a 7, what do I do?

Absolutely nothing.  Play the hand as is, and quit gambling entirely as soon as it is finished; your luck has evidently been depleted.

i have been dubbed the king of lame jokes by my grade

u dont want to hear them...... they cause heart failure

How do you make a venetian blind?

spoiler :  [hide]Poke him in the eye!

ok ok if u dont get it a venetian blind is thode blinds thingies

and a person from venice is called a venetian[/hide]

Why is 6 afraid of 7? Because 7 8 9. hahaha that was so lame.

Why was 9 afraid of 10? Because 4 5 6???? AHAHAHAHA. OK... I'll stop. :lol:

I'm sure Homo Erectus used fire, and I'm sure Neandertals (Notice the correct spelling) made fire.

Don't look at me, I'm just quoting Robin Williams from [u]Inside the Actors Studio[/u].

There were to ships: one carrying purple paint and one carrying red paint.  The ships collided.  What happened to the passengers?

Answer: They got marooned!  Ah ha ha ha ha....

Why did the chicken cross the Mobius Strip? To get to the same side! Aha hah heh um...

most of the jokes i no are very innapropriate. but.........theres sum jokes my choir teacher (mr. arnold) told us b4. (sunnyray should no these too)

a teacher was teaching the class about whales and she stated " it is impossible for a whale to swallow a human being whole" so a girl said "but i learned at my sunday school that jonas got swallowed by a whale" the teacher replied " no, thats impossible" so the girl said again "well, when i get to heaven, ill ask him" the teacher said "what if jonas isnt in heaven?" the girl said "then you ask him"

hahahahahahahaha...ha?

OH I HAV :varepsilon:   ON :varepsilon: !  HOW IVIANY SKINNY IVI :varepsilon: N   O :varepsilon: S   IT TAK :varepsilon:  TO I=ILL A BATHTUB?

I I)ON'T KNOW.  TH :varepsilon: Y  K :varepsilon:  :varepsilon: P SLIPING OWN TH :varepsilon:  I) :gamma: AIN!

THAT'S SOOOOOO I=UNNY,  :gamma: IGHT?

trying to type half leet half english or something?

try1|\|g  t0  typ3  h4L|=  l33t h4L|=  3|\|gL15h  0r  s0|\/|3th1ng?

um what did you say? and i posted:

how many skinny men does it take to fill a bathtub?

i don't know. they keep sliping down the drain :lol:

that's so funny! hahahahahahahahahahahahahahaha...right?

oh yeah that...its cured. i fixed it. i put it on another topic, i don't know which.

Hello, know any "Dumb and Dumber jokes"?
Here's a sampling of this type of silly humor . . .

"He's so dumb, he thinks a football coach has four wheels."
"That's really dumb!  How many wheels does it have?"
[Heard that one on the playground during recess ... a century ago.]

Two carpenters are working on a house.  One of them keeps hearing
a plink-sound every few seconds.  He turns and finds the other man
throwing away every other nail.
"What's going on?" he asks.
"Half these nails are no good.  The points are on the wrong end."
"You dummy!" says the first.  "They're for the other side of the house."

Bubba and Junior rented a boat and went fishing.  They caught many fish,
drank a lot of beer and had a great day.
While driving home, Bubba said, "I hope we remember that good fishing
spot the next time we come.
"Don't worry," said Junior.  "I marked it.  I put an X on the side of the boat."
"That's really dumb!" said Bubba.  "Suppose we don't get the same boat?"

Two men sre standing at the base of a flagpole, looking up.
A woman sees tham and askd what they're doing.
"We have to measure the height of the pole and we won't have a ladder."
The woman takes a wrench from her purse, loosens some bolts, and
lays the pole on the ground.  Then she uses her tape measure and
announces, "Twelve feet, 9 inches" and walks off.
The two men shake their heads.  "Ain't that just like a woman!" says one.
"We ask for the height ... and she gives us the length."

Bubba and Junior are told to get the mule into the barn. As they led the
mule up to the door, Bubba says,"Oh-oh!  His ears aren't gonna make it
through the door!"
So they sat and thunk for a while.
Then Junior says, "I've got it!  We'll dig a trench and let the mule walk in it."
"No, no!" said Bubba.  "It's his ears that're too long, not his legs!"

[quote="Scrambled"]cactus cacti
radius radi
octopus octopi
(there are a lot more im not thinking of)
so logically
bus bi[/quote]

Forgot one in particular; what's the singular form of "pi"?

wow.. this thread became so popular..
heres more from me..

What has 148 teeth and holds back the incredible hulk?
my zipper

Funny pick up line:
someone farted, lets get outta here..

[b]16 years later...[/b]Oh yeah! This thread is so popular. :rotfl:


[quote="Raja Oktovin"]Francesca and Giorgia play the following game. On a table there are initially coins piled up in some stacks, possibly in different numbers in each stack, but with at least one coin. In turn, each player chooses exactly one move between the following:
(i) she chooses a stack that has an even non-zero number of coins $ 2k$ and breaks it into two identical stacks of coins, i.e. each stack contains $ k$ coins;
(ii) she removes from the table the stacks with coins in an odd number, i.e. all such in odd number, not just those with a specific odd number.
At each turn, a player necessarily moves: if one choice is not available, the she must take the other.
Francesca moves first. The one who removes the last coin from the table wins.
1. If initially there is only one stack of coins on the table, and this stack contains $ 2008^{2008}$ coins, which of the players has a winning strategy?
2. For which initial configurations of stacks of coins does Francesca have a winning strategy?[/quote]
a) Easy to check that if $ n\equiv 1 (\mod 2)$ then the first player always wins.
We only need to consider the case $ n\equiv 0 (\mod 2)$
We will prove that the first player wins iff $ n\equiv 0 (\mod 4)$
The first step he divides in two stacks consist  2 and $ (n \minus{} 2)$ coins .
Second player must divides one of two stacks in two stacks consist an odd number of coins .
Next step the first player move two odd stacks .
This progress ends when the first player move last coins .
So he wins .
If $ n\equiv 2 (\mod 4)$ the first step the first player must divide in two ways :
1. Two stacks of the form $ 2t\plus{}1$ , this case the second player move two stacks then he wins .
2 . One stack of the form 4k and the other in the form $ 4t\plus{}2$
The second player will divide stack consist 4k coins in two stacks such that consist $ 2k$
If the first divide stack has 4t+2 coins , the second will move them in next steps.
This progress will end when all stacks have odd number and the second only need to move all of them . Therefore he will .
With $ n \equal{} 2008^{2008}$ the first player has a strategy to win .
b) I don't understand the question .Does he always win with number of stacks , it not depend on number of coins ?
If not he doesn't always win if the stacks of coins in the form : 2,2,2..,2  :)

You misunderstood the problem

[quote](i) she chooses a stack that has an even non-zero number of coins  and breaks it into two identical stacks of coins, i.e. each stack contains  coins; [/quote]


[hide="what i think"]
Yoda is the best!!!!!
[/hide]

Darth Vader is awesome.

Yoda is second.

Yoda! I think Mace Windu is cool too... as well as R2D2 and C3PO.

make a poll! I like YODA!!!
then R2D2
then Obi-Wan and Darth Vader
then ugly Palpatine
then Qui-Gon or whatever his name is
then everyone else lead by Mace Windu (I like the purple saber).

palpatine.
then darth vader

i hate the good guys :)

Mace Windu is my favorite, but Obi-Wan and Yoda rock also. I especially like Matthew Stover's vision for Mace and Obi.

[quote="juicybooty911"]padme hehehehe :D[/quote]hahaha Same here...natalie portman is...quite attractive.  :blush:

The way Yoda speaks is quite interesting.

Or

Interesting, the way Yoda speaks is.

Obi-Wan from the good ones and Dath Vader from the bad ones
(and yeah, i also like Padme (and Natalie Portman))

face it... the ewoks are the best characters ever made  :lol:

See avatar for details.

Second would be Mace Windu.

I like YODA.....especially the way he talks.........reminds me of...........*cough*

[quote=YODA]
Not if something to say about it, I have.
[/quote]

 :arrow: Dramatic moment.Dramatic sentence

[quote="eryaman"][quote="juicybooty911"]changed my mind, i like all of them, every single one that ever appeared. :)[/quote]Even Jar-Jar Binks?[/quote]

jar-jar is awesome :D

Darth Vader! But since I am mathjedi1, that's sorta biased

Guess what?  Go Clone Blaze Troppers!  Duh.  Everyone should know wat those r.
rofl.

----------------------------------------------------------------------------------------------------
[i]Even if you stay on the right track, youll still get run over.[/i]

................

Don't bump 5 year old threads.

lol

Now that this is bumped, we might as well keep it. YODA RULES!!! YODA CAN DEFEAT CHUCK NORRIS ANY DAY!!!


geometry
trigonometry
angle bisector
geometry solved

let $ABC$ be a triangle with incircle $(O)$. the incircle is tangent to the sides $AB$ and $BC$ at $P$ and $Q$ respectively . the internal angle bisector from vertex $A$ intersects the ray $PQ$ at $E$ .
prove that perpenticular $AC$ to $CE$.

[u]1. The syntetical method.[/u]

$m(\widehat {CIE} )=\dfrac 12 \cdot (A+C)=90^{\circ} -\dfrac 12 \cdot B,\ m(\widehat {CQE} )=m({\widehat {BQP})=\dfrac 12 \cdot (180^{\circ} - B)=90^{\circ}-\dfrac 12 \cdot B\Longrightarrow}$

$\widehat {CIE} \equiv \widehat {CQE} \Longrightarrow$ $CIQE\ -\ cyclic\Longrightarrow \widehat {CQI} \equiv \widehat {CEA},\ QC\perp QI \Longrightarrow EC\perp EA.$

[u]2. The first metric method.[/u]

I note $D\in BC\cap AI$ and I suppose WLOG that $b>c$. I observe that $DQ=DB-BQ=\dfrac{ac}{b+c} -(s-b)=\dfrac{b-c}{b+c}\cdot (s-a),\ PB=QB,\ PA=s-a$.

From the Menelaus's theorem applied to the transversal $\overline {PQE}$ in the triangle $ABD$ results the relation

$\dfrac{ED}{EA}\cdot \dfrac{PA}{PB}\cdot \dfrac{QB}{QD}=1,\ PB=QB\Longrightarrow \dfrac{ED}{EA}=\dfrac{QD}{PA}=\dfrac{b-c}{b+c}\Longrightarrow$ $\dfrac{ED}{b-c}=\dfrac{EA}{b+c}=\dfrac{AD}{2c}\Longrightarrow$

$AE=\dfrac{b+c}{2c}\cdot \dfrac{2bc}{b+c}\cdot \cos \dfrac A2\Longrightarrow AE=AC\cdot \cos \dfrac A2\Longrightarrow EA\perp EC.$

[u]3. The second metric method.[/u]

$CE\perp AE\Longleftrightarrow CE\perp AD \Longleftrightarrow CA^2-CD^2=EA^2-ED^2\Longleftrightarrow$

$b^2-\left( \dfrac{ab}{b+c}\right) ^2=\left[ \left( \dfrac{b+c}{2c}\right)^2-\left(\dfrac{b-c}{2c}\right)^2\right]\cdot AD^2$ $\Longleftrightarrow \dfrac{4b^2 s(s-a)}{(b+c)^2}=\dfrac{4bc}{4c^2}\cdot \dfrac{4bcs(s-a)}{(b+c)^2}\Longleftrightarrow True.$

[u]Remark.[/u] If $R\in AC\cap C(I,r),\ F\in QR\cap AI$, then $BF\perp AI$, i.e. $BF\parallel CE.$

The division $(A,D,F,E)$ is a harmonical division, i.e. $\dfrac{FA}{FD}=\dfrac{EA}{ED}.$

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=23677 post #7, extended problem.

  darij

I remind the extended problem mentioned by Darij, easily modified:

Let $ABC$ be a triangle and the midpoints $M,N,P$ of the sides $[AB],[BC],[CA]$ respectively.

The incircle $C(I,r)$of the triangle $ABC$ touches the sides $[AB],[BC],[CA]$ in the points $F,D,E$ respectively.

I note $X\in MN\cap DE,\ Y\in NP\cap DF$. Prove that $\{X,Y\}\subset AI,\ BX\perp AI,\ CY\perp AI.$

[u]My solution.[/u] I note the intersections $X_1,\ X_2,\ Y_1,\ Y_2,\ S$ of the line AI with the lines $MN,\ DE,\ NP,\ DF,\ BC$ respectively.

From the Menelaus's theorem apllied to the transversals $\overline {MX_1N},\ \overline {DX_2E},\ \overline {PNY_1},\ \overline {FDY_2}$ in the triangles $ABS,\ ACS,\ ACS,\ ABS$

results $\dfrac{X_1S}{X_1A}=\dfrac{X_2S}{X_2A}=\dfrac{Y_1S}{Y_1A}=\dfrac{Y_2S}{Y_2A}=\dfrac{|b-c|}{b+c}$, i.e $X_1\equiv X_2\equiv X,\ Y_1\equiv Y_2\equiv Y$.

I note $X_3\in AI,\ BX_3\perp AI,\ Y_3\in AI,\ CY_3\perp AI$.Thus,

$\dfrac{X_3S}{X_3A}=\dfrac{BS}{BA}\cdot \dfrac{\sin \widehat {X_3BS}}{\sin \widehat {X_3BA}}=\dfrac{ac}{c(b+c)}\cdot \dfrac{\left| \sin \dfrac{B-C}{2}\right| }{\cos \frac A2}=\dfrac{ac}{c(b+c)}\cdot \dfrac{\left| \sin \dfrac{B-C}{2}\right| \cos \dfrac{B+C}{2}}{\cos \frac A2\cdot \sin \frac A2}=$

$\dfrac{a}{b+c}\cdot \dfrac{\left| \sin B -\sin C\right| }{\sin A}=\dfrac{a}{b+c}\cdot \dfrac{|b-c|}{a}=\dfrac{|b-c|}{b+c}\Longrightarrow X_3\equiv X$.

$\dfrac{Y_3S}{Y_3A}=\dfrac{CS}{CA}\cdot \dfrac{\sin \widehat {Y_3CS}}{\sin \widehat {Y_3CA}}=\dfrac{ab}{b(b+c)}\cdot \dfrac{\left| \sin \dfrac{B-C}{2}\right|}{\cos \frac A2}=\cdots =\dfrac{|b-c|}{b+c}\Longrightarrow Y_3\equiv Y.$

[u]Thank you, [i]Darij[/i], for the address of this nice and interesting extended problem ! I think that you have a remarkable memory or you are very orderly (organized).[/u]


inequalities
real analysis
real analysis unsolved

My teacher told me that Kummer's test is more stronger than Raabe test in term of testing the convergent of  the series. And I am very curious to see the proof of Kummer's test.
Can anyone help me to find the proof for Kummer 's test for convergent please?
Kummer's test is in the website: http://mathworld.wolfram.com/KummersTest.html
  Many thanks! :)

It is very simple. If the limit is positive, then, for some $\delta>0$, $M\in\mathbb N$, we have $a_n\frac{u_n}{u_{n+1}}-a_{n+1}>\delta$ for all $n\ge M$. Rewrite it as $a_{n+1}u_{n+1}+\delta u_{n+1}<a_n u_n$ and add these inequalities up from $M$ to $N$. You'll get $\sum_{n=M}^Na_{n+1}u_{n+1}+\delta\sum_{n=M}^N u_{n+1}<\sum_{n=M}^N a_n u_n$. A lot of terms cancel out and you are left with $\delta\sum_{n=M}^N u_{n+1}<a_M u_M- a_{N+1}u_{N+1}<a_M u_M$. The rest should be clear :).


I hope I'm not bombarding this forum with programming posts, but I really need the help.

Well, anyways, here is the problem (I'm showing you this since I'm really bad at explaining things.):
[quote]
 N opaque rectangles (1 <= N <= 1000) of various colors are placed on a white sheet of paper whose size is A wide by B long. The rectangles are put with their sides parallel to the sheet's borders. All rectangles fall within the borders of the sheet so that different figures of different colors will be seen.

The coordinate system has its origin (0,0) at the sheet's lower left corner with axes parallel to the sheet's borders.
PROGRAM NAME: rect1
INPUT FORMAT

The order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle "on the bottom".
Line 1: 	A, B, and N, space separated (1 <= A,B <= 10,000)
Lines 2-N+1: 	Five integers: llx, lly, urx, ury, color: the lower left coordinates and upper right coordinates of the rectangle whose color is `color' (1 <= color <= 2500) to be placed on the white sheet. The color 1 is the same color of white as the sheet upon which the rectangles are placed.
SAMPLE INPUT (file rect1.in)

20 20 3
2 2 18 18 2
0 8 19 19 3
8 0 10 19 4

OUTPUT FORMAT
The output file should contain a list of all the colors that can be seen along with the total area of each color that can be seen (even if the regions of color are disjoint), ordered by increasing color. Do not display colors with no area.
SAMPLE OUTPUT (file rect1.out)

1 91
2 84
3 187
4 38[/quote]

Here are my three ways that I have figured out.  I'll list the pros and cons right here.

1.Use a two-dimensional array to represent the board, and for each square, make all of the spaces that it ocupies on the array equal to that color, since it's opaque.

pro - easily implemted, and somewhat easy to debug.
con - takes up too much space for the larger boards.

2.For each square, go through each little section that ocupies it and check whether it's in one one earlier squares area, starting from the last one layed down before it.  If it's in the area, then subtract 1 from an array of colors for tha color the square is.  Finally, add the total area of the square to whatever color the square is in to the color array.

pro - doesn't take up too much space, simple, and easy to debug
con - takes too much time when you have a whole bunch of squares.

3.For each square, check the overlap between it and the other squares.  Then add/subtract the amount of overlap to the square you need to.

pro - takes a very short amount of time
con - very very very very confusing to implement since there is a whole bunch of different ways there are overlap.

Does anyone know a better idea.  Right now, I'm using 2 and it might be the right way, just that there are probably plces where it could be cut down a little bit.  If it is 2, just tell me.  If not, then try guide me to the right way so I can try and learn a little bit oo.

Thanks to those who can help me.

hahahaha....wow..........im on the same section as well in the usaco site.  I'm stilll reading the text tho.  This problem seems challenging at first glance.................. hmmmm.......
Hey, since we're in the same boat, we could help each other out as we go through the problems.   We could use this portion of the site for questions.  word.  :P

[quote="jeffrey"]What do you know...I've been working on the same problem and have been trying the same things (and failing) that you have...[/quote]

I e-mailed Mr. Kolstad, the director of the USACO, and here's what he sent me:
[quote="Mathijs Vogelzang"]rect1:

Further Analysis by Mathijs Vogelzang

A straightforward approach to this problem would be to make an array
which represents the table, and then draw all the rectangles on it. In
the end, the program can just count the colors from the array and
output
them.  Unfortunately, the maximum dimensions of this problem are 10,000
x 10,000, which means the program uses 100 million integers. That's too
much, so we need another approach.

An approach that does work for such large cases (and it actually is
a lot faster too) is to keep track of the rectangles, and delete
portions
of them when they are covered by other rectangles.

Consider this input set:

0 0 10 10 5
5 0 15 10 10

The program first reads in the first rectangle and puts it in a
list.
When it reads a new rectangle it checks all items in the list if they
overlap with the new rectangle. This is the case, and then it deletes
the old rectangle from the list and adds all parts which aren't covered
to the list.  (So in this case, the program would delete the first
rectangle, add <tt>0 0 5 10 5</tt> to the list and then add the second
rectangle
to the list).
``

If you're unlucky, a new rectangle can create lots of new rectangles
(when the new rectangle entirely fits into another one, the program
creates four new rectangles which represent the leftover border:

+--------+      +-+--+--+
|        |      | |2 |  |
|        |      + +--+  |
|  +-+   |  --> | |  |  |
|  +-+   |      |1|  |3 |
|        |      | +--+  |
|        |      | | 4|  |
+--------+      +-+--+--+

This is not a problem however, because there can be only 2500
rectangles
and there is plenty of memory, so rectangles have to be cut very much
run out of memory.

Note that with this approach, the only thing that matters is how
many
rectangles there are and how often they overlap. The maximum dimensions
can be as large as you want, it doesn't matter for the running time.[/quote]

Did you turn that idea into a program yet? If so, I would like to see the code. My big problem is not that I don't know how to solve the problem, but I don't think I know too much about the details for C++, so I get errors that I don't know how to fix.

For example, my current version of this program gets the message "syntax error at the end of input," which doesn't really help me solve the problem.

Sorry, but I'm not really that good at finding out the compile-time errors.  Mabye if you put up your code I could help, but you would probably find a whole lot better help at a programming forum.

Did you have a working solution to this problem though?

i read in a different thread that you have progressed to the next section.
would it be possible for you to post (or pm) your code so that i can compare it with my own?

Yeah, I did but just a week ago.  I don't have acess to the code right now, but i just want to warn you - I have no comments, bad method and variable names that don't make any sense, and I code in C, not C++.


The figure consists of five congruent squares. The area of the figure is 180 square centimeters. How many centimeters are in its perimeter?
[asy]size(50); defaultpen(linewidth(0.8));
void drawSquare(int side, pair A){
 draw(A--(A.x + side,A.y)--(A.x + side, A.y + side)--(A.x, A.y + side)--cycle);
}

for(int i = 0; i < 3; ++i){
 drawSquare(1,(i,0));
}
for(int i = 1; i <= 2; ++i){
 drawSquare(1,(2,i));
}[/asy]

Call the side length of each square x.
So the area is 5x^2=180
therefore x=6.

there are 12 side lengths used to make the perimeter.
12*6=72 which is our answer


Are there some natural numbers $ n$ and $ k$ such that $ n\left(n\plus{}1\right)\left(n\plus{}2\right)\cdots\left(n\plus{}k\right)$ is a square?

[quote="SAPOSTO"]Are there some natural numbers $ n$ and $ k$ such that $ n\left(n \plus{} 1\right)\left(n \plus{} 2\right)\cdots\left(n \plus{} k\right)$ is a square?[/quote]

$ \prod_{i\equal{}0}^k(n\plus{}i)\equal{}q^m$ has no positive integers solution ($ n,k,q,m\in\mathbb N$) when $ m>1$ (see, I think, non elementary Erdos-Selfridge theorem)


geometry
parallelogram
ratio
analytic geometry
MATHCOUNTS
geometric transformation
reflection

$ ABCD$ is a parallelogram.$ E,F,G,H$  are the points on $ \overline {AB},\overline {BC},\overline {CD},\overline {DA}$ respectively.And $ \overline {AE}\equal{}\frac 1 2 \overline {AB},\quad \overline {BF}\equal{}\frac 1 2 \overline {BC},\quad \overline {CG}\equal{}\frac 1 3 \overline {CD},\quad \overline {DH}\equal{}\frac 1 4 \overline {DA}$.
If the area of the parallelogram $ ABCD$ is 1 ,then what is the area of $ PQRS$?

[hide="Solution; Generalizes Easily"]
The diagram is much messier than the solution actually is.

Start by putting all the ratio information down. Let $ [BFR]\equal{}a$, $ [CSG]\equal{}c$, $ [DPH]\equal{}d$, and $ [BQE]\equal{}b$. We can find $ [FRC]\equal{}a$, $ [GSD]\equal{}2c$, $ [HPA]\equal{}3d$, and $ [AQE]\equal{}b$ with the ratios we started out with. Now, note that $ [BEC]\equal{}\frac{BE}{BA}\cdot\frac12[ABCD]\equal{}\frac12\cdot\frac12\equal{}\frac14$, so $ [BQR]\equal{}\frac14\minus{}2a\minus{}b$. Similarly, we find that $ [QAP]\equal{}\frac38\minus{}2b\minus{}3d$, $ [PDS]\equal{}\frac13\minus{}2c\minus{}4d$, and $ [RSC]\equal{}\frac14\minus{}a\minus{}3c$. Now, it is not difficult to see that $ [PQRS]\equal{}1\minus{}\left(a\plus{}a\plus{}\frac14\minus{}2a\minus{}b\plus{}b\plus{}b\plus{}\frac38\minus{}2b\minus{}3d\plus{}3d\plus{}d\plus{}\frac13\minus{}2c\minus{}4d\plus{}2c\plus{}c\plus{}\frac14\minus{}a\minus{}3c\right)$
$ \equal{}a\plus{}b\plus{}2c\plus{}3d\minus{}\frac5{24}$. Note that $ [ABG]\equal{}[BCH]\equal{}[CDE]\equal{}[DAF]\equal{}\frac12[ABCD]\equal{}\frac12$, so from area-ratios, we have $ \frac{[CRB]}{[CEB]}\equal{}\frac{2a}{\frac14}\equal{}\frac{RC}{EC}\equal{}\frac{[CRD]}{[CED]}\equal{}\frac{\frac14\minus{}a}{\frac12}\implies a\equal{}\frac1{20}$. Similarly, $ \frac{2b}{\frac38}\equal{}\frac{\frac14\minus{}b}{\frac12}\implies b\equal{}\frac3{44}$, $ \frac{3c}{\frac14}\equal{}\frac{13\minus{}2c}{\frac12}\implies c\equal{}\frac1{24}$, and $ \frac{4d}{\frac13}\equal{}\frac{\frac38\minus{}3d}{\frac12}\implies d\equal{}\frac1{24}$, and $ [PQRS]\equal{}\frac1{20}\plus{}\frac3{44}\plus{}\frac1{12}\plus{}\frac18\minus{}\frac5{24}\equal{}\boxed{\frac{13}{110}}$.
[/hide]

[quote="PKMathew"]$ ABCD$ is a parallelogram.$ E,F,G,H$  are the points on $ \overline {AB},\overline {BC},\overline {CD},\overline {DA}$ respectively.And $ \overline {AE} \equal{} \frac 1 2 \overline {AB},\quad \overline {BF} \equal{} \frac 1 2 \overline {BC},\quad \overline {CG} \equal{} \frac 1 3 \overline {CD},\quad \overline {DH} \equal{} \frac 1 4 \overline {DA}$.
>> If the area of the parallelogram $ ABCD$ is 1 ,then what is the area of $ PQRS$? << [/quote]

From this question, it is predetermined that there is going to
be one answer, regardless of the parallelogram, so simplify
things and make it a square with area of 1 square unit.

In the xy-plane, let the vertices of the square be A(0,1), B(0,0),
C(1,0), and D(1,1).  Then the coordinates of the other points
would be:  E(0,1/2), F(1/2,0), G(1,1/3), and H(3/4,1).

By algebra, the equation for line AG is y = (-2/3)x + 1, the
equation for line BH is y = (4/3)x, the equation for line CE is
y = (-1/2)x + 1/2, and the equation for the line DF is
y = 2x - 1.

Set equations for the lines AG = BH, AG = DF, CE = BH, and CE = DF
to solve for the x and y coordinates in each intersection, respectively.

The internal points are:  P(1/2, 2/3), Q(3/11, 4/11), R(3/5, 1/5), and
S(3/4, 1/2).

Without regard to sign (clockwise or counterclockwise taking of
the order of the points), the area of this shape is:
< The following are supposed to be subscripts on x and y.** >

(1/2)[(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)]

When the eight values are substituted into this formula,
either -13/110 or 13/110 will be the result.

Area:  13/110 square units

** Ex.  x1y2 stands for the quantity (x sub one) multiplied by
the quantity (y sub two).

Better yet, you can assume that ABCD is a square with side length 1. That makes it a lot easier.

[quote="Arrange your tan"]

From this question, it is predetermined that there is going to
be one answer, regardless of the parallelogram, so simplify
things and make it a square with area of 1 square unit.

[/quote]
Ihatepie,

even if you were responding to math154,
I ALREADY had this equivalent content posted.

As a general rule, don't post a coordinate bashing solution unless an alternative can't be found (or if it's much faster). For example, in this case, you shouldn't post it because it's much faster with my method than with coordinate bashing (it's also much easier to mess up with coordinates). It's not good to assume anyway. :wink:

If you want to know a better way to bash without fractions, let the side lengths be $ 12$, and then divide the final area by $ 144$. :wink:

An instance in which coordinate bashing may have made calculations much more efficient and routine is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=77603#77603]this MATHCOUNTS problem[/url].

[quote="math154"]As a general rule, [I feel that you shouldn't] post a coordinate bashing solution unless an alternative can't be found (or if it's much faster) [but that is your choice]. For example, in this case, [I felt that] you shouldn't post it because it's much faster with my method than with coordinate bashing (it's also much easier to mess up with coordinates [in my opinion]). It's not good to assume anyway [I claim]. :wink:

If you want to know a better way to bash without fractions, let the side lengths be $ 12$, and then divide the final area by $ 144$. :wink:

An instance in which coordinate bashing may have made calculations much more efficient and routine is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=77603#77603]this MATHCOUNTS problem[/url].[/quote]

Oh, I will take EVERY opportunity to use coordinates to use to solve
and post them in solutions here as an experienced mathematician
who can show some students another way to solve certain problems.

Your insistence on limiting methods of solving to others is
ONLY opinion;  I amended your post with square brackets to
reflect this.  It broadens readers to different techniques
used within a method such as mine and/or others, AND it keeps
them from seeing the same method over and over by you.
(Also, I don't resort to typing smileys to punctuate
my comments.)


Let ABC trinagel with 0<A,B,C<pi/2. Prove: (sinA)^sinA . (sinB)^sinB .(sinC)^sinC > (2/3)^[( 3 . /sqrt[3](2) )/2]

[quote="ineq"]Let ABC triangle with $ 0<A,B,C<\frac{\pi}{2}$. Prove: $ (\sin A)^{\sin A}\cdot (\sin B)^{\sin B} \cdot(\sin C)^{\sin C} > \left(\frac{2}{3}\right)^{\frac{3\sqrt[3]{2}}{2}}$[/quote]
Is that what you mean?


probability
combinatorics unsolved
combinatorics

From a big box with 101 balls, numbered from 1 to 101, we pick 2 balls. What is the probability that the sum of the numbers of the 2 balls is 52?

The number of ways to pick up any two balls is $\dbinom{101}{2}$.
The ways that the sum is $52$ are $1,51$ $2,52$ $...$ $25,27$ and the permutations of those which are $50$.


logarithms
limit
trigonometry
calculus
factorial
Support
Pascal\u0027s Triangle

Hey, does anybody know why $0^{0}$ is undefined?  Can anyone tell me why $0!$ has no answer.  I was just wondering.

For example $\frac{0}{4}$ equals $0$,
but $\frac{4}{0}$ is also undefined.

Just curious as to why this is.  Does anyone have an explanation.  A scientific one would be best.  Not just like "because it is..." or something like that.

You know that $\frac{0}{0}$ is an indetermined value, then $\frac{0}{0}=0^{1-1}=0^{0}$, which is the same.

Nonmemory why $0!=1$

I think it's because $0^{0}=\frac{0^{1}}{0^{1}}$, which is undefined.
Doesn't $0!=1$?

Thank you, that makes a lot of sense.  But do you know WHY that it works like that.  Why does 0+0=0?  Why not 2 or 3.  You're helping me out a lot, but for every answer I get, I have about 20 more questions. (Like the game)

EDIT: @vamathletes

Huh?  I didn't know that 0! equaled 1.  Why, you're multiplying 0 by nothing else.  And even if you were, you're still multiplying by 0.  Why would it equal 1?

P.S. Does a question like this really belong on this forum or should I move it to the Intermediate?

[quote="vamathletes"]Doesn't $0!=1$?[/quote]
Yes but I don't remember how to prove it (when me they taught it said to me that it was a definition)

Yes - $0! = 1$ is given as a definition.

That was. Thanks for the culture bath  :rotfl:

[quote="rd5493"]Hey, does anybody know why $0^{0}$ is undefined?[/quote]

This actually depends on your definition of exponentiation, which depends on what math level you've gotten to so far.

[quote="rd5493"]Why does 0+0=0?  [/quote]

One way to define zero is the unique number such that for any other number $a$ we have $a+0 = a$.  So this follows by definition.

[quote="rd5493"]Huh?  I didn't know that 0! equaled 1.  Why, you're multiplying 0 by nothing else. [/quote]

This is a manifestation of the principle of the "empty product."  The empty product is the product of the elements of the empty set, which is a set with no elements.  Traditionally, and for convenience, this product is defined to be $1$.  In other words, the number of ways to permute the empty set is $1$ - the empty set itself.

[quote="vamathletes"]I think it's because $0^{0}=\frac{0^{1}}{0^{1}}$, which is undefined.
[/quote]

if so, then $0^{k}=\frac{0^{a+k}}{0^{a}}$ must be undefined as well.

But of course the identity $\frac{x^{a}}{x^{b}}= x^{a-b}$ is only valid when $x \neq 0$.

Again, it depends on which definition of exponentiation you are using.

[b]Definition:[/b]  $a^{n}= \underbrace{a \times a \times ... \times a}_{\text{n times}}$.  This is probably the most basic definition, and does not define what it means to take [i]any[/i] number to the power of $0$ except as the empty product, which is then $1$.

[b]Definition:[/b]  $a^{0}= 1$ and $a^{n+1}= a \cdot a^{n}$ for all integers $n$.  Here we have $0^{0}= 1$ by definition.

[b]Definition:[/b]  $a^{b}= e^{b \ln a}$.  This is the least basic definition, and forces $0^{0}$ to be undefined.

[quote="Teki-Teki"]Also, if one considers the limiting case and you use L'Hopital's rule (which is rigorous since 0^0 is a limiting case), then there are all sorts of fun ways to analyze this.  For example, what happens if $0^{0}= \lim_{n \rightarrow 0^{+}}\frac{\sin{x}}{x}$ with $x$ in radians? [hide=" :?: "]1[/hide]

but then again, calculus is not as rigorous as torajirou's last definition i believe.[/quote]

well this is a HSB topic, so special limits shouldn't be included =P

but shouldn't it be read:

$0^{0}= \lim_{x \rightarrow 0^{+}}\frac{\sin{x}}{x}$

(i changed the variable in which x approaches 0 from the right hand side instead of n)

and besides, isn't

$\lim_{x \to 0}\frac{\sin x}{x}=1$?

[quote="rd5493"][quote="nayel"]well, there was a problem in our BdMO this year:
Solve (x,y) in real numbers where $x^{x}=y$ and $y^{y}=x$.
some of the contestants got the pair $(0,0)$ a solution and some did not. my question is, is (0,0) a solution?
it doesn't matter to me now anymore, because i was the third top scorer in my category.  :winner_third: but yet, there's always a second time...[/quote]

UMMMM....wouldn't the solution be $(1,1)$?[/quote]

yes. and not only that, $(-1,-1)$ is a solution too (which i did not find). but the problem asks for all possible solutions. and i found $(0,0),(1,1)$ so my question is, will $(0,0)$ be counted as a solution?

I dont think it would be.  $0^{0}$, as far as i know, is undefined. period.

[hide="0^0"]It's 0/0, wich can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and so on. It can even be $\pi$.[/hide]

[hide="0!"]That's 1!/1=1. But some negative number, when put in a factorial, gives

$(-n)!=\frac{(-1)^{n+1}(n-1)!}{0}$

where n is natural.[/hide]

[quote="lkryptonicl"]Idk if I'm right but I think of 0! as how many ways can i arrange 0 things? the answer to that is just 1 no?[/quote]

That's right. Similarly, ${n\choose0}=1$ by the same reasoning (pascal's triangle supports this too.

[quote="lkryptonicl"]Idk if I'm right but I think of 0! as how many ways can i arrange 0 things? the answer to that is just 1 no?[/quote]
Yes, the answer is $1$ way which is empty set.


Princeton college geometry AMC AIME ARML calculus

I'm helping organize Pi Day at school.  Any ideas for what to do?  My school isnt crazy about math, but we do want to do something.  What are you doing at your school?

eat pie

none of that "pi is approximately 3.14" bs, just eat pie

Eating pie is good. Most of the cooler math teachers do it.

I think we are eating pie. But we haven't heard anything for like two months, so I don't know.

You can also talk about pi, but you get a lot of people to show up if you serve pie.

at my school we can sign up to be in the contest to recite digits of pi, then everybody eats pie
im not sure exactly about the pie eating part, maybe just the winners get pie  :maybe:
i used to know 66 i need to study more

These kind of things are one of many ways of showing nonmathematical nonsense as if it is of highest mathematical importance...

I'm not quite sure what we're doing at our school; I beleive u used to be able to throw pies at teachers, but I think they're planning to change that... I suggested a pi reciting competition. (I only know about 180 digits though.)

Well our school doesnt have pi day but you could ask all the math geeks to like write the digits of pi all through the school and have em stick the pi symbol onto lockers.

[quote="Xevarion"]Princeton Math Club has a pie eating contest. I got 4th place last year :(:(:([/quote]

how many pies did you eat?

I was going to post this!!!!! but oh well. My school is very untalented in the math section. Im a 7th grader at school and i'm in geometry: a class that does not exist. (cause its too high level)

Therefore we arn't gonna do anything.

By the way, Happy Early Pi day!!!!

[quote="#H34N1"]Memorizing pi up to 1000 digits :P[/quote]

Wow good luck..........

[quote]Pizza Party.[/quote] why not pi(e) party? $\pi$

[quote="bpms"][quote="PenguinIntegral"]I thought Pi day was on the 22nd of July?[/quote]
In Europe.[/quote]Indeed. Because instead of writing 3-14, they write 14-3, so it seems more logical for them to have 7-22 which is 22/7 over there.

[size=150][b]HAPPY PI DAY!!![/b][/size]

Our school's on spring break.

Happy birthday to all the $\pi 's$ out there.

Happy Pi Day!

Among other things, my school is having a raffle to throw pies at some teachers.

[quote="jhredsox"]what did you guys do at 1:59:26 pm/am?

jorian[/quote]

got yelled at by my spanish teacher forgot to check the time =)

oh wel the real pi day is in 8 years hahaha and i memorized i think 60 digits of pi =)

[quote="randomdragoon"]Why doesn't $e$ get any love?

Besides the fact that there aren't 71 days of February.[/quote]

hahaha true =)

e isn't appreciated because there is no significant value (to my classmates anyway) to their algebra 2 curriculm (yet) but once they start some calculus and logarithms they'll soon appreciate e...muwahahaha!

[quote="D3m0n Shad0w"]Lol i think pi day should be on 3/14

lol  :rotfl:[/quote]

oops pi day is on 3/14....3.14159265...lol thats all i know  :P

[quote="star99"][quote="InsanelyBored"]
Speaking of pi, theres this savant (saw it on 60 minutes) who memorized like 25k digits and can "sense" numbers. and he learned  polish i think in a week. some people are too smart.[/quote]

i saw that.  i dont know whether it was polish, but it was some crazyhard language that he learned in a week

he visualizes numbers as patterns and colors and emotions, so they just pop into his head[/quote]

Hey, that's how I remember numbers... :o

[hide="my pathetic pi memorization: 34 digits"]
3.141592653589793238462643383279504
[/hide]

I'm sure I got a number wrong and someone can do better...

If you want to write messages consisting of strings of digits of pi, go do it somewhere else.


Solve the equation [sqrt(x)]=sqrt[x], where [a] is the integer part of a real number a.

I don't think this post belongs here.
Anyway, $\sqrt{[x]}$ has to be an integer so $[x]$ has to be a perfect square.
So $x = p^2 + q$ where $p \in \mathbb{Z}$ and $0 \leq q < 1$.
All these numbers work.


An Australian bushman hunts kangaroos with the following weapon, a heavy rock tied to one end of a light vine of length 2 m.
He holds the other end above his head, at a point 2 m above ground level, and swings the rock in a horizontal circle.
The cunning kangaroo has observed that the vine always breaks when the angle (theta) (measured between the vine and the vertical) reaches 60 degrees.
At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit?

[quote]He holds the other end above his head, at a point 2 m above ground level, and swings the rock in a horizontal circle. [/quote]

Uhm... I don't get it. A vertical circle would make much more sense :huh:

I dont think ur problem is feasible...(unless I have misinterpreted the problem)...do you move to say that the vine moves such that the curve traced by the thread on revolution is a cone with vertex as ur head(I mean the bushman's!) and its base above the vertex.Well that means that both tension $T$ and the weight $mg$ act vertically downwards and hence the stone cannot be in equilibrium...In the othercase the initial height of release of the stone rmains indefinite.


[quote="ElChapin"]Prove that for all $ a$, $ b$, $ c$, $ d$ and $ e$ nonzero integers there are infinitely many integers $ k$ for wich the number $ (ad\minus{}bc)^{2}\minus{}4ek$ is a perfect square.

 :)[/quote]

Hint

[hide]Set the expression in terms of (ad-bc)^2-l^2=4ek, where l is a positive integer.[/hide]


What is the sum of the digits of the square of $ 111,111,111$?

$ \textbf{(A)}\ 18 \qquad
\textbf{(B)}\ 27 \qquad
\textbf{(C)}\ 45 \qquad
\textbf{(D)}\ 63 \qquad
\textbf{(E)}\ 81$

is there a good way to do this besides memorizing that mess or bashing it out?

I just did it like this:
There has to be a $ 1$ at the beginning, then a $ 2$ in the second spot all the way up to a $ 9$.  Then, however, our first number in the multiplication is used up of all its $ 1's$.  So, now we're just going back down to $ 1$, as slowly, each number in turn after we multiply it out is running out of its digits.  The sum is thus $ 1\plus{}2\plus{}3\plus{}...\plus{}9\plus{}8\plus{}7\plus{}..\plus{}1\equal{}\frac{90}{2}\plus{}\frac{72}{2}\equal{}\boxed{81}$.

Nevertheless, it is easier to just multiply them out.  It takes no more than $ 10$ seconds.

Alternatively, experiment with smaller numbers. I've already seen problems similar to this, so I knew what to do.

this is where ftw helped me...

theres a question on ftw asking for just that, the square of 111,111 or something like that... so the second i saw this, it clicked. also, $ 1^2\equal{}1$ $ 11^2\equal{}121$, $ 111^2\equal{}12321$, ... $ 111111111^2\equal{}12345678987654321$, and the sum is $ \frac{(8)(9)}{2}\cdot2\plus{}9\equal{}72\plus{}9\equal{}\boxed{81}$

The stupid stuff works, guys. Think multiplication algorithm.

A bit faster way to add those:

"going up 1" pairs with 'going down 8", "going up 2" pairs with "going down 7", etc., 8 pairs of nine + a nine, $ 9^2\equal{}81$.

this stuff fascinates little kids. this problem should bring back good childhood memories

[hide="interesting"]Consider $ (x^8 \plus{} x^7 \plus{} ... \plus{} x \plus{} 1)^2$. For $ k\leq 8$, $ x^k$ in expansion will have coefficient $ k \plus{} 1$ since $ a \plus{} b \equal{} k$ has $ k \plus{} 1$ solutions. Now, by symmetry, $ x^k$ has the same coefficient as $ x^{16 \minus{} k}$. Therefore, the expansion is $ 12345678987654321$ (since $ x \equal{} 10$).
[/hide]

there are 2 ways to do this problem.

the first involves knowing that 111,111,111 squared is equal to 12345678987654321.
(for a tip, when squaring any all 1 number such as 111 or 11111, u just take the amount of 1's there are and that becomes the middle digit making the number a palindrome. for 11 the answer would be 12321, for 11111 it would be 123454321)

ud then add 1,2,3,4,5,6,7,8,9,8,7,6,5,4,3,2, and 1 together thus giving u 81.

OR

you can add up the nine 1's, find that it equals 9, and then square it.
this would still give u 81

this was a cool problem in my opinion. hope i helped.   :D

the sum is $81$, we can write  $111,111,111^{2}$ as $(1 + 10 + 10^{2} + .... + 10^{8})^{2}$ it is enough to evaluate the sum of the coefficients of the above expression, $(1 + x + x^{2} + .... + x^{8})$ plugging $x = 1$ will give $81$


Consider an angle $ \angle DEF$, and the fixed points $ B$ and $ C$ on the semiline $ (EF$ and the variable point $ A$ on $ (ED$. Determine the position of $ A$ on $ (ED$ such that the sum $ AB\plus{}AC$ is minimum.

Let $G$ be the reflection of the point $F$ in the semiline $(ED$, and let $C'$ be the reflection of $C$ in the semiline $(ED$. It is clear that $C'$ must lie on the semiline $(EG$.

Then $AB + AC = AB + AC' \geq BC'$ by the triangle inequality, with equality when $A$ is the intersection of $BC'$ and the semiline $(ED$.


probability
vector
quadratics
linear algebra
matrix

Hello:

Can someone point out to me under what conditions one quadratic form
as shown below can be larger than another one

$q'Aq \geq  p'Ap$

where $A$ is a $n\times n$ symmetic matrix ($a_{ij}=a_{ji}$) with non-negative
elements ($a_{ij}\geq 0$) and $p$ and $q$ are probability vectors ($p_{i}\geq  0$ and $\sum p_{i}=1$)?

I need to know the conditions, as general as possible, on these
probability vectors for one has
a larger quadratic value than the other one.

Thanks for your help,

In particular, there also exist this relation between $p$ and $q$

$q_{i}= \frac{\sum_{j=1}^{n}p_{i}p_{j}a_{ij}}{\sum_{s=1}^{n}\sum_{t=1}^{n}p_{s}p_{t}a_{st}}$


function
induction
algebra unsolved
algebra

Let $S$ be a finite set. $f$ is a function defined on the subset-group $2^S$ of set $S$. $f$ is called $\textsl{monotonic decreasing}$ if when $X \subseteq Y\subseteq S$, then $f(X) \geq f(Y)$ holds. Prove that: $f(X \cup Y)+f(X \cap Y ) \leq f(X)+ f(Y)$ for $X, Y \subseteq S$ if and only if $g(X)=f(X \cup \{ a \}) - f(X)$ is a $\textsl{monotonic decreasing}$ funnction on the subset-group $2^{S \setminus \{a\}}$ of set $S \setminus \{a\}$ for any $a \in S$.

If I am not mistaken this is extraordinarily easy for a Chinese TST.  :huh:

We have to show:
(1) $\forall X,Y \subseteq S: f(X \cup Y) + f(X \cap Y) \le f(X) + f(Y)$ $\Leftrightarrow$
(2) $g_a$ is decreasing for any $a$.

For (1) $\Rightarrow$ (2) note that it suffices to show that
$g_a(X) \ge g_a(X \cup \{b\})$ for all $a \neq b \in S$, $X \subset S \setminus \{a,b\}$.
This however follows directly from (1) applied to the sets $X \cup \{a\}$ and
$X \cup \{b\}$.

For (2) $\Rightarrow$ (1) we use induction on the number $n$ of elements of $X \cup Y$.
If $n = 0$ (1) holds trivially. (1) also is trivial if $X=Y$.  So w.l.o.g. there is an $a \in S$
such that $a \in X$, $a \notin Y$. By induction hypothesis (1) holds for
the sets $Y$ and $X' : = X \setminus \{a\}$. Hence it suffices to show that
$f(X') - f(X' \cup Y) \le f(X) - f(X \cup Y)$.
But this is the same as $g_a(X' \cup Y) \le g_a(X')$, which holds by (2).

A curious note: Such an $f$ is called a [url=https://en.wikipedia.org/wiki/Submodular_set_function]submodular set function[/url], and at the forefront of discrete optimization theory these days.


What integers cannot be written as the difference of two squares?

suppose the integer is $n$ , then

$n=p^2-q^2$

consider mod 4 , then LHS$\equiv 0,1,3\mod 4$

so all $n\equiv 2\mod 4$ cannot be written as difference of square .

It remains to prove (a good exercise, in any case) that all integers $\equiv 0, 1, 3 \bmod 4$ [b]can[/b] be expressed as a [i]difference[/i] (edited) of two squares  ;)

[hide="Proof 1"]
$p^2 - q^2 = (p+q)(p-q)$, and clearly $p+q \equiv p-q \bmod 2$.  Hence any number $n$ that can be written in the form $n = rs$ where $r \equiv s \bmod 2$ can be written as $n = ( \frac{r+s}{2} )^2 - ( \frac{r-s}{2} )^2$.

If $r \equiv s \equiv 1 \bmod 2$ then we have all $n \equiv 1, 3 \bmod 4$.  If $r \equiv s \equiv 0 \bmod 2$ then we have all $n \equiv 0 \bmod 4$.
[/hide]

[hide="Proof 2"]
$k^2 = 1 + 3 + 5 + ... + (2k-1)$

Hence $p^2 - q^2$ is some sum of consecutive odd integers.  The sums of consecutive odd integers comprise all odd integers (clearly) and all even integers of the form $(2k-1) + (2k+1) = 4k$.

However, we cannot construct numbers $\equiv 2 \bmod 4$.  Consider the sequence of odd integers $\bmod 4$.  We have $1, -1, 1, -1...$.  Clearly any sum of consecutive members of this sequence can (starting from $1$) sum to $0, 1 \bmod 4$ or (starting from $-1$) sum to $0, -1 \bmod 4$, we can never have $\equiv 2 \bmod 4$.
[/hide]

$n\equiv 2\mod 4$ cannot be written as a difference of two squares.

[quote="t0rajir0u"]...all integers $\equiv 0, 1, 3 \bmod 4$ [b]can[/b] be expressed as a sum of two squares  ;) [/quote]

[b]Sum[/b] of two squares?  No number $\equiv 3 \bmod 4$ can be expressed as the sum of two squares.  Difference, yes, but not sum.

[hide="Proof of talking about sums"]All perfect squares are either $\equiv 0$ or $\equiv 1$ in mod 4.  Proof: Let $n$ be an integer.  So $2n$ is odd, $2n+1$ is even, and $n^2$ is an integer (this is pretty much common sense).  So $(2n)^2$, or $4n^2$, is $\equiv 0 \bmod 4$.  Also, $(2n+1)^2$, or $4n^2+4n+1$, is $\equiv 1 \bmod 4$.  By adding various possibilities (0+0, 0+1, 1+0, 1+1), we can only get 0, 1, and 2.[/hide]

Meh, it was probably a typo.


Let $ABC$ be a triangle,$\angle B=60^{\circ}$.Its Euler Line is $l$.$l\cap AB=M,l\cap CB=N$.
Prove that $\triangle BMN$ is an equilateral triangle.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16058 .

  darij


Proove that there exists a natural constant $ c$ such that every natural number $ n$ grater than $ 1$ can be represented as a sum of at most $ c$ prime number (not necesary distinct)

[url=http://en.wikipedia.org/wiki/Goldbach%27s_conjecture#Rigorous_results]The best known result[/url] is $ c \equal{} 6$, but I don't think it's elementary.

yes i know that but i'm interested in how to proove such a c-exist. it might be easier than prooving it for $ c\equal{}6$

[url=http://mathworld.wolfram.com/SchnirelmannsTheorem.html]You're right[/url], of course.  The key to the proof is that the set of sums of two primes has positive Schnirelmann density.

this is all seriously cool stuf:D tnx
i have one important question here. can u explain please?

i have problems with the notion $ \delta(A\plus{}B)$. how exactly is this defined? if we don't "allow" $ 0$ to be a natural number than $ A\plus{}B$ obviously doesn't iclude $ 1$ (no positive integer solutions to $ a\plus{}b\equal{}1$) which implies $ \delta(A\plus{}B)\equal{}0$ and this frankly isn't very useful
so i asume we "let" $ 0$ be a natural number. even more. we almost need $ 0$ in every sequence $ A$ or it's almost imposible to calculate with this sequence. so now arises the question what does than the notion $ A(n)$ mean. "the number of terms in the set $ A$ not exceeding $ n$"...do we count zero. well obviously from all the results on wiki and mathworld we dont... would you please explain the role of the $ 0$ here?

You just add some small numbers to the set under consideration (or $ 0$ to the initial set if you talk about sums of at most $ k$ numbers). The reason is simple: you cannot write e.g. $ 11$ as a sum of exactly $ 6$ primes.
And to define density, it's rather irrelevant where you start, you just need to look at the quotient "numbers of given type"/"numbers we consider".


geometry
3D geometry
icosahedron
symmetry

In a icosahedron we have in each edge a resistance $r$.
If we put a difference of potential in a vertece and on a another vertece, what is the equivalent resistance?

someone knows how to do this problem with graph?

Note: The difference of potential is applyed in 2 vertix of the same edge

how can we use symmetry and graphs to solve this in the most easy way?

thanks everybody

By symmetry, the current in the edge opposite the edge with the potential difference applied is zero.  (that is, if the potential is applied across $AB$, then the current in $CD$ is zero.)  That should help finish the problem.

Once you manage to draw a picture it's trivial ;)

This problem was on BAUPC some years ago, and all contestants were given small paper models of an icosahedron...

Hi everybody

if we make a 3D picture the problem it's easy to solve.

but if we make a graph of this i confuse a lot of thinks

we can generalize this problem to a polygon with n sides. if we make a graph of this we can solve. etc.

but using a graph. what is the strategy to solve?

sorry... im bad with graphs... i just know this because wikipedia

sorry for this...


logarithms
trigonometry
limit
complex numbers
algebra unsolved
algebra

Let $k \ge 2$ be a fixed integer.  For $n = 1, 2, 3, ...$ define

$a_{n}= \begin{cases}1 & \text{if }k \not | n\\-(k-1) &\text{if }k | n \end{cases}$

Evaluate

$\sum_{n=1}^{\infty}\frac{a_{n}}{n}$

The final answer has an even neater form  :wink:

Isn't it $-\infty$? Am I missing something :maybe:

[hide]
We can rewrite it as

\begin{eqnarray*}\sum_{n=1}^{\infty}\frac{a_{n}}{n}&=& \sum_{k \not| n}\frac{1}{n}-\sum_{k|n}\frac{k-1}{n}\\ &=& \sum_{n=1}^{\infty}\frac{1}{n}-\sum_{k|n}\frac{k}{n}\end{eqnarray*}.

In the last summation, replace $n$ by $kn'$ and we see that

$\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{k|n}\frac{k}{n}= \sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n'=1}^{\infty}\frac{1}{n'}= 0$.

Correct? [/hide]

Incorrect. You cannot freely rearrange a series which does not converge abolutely.

Write

$\sum_{n=1}^{\infty}\frac{a_{n}}n= \sum_{m=0}^{\infty}\sum_{n=mk+1}^{(m+1)k}\frac{a_{n}}n$

$=\sum_{m=0}^{\infty}\left(\sum_{n=mk+1}^{(m+1)k-1}\frac1n-\frac{k-1}{(m+1)k}\right)$

$=\sum_{m=0}^{\infty}\left(\sum_{j=1}^{k-1}\frac1{m+1}\cdot\frac{m+1}{mk+j}-\frac{k-1}{(m+1)k}\right)$

$=\sum_{m=0}^{\infty}\frac1{(m+1)k}\left( \sum_{j-1}^{k-1}\left(1+\frac{1-\frac{j}{k}}{m+\frac{j}{k}}\right)-(k-1)\right)$

$=\sum_{m=0}^{\infty}\frac1{(m+1)k}\sum_{j=1}^{k-1}\frac{1-\frac{j}{k}}{m+\frac{j}{k}}$

This does show that the sum converges, since the inner sum is $O(m^{-1}),$ making the whole sum over $m$ a sum of terms that are $O(m^{-2}).$ However, contrary to paladin8's claim, the sum is not zero. This calculation has demonstrated that it is positive.

My grouped series turned out to be absolutely convergent, but that is not the case with the original series. The original series is conditionally convergent and hence may be rearranged to any sum, including zero (paladin8) or $-\infty$ (perfect_radio). But if we preserve the original order of the series, it converges to a positive sum.

Rust has a very interesting computation, but I find his notation hard to follow. Here is a "cleaned up" version of what Rust said:

Let $\omega=e^{2\pi i/k}$ be the first nontrivial $k$-th root of unity.

We claim that $a_{n}=-\sum_{j=1}^{k-1}\omega^{nj}.$

(Note that this sum is usually presented with one more term, the $j=0$ term. By leaving that term out, we have shifted the answer by 1, as desired.) Then

$\sum_{n=1}^{\infty}\frac{a_{n}}n=-\sum_{n=1}^{\infty}\sum_{j=1}^{k-1}\frac{\omega^{nj}}n$

$=\sum_{j=1}^{k-1}\left(-\sum_{n=1}^{\infty}\frac{\omega^{nj}}n\right)= \sum_{j=1}^{k-1}\log(1-\omega^{j}).$

This is the complex logarithm, and this is a series with both real and imaginary parts. But the orginal series was real, so the imaginary parts must add to zero. Hence we need only concern ourselves with the sum of the real parts, and the real part of $\log w$ is $\ln|w|.$ Our original sum is now:

$=\sum_{j=1}^{k-1}\ln\left|1-\omega^{j}\right|$

But (draw the picture!!)

$\left|1-\omega^{j}\right|=\left|1-e^{2\pi ij/k}\right|= 2\sin\left(\frac{\pi j}k\right).$

So our original sum is $\sum_{j=1}^{k-1}\ln\left|2\sin\left(\frac{\pi j}k\right) \right|.$

[Which, the problem poser hints, can be further simplified.]

[quote="Kent Merryfield"]My grouped series turned out to be absolutely convergent, but that is not the case with the original series. The original series is conditionally convergent and hence may be rearranged to any sum, including zero (paladin8) or $-\infty$ (perfect_radio). But if we preserve the original order of the series, it converges to a positive sum.[/quote]
My mistake was another one. I incorrectly supposed that $a_{n}=-(n-1), \, \forall k | n$, in which case you do get $-\infty$.... :(

Anyway, I think that we have that $a_{nk}= 1+\frac12+\ldots+\frac1{nk}-1-\frac12-\ldots-\frac1{n}, \, \forall n \geq 1$, so if the sequence has any limit, then its $\ln k$ (I think that we can modify this to give an actual proof).

[hide="The simple answer."]Since the series converges, its sum is the limit of any subsequence of the partial sums. Let us compute

$\lim_{M\to\infty}\sum_{n=1}^{Mk}\frac{a_{n}}n= \lim_{M\to\infty}\left(\sum_{n=1}^{Mk}\frac1n-\sum_{j=1}^{M}\frac{k}{jk}\right)$

$=\lim_{M\to\infty}\left(\sum_{n=1}^{Mk}\frac1n-\sum_{j=1}^{M}\frac{1}{j}\right)$

Now use the fact that as $N\to\infty,\ \sum_{n=1}^{N}\frac1n=\ln N+\gamma+o(1)$ for a certain fixed constant $\gamma.$ (Note that we will not need to know the value of $\gamma$ in the argument that follow.)

$\lim_{M\to\infty}\sum_{n=1}^{Mk}\frac{a_{n}}n= \lim_{M\to\infty}\left[(\ln(Mk)+\gamma+o(1))-(\ln(M)+\gamma+o(1))\right]$

$=\lim_{M\to\infty}\left(\ln\left(\frac{Mk}M\right)+o(1)\right) =\ln k.$[/hide]

And perfect_radio's most recent conjecture is correct.

Very nice :)  The book the solution gives relies on dealing with the quotient and remainder when the $n^{th}$ partial sum is divided by $k$, which I found an interesting approach.


Hi,

I'm planning to write another Mock AMC this year. I only have #1-10 but if many people are interested in taking this, I'll try to finish around Mar. 17th. Unlike previous AMCs, this one will be based on the new scoring scale (6 pt correct, 1.5 pt blank, and 0 pt incorrect) so you can guess where you're at for next year's AMC. I'm not sure how close the difficulty of the test will be since no one has taken the new test yet but let's see how it goes.

Anyone interested? Post here. I'll only do it if there are at least 10 people.

Let's get some AMC action going, when is this?

[quote="Adunakhor"]I'm interested. Will it be a 12 or 10?[/quote]

This will be 12. 6 people interested so far. Yay!  :D

[quote] This will be 12. [/quote]

Oh ok, good   :D I'm sure there will be more than 10 people interested

interested.  The 17th is right during spring break, so I shouldn't have scheduling issues either.

[quote="scorpius119"]i'm interested. Time to see how quickly I can improve to the point where I get 150s  :rotfl: (like that's ever going to happen)[/quote]

Remembering your score on this year's Lehigh, I'm pretty sure you have chance to get 150!  :)

P.S. So far 8

I'll do it, I really need to work on AMC.

Count Me in By the Way How does it work? I have never done a mock test on here and just wondering how it went.  Thanks

basically someone posts the test at a designated time.  At the end of 75 minutes you send the person your answers, and he/she grades them.  Then a discussion thread is created to discuss problems/results.

It seems like a good deal amount of people are interested so I'll finish this Mock AMC. I wrote #1-19 and I'll probably finish #20-25 sometime today or tomorrow. Hopefully, by sometime in next weekdays, I should be able to make them in PDF and post it on say, Friday I think (or earlier).


linear algebra
matrix
superior algebra
superior algebra unsolved

give an example of $ C^*$-algebra and positive elements $ A,B,C,D$ such that $ A\minus{}c$ and $ B\minus{}D$ are positive and $ AB\minus{}CD$ isn't positive...

$ AB \minus{} CD \equal{} (A \minus{} C)B \plus{} C(B \minus{} D)$ is a sum of products of positive elements. so in the commutative situation (where the product of self-adjoint elements is again self-adjoint) this is positive. a counterexample has to be noncommutative. take $ C \equal{} 0$, then we only have to find positive elements $ A,B$ whose product $ A*B$ is not self-adjoint. consider the $ C^*$-algebra $ M_2(\mathbb{C})$ and $ A \equal{} \begin{pmatrix} 1 & 0 \\
0 & 0 \end{pmatrix} , B \equal{} \begin{pmatrix} 1 & \minus{} 1 \\
\minus{} 1 & 1 \end{pmatrix}$.


Let ABC be a triangle. Prove that:

$R\cdot a^2 + R\cdot b^2 + R\cdot c^2 \leq 3abc$

Hey there!!

you replace $R=\frac{abc}{4S}$. then divide with abc and you get that:

$a^2+b^2+c^2\leq 12S$ which was solved somewhere on this forum!!! cheers lads!!!!!!!

:D


$ ABC$ is triangle with $ \angle C\equal{}90$. $ CD$ is the altitude from $ C$. $ I,I_1,I_2$ are the incircle of $ ABC,ACD,CDB$ respectively. Prove that $ CI\perp I_1I_2$.

It's a classical result , moreover $ I$ is the orthocenter of $ \Delta CI_1I_2$.  :) It's a  pure angle chasing problem

 [geogebra]0d821edcc68072f6db4b56a254b6df63e742fc0e[/geogebra]


Is there exist such a $ "n"$ which has exactly 2000 prime divisors and $ n$ can divide $ 2^n \plus{} 1$
Help me please  :blush:

It's IMO 2000.  :wink:
http://www.mathlinks.ro/resources.php?c=1&cid=16&year=2000


function
integration
limit
Functional Analysis
advanced fields
advanced fields unsolved

A question which I would like some help with please:

Let $ 1\leq p<\infty$ and let $ X$ denote the space of complex-valued continuous functions defined on the interval [0,1] equipped with the $ L^p$ norm. Is this space a Banach space with the norm defined?

So really I need to show if the space is complete or not, i.e. that every cauchy sequence converges to a limit in $ X$, or that I can find one which converges to something outside $ X$. Can anyone give me some pointers on where to start?

Thanks

We know that the uniform limit of continuous functions is continuous. But a sequence can converge in the $ p$ norm without being uniformly convergent - so perhaps we should be looking for a counterexample.

Pick a target function $ f$ to converge to that's not continuous - a step function, perhaps, with a jump discontinuity.

Choose a sequence $ f_n$ of continuous functions that converge as simply or nicely as you can manage to $ f.$ (Personally, I'd probably make the $ f_n$ piecewise linear and continuous - they're pretty easy to explain and draw.)

Compute $ \|f_n\minus{}f\|_p$ for your example. Does it tend to zero? If it does, then it would follow that the sequence $ (f_n)$ would be Cauchy in the $ p$ norm.

Did you try the sequence $ \{x^n\}_{n\in\mathbb N}$?

No I haven't tried that sequence. I don't understand how that would work... actually I don't see how its a Cauchy sequence and how you explain what it tends too... Sorry I'm a bit stupid.

Well, with the functions mentioned above you can build up a counterexample: Take $ f_n(x) \equal{} \min\{1,(2x)^n\}$. Then use the fact that the $ L^p$-norms of $ \{f_n\}$ restricted to $ [0,\frac {1}{2}]$ tend to zero to show that $ \{f_n\}$ is Cauchy in your function space. Now consider the pointwise limit of $ \{f_n\}$ to get your result.

I'll go with my example rather than Third Edition's because I like piecewise linear continuous functions - but the essence of my example is the same as his.

For $ n>2,$ let $ f_n(0)\equal{}0, f_n\left(\frac12\minus{}\frac1n\right)\equal{}0,f_n\left(\frac12\right)\equal{}1,$ and $ f_n(1)\equal{}1,$ and connect those dots by line segments.

The pointwise limit is $ f(x)\equal{}\begin{cases}0&0\le x<\frac12\\1&\frac12\le x\le 1\end{cases}.$ Note that $ f$ is not continuous.

$ f_n(x)\minus{}f(x)$ is zero except on $ \left(\frac12\minus{}\frac1n,\frac12\right),$ where it takes on the value $ n\left(x\minus{}\frac12\plus{}\frac1n\right)$

With the appropriate change of variables, we get that

$ \|f_n\minus{}f\|_p^p\equal{}\int_0^{\frac1n}n^pt^p\,dt\equal{}\frac1{n(p\plus{}1)},$ or $ \|f_n\minus{}f\|_p\equal{}\frac1{\sqrt[p]{n(p\plus{}1)}}.$

So $ \lim_{n\to\infty}\|f_n\minus{}f\|_p\equal{}0,$ from which it follows that $ \|f_n\minus{}f_m\|_p\le\|f_n\minus{}f\|_p\plus{}\|f_m\minus{}f\|_p$ can be made arbitrarily small for sufficiently large $ n$ and $ m.$

So this is a Cauchy sequence. If there were any continuous function $ g$ that it converged to, we would have to have $ \|f\minus{}g\|_p\equal{}0,$ which can't happen, not for a continuous $ g.$


John writes a number with 2187 digits on the blackboard, each being a 1 or 2. Judith creates a new number from John's number by reading his number from left to right and wherever she see a 1 writing 112 and wherever she sees a 2 writing 111. (for example, if John's number begins 2112, then Judith's number would begin 111112112111). After Judith finishes writing her number, she notices that the leftmost 2187 digits in her number and in John's number are the same. How many times do five 1's occur consecutively in John's number?

(a) 182     (b) 183     (c) 184     (d) 185     (e) 186

er...wow...

well, let me state the obvious.  for it to be five 1's in Judith's number, it must be the sequence 2-1 in John's number.

[b]a=182.[/b]
My solution is too long to go into here.
I just say it has something to do with $2187=3^{7}$, repetition x3 of strings, and changing the last digit of a string.
Here is a table of repeats for if John's number is $3^{n}$, and m is the number of strings of 5.
n 0  1  2  3  4  5  6  7   8    9
m 0  0  0  2  6  20 60 182 546 1640
For m, starting from n=4, it alternates x3 and x3+2.

Scary, and I probably got it wrong too!

EDIT: Here's a link I found for the pattern!
[url=http://www.research.att.com/~njas/sequences/?q=2%2C6%2C20%2C60%2C182&sort=0&fmt=0&language=english&go=Search]Link[/url]

It says the formula for term a (counting n=3, m=2 as the 1st term) is
$\lfloor\frac{3^{a+1}}{4}\rfloor$

[hide]
The answer is A.
By brute force
John $1 \ 1 \ 2 \ 1 \ 1 \ 2 \ 1 \ 1 \ 1 \ 1 \ 1 \ 2 \cdots$
Judith $112 \ 112 \ 111 \ 112 \cdots$
There are $5$ ones per $12$.
Divide and floor $\frac{2187}{12}= 182$
[/hide]

Umm waaait ...
@ AstroPhys:
Is it just me, or does your solution imply that John's number repeats every 12 digits or so?  ... Because I wrote out the first 81 digits (while finding my pattern) and it  doesn't seem to be 5 ones for every 12 digits.

Still, we seem to have the same answer, that has to count for something.

Isn't this a problem from some year's AMC? Or was it a Canadian math competition?
Anyways, I tried it and it made my head hurt for like 3 hours afterwards.

This was part of a Canadian Grade 9 Pascal competition.
It took me a whole entire day to understand everything.
Still a little confused today as well.

CircleSquared
 :huh: I assumed so, though it appears not to be so.


blogs
Support
geometry
geometric transformation
reflection

Don Imus called the rutgers girls basketball team "nappy-headed hoes" on his radio talk show. What do you think should be done about it?

My opinion:

What Imus said was very offensive, and although there is freedom of speech, he shouldn't have said it. However; rappers(see signature) say that kind of thing about women all the time. Imus' show should be suspended, but not cancelled. He should also apologize for his comment.

see soxfan14's signature
case dismissed

anyways...
rappers shouldn't be allowed to say these things anyways
i think he should be fired, this isn't the kind of thing a civilized person wants to hear on the radio  :P
and they should ban from the radio/whatever they can ban from all rappers that are inappropro
its not even good music anyways, people just like it because its "cool"
yes, i know thats probably a gross generalization and only my opinion, so dont get mad at me for saying it, just trying to prove a point

I absolutely hate rap. However, I'm one of the biggest fans of free speech. People should be free to say anything they want, even if it's slanderous, objectionable, or just plain rude. Thus I say it was really distasteful what he said, but I also believe that he shouldn't be forced not to say it. However, if MSNBC wants to drop him, that's their decision; similarly, if someone else offers him a place, that's fine.

Consider the internet, currently a model of free speech. You can be paid to blog on various sites. However, if you decide one day to post pr0n, your blog provider may decide to fire you. But you can proceed to go get your own blog and go into business for yourself, and you can post the vilest pr0n you want and make money off advertising. That's the beauty of free speech and a free market: anyone can say anything they want, but you don't have to listen it.

Here's a really, really, really simple question:

What does Don Imus have to do with rappers?

Because I really, really don't understand why a discussion of Imus being a misogynist, racist jerk ends up as an entirely different discussion.  Imus is a misogynist, racist jerk -- not only did he have this firing coming, but it's remarkable that anyone (politicians, journalists, etc.) ever thought that he was a sufficiently respectable person that they should go on his show.  Now, tell me again what this has to do with rap music?

Also, can anyone tell me what this means: [quote]yes, i know thats probably a gross generalization and only my opinion, so dont get mad at me for saying it, just trying to prove a point[/quote] Huh?  I mean, really, what is the purpose of that sentence?  The reason I shouldn't be annoyed by your comments is because you're making an argument which you can't actually support ("probably a gross generalization," "only my opinion")?  Maybe you just shouldn't make the argument, if you have that little faith in it and nothing to support it with.

There is some connection of rappers and Don Imus:

Rappers degrade women
Don Imus degraded a group of women

Rappers don't get fired
Don Imus did

There is the connection for you

Re: Ignite: racism, offensiveness.

[quote]Rappers degrade women
...
Do research on topics before you post ignorant posts[/quote]
Uh-huh.

In a discussion about misogyny [i]in popular culture and the media[/i], a discussion of rap music would be relevant.  In a discussion about the misogyny and racism [i]of Don Imus[/i], the misogyny and racism of anyone else is irrelevant.  The fact that people seem to have trouble grasping this is somewhat embarrassing.  Indeed, the only purpose this analogy appears to serve is to divert attention from the fact that Imus is a misogynist, racist jerk.
Of course, the general weakness of the rapper-Imus analogy is itself really remarkable.
(1) It requires me to accept the equivalence of an individual and a particular comment of his to an entire class of performers, encompassing thousands of people and hundreds of thousands of songs.*
(2) It requires me to accept the equivalence between the political commentary of someone paid to make political commentary (indeed, someone who regularly hosts prominent politicians and others on his show) and the political commentary of people paid to make music.
(3) It requires me to ignore the clear racial undertones of Imus' comments.

*Why is this important?  Well, because to say, "But, Ludacris is a misogynist, too!" sounds totally absurd -- it's obviously a non sequitur.  The only way to make this not sound absurd is to associate it with the entire class of rappers.  Of course, this requires the audience to be willing to accept such a blanket assertation.  This has a lot to do with why I think this analogy reflects so badly on those who present or defend it.

Also, a general word of advice: calling moderators names is not a good idea.

[quote="ZetaX"]Aznwithabrain, would you please calm down and shut up¿
If you aren't able to talk like an adult or accept something an adult says: leave this subforum and look out for a kindergarten. And be happy that it's not me who decides on bannings (nonetheless, the admins will be informed about this).
JBL is moderator here and I don't think he's done anything wrong (I haven't seen Ignite's post, but already the fact that it's one of Ignite tells me everything I've to know ;) ).[/quote]

It's people who stereotype things like you that result in Don Imus incidents.

A quick paraphrase/summary of my post: I don't see why people immediately accused Don Imus of being racist. He mentioned nothing about race specifically, and if people are quick to associate "nappy headed hos" with African Americans, then it shows that it is them who are the racist and making connections of traits (especially derogatory ones) on the basis of race and not Don Imus.

Obviously, that post disagrees with JBL's viewpoint that Don Imus is racist and I'm guessing that was why it was deleted.

[quote]
IMUS: So, I watched the basketball game last night between -- a little bit of Rutgers and Tennessee, the women's final.

ROSENBERG: Yeah, Tennessee won last night -- seventh championship for [Tennessee coach] Pat Summitt, I-Man. They beat Rutgers by 13 points.

IMUS: That's some rough girls from Rutgers. Man, they got tattoos and --

McGUIRK: Some hard-core hos.

IMUS: That's some nappy-headed hos there. I'm gonna tell you that now, man, that's some -- woo. And the girls from Tennessee, they all look cute, you know, so, like -- kinda like -- I don't know.

McGUIRK: A Spike Lee thing.

IMUS: Yeah.

McGUIRK: The Jigaboos vs. the Wannabes -- that movie that he had.[/quote]

It is true that there was no mention of race... :huh:
I should've read more carefully
well now I can see why some people feel that Imus really didn't do anything that was very racially offensive
but
I still think Don Imus should be fired :lol:

[i]Moderator says: the remainder of this post is in reference to posts which were part of a pointless flame war and which have been deleted.[/i]

oh and

Zetax, how was I not calm (I was laughing for godsake)
Ignite, yeah I kinda messed up on the "hypocrisy" part; I take that part back

Edit:
Is nobody paying any attention to Treething's post?
Same as my posts
They're both attacks on jbl and imus

Edit2:
If you agree on not firing Imus
You're agreeing with Rosie O' Donnell (evidence on Wikipedia)
 :rotfl:

[quote="Ignite168"]A quick paraphrase/summary of my post: I don't see why people immediately accused Don Imus of being racist. He mentioned nothing about race specifically, and if people are quick to associate "nappy headed hos" with African Americans, then it shows that it is them who are the racist and making connections of traits (especially derogatory ones) on the basis of race and not Don Imus.[/quote]
This is an interesting point, but in my opinion it is not a legitimate one. He's a professional entertainer. It's his job to know his audience. It may not [i]necessarily[/i] be true that "nappy-headed hos" must refer to African Americans, but obviously a vast majority of people do make that association, so even if Imus didn't intend to make a racial slur, he should still have been aware that the comment would likely be construed as a racial slur.

Let's be clear: the fact that you can pretend that the words "nappy headed" aren't a racial slur does not, in fact, mean that they are not a racial slur.  The word "nappy" is invariably used in reference to the hair of African-Americans.  See, for example, http://www.urbandictionary.com/define.php?term=nappy .  (Why am I citing urbandictionary?  Because this is a question of common usage, not of formal meaning.)  If the Rutgers team had five white starters with curly hair, I promise you 100% that the word "nappy" would never have arisen.  (Or, for that matter, the word "ho.")  Your own naivete about racism is unsurprising, but you should know something about subjects like this before you jump into discussions where such knowledge is relevant.

Sorry to contradict you JBL, but my sister uses "nappy" to refer to white people's hair fairly frequently. (I consider her to be a fairly good indicator of pop culture trends.) She also occasionally calls white girls "hos". But again, what's important is not how he meant it himself but how the rest of the world would perceive it. (Also, I see no reason to believe that he didn't mean it as a racial slur, considering other things he's said in the past.)

[quote="JBL"]Let's be clear: the fact that you can pretend that the words "nappy headed" aren't a racial slur does not, in fact, mean that they are not a racial slur.  The word "nappy" is invariably used in reference to the hair of African-Americans.  See, for example, http://www.urbandictionary.com/define.php?term=nappy .  (Why am I citing urbandictionary?  Because this is a question of common usage, not of formal meaning.)[/quote]

Nappy, is NOT a racial slur for the refernece of black people. I have heard it used to describe whites and blacks many times with no real trend to one or the other.

The fact that you're citing urbandictionary is laughable. Urbandictionary is not authoritative at all, even on slang, nappy is a perfectly acceptable word around here racially. While it still is mostly an insult, there are no racial grounds. As for the credibility of urbandictionary of "female" let's look at their definition. http://www.urbandictionary.com/define.php?term=female
Well, I'm certaintly sure that everytime someone says female they are referring to soul sucking evil counterparts of males who **** with your emotions etc.

[quote="JBL"]If the Rutgers team had five white starters with curly hair, I promise you 100% that the word "nappy" would never have arisen.[/quote]

This argument holds no weight whatsoever seeing as it's hypothetical and offers no justification whatsoever. Heck I could say that if the Rutgers basketball team was all white, Don Imus would've 100% used even worse words because he hates white females.

[quote="JBL"](Or, for that matter, the word "ho.")  Your own naivete about racism is unsurprising, but you should know something about subjects like this before you jump into discussions where such knowledge is relevant.[/quote]

LMAO. First of all, nice ad hom, I notice you deleted your flames to other people about them being jerks and such after it being pointed out but you're still at it. Next, you accuse me of knowing nothing about these words when you just obviously proved that it is you who knows nothing about the word "ho." The word "ho" means prostitute, period. It has NEVER had any racial charges behind it and still does not. Heck, even your beloved urbandictionary agrees http://www.urbandictionary.com/define.php?term=ho with the entire first page mentioning prostitute but not black at all. Ho was most probably used on the basis of them being females and you trying to assert it being used on the grounds of race is ludacris (I like his songs) and ridicolous.

Maybe you should actually get some credible knowledge aka one not from urbandictionary before you try and discuss his words.

Ignite, you seem to be fairly talented at constructing good arguments and your writing is pretty decent too. So why do you waste those skills arguing ridiculous points? As I said before, it doesn't matter that some people (such as yourself or my sister or possibly Imus) use "nappy" without racial connotation. The point is that enough people do use it with that connotation for there to have been public uproar after Imus' comment. The fact that we here are arguing about it is sufficient evidence for that in my opinion. And as an entertainer it is Imus' job to be aware of potential public response to what he says on air.

(Just a note: I don't personally think that there's anything intrinsically wrong with what he said, on freedom of speech grounds. I don't care if this means that I agree with Rosie O'Donnell. But it was still a mistake for him to say it, because it was inappropriate for the medium and he should have known that many people would be upset by it. The purpose of regulation of this kind of stuff on the air is to prevent too many people from getting pissed off at the broadcasting companies. Imus screwed up on this front and therefore it makes sense that he be punished for it. But I think it is wrong to see this as a punishment for his racism. He's free to be a racist jerk if he wants.)

[quote="Ignite168"]The word "ho" means prostitute, period. It has NEVER had any racial charges behind it and still does not.[/quote]

And I didn't say it does.:roll:  I did say that Imus would never have called a team of white players a team of hoes, but that doesn't mean he wouldn't have done it because the term ho has a racial connotation.

[quote]The purpose of regulation of this kind of stuff on the air is to prevent too many people from getting pissed off at the broadcasting companies. [/quote]

Incidentally, I don't think there should be a law or regulation preventing Imus from being a racist, misogynist jerk on the air -- I just think it's absurd that anyone pays him to do it.

Instances in the past with Imus:

[quote]As reported by New York Times columnist Bob Herbert,[36] in the course of a 1998 interview with Mike Wallace on 60 Minutes, Imus admitted telling a producer off-camera that McGuirk was hired to perform "n***** jokes."[/quote]

[quote]Imus has also repeatedly referred to Arabs as "ragheads."[/quote]

[quote]He has berated many female newsreaders, most recently Contessa Brewer, which caused her to leave the show. After she left the show, Imus went on a tirade, saying, "With that fat ass she's got, she wouldn't be one of 'em," [a beautiful woman]. Imus said on the air, "That skank has to spend three hours with makeup in the morning." The tirade was also tied to comments that were overheard of Contessa's calling Imus "a cantankerous old fool" at a dinner in a restaurant in 2005, when she was still newsreader.[38] During Imus' show a producer also poked fun at poet Maya Angelou.[/quote]

[quote]On his December 15, 2004, show, Imus referred to publishers Simon & Schuster as "thieving Jews", and later in the show issued a mock apology, saying the phrase was "redundant". In October 1998 he described media critic Howard Kurtz as "that boner-nosed . . . beanie-wearing little Jew boy".[/quote]

[quote]Imus' routines were full of homophobic epithets, including "faggot", "lesbo", and various terms for gay sex.[/quote]

[quote]Imus has also made fun of the Irish, Roman Catholics, Italians, and other assorted peoples.[/quote]

Brief History:

[quote]Originally considered a shock jock, Imus turned his show more towards politics and news and gained public influence as a result. During the 1990s, Imus in the Morning became increasingly important as a useful, non-traditional platform for politicians to express their views and gain exposure. Former U.S. Sen. Al D'Amato (R-N.Y.) was one of the first officeholders to ingratiate himself with "The I-man", as Imus was frequently called by cast and guests. The show's many guests included prominent politicians such as Barack Obama, Joe Biden, Mitt Romney, John McCain, John Kerry, Vice President Dick Cheney, J.D. Hayworth and Harold Ford, Jr., as well as reporters and columnists from Newsweek, NBC, MSNBC, and other media outlets. In 1997, Imus was named as one of Time magazine's "25 most influential people in America"; he was also on the cover of Newsweek in 1999.

In 2005, Imus in the Morning was carried by about 90 radio stations across the United States[1], although at the time of its cancellation it was carried by 61 stations.[2] Arbitron analysis put the program among the most listened to radio talk shows in the United States,[3] with about 1.6 million radio listeners per week[4].

In an April 26, 2005, article, David Kiley wrote in Business Week magazine that Imus in the Morning commanded higher advertising rates compared to competing talk shows with larger audiences, because the Imus show was less politically partisan and its demographics were "affluent, educated and influential men". Kiley also quoted former CBS News anchor and Face the Nation moderator Bob Schieffer as saying, "I don’t know anyone in Washington who doesn’t listen to Imus or watch him on TV. I get more feedback off my spots on Imus than from my own shows."[5] (This included the former Howard Stern terrestrial radio broadcast. In comparison, Stern attracted more than 8.25 million listeners on half as many stations and Rush Limbaugh attracts 13.5 million listeners).

At the time of its television cancellation, Imus in the Morning averaged 361,000 viewers and was up 39 percent, ranking third among cable morning news programs in the Nielsen ratings, compared to 769,000 viewers of Fox and Friends on Fox News Channel and 372,000 viewers of CNN's American Morning.[6] The program was reported to generate $20 million in annual revenue for WFAN, representing approximately 25 percent of the New York anchor station's revenue. Total revenue, including affiliate advertising and MSNBC, was said to be$50 million.[7]

Imus generally selected country and western songs as bumper music, often playing them at substantial length, and he often promoted artists he liked such as Delbert McClinton, The Dixie Chicks, and The Flatlanders. Sometimes when soundbites were played as part of the news, such as a dissembling politician, a shrieking "SHUT UP!" sound effect was played, abruptly ending the sound bite.[/quote]

After reading all this on Wikipedia, I do believe Don Imus should have been fired.


algebra
polynomial
Vieta
logarithms
function
domain
complex numbers

If $ 1,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are $ 5^{th}$ roots of unity.Evaluate
$ \sum_{i\equal{}1}^{4}\frac{1}{2\minus{}\alpha_{i}}$

[hide="Solution"]In the solution below, all 'polynomial' will mean 'fourth-degree polynomial'.

A polynomial that has $ \alpha_1,\alpha_2,\alpha_3,\alpha_4$ as its four roots is $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$.

Now, we can construct a polynomial that has $ \frac {1}{2 \minus{} \alpha_1},\frac {1}{2 \minus{} \alpha_2},\frac {1}{2 \minus{} \alpha_3},\frac {1}{2 \minus{} \alpha_4}$ as its roots and use Vieta's relations.

A polynomial that has $ 2 \minus{} \alpha_1, 2 \minus{} \alpha_2, 2 \minus{} \alpha_3, 2 \minus{} \alpha_4$ as its roots is $ (2 \minus{} x)^4 \plus{} (2 \minus{} x)^3 \plus{} (2 \minus{} x)^2 \plus{} (2 \minus{} x) \plus{} 1 \equal{} x^4 \minus{} 9x^3 \plus{} 31x^2 \minus{} 49x \plus{} 31$.

A polynomial that has $ \frac {1}{2 \minus{} \alpha_1},\frac {1}{2 \minus{} \alpha_2},\frac {1}{2 \minus{} \alpha_3},\frac {1}{2 \minus{} \alpha_4}$ as its roots is $ 31x^4 \minus{} 49x^3 \plus{} 31x^2 \minus{} 9x \plus{} 1$.

The sum of the roots of the polynomial above is $ \frac {49}{31}$.

Answer: $ \frac {49}{31}$[/hide]

Nice :)
Let others also post their different solutions,I will post my totally different solution at the end.

Let $ \alpha _0 \equal{} 1$, remark that $ \alpha _3 \equal{} \overline{\alpha _2},\ \alpha _4 \equal{}\overline{ \alpha _1}$.

[hide="My Most Elegant Solution That I Have Ever Found"]
Since $ 1,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are the roots so we can write it as $ x^{5}\minus{}1\equal{}(x\minus{}1)(x\minus{}\alpha_{1})(x\minus{}\alpha_{2})(x\minus{}\alpha_{3})(x\minus{}\alpha_{4})$

Taking log both sides
$ \ln( x^{5}\minus{}1)\equal{}\ln (x\minus{}1)(x\minus{}\alpha_{1})(x\minus{}\alpha_{2})(x\minus{}\alpha_{3})(x\minus{}\alpha_{4})$
which can be written as
$ \ln( x^{5}\minus{}1)\equal{}\ln (x\minus{}1)\plus{}\ln (x\minus{}\alpha_{1})\plus{}\ln (x\minus{}\alpha_{2})\plus{}\ln (x\minus{}\alpha_{3})\plus{}\ln(x\minus{}\alpha_{4})$
Diferenciating both sides we get $ \frac{5x^{4}}{x^{5}\minus{}1}\equal{}\frac{1}{x\minus{}1}\plus{}\frac{1}{x\minus{}\alpha_{1}}\plus{}\frac{1}{x\minus{}\alpha_{2}}\plus{}\frac{1}{x\minus{}\alpha_{3}}\plus{}\frac{1}{x\minus{}\alpha_{4}}$
Now substitute $ x\equal{}2$ so the answer will be $ \frac{80}{31}\minus{}1\equal{}\frac{49}{31}$

[/hide]

@ GROVER:

cool solution, an elegant one indeed.  however, what can we do for the same problem if we want to evaluate $ \sum_{i\equal{}1}^{4}\frac{1}{1\minus{}\alpha_{i}}$  instead ?  maybe we should eliminate the $ (x\minus{}1)$ factor first ?  Also, how can we use this technique if we want $ \sum_{i\equal{}1}^{4}\frac{1}{\minus{}1\minus{}\alpha_{i}}$. ?  Will the technique still be valid ?

As far as I can remember, the $ ln(x)$ function is only differentiable on its domain, which is $ x>0$ so maybe we can't substitute $ x\equal{}\minus{}1$ for this case...  am i right ?    :maybe:

[quote="RAGHAV GROVER"][hide="My Most Elegant Solution That I Have Ever Found"]
Since $ 1,\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ are the roots so we can write it as $ x^{5} \minus{} 1 \equal{} (x \minus{} 1)(x \minus{} \alpha_{1})(x \minus{} \alpha_{2})(x \minus{} \alpha_{3})(x \minus{} \alpha_{4})$

Taking log both sides
$ \ln( x^{5} \minus{} 1) \equal{} \ln (x \minus{} 1)(x \minus{} \alpha_{1})(x \minus{} \alpha_{2})(x \minus{} \alpha_{3})(x \minus{} \alpha_{4})$
which can be written as
$ \ln( x^{5} \minus{} 1) \equal{} \ln (x \minus{} 1) \plus{} \ln (x \minus{} \alpha_{1}) \plus{} \ln (x \minus{} \alpha_{2}) \plus{} \ln (x \minus{} \alpha_{3}) \plus{} \ln(x \minus{} \alpha_{4})$
Diferenciating both sides we get $ \frac {5x^{4}}{x^{5} \minus{} 1} \equal{} \frac {1}{x \minus{} 1} \plus{} \frac {1}{x \minus{} \alpha_{1}} \plus{} \frac {1}{x \minus{} \alpha_{2}} \plus{} \frac {1}{x \minus{} \alpha_{3}} \plus{} \frac {1}{x \minus{} \alpha_{4}}$
Now substitute $ x \equal{} 2$ so the answer will be $ \frac {80}{31} \minus{} 1 \equal{} \frac {49}{31}$

[/hide][/quote]

You have to explain how to define the natural logarithm of a complex number.  Luckily this can be done by "extending" the definition from the positive reals to all complex numbers except nonpositive reals.  I do not know the specifics, but it is possible.

Kunny you are a genius man.The solution referred at your link is same as mine. :lol:

@ kunny:

i think your solution at the link fixes all wrinkles and clarified my doubts about GROVER's technique.  thanks.    :coolspeak:

To Maths 486 post#6

You can solve the original problem and the additional problems posed by you without logarithmic differentiation.

 $ 1+Z+Z^2 +Z^3 +Z^4 =(Z-a_1)(Z-a_2)(Z-a_3)(Z-a_4)$

Denote the left side of this equation by f(Z) and the right side by g(Z).  Differentiating both sides with respect to Z gives

$ f^1(Z)=1+2Z+3Z^2 +4Z^3 =(Z-a_2)(Z-a_3)(Z-a_4) +(Z-a_1)(Z-a_3)(Z-a_4) +(Z-a_1)(Z-a_2)(Z-a_4) +(Z-a_1)(Z-a_2)(Z-a_3)$

So $ \sum{\frac{1}{Z-a}}$ =$ \frac{f^1(Z)}_{g(Z)}$

Finding $ f^1(Z)$ will give no trouble but finding g(Z) will require more work.

For example,when Z=1 (your first question) the main difficulty is finding $ (1-W)(1-W^2)(1-W^3)(1-W^4)$  where W is a complex 5th root of unity. Start by regrouping this  as $ [(1-W)(1-W^4)].[(1-W^2)(1-W^3)]$ and simplify as you go using $ W^5$=1 and $ 1+W+W^2 +W^3 +W^4$=0.(my four lines of working gave the value 5).


Does anyone know any useful math "tricks" that are not taught in high schools (usually because they're too complicated and difficult to explain) ? For example, I found matrixes to be a very powerful tool when dealing with vectors. Also, not very useful, but it's cool knowing how to develop the Taylor series of sin(x), cos(x), etc. and calculating this stuff without the help of a calculator.

So, anything?

Tricks used for solving tougher math problems can be found in specialized textbooks, such as Art of Problem Solving. More advanced concepts that are taught in college and beyond are generally not needed for contest math.

i agree, get the AOPS books...there are so many *tricks* to solving math problems really fast that there's really no point listing them all here. Plus, you'll get a better understanding through those books.

Also, you can go through the contest directory to see solutions to AMC and AIME problems. I'm sure the solutions especially from AOPSers are the simplest and most efficient ways of doing each problem :)

also, i hope i don't get banned for this, but AoPS isn't hte only resource to get cool math tricks. In fact, grab any math contest book, grab its solution manual as well, and read through the solutions. Tons of tricks in there.


geometry
angle bisector
geometry proposed

Let ABC be a triangle with $ AB \equal{} BC \equal{} a$ and the angle $ ABC \equal{} 140^o$. Let AN and AH respectively be the angle-bisector and the altitude from A. Prove that $ 2BH.CN \equal{} a^2$

It is a problem in Youth and math magazine, don't post here till the deadline.


superior algebra
superior algebra theorems

Could you elaborate on what you have tried, and where you are stuck?

For example, to show that something is a group there is some small number of axioms to verify--do you need help with all of them, or just a couple?

Im a little confuse what does $ R^n$ means. Does it mean a single base, such as {${ ...,a^{ - 2},a^{ - 1},1,a,a^2},...$}?

[quote="t0rajir0u"]$ \mathbb{R}^n$ is the set of $ n$-tuplets $ (x_1, x_2, ... x_n), x_i \in \mathbb{R}$.  Presumably the addition is componentwise.[/quote]'

Oh, thanks, now I remember.


A palindrome is a number that reads the same forwards and backwards, such as 3003. How many positive four-digit integers are palindromes?

The number will be in the form abba  (1001a+110b) where a and b are digits. There are 9 choices for a, since 0 doesn't count, and 10 for b. The are no overlaps so we have 9*10 possiblities or $ \boxed {90\text{ solutions}}$

umm, couldnt you say that  given the first 2 digits of the number, you can find the last 2 digits so just find the number of 2 digit numbers.