Everyone comes to the forum with an username,
the username must be more or less having some significance to you.

Let's talk about the reason why you choose your this specific username in this forum

My username is simply my real name... :P

well, when i created a username, i didn't think i would like AoPS.  I ended up liking it. :D  I chose math becuase thats what this forum is about and i chose 92 because i was born in 1992. :)

my screen name is furious as i was fed up with having to change my name (this was about a year ago). i was genius1601 until for some reason that name wouldn't login. then i used a name that was purposely very similar to someone else's until and an admin changed it to mathguy, as that was causing confussion. so now i'm furous (pun intended). :)

you can guess mine...

I'M ASIAN! (Chinese)
and i was born on the year of the monkey...

This is really complicated, but I really made this up when I was 10, and then I stared using that username for everything else, including AoPS.  And since you can't change it, I'm kind of stuck with this one.

The h and the s are for the first two letters of my name, shravas.
The potter is for harry potter.
And the 2002 is when I first made this for something, yahoo mail.

mine is just my name which and it is also easier to remember by using my own name instead of creating different kind of username which difficult you to login in . Well , i used to forget what username I register for other website so I think i better stick to the easy one rather than register the whole thing again just because you have too much username .  :P

My username was created back in the days when I didn't have a shred of creativity. Math is for math, nerd is for nerd, and 314 is for pi. But now that it sounds kinda cool, I'm using it for everything.

[quote="ToBe"]truthfully, i have NO idea why i chose this user name. It's not my real name or anything...[/quote]
or NotToBe?  :P

I like calculus$\rightarrow$it rulz (rules)

That looks very interesting in light of your signature...

[quote="shadysaysurspammed"]the rest was plain stupid of me...[/quote]
"Think twice before you speak."

[quote="shadysaysurspammed"]and now i can't even change my username to something cool like CREED or Shady49 :mad:[/quote]
"Where there's a will, there's a way."

[quote="shadysaysurspammed"]I was pretty enthusiastic(Sorry for bad spellings) when riddler told me about this site...so i did not waste anytime and i really like eminem(the singer)  who is known as Slim Shady...so that's where shady came from...the rest was plain stupid of me...and now i can't even change my username to something cool like CREED or Shady49 :mad:[/quote]

You can get your username changed, just PM one of the administrators.

[quote=Sparta07]i made this account a few years ago when i didnt realize how absolutely brutal and horrific the spartans were and i just thought they were cool, but i dont want to change it because its kind of nostalgic[/quote]

You cannot change your username either.

[quote=Ihlathi]I found the name Ihlathi through google translate (just trying things in random languages until I liked something) and have used it basically everywhere since. It kinda feels like a second name to me now![/quote]

The forest?

[quote=Ihlathi]Yes. Forest.[/quote]

Nice (google translate :P )

[quote=megarnie][quote=ihatemath123]i hate math (123)[/quote]

then why are you on aops?[/quote]

to find balance, to complete that missing part of ourselves...

two worlds I really like: magic and math

I am a crab. I live in maryland a state known for crabs. In chess vishy anand is known as the tiger of madras. I needed an online username for lichess. there is also a crab opening that i use to troll. I also am knowingly and intentionally a hermit. Which would make me a hermit crab

[quote=aqua2026]i was reading the drug facts of my hand soap when i was creating my account.
the first ingredient was "aqua" and my year of grad is 2026 (im in 8th grade) so i combined the two!

i also like the color aqua/teal/blue so i guess that makes sense too.
funny backstory :D[/quote]

aqua means water XD

My dad made this username.

mega arnie(a nickname that hasn't been used for 8-9 years)

oh lol i always prounounced it as:

may - jar - nee

didnt even see the word mega!!

i thought it was meg+arnie (like a name) lol


calculus
integration
logarithms
Putnam
calculus computations

Not quite the same problem as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5[/url].

The $ \int_1^\infty$ part can be computed using the substitution $ y \equal{} \frac1{x}$. We are then left with $ \int_0^1 \frac { \minus{} \ln x}{x^2 \plus{} 1} \, dx$, which, after writing $ \frac1{1 \plus{} x^2}$ as a series, turns out to be equal to Catalan's constant.


The sector $OAB$ of a circle with center $O$, has a perimeter of $12$. Determine the radius of the circle which maximizes the area of the sector.

Looks simple, but I'm not getting it :( Anyone know how? Full solution please.

[hide]Let $r$ be the radius, $\theta$ be the angle in [i]radians[/i]. The perimeter is given by $r(2+\theta )=12$, and the area is $\frac{1}{2}\theta r^2$. Substitute $\theta=\frac{12}{r}-2$, so that the area is $6r-r^2=9-(r-3)^2$. Therefore area is maximized at $r=3$, $\theta=2$.[/hide]


inequalities
function
triangle inequality
inequalities proposed

For $m_a,m_b,m_c$ medians of a triangle of sides $a,b,c$ prove

$\sum \frac{a^2}{m_b^2+m_c^2} \leq 2$

Trivial!

Since $\displaystyle a^2 = \frac49 \left(2m_b^2 + 2m_c^2 - m_a^2\right)$, etc., the inequality is equivalent to:

$\displaystyle \sum_{cyc} \frac{4}{9}\left(2 - \frac{m_a^2}{m_b^2+m_c^2} \right) \leq 2$  which turns into Nesbitt.

Using  $m_b^2\geq s(s-b) , m_c^2\geq s(s-c)  $  we  get
$\Sigma\frac{a^2}{m_b^2+m_c^2}\leq\Sigma\frac{a}{s}=2 $


ARML
geometry
circumcircle
3D geometry
Mafia

Pwned everyone.  Actually not quite.
None of the Chicago teams ranked nationally, none made it to the tiebreaker round, none won candy from super relays, and we didn't beat Minnesota in Ultimate.
To make things worse, D team didn't even beat C team lol. :/

However, I did eat a lot of food...which I most certainly enjoyed.

Hm I wonder what the A team scores were...

mreow we beat you for the first time in like EVER. this is a historic moment. xD

In honor of that moment, here's a "medal count" of placements at the Iowa site since 1995:
$ \begin{tabular}{crrr}Team&1&2&3\\Chicago&10&4&1\\Texas&3&5&2\\Minnesota&1&3&4\\Indiana&1&0&1\\Michigan&0&3&4\\Missouri&0&1&0\\Wisconsin&0&0&2\end{tabular}$

This year's tie is counted as second place for both teams.

Point P is in the plane of triangle ABC. Prove the circumcircles of the medial triangles of ABC, PAB, PBC, and PCA concur

Sound familiar?   :wink:

Hm one more thing... Arjun messed up #9 cause he thought it said cone. I think some other kid (was it Ravi?) messed up #9 for the same reason. At first, i also thought it said cone....
So I have a theory -- Perhaps the proctor/chi arml coach who read the questions out loud said "cone" instead of "cylinder" =0

I was proctoring Nebraska so I was able to see the problems real time and do them in my head.  I do not remember what Mr. Goodman read, although that is a great theory.

I also thought "cone".  I started to do some work in my head and thought "Too much work, I wish it said cylinder" and cylinder all of a sudden appeared on the page.  I then wondered why the same problem was on the INDYs twice  -- #4 and #9.

It would be interesting to know how many kids misread "cone" at the other sites.  I suspect this was a matter of quick reading by the brain and realizing cylinder made it trivial and that can't be the case for a #9.

Too bad for Arjun as he was 8/8 to that point. And with Ravi getting #10...missing #9 seems crazy.

If you look at the team scores on #9, the C-team dominated Chicago with 13.  Not sure if that's from me always reminding them to make sure they get the first question or whether they did the right thing and said "Hmm, #10 looks ridiculous.  I better spend all my time on #9."

coach fritz

Great theory?  No way!

One year I misread something and the whole room yelled at me.  That didn't happen this year.  I'm quite sure I said cylinder, as I read the question to myself before hand to make sure.

Thanks for the conspiracy theory.

Mr. G.

hey Mr. Goodman!

Did anyone on chicago arml get special awards, btw? And did ravi get anything special for being a middle schooler?

There are no special awards for younger participants. With a sixth grader winning and a seventh grader sixth, they don't need it.

Hi, Megan.

Kate Rudolph won the Douglas Cameron Baker Memorial Award (1 of 2 awarded).
Ravi got special mention for being the only one at Iowa to answer #10.
The whole Chicago A-team was the Iowa site winner for being the highest score from Iowa, not in the top 9 nationally.

It was a fun weekend, even though the Chicago score was down a bit this year.

[/quote]

#9...uh well I don't really listen to he says, I usually read ahead on the problem and try to think about how to do the problem instead of actually listening haha.
And if Ravi keeps pwning he might be one of the few (or the first? I don't know) that will have been on Chicago A for 6 years straight o_o

[quote="Aryth"]Mr. Kim was also nominated "Most Likely to be in the Mafia", but we managed to eliminate him before he could be honored. I wasn't honored for this, but I was by far Most Likely to be Killed by Chicago ARML Members. That includes coaches.[/quote]

Rofl why wouldn't we kill you first?  :O

=P

Hmm it seems that not every individual award was mentioned... I recall that Kevin gave himself the award of "Highest Individual Score that is not an 8, 9, or 10" or something like that


Here are some interesting Harvard problems I found (geometry only), I think they're great practice for contests such as the AMC/AIME

1) A circle has parallel chords of length $x$ that are $x$ units apart. The area of the circle between these two chords is $2+\pi$. Compute $x$

2) A point on a circle inscribed in a square is $1$ and $2$ units from the two closest sides of the square. Find the area of the square.

1.
[hide]The given information means that a square of side length x is inscribed in the circle. Therefore, the radius is $\frac{x\sqrt 2}2$. The area is comprised of two quarter-circles, which make one semicircle of area $\frac{\pi x^2}4$ and two triangles which can be seen to be 45-45-90. Their total area is $\frac{x^2}2$. So we have $\frac{x^2}2=2$ and $\frac{x^2}4=1$, meaning that x=2.[/hide]


How many different 10-letter arrangements of the letters in the "word" GOPALADINS are there?

I don't know how to do this when there are more than one of the same letter.

If each letter was unique, we would have $10!$ permutations.  Obviously when there is a repeated letter we have to divide that $10!$ by some number.

Since there are 2 A's in Gopaladins, try dividing $10!/2!$

In problems like this, when you have $n$ objects, $n_1$ indistinguishable of the first type, $n_2$ indistinguishable of the 2nd type, etc..

where $n_1+n_2+......+n_r=n$, there are $n!/n_1!n_2!....n_r!$ arrangements.


The content in the link given from user enescu

contains some of these errors/errors in judgment:

"m/m" should instead be "m/n"

The phrase "obvious step" is ambiguous.  To whom?  It should not have been included.  The step, IT REFERRED TO, would still have been included under different phrasing.

"1/(n(n+1)" should be "1/(n(n+1))"

660.1319, 661.1318 are not the products of  660 and 1319, and
661 and 1318, respectively.  The period should be reserved for
decimal points.  Instead, 660*1319 and 661*1318 can do it.  Or
use the raised (centered) dot for multiplication between the numbers.  These are two ways out of a larger choice of legitimate/non-ambiguous ways for multiplication symbols.


real analysis
advanced fields
advanced fields unsolved

We have already defined Lebesgue measure for two kinds of sets: finite unions of intervals and sets of Lebesgue measure zero. Show that ese two definitions are not contradictory; that is, show that the interval $[a,b]$, $a<b$, is not a set of measure zero. (Hint: Use the Heine-Borel property of compact sets.)

On the problem set, I felt like I didn't properly use the Heine-Borel theorem. Some other notes:

The set $[a,b]$, $a<b$, has measure $b-a$. The measure of a union of disjoint intervals is just the sum of their individual measures. A set has measure zero if for every $\epsilon>0$, there exists a covering of the set with measure less than $\epsilon$.


factorial
function
algebra
floor function

I got 500 as well. You have 401 multiples of 5, 80 multiples of 25, 16 multiples of 125 and 3 multiples of 625. $401+80+16+3=500$

[quote="236factorial"]How many zeros end 2005! when it is multiplied out as an integer?

500?[/quote]

[hide]there are 401 divisible by 5
there are 80 divisible by 25
there are 16 divisible by 125
there are 3 dvisible by 625

add them and get 500[/hide]

How does dividing by 5, 25, 125, and 625, help find out how many zeros there are?  PM me plz

each 0 means that we have a 10.  Each 10 is made up of 5*2
their are a lot more 2's then 5's, so just figure out how many factors of 5.

I just think real samples are more explanatory than boring formulas

[quote="Hexahedron"]well here's the formula anyway: for 2005! answer is [2005/5] +[2005/25] + [2005/125] + [2005/625] =500

*[x] is the floor function for x*[/quote]

Thanks for your first response to my question.  That kinda clears things up.  And as for the formula, do you mind posting it on my thread, "What we really need to know?"

Thanks again!!! :D

I was wondering about this because I saw a state countdown round with a similar problem, except with the number 236 factorial (where my name came from), and later a problem from 1997 about zeros in 1997!.

Thanks for the replies.


function integration trigonometry limit LaTeX

let's see if this works [tex] \frac 12 \sqrt{41} - x^2 + a_1 + a_2 +\cdots +a_n [/tex]

let $f:\mathbb{R}\to\mathbb{R}$ be a variable function such that [tex]\int^1_0 f(x)dx= 1 [/tex]. Prove that ...

[tex] \displaystyle \int^1_0 f(x)dx [/tex] ...

let's also try this \[ f(x) d x \frac 12 \]

[quote="Valentin Vornicu"]trying again [tex]\int^1_{a-b}f(x)-\frac 12^3[/tex][/quote] and again :)  [tex]\int^1_{a-b}f(x)-\frac 12^3[/tex]

[quote="Valentin Vornicu"][tex] testing\ this \ forumla \sum^n_{k=1} \frac {n!}{n-k!} = 2m [/tex] is it ok ? [tex] f(x) = g(x) \Rightarrow f^2(x) = g^2(x) [/tex][/quote]
You can use \text to include text in a formula
[tex]\text{testing this formula} \sum^n_{k=1} \frac {n!}{n-k!} = 2m [/tex] is it ok ? [tex] f(x) = g(x) \Rightarrow f^2(x) = g^2(x) [/tex]

or use \mbox to change font size
[tex]\mbox{\tiny testing this formula} \sum^n_{k=1} \frac {n!}{n-k!} = 2m [/tex] is it ok ? [tex] f(x) = g(x) \Rightarrow f^2(x) = g^2(x) [/tex]

[tex]\boxed{\displaystyle\int_{a}^{b}f\left( x\right) dx=\lim\limits_{\delta x\rightarrow 0}^{{}}\sum\limits_{x=a}^{x=b}f\left( x\right) \delta x}}[/tex]

[tex]\kappa =\dfrac{d\psi }{ds}=\dfrac{\dfrac{d^{2}y}{dx^{2}}}{\left( 1+\left( \dfrac{dy}{dx}\right) ^{2}\right) ^{\frac{3}{2}}}=\dfrac{\overset{.}{x}\overset{..}{y}-\overset{..}{x}\overset{.}{y}}{\left( \overset{.}{x}^{2}+\overset{.}{y}^{2}\right) ^{\frac{3}{2}}}[/tex]

Wonderful!

[tex]\dfrac{\sin x}{n}=\dfrac{\text{si}\hspace{-1mm}\not{n}\hspace{1mm} x}{\not n}=$ six$[/tex]  :)

[tex]\displaystyle \pi = \sum _{k=0} ^{\infty}  \dfrac{1}{16^k} \left[\dfrac{4}{8k+1}-\dfrac{2}{8k+4}-\dfrac{1}{8k+5}-\dfrac{1}{8k+6}\right] [/tex]

[tex]f(x)=\left\{\begin{array}{cc}0,&\mbox{ if } x\leq 0\\1, & \mbox{ if } x>0\end{array}\right[/tex]

This is addictive!

Let [tex]M(n)=\sum^n_{i=1}\mu(i)[/tex], Prove that [tex]M(n)=O(n^{\frac12+\epsilon}), \forall\epsilon>0.[/tex]

[quote="Valentin Vornicu"][quote="stevem"][quote="Valentin Vornicu"]however, for the page to be faster to load I strongly recommend not using the text into the pictures, only the formulaes that are necesary.[/quote]OK like this
The highest power of a prime [i]p[/i] dividing [i]n![/i] is [tex]\sum_{k=1}^\infty \left[\frac{n}{p^k}\right][/tex][/quote] nice idea using italics tags for the variable inline :D:D[/quote]
don't italicize "!"


Factor $30x^5 - 166x^4 -542x^3 + 2838x^2 + 1520x - 800$ use synthetic division not long! :)

Uhhh.. there are way too many possible rational roots...

[hide="Answer with Ti 89"]$2(x+4)(3x-1)(5x+4)(x-5)(x-5)$[/hide]

And yeah, too many possible roots according to factor theorem.


I am curious to know how much time do you guys spend on solving problems vs studying theory.

I personally think that actual problem solving is much more important than theory.  It is quite common to be able to solve a problem without understanding the theory behind it; for non-proof contests this is all that you need.  However, perfectly understanding a concept without being able to apply it to the problem is a virtually useless skill.  I also feel that doing many problems gives you a chance to discover the theory on your own.

probably about 40%.  I have to learn material before i can apply it.


Ordinarily we don't post job listings on our site, but we're a big fan of Mathcamp and they're looking for a very responsible person to do part-time work from home.  The details can be found here:

[url]http://mathcamp.org/admincoordinator.php[/url]


[quote="cincodemayo5590"][quote="Quevvy"][hide]The last side will be $\sqrt{x^{2}+x^{2}}=\sqrt{2x^{2}}=x\sqrt{2}$ by the Pythagorean theorem (reject the negative value)[/hide][/quote]

Lol, hooray for reading the problem!  :D

[hide="AREA"]The area will be $\frac{\text{base}\cdot\text{height}}2 =\frac{s \cdot s}{2}=\frac{s^{2}}2$[/hide][/quote]
Oh, I'm stupid... but my way is how to find the hypotenuse of an isosceles right triangle :P

Edit: Here's a good problem that combines both things! Express the area of an isoscles right triangle in terms of x if x is the hypotenuse.

To Quevvy's question:
[hide] Say that the length of each congruent side is y.  Using cincodemayo5590's idea, we have that the area is $\frac{y^{2}}{2}$ and using Quevvy's way we find that $y\sqrt{2}=x\Rightarrow y=\frac{x}{\sqrt{2}}=\frac{x\sqrt{2}}{2}$.  Plugging in this value for y, we find that $\frac{y^{2}}{2}=\frac{(\frac{x\sqrt{2}}{2})^{2}}{2}=\frac{x^{2}}{4}.$ [/hide]

[quote="Interval"]Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides.[/quote]

[hide]

You have triangle $\triangle XYZ$, where $XY=XZ=x$. You find $[XYZ]$. So, you use the triangle formula, and get $\frac{bh}{2}=\frac{s^{2}}{2}$. I hope I read the problem right.
[/hide]

yay! I remembered to close the hide :!:


Hmmm, any ideas for how I could recognize and meet up with AoPSers at State Mathcounts?

Especially this deals with those who are new to Mathcounts or I haven't seen quite much (e.g. suma_milli, RussianRocket)

This may sound a bit silly, but if you are eager to meet with other AoPSers, why don't you just wear an AoPS T-shirt? It won't allow [i]you[/i] to recognize other AoPSers, of course, but, at least, it will allow [i]them[/i] to recognize you as an AoPS member. :P.

That's what I always knew and thought about at first, but I wanted the other IL kids (e.g. RussianRocket, suma_milli, avec_une_h, etc.) to help brainstorm ideas. And this especially goes for the mathcounts kids from IL.

you guys could arrange to meet at a certain spot... make a giant sign saying "AoPSers" and hold it up.

Hmmm, that may be great but doesn't that mean some non-AoPSers would be clueless and they're like "huh"?

Or the aopsers could wear a nametag with their aops username and stick it on their shirt :D

Haha I like Arnyth's idea.  I'm rather weird and I think my teammates have gotten used to it by now (:P),  hopefully they won't say anything when I hold up a huge sign that says "AOPSers".

Quick questions about mc:
-To make it to state/nats do you have to have your whole team pass chapter?  If so I'm going to be worried..
-Countdown is not official in Illinois, right?

[quote="Aryth"]You can qualify for State in one of two ways:

1. Be on the top team in your chapter
2. Be one of the top two individuals who are not on the top team in your chapter.

To qualify for nationals, you must be one of the top four individuals. The winning team's coach becomes the coach of the state team. It is possible (but probably unlikely) that the state coach is not from the same school as any of the national competitors.

Countdown is not official in Illinois.[/quote]
some chapters send more than two top individuals...mine actually sends the top 4 teams

and his name is Aryth, not Arnav, suma. :wink:

Also, suma, are you in the North Suburban chapter? I am. :D

Oops sorry Aryth!  :blush:

Nope.  North suburban would be around chicago, right?  That's definitely not me.

Then where are you? Are you around the Champaign area?

Yeah I'm around the Champaign area.

Edit:  Does individual include the people that participate in the team round and the people who don't?

Yeah.

Btw suma are you better at written or countdown?

My FTW rating is 1308. So Im bad at countdown. Epic Phail. :(

Mine is even lower (super low).  Even more epic phail

What is it?

Btw don't u think its hard for ppl to distinguish us because we have the same avatar?

And i find it funny and fun :D

Go  :D !

It is < 1100.  Ultimate phail.

And yeah. :D :P

lolz that sounds more like a smiley truce :P  :D

Oh that just made me think of something!

How many ways can you seat Mr. Green, Happy Smiley Face, Frowney Smiley Face, Gasp Smiley Face,  Oops Smiley Face,  and Razz Smiley Face in a row if Razz and Mr.Green must sit together as a sign of their truce and happy smiley and frowny refuse to sit next to each other?

Oh darn:
 Too many smilies in your post: 8.
You can have at most 3 smilies in this message. Please remove some smilies before submitting this message.

That is why this post now does not include smiley faces.  :( (except for that one)

hahaha...well ok i suck at combinatorics but I'd at least try:

So there are 6 smilies. There are 6!=720 ways to seat them in a row with no restrictions. If Mr. Green and Razz were to sit together, there would be 2*5! ways for them to do so. Also, if happy and frowny smiley were to sit together, along with Mr Green and Razz being together, there would be 4*4! ways for the smilies to sit in the row. All the rest would have frowny and happy separate, so 2*5!-4*4!=144 ways for the smilies to sit together.

And you got the smilies' names wrong! ZOMG their names are Mr. Green, Very Happy, Sad, Shocked, Embarrassed, and Razz!

LOLOLOLOLOLOLOLOL......................................................................

look what im about to put.....

 :lol:  :D  :(  :oops:  :o  :P

hehe!!!!!!

[quote="suma_milli"]Correct!

Here's how I did it:

Consider Mr. Green and Razz to be one.

We have 5 * (2!) * 4! = 240 ways to seat Mr. Green and Razz together.  Then we just have to subtract the number of ways included in those in which happy smiley and frowny are sitting next to each other.  Now we consider happy smiley and frowny to be one.

(2!) * 4 * (2!) * 3 * 2! = 96

240 - 96 = 144 :D

By the way I was refering to :) by happy smiley and :oops: by oops smiley.

...[img]http://www.artofproblemsolving.com/Forum/images/smiles/tongue.gif[/img][/quote]
 :blush: can also indicate oops

Oops, that's the smiley I meant when I said oops smiley. ... :blush:

lol what an ironic post :P

Let's say it's an oops smiley in my own world..

Let's say it's a BLUSH smiley in MY own world... :rotfl:  :D

Whatev.  :P  :rotfl:

Let's just say in my world chocolate grows on trees. :P

:(  :o  :D

Wow that TOTALLY didn't make sense at all...

Let's say in MY WORLD, all human faces are green.
And that DOES include avec_une_h.

[quote="avec_une_h"]What?

Well, in my world, you all speak Latin![/quote]
I mean, wen i saw u at states, I think your face looks like  :D


Let $ f: \mathbb R\to\mathbb R$ defined by $ f(x)\equal{}3x^2\plus{}4.$ Consider $ \epsilon,k>0.$ Prove that if $ |x\minus{}k|<\epsilon\implies|f(x)\minus{}f(k)|<3\epsilon(\epsilon\plus{}2k).$

The hardest part is maybe showing a triangle inequality. You can do it in your head if you work backwards.

$ \frac {f(x) \minus{} f(k)}{x \minus{} k} \equal{} 3(x \plus{} k)$.  What can you do from here?

I don't know how you got that, but I was thinking on $ |3x^2 \plus{} 4 \minus{} 3k^2 \minus{} 4| \equal{} 3|x \minus{} k\parallel{}x \plus{} k|,$ then $ |x \plus{} k| \equal{} |(x \minus{} k) \plus{} 2k|\le|x \minus{} k| \plus{} 2k < \epsilon \plus{} 2k,$ so the conclusion follows?

That's exactly what I wrote.  Where is the confusion?


calculus
derivative
function
calculus computations

Anyone please explain to me why :

if the power series representation of f(x) has a radius of convergence of R then the term by term differentiation of the power series will also have a radius of convergence of R .

i'm kinda new to this ..

[b]Theorem[/b] Suppose that the series $ f(x_{0})\equal{}\sum_{n\equal{}0}^{\infty}a_{n}x_{0}^{n}$converges and let $ a$ be any number with $ 0<a<\left|x_{0}\right|$.  Then on $ [\minus{}a,a]$ the series $ f(x)\equal{}\sum_{n\equal{}0}^{\infty}a_{n}x^{n}$ converges uniformly and absolutely.  Moreover, the same is true for the series $ \displaystyle g(x)\equal{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$.  Finally, $ f$ is differentiable and $ f'(x)\equal{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$ for all $ x$ with $ \left|x\right|<\left|x_{0}\right|$.

[i]Proof[/i]
Since $ \sum_{n\equal{}0}^{\infty}a_{n}x_{0}^{n}$ converges, the terms $ a_{n}x_{0}^{n}$ approach $ 0$.  Hence, they are bounded: there is some number $ M$ such that
$ \left|a_{n}x_{0}^{n}\right|\equal{}\left|a_{n}\right|\cdot\left|x_{0}^{n}\right|\leq M$ for all $ n$.  Now if $ x$ is in $ [\minus{}a,a]$, then $ \left|x\right|\leq\left|a\right|$, so
$ \left|a_{n}x^{n}\right| \equal{} \left|a_{n}\right|\cdot\left|x^{n}\right|\leq\left|a_{n}\right|\cdot\left|a^{n}\right|\equal{}\left|a_{n}\right|\cdot\left|x_{0}\right|^{n}\cdot\left|\frac{a}{x_{0}}\right|^{n}\leq M\left|\frac{a}{x_{0}}\right|^{n}$
But $ \left|a/x_{0}\right|<1$, so the series $ \displaystyle\sum_{n\equal{}0}^{\infty}M\left|\frac{a}{x_{0}}\right|^{n}\equal{}M\sum_{n\equal{}0}^{\infty}\left|\frac{a}{x_{0}}\right|^{n}$
converges.  Choosing $ M\cdot\left|a/x_{0}\right|^{n}$ as the number $ M_{n}$ in the Weierstrass M-test, it follows that $ \sum_{n\equal{}0}^{\infty}a_{n}x^{n}$ converges uniformly on $ [\minus{}a,a]$.  To prove the same assertion for $ g(x)$, notice that
$ \left|na_{n}x^{n\minus{}1}\right| \equal{} n\left|a_{n}\right|\cdot\left|x^{n\minus{}1}\right|\leq n\left|a_{n}\right|\cdot\left|a^{n\minus{}1}\right|\equal{} \frac{\left|a_{n}\right|}{\left|a\right|}\cdot\left|x_{0}\right|^{n}n\left|\frac{a}{x_{0}}\right|^{n}\leq\frac{M}{\left|a\right|}n\left|\frac{a}{x_{0}}\right|^{n}$
Since $ \left|a/x_{0}\right|<1$, the series $ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{M}{\left|a\right|}n\left|\frac{a}{x_{0}}\right|^{n}\equal{}\frac{M}{\left|a\right|}\sum_{n\equal{}1}^{\infty}n\left|\frac{a}{x_{0}}\right|^{n}$
converges.  Another appeal to the Weierstrass M-test proves that $ \sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$ converges uniformly on $ [\minus{}a,a]$.  Finally, a previous corollary proves, first that $ g$ is continuous, and then that
$ \displaystyle f'(x)\equal{}g(x)\equal{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$ for $ x$ in $ [\minus{}a,a]$.
Since we could have chosen any number $ a$ with $ 0<a<\left|x_{0}\right|$, this result holds for all $ x$ with $ \left|x\right|<\left|x_{0}\right|$.

JRav: what is this "previous corollary" that you refer to?

My mistake.  It refers to corollary that justifies the continuity of individual terms.

what exactly is this corollary? and how is it proved?

The statement is that if $ \sum_{n\equal{}1}^{\infty}f_{n}$ converges pointwise to $ f$ on $ [a,b]$, each $ f_{n}$ has an integrable derivative $ f_{n}'$ and $ \sum_{n\equal{}1}^{\infty}f_{n}'$ converges uniformly on $ [a,b]$ to some continuous function, then $ f'(x)\equal{}\sum_{n\equal{}1}^{\infty}f_{n}'(x)$.

The proof requires you to take the limit of partial sums.


limit
real analysis
real analysis unsolved

Prove that $ f: [a,b]\to\mathbb R$ is continuous if and only if the following conditions hold:

(a) For any $ c, d\in[a,b]$ and any $ u$ between $ f(c)$ and $ f(d)$, there exists $ \xi$ between $ c$ and $ d$ such that $ f(\xi)\equal{}u$.

(b) For any $ x_0\in[a,b]$, $ \lim_{x\to x_0} f(x)$ exists.

($ \Rightarrow$) Obvious.

($ \Leftarrow$) Assume the contrary that there is $ x_0\in[a,b]$ such that $ f$ is not continuous at $ x_0$. As $ \lim_{x\to x_0}f(x)$ exists, we have $ \lim_{x\to x_0} f(x)\neq f(x_0)$. Denote the limit by $ L$. WLOG, suppose $ L$ is greater than $ f(x_0)$, i.e.

$ L \minus{} f(x_0) \equal{} \lim_{x\to x_0} [f(x) \minus{} f(x_0)] > 0.$

Then, there exists $ \delta > 0$ such that

$ f(x) \minus{} f(x_0) > \frac {L \minus{} f(x_0)}2 > 0$ whenever $ x\in (x_0 \minus{} \delta, x_0 \plus{} \delta)\cap([a,b]\backslash \{x_0\}).$

Now, we choose $ c \equal{} x_0$ and $ d\in(x_0, x \plus{} \delta)\cap[a,b]$. Then, for all $ \xi\in (c,d]$, we have

$ f(\xi) \minus{} f(c) > \frac {L \minus{} f(c)}2$,

i.e. $ f(\xi) > \frac {L \plus{} f(c)}2.$

Hence, if we choose $ u\in\left(f(c), \frac {L \plus{} f(c)}2\right)\subset(f(c), f(d)]$ , we have

$ f(\xi)\neq u$ for all $ \xi\in[c,d],$

which contradicts the condition (a). Thus $ f$ is continuous on $ [a,b]$.


A quick easy question  :)

Consider the sequence $8,32,128,...$ obtained by $a_n=2^n.2^{n+1}$

Now we create a new sequence $b_1, b_2, ...$ by changing $\left\{a_1, a_2, ...\right\}=\left\{2.4, 4.8, 8.16, 16.32, ...\right\}$ to $\left\{2.3, 4.7, 8.15, 16.31,...\right\}=\left\{b_1, b_2, b_3, ... \right\}$

Prove that $b_n$ is a perfect number if $2^{n+1}-1$ is prime.

[i]A perfect number is defined as a number that equals the sum of all his divisors exept the number itself.[/i]

Proof: Suppose that $2^{n+1}-1$ is prime. It is easy to see that $b_n=2^n(2^{n+1}-1)$. If $S(x)$ denotes the sum of the divisors of $x$ (including $x$ and $1$), then $x$ is a perfect number iff $S(x)-x=x$, or $S(x)=2x$.

The sum of the divisors of $b_n$ is given by $S(b_n) =(1+2+2^2+...+2^n)(1+2^{n+1}-1)=(2^{n+1}-1)(2^{n+1})=2(2^n(2^{n+1}-1))=2b_n$, which completes the proof.


trigonometry
ratio
geometry
trig identities
Law of Sines

Well the law of sines is
               a/Sin A = b/Sin B= c/Sin C
Well my problem is last time I attempted to solve I had    a and Sin A i didn't have Sin of B but I had b
When I went to solve it I blanked out
Can someone give me a really easy example of how to solve it when you ever everything but one of the Sins so I can see a step by step process of it? I have my Math B regents in a matter of hours and wanted to make sure i understood that

Well, it is basic arithmetics to find the unknown sine:

$ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}$ gives you $ \sin B\equal{}\frac{b\sin A}{a}$.

Now, there you have a number between 0 and 1 as a result. The question arises: what is the angle B? If you get a well-known value for $ \sin B$ (i.e. $ 1,\frac{1}{2},\frac{\sqrt {3}}{2},\frac{\sqrt {2}}{2}$) you can have the solution right away, from your head. If not, then calculator is your friend... And $ \sin^{\minus{}1}$ function.

Generally if you have three pieces of information in the Law of Sines ratio, you can find the fourth by manipulation of proportions.

I agree with hsiljak. Solve like a regular proportion at first. Then when you have sin B isolated, just use inverse sine.

On most calcs it's [2nd][sin].


Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the opposite direction at an average speed of six miles per hour. It took Calvin 30 minutes less than it took Louisa to run the full path once. How many miles did Louisa run when she completed one circular path?

[quote="vallon22"][quote="dragon96"]Let x be the amount of time it took Louisa to run around the track. The time and rates are inversely proportional, since the distance is same. We have 6(x-30)=5x, and simplifying gives us x=180. Therefore, it took Louisa 3 hours to run around the path.[/quote]

that's not the answer to the question....
the question askes how many miles she ran when she finished the circular path
u gave us 3 hours which isn't how many miles[/quote]

Oh right...

Let me finish:

$ 3\cdot 5 \equal{}15$ miles.
I'm not sure if thats right because it is not likely a park path is 15 miles. Tell me if I'm wrong.

[quote="ernie"]Sounds like you're right, because the method is.[/quote]

Yeah, I'm positive he's right.


analytic geometry
combinatorics unsolved
combinatorics

100 unit squares of an infinite squared plane form a $ 10\times 10$ square.  Unit segments forming these squares are coloured in several colours.  It is known that the border of every square with sides on grid lines contains segments of at most two colours. (Such square is not necessarily contained in the original $ 10\times 10$ square.) What maximum number of colours may appear in this colouring?

[i]Author: S. Berlov[/i]

Yes.

You can achieve $ 11$ colors, and no more. In general for an $ n \times n$ board the answer is $ n\plus{}1$. Fun problem - thanks for "Problem of the Day" for floating it up.

Assume the board stretches from $ (0,0)$ to $ (10,10)$ on the Cartesian plane. It'll be convenient to refer to the segments like this: $ R(i,j)$ is the segment that runs to the right from $ (i,j)$, namely it connects $ (i,j)$ with $ (i\plus{}1,j)$, an similarly for $ L(i,j)$ ("left"), $ U(i,j)$ ("up") and $ D(i,j)$ ("down"). So the segments on the board can be described as $ R(i,j)$ for $ 0 \leq i \leq 9$, $ 0 \leq j \leq 10$ and $ U(i,j)$ for $ 0 \leq i \leq 10$, $ 0 \leq j \leq 9$.

How to use 11 colors: Make all the segments black (let's call this "Color 0"), except that $ R(i,i)$ has color $ i\plus{}1$ for $ 0 \leq i \leq 9$. It is obvious that no square contains $ R(i,i)$ and $ R(k,k)$ for $ i < k$: these segments are both horizontal, so if a square contained both of them it would have to have horizontal edges along row $ i$ and row $ k$, so it's height must be $ k\minus{}i$. But then its width is $ k\minus{}i$ as well, and since it contains the point $ (i,i)$ it cannot contain $ (k\plus{}1,k)$ which is the rightmost end of $ R(k,k)$. So this is a legal coloring of the segments on the board.

Now the proof that $ 11$ is the highest number possible:

First, let's suppose that all the horizontal segments are the same color - say, black. Now two vertical segments at the same height, say $ U(i,z)$ and $ U(j,z)$, cannot have two different colors that aren't black, since there's a square with edges along both segments as well as a black horizontal line. Thus for every row $ z$ there's a maximum of one non-black color among the vertical segments at this height. Overall this gives $ 11$ colors at most.

Next, let's suppose that each row (the horizontal segments with the same $ y$ coordinate) is monochromatic (single-colored), but different rows may have different colors. Note that there can be at most two different row colors (proof: exercise  :) ). So all rows are either all black or all white. There must be two adjacent rows of different colors, say rows $ z$ and $ z\plus{}1$. Now the vertical segments $ U(i,z)$ cannot have any color other than black/white, and the verticals at every other height $ z'$ can have only one additional color as before - so once again we cannot exceed $ 11$ colors (black, white, and at most 9 new colors from each height other than $ z$).

Those arguments obviously apply to columns as well, so we can assume that the rows are not all monochromatic and neither are all columns. Suppose then that there's a row that contains both black and white horizontal segments. Then there's an adjacent pair $ L(i,j)$, $ R(i,j)$ of segments one of which is black and the other is white. This implies that every vertical segment $ U(x,y)$ where $ x \neq i$ must be either black or white, since for each such segment we can find a square that contains it and also contains $ L(i,j)$ and $ R(i,j)$. Thus all columns except for column $ i$ are black/white. By the same argument, since we have a non-monochromatic column, all rows must have colors $ C/D$ for some colors $ C$ and $ D$, except for maybe one row. Overall this enables $ 8$ colors at most (black, white, $ C$, $ D$ two colors at column $ i$ and two colors at the exempt row). In fact much more can be said here but for our purposes this is enough: there cannot be more than $ 11$ colors.

Yes.

just an outline: assume that there are two side-by-side segments with colors 0,1;call WLOG R(i,j) and R(i+1,j). it follows that the only segments which can have colors different from 0,1 are :

R(k,j+1)&R(k,j-1) with 0=<k<10

R(h,j+2)&R(h,j-2) with 0=<h<4 ,6=<h<10

R(l,j+3)&R(l,j-3) with 0=<l<3 , 7=<l<10

R(m,j+4)&R(m,j-4) with 0=<m<2 , 8=<m<10

R(n,j+5)&R(n,j-5) with 0=<n<1 , 9=<n<10

you can do some little case work and get 11 as the most number of colors one can arrives. for an n*n grid one can do the same and got n+1 the maximum number of colors for coloring the grid.

A nice problem,it is better than this years third on Tuymaada(Combo),but not hard.Firs,an example for $11$.Color all segments with color $1$ expect the first vertical one in the first row,the second vertical in the second row...the tenth vertical in the tenth row.

Now,the proof that it can't be more.Consider all lines expect the border lines(which build the border of $10*10$).Now,we know that we color them with at most two colors.Now,let all colors that are not those be special colors.Note that we can't have $1$ vertical and $1$ horizontal line which are colored with distinct special colors.Now,from this we can assume that all special colors are all horizontal or vertical,so now it is easy that if we can't have two segments which are colored with distinct special colors and are in the same row or column,so from this we are finished.

I think there is easy solution with induction.


topology
advanced fields
advanced fields unsolved

Let $\Bbb{N}$ be set of natural number, contruct the metric
$d(m,n)=\left\{\begin{array}{c}-1&n=m\1+\frac{1}{m+n}&n\not=m\end{array}$
prove that $(\Bbb{N},d)$ is complete metric space

I think you mean $d(n,n)=0$. Just remark that if $d(n,m)<\frac{1}{2}$ then $m=n$. That's all.

that's right it is $0$, but we need complete, can you show clearly, your remark only need for metric?  :maybe:

alekk's observation implies that any Cauchy sequence is eventually constant (hence convergent).

By the way, gemath, your question probably belongs in the analysis forum, not that we don't like having you here  :)


[size=109]If $ x \plus{} y \plus{} z \equal{} \pi$,

prove that:

$ \tan(x) \plus{} \tan(y) \plus{} \tan(z) \equal{} 4 \tan{\frac {x}{2}} \tan{\frac {y}{2}} \tan{\frac {z}{2}}$[/size]

[hide="PS:"]I edited it Sonnhard,thnx.[/hide]

hello, or do you mean $ x\plus{}y\plus{}z\equal{}\pi$?
Sonnhard.

I will prove that the given equality is true only when one of the angles is $ \pi$ and the others are $ 0$.

[hide="Proof"]We will assume that the given is true for all $ x, y, z$ such that $ x\plus{}y\plus{}z\equal{}\pi$ and lead a contradiction.

Since $ x\plus{}y\plus{}z\equal{}\pi$, we have $ \tan x\plus{}\tan y\plus{}\tan z\equal{}\tan x\tan y\tan z$.

Hence, we have $ \tan x\tan y\tan z\equal{} 4\tan \frac{x}{2}\tan \frac{y}{2}\tan \frac{z}{2}$.

Since $ \tan \frac{x}{2}\equal{}\frac{\sin x}{1\plus{}\cos x}$, we have
\[ \frac{\sin x\sin y\sin z}{\cos x\cos y\cos z}\equal{}\frac{4\sin x\sin y\sin z}{(1\plus{}\cos x)(1\plus{}\cos y)(1\plus{}\cos z)}\\ \\ \implies (1\plus{}\cos x)(1\plus{}\cos y)(1\plus{}\cos z)\equal{}4\cos x\cos y\cos z\]
Here, assume that none of the angles are $ \pi$.

Since none of the angles are $ \pi$, all of $ 1\plus{}\cos x$ and cyclic are positive.

Applying AM-GM on $ 1\plus{}\cos x$ and cyclic, we have $ (1\plus{}\cos x)(1\plus{}\cos y)(1\plus{}\cos z)\ge 8\sqrt{\cos x\cos y\cos z}$.

Since $ (1\plus{}\cos x)(1\plus{}\cos y)(1\plus{}\cos z)\equal{}4\cos x\cos y\cos z$, we have \[ 4\cos x\cos y\cos z\ge 8\sqrt{\cos x\cos y\cos z}\implies \cos x\cos y\cos z\ge 4,\]which is clearly false.

Therefore, the equality given is true only when one of the angles are $ \pi$, and the others are 0. $ \blacksquare$[/hide]


The odds in favour of a Pythag-Air-US airlines flight being on time are 3:1.
a) What is the probability that this airline's next eight flights will be on time?
b) What is the expected waiting time before a flight delay?

For the first 1, the probability is $\left(\frac{3}{4}\right)^8$, which I'm two lazy to calculate out.

Is the second one supposed to be a trifck question?  Like, there is no waiting before a delay?  Otherwise, it's totally irrelevant.

once a flight is delayed, can the other flights still be on time?

what's with the second question?


probability
inequalities
calculus
integration
geometry
triangle inequality

You break a stick at two random places. What is the probability that a triangle is formed?

[hide] I just did it geometrically, let the three sides be x y and z.  so let the stick have length 1, and we know x+y+z = 1 and since they are all positive we can just limit it to the triangle that has them all positive.  then we also know that x<1/2, y<1/2, z<1/2 because otherwise the triangle inequality would be broken.  So after drawing those in we get the intersection to be just the small equilateral triangle made by connecting the midpoints of the sides. So it is well known that this is 1/4 the area. and thats our answer[/hide]

[quote="pianoforte"][hide="answer"]I got 3/4 ?[/hide]
[hide]Let the segment have length 1, and two of the resulting pieces have length a and b, the last piece have length 1-a-b. From triangle inequatility, we get
$a+b>1-a-b \implies 2a+2b>1$
$a+1-a-b>b \implies 1>2b$
$b+1-a-b>a \implies 1>2a$, so that
$1>a+b>\frac{1}{2}$. Graphing a on the x-axis, and b on the y-axis, on the triangle (0,0), (0,1) and (1,0), I get 3/4 ?[/hide][/quote]

[hide] you forgot to limit a and b to be less than 1/2, otherwise the other 2 must add up to less then that. so once you add that in i get 1/4[/hide]

My teacher actually solved this by using three different methods.

1. Random Numbers, generated from a calculator.
2. Tree Diagram
3. Integration, very impressive!

[quote="Farenhajt"]I must say I'm getting $P={1\over 8}$: $a<{1\over 2}$ and $b<{1\over 2}$ define a square of the area ${1\over 4}$, and the line $a+b>{1\over 2}$ cuts that square in half, being its diagonal. Hence we must take into account only the area that satisfies all three inequalities, and that's the triangle $({1\over 2},0),(0,{1\over 2}),({1\over 2},{1\over 2})$.[/quote]

I am pretty sure the answer is $\frac{1}{4}$  :maybe:

Because I have now seen 5 methods of solving this problems.

Oh it IS my bad after all, I just got that: the area of all the possible events isn't the square of side $1$ but the half of it, lying under a diagonal, hence the probability is twice my result, i.e. ${1\over 4}$.

Sorry everybody  :!:

Expansion:

A stick is broken into four pieces. If three of these pieces are chosen at random, what is the probability that they form a triangle?

A stick is broken into $n$ pieces. If three of these pieces are chosen at random, what is the probability that they form a triangle?


Prove:
The necessary and sufficient conditions of $ f(x)\equal{}cx$ is $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(1)\equal{}c$.$ c$ is a constant.

[quote="thomaszhuk"]Prove:
The necessary and sufficient conditions of $ f(x) \equal{} cx$ is $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and $ f(1) \equal{} c$.$ c$ is a constant.[/quote]

It's well known that (at least with  axiom of choice) it exists some non linear functions (so different from $ cx$)  from $ \mathbb R\to\mathbb R$ such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(1)\equal{}c$

If you want to be sure to have only linear solutions, you need to add some constraints :
$ f(x)$ monotonous
$ f(x)$ continuous
$ f(x)$ bounded (either below or above) on some non empty interval $ (a,b)$
...


Can someone post solutions for the following questions?
1.  A student bought x pencils for y one dollar bills.  The seller advised him to buy 10 more pencils so that he would get all for $2 and save 80 cents per dozen.  Find x.
[hide="Answer"]5[/hide]

2.  The tide level follows a sinusoidal curve.  In the morning of March 1, 2003, low tide was at 1:04 a.m. and the next high tide after that was at 7:01 a.m..  When is the next day that has 3 high tides (in the same day)?
[hide="Answer"]April 5, 2003[/hide]

3.  If (x^2+y^2)/(x+y)=4, find all possible values for x-y.
[hide="Answer"][-4,4][/hide]

[edit]

solution (thanks to treething)
[hide]
u kno $xk=y$ for some $k$ which is how much a pencil costs
since $y$ is the number of one doller bills, and you know that for $x+10$ pencils you pay with two one dollar bills, $y = 1$
so $k= \displaystyle\frac{1}{x}$
then you know that
$(x+10)k - (\displaystyle\frac{x+10}{12}) (\frac{4}{5}) = 2$
$(x+10)(k-\displaystyle\frac{1}{15}) = 2$
substitute the $k= \displaystyle\frac{1}{x}$ into the equation and simplify that to get
\[ x^2+25x-150 = 0\]
\[x=5 or -30\]
and you can ignore the negative answer so you get $x=5$
[/hide]

Thanks for the solution.  Does anyone have a solution for #2 and 3?

[hide]The first high tide on March 1 was on 7:01 AM. Since it was 5 hours and 57 minutes away from the first low tide, the next high tide is 11 hours and 54 minutes from 7:01 which is 6:55 PM. The next high tide is also 11 hours and 54 minutes from that which is 6:49 AM. For all of the high tides to be in one day, then, they must all lose 6 hours and 50 minutes which is also 410 minutes. Since everyday each high tide loses 12 minutes, it will take $\frac {410}{12} = 34 \frac {1}{6}$ days for all three high tides to be on the same day. But, since it's not quite there on the 34th day, then it will take another day. 35 days more than March 1st lands on April 5th.[/hide]

[quote="jen7"]
3.  If (x^2+y^2)/(x+y)=4, find all possible values for x-y.
[hide="Answer"][-4,4][/hide][/quote]

So we know x != -y.  Assuming this, we multiply through by x+y to get

$x^2 + y^2 = 4x + 4y$

$x^2 - 4x + 4 + y^2 - 4y + 4 = 8$

$(x-2)^2 + (y-2)^2 = (2\sqrt{2})^2$

which is a circle of radius $2 \sqrt{2}$ centered at 2, 2.  The only point that is excluded is (0, 0).  Now, let a line of slope one go through the point (a, b).  Then its equation is y - b = x - a --> y + a - b = x, so the x intercept is a - b.  Therefore, the range of the x intercepts of lines of slope 1 is what we're looking for.  The bounds of these obviously occur at tangency between the line and the circle.  The rest you should be able to figure out; I can't really explain without a diagram :(.

i think i have better solutions but theres a problem

im just a begginer here and i think i have a good solutions for these problems :lol: . but i also have a problem: i dont know were i cant write the solutions!:?:

well if someone can teach me where i can write the solutions maybe i could give it to you.  :D

sorry just a begginer :oops: [/u]

You just click "Post Reply" and type up the solutions in the message body and click "Submit."  :) It doesn't have to be in LaTeX.

WOW. For #3, they made a mistake in finding solutions.

[hide]We multiply x+y to each side and we get x^2+y^=4x+4y, we can see clearly that x and y can both be =2. Therefore 4-4=0, and that wasn't included...[/hide]

  ;)

It was included, but he just didn't list out all the possible points..
Just put four and four in the circle equation and it will come out eight.

Thanks for all the help!  But for #3, why wouldn't it be [-4,0) (0,4] since (0,0) isn't part of the graph?

Well by giving you the equation he gave you every single possible value of $x$ and $y$ so really he did give you all those answers.

but since 0 isn't a possible x-intercept, wouldn't 0 be excluded from the possible values?

so then why wouldn't the answer be [-4,0)U(0,4] instead of [-4,4]? :?


[hide="2."]

Let it be $ (r^2\plus{}ar\plus{}b)(r^2\plus{}cr\plus{}d)$

Expanding and setting like terms equal,

a+c=0
ac+b+d=0
ad+bc=12
bd=323

Note that $ 323\equal{}18^2\minus{}1\equal{}17\cdot19$

On a hunch, let $ b\equal{}17$ and $ d\equal{}19$

Now we have

a+c=0
ac=-36
19a+17c=12

From the first two equations, a and c are $ \pm 6$

From the second, a=6 and c=-6.

Thus:

 $ (r^2\plus{}6r\plus{}17)(r^2\minus{}6r\plus{}19)\equal{}r^4\plus{}12r\plus{}323$[/hide]


function
trigonometry
logarithms
algorithm
algebra
functional equation
system of equations

Show that there exists a function f : $ \mathbb{R}\longrightarrow\mathbb{R}$
 which is continuous, and satisfies the equation f(x) + f(2 x) + f(3 x) = 0 for all x, but which is not the zero function.
-good luck. :)

Let $ r\in \mathbb{C}$ satisfy the equality
\[ 1 \plus{} 2^r \plus{} 3^r \equal{} 0
\]
[hide="remark"]
Need to verify such an $ r$ exists, but it's not too hard and a numerical approximation gives one solution $ r\approx \minus{} 0.454397 \plus{} 3.59817i$[/hide]and let $ g(x) \equal{} \begin{cases} Cx^r & x\neq 0 \\
0 & x \equal{} 0 \end{cases}$ for some constant $ C\in \mathbb{C}$.  Then we have
\[ g(x) \plus{} g(2x) \plus{} g(3x) \equal{} Cx^r(1 \plus{} 2^r \plus{} 3^r) \equal{} 0
\]
and hence $ f_1 \equal{} \Re g$ and $ f_2 \equal{} \Im g$ are real functions that satisfy the functional equation.
More generally, the functional equation
$ \sum_{k \equal{} 1}^n{\alpha_k f(\beta_k x)} \equal{} 0$ has solution $ f(x) \equal{} \Re \sum_{j \equal{} 1}^d C_jx^{r_j}$ where $ \{ r_j \}_{j\in \{1,2,...,d\}}$ are solutions of $ \sum_{k \equal{} 1}^n{\alpha_k \beta_k^r} \equal{} 0$.

Edit: On second thought, since $ \Re r<0$ the function constructed will have a nasty discontinuity at the point $ x \equal{} 0$.  Maybe a properly weighted linear combination of particular solutions can get rid of the discontinuity, but I've no good ideas right now.

many thanks JoeBlow for you'r nice approche to the answer but i am asking how to find such $ r$ and is-it unique?

If $ r \approx 0.9405921832 \minus{} 8.230709548\mathrm{i}$, then $ 1 \plus{} 2^r \plus{} 3^r \equal{} 0$.
Also, $ f(x) \equal{} |x|^r$ is what we want.

It is not unique, there are infinitely many families of solutions (but countably so).  Unfortunately there's not much hope of a closed form for exact solutions to a transcendental equation like $ 1 + 2^r + 3^r = 0$, but there are fairly precise and reliable ways to generate some approximate solutions.  One direct approach would be to write $ r = x + iy$, $ x,y\in \mathbb{R}$, and solve the resulting system of equations by taking real and imaginary parts:
\[ 1 + 2^x\cos (y\log 2) + 3^x \cos (y\log 3) = 0 \\
2^x \sin (y\log 2) + 3^x \sin (y\log 3) = 0
\]
The second equation gives you an expression for $ x$:
\[ x = \log_{\frac {3}{2}}\left( - \frac {\sin (y \log 2)}{\sin (y \log 3)\right)}
\]
Now substitute that into the first equation and you have an expression purely in terms of $ y$, for which you can find roots using your favorite root-finding algorithm (e.g. some variant of Newton's method).  Note that this approach also has the nice theoretical feature of giving you a good way to actually [i]prove[/i] the existence of solutions, since you need only show that the continuous expression in terms of $ y$ switches signs somewhere.

Using this method you can find better particular solution than the one I posted originally.  For example: $ r\approx 0.940592 + 8.23071i$ (cheers to Batominovski for beating me to it by a few minutes :)), which since $ r > 0$ has the advantage of yielding continuous solutions to the functional equation:
\[ f(x)\approx c_1|x|^{0.940592}\cos(8.23071 \log |x|) + c_2|x|^{0.940592}\sin(8.23071 \log |x|)
\]
with $ c_1,c_2\in \mathbb{R}$.  As I mentioned before, it's most certainly not unique, there are infinitely many such $ r$.  Some other solutions are given by, for example: $ r \approx 0.132624 \pm 20.9796i, r \approx 0.456505 \pm 24.84572i$, etc.  Furthermore, any linear combination of solutions gives another solution, so you have in principle infinitely many degrees of freedom to construct a wide variety of solutions.

An interesting follow-up: can there exist a continuously differentiable such function?  (can we have an $ r$ with $ \Re r > 1$?)


[quote="ttt777"]Hi i have the question as the following:

(1.08)^t - 1 = (0.8)(1.08) ^ t

and the next step is (0.2)(1.08)^t = 1

But my question is how do i get to the next step, the teacher gives me the answer, but i really don't know how she does it.  Please help me, thank you :oops:[/quote]
$ (1.08)^t\equal{}5$
$ t\equal{}\text{log}_{1.08}5$


Given that $ m$ and $ n$ are the least positive integers that satisfy the equation $ 8m\equal{}10n$, what is the value of $ m\plus{}n$?

First, we divide the equation by 2.
8m=10n is the same as 4m=5n.
We also know that both m and n are positive integers.
Clearly, 5n is a multiple of 5, and since 4m=5n, 4m must also be a multiple of 5.
In addition, since 4 is not a multiple of 5, m must be (a multiple of 5).
The smallest positive integer that is a multiple of 5 is 5.
Thus, m=5.
Using 4m=5n, we solve and get n=4.
Therefore, the smallest sum of m+n=5+4=[b]9[/b]


The number of digits in an integer, n, is $ \lfloor\log_{10}(n) \plus{} 1\rfloor$

[hide="Proof"]$ n \equal{} a \times 10^b$ for some $ a$ where $ 1 \le a < 10$ and $ b$ is a nonnegative integer.

We know that $ 10^b$ has $ b\plus{}1$ digits. Therefore, $ a \times 10^b$ also has $ b\plus{}1$ digits. Therefore, $ n$ has $ b\plus{}1$ digits.

Note that $ 1\plus{} \log_{10} n \equal{} 1\plus{} b \plus{} \log_{10} a$.

$ \lfloor 1\plus{} \log_{10} n \rfloor \equal{} \lfloor 1\plus{} b \plus{} \log_{10} a \rfloor$.

$ \log_{10} a < 1$, so $ \lfloor 1\plus{} \log_{10} n \rfloor \equal{} b\plus{}1$.

So $ \lfloor 1\plus{} \log_{10} n\rfloor$ is the number of digits of $ n$.

[/hide]

Sorry about the off topicness, but what do the $ \lfloor \rfloor$ brackets mean?

They denote the floor function.

If $ x \equal{} n\plus{}k$ for an integer $ n$ and a nonnegative decimal $ k$ where $ 0 \le k < 1$, then $ \lfloor x \rfloor \equal{} n$.

Basically, it rounds the number down if it is a decimal, and does nothing if it's an integer.


Hi guys,
I'm doing a project on tessellations. Does anyone have a good idea what I should do, or how I should present the project, to make it look nice?

What is the project for ? { I'm basically asking this question to understand how detailed you need to be while doing your project }

You may find some useful resources at [url=http://dmoz.org/Science/Math/Recreations/Tessellations/]dmoz.org/Science/Math/Recreations/Tessellations/[/url].

There are many different tesselations...


find all $f$ such that for all real number $x,y$:$f(xf(x)+f(y))=f^{2}(x)+y$

[quote="zhaobin"]for all real number $x,y$:$f(xf(x)+y)=f^{2}(x)+y$[/quote]

Maybe $x,y>0$ is required?
Otherwise, let $x=0$, then $f(y) = y + C$ and solve for $C$.

my mistake.and now I have edited it.Slove it guys. :)

[quote="Armo"]Nice one.
Clearly $f$ is sjurective. Assume that $f(t)=0$, take $x=t$ to obtain: $f(f(y))=y$ for all $y$.Put $f(x)$ instead of $x$, thus $f(xf(x)+f(y))=x^{2}+y=f^{2}(x)+y\Rightarrow f^{2}(x)=x^{2}$ (*).After this just square up the original equation  and use (*) to have $f(x)f(y)=xy$ for all $x,y$. Put $y=1$, so $f(x)=f(1)x$.But from (*) we have that $f(1)=1$ or $f(1)=-1$, so the answer is $f(x)=x$ and $f(x)=-x$.And it is easy to verify that this two functions satisfy the original relation.[/quote]
can you explain why there exist a $\"t\"$ such that $f(t)=0$?

can you explain why there exist a $\"t\"$ such that $f(t)=0$?

x=0 leads to $f(f(y))=f^2(0)+y$ choose $y=-f^2(0)$ you get

$f(f(-f^2(0)))=0$ and $t=f(-f^2(0))$

Because $f(x)$ is surjective because in original equation putting $x=0$ we get $f(f(y))=c+y$ that can take any value hence for some $y$, $f(y)=0$.Sorry I wrote the same as Moubinool wrote :blush: .


Matt has exactly $ \$1.60$ in his pocket. He only has nickels and dimes. There are twice as many nickels as dimes. What is the number of coins in Matt's pocket?

Let $ x$ be the number of nickels, and $ y$ the number of dimes. Then,

$ 5x\plus{}10y\equal{}160$
$ x\equal{}2y$.

Substituting the second equation into the first, we have

$ 20y\equal{}160 \implies y\equal{}8$. Plugging this in yields $ x\equal{}16$, so $ 16\plus{}8\equal{}\boxed{24}$.


analytic geometry
geometry
geometric transformation
rotation
trigonometry
group theory
abstract algebra

The coordinates of the three vertices are :
$(x_0, y_0), (x_1 = x_0 + c \cos(t), y_1 = y_0 + c \sin(t))$ and $(x_2 = x_0 + c \cos(t+ \frac {\pi}3), y_2 = y_0 + c \sin(t+\frac {\pi}3)).$

Then :
If $y_1 \neq y_0 : \sqrt{3} = \frac {x_0 + x_1 - 2x_2}{y_1 - y_0} \implies$ impossible;
If $y_1 = y_0 : x_1 \neq x_0$ and $\sqrt{3} = 2 \frac {y_2 - y_0}{x_1 - x_0} \implies$ impossible as well.

So we're done.

[quote="Peter"]Prove no three lattice points in the plane form an equilateral triangle.[/quote]

 We generalize it slightly.

 Theorem.  No three rational points in the plane form an equilateral triangle.

 First Proof.  Assume on the contrary that three vertices of an equilateral triangle $ ABC$ are rational points. Write $ A(a_{1}, b_{1})$, $ B(a_{2}, b_{2})$, $ C(a_{3}, b_{3})$, where $ a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}\in\mathbb{Q}$. Since $ ABC$ is equilateral, we obtain
 \[ \left( a_{1}\minus{}a_{2}\right)^{2}\plus{}\left( b_{1}\minus{}b_{2}\right)^{2}\equal{}\left( a_{2}\minus{}a_{3}\right)^{2}\plus{}\left( b_{2}\minus{}b_{3}\right)^{2}\equal{}\left( a_{1}\minus{}a_{3}\right)^{2}\plus{}\left( b_{1}\minus{}b_{3}\right)^{2}\equal{}\lambda.\]
 for some $ \lambda\in\mathbb{Q}$. After setting $ p\equal{}a_{1}\minus{}a_{2}$, $ q\equal{}a_{2}\minus{}a_{3}$, $ r\equal{}b_{1}\minus{}b_{2}$ and $ s\equal{}b_{2}\minus{}b_{3}$, it becomes
 \[ p^{2}\plus{}r^{2}\equal{} q^{2}\plus{}s^{2}\equal{}\left( p\plus{}q\right)^{2}\plus{}\left( r\plus{}s\right)^{2}\equal{}\lambda.\]
 We obtain
 \[ 2 ( pq\plus{}rs ) \equal{}\left( p\plus{}q\right)^{2}\plus{}\left( r\plus{}s\right)^{2}\minus{}\left( p^{2}\plus{}r^{2}\right)\minus{}\left( q^{2}\plus{}s^{2}\right) \equal{}\minus{}\lambda.\]
 or
 \[ pq\plus{}rs \equal{}\minus{}\frac{1}{2}\lambda.\]
 It follows that
 \[ (ps\minus{}qr)^{2}\equal{}\left( p^{2}\plus{}r^{2}\right)\left( q^{2}\plus{}s^{2}\right)\minus{}( pq\plus{}rs)^{2}\equal{}\frac{3}{4}{\lambda}^{2}\equal{} 3\left(\minus{}\frac{\lambda}{2}\right)^{2}\]
 or
 \[ \vert ps\minus{}qr\vert \equal{}\sqrt{3}\,\vert pq\plus{}rs\vert.\]
 Notice that both $ ps\minus{}qr$ and $ pq\plus{}rs$ are rational numbers. If $ pq\plus{}rs\neq 0$, then $ \sqrt{3}\equal{}\left\vert\frac{ps\minus{}qr}{pq\plus{}rs}\right\vert\in\mathbb{Q}$, which contradicts for the irrationality of $ \sqrt{3}$. Hence, $ pq\plus{}rs\equal{}0$. This implies that $ ps\minus{}qr\equal{}0$ so that
 \[ \left( p^{2}\plus{}r^{2}\right)\left( q^{2}\plus{}s^{2}\right) \equal{} (pq\plus{}rs)^{2}\plus{}(ps\minus{}qr)^{2}\equal{}0.\]
 Hence, we have $ (p,r)\equal{}(0,0)$ or $ (q,s)\equal{}(0,0)$. This means that
 \[ A\left( a_{1}, b_{1}\right) \equal{} B\left( a_{2}, b_{2}\right)\;\; or\;\; B\left( a_{2}, b_{2}\right) \equal{} C\left( a_{3}, b_{3}\right).\]
 Since $ ABC$ forms a triangle, this is a contradiction.

 Second Proof. We need the following well-known result.

 Lemma. Any triangle whose vertices have rational coordinates has rational area.

  Proof of Lemma. Suppose that rational points $ P(a_{1}, b_{1})$, $ Q(a_{2}, b_{2})$, $ R(a_{3}, b_{3})$ form a triangle.  By the well-known formula, we obtain
  \[ [\triangle PQR ]\equal{}\frac{1}{2}\vert a_{1}b_{2}\plus{}a_{2}b_{3}\plus{}a_{3}b_{1}\minus{}a_{2}b_{1}\minus{}a_{3}b_{2}\minus{}a_{1}b_{2}\vert\in\mathbb{Q}.\]
  Assume on the contrary that three rational points form an equilateral triangle with the length $ l$. Since vertices are rational points, we find that $ l^{2}$ is also positive rational. On the one hand, the above proposition means that the area of the triangle is rational. On the other hand, the area of triangle is equal to $ \frac{\sqrt{3}}{4}l^{2}$, which is irrational. This is a contradiction.

 Third Proof. We present another alternative proof of Theorem 10.2. Assume on the contrary that three rational points form an equilateral triangle $ ABC$. Then, we may translate $ ABC$ to $ OB'C'$ by $ \minus{}\overrightarrow{OA}$, where $ O$ is the origin. Then, three rational points $ O$, $ B'(p,q)$, $ C'(r,s)$ also form an equilateral triangle. By $ \frac{\pi}{3}$ rotation around the origin, $ B'$ corresponds to $ C'$. In case of $ \frac{\pi}{3}$ \textit{counterclockwise} rotation, we obtain
 \[ \left(\begin{array}{c}r\\ s\end{array}\right) \equal{}\left(\begin{array}{cc}\cos\frac{\pi}{3}&\minus{}\sin\frac{\pi}{3}\\ \sin\frac{\pi}{3}&\cos\frac{\pi}{3}\end{array}\right)\left(\begin{array}{c}p\\ q\end{array}\right)\]
 or
 \[ r \equal{}\frac{1}{2}p\minus{}\frac{\sqrt{3}}{2}q,\; s\equal{}\frac{\sqrt{3}}{2}p\plus{}\frac{1}{2}q.\]
 Since $ r$ and $ p$ are rational numbers, we see that $ \frac{\sqrt{3}}{2}q\equal{}r\minus{}\frac{1}{2}p$ is rational. Since $ \frac{\sqrt{3}}{2}$ is irrational and since $ q$ is rational, this means that $ q\equal{}0$. From $ s\equal{}\frac{\sqrt{3}}{2}p\plus{}\frac{1}{2}q$, we also get $ p\equal{}0$. This means that $ B'$ coincides with the origin, which is a contradiction.
 In case of $ \frac{\pi}{3}$ \textit{clockwise} rotation, you may change the role of $ B'$ and $ C'$ to reach the contradiction.

Here is a different approach leading to another generalization:
Assume that triangle $ ABC$ is a lattice equilateral triangle. It is known that if $ \omega$ is the Brocard angle of a triangle $ ABC$, then $ \cot{\omega} \equal{} \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{4S}$. By Pick's theorem, the area of triangle $ ABC$ is (at least) rational. Thus, $ \cot{\omega} \in \mathbb{Q}$. But since $ ABC$ is equilateral, its Brocard angle is $ \frac {\pi}{6}$, and therefore $ \cot{\omega} \equal{} \cot{\frac {\pi}{6}} \equal{} \sqrt {3} \not\in \mathbb{Q}$. Contradiction.

[color=darkblue][b]Generalization[/b][/color] (communicated by Cezar Lupu): [color=blue]The Brocard angle of a lattice triangle is not a rational multiple of $ \pi$.[/color]
This now follows from the fact that $ \cot{\mathbb{Q}\pi} \in \left\{ \minus{} 1, 0, 1\right\}$, combined with [b]Johnson's theorem[/b] (which says that the Brocard angle of a triangle is less or equal $ \pi/6$, with equality when $ ABC$ is equilateral - see here: [url]http://mathworld.wolfram.com/BrocardAngle.html[/url]).

[quote="ideahitme"] Theorem.  No three rational points in the plane form an equilateral triangle.[/quote]
See [url=http://www.mathlinks.ro/viewtopic.php?t=201363]here[/url] for a possible generalization (in case that the second part is true).

There is also an n-dimensional version:

The vertices of an n-dimensional regular simplex can be realised as subset of $ \mathbb Q^n$ (respectively $ \mathbb Z^n$) iff $ n$ fulfills one of the following conditions:
- $ n \equiv 1 \mod 4$ and $ n \plus{} 1$ is a sum of two perfect squares.
- $ n \equiv 3 \mod 4$
- $ n$ is even and $ n \plus{} 1$ is a perfect square.

Nice! Though I have no idea how to prove it. Can you please tell me a reference or something? (It seems that I'm losing my Googling skills).

Well, there is one more beautiful problem related to the topic.
Prove that for $ n\geq 3$ there is no n-gon with all sides equal where each of the vertex lies in the lattice points. (of course without n=4)
I'd like to see your ideas :wink:

[hide="Small hint "]Infinite descent :wink: [/hide]

In fact, the only $ n$-gons contained in any lattice are $ n\equal{}1,2,3,4,6$.

@pohoatza: it was proven by Schoenberg, you may be able to find something about this on the net.

[quote="ZetaX"]In fact, the only $ n$-gons contained in any lattice are $ n \equal{} 1,2,3,4,6$.
[/quote]
I don't quite understand what you meant by that  :huh:

The only $ n$ such that the regular $ n$-gon can be realized by points of a lattice (of any kind) are $ n\equal{}1,2,3,4,6$.

[quote="ZetaX"]The only $ n$ such that the regular $ n$-gon can be realized by points of a lattice (of any kind) are $ n \equal{} 1,2,3,4,6$.[/quote]
Actually (in the spirit of this topic) $ n \neq 3$.  :D

A lattice is not necessarily 2-dimensional.
Another thing is that in my post, "lattice" can be a general lattice (a discrete subgroup of some $ \mathbb R^m$); e.g. we could have the hexagonal lattice.


What's the number of different roads from $ A$ to $ B$ if one can walk only down(verticaly or on the diagonals) or to the right(as shown by the arrows)?

[url=http://mathworld.wolfram.com/SchroederNumber.html]Schröder Number[/url]
[url=http://www.research.att.com/~njas/sequences/A006318]A006318 in OEIS[/url]

Ok....seems that there were a formula already....but what if you are given this figure? Is there any elegant way to find the number of different roads by walking only up or to the right?(I am intereseted in the way of counting not in the numeric answer  :)  )

$ \left( \binom{4}{2}\minus{}1 \right) \cdot \binom{4}{2} \equal{} 30$

For your second figure, given that it's a relatively small diagram, you should just do it out. Let $ f(P)$ be the number of ways to get to point $ P$. Say you can get to point $ P$ from $ Q_1$, $ Q_2$, $ \ldots Q_n$.

Then, $ f(P)\equal{}f(Q_1)\plus{}f(Q_2)\plus{} \ldots f(Q_n)$

Notice that if there are $ j$ ways to get to point $ (x\minus{}1,y)$ and $ k$ ways to get to $ (x,y\minus{}1)$, there are $ j\plus{}k$ ways to get to $ (x,y)$ given that there is a path leading from both aforementioned points to $ (x,y)$.

[quote="Lazarus"]Notice that if there are $ j$ ways to get to point $ (x \minus{} 1,y)$ and $ k$ ways to get to $ (x,y \minus{} 1)$, there are $ j \plus{} k$ ways to get to $ (x,y)$ given that there is a path leading from both aforementioned points to $ (x,y)$.[/quote]so how does he get $ \binom{4}{2}$?

[quote="123456789"][quote="Kouichi Nakagawa"]$ \left( \binom{4}{2} \minus{} 1 \right) \cdot \binom{4}{2} \equal{} 30$[/quote]How did you determine $ \binom{4}{2}$ was the way to count?[/quote]

R : advance to the right
U : advance to the up
The route supports that I arrange two R and two U to one line.
Therefore,
$ \dfrac{(2\plus{}2)!}{2! \ 2!}\equal{}\binom{4}{2}$

Thanks for your idea of counting backwards, it's calculatorious but it works :)

[quote="Kouichi Nakagawa"][quote="123456789"][quote="Kouichi Nakagawa"]$ \left( \binom{4}{2} \minus{} 1 \right) \cdot \binom{4}{2} \equal{} 30$[/quote]How did you determine $ \binom{4}{2}$ was the way to count?[/quote]

R : advance to the right
U : advance to the up
The route supports that I arrange two R and two U to one line.
Therefore,
$ \dfrac{(2 \plus{} 2)!}{2! \ 2!} \equal{} \binom{4}{2}$[/quote]

Sorry but I didn't understand your solution :( can you please explain it with more details? :oops:

And I was just thinking...what if we are allowed to walk in any direction but without passing twice through the same vertex?  :?:

[quote="sunnykim"]And I was just thinking...what if we are allowed to walk in any direction but without passing twice through the same vertex?  :?:[/quote]

as far as I know, that's a much harder problem

I don't think there is an elegant combinatoric solution to that, but if anybody wants to enlighten me, go for it

[quote="sunnykim"]Sorry but I didn't understand your solution :( can you please explain it with more details? :oops:[/quote]First, let's concentrate on the 2 by 2 square grid on the top right. Label the bottom left corner of that 2 by 2 grid $ C$. Now, every path going from $ C$ to $ B$ must contain two UPs and two RIGHTs, in any order. For example, RRUU or RURU. There are $ \frac {4!}{2! \cdot 2!}$ ways to permutate two indistinguishable $ R$s and $ U$s, hence $ \binom{4}{2}$.


A bag contains 20 black balls and 30 white balls. Four balls are randomly drawn without replacement. What is the probability of selecting a white ball followed by a black ball, and then selecting a black ball followed by a white ball? express your answer as a common fraction.

It's simply four independent probabilities. Just multiply the four together after the balls are drawn sequentially.

$ (\frac{30}{50})(\frac{20}{49})(\frac{19}{48})(\frac{29}{47})\equal{}\boxed{\frac{551}{9212}}$.


geometry
perimeter
circumcircle
inradius
trigonometry
geometry proposed

Let $\triangle ABC$ be an acute-angled triangle with perimeter $2s.$ $r_a,r_b,r_c$ denote the radii of its excircles against $A,B,C.$ $H$ is the orthocenter of $\triangle ABC$ and $\sigma_a,\sigma_b,\sigma_c$ denote the radii of the incircles of $\triangle HBC,$ $\triangle HCA,$ $\triangle HAB.$ Prove synthetically the relation

\[\sigma_a+\sigma_b+\sigma_c+r_a+r_b+r_c=2s \]

$ a \equal{} BC, b \equal{} CA, c \equal{} AB.$ $ R$ is circumradius, $ r$ inradius, $ r_a, r_b, r_c$ the A-, B-, C-exradii, and $ s$ semiperimeter of $ \triangle ABC$. $ \varrho_a, \varrho_b, \varrho_c$ are inradii and $ \sigma_a, \sigma_b, \sigma_c$ semiperimeters of $ \triangle HBC, \triangle HCA, \triangle HAB.$

$ R(\cos A \plus{} \cos B \plus{} \cos C) \equal{} R \plus{} r,$
$ R(\cos A \minus{} \cos B \minus{} \cos C) \equal{} R \minus{} r_a.$

Using these equations yields

$ \sigma_a \equal{} \frac {2R(\cos B \plus{} \cos C) \plus{} a}{2} \equal{} \frac {r_a \plus{} r \plus{} a}{2},$
$ \varrho_a \equal{} (\sigma_a \minus{} a) \tan \frac {\widehat{CHB}}{2} \equal{} \frac {r_a \plus{} r \minus{} a}{2 \tan \frac {A}{2}} \equal{}$ $ \frac {2s \minus{} a \minus{} 4R \cos^2 \frac {A}{2}}{2} \equal{} s \minus{} \frac{a}{2} \minus{} R(1 \plus{} \cos A).$

Cyclic sum is

$ \sum_{\odot} \varrho_a \equal{} 3s \minus{} s \minus{} 3R \minus{} R \sum_{\odot} \cos A \equal{} 2s \minus{} \sum_{\odot} r_a.$


number theory
greatest common divisor
summer program
MathPath

Show that if the positive integers a, b, c, d satisfy ab = cd, then we have gcd(a,c) gcd(a,d) = a gcd(a,b,c,d).

dudes someone answer i don't know the answer  :(
seriously or could a mod move this to a forum where people would try it?

[quote="rd5493"]where r u getting all of these foreign tests.  u r from the states arent u?  to be honest, i dont feel like trying to work it out right now, got 2 practice for nats and try to figure our mathpath qualifying exam[/quote]
dude am i from the states?
yeah
just look for foreign tests and you find them...some without answers...or solutions which do help.


quadratics
algebra
polynomial
LaTeX
geometric series

Ok, our summer class just proved and derived a formula to find the Nth fibonacci number.  See if you can do it too!

Specific rules:

1) State the formula
2) Tell how you found the formula
3) Prove the formula!

I'd like to see if there are any other ways than the way we did it.

Extra challenge:

Derive and prove a formula for the Nth term in the following sequence:

$1, 1, 3, 7, 17, 41, 99, 239 ...$

Despite the fact that this series is probably named by now, they will henceforth be called the Dilnacci numbers (named after the teacher who told us to find the explicit formula).

I've done this before but I've never looked at the theory behind it. Here is a general idea.
[hide]You define the sequence recursively, then find a corresponding equation (quadratic in this case). The roots of this corresponding equation are the ?characteristic roots? of the original sequence. The formula for the nth term is the sum of each root to the nth power, with a constant in front of each root to the nth power. The explicit formula for the nth term of the Fibonnacci Sequence is Binnet's Formula.

Supposively a recursive sequence like the Fibonacci Sequence can be approximated as the sum of two geometric series because its defining recursive relation involves the two previous terms right? Also, I know that this method does not always work. Can anyone exactly what's behind this method?[/hide]

Sure you can, why not?

[hide]Your sequence is represented recursively by

$\\ a_1 = 1 \\ a_2 = 1 \\ a_{n+1} = 2a_{n} + a_{n-1} \ \mbox{for} \ n > 2$

You've got

$a_{n+1} - 2a_n - a_{n-1} = 0$

The characteristic polynomial for the reccurence above is

$x^2 - 2x - 1 = 0$

Lets find zeros for the equation above

$\Delta = 4 - 4(-1) = 8$
$\sqrt{\Delta} = \sqrt{8} = 2 \sqrt{2}$

$x_1 = \frac{2 + 2\sqrt{2}}{2a} = 1 + \sqrt{2}$

$x_2 = \frac{2 + 2\sqrt{2}}{2a} = 1 - \sqrt{2}$

So the formula for $n$-th number is given by

$a_n = A (1 + \sqrt{2})^n + B(1 - \sqrt{2})^n \ \mbox{for some} \ A, B \in \mathbb{R}$

We've got to find $A \ \mbox{and} \ B$

$a_1 = A(1 + \sqrt{2}) + B(1-\sqrt{2}) = 1$
$a_2 = A(1 + \sqrt{2})^2 + B(1-\sqrt{2})^2 = 1$

$a_1 = A + A \sqrt{2} + B - B \sqrt{2} = 1$
$a_2 = A(1 + 4\sqrt{2} + 2) + B(1 - 4\sqrt{2} + 2) = 1$

After finding roots of two linear equations above you have

$A = \frac{\sqrt{2}-1}{2} \ \mbox{and} \ B = \frac{\sqrt{2}+1}{2}$

Eventually we have a compleate formula

$a_n = \left(\frac{\sqrt{2}-1}{2}\right)(1 + \sqrt{2})^n + \left(\frac{\sqrt{2}+1}{2}\right)(1 - \sqrt{2})^n$[/hide]

[edit]

What's the point in hiding a solutions?
I haven't heard of that rule before.

Drunner, you have the basic idea, just put it into math.

Sage, would you mind hiding your solution-ish thing (I guess it's more of a derivation).

Excuse my Latex if it is not correct. I haven't used extensive amounts of it for months now, and I'm learning new things (at least trying to).

[hide]Binet's formula for the Fibonacci sequence states that the nth term of the Fibonnaci sequence is $\frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^{n+1}-\left(\frac{1-\sqrt5}{2}\right)^{n+1}\right]$.

A derivation of this formula is given in the AoPS book.

The main idea is to guess that the Fibonacci sequence can be written as the sum of two different geometric sequences (let them be $ar^n$ and $bs^n$) to that

$F_n=ar^n+bs^n$.

By the definition of the sequence, $ar^{n-1}+bs^{n-1}+ar^n+ bs^n =ar^{n+1}+bs^{n+1}$.

Grouping the a's together and factoring, we get

$ar^{n-1}(1+r-r^2)=-bs^{n-1}(1+s-s^2)$

Then solve for r and s, and using that, a and b can be found. The formula could then be derived.
[/hide]

The point would be that people don't accidentally stumble upon a way to do it, they choose to by clicking (I usually assume something not hidden is not a solution and thus I can read it even if I still want to do the problem myself).  It's just kind of a courtesy, I guess.  If you don't want to, that's fine, I guess.  It's not that big of a deal at all :).

[quote]Excuse my Latex if it is not correct. I haven't used extensive amounts of it for months now, and I'm learning new things (at least trying to).

[hide]Binet's formula for the Fibonacci sequence states that the nth term of the Fibonnaci sequence is .

A derivation of this formula is given in the AoPS book.

The main idea is to guess that the Fibonacci sequence can be written as the sum of two different geometric sequences (let them be  and ) to that

.

By the definition of the sequence, .

Grouping the a's together and factoring, we get

Then solve for r and s, and using that, a and b can be found. The formula could then be derived. [/hide]
[/quote]

This is the equation we used in my class!  If you want, feel free to state what the AoPS book has for the derivation, I'd love to see if it's the same as my coach's (I think he got his possibly from it as well, though..._).  Does it use $\alpha$ instead of $\frac{1 + \sqrt{5}}{2}$ and $\beta$ instead of $\frac{1-\sqrt{5}}{2}$?  That at least is how we simplified it... fascinating though how they're both technically roots of $x = 1 + \frac{1}{1 +\frac{1}{1+\frac{1}{1 + ...}}}$!  $\phi$ is so cool!

[quote="futuremontrealian"]
fascinating though how they're both technically roots of $x = 1 + \frac{1}{1 +\frac{1}{1+\frac{1}{1 + ...}}}$!  $\phi$ is so cool![/quote]

careful, that expression is not negative so it cannot be the negative root

well, I said "technically" for a reason.  I understand that it's obviously positive :).

[quote="Sage!"]What's the point in hiding a solutions?
I haven't heard of that rule before.[/quote]

Solutions are hidden in case someone who has not yet solved the problem does not wish to see the solution.

All of these can be solved using characteristic equations.  For the second one, use the fact that $a_{n+1}=2a_n+a_{n-1}$.  The rest is just simple.

EDIT: I just noticed it was already answered.  :oops:

[quote="futuremontrealian"]This is the equation we used in my class!  If you want, feel free to state what the AoPS book has for the derivation, I'd love to see if it's the same as my coach's (I think he got his possibly from it as well, though..._). [/quote]

The entire thing would take too long to type, and the amount of Latex is enormous. Do you have the book? [It's in the second volume, chapter 16.]

I'm pretty sure my coach does.  I can understand why it would be hard to latex :).  But if it uses $\alpha$ and $\beta$ chances are it's the same one.

And don't be afraid to post your ideas about how to solve it even if it's already been done, I'd like to see how many people can find a good equation :)

[quote="Sage!"]
[hide]$a_n = \left(\frac{\sqrt{2}-1}{2}\right)(1 + \sqrt{2})^n + \left(\frac{\sqrt{2}+1}{2}\right)(1 - \sqrt{2})^n$[/hide]
[/quote]

[hide]I can't help but notice that this can be simplified further to

$a_n = \frac{\left(1+\sqrt{2}\right)^{n-1}-\left(1-\sqrt{2}\right)^{n-1}}{2}$
[/hide]

[quote="TheAmazingOne"][hide]I can't help but notice that this can be simplified further to

$a_n = \frac{\left(1+\sqrt{2}\right)^{n-1}-\left(1-\sqrt{2}\right)^{n-1}}{2}$
[/hide][/quote]
Not quite.  Check your signs - $\sqrt2$ terms have to cancel.

Are you sure? At least according to my calculator it gives the right sequence.

On mine it doesn't.  For $a_2$ it gives me $\sqrt{2}$... I'm pretty sure that's not in the sequence  :lol:

Oh, I see. The problem is that Sage's answer had a sign error in it, so although my equation was equivalent to his, it also had a sign error in it. The terms in the numerator should be added rather than subtracted.

I agree that it was his error (no offense at all to him, I probably couldn't do it on my own)... I put his onto my calculator as well and it yielded the same answers as yours.

What equation is on your calculator giving you the right sequence?  It must be different or something...

The correct equation is

[hide]
$a_n = \frac{\left(1+\sqrt{2}\right)^{n-1}+\left(1-\sqrt{2}\right)^{n-1}}{2}$
[/hide]


Find the integer values of x for which the following function takes integer values

f(x)=x^2/(x+3)


For any positive integer n the sum 1+1/2+...+1/n is written in the form P(n)/Q(n), where P(n) and Q(n) are relatively prime.
a) Prove that P(67) is not divisible by 3;
b) Find all possible n, for which P(n) is divisible by 3.

I think the problem has been discussed at here last summer: http://www.mathlinks.ro/Forum/viewtopic.php?t=5762

[quote="iura"]I agree with you, grobber.
Define S[n] the residues mod 3^n and coprime to it.
then for every a there is a' s.t.aa'/3^n=1(mod 3^n)
Also if 3^k|a, define a' s.t. aa'/3^(n-k)=1(mod 3^(n-k)).
Suppose 3^k<n<3^(k+1) we must have
sum a' (a=1..n) divisible by 3^(k+1) (*)
 We group 1..n into A[1]..A[k] by the power of 3 dividing them
By (*) we must have 3|sum(a')  ai n A[k].
Hence both 3^k and 2*3^k belong to A[k]. So 2*3^k<=n.(1)
If we have k=1 then this yields n=2.
Now consider A[k-1] By (1) we can show 1,2,4,5 belong to A[k-1] and maybe 7 or 7, 8.
But we use (*) and (1) to get 3|sum(a') a' in A[k-1] + 1.
Checking we get just 7 belong to A[k-1].
So 7*3^(k-2)<=n<8*3^k-2. If k=0 we get n=7.
Continuing like this we get 22/27<=n<23/27 yielding n=22 for k=3.
But this procedure gives no solution further (we have 2 possible residues to choose, and the desired residue in luckily not among them).
Hence we stuck and we get grobber's result.
I know my solution is hard to understand. Sorry I can't explain it.
To solve it you must draw the tables of a' for S[1],S[2],S[3],S[4]. 81 numbers.
Elegant, huh?
If this is the only solution they've got, this is a horrible problem.
Maybe it involves a beatiful idea, but counting 81 numbers..[/quote]
can someone explain it because I can't understand it?please write the thought and the conclusion.
I also have a thought.for P(k) is not divisiable by 3,we can prove P(3k),P(3k+1),P(3k+2) is not ,either.

[quote="zhaobin"][quote="iura"]I agree with you, grobber.
Define S[n] the residues mod 3^n and coprime to it.
[..............]
Maybe it involves a beatiful idea, but counting 81 numbers..[/quote]
can someone explain it because I can't understand it?please write the thought and the conclusion.
I also have a thought.for P(k) is not divisiable by 3,we can prove P(3k),P(3k+1),P(3k+2) is not ,either.[/quote]Could you please continue in that topic this discussion? Thanks!


For $a,b,c$ positive reals such that $ab+bc+ca=1$ prove that

$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \sqrt{3} +\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}$

Replace each $1$ from the numerators with $ab+bc+ca$, then subtract $\sum \frac{ab}{a+b}$ from both sides. We get $a+b+c\ge \sqrt 3$, which is obvious from $(a+b+c)^2\ge 3(ab+bc+ca)=3$.

It was proposed form some Romanian NMO (I think it was in the contest).

This one was proposed for the national stage in 2002, 9th grade, it was in the contest, author: Dinu Teodorescu.

I think i hv another sol.
Put x=tanA, y=tanB, z=tanC s.t. A+B+C=90
it reduces the ineq to tanA+tanB+tanC>=sqrt3 which is trivial as
tanA+tanB+tanC >= sqrt(3(tanAtanB+tanBtanC+tanCtanA))=sqrt3


Let $a, b, c$ are positive real numbers such that $abc = 1$. Then show that
$a^{b+c}b^{a+c}c^{a+b}\leq 1$.

Inmo 2001 here [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=a%5E%7Bb%2Bc%7D&t=55430[/url]

Taking logarithms, we want to show that given $\sum \log a = 0$, we have

$\sum (a+b)\log c \leq 0$

If $a\leq b\leq c$, then $\log a \leq \log b \leq \log c$, so that by Chebyshev,

$\sum (a+b) \log c \leq \frac{1}{3}\left(\sum (a+b)\right)\left(\sum \log a \right)=0$

we only need to use $a^ab^b\geq a^bb^a$ for all positive $a,b$ and notice that the inequality is equivalent to

$a^{b+c}b^{a+c}c^{a+b}\leq (abc)^{\frac{2(a+b+c)}{3}}$


What is the number of square units in the area of this rectangle if the radius of each circle is 5' units and the circles are tangent to each other and the rectangle as shown?
[asy]for(int i = 0; i < 15; ++i){draw(Circle((1+ 2*(i % 5), 1+ 2 *(int) (i/5)),1));}
draw((0,0)--(10,0)--(10,6)--(0,6)--(0,0));[/asy]

if the radius is 5 then the diameter of each circle is 10.

we see that there are 5 circles going across so 5*10=50
we see that there are 3 circles going down so 5*3=15

15*50=750 in area

actually, you made a mistake yaofan :(

The longer dimension is correct: $ 50$, but the shorter dimension is not. You stated that each circle has diameter $ 10$, and there are $ 3$ circles going down, so the shorter dimension is $ 30$. This results in an area of $ 30\times50\equal{}\boxed{1500}$.
I believe you may have multiplied the radius by $ 3$ instead of the diameter ;)


I've been doing a small Middle School contest for a while and thought people might be interested.
You can find everything (including rules) on [url=http://ca.geocities.com/dmmpsolved]ca.geocities.com/dmmpsolved[/url],
but ignore the part about Don Mills and you don't need to write the name of your math teacher
if you're not from Don Mills  :D
Anyways here is problem set 5 (Please send your solutions to dmmpsolved@yahoo.ca until April 30, 2005
-> tons of time :))

[size=150]DMMP (Don Mills Mathematics Program)[/size]

5.1 A positive integer is called [i]lucky[/i] if $7$ divides the sum of its digits. A [i]lucky pair[/i]
      is a pair of two consecutive positive integers that are both lucky.
      [4] (a) Give an example of a [i]lucky pair[/i]
      [3] (b) Prove that there are infinitely many [i]lucky pairs[/i]

5.2 We place a “$+$” in one corner cell of a $4$×$4$ table and “$-$“ in all the
      other cells. It is allowed to flip the signs (replace a $+$ by a $-$ or
      a $-$ by a $+$) either in a column or in a row. Can one get a table with all
      pluses in a finite number of steps? If yes, how? If no, why not?

5.3 Let $ABC$ be an isosceles triangle with $BC=AC=7$ and $AB=2$ . Let $D$
      be a point on $AB$ produced so that $B$ lies between $A$ and $D$, and
      $CD=8$ . Determine (with proof) the length of $BD$.

5.4 Does there exist a triangle with
     [3](a) Area greater than $100cm$2 , but all altitudes smaller than $1cm$ ?
     [4](b) Two of the altitudes greater than $100cm$ , but area smaller than $1cm$2 ?
        For both parts, justify your answer.

5.5 Let $n>1$ be a positive integer.
Prove that $n<1+\sqrt[2]{2}+\sqrt[3]{3}+...+\sqrt[n]{n}<2n$

Good luck with these  :D

[quote="hiitsme"]April 30?!!???!?!!??!?!!?
Is that when dmmp 6 is coming?[/quote]
Yes, I don't expect people to do math during their marchbreak, as they might be on vacation :D
I also want to give people more time, so there is no time pressure. :P

[quote="mathgeek2006"]Then people will procrastinate and forget, then it won't work out.[/quote]
Who says you have to hand it in on the last day?  ;)

Well, people love to procrastinate, and also do extreme things, so they'll wait 'til the last minute, and then they'll fail! ;)

It's funny you say that on my last day of march break, as I haven't finished my homework  ;) .

Our school has two more days left of the break. *so lucky*


MIT college AMC USA(J)MO USAMO calculus geometry

Dude, just make like Physics Team or MOSP, and then you can get a few Bs and still get into a good college.

[quote="jhredsox"]next year i will be a freshman. i am worrying that if i dont get straight a's (won't be easy! - seriously! :mad: )...what should my strategy be

how much time should i study on english, science, etc. obviously i don't want to spend ALL of my time, but mit is worth doing a lot of work (when you think about 4.5 years from now)

-jorian[/quote]

you really oughtn't have a problem with grades, really, until the heavier coursework years, like junior or senior year. I can't say for your school, but freshman year will be the easiest of your four years, and you ought to take the time to study as much math/science for competitions as you can, while you still have the luxury of doing so -- I regret not studying more math freshman/sophomore year, and I never made it past AIME.

[quote="jhredsox"]you want me to read off the validctorian's scores:

Here it is
(***names have been changed***)
______________________________________________________
[quote]_________, daughter of ____ and _________________ is the top-ranked senior. She will attend Tufts University to major in biology. ________ was inducted into the National Honor Society, Science National Honor Society and Foreign Language National Honor Society for French and Latin. She was designated a national Merit Commended Scholar, Governor's Scholar and AP Scholar. ________ was the recipient of the Governor's Academic Achievement Award, the University of Pennsylvania Book Award, the CAS-CIAC Scholar athlete award and the Hartford Courant All Academic Award. _______ achieved summa cum laude in the national Latin Exam and first place in the National French Essay Contest. ________'s activities included the varsity swim team, where she was captain, treasurer of the Key Club, secretary of the French Club and sports editor for the Hawk's Eye. She participated in the Eastern Region Music Festival and New England Music Festival, the French Exchange Program in Dinard and Camp Sunrise.[/quote]
-jorian[/quote]

What does this have to do with the topic? So there exists some academic/extracurricular beast. And that means... you need to get straight A's?

Returning to the topic at hand, there's not a surefire studying strategy to maintain a 4.0 GPA through your high school years. It is probably true, however, that your freshman year will be the easiest. You can probably keep your grades up just by doing the work and paying attention in class; it's a lot easier than doing something like making the USAMO/MOP. And of course MIT knows that, so if you're weighing a 4.0 against USAMO qualification, I think you know which way the scale will tilt.

[quote="paladin8"]You can probably keep your grades up just by doing the work and paying attention in class; it's a lot easier than doing something like making the USAMO/MOP.[/quote]

True, but trying to qualify for MOP is much more fun :P

[quote="JSteinhardt"]And at the same time Freshman year is your opportunity to explore, well, opportunities. Activities. Things you might like that you did not previously know about. Don't waste it worrying incessantly about grades.

Though for English, maybe a start would be to start using correct grammar on this forum  :wink: .

Oh and I'm not sure that Valedictorian's list of achievements is really that impressive except that she can apparently study well, presumably likes Latin, and is quite good at inflating her resume. I think most colleges can recognize when someone is inflating their resume.

And I'm not sure you would be particularly happy at MIT if you have to work non-stop for 4 years just to get in. Not that obsessively caring about a 4.0 GPA will particularly do anything positive for you, at least as far as MIT is concerned:

[quote="http://www.mitadmissions.org/topics/before/recommended_high_school_preparation/index.shtml"]
Some applicants struggle to turn themselves into clones of the "ideal" MIT student - you know, the one who gets triple 800s on the SAT. Fortunately, cloning is still for sheep. What we really want to see on your application is you being you - pursuing the things you love, growing, changing, taking risks, learning from your mistakes, all in your own distinctive way. College is not a costume party; you're not supposed to come dressed as someone else. Instead, college is an intense, irreplaceable four-year opportunity to become more yourself than you've ever been. What you need to show us is that you're ready to try.
[/quote]

With that in mind, it is good to try in your classes, and to be learning, but in some senses, provided that you are taking sufficiently challenging classes, all that getting below an A shows is that you are being challenged enough.[/quote]

Yes, in English I have trouble getting good grades on compositions. They are not to my liking. I hope I do get placed in Level 1 for English next year.

-jorian

wtf, dude i had like 4 Cs in high school and an unweighted GPA of about 3.5 (B average) and I ended up getting accepted into Brown (still waiting on my other colleges)...grades are not THAT big of a deal. It's really all about your personality in your essays and your ECs.

On the subject of this thread, what does a ninth grade class schedule look like in your school? I'd love to hear lots of examples.

Here's one schedule example I know of:

1) honors English

2) honors science

3) algebra I

4) general business

5) physical education

6) social studies (American history)

Here's another, for ninth grade at a different school:

1) calculus C and linear algebra

2) AP United States history

3) AP physics B

4) C programming III

5) 10th grade honors English

6) seminar: democracy, freedom, rule of law

Note that both of these schedules lack ninth grade foreign language courses, because of limitations of course offerings at the schools. I think, because of a different place where I lived while in seventh and eighth grade, that it is perfectly natural, and desirable, to begin foreign language study at a younger age.

What is the schedule for ninth grades like at your school? What is/was your schedule like?

not sure for everyone but java seems to be the programming language to learn(not c or c- or c..)
and Latin and Greek are foreign languages(i will be only taking Latin in 9th)

tokenadult, what kind of school did that second kid go to? That's outrageous... (I'm jealous)

here's what I took as a 9th grader:
1) geometry
2) physical science (1 semester chem, 1 sem physics)
3) Spanish 1
4) English 1
5) world history
6) web page (intro to html, frontpage, dreamweaver)
7) driver's ed (1 quarter)/PE (3 quarters)

that 2nd kid looks nuts

my freshman schedule was like, english/math/history/science, PE, lunch, german (only two classes were honors)

uh

ap calc
ap stat
"gifted and talented" (gt) biology
gt american government
gt french 3
honors english
P.E.

i think that's it

Mine was:

Honors English I
Geometry Honors
Latin I
Physical Science Honors
Intro to Keyboarding
Gate/Honors Global Studies I
PE
Lunch

Wow, my 9th grade schedule was lame.

I'm sure we'd all love to not need to sleep, but it's kinda a necessity...

I've tried getting my counselor to give me harder classes, but apparently if I don't take the required classes, I can't graduate. (I suppose that doesn't matter now that I'm going to Early Entrance.) But this is my schedule (as a 9th grader):

AP Calc (and it took a long time to convince the head of the department to let me in)
Latin I
Orchestra
World History
9th grade int. science
9th grade Language Arts

hmm i've had some different experiences...
i've had to deal with 2 dept heads in my life for math, and they were both very accomadating. in MA, the dept head just said that, if i wanted to take BC Calc as a 10th grader, it already shows the resolve and ability...

and my dept head here was slightly skeptical at first, but i told her about my competition history, and she let me into the program stanford has with our high school :D  (we take classes with the undergrads, but theres no official record other than independent study credits and possibly a grade, not even sure if we get a letter grade...)

I had world history this year.

Honors Alg II/Language Arts/ Biology/History/Math Team/Spanish I/Cross Country-Track

only math at our school, unless it is Advanced Placement, is weighted out of 5.

[quote="13375P34K43V312"]This will be my schedule for 9th grade.

Honors Precalc
English (Unleveled)
World History I (Unleveled)
Honors Earth Science
Honors Spanish II
PE
Health
Maybe Intro to C++ who knows[/quote]
That's a little odd that your english is unleveled. By the way, how do you know your schuedule already??? In your school, is Honors the same as AP? (Or is it just level 1) I should try to get into AP Bio freshman year, geometry (i know, i'm dumb), French..., English lvl. 1......
-jorian

You guys are lucky. I just transfered to a school that doesn't have an Honors sequence or any AP classes. They claim that all classes are Honors. Pshhh. Plus, I am now in the middle of nowhere.

This was my freshman schedule:

Algebra II
Biology
Technology
English 9
Computer Science
Japanese 1
PE/health

(All classes at TJ default honors)

your school offers japanese? no fair

haha we only have spanish, german, french, and latin  :mad:

Dude, he goes to TJ.

Those are the four languages my school offered as well.

[quote="now a ranger"]

haha we only have spanish, german, french, and latin  :mad:[/quote]

Learn the languages in this priority: German, Latin, Spanish, French.  :)

Our school has spanish, french, latin, german, italian, mandarin, and asl (american sign language). I'm planning to only do one, Spanish.

I'm Taking spanish

I'm chinese myself, so Mandarin would be really fun hearing all the other non-satisfactory accents  :lol:

they should have a taiwanese class too someday

ASL rocks, when teachers tell you to be quiet you can use signs to talk

The high school in my town has
______________________________________________________[list]
1.American Sign Language
2.Arabic
3.Chinese
4.French
5.German
6.Japanese
7.Latin
8.Russian
9.Spanish
10.English Language Learners
[/list]
[url=http://www.foreignlanguage.org/languageresources/index.htm]Click here for details.[/url]

Guess which two I'm taking!

-jorian


To walk on water, all you need is a pair of water-walking boots shaped like boats. If each boot is 26 cm high and 32 cm wide, how long must they be to support a 78 kg person?

Shouldn't we know the weight of the wood?