Why is the sum of the exterior angles for any polygon $ 360^{\circ}$?

Does a circle have exterior angles?

Sum of the interior angle is $ 180(n \minus{} 2)$ where n is the number of sides of a polygon

So EACH angle is $ \boxed{\frac {180(n \minus{} 2)}{n}}$

draw a hexagon or polygon any regular polygon shape for that matter and extend the line.

[img]http://i21.tinypic.com/30cuflw.jpg[/img]

Notice that each sides is supplementary meaning that the interior + exterior angles add up to $ 180^o$

therefore each exterior angle = $ 180 \minus{}  \frac {180(n \minus{} 2)}{n}$

so sum of $ n$ interior angles would be $ n[180 \minus{} \frac {180(n \minus{} 2)}{n}]$

expand and implify and you will get $ 360^o$

-------------

you should always ask good quesitons like this. Reply back if you don't understand any step

The measure of each angle you give is true only for regular polygons.

What if the polygon is irregular. How can one prove the sum is still $ 360^{\circ}$

Is a circle a regular polygon? Does it have infinite exterior angles (tangents)?

[quote="Fraenkel"]
Is a circle a regular polygon? Does it have infinite exterior angles (tangents)?[/quote]

Yes, a circle is like a regular polygon with an infinit number of sides (and tangents/exterior angles)

[quote="Fraenkel"]The measure of each angle you give is true only for regular polygons.

What if the polygon is irregular. How can one prove the sum is still $ 360^{\circ}$

[/quote]

Here's a picture that I drew..

The answer to your question will be yes if a polygon is not concave.

(There's supposed to be a line at the bottom..)

Yeah I was able to prove by induction that for any polygon the exterior angles' sum is $ 360^{\circ}$.

Does anyone know a proof for this without using induction.

[hide="proof"]
Let an n-gon have interior angle measures $ a_{1}$, $ a_{2}$, $ a_{3}$...$ a_{n}$.  Then, the exterior angle measures are $ 180\minus{}a_{1}$, $ 180\minus{}a_{2}$, $ 180\minus{}a_{3}$...$ 180\minus{}a_{n}$.  Summing these terms gives us $ 180n\minus{}a_{1}\minus{}a_{2}\minus{}a_{3}\minus{}...\minus{}a_{n}\equal{}180n\minus{}(a_{1}\plus{}a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n}$.  However, the sum of the interior angles of an n-gon is $ 180(n\minus{}2)$.  So $ a_{1}\plus{}a_{2}\plus{}a_{3}\plus{}...\plus{}a_{n}\equal{}180n\minus{}360$.  Plugging this in gives us $ 180n\minus{}(180n\minus{}360)\equal{}180n\minus{}180n\plus{}360$ which is 360
[/hide]

[quote="alanchou"][hide="proof"]
Let an n-gon have interior angle measures $ a_{1}$, $ a_{2}$, $ a_{3}$...$ a_{n}$.  Then, the exterior angle measures are $ 180 \minus{} a_{1}$, $ 180 \minus{} a_{2}$, $ 180 \minus{} a_{3}$...$ 180 \minus{} a_{n}$.  Summing these terms gives us $ 180n \minus{} a_{1} \minus{} a_{2} \minus{} a_{3} \minus{} ... \minus{} a_{n} \equal{} 180n \minus{} (a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n}$.  However, the sum of the interior angles of an n-gon is $ 180(n \minus{} 2)$.  So $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} ... \plus{} a_{n} \equal{} 180n \minus{} 360$.  Plugging this in gives us $ 180n \minus{} (180n \minus{} 360) \equal{} 180n \minus{} 180n \plus{} 360$ which is 360
[/hide][/quote]

That's exactly how I did it, but I had to prove $ 180(n\minus{}2)$ is the sum of the interior angles by induction.
At the inductive step I split the polygon into triangles and so forth came to the conclusion.

I was just curious whether we can do this step without induction.


For natural numbers $ p,q$ such that $ p^2>4q$, let $ a\equal{}\frac{p\plus{}\sqrt{p^2\minus{}4q}}{2},b\equal{}\frac{p\minus{}\sqrt{p^2\minus{}4q}}{2}$.

Prove that $ a^m\minus{}b^m$ divides $ a^{km}\minus{}b^{km}$ for all natural numbers $ m,k$.

let x,y denote $ a^m,b^m$.  obvious that we can factor out $ x\minus{}y$ from $ x^k\minus{}y^k$. to show the other factor is an integer, note that its symmetric in a,-b.  so if a-b and -ab are integers we are done.

[quote="Singular"]let x,y denote $ a^m,b^m$.  obvious that we can factor out $ x \minus{} y$ from $ x^k \minus{} y^k$. to show the other factor is an integer, note that its symmetric in a,-b.  so if a-b and -ab are integers we are done.[/quote]

How can you be sure it can be written in form of $ a\minus{}b$ and $ \minus{}ab$?

The problem assumes implicitly that $ a^m \minus{} b^m$ is an integer, but this is plainly false when $ m \equal{} 1$.  Did you mean $ b \equal{} \frac{\minus{}p \plus{} \sqrt{p^2 \minus{} 4q}}{2}$ or $ a^m \plus{} b^m$?

Terribly sorry... actually I wanted to mean $ \frac{a^m\minus{}b^m}{\sqrt{p^2\minus{}4q}}$  :blush:

Note that $ a, b$ are roots of the quadratic equation $ x^2 \equal{} px \minus{} q$, hence $ s_m \equal{} \frac {a^m \minus{} b^m}{a \minus{} b}$ satisfies the recurrence $ s_{m \plus{} 2} \equal{} ps_{m \plus{} 1} \minus{} qs_m$.  We also have $ s_0 \equal{} 0, s_1 \equal{} 1$.  The result is a generalization of the similar result for Fibonacci numbers ($ p \equal{} 1, q \equal{} \minus{}1$) and can be proven similarly:  calculate the sequence $ \bmod s_m$.


[quote="Zenith Xan"][quote="The QuattoMaster 6000"][hide] Firstly, note that $\frac{x-5}{x+3}=\frac{2(x-5)}{2(x+3)}=\frac{2x-10}{2x+6}$.  Now, we have that $\frac{7x}{2x-6}-\frac{x-5}{x+3}=\frac{7x}{2x+6}-\frac{2x-10}{2x+6}$ because of our first equation.  Now, we can say that $\frac{7x}{2x+6}-\frac{2x-10}{2x+6}=\frac{7x-(2x-10)}{2x+6}=\boxed{\frac{5x+10}{2x+6}}$. [/hide][/quote]

You didn't actually finish. You know your boxed equation is equal to 8, right?[/quote]
Oh, whoops!  I thought it said simplify!   :blush:
[hide] Alright, so we have that $\frac{5x+10}{2x+6}=8$.  We can multiply $2x+6$ on both sides to obtain that $5x+10=8(2x+6)=16x+48$.  Now, we can simplify to obtain that $11x+38=0\Rightarrow 11x=-38\Rightarrow \boxed{x=\frac{-38}{11}}$ [/hide]


function
trigonometry
algebra
partial fractions
rational function
calculus
calculus computations

I have a problem with this: Determine function $f: \mathbb{R}\to \mathbb{R}$ such that its Laplace transformation is $F(s)=\frac{s}{s^2+4}-\frac{2s}{e^s(s^2-4)}$ for every $s\in\mathbb{R}$. I found that $\frac{s}{s^2+4}$ is the Laplace transformation of $\cos 2x$ but what about the latter term?

Start with finding the inverse Laplace transform of $\frac{2s}{s^2-4}$ (without the exponent). Using partial fractions, we see that $\frac{2s}{s^2-4}=\frac 1{s-2}+\frac1{s+2}$. Hence the inverse Laplace transform of this rational function is $e^{2t}+e^{-2t}$. Now recall that $\mathcal L^{-1}[f(s)e^{-as}]= u_a(t)[\mathcal L^{-1}f](t-a)$ where $u_a$ is the step function jumping from $0$ to $1$ at $a>0$.


Triangle is acute  and non-isosceles
H denotes the orthocentre, I the incentre and O the circumcentre

If one vertex of the triangle lies on the circle through H,I and O prove that at least two vertices lie on this circle

Yetti commented on this problem here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43999]www.mathlinks.ro/Forum/viewtopic.php?t=43999[/url]


Find the number of roots of $ 2x = (2n+1) \pi ( 1 - \cos{x} ) $ , where $n$ is a positive integer. Find an estimate of the domain of the roots.

Won't that be somewhat crude? Is there some sort of algebraic approach??

Yes there is a algebraic method. The answer is [hide] $ 2 n + 3 $ [/hide] . I'll post the solution later!

Try considering the sign of LHS - RHS at $k \pi$ and $(k+1) \pi$ (both even and odd) . The answer comes straight away after including $0$ and $(2n+1) \pi$


Hi all,

I have a friend who created a thesis.cls template. It works great. He found out that the first page of each chapter should start 2 inches from the top rather than 1.5. He gave me the code he used for the .cls template.

\def\ps@myheadings{\ps@empty \vspace{1in}
  \def\@evenhead{
    \setTrue{runhead}
    \normalfont\normalsize
}
  \def\@oddhead{
    \setTrue{runhead}
    \normalfont\normalsize
}
}

To call it up at the proper time in the tex document, this is the code used:
\chapter{INTRODUCTION}
\thispagestyle{myheadings}

It works great except for some reason, the page number is no longer on that page. The words are type set all the way to the bottom with no page number. On the following pages, the page numbers correctly resume.  This is happening throughout the document wherever  there is a chapter page. So, if the chapter heading page is #3, then it no longer shows up, but the next page shows the page number correctly as  #4.  Then the next chapter page could be 16 and won't show up, but the following page is 17 and it resumes properly, etc...
If there is anyone with any idea on how I  can get the .cls code to recognize the page number, please let me know! Thanks again - my buddy is away for the summer so he can't help with this.

I've been fiddling with the .cls template code and I finally got the chapter page to actually show the page number. However, the page number is located in the top right hand corner of the page instead of the center of the page at the bottom. This is occuring for each Chapter page. The rest of the pages are not affected.

Here is the revised code, which can be compared with the earlier code I sent which resulted in no page number being shown for the chapter page:

\def\ps@myheadings{\ps@empty \vspace{1in}
  \def\@evenhead{
    \setTrue{runhead}
    \normalfont\normalsize
     \rlap{\thepage}\hfil\leftmark{}{}\hfil}
   \def\@oddhead{
    \setTrue{runhead}
    \normalfont\normalsize
      \rightmark{}{}\hfil \llap{\thepage}}
      \let\@mkboth\markboth

}
}

If anyone can decipher this and show me how to get the page number to show at the bottom center of the chapter page, please help. I would greatly appreciate it!
Thanks again - I'm hoping to hear from a LaTeX wiz!!


The energy that a fish uses swimming in a river is determined by the following formula:

$ E(v) \equal{} 2.73v^3(\frac{10}{v\minus{}x})$

Being x the speed of the current.

What would be the value of $ v$ that would minimize the energy used when the current is at $ 5$ miles per hour?

hello, the first derivative with respect to $ v$ ís given by
$ E'(v)\equal{}\frac{54.6v^2(v\minus{}1.5x)}{(v\minus{}x)^2}$
solve the equation $ E'(v)\equal{}0$.
Sonnhard.

Yes as Sonnhard said. I will show steps for the differentiation:

$ E'(v) \equal{} \frac {10}{v \minus{} 5} \cdot \frac {d}{dv} 2.73v^3 \plus{} 2.73v^3 \cdot \frac {d}{dv} \frac {10}{v \minus{} 5}$

$ E'(v) \equal{} \frac {81.9 v^2}{v \minus{} 5} \plus{} 2.73v^3 \big( \frac {(v \minus{} 5) \frac {d}{dv} 10 \minus{} 10 \frac {d}{dv} (v \minus{} 5)}{(v \minus{} 5)^2} \big)$

$ E'(v) \equal{} \frac {81.9 v^2}{v \minus{} 5} \plus{} 2.73v^3 \cdot \frac { \minus{} 10}{(v \minus{} 5)^2}$

$ E'(v) \equal{} \frac {81.9v^2(v \minus{} 5) \minus{} 27.3v^3}{(v \minus{} 5)^2}$

$ E'(v) \equal{} \frac {54.6v^3 \minus{} 409.5v^2}{(v \minus{} 5)^2}$

Step 1: Product Rule
Step 2: Quotient Rule and simplified left term
Step 3: Simplified right term
Step 4: Common denominators and simplify

So now we want to know when $ E'(v) \equal{} 0$. I'll let you try it, and if you need any help, just ask.


Hello,

   I have tried sending an e-mail to the AMC and the e-mail bounced. I sent it to multiple addresses under the maa.org domain and also had a friend attempt to send it as well, but all of the messages were bounced with the explanation that the address of the recipient did not exist. Is this intentional to avoid being flooded with e-mails, or is the server down, or is there a problem on our end (we do both use gmail, so it's conceivable that gmail is failing somehow).

Thanks,
~Jacob Steinhardt

The page of [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2008-12a/spry_perfect/2008-HSperfect.shtml]AMC perfect scorers[/url] says that there is sometimes a problem with the maa.org e-mail address, and suggests a unl.edu address.  Maybe you can try that address until maa.org seems okay again?

I've also sent four emails (using a yahoo account) to the maa.org address in the past three months regarding the AMCs but haven't gotten anything back (not even a bounce).

At first I thought they had a filter that got rid of emails that were too long, so I kept truncating my emails.. which didn't work.

[i]deleted by Admin -- see next post[/i]

Please do not use Steve Dunbar's (or anyone else's) non-AMC email address to contact about AMC matters.  Many people like to keep separate emails for separate purposes, and if Prof. Dunbar wanted that email to be available for AMC purposes, he would have put it on the AMC webpage.

If you need to contact the AMC urgently, their phone number is on [url=http://www.unl.edu/amc]their website[/url].

First of all, I apologize for that response. I didn't think of it that way (I simply thought about any other way that the email may be able to go through).

Anyway, yeah, ignore my earlier response to those who have read it.  :)


Answer the poll question, and add any other comments you would like.

I think Pittsburg defense, in particular their secondary is just too physical for Arizona's receviers.  Pittsburgh doesn't have a great offense but then again, Arizona doesn't have much of a defense.  I think Pittsburgh will be able to control the tempo of the game enough to pull out a win, although anything is possible.

I disagree. Arizona has more experience playing in conditions like they will be in Tampa. Both of Pittsburgh's games were in the snow, so they might have trouble adjusting to Tampa's ground. Also, Larry Fitzgerald and Anquan Boldin pose much more of a threat and are much better playmakers than their counterparts on the opposing side; Nate Washington and Hines Ward (who may not play). In fact, the only dynamic playmaker they have on offense is Santonio Holmes. I think Pittsburgh will end up double-teaming Fitzgerald and maybe even Boldin, opening up many 1 on 1 opportunities for the other receivers. and dont forget Kurt Warner is superb under pressure; the Eagles blitzed him almost 15 times and sacked him maybe 1 time, and blitzing is really what gives Pittsburgh its dynamite defense. Look for Kurt Warner to foil the blitzes and for Arizona to pull out an upset.

[quote="mz94"]Arizona has more experience playing in conditions like they will be in Tampa. Both of Pittsburgh's games were in the snow, so they might have trouble adjusting to Tampa's ground.[/quote]
That's a pretty strange argument.  Pittsburg is not a team that relies too heavily on its offense to win.  Pittsburgh total defense, rush defense, and pass defense were all first in the league, and Arizona has a minimal running game.  The secondary can just play the pass and force Arizona to win on the ground, which would be difficult.

[quote="JRav"]...and Arizona has a minimal running game. The secondary can just play the pass and force Arizona to win on the ground, which would be difficult.[/quote]

Arizona's running game had underperformed for much of the regular season, but it has improved greatly.  I think Arizona can be effective with the run if it becomes necessary.  However, I don't think it will be as easy for Pittsburgh to "play the pass" as you suggest it will be.

w00t! I live 60 miles away from Tampa!!! But, I don't understand football.

Does this belong in the Round Table instead?

i could really care less about sports. but i DO know who NOT to like, so i chose Stealers.

I think it will be a good game.  And in games like this where I don't really like or dislike either team, I root for the underdog, which in this case is Arizona.  And I think they're going to win.

Gah congrats to the Steelers... Cardinal's would've won if they didn't mess up at those two crucial points.

That 100 yard return was quite beast.

San Antanio Holmes was pretty awesome too.  w00t Steelers!! 6 Superbowl wins!!

[quote="1=2"]BOBBY SHEN IS GONNA BEAST THIS YEAR![/quote]

Um...what is this?

That would be spam in my thread.  Shame on you, 1=2!

Anyway, my opinion of the game, taken from my blog:
[quote]I just saw a video of the so-called "fumble" that ended the game (I didn't see the game on TV), and it was unquestionably an incomplete pass. The other major plays of the game (the interception return and the game-winning touchdown) were called correctly, but that one certainly wasn't. I'm not saying the Cardinals would've won the game given another opportunity, but they definitely had the chance taken away from them with a bad call from the officials.[/quote]

The refs got way too involved in the game and although Fitzgerald made a great play for the touchdown in the 4th, he left 2:37 left on the clock, which is far too much time.  Recall last year, the Patriots handed the ball over to the Giants with 2:39, and we all know what happened.

The patriots should have on last year. There were many penaltys on the giants that the refs didn't call. Such as a holding on the miracle catch by david tyree. Also on a huge gain on plaxico buress, there was learly an offensive pass interferene.

In that game, Mike Carey and his crew didn't get involved in the game for the most part, which is a good thing.  It's one thing if you consistenly call penalties on one team but not the other, but if you're not calling penalties on anyone, you can't complain.  I thought Carey and his team did a really good job last year.

Yeah, but still you can't be too leniant. There were some big plays that should have obviously been penalities. It could have changed the outcome of the game.

[quote="JRav"]the game[/quote]

GAAAH

I just looked football up on wikipedia, and it doesn't say how to play. How do you play?

[quote="1=2"]I just looked football up on wikipedia, and it doesn't say how to play.[/quote]
If you're being serious, actually, [url=http://en.wikipedia.org/wiki/American_football_rules]it does.[/url]

I don't care if a lot of penalties are called or if very few penalties are called, as long as it is consistent on both sides.  jjx1, I don't remember the exact call you're referring to (I remember the play, of course, just not the claimed missed holding call), but I can say that holding is often a tough judgment call to make.  It's also possible the referees simply failed to see it.

[quote="JRav"]...although Fitzgerald made a great play for the touchdown in the 4th, he left 2:37 left on the clock, which is far too much time.[/quote]
Good point.  I was thinking the same thing, but I hoped Arizona's defense could step up one more time.

[quote="cf249"]Good point. I was thinking the same thing, but I hoped Arizona's defense could step up one more time.[/quote]
And the thing is, from the third quarter until that drive, Arizona's defense was great.  Recall that they stopped Pittsburgh on 1st, 2nd, and 3rd and goal twice on the same drive.  And even on that last drive, they got Pittsburgh to third and long.  But when it really mattered, they just couldn't hold.

[quote="1=2"]How do you play?[/quote]
Here's the basic premise: get from one side of the field to the other.  You have 4 chances to advance a certain distance and if you can't the other team gets the ball.  If you can, you get another 4 chances to advance more, and so on.

It's Santonio Holmes, not San Antonio Holmes.

He went to Ohio State.  I remember watching him when they played Penn State (on tv...i'm not at college for 3 more years).
But yeah, I think pro football is a joke.  College football is sooooooooooooooooooooooooooooooooooo much better.


Distinct integers are given. Is it posible that sum of any two divides the sum of all given integers?

Ooo, i forgot, these are positive integers  :roll:

That's all you forgot?  :lol:

Let a and b be distinct positive integers. Then any sum of two elements of M={a,b} divides the sum of all elements of M.

:lol: Ok, and if that set has $2006$ or $2007$ elements. I think that for $2006$ elements it was on some Romanian competition...


probability
geometry
3D geometry
analytic geometry
pyramid
calculus
integration

Let $ N$ be an integer.

Consider the range $ [0, N]$.

Choose $ N$ real numbers randomly from within this range.

What is the probability that the sum of these $ N$ real numbers is less than $ N$?

[quote="atomicwedgie"][hide]1/n![/hide][/quote]

how do you prove this>?

The total space that we have to consider is an $ n$-dimensional cube of side length $ n$. The subset of this space that satisfies the condition is all points below the plane of points with sum of coordinates equal to $ n$. This region is an $ n$-dimensional pyramid. The easiest way to determine that this $ n$-dimensional pyramid has volume $ n^n/(n!)$ is with integrals, though I think there are other ways. (start by proving that the area of an $ n$-dimensional pyramid under the plane of poitns with sum of coordinates $ 1$ is $ 1/(n!)$) Dividing by the volume of the cube, $ n^n$, gives a probability of $ 1/(n!)$.

[quote="atomicwedgie"][hide]1/n![/hide][/quote]

[hide="I second that"]
Let's take a random [b]average[/b] for the n numbers.  If it is less than 1, then the condition is true.  So we have $ \frac{1}{n}$

[/hide]

Sorry, not quite right. For one thing you misread his answer: it's $ 1/n!$ with a factorial, not $ 1/n$. The reason that method doesn't work is because the probability distribution over all possible averages isn't uniform. That is, there are some averages which are more likely to occur than others.


limit
trigonometry
logarithms
calculus
calculus computations

$ \lim_{x\to0}\frac{\sqrt{1\plus{}x\plus{}x^2}\minus{}1}{x}\minus{}\lim_{x\to\frac{\pi}{4}}\frac{\sqrt{2}\minus{}2\cos{x}}{\sin(x\minus{}\frac{\pi}{4})} > \log_{10} 0.005$

Please tell it is true or false with explanation. Thank you so much for your help.

Sure, it's true. However, this is really easy - maybe you should try solving it yourself and just ask when/if you run into any problems. Let me just give you some hints and partial results:

$ \lim_{x\to 0}\frac {\sqrt {1 \plus{} x \plus{} x^{2}} \minus{} 1}{x} \equal{} \frac {1}{2}$, you can use L'Hospital or, if you prefer something more natural, so to speak, you can try using the old trick $ \sqrt {a} \minus{} 1 \equal{} \frac {a \minus{} 1}{1 \plus{} \sqrt {a}}$ to prove this.

$ \lim_{x\to\frac {\pi}{4}}\frac {\sqrt {2} \minus{} 2\cos{x}}{\sin(x \minus{} \frac {\pi}{4})} \equal{} \sqrt {2}$, I guess L'Hospital would be the easiest way here.

At the end, simply prove (it's easy!) that $ \log_{10}{0.005} < \minus{} 1 < \frac {1}{2} \minus{} \sqrt {2}$ and you're good to go.

hello, we have
$ \lim_{x \to 0}\frac{\sqrt{1\plus{}x\plus{}x^2}\minus{}1}{x}\equal{}\lim_{x\to 0}\frac{1\plus{}x}{\sqrt{1\plus{}x\plus{}x^2}\plus{}1}\equal{}1/2$
Sonnhard.

hello, and by the rule of L'Hospitale we can prove that
$ \lim_{x \to \frac{\pi}{4}}\frac{\sqrt{2}\minus{}2\cos(x)}{\sin(x\minus{}\frac{\pi}{4})}\equal{}\lim_{x \to \frac{\pi}{4}}\frac{2\sin(x)}{\cos(x\minus{}\frac{\pi}{4})}\equal{}\sqrt{2}$.
Sonnhard.

hello, ok, the final step ist posted already.
Sonnhard.


Valentin, you really shouldn't do any "tests" with admin privileges...some things may work for you but they may not work for us.

That is, assuming that your test is related to the board technically.

:D thanks for the hidden compliment ;) (new emoticons :))

oh that was what you were testing

so it was VERY recently indeed :P (oops the razz isn't working yet :wink: )

you can use the classical ; ) instead of :wink: ;)

Oh :) nice, I usually use ; ) but I'm used to getting on other boards where they kept the :wink: :D

Great ;)


$ \tan(\dfrac{3\pi}{11})\plus{}4\sin(\dfrac{2\pi}{11}) \equal{} ?$

good luck!

[hide="Solution"]This calculus is very hard.
First let's prove that $ \tan\frac {3\pi}{11} \plus{} 4\sin\frac {2\pi}{11} \equal{} \cot\frac {\pi}{22} \minus{} 4\sin\frac {4\pi}{11}$.
This is equivalent with $ \cot\frac {5\pi}{22} \plus{} 4\sin\frac {2\pi}{11} \equal{} \cot\frac {\pi}{22} \minus{} 4\sin\frac {4\pi}{11}\ (1)$
Now let $ a \equal{} \frac {\pi}{22}$
So $ (1)$ becomes $ \cot 5a \plus{} 4\sin 4a \equal{} \cot a \minus{} 4\sin 8a$
$ \Longleftrightarrow 8\sin 6a\cdot\cos 2a \equal{} \frac {\sin 4a}{\sin a\cdot\sin 5a}\ \Longleftrightarrow \sin 4a \equal{} 8\sin a\cos 2a\sin 5a\sin 6a$

$ \Longleftrightarrow 2\sin 2a\cos 2a \equal{} 8\sin a\cos 2a\sin 5a\sin 6a$ $ \Longleftrightarrow 2\sin a\cos a \equal{} 4\sin a\sin 5a\sin 6a$

$ \Longleftrightarrow \cos a \equal{} 2\sin 5a\sin 6a \equal{} \cos a \minus{} \cos 11a \equal{} \cos a$, because $ \cos 11a \equal{} \cos\frac {\pi}{2} \equal{} 0$.

Now let's show that $ \cot\frac {\pi}{22} \minus{} 4\sin\frac {4\pi}{11} \equal{} \sqrt {11}$.

Because $ \sum_{k \equal{} 1}^n\sin kx \equal{} \frac {\sin\frac {(n \plus{} 1)x}{2}\sin\frac {nx}{2}}{\sin\frac x2}$, for $ x\in\mathbb{R} \minus{} \{2k\pi|k\in\mathbb{Z}\}$, for $ n \equal{} 5$ and $ x \equal{} \frac {\pi}{11}$ we obtain:

$ A \equal{} \sin\frac {\pi}{11} \plus{} \sin\frac {2\pi}{11} \plus{} \sin\frac {3\pi}{11} \plus{} \sin\frac {4\pi}{11} \plus{} \sin\frac {5\pi}{11} \equal{} \frac {\sin\frac {3\pi}{11}\sin\frac {5\pi}{22}}{\sin\frac {\pi}{22}} \equal{} \frac 12\cot\frac {\pi}{22}\ (2)$

Let $ B \equal{} \cos\frac {\pi}{11} \minus{} \cos\frac {2\pi}{11} \plus{} \cos\frac {3\pi}{11} \minus{} \cos\frac {4\pi}{11} \plus{} \cos\frac {5\pi}{11}$ and

     $ C \equal{} \sin\frac {\pi}{11} \minus{} \sin\frac {2\pi}{11} \plus{} \sin\frac {3\pi}{11} \minus{} \sin\frac {4\pi}{11} \plus{} \sin\frac {5\pi}{11}$,

     $ D \equal{} \sin\frac {\pi}{11} \plus{} \sin\frac {2\pi}{11} \plus{} \sin\frac {3\pi}{11} \minus{} \sin\frac {4\pi}{11} \plus{} \sin\frac {5\pi}{11}$.

Because $ B \plus{} iC \equal{} \frac {\cos\frac {5\pi}{22}}{\cos\frac {\pi}{22}}\left(\cos \frac {3\pi}{11} \plus{} i\sin\frac {3\pi}{11}\right)$

$ \Longrightarrow B \equal{} \frac {\cos\frac {5\pi}{22}\cos\frac {3\pi}{11}}{\cos\frac {\pi}{22}} \equal{} \frac 12$

Calculating $ D^2$ and transforming the products that appear in sum we obtain $ D^2 \equal{} \frac {B \plus{} 5}{2} \equal{} \frac {11}{4}$ and how $ D > 0$ $ \Longrightarrow D \equal{} \frac {\sqrt {11}}{2},\ (3)$.

From $ (2)$ and $ (3)$ $ \Longrightarrow A \minus{} D \equal{} \frac 12 \cot\frac {\pi}{22} \minus{} \frac {\sqrt {11}}{2}$, so $ \cot\frac {\pi}{22} \minus{} 4\sin\frac {4\pi}{11} \equal{} \sqrt {11}$.

$ \Longrightarrow \tan\frac {3\pi}{11} \plus{} 4\sin\frac {2\pi}{11} \equal{} \sqrt {11}$.[/hide]


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AIME I
\/closed

Nothing can "guarantee" your score to be $\ge k$ unless $k=0$ :roll: but it can definitely help you increase your score. If you study hard enough, you can improve a LOT over a year. I got a 18/46 on MATHCOUNTS states last year and now my lowest recent AIME practice score is a 7. I've only taken 2 AoPS classes, you've taken much more. You should also aim HIGH. Last year after MC states (when I finished crying) I decided my new goal would be to make it to USAMO. Pretty high goal, right? Well, look how much I've improved...

But anyway, about the problem series, I think that it would obviously help but it's not necessary. You can just do lots of AIME questions on your own, if you have the motivation. But others who have taken the Problem Series can probably give better advice in this area.

Any more comments? I like to hear some from instructors as well.

[quote="rrusczyk"]I'm not sure I entirely understand your question, but I'll say that the AIME Problem Series are in general better designed for the student who is already in the 4-8 range and looking to move into the 8-12 range than for the student who is in the 1-3 range looking to move to the 4-8 range.  I'm not sure if that answers your question.  If it doesn't, feel free to ask again :)[/quote]

Sorry if my question was unclear. Let me rephrase it.

About 3-4 months, I got 5 on one of the recent AIMEs, and happened to be my first AIME. So, I'm guessing my score is bit higher than 5 now but it's around there (I'm not entirely sure.. I never took full AIME test after it). My question was that since I'll be junior next year, I'd like to bring my score to at least 10-12.

Your comment answered the question since my situation was exactly what you described. BTW, can 10 on AIME and 120 on AMC-12 (which I need to work on :huh: ) get me USAMO as a junior?

There is no way of knowing what scores will make the USAMO.

However, it should be noted that there have been gold medalists at the IMO who didn't break 10 on the AIME.  Though AIME scores at the top end are likely getting higher, the USAMO expansion will hopefully keep the necessary scores relatively constant.

[quote="kyyuanmathcount"]Mr. Crawford,

The AIME Problem Series class will begin in June, but that is too early for me.  I'm not best prepared for that class yet. I'd like to take it in the fall or winter.  But if this is the last class before next year's AIME, I might have to take it.  Please let me know if there'll be another one offered later in the year.

Thanks.[/quote]

In the past, we've typically offered an AIME Problem Series both during the summer and during the school year (leading up to the AIME).

[quote="filletwho"][quote="rrusczyk"]I'm not sure I entirely understand your question, but I'll say that the AIME Problem Series are in general better designed for the student who is already in the 4-8 range and looking to move into the 8-12 range than for the student who is in the 1-3 range looking to move to the 4-8 range.  I'm not sure if that answers your question.  If it doesn't, feel free to ask again :)[/quote]

What about students in the 8-12 range?[/quote]

There is a lot in the AIME Problem Series for students in the 8-12 range.  Problems range all over the AIME, which includes a fair number of 10-15 range.  Students in the 8-12 range might also learn how to find more elegant solutions to those they can already solve.  I personally think this kind of class serves as a good "in" to Olympiad math for many students who can qualify for the USAMO, but aren't yet at the point of solving USAMO problems very often.  To me, the most interesting part of the AIME is how much math is behind the problem writing.  Proof techniques often shorten the time needed to solve or be sure of an AIME, and we try to make that plain in many of the solutions discussed in class.

I will add that the right place for students in the 8-12 range next school year is WOOT...

[quote="chess64"]Do MOPpers get in for free again next year?[/quote]Yes, that's the plan.  Assuming that MOP does not have a huge unexpected increase in enrollment, all non-graduating MOP students will be invited into WOOT for free.

This is made possible by our sponsorship from [url=http://www.deshaw.com]D. E. Shaw group[/url].

[quote="ffdbzathf"]where is the right place for students in the 5-8 AIME range?[/quote]

The larger, school-year-long program for students who can score 5 or above on the AIME is WOOT.  The AIME Problem Series are good classes on a smaller scale and don't have much of a focus on proof writing.

[quote="calc rulz"]So does this mean WOOT is also helpful for AIME? Or is it just focused on olympiads?[/quote]

It covers material needed for both.

[quote="Silverfalcon"]Hi

For Mr.Crawford and Mr.Rusczyk, I'm only going to consider taking AIME Problem Series A if it can guarantee my score to be around 11-15 in next year's AIME. Currently, I'm not exactly sure what I'll get but I think it's around 5-9. I only did one practice (3-4 months ago) so this is really hard for me to know what kind of score I'll receive. My goal in high school for math is USAMO (MOP is in list but I consider USAMO to be more big thing) and I heard when you're junior or senior, you need to score pretty high to get into USAMO. So, will AIME Problem Series A good for me on achieving this goal? I'll be busy during the summer with SAT studying so I may not be able to actually attend classes in time (but I'll review transcripts and work on class problems). I'll appreciate this very much. AoPS got me to qualify to AIME this year and again, I'm asking for help to USAMO.  :D[/quote]

Silverfalcon, you should already know from being around these forums for so long there nothing you can do, no training that will "ensure" you get a grade of X on test Y.  The instructors can structure a course toward a certain ability level but they can by no means guarantee you will perform at a certain level on the real competition.

I think I undertand what you're asking but honestly you need to work on articulating your questions correctly so you get the responses you want.  I think that the AIME Problem Series offer a curriculum which, in theory, can get your AIME score high enough to make USAMO.

WOOT is mostly focused on Olympiad material but I definitely believe that it can boost the AIME performance of students who are just below USAMO qualifying level or at a higher level.  First, olympiad problems will toughen you up big time.  You won't quit after your first or even second approach fail and you'll be able to attack these problems for longer periods of time.  It's this shift from speed-based, 30 second math to more methodical, long term problem solving which can boost a borderline student over the edge and get them into USAMO.  Additionally, an appreciation for proof and justification usually grows in the student as (s)he delves into the olympiad mathematics world.  This can actually help out on the AIME as one will be less susceptible to be fooled by "intuitive" notions which are actually wrong.


number theory
greatest common divisor
combinatorics unsolved
combinatorics

Some cards each have a pair of numbers written on them. There is just one card for each pair $(a,b)$ with $1 \leq a < b \leq 2003$. Two players play the following game. Each removes a card in turn and writes the product $ab$ of its numbers on the blackboard. The first player who causes the greatest common divisor of the numbers on the blackboard to fall to $1$ loses. Which player has a winning strategy?

May be i am missing something because it looks too easy.
There is some prime p between 1002 and 2003 by Chebyshev's theorem. So if the first player pick a card with numbers 1 and p he will write p on the blackboard and whatever the second player chooses the gcd will be 1.

Yeah, those primes do not work.
For example, 2003 is prime and the numbers of cards whit a 2003 is even (note that a<b), so you would lose


trigonometry
integration
geometry unsolved
geometry

I try solve problem where I should evaluate a average value of sin(x).

For sin(x) in interval 0 to Pi/2 aj still obtain 2/Pi. But results should be 1/sqrt(2).

Any idea what it can be?

You are right. $ \int_{0}^{\pi/2} \sin x \textrm{d}x \equal{} [\minus{}\cos x]_{0}^{\pi/2} \equal{} 1$, therefore the average value of $ \sin x$ over that interval is $ \frac {1} {\pi/2 \minus{} 0} \equal{} \frac {2} {\pi}$.


Use rational root theorem.  After some checking, we factor:

(x+1)(3x-1)(12x^2+4x-1)

[quote="Arne"]Solve for $x$, given $36x^4 + 36x^3 - 7x^2 - 6x + 1$.[/quote]

I suppose we want that to equal zero? :D


function
algebra
polynomial
calculus
integration
rational function
calculus computations

how would you go about finding the primitive of (x^2/(1+x^2)dx)

I know the primitive of 1/(x^1+1) is arctan(x) but i dont know how to get rid of the x^2

Long division.

This is a rational function (meaning a quotient of polynomials), and it is an improper fraction (meaning the degree of the numerator is at least as large as the degree of the denominator). Just as a numerical proper fraction such as $\frac{17}5$ can be rewritten as $3+\frac25,$ an integer plus a proper fraction, then an improper rational function can be written as a polynomial plus a proper fraction.

That's the first step, anyway.

thanks a lot man.  I was almost about to shoot myself because this was so confusing.  I tried doing integration by parts and it took so much space and time and i wasn't getting anywhere lol

it didn't work...  integration by parts is i think what needs to be done but i dont know how to do it

the original function was integral of (ln(1+2^2)dx)

If anyone can show me how to integrate that step by step i would love u forever because i've been trying for 2 hours now...


LaTeX
number theory unsolved
number theory

[quote="nmt_2006"][img]http://i102.photobucket.com/albums/m86/nguyenminhtrang_2006/T8-Algorithm1.jpg[/img]
So hard![/quote]
may be not very hard.if$k_{0}\in A$
let $u=[\frac{k_{0}}{a}]$then $u\in A$
and $k_{0}+2la\in A (l\geq 0)$so $u+2l\in A (l\ge 0)$
there exist a positve integer $l'$ such that $u+2l'-2<a(u-1)\leq u+2l'$
then $[\frac{u+2l'}{a}]=u-1 \in A$ then every integer $t\geq u-1 \in A$
the rest is easy.

[quote="gerrard"]Continue
[img]http://i102.photobucket.com/albums/m86/nguyenminhtrang_2006/T8-Algorithm.jpg[/img][/quote]

Simply, $|A|=\varphi(p_{1}p_{2}p_{3})=(p_{1}-1)(p_{2}-1)(p_{3}-1)\geqslant (2-1)(3-1)(5-1)=8$. The equality occurs for $p_{1}=2,p_{2}=3,p_{3}=5$.

Type it down with $\LaTeX$ on the forum please.


Let $\ds E(a,b,c)=\sum_{cyclic}\frac{a^n}{1+a+ab}$. Show that if there is $n\geq 2$, for which $E(a,b,c)=1$, this happens for all $n\geq 2$.

I have some doubts. I have expanded it for $n=2$ and $n=3$. MathCad doesn't show any relations. The expressions cannot be factorized.


logarithms
function
algebra unsolved
algebra

[color=darkblue]Slove this equation: $ (1 \plus{} cosx)(2 \plus{} 4^{cosx}) \equal{} 3.4^{cosx}$[/color]

[quote="loup blanc"]$ \cos(x)\in\{0,0.5,1\}$.[/quote]

[color=darkblue]Can you post solution?[/color]

Let $ a\equal{}\cos(x)$ and $ u\equal{}2a\minus{}1,u\in[\minus{}1,1]$. The equation becomes:
$ f(u)\equal{}u\log(2)\minus{}\log(\dfrac{3\plus{}u}{3\minus{}u})\equal{}0$.
$ f$ is odd and $ f(0)\equal{}0,f(1)\equal{}0$; then we may assume $ u\in[0,1]$.
$ f''(u)\equal{}\dfrac{\minus{}12u}{(9\minus{}u^2)^2}<0$; then $ \minus{}f$ is a strictly convex function and has no zero in $ ]0,1[$.

I have forgotten the case $ u\in]\minus{}3,\minus{}1]$.
 Now $ f$ is a convex function and $ f'(\minus{}1)<0$; then $ f>0$ on $ ]\minus{}3,\minus{}1[$.


AMC
AIME
analytic geometry
geometry
trigonometry

Two circles touch the x-axis and the line $y = mx (m > 0)$. They meet at $(9,6)$ and another point and the product of their radii is $68$. Find $m$.

any nice solutions?

Sorry if Im dumb but is this the figure?

yea id think so
thats wat i got at least

really too hard or something of coordinate geometry can somebody post the solution?????

[hide]Tangents to a circle are equal. The lines $y=mx$ and the line $y=0$ are tangents to the circles. Thus, the bisector of the angle they form at $(0,0)$ goes thru the centers of the two circles. This line has equation $y=cx$. Call the angle this line makes with the $x$ axis $\theta$. Then we have $c = \tan \theta$ and $m = \tan 2\theta$. Since

$tan 2\theta = (2 \tan \theta)/(1-\tan ^2 \theta)$

we have that $m = (2c)/(1-c^2).$

Now, assume the center of one of the circles is $a, ca.$ Then the equation for the circle will be:

$(x-a)^2 + (y-ca)^2 = (ca)^2$

Since it goes through $(9,6)$, we have that

$(9-a)^2 + (6-ca)^2 = (ca)^2$ or
$117 + a^2 -18a - 12ca = 0$ or

$a^2 - a(18+12c) + 117=0$.

This is an equation with two solutions which correspond to the equations of the two circles. The product of the solutions is $117$. The two product of the two radii is then $117(c^2)$, since the radii are $ca$ and thus

$117c^2 = 68$, or

$c = \sqrt{68/117}.$

So $m = 2\sqrt{68/117}/ (1-68/117)$.[/hide]


calculus
integration
inequalities
number theory proposed
number theory

The numbers from $ 51$ to $ 150$ are arranged in a $ 10\times 10$ array. Can this be done in such a way that, for any two horizontally or vertically adjacent numbers $ a$ and $ b$, at least one of the equations $ x^2 \minus{} ax \plus{} b \equal{} 0$ and $ x^2 \minus{} bx \plus{} a \equal{} 0$ has two integral roots?

Suppose to the contrary that such arrangement is possible.
If number placed in corner cell of our board,we will call it "corner" number.
Consider numbers $ 149$:
[b]-[/b]Cell in which number $ 149$ is placed,has at least two adjacent cells.Let $ a$ be one of the numbers written on them.
If for some integer $ d$ one of the following equalities holds:
$ a^2 \minus{} 4\cdot 149 \equal{} d^2$ or $ 149^2 \minus{} 4a \equal{} d^2$,where $ 51\leq a\leq 150$,it follows that one of the equations $ x^2 \minus{} ax \plus{} 149$ or $ x^2 \minus{} 149x \plus{} a$ has two integer roots.Since $ 149$ has at least two adjacent numbers,we conclude that these two equations must have at least two roots in common.
Let's consider first equation:
If $ (a \minus{} b) \equal{} k$ and $ (a \plus{} b) \equal{} \frac {4\cdot 149}{k}$,the following inequalities must hold:
$ 102\leq 2a \equal{} k \plus{} \frac {4\cdot 149}{k}\leq 300$.But it holds only for $ k \equal{} 2,3,4,5$.
For $ k \equal{} 3,4,5$ we have that $ a$ is not integer,hence the only possible solution is $ k \equal{} 2$,and $ a \equal{} 150$.
Consider second equation:
$ 149^2 \minus{} 4a \equal{} d^2$.It is clear that $ d$ is odd.And $ \sqrt {149^2 \minus{} 4\cdot 150}\leq d\leq \sqrt {149^2 \minus{} 4\cdot 51}$.Or
$ 146 < d < 148$,hence $ d \equal{} 147$,but if $ d \equal{} 147$,it follows that $ a \equal{} 148$.
Therefore,cell in which $ 149$ is placed has only two adjacent cells,with number $ 148,150$ written on them,i.e $ 149$ is corner number.
Now let's try something similar with numbers $ 148,150$.
[b]1.[/b]Consider equation:
$ a^2 \minus{} 4\cdot 148 \equal{} d^2$,it has two integer solutions:$ a \equal{} 76,d \equal{} 72$ and $ a \equal{} 149$,$ d \equal{} 147$.
Equation $ 148^2 \minus{} d^2 \equal{} 4a$,has only one solution $ a \equal{} 146$.
[b]2.[/b]Consider equation:
$ a^2 \minus{} 4\cdot 150 \equal{} d^2$ it has two solution $ a \equal{} 77,d \equal{} 73$ and $ a \equal{} 53$,$ d \equal{} 47$.
But since cells with numbers $ 148$ and $ 150$ have common adjacent cells,contradiction follows,as $ 77,53\neq 76,146$.
[b]P.S:[/b]
I've tried only few minutes to solve this problem,and the solution above is the first idea that came to mind,so I'm pretty sure that there exist more elegant solution,and I'm waiting for the moment when some of you will find it...
Best regards,Erken.

Why can't we just say 51 can be only adjacent to one number.

If $ x^2\minus{}51x\plus{}b$ has two integral roots, then $ b\le 50$. If $ x^2\minus{}ax\plus{}51$ has two integral roots, then $ a\equal{}52$ is the only possibility. So 51 can't be on the board? I'm not sure whether I'm missing something here.

$ (x\minus{}49)(x\minus{}2) \equal{} x^2 \minus{} 51x \plus{} 98, (x\minus{}48)(x\minus{}3) \equal{} x^2 \minus{} 51x \plus{} 144$  :wink:

On the other hand, doesn't it suffice to pick $ 5$ primes $ > 78$? For any such prime $ p$, we have either $ x^2 \minus{} ax \plus{} p \equal{} 0$ or $ x^2 \minus{} px \plus{} b \equal{} 0$ has integer roots; in the former, the product of the two (positive) integer roots is $ p$, so they must be $ (1,p) \Longrightarrow a \equal{} 1\plus{}p$. In the latter, the sum of the two integer roots is $ p$; let them be $ k$, $ p\minus{}k$, wlog $ k < p/2$, so $ 51 \le b \equal{} k(p\minus{}k) \le 150$. But $ k(p\minus{}k)$ is strictly increasing on the interval, and for $ k\equal{}2$ we have $ 2(p\minus{}2) > 2 \cdot 76 > 150$, and for $ k \equal{} 0, k(p\minus{}k) \equal{} 0 < 51$, so we must have $ k \equal{} 1$, giving $ b \equal{} p\minus{}1$. So there are only two possible numbers adjacent to each of those primes, contradiction.

Sorry, that was indeed wrong.

What about the numbers between 54 and 97 can only have at most 3 adjacent numbers? But we can only have 36 numbers on the periphery.


If the following three numbers are placed in ascending order, which number would be second? $ 5^{56}$, $ 7^{49}$, $ 13^{35}$

This is a Workout so calculators should be allowed...

There isn't a fast non-calc method to solve this because the bases are all prime numbers.

After typing these 3 numbers into a calculator, you find that the answer is $ \boxed{5^{56}}$

What if we're using a scientific calculator, like me? :(


number theory
prime numbers
complex numbers
\/closed

I'm 17 and I can't do half the beginners problems. I thought I was pretty good at maths and I've always found it very east in college, but how can someone the age of my sister know all this stuff?

Do some people have tutoring or something? I never got any of that, all I did was do what I was taught in school and ended up with decent grades.

It's a shame because now I feel like I've wasted something I could have had, since I've always liked maths! I even did some extra maths, but our syllabus in the UK is a bit weird, I find myself not really knowing about prime numbers and that but from further reading I can do complex numbers in exponential forms and weird things like that. silly!

So anyway, my question is how does someone like me get better at maths? I don't even know what some notation means, like {} and the U shaped things. (and upside down versions). :cap: is one (I just spotted it on the left panel!)

Thanks for any suggestions, I really want to do well since I'm entering into the SMC (senior maths competition... the top so many go into the British maths olympiad round 1 then the top ones go into round 2 for the IMO). I just want to get into the BMO, I've seen the questions and they're really hard but I set it as a target.

So the SMC is pretty easy by the standards on here. The first 15 questions are mostly straightforward, and the last 10 are harder (and you lose marks for getting them wrong). How have you all become so good at maths? :lol:

Thanks for any adive, the test is next month I think! :cool:

The titles of AoPS forums are not representative of all math, but are representative of competition math starting from middle school on up.  Certainly there are some topics that achieving math students like yourself get more exposure to such as complex numbers, but as you learn more about number theory you will find that some problems you didn't previously know how to solve are not really any harder than some that you already do, but that they simply take learning a few ideas that aren't typically taught in middle or high school curriculum.

We offer classes here for students who are interested in learning nearly anything that would come up on exams used to select IMO teams.  You might also spend your time working through the Beginners forum until you are confident you can solve every problem there.  The Intermediate forum is a nice step toward Olympiad problems, though proofs are usually relegated to the Pre-Olympiad forum and up.


geometry
HMMT
incenter
trigonometry
trig identities
Law of Sines
Law of Cosines

Source: HMMT 2005 Geometry #4

Let $ XYZ$ be a triangle with $ \angle X \equal{} 60^\circ$ and $ \angle Y \equal{} 45^\circ$. A circle with center $ P$ passes through points $ A$ and $ B$ on side $ XY, C$ and $ D$ on side $ YZ,$ and $ E$ and $ F$ on side $ ZX$. Suppose $ AB \equal{} CD \equal{} EF.$ Find $ \angle XPY$ in degrees.

Source: HMMT 2006 Geometry #8

Triangle $ ABC$ has a right angle at $ B$. Point $ D$ lies on side $ BC$ such that $ 3 \angle BAD \equal{} \angle BAC$. Given that $ AC \equal{} 2$ and $ CD \equal{} 1,$ compute $ BD$.

[hide="hint for 4"]show that P is the incenter[/hide]

[hide="8"] Let $ \angle{BAD} \equal{} \theta$. Then, $ \angle{ADC} \equal{} 90 \plus{} \theta$, $ \angle{DAC} \equal{} 2\theta$, and $ angle{ACD} \equal{} 90 \minus{} 3\theta$.

By the law of sines, $ \dfrac{2}{\sin(90 \plus{} \theta)} \equal{} \dfrac{1}{\sin{2\theta}}$. By the fact that $ \sin(90 \plus{} \theta) \equal{} \cos{\theta}$ and $ \sin{2\theta} \equal{} 2\sin{\theta}\cos{\theta}$, we find that $ \sin{\theta} \equal{} \dfrac{1}{4}$.

Note that $ \cos(90 \minus{} 3\theta) \equal{} \sin{3\theta} \equal{} 3\sin{\theta} \minus{} 4\sin^{3}\theta \equal{} \dfrac{11}{16}$. Now, by the law of cosines, $ AD^{2} \equal{} 4 \plus{} 1 \minus{} \dfrac{11}{4} \equal{} \dfrac{9}{4}$. Thus, $ AD \equal{} \dfrac{3}{2}$. Now, $ \dfrac{BD}{AD} \equal{} \sin{\theta} \equal{} \dfrac{1}{4}$, so $ BD \equal{} \boxed{\dfrac{3}{2}}$.[/hide]

So we have that cos(90-3a)=x/2=sin(3a)=sin(2a)cos(a)+sin(a)cos(2a),

if you solved for sin(a) earlier it was 1/4, so sin(2a)=sqrt15/8, cos(2a)=7/8, cos(a)=sqrt15/4,

plugging each one in we get x/2=11/16, so x=11/8.

11/8-1=BD=3/8.  x=BC


Ten people are attending a convention. They can choose any one of four talks to attend. These talks occur simultaneously in four different rooms. Consider only the numerical compositions of attendees in any particular room. For example, Jane and John in Room A is the same numerical composition as Jane and Jerry in Room A or as Julia and James in Room A, etc. However, the composition of 4 people in Room A, 4 people in Room B, 2 people in Room C, and 2 people in Room D, would be a distinct composition from that of 4 people in Room A, 2 people in Room B, 4 people in Room C, 2 people in Room D because of the differing numbers of people in particular rooms. Under these conditions, in how many distinct ways can these 10 people be divided into numerically composed audiences for the individual talks? Give your answer as an integer.

ok we just have i) indistinguishable balls
ii)distinguishable urns

i think it is clear how do to now


Prove that: there exits positive integer [i]n[/i] satisfying the following condition:
$ 18^n \plus{} 45^n \plus{} 50^n \plus{} 125^n  \vdots 2006$

$ 2006\equal{}2*17*59$.
Let $ a_n\equal{}18^n\plus{}45^n\plus{}50^n\plus{}125^n$. Obviosly $ 2|a_n$ for all n.
$ 18\equal{}1\mod 17, \ 50\equal{}\minus{}1\mod 17, \ 45\equal{}\minus{}6\mod 17, \ 125\equal{}6\mod 17$, therefore $ 17|a_n$ if n odd. If n=2k, then $ a_n\equal{}2(1\plus{}2^k)\mod 17$. Therefore $ 17|a_n$ if and only if $ n \minus{}odd$ or $ n\equal{}8\mod 16$.
$ \binom{18}{59}\equal{}\minus{}1,\binom{45}{59}\equal{}1,\binom{50}{59}\equal{}\minus{}1,\binom{125}{59}\equal{}1$
Therefore $ n\equal{}29\mod 58$ work.


inequalities
linear algebra
matrix
LaTeX
inequalities theorems

In order to find the "most important"  work on [b]Majorization theory [/b] , please see the :

http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?did=D39883

You find the [b]Schur's thesis[/b] :

[color=red]  [b] " Ueber eine  Klasse von Matrizen  die sich einer gegebenen Matrix  zuorden lassen "  [/b]
[/color]
[color=green]        Inaugural -Dissertation    (Firedrich-Wilhelms-Universitaet zu Berlin)   am 27.November  [b]1901[/b]
[/color]
 von  [color=red] [b]  Issai Schur[/b][/color]     (1875--1941)
[b](aus Russland, geboren am 10.Januar 1875 zu Mohilev (=Moghilev) am Dniepr) [/b]
Opponenten: L.Lewent , R. Ziegel , Edmund Landau

Sorry I cannot find nothing about  Schur's thesis and this majorization theory. Could you post it in Latex or pdf ?Thanks

Silouan, it is here : http://dz-srv1.sub.uni-goettingen.de/cache/toc/D39883.html

(however, this is in German.. (unless I'm mistaken  :oops: ))


If I have asymptote code within a LaTeX file, how do I force TeXnicCenter to center the image?

You do it in exactly the same way as you centre any image (as the latex code just includes the image and doesn't care that it was produced by Asymptote or anything else). So use [code]\begin{center}
\begin{asy}
...
\end{asy}
\end{center}[/code]or, if you are using \begin{figure}...\end{figure} you can write instead[code]\begin{figure}
\centering
\begin{asy}
...
\end{asy}
end{figure}[/code]

Sorry, that's not quite the question I meant to ask (hence the mismatch between the title and the text).  Is there any way to make the text wrap around the image?

Again, the question is the same as wrapping text around any image. There are 3 packages that aim to do this, [i]floatflt[/i], [i]picins[/i] and [i]wrapfig[/i] all available in MiKTeX. [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=textflow]Flowing text around figures in LaTeX[/url] gives the pros and cons of each package. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=760687#760687]DPatrick[/url] says picins is used for typsetting AoPS textbooks.

If you try these packages feedback would be welcome as to how you got on with them as wrapping text is sometimes asked about.

I'm trying to use picins, but what is the correct formatting with \parpic{} when I am using asymptote code that is written into the document (as opposed to a separate file accessed with \includegraphics)?

It looks as if \parpic won't work with asymptote code, probably because \parpic requires { and } instead of an environment with \begin and \end.
However, I have just tried floatflt and wrapfig which use \begin...\end and they seem to work so try one of those.


Support
geometry
rectangle
function
probability
calculus
calculus computations

Suppose the p.d.f of the two-dimensional r.v (X,Y) is given by   F(x,y)=(3/2)xy^2, o<x<2, 0<y<1. O, elsewhere

a.  What is the marginal pdf of X and Y
b. Find P{1<X<2}
c. Find P{Y<X)
d. Find E[X/Y^2]
e. is X and Y independent? Why?

Notice that the support of this p.d.f. is the rectangle $[0,2]\times[0,1]$ (a cartesian product) and note also that we can completely factor $f(x,y)$ as a function of $x$ only times a function of $y$ only. From this it follows that the answer to (e) is that $X$ and $Y$ are independent. Furthermore, the marginal density for $X$ is $c_{1}x$ on $[0,2]$ and the marginal density for $Y$ is $c_{2}y^{2}$ on $[0,1].$ All you have to do now is figure out what $c_{1}$ and $c_{2}$ have to be in order to make those proper probability densities.


In the plane there are given the lines $\ell_1$, $\ell_2$, the circle $\mathcal{C}$ with its center on the line $\ell_1$ and a second circle $\mathcal{C}_1$ which is tangent to $\ell_1$, $\ell_2$ and $\mathcal{C}$. Find the locus of the tangent point between $\mathcal{C}$ and $\mathcal{C}_1$ while the center of $\mathcal{C}$ is variable on $\ell_1$.

[i]Mircea Becheanu[/i]

My opinion (generally):
Let O be the section point of  $\ell _1 ,\ell _2 ,OX$
the bisector of the angle of
$\ell _1 ,\ell _2 .$
Consider
$C_1 $
instantaneously constant witch tangents
$\ell _1 \;at\;T$
and intersects the bisector at P, every point of the minor arcTP is a tangent point  one circle 'type'
$C $ (a family of circles...)   with $C_1$.
Finally the locus is  the union of the angle  <XOD with D point of $l_1$ and the symmetric of <XOD with respect to the point O (and the point O of course).

S.E.Louridas

Sorry, I forgot to mention that we work the same way for the other (π-φ) of the angle is being formed by
$\ell _1 ,\ell _2 \;also..$
so we have an other set (respectively) of points of focus

S.E.Louridas

[quote="S.E.Louridas"]Sorry, I forgot to mention that we work the same way for the other (π-φ) of the angle is being formed by
$\ell _1 ,\ell _2 \;also..$
so we have an other set (respectively) of points of focus.
Finally finally
The locus is the whole plane while it is sufficient to consider the point us a circle with radius 0.

S.E.Louridas[/quote]


I just finished writing the first problem's solution. It is 2 pages and before I move on, I have one question.

How much work should be shown?

Mine for the first one was quite simple. I gave one example and said, "Using this logic, we can determine...." Is this okay? Like do I have to show the process of working on every equations I have?

Sorry if this is unclear. I don't know how I can express this very clearly without actually giving away the answers.

Thanks!

I don't think we're supposed to discuss questions like this until after the round is over, in fairness to all competing.

Okay, thanks.

[quote]Also, you can never be penalized for showing too much work and let's leave it at that.[/quote]

This part was what I'm unsure about. Thanks.

[quote="DPopov"]Just use your best judgement, it's part of the competition. Also, you can never be penalized for showing too much work and let's leave it at that.[/quote]

Though whenever we graders find a ten-page solution for a single problem, we are tempted to invent such a penalty.

We mathematicians use "Likewise" to represent the phrase "Using the same logic over again but with different numbers". The caveat is that it has to be the same logic and not just similar logic. Though we may modify the "likewise" with clauses, such as "Likewise, except that we do not have to check whether X is negative in this case,...."

Erin Schram
USAMTS grader

I'm not sure what you mean by "using the same logic and not just logic."

So, if I solved this problem:

\[x+5 = 2\]

[hide="Solution"]

Subtract 5 from both sides. Since $x$ is isolated, it equals to LHS which is 2-5 or -3. [/hide]

And I have two other problems that look like $y+2 = 7, z+5 = 9$.

So, for the remaining two problems, can I say the term "likewise" since I'll be basically doing the same thing? (I'm not necessarily subtracting same number but I'm subtracting number to isolate the $x$)

[quote="Silverfalcon"]I'm not sure what you mean by "using the same logic and not just logic."[/quote] I said "not just similar logic."

[quote="Silverfalcon"]So, if I solved this problem:
\[x+5 = 2\]
Subtract 5 from both sides. Since $x$ is isolated, it equals to LHS which is 2-5 or -3.

And I have two other problems that look like $y+2 = 7, z+5 = 9$.

So, for the remaining two problems, can I say the term "likewise" since I'll be basically doing the same thing? (I'm not necessarily subtracting same number but I'm subtracting number to isolate the $x$)[/quote]
Yes, you could say, "Likewise, $y+2=7$ has the solution $y=5$." However, if you had the problem $2y=6$, you would not be able to use likewise, because while division is similar to subtraction, it is only similar logic, not the same logic.

Erin Schram

Are those just examples...?

Would I really have to explain all the steps to get $x$ from

$x + 5 = 2$
to
$x = -3$

or can I just assume that the reader can figure this out themself?

Don't worry, those are just examples - steps that are obvious can be omitted (of course, sometimes it can be difficult to determine exactly when that cut-off is; it's certainly more advanced than at addition and multiplication, though :-P)

Excessively long-winded solutions can hurt your chance at a commended solution, but that's about all. They do annoy the graders, and might not even be fully read; if the grader sees enough for a 5, he or she might just stop there.

On the other hand, a complete and very long solution will get the 5 points.

Guess-and-check is a perfectly respectable way to prove something, though it's not very insightful mathematically.  For example, if a problem wanted you to find a solution to $2^x = 16$, your proof could consist of "testing $x = 1,2,...$, we find that $x = 4$ gives $2^4 = 16$."  Or, even more simply, you could skip the explanation of your guess and check and say "note that $2^4 = 16$".

[quote="JesusFreak197"]On this same note, do we have to prove something that, without extremely complicated methods, just requires some guess-and-test? Should we include some of the easier parts that eliminate parts of the guess-and-test? I've tried to understand Mathworld's method for what I'm stuck on, but I have no idea about half of what they're saying; I got the answer a different way that isn't very rigorous, but isn't quite directly related to the problem.

Sorry if the explanation is a little unclear, I'm trying to express my problem without giving anything away about what problem it is or how to solve it.[/quote]

On most problems, we simply say, "with proof," to remind solution writers to show their reasoning rigorously. In problems where there are many possibilities about what could be proven and proving everything would be exhausting, we try to be explicit about what we want proven. For example, the phrasing "Find five answers and prove that they are valid" means that we don't care how you found the answers, we just care that they are the right answers. (Nonetheless, explaining a good method for finding the answers could earn a commendation.) The phrasing "Find the answers and prove you found all of them" requires an explanation of a method for finding all possible answers.

Occasionally we stick to the original wording of the problem that a mathematician submitted to us for the USAMTS, but most often we are very precise about our grammar.

Erin Schram
USAMTS grader

Okay, I think that what I was thinking of putting down would be okay. I'll stick with that and not make stuff too complicated.


geometry perimeter FTW MATHCOUNTS function Asymptote calculus

I decided that I can't wait to let others take my mock test! So here it is, and it will also be posted in my blog. Instructions are on the first page of the test. READ IT CAREFULLY!!! :D

Technically, this doesn't start until after AIME15's competition is over, but I've decided to be nice to you and let you take it until...

[size=200]August 13th!![/size][size=59]This deadline is not strict. I may update it depending on others' schedules.[/size]

My test is posted below. No need to say "I sign up" or whatever. Just download this test and pm me, and I will automatically include you in my scoring.

Just a final word: [b]THIS IS A NATIONAL TEST.[/b] Do NOT, I repeat, DO NOT feel like you are stupid if you get below an $ x$ on this (where $ x$ is the score you normally get on school/chapter rounds). This is designed to be challenging and fun at the same time. Have fun!

The posts in this topic will be for any questions about any problems on my mock test (and/or spam, although please refrain from posting that :D ).

Meanwhile, I will update this list with people who have taken my test and who have been scored by Yongyi781.

1. tinytim

Er...on your target #8, I'm not quite sure how the dog travels. Does it always stay right next to the army...or like...was it something that I was too stupid to think of??? :blush:

[quote="Yongyi781"]Think of it as tracing the perimeter of the square. However, the thing is that the army is in MOTION while the dog is moving as well. An example of how this affects the dog's movement is that during its horizontal travels, the dog actually travels "diagonally" relative to a stationary object, and it takes longer going up the length of the army than down.[/quote]

OK that's what I thought it meant...just wasn't sure...thanks

[quote="Poincare"]Dumb question :
Does general relativity apply?[/quote]

You may assume it does, but then it only makes my mock test harder! :D

Oh, and another thing: #6 is sort of a trick; You may wonder whether 0 has a prime factor greater than 5 (and I can't blame you), but wait, I said POSITIVE INTEGERS!! :D

Updated list of people who sent in their scores:

1. tinytim
2. isabella2296

HURRY! You have until August 13th to send in your answers! :!:

Oh, and gauss1181 wants to register too. :D

only two people? *reminds me of how many people took my mock test* :(  :(  :(

I'm not from USA, can I be participant?  :)

Yes, anyone can participate! :D

Updated list (order submitted):
1. tinytim
2. isabella
3. chenhsi
4. shentang

Waiting list:
5. gauss1181

Hello, I just came back from my 4-day trip to Cape Cod!! OK, now here is the updated list of people I have scored in order submitted:

1. tinytim
2. isabella2296
3. chenhsi
4. shentang
5. gauss1181
6. Bachukas

can u publicly release the scores and current ranks?

Um, say a certain person did really badly and got a score a lot lower than anyone else, could you just post the rankings?  :oops:

Sure, isabella2296.

[b]In first place[/b]: tinytim
[b]Second place[/b]: Bachukas
[b]Third place[/b]: gauss1181
[b]Fourth place[/b]: shentang
[b]Fifth place[/b]: chenhsi
[b]Sixth place[/b]: isabella2296

Keep in mind that the registration is still open, so send in your answers and beat tinytim!!
Also, at the end of the period, I will publicly release an Excel file with all the contestants and their scores, as well as which questions they got correct or incorrect. I will most likely analyze these and post some fun facts! :lol:

I will also publicly release the answer key. If enough people are interested, I will also write solutions to all the problems.

Not only is my printer still broken, but my dad is too lazy to go buy ink.  :rotfl:

Meh, now my printer is having some color mismatch problems which nobody is going to fix for a while... :mad:

[quote="tinytim"]I love how all of a sudden we are talking about printers now! :rotfl:[/quote]

Hint, hint... ;) ;) ;)

[size=200]A reminder to all that mock test answers are due tomorrow!![/size]

Updated list (in order submitted):

1. tinytim
2. isabella2296
3. chenhsi
4. shentang
5. gauss1181
6. Bachukas
7. 27r
8. andersonw
9. math154
10. xpmath
11. FantasyLover

Waiting list:
12. Nexmus
13. PI-Dimension

geez why isn't anyone doing MY MOCK TEST???????

it is due [size=150][b]THURSDAY AUGUST 21 AT 10 PM CDT!!!!!![/b][/size]

MY mock test is here in this link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219024[/url]

oh and i hope i don't phail too badly on the yongyi test :o

Only 2 people did my mock test. Join the club!

And only 4 people did mine, so I get to join the club. What's the club's name?

Only 12 people did mine! I get to join the club as well! :D

I only do mock tests if the person who created it did my mock test!

We should call the club "rare mock test club"

guys I submitted mine.

I ran out of time on 25 on sprint or something (i was doing that one last)

I must say it used too much AIME knowledge, even though its trivial after that

Definitely not the same as nats, when many were multi-step, but not so abstract knowledge

[quote="nick42"]anyway...WHO WON???? and i want to know if you can post the answers.

was it tinytim?[/quote]
i think andersonw won. :D

nick42 are u gonna do my mock test? WE NEED MORE PPL DOING IT, TARGET IS DUE THURSDAY

This test is now officially over. Results can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221270]here[/url].


There seem to be a lot of problems from 1985 posted, so here is another.

[b]Problem: [/b] Evaluate \[\int_{0}^{\infty}\frac{1}{x^{1/2}}e^{-k(x+\frac{1}{x})}dx\] and you may assume that $\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$.

[b]Solution: [/b] The funny thing is that this Putnam problem is easily solved with another Putnam problem: [url]http://www.kalva.demon.co.uk/putnam/psoln/psol6810.html[/url]

\begin{eqnarray*}\int_{0}^{\infty}\frac{1}{x^{1/2}}e^{-k(x+\frac{1}{x})}dx&=& 2\int_{0}^{\infty}e^{-k(u^{2}+\frac{1}{u^{2}})}du~~~~~~~~~~~~~~~~~~~~~~~~~(x=u^{2})\\ \ &=&\int_{-\infty}^{\infty}e^{-k(u^{2}+\frac{1}{u^{2}})}du\\ \ &=&\int_{-\infty}^{\infty}e^{-k((u-\frac{1}{u})^{2}+2)}du\\ \ &=&\int_{-\infty}^{\infty}e^{-k(u^{2}+2)}du \end{eqnarray*}

The last step was obtained using the other Putnam problem mentioned above, and at this point we are pretty much done.  The solution at Kalva involves differential equations, and the soultion in the 1985-2000 book is slick, but not easy like this one.


trigonometry geometry calculus derivative inequalities function ARML

I've done two old AMC 10s and got around 110-115 on both of them.  However, yesterday I did the 2001 AMC 12 and got like a 70!  I only got to (read) about 12 problems before my time ran out, and the problems seemed a lot harder (than AMC 10's).  I know that the AMC 12 covers material not included on the AMC 10, but it wasn't like the questions were on a subject I didn't know - rather they just seemed really hard.

 Is the AMC 12 really that much harder?

This sort of has to do with this topic:

If a person scores an average of about 135 on the AMC 10, how well will he/she be expected to do on an AMC 12?

Wellllll, I don't think that's enough to determine because I score about 135-140 on AMC 10 but 115-120 AMC 12. (This is not intended as an arrogant message; apoligies to AIME15USAMO)

It depends on your knowledge of more "advanced" topics such as trig and logarithms...if you know those, your AMC 12 score will be a bit higher than the score of one who doesn't know them.

Well, he said that he knew how to do everything, but just didn't have enough time, so he should already have that knowledge.

I'm really weak at geometry, so I thought I'd look and see if I could find out how many geometry questions typically occur on the AMC 12 - I found this [url]http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012[/url]

If you go to AMC 12 topic breakdown it says that there is an average of 4 questions per test that are "Calculus & Analysis".  However, I've never seen a "calculus question", and the website says that every question can be solved using pre-cal methods.  Does this mean that some of the questions have both calculus and pre-cal solutions?

The calculus solutions are faster, but the problem does not have to be solved using calc, and can be solved using pre-calc.

A few of those 4 may also be in the "Analysis" part, but I don't know what that is.

The main subjects are Geometry and Number Theory.

Well, if there ever is a min./ max. problem, it is usually easy enough to be done with straight up derivatives  :lol:

Or, you need to manipulate a bit. Either way though, calc is NECESSARY, although helpful.

Perhaps you meant "not NECESSARY?"

Also, many optimization contest problems have three or more variables, requiring the use of things more tricky than a direct application of derivatives. I mean, there are partial derivatives, but those require calculations which take much longer than, say, a simple application of AM-GM. If all optimization problems on these contests could be solved by just derivatives, then the problems should not be on the contest at all.

There are no calculus questions on the AMC. There are no questions where calculus is easier or faster than the official solution. The AMC is aware that people sometimes try to solve problems with calculus - they're not going to write problems that give an advantage to people who do that.

All these optimization problems that look easy to solve with calculus are much easier to solve with basic inequalities like AM-GM.

As for that table, I'm not quite sure what they're talking about. I just looked through the 2008 AMC 12A and I can assure you that there are not five problems which even be vaguely considered to be calculus problems. My best guess is that some of the sequences/series problems and function problems, or maybe the distance/rate problems, are ending up in that category.

[quote="topofmath"]... about 135-140 on AMC 10 but 115-120 AMC 12. ...[/quote]
This seems pretty good to me. For the purposes of choosing an ARML team, I collect AMC/AIME scores. I've always discounted AMC10 scores by 20 points, and in most years, that seems quite fair.

[quote="AIME15USAMO"]Well, I average around a 140 on the AMC 10 and my AMC 12 scoring range is like 102-114. I'm assuming someone who averages a 135 on the AMC 10 would be similar to this.[/quote]

This totally depends.
I remember in my sophomore year, I got a 126 on the 10 (with the help of a few careless mistakes) and a 111 on the 12 (and I really could have done more)

It just depends on whether you're ready for the harder problems or not, imo.

[quote="RaviB"]The 2001 AMC 12 was the hardest AMC 12 ever. All you needed was an 84 on that test to make AIME.[/quote]

good point

[quote]If you go to AMC 12 topic breakdown it says that there is an average of 4 questions per test that are "Calculus & Analysis".[/quote]

Please state precisely where you found this breakdown.  It will help clarify how to answer your question.

If you study the AMC contest problems that are in the AMC 10/AMC 12 Math Club Package for example, you may note that in fine print at the bottom of each problem page are two lines.   One is labeled in a boldface font "NCTM Standards" and the other is labeled "MathWorld.com Classification".   We use those to try to classify and categorize the problems, in order to index them.  The   NCTM Standards are from the now well-known "Principles and Standards" document from the National Council of Teachers of Mathematics.  This document is excerpted at the NCTM website, nctm.org.
The other is a hierarchical classification of mathematics subjects at the on-line encyclopedia mathworld.com.  If you go to mathworld.com you will see the hierarchy and subjects classified under Calculus&Analysis, e.g Absolute Value is
Calculus&Analysis > Functions > Absolute Value.   We then use these classifications to index problems, so that students and teachers can consult the topic index to find problems of a certain type, or meeting a certain NCTM Standard.  The MathWorld.com classification also gives someone unfamiliar with the topic at least one place to make a first look-up.

However, neither classification system is perfect, that's why we use two.  Furthermore, many AMC problems do not neatly fit into any category.   It's a judgment call about how to classify any particular problem, and different people may make different classifications.  Finally, I do not always agree with the way the MathWorld hierarchy is organized, for instance putting Absolute Value in the Calculus&Analysis section makes some sense from an advanced mathematical viewpoint, but from a student's viewpoint it might not.

As a previous commeter wrote,
[quote]There are no calculus questions on the AMC. There are no questions where calculus is easier or faster than the official solution. The AMC is aware that people sometimes try to solve problems with calculus - they're not going to write problems that give an advantage to people who do that. [/quote]

That is a correct statement.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="AMCDirector"]Please state precisely where you found this breakdown.  It will help clarify how to answer your question.[/quote]
[quote="Evolved"]I'm really weak at geometry, so I thought I'd look and see if I could find out how many geometry questions typically occur on the AMC 12 - I found this http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012 [/quote]
But then why is there a Calculus & Analysis in the first place, when none of the questions really fit the category (they are all pre-calculus)?

Compare the topic titles in that AMC classification table with the topics titles in the top-level of the hierarchy at [url]mathworld.wolfram.com[/url].  You will see they are the same.

We use the MathWorld classification hierarchy because I have not found an alternative that is as complete or as useful.

If you want a more detailed or fine-grained classification or an alternative classification using for example the NCTM Standards, there are several in the AMC 10/AMC 12 Math Club Package.

Steve Dunbar
MAA Director, American Mathematics Competitions


probability
limit
conditional probability

some1 spamming the British MO...
plz help!

[hide="number 2"]There is a $1/6$ chance of rolling 3 on first roll.
There is a $(4/6)(1/6)$ chance of rolling 3 on second roll, with first roll not 1 or 2.
...
$(1/6)+(4/6)(1/6)+(4/6)^{2}(1/6)+...=\frac{(1/6)}{1-(4/6)}=1/2$?.[/hide]

[quote="8parks11"]1. What is the coefficient of $wx^{2}y^{3}z^{4}$ in the expansion of $(8w+\frac{1}{3}x+\frac{1}{2}y+z)^{10}$?

2. What is the probability that you will roll a 3 before you roll both a 1 and a 2 if we roll a fair 6-sided die repeatedly?[/quote]
[hide="1"]If the coefficients of w, x, y, and z were 1 in the factored form then it's coefficient would be $\frac{10!}{1!2!3!4!}=12600$. Now we multiply this by what the coefficients are, $12600*8*\frac{1}{9}*\frac{1}{8}*1=1400$[/hide]

2differentanswers?
letme try it later and ill post

no and no TIM- > although the setup for #1 is right (calculated wrong -> 1400

#2 is wrong

Bad with probability, but here's a shot:

[hide]Rolls of 4, 5, or 6 are inconsequential, and if they occur, will be ignored (re-roll).

Therefore, the first legitimate roll can be either a 1, 2, or a 3, with a $1/3$ chance for each.

If we roll a 3 first, then we win. $1/3$ chance of this.

If we roll a 2 first ($1/3$ chance), then we win if we roll a 3 before we roll a 1. Rolls of 2 are inconsequential, and will be ignored (re-roll). We want the first legitimate roll to be a 3 but not a 1, so there is a $1/2$ chance of this. Multiply by the $1/3$ chance of rolling the first 2 gives a $1/6$ of winning in this scenario.

The math is the same if we roll a 1 first: $1/3$ chance of rolling the first 1, and $1/2$ chance of winning given that the first 1 was rolled, so there is a $1/6$ chance of winning in this scenario.

The total probability of winning is $1/3+1/6+1/6=\boxed{2/3}$

 :maybe:
[/hide]

i don't actually know the answer but know the first one is wrong...
my guess

the question asks for the P of getting the pattern 3 then 1 and 2 in which 1 and 2 are switchable. Ok.

To get a $3$: 1/6

Here's what i think is the tricky part.

Getting a 1 and a 2 is actually quite simple, for rolling another 3,4,5,and 6 are will not affect the probability. If i simply roll another 3, well, then i'm still at step one.

Rolling a 3 then a 3 then a 3 then a 3 then a 3 is just resetting the experiment. So! Getting a probability of 1 or 2 would be 2/6. The final die roll MUST roll the leftover number, so 1/6.

$1/6*2/6*1/6 = 2/216$

my approach was that I focus on the probability of getting exactly those numbers without any conditions or affects on my P. Simply target the prob of the die rolls, and shoot. Is my answer right?

[quote="8parks11"]I focus on the probability of getting exactly those numbers without any conditions or affects on my P. Simply target the prob of the die rolls, and shoot. Is my answer right?[/quote]

Your approach is unsound.  Rolling a $1$ and a $2$ is a very different kind of event from rolling a $3$ - for one thing, it requires two rolls rather than one.

You also totally ignore the case where you roll a $3$ not on the first but the second, or third, or fourth, roll (without rolling both a $1$ or a $2$).

[quote="tim1234133"][hide="I would deny that I was spamming... heres no 1."]
There are $\binom{10}{4}$ ways that the $z$s can be chosen, $\binom{6}{3}$ ways that the $y$s can be chosen, $\binom{3}{2}$ ways that the $x$s can be chosen:

$\binom{10}{4}\cdot\binom{6}{3}\cdot\binom{3}{2}\cdot\frac{1}{8}\cdot\frac{1}{9}\cdot8=8400$ [/hide][/quote]You did the calculation wrong. 1400 is the correct answer, which is equivalent to the left hand side of your equality.

For 2

You have the solution of the  above problem here but as I misread
[hide="it solves  the  wrong problem first"][hide]Let the probablity be $p$ that you do this (I call it a "win")
Then what can happen -
Case 1: You throw a '3' (Probablity 1/6)   and you are done. (You win)
Case 2: You throw a '1' or a '2'  and you are done too (or should it be called "undone" :)   - you lose)
Case 3: You throw some thing else (Proablity 3/6 = 1/2): You are back to the begining ... and the chances (to win) now is, of course $p$

So you have $p = \frac{1}{6}+\frac{p}{2}$
This gives $p = \frac{1}{3}$ [/hide]

Hope this helps.

Pianoforte - you were almost correct, except for the typo, you should have (3/6) (instead of 4/6)  etc...)
[hide]
(Intuitively one can see that "winning" of 1 or 2 is twice as likely as "winning" by 3  ...  that again gives the  answer) [/hide]

Edited later: Didn't see the orignal problem carefully, the answer above is rolling 1 or 2, if it  1 and 2 the answer is different.. but can be done as above.[/hide]

(Edited again:I see The Zuton Force has already posted the answer... Keeping the post any way - hope its helpful.

. [hide]The answer in this case would be (2/3)
Again:
[hide]Throw a 3 - You are done:
Throw  a (1 or 2 ) - Now you win 1/2 the time (symetry)
Throw  other - back to  the begining.. so
$p = 1/6+(1/6+1/6)(1/2)+(3/6) p$ gives $p$ [/hide][/hide]

[quote="Gyan"]For 2

[hide]Case 2: You throw a '1' or a '2'  and you are done too (or should it be called "undone" :)   - you loose) [\hide]
[/quote]
But don't you have to roll a 1 [b]and [/b]a 2 to lose?

[quote="bpms"][quote="Gyan"]For 2

[hide]Case 2: You throw a '1' or a '2'  and you are done too (or should it be called "undone" :)   - you lose) [\hide]
[/quote]
But don't you have to roll a 1 [b]and [/b]a 2 to lose?[/quote]
Yep! tha's why I have [quote]Edited later: Didn't see the orignal problem carefully, the answer above is rolling 1 OR  2, if it 1  AND  2 the answer is different.. but can be done as above.
[/quote] in the quotes above !Sorry for the confusion, as that part was when I was solving (wrong) problem (you lose when you get (1) or (2))
You must see the  edited part to get the correct answer for the 1 AND 2.

I will edit the answer any way to remove the confusion.

[hide="#1"]Number 1 can be solved by inspection:

Splitting $\left(8w+\frac13 x+\frac12 y+z\right)^{10}$ into a sum of binomials,

$a = 8w+\frac13 x$ and $b= \frac12 y+z$, it is evident that the $wx^{2}y^{3}z^{4}$ term is contained in the $a^{3}b^{7}$ term of $\left(a+b\right)^{10}$ for which the coefficient is $\binom{10}{7}.$

The coefficient of the $wx^{2}$ term in $a^{3}$ is $\binom{3}{2}\cdot 8\cdot\left(\frac13\right)^{2}.$

The coefficient of the $y^{3}z^{4}$ term in $b^{7}$ is $\binom{7}{4}\cdot\left(\frac12\right)^{3}.$

Altogether the coefficient of the $wx^{2}y^{3}z^{4}$ term in the original expression is:

$\binom{10}{7}\binom{3}{2}\cdot 8\cdot\left(\frac13\right)^{2}\binom{7}{4}\cdot\left(\frac12\right)^{3}.$
[/hide]

[hide="#2"]A couple of you have had the right idea, but I don't believe anyone has given a complete explanation yet.

You either roll a 3 on the first roll or you don't.

The probability of doing so is: $\frac16.$

If you don't, you continue playing until you roll a 3 and win or until you roll a 1 or 2, in which case you lose. If we call $q=1-p$ the probability of losing then the probability that you may continue to play for $n$ passes is:

$\sum^{n}_{i=1}p^{i}q =\frac{pq\left(1-p^{n}\right)}{1-p}.$

The probability that may continue to play indefinitely is:

$\lim_{n\to\infty}\sum^{n}_{i=1}p^{i}q = \lim_{n\to\infty}\frac{pq\left(1-p^{n}\right)}{1-p}=\frac{pq}{1-p}=\frac{pq}{q}= p.$

If $q = P\left(1\text{ or }2\right) =\frac16+\frac16 =\frac13$ then $p = P\left(3,4,5,\text{ or }6\right) = 1-q = \frac23.$

The conditional probability that you win, by rolling a 3, given that you continue to play indefinitely is: $\frac14.$ (Equal to the probability of obtaining $3$ from $\lbrace 3,4,5,6\rbrace.$)

Thus the probability that you continue to play indefinitely and win is: $p\times\frac14 =\frac23\times\frac14 = \frac16.$

Conclude: The probability of winning, i.e. rolling  3 before rolling 1 or 2, is the probability of rolling once and winning OR rolling indefinitely and winning $= \frac16+\frac16 =\frac13.$[/hide]

[quote="8parks11"]its $\frac{2}{3}$ The Zuton Force was right[/quote]
Yep.. that's what I posted too.

1 or 2? I thought the question said that you must have both of them.

If I roll a 2 OR a 1, I must not roll the other or I lose. I need a 3. So it's (1/3) which is the prob of rolling a 2 or 1, multiply 1/6 which is the prob of getting a 3.

So 1/6 + 1/18 = 2/9...

whats wrong

i love it when people discuss such basic questions to no end  :wink:


probability
inequalities
triangle inequality

The lengths of two sides of a triangle are 7 centimeters and 12 centimeters. If a number is chosen randomly from the set $ \{1,2,3,\ldots,20\}$, what is the probability that the number chosen could be the length in centimeters of the third side? Express your answer as a common fraction.

From the [url=http://www.artofproblemsolving.com/Wiki/index.php/Triangle_inequality]triangle inequality[/url], we get the number must be in the interval $ 6\le n\le 8$ which has $ 13$ solutions, so the probability is $ \boxed{\frac{13}{20}}$.

The interval would actually be $ 6\le n\le 18 $

Actually, you don't have to solve the inequality.

Let x = the smaller triangle side given
      y = the larger triangle side given

The largest value of the third side is $x+y-1.$

The smallest is $y-x+1.$

The number of values in between is $(x+y-1)-(y-x+1) + 1 = x+y-1-y+x-1+1 = 2x-1.$

Thus $7\cdot2-1 = 13.$
Also, the values are all under 20 because $7+12$ is under 20.

$\boxed{\frac{13}{20}}.$

$2x-1$ works for all cases.


EDITED: If $ a, b, c$ are positive real numbers such that $ abc \equal{} 1$, then
\[ (a \plus{} b \plus{} c)^2 \plus{} a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {c^2\minus{}a^2}{a^2} \plus{} \frac {a^2\minus{}b^2}{b^2} \plus{} \frac {b^2\minus{}c^2}{c^2} \ge 2(\frac {ba \plus{} 1}{a^2} \plus{} \frac {cb \plus{} 1}{b^2} \plus{} \frac {ac \plus{} 1}{c^2})
\]

[quote="pythag011"]If $ a, b, c$ are positive real numbers such that $ abc \equal{} 1$, then
\[ (a \plus{} b \plus{} c)^2 \plus{} \frac {c^2}{a^2} \plus{} \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \ge 2(\frac {ba \plus{} 1}{a^2} \plus{} \frac {cb \plus{} 1}{b^2} \plus{} \frac {ac \plus{} 1}{c^2})
\]
[/quote]

If the condintion is $ abc\equal{}1$ then try $ a\equal{}10,b\equal{}1$ and $ c\equal{}0,1$.


superior algebra
superior algebra unsolved

Let $A=\mathbb{Z}[X]$. Prove that

   i) $(4, X)$ is a primary ideal, $\sqrt{(4,X)}=(2,X)$, but $(4,X)$ is not a power of $(2,X)$.

  ii) $(X^{2}, 2X)$ is not a primary ideal.

Sometimes notation varies, are you considering a power of $(2,X)$ to be (just in the case of 2) $(2,X)(2,X)=\{\sum a_{i}a_{j}: a_{i},a_{j}\in (2,X)\}$, or $(2,X)^{2}=\{\sum a_{i}^{2}: a_{i}\in (2,X)\}$?


calculus derivative algebra function domain

Q: Take the derivative.
f(x) = (x^4 + x^2 ) ^ (1/2)

A: (2x^2 +1)/(x+1)^(1/2)

The above answer is from the book.
However, isn't it (2x^3 + x)/(x^4 + x^2)^(1/2) since x can also be negative thus preventing it from being simplified?

In the future, calc problems would go better in the calc forums ;)

You dropped a ^2 in the book's answers.  Intro calc textbooks tend to be really awful about this sort of issue (implicitly restricting domains arbitrarily in order get $ \sqrt{x^2} \equal{} x$, for example).  You are correct that this function is defined for all of $ \mathbb R$ and so they should have pulled out a $ |x|$ instead of an $ x$, which would prevent cancellation.  Other ways to see that the book is wrong are to note that they're claiming the derivative of an even function is even when in fact derivatives of even functions are always odd, or that they're claiming the derivative at 0 exists when in fact it doesn't (the graph has a cusp there).

Same exact topic in another subforum (HSB):

[url]http://www.mathlinks.ro/viewtopic.php?t=258904[/url]


induction
arithmetic sequence
combinatorics unsolved
combinatorics

find the greatest integer $k$ such that we can find $k$element subset of a set $\{1,2,3,...,3^8\}$, in which there doesn't exist 3-element arithmetic progression

write these elements in base 3,and subtract 1 from each of them.Then we need to find a largest possible subset of $\{0,1,2...3^8-1\}$
i.e among the numbers denoted by strings of length $n$ consisting of $0,1,2$.Take all the strings with no 2's among them.clearly they have no 3 elt AP among them.thus you can get a set of size $2^8$.

actually the $2^8$ element set described is obtained by choosing greedily.i think it must be the best possible size too(of course,one can get other sets of the same size too)....however i don't have a proof of that now.

hmm...i conjecture that the answer is 2^n from the set {1, 2, ... , 3^n} for all positive integers n. :D induction ought to work, but i haven't quite made the bridge from the hypothesis yet.... :(

Anyone agree with me? or already proved it? :lol:

I also believe it to be true, but the only cases I could  check directly are $n=1,2$ :).

[quote="al.M.V."]i conjecture that the answer is 2^n from the set {1, 2, ... , 3^n}[/quote]
Sorry, the conjecture is false.  Behrend (1946) found much denser subsets without 3-progressions.  See
[url]http://planetmath.org/encyclopedia/BehrendsConstruction.html[/url].

Thank you all, but can someone tell me what is the final answer??

As I understood, final answer is unknown.  :?


limit
real analysis
real analysis unsolved

x_1, x_2, x_3, ... is a sequence of positive reals such that x_n < x_(n+1) + (x_n)^2. Prove that sum of  x_n diverges.

So we've got: $ x_{n+1}>x_{n}-x_{n}^{2}>x_{n}>0$, so $ \{x_{n}\}$ is increasing with $ x_{1}>0$, so $ \lim_{n}x_{n}$ can't be a $ 0$, so $ \sum_{n=1}^{\infty}x_{n}$ can't be convergent.


I wish to be better than I was last year ...

AND i wanna get to state mathcounts, and top 4 at states

[quote="BanishedTraitor"]-Become eligible for Mathematics and English Extension 2
-Make it into the Australian Chemistry Olympiad Team
-Get a decent score on the Australian Mathematics Olympiad
-Prove the Riemann Hypothesis
-Disprove Fermat's Last Theorem
-Prove Poincare Conjecture/Theorem in one line

I am sure you can tell the real ones from the not so real ones.[/quote]

I believe it is the hidden one is the real wish.

I wish I made Black MOP.

I wish...for....nothing. I don't need wishes :P

I want to make USAICO and MOP. I don't really care whether I make blue or black MOP though. Eventually I want to make black...but it doesn't have to be this year.

Make USAMO, top 12 Nats Mathcounts, have someone end world hunger, make someone invent solar panels made of plastic that have 30% efficiency. :)

[quote="randomdragoon"]While everyone is wishing for stuff, I might as well ask for a Pikachu.

/Ponies? Bah, who needs em[/quote]

 :rotfl:  EPIC.

I wished that 2008 gave us cool quotes just like 2007 (Which includes - "DONT TAZE ME BRO!" and "CAN I HAS RECON>??!!")

I wish I will be better in math. :P

Happy year to everyone!  :)

[quote="sylow_theory"]I wish for 2 more wishes  :P
Is that against the rules :maybe:[/quote]

Why not. I wish for infinite wishes that I can use throughout this year.

Actually, numerous fairies have informed me that it is actually against the rules to wish for more wishes. There is actually a pretty extensive code of things that you can't wish for. For example, the fairies must still obey conservation of matter & energy, so just wishing for a pony won't work (can't materialize a pony out of thin air).

Silly, you know Gamp's law is make-believe.

[quote="CatalystOfNostalgia"]guys see Gamp's Law of Universal Transfiguration (I think that's what it's called...)[/quote]

Gamp's Law of Elemental Transfiguration.

[b][size=200]HAPPY NEW YEAR!!!!!!!!!!![/size][/b]

i want to be smarter!

[quote="shaggy75"]i want to be smarter![/quote]

So Do I!

I wish that global warming will not cause humans to go extinct.
I wish that I could visit AoPS 24hrs a day, 7days a week, 52 or so weeks a year, 100years a century, etc.

I just hope I can work harder;I would be so much better if I could actually sit down and work on stuff.

[quote="ZhangPeijin"]I wish that I had unlimited money.
 :lol:[/quote]

WG. Counterfeit money.

OOPS, wrong topic. :blush:

You make a good point. I'll get a Magikarp instead and give it an Everstone.

i wish that i can get a wii... :(
all my friends have one, and they pwn me in wii tennis...
but i still beat my best friend at rayman raving rabbids.. :D

[quote="kimmystar94"]
all my friends have one, and they pwn me in wii tennis...
[/quote]

If that ever happened to me I'll just grab my racquet and crush them in real life tennis.

Seriously, even though Wii Tennis is fun, it's NOTHING like real tennis.

well.. the thing is, my friend's amazing at real tennis.. more than she is at wii tennis...
:D
and i play real tennis, too..
oh.
i wish to become better at tennis..

i wish i made USAMO and nats MC. i wish our science olympiad team would win nats. i wish i could get more wishes

[quote="randomdragoon"][quote="erabel"][quote="randomdragoon"]While everyone is wishing for stuff, I might as well ask for a Pikachu.

/Ponies? Bah, who needs em[/quote]

I would rather vote for a Munchlax!!![/quote]

Real life isn't like the game, you would have to consider very much how much you spend feeding your munchlax and where you would house it if it were to evolve.

See, a Pikachu doesn't eat much food and even pays for itself in reduced home electricity bills.[/quote]

I would get an mudkip and get my free water.
So, I herd joo liek mudkipz?

nooo.. don't even go there, my friends...
"joo liek mudkipz?"
noo...
please don't..
it shall haunt me in my dreams......

For tennis, I don't really need the lessons. Not that I am amazing or anything, I just don't think they help at all. All you need is constant practice  :)

[quote=7h3.D3m0n.711"]All you need is constant practice
[/quote]

This usually goes for all sports.

I wish I will make it to Black MOP. Now for a sane wish. I wish I made it Nats Math and AIME.

I also wish for a Gengar as my bodyguard. It's a ghost, so it doesn't have to eat, and therefore is a cheap pet.


trigonometry
calculus
integration
real analysis
real analysis solved

let us recall that $ \frac{d^2}{dx^2} (fg) = f''g + 2f'g' + fg''$, and that we have $ \frac{d}{dx}\tan x = 1+\tan^2 x$, $ \frac{d^2}{dx^2}\tan x = 2\tan x (1+\tan^2 x)$.
we can then write:
\[ \frac{d^2}{dx^2}(y \tan x) = (\tan x )y\"+2(1+\tan^2x)y'+2 \tan x(1+\tan^2x)y = y\tan x \]
setting $ z = y \tan x $ we should be able to get $ z = Ae^x + Be^{-x} $ for arbitrary $A$ and $B$, so
\[ y = (Ae^x + Be^{-x}) \tan^{-1} x \]
maybe i made some mistake in computations, but i am really not in the mood for deriving!

[edit: the error was in integration :|]

$y = \frac{Ae^x + Be^{-x}}{\tan x}$

Congratulation ma_go


inequalities
integration
logarithms
calculus
function
real analysis
calculus computations

Man, that one looks a lot like [b]Stirling's Formula[/b]...

$ n!=\sqrt{2\pi n}\cdot (\frac{n}{e})^{n}\cdot (1+O(\frac{1}{n})).$
So if we put this it will be easy to prove the inequlities.  :P

Of course it won't: an asymptotic expression alone can't provide proof for an inequality at any given value of the variable.  Now, the [i]proof[/i] of the asymptotic expression might be useful for proving the inequalities.  For example,
[hide="LHS"]Estimate $ \int_{1}^{n+1}\ln x \; dx$ by the sum $ \sum_{i = 1}^{n+1}\ln i$.  The latter is the right-endpoint sums for the Riemann sum of the integral of a concave function, so is larger than the value of the integral.  Evaluate the integral and exponentiate to get the left-hand inequality.[/hide]

The idea for the right-hand side is almost exactly the same.

[quote]Of course it won't: an asymptotic expression alone can't provide proof for an inequality at any given value of the variable.[/quote]

This is really just a side note to JBL, but it is an important point to make.  Technically JBL has said nothing wrong, but I should point out that Stirling's approximation takes an equality form as follows:

$ n!=\sqrt{2 \pi n}\left( \frac{n}{e}\right)^{n}e^{\lambda_{n}}$ where $ \frac{1}{12n+1}< \lambda_{n}< \frac{1}{12 n}$.

Sometimes this form is useful for proving limits in a rigorous way.

Getting back to the original topic...

If $ f$ is an increasing function on $ [1,\infty]$, then

$ f(1)+\ldots+f(n-1) < \int_{1}^{n}f(x) dx < f(2)+\ldots+f(n).$

Let us make $ f(x) = \log x$ in the above inequalities.. [b]ac1072[/b]'s inequality follows at once from this substitution.

[quote="Yuriy Solovyov"]$ n!=\sqrt{2\pi n}\cdot (\frac{n}{e})^{n}\cdot (1+O(\frac{1}{n})).$
[b]So if we put this it will be easy to prove the inequlities.[/b]  :P[/quote]

[b]Of course, I never meant that...  :(  All I said was that the purported inequality reminded me a lot of the way I usually write Stirling's Formula. Anyways, I've already posted a somehow complete solution to the problem...[/b]