inequalities trigonometry logarithms geometry

Let $ ABC$ be a triangle. Prove that
\[ \tan\frac {\alpha}{2}\cdot \tan\frac {\beta}{2} \cdot\tan\frac {\gamma}{2}\leq \frac {\sqrt {3}}{9}\]

[hide="Solution"]
Let $ f(x)=\ln \tan x$,

$ f(x)$ is concave down, therefore by [url=http://www.artofproblemsolving.com/Wiki/index.php/Jensen%27s_Inequality]Jensen's Inequality[/url],

\begin{eqnarray*}
\dfrac{f(\dfrac{\alpha}{2})+f(\dfrac{\beta}{2})+f(\dfrac{\gamma}{2})}{3}&\le& f\left(\dfrac{\alpha +\beta+\gamma}{6}\right)\\
\dfrac{\ln (\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2})}{3}&\le& \ln \tan \dfrac{\alpha+\beta+\gamma}{6}\\
\ln  (\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2})^{\dfrac{1}{3}}&\le&\ln \tan \dfrac{\pi}{6} \\
\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2}&\le& \tan^3 \dfrac{\pi}{6}\\
\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2}&\le& \left(\dfrac{1}{\sqrt{3}}\right)^3\\
\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2}&\le&\dfrac{1}{3\sqrt{3}}\\
\tan \dfrac{\alpha}{2}\cdot \tan \dfrac{\beta}{2}\cdot \tan \dfrac{\gamma}{2}&\le&\dfrac{\sqrt{3}}{9}
\end{eqnarray*}
[/hide]

[hide="Solution 2"]

$ \tan\frac{\alpha}{2}\cdot\tan\frac{\beta}{2}\cdot\tan\frac{\gamma}{2}\equal{}\prod_{cyc}\sqrt{\frac{(s\minus{}b)(s\minus{}c)}{s(s\minus{}a)}}\equal{}\frac{r}{s}$ .

So we need to prove that $ \frac{r}{s}\le \frac{\sqrt 3}{9}$ or $ 3\sqrt 3 r\le s$ .

Now, by $ AM\minus{}GM$ ineq. applied to $ s\minus{}a,s\minus{}b,s\minus{}c\ \implies s\ge 3\sqrt[3]{(s\minus{}a)(s\minus{}b)(s\minus{}c)} \Longleftrightarrow\ s^4\ge 27S^2$ , $ S$-area of $ \triangle ABC$ .

From $ S\equal{}sr\ \implies s\ge 3\sqrt 3 r$ so we are done.

[/hide]


Let $n$ be a positive integer. Denote $M = \{(x, y)|x, y \text{ are integers }, 1 \leq  x, y \leq  n\}$. Define function $f$ on $M$ with the following properties:

[b]a.)[/b] $f(x, y)$ takes non-negative integer value; [b]
b.)[/b] $\sum^n_{y=1} f(x, y) = n - 1$ for $1 \eq  x \leq n$;
[b]c.)[/b] If $f(x_1, y_1)f(x2, y2) > 0$, then $(x_1 -  x_2)(y_1 -  y_2) \geq 0.$

Find $N(n)$, the number of functions $f$ that satisfy all the conditions. Give the explicit value of $N(4)$.

I know that N(n)=(n^2-1)C(n-1)  so N(4)=15C3=455
Because of my poor English and poor LaTex skills, I can't give the whole solution.

Here is the official solution which is much more elegant than the previous one :)

Good solution!And I think we can use generating function too.


abstract algebra
superior algebra
superior algebra unsolved

let $ G$ be a non-abelian group and $ |G|\equal{}p^3$ (p is prime),find a known group isomorphic to $ G'$ .

I don't think I understand your question. Do you mean the following?

Since $ G$ is a $ p$-group, it is solvable. Hence $ G'$ has order $ 1,\, p,\, p^2$. Since it is given that it is not $ 1$, we have that it is $ p$ or $ p^2$, hence $ G'$ is isomorphic to one of the following:

$ C_p,\ C_{p^2},\ C_p \times C_p$.


If $ \frac{a}{b}$ and $ \frac{c}{d}$ are neighbor fractions, their difference has a numerator of plus-minus 1. Prove that $ \frac{a\plus{}b}{c\plus{}d}$ is between $ \frac{a}{b}$ and $ \frac{c}{d}$ and $ \frac{a\plus{}b}{c\plus{}d}$ is a neighbor fractions to both $ \frac{a}{b}$ and $ \frac{c}{d}$

thanks :)

Two keywords:  [url=http://mathworld.wolfram.com/FareySequence.html]Farey sequence[/url] and [url=http://en.wikipedia.org/wiki/Pick_theorem]Pick's theorem[/url].

this is supposed to be a very easy question supposed to be solved with very basic tools so...

As mentioned in the other thread, you've got the fraction wrong; it's $ \frac{a \plus{} c}{b \plus{} d}$.  You might find the problem easier now (since it was false before).  I mention those two key words not as a solution method (although Pick's is very nice and not exactly advanced mathematics) but as an application of the notion of neighbor fractions.


inequalities
search
inequalities proposed

Note that, using this inequality, we can prove
\[\sqrt[5]{\frac{x}{y+k}}+\sqrt[5]{\frac{y}{z+k}}+\sqrt[5]{\frac{z}{x+k}}\ge \frac{3}{\sqrt[5]{k+1}}\]
for any $x,y,z,k>0$ such that $xyz=1$.

@Dear Hung: Have you got any nice solution for it?

[quote="silouan"]I think the following is stronger (it belongs to Arqady)
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq\sqrt[4]{27(a^{4}+b^{4}+c^{4})}$

and its solution is only AM-GM !  :wink:[/quote]
No, since
\[\sqrt[6]{243(a^{6}+b^{6}+c^{6})}\ge \sqrt[4]{27(a^{4}+b^{4}+c^{4})}\]
by Cauchy Schwarz. :P
My inequality is stronger!

[quote="toanhocmuonmau"]
My inequality is stronger![/quote]
Yes, toanhocmuonmau! And the following inequality is wrong already:
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\cdot\sqrt[7]{\frac{a^{7}+b^{7}+c^{7}}{3}}.$

I only can prove that $LHS \ge 3\sqrt[5]{\frac{a^{5}+b^{5}+c^{5}}{3}}$, can you post your solution, toanhocmuonmau ?

This is in "Sáng tạo Bất đẳng thức". You yourself can check it, 10maths. It is also discussed here. Search with Arqady or Hungkhtn.

[quote="10maths_tp0609"][quote="silouan"]I think the following is stronger (it belongs to Arqady)
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq\sqrt[4]{27(a^{4}+b^{4}+c^{4})}$

and its solution is only AM-GM !  :wink:[/quote]
can you post that solution?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38369
It's Gladkich's proof. :wink:

[quote="toanhocmuonmau"]Prove that for any positive real numbers $a,b,c,$ we have
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\sqrt[6]{\frac{a^{6}+b^{6}+c^{6}}{3}}\]
:)[/quote]
Any nice solution for it?  :?:

[quote="toanhocmuonmau"]Prove that for any positive real numbers $a,b,c,$ we have
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\sqrt[6]{\frac{a^{6}+b^{6}+c^{6}}{3}}\]
:)[/quote]

To prove this inequality,we can use a lemma of can_hang2007:

\[ \dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a} \geq \dfrac{3\sum a^4+13\sum a^3(b+c)-\sum a^2b^2-abc\sum a}{3\left(\sum a\right)\left(\sum ab \right)} \]
Then use $ p,q,r $ or $ u,v,w $ theorem.

[quote=quykhtn-qa1]To prove this inequality,we can use a lemma of can_hang2007:

\[ \dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a} \geq \dfrac{3\sum a^4+13\sum a^3(b+c)-\sum a^2b^2-abc\sum a}{3\left(\sum a\right)\left(\sum ab \right)} \]
Then use $ p,q,r $ or $ u,v,w $ theorem.[/quote]
It's wrong if $a=\frac{1}{7},\,b=\frac{1}{62},\,c= 1.$


Determine all injective functions f: N -> N such that

f(f(n)) $\leq$ (n+f(n))/2.

I think that the answer is [color=white] f(n) = n [/color] and have a hazy proof which i havn't rigorised yet...

Have a look at :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5449

Pierre.

[quote="Peter VDD"]I think any funtion saying:
$f(0\rightarrow n)=0$
$f(n+1)=1$
$f(n+i)=i$
fulfills the requirements :)[/quote]

We can't have this I think because the function has to be injective.

Thanks for the solution Pierre, it was basically the same as mine, just but a little (well a lot!) neater


geometry
circumcircle
power of a point
radical axis
geometry proposed

Triangle ABC isnt isosceles. O is the circumcenter of ABC. The incircle of triangle ABC touches sides BC,CA,AB at D,E,F. $ P \equal{} DE\cap AB, Q \equal{} DF\cap AC$. M,N are the midpoints of QE,PF. Show that three perp lines from O,P,Q to MN,BE,CF resp are concurrent.

[quote="mr.danh"]
Let $ \triangle ABC$ be non-isosceles triangle with circumcenter $ O$.Let $ D,E,F$ be the tangency points of incircle $ \omega(I)$ with $ BC,CA,AB$,respectively.Denote $ P\in\DE\cap AB$,$ Q\in DF\cap AC$.Let $ M,N$ be the midpoints of $ QE,PF$,respectively.
Prove that perpendicular lines from $ O,P,Q$ to $ MN,CF,BE$,respectively,are concurrent.
[/quote]
Proof:
Lemma:
1.$ CF$ is the polar of $ P$ w.r.t $ \omega$.
Proof of lemma:
See here http://www.mathlinks.ro/weblog.php?w=1027
Now,using lemma,we know that $ PI\perp CF$ and $ QI\perp BE$,thus our goal is to show that $ OI\perp MN$,
As we know from Pythogorean theorem,$ OI\perp MN\Longleftrightarrow IM^2 - IN^2 = OM^2 - ON^2$.
Since $ FM = MP$ and $ EN = NQ$,we conclude that $ IM^2 - IN^2 = \frac {FP^2 - QE^2}{4}$.
On the other hand,$ OM^2 - ON^2 = (OX^2 - ON^2) + (XM^2 - NY^2)$,where $ X,Y$ are midpoints of $ AB$,$ AC$,respectively.
Since $ OX^2 - OY^2 = AY^2 - AX^2$,we conclude that $ OM^2 - ON^2 = AY^2 - NY^2 + XM^2 - AX^2 = AM\cdot MB - AN\cdot NC$.
So we need to prove that $ 4(AM\cdot MB - AN\cdot NC) = FP^2 - QE^2$.
$ AM\cdot MB = (MF - AF)(MF + BF) = MF^2 + MF(BF - AF) - AF\cdot BF$,so all we need to prove is that
$ MF(BF - AF) - AF\cdot BF = NE(CE - AE) - AE\cdot EC$,
But as we know $ (PF,AB) = - 1$,hence $ \frac {PB}{PA} = \frac {BF}{FA}$,thus $ 1 + \frac {c}{PA} = \frac {p - b}{p - a}$,so
$ \frac {c}{PA} = \frac {a - b}{p - a}$,where $ AB = c,AC = b,BC = a,p = \frac {a + b + c}{2}$,
Since $ M$ is the midpoint of $ PF$ and $ PF = PA + AF = \frac {(p - a)c}{a - b} + p - a = \frac {2(p - a)(p - b)}{a - b}$,we conclude that
$ MF = \frac {(p - a)(p - b)}{a - b}$,hence $ MF(BF - AF) - AF\cdot BF = \frac {(p - a)(p - b)}{a - b}(a - b) - (p - a)(p - b) = 0$,similarly $ NE(CE - AE) - AE\cdot EC = 0$,so we are done.

It's clear that $CF,BE$ are the polars of $P,Q$ WRT the incircle $(I)$ of $\triangle ABC,$ thus the prependicular lines dropped from $P,Q$ to $CF,BE$ pass through $I.$ Since $(A,B,F,P)=-1$ and $(A,C,E,Q)=-1,$  by Newton's theorem it follows that $NF^2=NA \cdot NB$ and $ME^2=MA \cdot MC$ $\Longrightarrow$ $M,N$ have equal power WRT circumcircle $(O)$ and incircle $(I)$ $\Longrightarrow$ $MN$ is the radical axis of $(O),(I)$ $\Longrightarrow$ $MN \perp OI.$ Hence, perpendicular lines dropped from $O,P,Q$ to $MN,CF,BE$ concur at $I.$


linear algebra
matrix
linear algebra unsolved

$M\in M(n,R)$
k>0 fixed integer such that $(k+2)M^{k+1}=(k+1)M^{k+2}$

Find the inverse  of $M^2-2M+I_n$

I´ll prove that $B=(-I_n-M-...-M^k+(k+1)M^{k+1})^2$ is the inverse of $(M-I_n)^2$. It´s just multiplicate and we´ll obtain: $(M-I_n)^2B^2=(M-I_n)(M-I_n)BB$ and now we can observe that $(M-I_n)B=-M-M^2-...-M^{k+1}+(k+1)M^{k+2}+I_n+M+...+M^k-(k+1)M^{k+1}=I_n-M^{k+1}-(k+1)M^{k+1}+(k+2)M^{k+1}=I_n$ and so it´s proved that $(M-I_n)^2B^2=(M-I_n)(M-I_n)BB=I_n$


now its coming on both the problem and lessons....i cant see any of my areas in my report either...

EDIT: if i click badges, i get this:

Fatal error: Cannot use string offset as an array in /var/www/Alcumus/Badges.php on line 76

ouch...... but, did u report it to the admins?

try that......

but srry, i dnt know wat 2 do......

Now when trying to get a probelm, it says "Cannot contact server...".  The internal errors are gone, but still nothing shows up.

It appears that someone has been working on Alcumus and temporarily took it down.

When AoPS crashed yesterday, all the forums went down and thus I suppose that made the Alcumus system confused and thus did not display anything. Yeah, when I saw that, I was shocked too, because I didn't think I finished all of the videos and problems. But I think they have got it fixed, so it should work now, I guess.

Our server developer is still working on the problem.


what can a movie like Good Night and Good Luck, describing historical events set in 1953, tell us about journalism, politics and civil liberties today, especially in the post 9/11 era?

Good Night and Goood Luck:

Good Night, and Good Luck takes on an historical moment in American history. It was a period at the height of the cold war, where the fear on nuclear annihilation by the Soviets was high, and patriotism often meant that dissenting ideas were not tolerated.   Edward R. Murrow, among others, did take on Senator McCarthy, who was leading a crusade against subversives through the House UnAmerican Activities Committee (HUAC).  The clips of Senator McCarthy and the Committee in the film are not acting, but documentary footage. As the Good Night, and Good Luck webpage notes:

He [Murrow] manages to highlight the serious issues involved in the McCarthy hearings: namely, the line between investigation and 'persecution'; that dissent is not disloyalty; accusation is not proof; conviction depends upon evidence and due process of law; and finally, as defenders of freedom abroad, the US cannot desert it at home.

What do you think?

[quote="JBL"]I think you should do your own essays for class.[/quote]

 :rotfl:  nice.

[hide="Hint"]
What's the point of studying history?
[/hide]

[quote="mdk"]Good Night and Goood Luck[/quote]That's what the vampires said in 30 Days of Night when they started the hunt  :D

[quote="mdk"]This is not homework. It was a prompt given to us due a couple of days ago, and I wanted to know what you guys thought?[/quote]

i generally avoid thinking if i can get away with ti


Let $a_1,a_2,..,a_n (n\geq2)$ be positive real numbers, such that $\sum\limits_{1\leq i\leq j\leq n} \frac{1}{a_ia_j}\geq 1$. Prove that

$\frac{a_1}{a_2^2}+\frac{a_2}{a_3^2}+..+\frac{a_{n-1}}{a_n^2}+\frac{a_n}{a_1^2}\geq \sqrt\frac{2n}{n-1}$


The problem maybe is easy for you.
Let $a_1,a_2,...a_{16}$ be integers such that $0<a_1<a_2<...<a_{16}<140$,prove that there exist four positive integers $p,q,r,s$ s.t. $0<p<q<r<s<16$,and $a_p+a_s=a_q+a_r$.

hmm,why still no one? :( please help me solve the problem as quickly as possible,thanks! :)

[quote="shyong"]I think it is very obvious that it exist (if i didnt misintepret the question)

Since $a_p<a_q<a_r<a_s$

Then we have $a_q+a_r=a_p+a_s>a_p+a_r$  :arrow: $a_q>a_p$ which is always true .

i.e $a_p=1$ , $a_q=2$ , $a_r=3$ , $a_s=4$ .[/quote]
strangely, :roll: what do you mean?I can understand nothing!

[quote="shyong"]I think it is very obvious that it exist (if i didnt misintepret the question)

Since $a_p<a_q<a_r<a_s$

Then we have $a_q+a_r=a_p+a_s>a_p+a_r$  :arrow: $a_q>a_p$ which is always true .

i.e $a_p=1$ , $a_q=2$ , $a_r=3$ , $a_s=4$ .[/quote]sigh. The 16 numbers are given. You have to prove that for each 16 such numbers, there exist 4 indices p,q,r,s. So you cannot put $a_p=1$ , $a_q=2$ , $a_r=3$ , $a_s=4$.

But you can put $p=1$ , $q=2$ , $r=3$ , $s=4$, which doesn't solve the problem.

still no one, :( I am sure you can solve it! :)


When and how the results of tomorrow's AMC 8 will be available?
Thank you.

After the AMC 8 is administered, the Contest Manager, (typically your math teacher) mails the contests to us.  That takes anywhere from 2 days to 8 days.  The AMC office sorts and organizes the incoming packages, that usually takes 2 to 3 days.  Then the scanning can begin, and we can scan approximately 40,000 answer forms per day, so it will take about 5-7 days to scan all the results.  Then the data has to be collated, and tabulated, and that usually takes about 2-4 days.  Then the reports can start being emailed, along with statistical results from the tabulations.  Don't forget that (US) Thanksgiving holiday falls in there, so we won't be working the two days of  November 23rd and 24th.  So expect the first results to be emailed to the Contest Manager, that is usually your math teacher, about 3 weeks after the test is given.  That would be about December 5.

However,  a lot depends on the US mail, and the timing depends on the scanners working well.  It also depends on you the student writing clearly and filling in all circles darkly and completely.

Steve Dunbar
AMC Director

When will the perfect scores be shown on the AMC8 website? Are any other names mentioned?

-jorian

I wish they had a section for high scorers with 23 or 24. That would really tone down the pressure for getting a high score. One problem. wrong.. yeah . That way, someone who worked hard but ended up just short of the score could be recognized. Yeah, scores , especially mail ins take a while. Just like the Fall Startup- which took like 3 weeks.

  :)

At least January, and probably February because:
(1) It is the last thing completed, and requires all scores to be tabulated in order to get a complete list of perfect scores.  We will not put up a partial list of perfect scores.
(2) It takes that long to collect the photos.

Steve Dunbar
MAA Director of AMC

You people have to understand how much it takes to do all this...

Will we be allowed to post answers the day after the test?

EDITED:

SEE BELOW:

WAIT UNTIL AFTER NOVEMBER 21 TO DISCUSS IT!

When will the AMC post the answers so we can unofficially see how we do? Rather, when is discussion going to be allowed?

well obviously after the 21st we can discuss it, because no one will be taking it
-jorian

[quote="easyas3.14159..."]Don't teachers usually have an answer key?[/quote]

I don't think so.

[quote="mysmartmouth"]Yes teachers do get an answer key.  It seems to me that severely jeopardizes the integrity of the results: it takes just 1 dishonest teacher to add a perfect score to the list.  But the solution manual is sealed with instructions not to open until x/y/2006.

(At least, that's how I remember it...)[/quote]
You are right, but one "dishonest teacher", in the very very unlikely event that that happens, would probably add two or three perfect scores, and wouldn't need an answer key to do so (they have a copy of the test and most math teachers should be able to do those problems). I think that if a student were in that situation, it is their honesty at stake. As much as I want to do well, it really annoys me when I see threads littered with stuff like "I would give up a kidney to win national mathcounts". If you don't win, you probably didn't deserve to, so you stole the victory from someone. That is how I feel.

There is plenty of recognition for various scoring levels.

Good luck on the test. Remember, NO discussion of results or content of the test until after the 21st.

are there regional and state winners? like awards for amc 8

and is it graded by the amc directors and staff?

Yes if win for your state, you get a shiny plaque that says so. It's really nice. Some states have you take a photo with your governor/senator/whatever or something along that lines, but that's all.

Last year I made a bunch of careless mistakes and ended up with a 22.  It was still (tied for) the highest score in SC.  So all 3 high scorers got a plaque reading:

[Your Name]
[Your School]
1st Place in [Your State]

AMC DETAILS HERE

It was a nice plaque, too.

Answers to every question posted in this thread can be found at the AMC website, or by inquiring with your math teacher.  Read the Teachers' Manual thoroughly.

See Section 5, middle inside panel of the 2006 AMC 8 Brochure
http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/2006-AMC8Bro.pdf
for a description of state and national awards.

For a description of within-school awards, see Section X. Page 6, of the 2006 AMC 8 Teachers' Manual:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/06-AMC8TM/2006-amc8TM.pdf

Steve Dunbar
AMC Director

[quote="now a ranger"]but do you have to have a perfect score? or does it depend on state. I don't think I will get anything with a 24 lol. so i basically didnt do well[/quote]

I got a 24 last year, and I got a nice plaque for being first in Wisconsin. 24 is a respectable thing, because even if you had no clue on the 25th problem, you can say "Ugh. I made one stupid mistake and that stopped me from being perfect!".

Unless you live in a state with someone really smart in it, you'll probably get a plaque that says "1st in ____"

P.S. I really did make a stupid mistake though.

Did you guys all take it today?

my school isn't taking it until tomorrow... :maybe:

[quote="nebula42"]Please DON'T discuss the AMC 8 yet. This includes discussing problems and discussing scores (as you just did).

Also, this was already posted in several places, including:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119364[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119172[/url][/quote]

Do you guys not understand this?

[i]Edited by a moderator - don't discuss the AMC 8 until you are allowed to do so.[/i]

Please do not discuss the AMC 8 until November 21, in terms of problems or scores.


Hi all,
I have an interesting observation which I made while I was browsing throught the Combinatorics unsolved problems section of the Forum.
Consider the following statement:
"If there are two posts A and B, and if post A is posted before post B, on an average post A should have more views than post B".

But, infact check for yourself, sometimes this statement is far from the truth. In a few cases in the same Combinatorics forum ( I plan to check in more sub-forums), posts which are posted a day after have about 50 more views. But in some cases it is round about the same.
Comment, on your views on this freaky observation.

Well if a thread gets more replies it gets pushed to the top again, so people are more likely to view it... also factors like attractive title, difficulty, source, poster, etc. influence what threads people will view.


For research papers, is it more appropriate to use "we" or "I" when saying something like Here, we (I) introduce this new formula...

I am writing this by myself with no help from a professor, so do I still use we or should I use I, which seems a lot stronger.

Thanks

"We" in sentences like "[i]Now we derive the corollary from the theorem[/i]" usually means "we, the author and the reader" rather than "we, the authors", so I usually use "we" in all cases except in sentences like "[i]I learnt this technique from...[/i]". Even in those you can avoid the first person by using "the author" instead of "I" if you want but I do not see a big difference. "I" repeated too many times usually leaves an impression of the author giving a lecture to the reader rather than sharing his thoughts with him.

Anyway, there is no great difference and you may be sure that this "I" versus "we" issue will be the last one anybody will take into account when forming his opinion of who really did the job :).

Alright thanks.  Compared to the actual research, this is a very small thing, but I wanted to be stylistically correct.


analytic geometry
geometry
geometric transformation
dilation
trigonometry
quadratics
linear algebra

Hi,
We can post USAPhO problems in this thread. We don't necessarily have to have 1 a day though. Multiple choice and open response are ok, but this is mainly for open response since there are no solutions avalible to them. Problems can be found at http://www.compadre.org/psrc/evals/olympiad.cfm .

OK Let's start off with one not too hard!

Suppose that Galileo is in the Leaning Tower of Pisa. He throws a cannonball straight down from height $ H$ above the ground. The ball leaves his hand with inital speed $ v_0$. At the same instant, his friend throws a cannonball straight up from ground level with an intial speed of $ 2v_0$. When the cannonballs collide, they are traveling in the same direction, and Galileo's cannonball is traveling with 7 times the speed of his friend's. Ignore air resistance.

1. At what time does the collisions occur? Express in terms of $ H$ and $ v_0$

Well, I don't believe the first part of the question really matters.  Both of them are accelerating downwards at $g$, so it doesn't really matter- all you need to know is that the balls are approaching each other with velocity $3 v_0$.  So their separation is $H$, and its just $\frac{H}{3v_0}$

Next question(1999):

A car accelerates uniformly from rest.  Initially, its door is slightly ajar.  Calculate how far the car travels before the door slams shut.  Assume the door has a frictionless hinge, a uniform mass distribution, and a length $L$ from front to back.

Are you sure that is correct? What is wrong with what I did:

$x=1/2 gt^2+v_0t$
$H-x=1/2 gt^2-2v_0t$

Add both to get $H=gt^2+v_0t$
Solve for t to get $t=\frac{\sqrt{4gh+{v_0}^2}+v_0}{2g}$

daveed misread the problem, they collide when they are traveling in the same direction

For one, why are you setting two $x$'s in the two equations equal to each other?  Doesn't that imply that the displacement(vertically) of the first ball is equivalent to that of the second ball?

in response to the post above:
I don't believe I misread the question.  They might be moving in the same direction, but their relative velocity is still $3v_0$, because they're in the same accelerated frame of reference.  Gravity accelerates both objects at $g$- so the relative velocity stays the same: they are always moving towards each other at that rate.

No, they are not equal to each other; first displacement is taken to be x, and the second is H-x
This is because when the collision occurs, the sum of the distance traveled by the two balls is H.

We'll do this with your method.

Draw out your little picture/graph/chart/whatever you call it.  You have two objects going toward each other, one going down at $v$, one going up at $2v$.

Their separation is $H$

Now, going by your method, you set one of the points as your origin- your "0-distance", if you will.  In your case, you did that with Galileo's ball, the one going at $v$

It's accelerating in the same direction as it's velocity: therefor, it's displacement is, as you say, $x=1/2 gt^2+v_0t$

Now, let's look at the second ball. [b]It[/b]'s displacement in space(from the starting point) is $x_1=2v_0t - 1/2gt^2$ (this differs from your's because I am defining the direction of positive-displacement as towards galileo's ball: this comes into factor when we subtract $H-x_1$ later on.

However, the tricky part here is that $x$ and $x_1$ cannot be the same thing.  They are defined as the displacements of each respective particle after a given period of time, which are most definitely not equal.

Now, because we anchored ourselves at Galileo, we then know that the coordinate of the other ball is then $H-x_1=H+1/2 gt^2-2v_0t$

Then, you want to find out when the coordinates of those are equal.
So you set $x=H-x_1$
and have
$H+1/2 gt^2-2v_0t = 1/2 gt^2+v_0t$
And then the $1/2gt^2$'s cancel out,
and you have
$H=3v_0t$
and, so
$t=\frac{H}{3v_0}$

i still think H/(3v) would give the time IF they collide while one is dropping and the other is going up..

there must be more to this problem.. it says they collide when both are falling

===================================

edit:
i did the problem and got H/(3v_0) as well
amazing

They can still collide when both are falling.  It just depends what the values of $H$ and $v$ are -- I believe you can play around to get the right relation to get the specifications of the problem.  However, the first point I put forward should be correct- the relative velocity stays the same.

yes you are correct.

I know the balls can collide going in the same direction -- just that t= H/(2v+v) makes me think that gives the collsion time such that v is large enough, so the collision occurs before the 2v ball starts falling.

I wrote equation in a way that the bottom ball reaches max height, and falls again --- reached the same answer ---  H/(3v)

i guess as long as H/(3v) > 2v/g then its fine

Why dont we do the 2nd part of the problem first, before working yours

2. At what height above the ground does the collision occur? Express your anwer in terms of H only.

Well, that's not hard.

Look at the bottom ball- it's height above the ground is modelled by $x_1= 2v_0t-1/2gt^2$
And... since we just figured out what $t$ was... plug and chug

Edit: Whoops I missed the part about in terms of H only.

Well, then I suppose you use the first part of the problem after all.  I guess, energy conservation and the above is what we need.

I agree with daveed; they are going the same direction when they collide, but in the accelerated frame, there is no acceleration.  So the ball tossed upward keeps on going upward at constant speed and the ball tossed downward also keeps going at the same speed.  This is mathematically represented by the cancelling of the (1/2)gt^2 terms.
For the 1999 problem,

the cm of the door experiences an inertial force Ma from the acceleration of the car.  This torque on the hinge should be set to the moment of inertia of the door times its angular acceleration.  After using the small angle approximation (b for sin(b)), we get a differential equation and so we can find the quarter period it takes for the door to swing shut.  This can be used to find the distance with (1/2)(a)(t^2).  I got (pi^2)L/12.[/u][/list][/quote]

You use the relationship between velocities when the collision occurs. I got H/4

(can we do more than 1 a day? this is way to exciting to stop)

[quote="qweretyq"]I suppose we can do more than one if it is multiple choice,

Here's one from 2005 exam (without the choices):

Two inertial observers, A and B, are traveling with respect to one another with velocity $v=0.6c$. Observer A has a sample of radioactive phosphorus $^{32}P$ and measures its half life to be $16$ days. Observer B will measure the sample life to be __?

As soon as someone answers, post the next one.[/quote]

um, 20 days?

daveed posted a question in the 2nd post which I don't think has been answered yet

[quote="MysticTerminator"]Let the positions of the two balls be $ x_1, x_2$, respectively. Then if $ L$ is the normal length of the spring, we have from Hooke's Law that $ \frac {d^2}{dt^2}x_1 \equal{} \frac {k}{m_1} (x_2 \minus{} x_1 \minus{} L)$, $ \frac {d^2}{dt^2} x_2 \equal{} \frac {k}{m_2} (L \plus{} x_1 \minus{} x_2)$, so we have a system of linear differential equations. I am now going to try to remember some of that stuff...so let $ y_1, y_2 \equal{} \frac {d}{dt}x_1, \frac {d}{dt}x_2$. Let $ X$ represent the matrix $ \begin{pmatrix} x_1 \\
y_1 \\
x_2 \\
y_2 \end{pmatrix}$ so that $ X' \equal{}$ OK THIS IS STUPID NEW METHOD

Place the center of mass at the origin, and let the normal (equilibrium) position of $ m_2$ be $ x$, so that the equilibrium position of $ m_1$ should be $ \minus{} \frac {xm_2}{m_1}$. In fact, this last equation should always be true as the center of mass remains at rest (at the origin). Now, we have $ \frac {d^2}{dt^2} x \equal{} \frac {k}{m_2} (L \plus{} x_1 \minus{} x_2) \equal{} \frac {k}{m_2} (L \minus{} \frac {xm_2}{m_1} \minus{} x) \equal{} k (\frac {L}{m_2} \minus{} \frac {x}{m_1} \minus{} \frac {x}{m_2})$. Using my psychic powers, it looks like the frequency will be $ \sqrt {k(\frac {1}{m_1} \plus{} \frac {1}{m_2})}$ so the period will be $ \frac {2 \pi}{\sqrt {k (\frac {1}{m_1} \plus{} \frac {1}{m_2})}}$.[/quote]

Why do you know $ \frac {d^2}{dt^2} x \equal{} \frac {k}{m_2} (L \plus{} x_1 \minus{} x_2)$ is true? i.e. why is $ \Delta x \equal{} L \plus{} x_1 \minus{} x_2$? After that, how did you get rid fo the L from the equation?

You could alternatively do the above problem in like three seconds by transforming it into a reduced mass problem, so $ T\equal{}2\pi \sqrt{\dfrac{k}{\mu}}$ which gives you the same thing

@123456789: That comes directly from $ m_2 \ddot{x} \equal{} kx$. Clearly the displacement from arnav's origin is $ L\plus{}|x_1\minus{}x_2|$

[img]http://img413.imageshack.us/img413/4724/cuberesistor.gif[/img]

too lazy to type upXD

Dang, I was thinking about something like A) yesterday night.  But anyway...

Due to the great amount of symmetry in this problem, we basically see that the points [i]b[/i], [i]c[/i] and [i]e[/i] have the same potential, and the same can be said about [i]d[/i], [i]g[/i], and [i]f[/i].  Then we can group each of these equipotential trios as one point.  Then, this cube resistor is equivalent to the following resistors in series:

* 3 "3Ohm" resistors in parallel
* 3 "8Ohm" resistors and 3 "12Ohm" resistors in parallel
* 3 "1Ohm" resistors in parallel

Doing the calculations, these sub-groups have equivalent resistances of

* 1 Ohm
* 1.6 Ohm
* 1/3 Ohm

And since these are all in series, the total equivalent resistance is $ 1\plus{}1.6\plus{}1/3\equal{}\boxed{44/15}$ Ohms

And maybe I'll do the second part later.  Or someone else could do it  :)  ...

[hide="A1 (b)"]
After a long time the capacitors act as open-circuit i.e, no current flows through them. Under these conditions, removing the parts $ CD$, $ BF$ and $ EG$, we've equivalent circuit as
3 circuit branches of the form "$ 3 \Omega$ , $ 8 \Omega$ and $ 1 \Omega$ in series"  in parallel
So equivalent resistance now is $ \frac{3\plus{}8\plus{}1}{3}\equal{}4 \Omega$
Current $ I\equal{}3 \ A$
So potential difference between $ C$ and $ D$ is $ V_{cd}\equal{}V_{cg}\equal{} (8 \Omega) \left( \frac{3A}{3}\right)\equal{}8 V$
$ Q_{cd}\equal{}CV_{cd}\equal{}120 \mu C$
[/hide]

[img]http://stanford.edu/~yckato/aops/general/p13.png[/img]

This problem has quite a few variables but it's not too hard I think.

[hide="A2 (a)"]
[hide="(i)"]
Volume is constant. $ \implies  \ \ \frac{P}{T}\equal{} \text{constant}$
$ T_{cr}\equal{} \left(\frac{P_{cr}}{P_{0}} \right) T_{0}$
[/hide]

[hide="(ii)"]
Adiabatic process
$ PV^{\gamma}\equal{} \text{constant}$
$ L_{cr}\equal{} \left (\frac{P_{0}}{P_{cr}} \right)^{\frac{5}{3}} L_{0}$
[/hide]
[/hide]

[hide="A2 (b)"]
Atleast force due to pressure of gas on bullet should be equal to Max. Static friction
$ P_{cr}( \pi r_{c}^2)\equal{} \mu N$

$ N\equal{}P' \times (2 \pi r_{c} h)$

$ P'\equal{} \left(\frac{\Delta r}{r_{c}} \right) E$

Finally
$ P_{cr}\equal{} \frac{2 \mu Eh \Delta r}{r_{c}^2}$
[/hide]

[url=http://www.mathlinks.ro/viewtopic.php?p=1785996#1785996]Next question on Electrostatics :D[/url]

Has anyone any idea on how to do A3 on the 2003 Semifinal Exam? I've been trying it for over an hour and can't seem to get anywhere. It involves the magnetic field from an infinitely large conducting slab.

So...some hints, I suppose.

* Consider a wire with the current in the same direction as in the problem - what does the B-field look like?  Now stack a few of those in the x-direction, what does the B-field look like then?  y-direction?

* Make some assumptions based on symmetry and/or infiniteness - where will the B-field be zero (if at all)?  Which direction will it point in general?

* Use Ampere's Law with a nice rectangular loop.  Choose wisely  :)

Once you get part a), b) and c) shouldn't be too hard I think.

For part (a), I got [hide]$ \|B\|\equal{} \frac{\mu_0JL}{2}$ for $ |z| \geq \frac{L}{2}$
$ \|B\| \equal{} \frac{\mu_0J|z|}{2}$ for $ |z| < \frac{L}{2}$ where $ z$ is the z-coordinate of the point.
$ B$ points to the left for $ z > 0$ and to the right for $ z<0$.
[/hide]

For part (b), I got [hide] $ F \equal{} 0$ and $ \tau \equal{} BIa^2sin\theta$.[/hide]

Can anyone confirm these? Thanks!

[hide="Part A"]I'm a little concerned about the fact that the B-field isn't continuous at the border, given the simplicity of the system.  Perhaps the second one shouldn't have a "$ \frac{1}{2}$" in it?...[/hide]

[hide="Part B"]Your final torque shouldn't be expressed in terms of $ B$, I think  :wink: ... and while $ sin \theta$ probably refers to the sine of the angle between the normal and the B-field, they already have a different $ \theta$ defined in the problem...[/hide]

Lyttleton - Bondi Model for the Expansion of the Universe
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained on
the basis of Newtonian if matter carried a net electric charge. Imagine a spherical volume of
astronomical size and radius R containing un-ionized atomic hydrogen gas of uniform density 
(= 1026 kg·m3) and assume that the proton charge ep = 1(1 + y)e, where e is the electron charge.
(a) Obtain the value of y for which the electrostatic repulsion becomes larger than the gravitational
attraction and the gas expands.
(b) Obtain an expression for the force of repulsion on an atom which is at a distance R from the
centre of the spherical volume. Hence show that the radial velocity is proportional to R. Let us
label the proportionality constant as H. Assume that the density is maintained constant by the
continuous creation of matter in space. Assume that the value of y is larger than the equilibrium
value calculated in part (a) above and hence ignore gravity.
(c) Calculate the numerical value of H. Take the value of y to be one order of magnitude larger than
the equilibrium value calculated in part (a) above.
(d) Given that at time t = 0, the volume of the Universe was V0, obtain an expression for the volume
expansion of the Universe.
(e) Why do you think the Lyttleton - Bondi model has been largely discarded by the scientific
community?

A problem from 2009 semi-finals... [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=332&t=586715]Problem Thread[/url]

[b]There is no answer from AAPT![/b]


quadratics
geometry
geometric transformation
reflection
conics
parabola

The equation x^2 - 6x +13 = 0 has roots 3 + 2i and 3 - 2i. Is it a coincidence that if you reflect the graph of y = x^2 - 6x +13 in the horizontal line through its lowest point , these roots can be read off from the x-axis (doing the reflection algebraically, they are the roots of x^2 - 6x + 5 =0, i.e. x= 1 and x = 5, interpreted as 3 - 2i and 3+ 2i).

No, it's not a (one-time) coincidence.

Ex.)
Shift the parabola one unit to the right:

y = (x - 1)^2  - 6(x - 1) + 13

y = x^2 - 8x + 20

This has a vertex of (4,4).

Its roots are x = 4 - 2i, 4 + 2i.

Drop the parabola down 8 to get:

y = x^2 - 8x + 12

This parabola has roots of x = 2, 6.  <--> Compare to the complex roots.


function
floor function
inequalities
algebra

Source: 360 PMC
Solve the equation
$\left\lfloor\frac{25x-2}{4}\right\rfloor=\frac{13x+4}{3}$.

I don't get your hint, lotrgreenapple. :(  :?:

[quote="lotrgreengrapes7926"]Can't you just solve the equation without the floor function and check numbers close to it?
[hide="Hint"] And you know it's 2 mod 3. [/hide][/quote]

Why is it 2 mod 3?

[quote] And you know it's 2 mod 3.[/quote]

What if x is not even an integer? :)
Which happens to be the case here, By the way!

Unless you consider the following ... BUT
Don't peek, unless you know the answer,   as this will give away an answer

[hide]
I guess $14 \equiv 2 \bmod 3$ and $13 \equiv 1 \bmod 3$ so you may
consider  $\frac{14} {13} \equiv (2/1) \equiv 2 \bmod 3$
and you get  $\frac{14} {13}$  as 2 mod 3 <smile>
and $x=\frac{14} {13}$ happens to be one of the answers  :rotfl:

BTW, the method to find the answer is fairy straight forward...
[/hide]

[quote="lotrgreengrapes7926"]Can't you just solve the equation without the floor function and check numbers close to it?
[hide="Hint"] And you know it's 2 mod 3. [/hide][/quote]
I don't think that works.....
BTW there are two answers.
[hide="hint"]
basically, when i see those floor functions, i set up an inequality:
$a\leq \lfloor x \rfloor < a+1$, for some integer $a$.
Try to apply this to the given equation.[/hide]

[hide="solution"]
by floor function identity, we have $p\le \frac{25x-2}{4}<p+1$ for some integer $p$, and we know that $p = \frac{13x+4}{3}$.  Thus we can solve for $x$ in $p$ and plug in the inequality, solving we get $5 < p < 8$.  Plug back in we have $x = \frac{17}{13}, \frac{14}{13}$.
[/hide]

Take y which is in the geatest integer function. then solve it you will get the trur answer

[quote="kimby_102"][quote="pkerichang"][hide="solution"]
by floor function identity, we have $p\le \frac{25x-2}{4}<p+1$ for some integer $p$, and we know that $p = \frac{13x+4}{3}$.  Thus we can solve for $x$ in $p$ and plug in the inequality, solving we get $5 < p < 8$.  Plug back in we have $x = \frac{17}{11}, \frac{14}{11}$.
[/hide][/quote]
You mean $\frac{14}{13}$ and $\frac{17}{13}$? :D[/quote]

You got me there....
Man, I ALWAYS mess up on the last step!!! :(


[quote="mr. math"]how would u do this problem?

what is the probability of drawing a heart or a 7 from a regular deck of 52 cards?[/quote]

Answer=P(Hearts)+P(7)-P(7 of heart)=$13/52+4/52-1/52=16/52=\boxed{4/13}$

This type of problem uses the principle of inclusion and exclusion (PIE).  (look this up on AoPSWiki)

First, find the number of cards that are hearts.  There are 13.  Then, there are 4 hearts.  However, one of those cards is the 7 of hearts, and it was counted twice, so you must subtract 1 so it is only counted once.

13+4-1=16

There are 52 total cards, and 16 are "good" cards, making the probability $\frac{16}{52}=\frac{4}{13}$.

This can also be expressed as $P(\text{ Heart }\cup\text{ 7 })=P(\text{ Heart })+P(\text{ 7 })-P(\text{ Heart }\cap\text{ 7 })$

[quote="vishalarul"]Why can't we include the 7 of hearts?[/quote]

Well, since you counted the number ways to get a 7 and the number of ways to get a heart. You over-counted 7 of hearts


function calculus integration probability geometry conics AMC

The AMC is the only of the American Mathematics Competitions which allows the use of a calculator, and so it should be used to its full advantage. This topic is for discussion of various things you can do with your calculator that will make medium or hard problems relatively trivial. Feel free to post any methods that you've come across as useful, or any problems that you think can be solved easily with a calculator.

I'll start out with a bunch of topics:
- Graph given functions or geometrical constructions when you can, especially in questions that ask for tangent lines or intercepts.
- In problems that ask you to count something, if you don't know how to count the topic in question, you could make a calculator program to check which numbers satisfy the criteria.
- The numerical integral of a function can be used to determine the probability that something holds in a given region, or the area under any curve that you can find a function for.

Care to add anything? Comments? Better explanations?

[quote="qqwref"]The AMC is the only of the American Mathematics Competitions which allows the use of a calculator, and so it should be used to its full advantage. [/quote]

Not quite getting that logic there...
As the booklet says, or said, the problems aren't designed with the use of a calculator in mind, and all the problems can be solved without the use of one, in ingenius methods.

Probably because my knowledge of how to use my TI-83+'s more advanced functions is not so deep, but I have almost never benefited from having a calculator during the AMC's.

Though it isn't neccessary, I think this topic should go on.  It could provide some fruitful results.

I personally love graphing to count the solutions.

Anyone have good ideas in relations with conics?

Gah, this is a terrible topic. Not only does using your calculator often provide solutions that may even take more time than searching for an elegant method, it indisputably encourages you to rely on this machine of brute-force and will thus prevent you from doing as well on future exams, such as the AIME.

[i]As the booklet says, or said, the problems aren't designed with the use of a calculator in mind, and all the problems can be solved without the use of one, in ingenius methods.[/i]

That doesn't mean you'll think of them during the test! Unless you're sure to get above 140 anyway - in which case you shouldn't be hanging out in AMC preparation topics - you have a good chance that there will be a problem or two (or more) that you can't get within the time for some reason. The purpose of this topic is to provide advice on how to effectively use a calculator on the AMC in order to get an extra 5-10 points or so. Although a calculator is not necessary, if you know how to use it it can be very useful.

[i]Having a TI-89 certainly helps[/i]

Yes, it does, but other calculators are useful too. I personally prefer the TI-83+ :o because it automatically evaluates everything numerically and because the programming engine is less complex (and therefore easier to do rapidly).

[i]I personally love graphing to count the solutions.[/i]

This is definately a good use!

[i]Anyone have good ideas in relations with conics?[/i]

What kind of ideas are you looking for?

[i]Not only does using your calculator often provide solutions that may even take more time than searching for an elegant method, it indisputably encourages you to rely on this machine of brute-force and will thus prevent you from doing as well on future exams, such as the AIME.[/i]

I'm not encouraging brute force as a way to solve problems. That isn't a worthwhile approach. I'm just saying that, for problems 19-25 especially, it is quite within the realm of reason that you can't (or will take at least 5-10 minutes to) solve a problem and that you can do it faster by typing some kind of graph, summation, or program into your calculator than by trying (possibly fruitlessly) to find an elegant solution.

The AMC and AIME are very different because of their time limits. In this test, you don't have a lot of time, and many of the questions are quite difficult near the end. Although you should be able to think of a nice solution for every problem, practically that isn't always the case, especially if there isn't much time left in the test. If you try a problem and can't find a good solution, I'm just saying that it could be useful to quickly try solving the problem on the calculator.

The calculator shouldn't be used as a first attempt, but it's a very good last resort.

Use programs to add an infinite or finite sums.

Such as:

$\sum_{n=0}^{\infty} \cot^{-1}(k^{2}+k+1)$

Or (If you don't get the pattern):
$\sum_{n=1}^{2006} (-1)^n$

On 2006AMC10B#19, I used my calculator to find the area of triangle ODE.
Even though all the trig functions that were involved used $\sqrt{6}$ and $\sqrt{2}$, The area of the triangle was a nice integer.

A forloop for a very large value, or use labels and stop it after a while.  It'll be approximate, but close enough.

Are there any helpful calculator programs for the AMC?

[quote="drummergirl249"]How would you write a program to find sums? - I use a TI 83+.[/quote]
Actually, if you just want to evaluate sums, you can use "sum(seq([i]expression, variable, lower bound, upper bound[/i]))" command. (See the manual for more info)

For instance, if you want to evaluate $\sum_{n=1}^{100} n^2$, you can type in "sum(seq(n^2,n,1,100))", and TI-83 gives you the answer :lol:

[quote="drummergirl249"]How would you write a program to find sums? - I use a TI 83+.[/quote]
for $\sum_{x=m}^n{y_1(x)}$ you could do this:

:Input "Y1=",Strg1  <- (that where you write "x" or "sin(x)" (w/o quotes of course))
:Strg1 -> Y1
:Prompt M,N
:0 -> S
:For(X,M,N)
:S+Y1 -> S
:End
:Disp S

Edit: Or that ^^ :roll:

[quote="qqwref"]I'm not encouraging brute force as a way to solve problems. That isn't a worthwhile approach. I'm just saying that, for problems 19-25 especially, it is quite within the realm of reason that you can't (or will take at least 5-10 minutes to) solve a problem and that you can do it faster by typing some kind of graph, summation, or program into your calculator than by trying (possibly fruitlessly) to find an elegant solution.[/quote]

Agreed that this is a possibility, but IMHO, a well-written problem should not allow this to happen.  This is especially true on the last few problems on the test, which should be separating out the top students from each other, not the students who know how to use their calculator to maximum advantage.

In general, it's my belief that:
1. Students tend to rely too much on calculators where they are allowed, and solve problems with a calculator that were intended to be solved by hand.
2. A better test of mathematical ability is one's ability to work without a calculator.
3. It's more worth your time to practice doing problems the "real" way than learning to do quick calculator solutions.
4. There should be more competitions in which calculators are not allowed.

I dunno, it really seems like most competitions seem to be no calculator, at least outside of the world of Mu Alpha Theta.

I generally think that almost all the questions in the last 5 are undoable using amazing calculator skills for the most part.  Geometry isn't something that you can really do with a calculator, and thats almost assured to be one of the last few, along with probability... though it would be impressive to write a program for that... but even then, you'd have to know alot of math involved with it, in which case you could probably do it by hand anyway faster.  However, it can save you on some things... you could probably store information on there if you wanted to really.

[quote="Xantos C. Guin"]Use programs to add an infinite or finite sums.

Such as:

$\sum_{n=0}^{\infty} \cot^{-1}(k^{2}+k+1)$
[/quote]

Grr... seems familiar, eh? (hem hem gctm state) :lol:

If you all want to see an example in place, look at the Log Probability question on the AMC. I'm too lazy to put up the link, but it's on this first page of topics (i think).

Another helpful calculator trick for the TI-89 is to
[u]
[i][b]solve equations with complex numbers and real numbers[/b][/i][/u]

Pwnage right there  :D

But remember, use the calculator sparingly. Take that extra 5 seconds to read the problem carefully and see if there are any previous tricks you can use before quickly going to the calculator and typing away rapidly... ten minutes later, you may still be typing away, when if you had just spent those 5 seconds, you could've gotten the problem done in 30 seconds.  :)

Even thought i don't use a calculator that much in AMC, i like making programs.  Here's one for the formula in my signature.
(Put the points $a_1,...,a_{n+1}$ in $\text{L}_1$ and ${f(a_1},...,f(a_{n+1})$ in $\text{L}_2$)

\begin{eqnarray*} &: & \text{Prompt X} \\ &: & 0 \rightarrow \text{S} \\ &: & \text{For}(\text{K},1,\text{dim}(\text{L}_1)) \\ &: & \text{L}_2(\tex{K}) \rightarrow P \\ &: & \text{For}(\text{I},1,\text{dim}(\text{L}_1)) \\ &: & \text{If I}\neq\text{K} \\ &: & \text{Then} \\ &: & \text{P*}(\text{X}-\text{L}_1(\text{I}))\text{/}(\text{L}_1(\text{K})-\text{L}_1(\text{I})) \rightarrow \text{P} \\ &: & \text{End} \\ &: & \text{End} \\ &: &\text{S}+\text{P}\rightarrow\text{S} \\ &: & \text{End} \\ &: & \text{Disp S} \end{eqnarray*}

you mean polynomial instead of function?

[quote="nat mc"]Even thought i don't use a calculator that much in AMC, i like making programs.  Here's one for the formula in my signature.
(Put the points $a_1,...,a_{n+1}$ in $\text{L}_1$ and ${f(a_1},...,f(a_{n+1})$ in $\text{L}_2$)

[/quote]

What is X?


geometry
parallelogram
geometry proposed

Two circles $ \omega_{1}$ and $ \omega_{2}$ intersect in points $ A$ and $ B$. Let $ PQ$ and $ RS$ be segments of common tangents to these circles (points $ P$ and $ R$ lie on  $ \omega_{1}$, points $ Q$ and $ S$ lie on $ \omega_{2}$). It appears that $ RB\parallel PQ$. Ray $ RB$ intersects $ \omega_{2}$  in a point $ W\ne B$. Find $ RB/BW$.

[i]S. Berlov [/i]

Let X the intersection of AB and PQ. We know that AB||PR and RB||PX, so, XBRP is a parallelogram,
and BR = PX = PQ/2 = RS/2.

We know that $RB.RW = RS^{2}= 4PX^{2}\Longrightarrow RW = 4PX \Longrightarrow BW = 3PX$.

With this, we find our razon: 1/3

Dear N. T. TUan and Mathlinkers,
is there a typos in the formulation of the problème?
Sincerely
Jean-Louis


[quote]Justin earned scores of 85, 92, and 95 on his science tests. What does he need
to earn on his next science test to have an average (arithmetic mean) of 93%?[/quote]

What is the formula to find the score he needs to earn on his next test?

If $ x$ is the score he needs, than:

$ \frac{85\plus{}92\plus{}95\plus{}x}{4}\equal{}93$. Then $ x\equal{}93\cdot 4\minus{}(85\plus{}92\plus{}95)\equal{}100$.

If you have the results of $ n$ tests $ a_1,a_2,...,a_n$, and if there are $ n\plus{}1$ tests, and $ k$ is the average of all tests, and if he has unknown score $ x$ on $ n\plus{}1$-th test, then:
$ \frac{a_1\plus{}a_2\plus{}...\plus{}a_n\plus{}x}{n\plus{}1}\equal{}k$ and $ x\equal{}(n\plus{}1)k\minus{}(a_1\plus{}a_2\plus{}...\plus{}a_n)$

The reasoning behind this is that the average for the following set of numbers:

$ \{a_1, a_2, a_3, a_4, \dots, a_n\}$

is

$ \frac{a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} \dots \plus{} a_n}{n}$

So this would mean the sum of $ a_1$ through $ a_n$ is the average times the number of terms. Thus, if you had 4 items, and you know 3 of them, and the average of the numbers, you could take the sum and subtract the 3 numbers from this, getting the number you are looking for.


modular arithmetic
search
calculus
integration
function
number theory unsolved
number theory

find all $a,b,c \in \mathbb{N}$ s.t for every natural $n$ we have:
$2^{n}a+b \mid c^{n}+1$

[quote="BaBaK Ghalebi"]for every [b][u]natural[/u] [/b]$n$ [/quote]
trust me,this problem is very harder than it looks like...  :wink:

If $2\nmid (a,b)$:

1)$a+b=2^{k}$ for some positive integer k

Let's suppose there exist an odd prime p such that $p|a+b$

Than $2^{p-1}a+b\equiv a+b\equiv 0 \pmod p$ but $c^{p-1}+1\equiv 1+1\not\equiv 0 \pmod p$, absurd.

2) $b=1$

b is odd, for a+b is even and $2\nmid(a,b)$. Let's suppose $b\neq 1$ Then there is a positive integer k such that $2^{k}\nmid b-1$

$2^{2^{k}}a+b|c^{2^{k}}+1$ but all odd divisors of  $c^{2^{k}}+1$ are $\equiv 1 \pmod{2^{k+1}}$ while $2^{2^{k}}a+b$ is not. Absurd

3) $a=1$

Let $p>c+1$ be a prime. Then all the divisors of $c^{p}+1$ either are divisors of $c+1$ or are $\equiv 1 \pmod p$

$2^{p}a+1|c^{p}+1$ so $(2^{p}a+1,c+1)\equiv2^{p}a+1\equiv 2a+1 \pmod p$

But $p>c+1$ so $(2^{p}a+1,c+1)=2a+1$, hence $2a+1|2^{p}a+1$ that is to say $2a+1|2^{p-1}-1$ for every prime p>c+1, so $2a+1=3$

[b]Now we have to solve $2^{n}+1|c^{n}+1$ for every natural n[/b]

If $ord_{p}(2)=2n$, then $ord_{p}(c)|2n$

So the equation $2^{x}\equiv c \pmod p$ is solvable for every prime p such that $2|ord_{p}(2)$

[b]This implies $c=2^{k}$ for some integer k[/b]

[size=75]I shall prove this claim in another post, for this one is already too long and messy.[/size]

if k is even, then $3\nmid 2^{k}+1$, if k is odd then we can easily see that $2^{n}|c^{n}+1$ for every n.

NB: now we see that $2\nmid (a,b)$ is a necessary condition because c is even.

In conclusion all the solutions are:

$a=1$, $b=1$, $c=2^{2m-1}$ where m is a positive integer.

[quote="piever"]So the equation $2^{x}\equiv c \pmod p$ is solvable for every prime p such that $2|ord_{p}(2)$

[b]This implies $c=2^{k}$ for some integer k[/b]

[size=75]I shall prove this claim in another post, for this one is already too long and messy.[/size][/quote]

Also, if the equation $2^{x}\equiv c \pmod p$ is solvable for every prime p such that $2|ord_{p}(2)$, it is true for composite numbers too.

Let's write again the claim:

Hp: if for a positive integer n $2^{x}\equiv 1 \pmod n$ is solvable, than both $2^{x}\equiv c \pmod n$ and $2^{x}\equiv-c \pmod n$ are solvable

Th: $c=2^{k}$

Let's suppose, by absurd, $c\neq 2^{k}$.

Obviously, one of the two last equation have a nonnegative integer solution x smaller than the smallest positive integer solution to the first equation.

Let's say $n=2^{d}+1$ and $n>c$

$x=d$ is a solution to the first equation, so there is a solution with $x=j< d$ to $2^{x}+c\equiv 0\pmod n$

But then, since $2^{j}-c=0$ is impossible, we have: $c=2^{d}+1-2^{j}$ so $v_{2}(c-1)=j$

Now let's say $n=2^{e}+1$ and $e>d$

$x=e$ is a solution to the first equation, so there is a solution with $x=l< e$ to $2^{x}+c\equiv 0\pmod n$

But then, since $2^{l}-c=0$ is impossible, we have: $c=2^{e}+1-2^{l}$ so $v_{2}(c-1)=l>j$, ABSURD.

Such questions have been abounding on the forum in the past two years (from AMM, Polish Olympiad, TST's, a recent Shortlist). To name the most prominent threads:

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=487801093&t=4556 ("A masterpiece" by harazi);

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=487801093&t=136198 ("(f(z)^n - 1)/(g(z)^n - 1) is an integral function for each n" by maxal);

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1531670334&t=27052 ("If the quotient of two power sums is integer-valued, then..." by vess), of which THIS problem (in this thread) is a very special case.

Thanks for the links Vess.

It looks like "the masterpiece" is equivalent to this problem, I have posted a solution that is nearly equal to this one.

The third link is a very interesting fact, I'll try to study your proof.


function
inequalities unsolved
inequalities

Why use $\tanh$ if we can simply say that $a_{1},a_{2},...,a_{n}$ are positive numbers with $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=1$. Prove that $\sum_{k=1}^{n}\sqrt{\frac{1-a_{k}^{2}}{a_{k}^{2}}}\geq n \sum_{k=1}^{n}a_{k}$.

[i]Edit[/i] I edited the ineq, but it seems it is wrong anyway.

[quote="Myth"]Why use $\tanh$ if we can simply say that $a_{1},a_{2},...,a_{n}$ are positive numbers with $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=1$. Prove that $\sum_{k=1}^{n}\sqrt{\frac{a_{k}^{2}}{1-a_{k}^{2}}}\geq n \sum_{k=1}^{n}a_{k}$.[/quote]
just chebhev(i forgot how to spell)
$\sum_{k=1}^{n}\sqrt{\frac{a_{k}^{2}}{1-a_{k}^{2}}}\geq \frac1 n \sum a_{k}\sum \frac{1}{\sqrt{1-a_{k}^{2}}}\ge n \sum_{k=1}^{n}a_{k}$

[quote="Minh Can"]the reformulated ineq is [b]not [/b]true. Take a(1) = a(2) =...= a(n) = (1/n)^(1/2) and n = 3 then the lhs < rhs.[/quote]
Check it again, I misplaced denominator and numerator, but anyway it is wrong. :maybe:

Sorry,I didn't check the coeffient.but any way, we can get a best coeffient $t$ such that $\sum_{k=1}^{n}\sqrt{\frac{a_{k}^{2}}{1-a_{k}^{2}}}\geq \frac1 n \sum a_{k}\sum \frac{1}{\sqrt{1-a_{k}^{2}}}\ge t \sum_{k=1}^{n}a_{k}$
just use power mean.I am so lazy to compute the coeffient.


algorithm
superior algebra
superior algebra unsolved

Let $ F$ a free group, and $ a,b \in F$ such that $ a^n\equal{}b^n$ for a $ n \in \mathbb{Z}$.Prove that $ a\equal{}b$.

I consider all the elements written in the shortest form in terms of the free generators.

First suppose that $ a$ can be written $ a\equal{}xa'x^{\minus{}1}$ for some $ x$ in the free generators set.
Then the first letter of $ b$ must be $ x$ and the last letter of $ b$ must be $ x^{\minus{}1}$. Hence $ b\equal{}xb'x^{\minus{}1}$.
Now we get that $ a'^{n}\equal{}b'^{n}$.
We continue this process until we can.

Now we can suppose that $ l(a^n)\equal{}nl(a), l(b^n)\equal{}nl(b)$ (else apply the algorithm above and eventually get this). From this results that $ l(a)\equal{}l(b)$ and now since the first $ l(a)$ letters of $ b$ must be identically with the letters of $ a$ it follows that $ a\equal{}b$.


Evaluate $\tan 20 \tan 40 \tan 60 \tan 80$ [b]without [/b] a calculator.

[hide="Hint"]Write in radians... do the 9's mean something?[/hide]
[hide="Bigger Hint"]$\tan 9\theta = 0$... what are the roots of this?[/hide]
[hide="Biggest Hint"]Let $x = \tan\theta$... write the 'bigger hint' in terms of $x...$[/hide]
[hide="Turnips... aka giveaway hint"]What does $\tan n\theta$ equal in terms of $\tan \theta$?[/hide]

[quote="puuhikki"]By Taylor's formula, $\tan 20 \tan 40 \tan 60 \tan 80\approx-7.2021$[/quote]

Hmmm, my calculator tells me the answer is actually 3.

[quote="Xaero"]Hmmm, my calculator tells me the answer is actually 3.[/quote]
Your calculator is in degree mode. The question was to evaluate
$\tan 20 \tan 40 \tan 60 \tan 80$, not $\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ$. That is easy to prove to be 3.

[quote="Benjamin Hu"][quote="Benjamin Hu"]I have a somewhat related problem.
There are an infinite number of solutions (a,b) to the equation a+b=ab. Write a and b in terms of sin(theta) and cos(theta).

Biggest hint: use the Pythagorean theorem.[/img][/quote]
For harder challenge, try to write a and b in terms of e^x and e^-x.

*additional challenge*
There are also an infinite number of solutions (m,n) to the equation m-n=mn. Write m and n in terms of sinh(theta) and cosh(theta).[/quote]
Revise the above "harder challenge" to say "write a and b in terms of e^ix and e^-ix". Also, challenge for the "additional challenge": Write m and n in terms of e^y and e^-y.

[quote="Benjamin Hu"][quote="Benjamin Hu"]
Revise the above "harder challenge" to say "write a and b in terms of e^ix and e^-ix". Also, challenge for the "additional challenge": Write m and n in terms of e^y and e^-y.[/quote]
Hint: use the hyperbolic trigonometric functions to hellp you here. I also reccomend reading the book [u]e: the Story of a Number[/u].[/quote]

And that is by...who?

Eli Maor

Here's the link at Amazon:

http://www.amazon.com/e-Story-Number-Eli-Maor/dp/0691058547/ref=pd_bbs_sr_1/102-5133942-6659347?ie=UTF8&s=books&qid=1176575825&sr=1-1

Good book.
Fulton

Wow. I spelled "recommend" wrong. Stupid me.

Even more appalling, I spelled "help" wrong, too.


I'm having trouble finding an approach to start most problems (including this one), so if you could share why you attacked it a certain way that would be great!

$ f(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$

Where $ a, b, c,$ and $ d$ are real numbers.

Suppose the graph of $ y \equal{} f(x)$ intersects the graph of $ y \equal{} 2x \minus{}1$ at $ x \equal{} 1, 2,$ and $ 3$

What's the value of $ f(0) \plus{} f(4)$?

I started this problem by plugging in $ f(1) \equal{} 1$ , $ f(2) \equal{} 3$ , and $ f(3) \equal{} 5$ and then I tried to rewrite the variables in terms of each other, but they always completely canceled each other, so I don't think that's the right approach.

What happens if you change the coordinates to $ x' \equal{} x, y' \equal{} y \minus{} 2x \plus{} 1$?

[hide="solution"]
$ f(x)\minus{}(2x\minus{}1)\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}a)$

So $ [f(0)\minus{}(2 \times 0\minus{}1)]\plus{}[f(4)\minus{}(2 \times 4\minus{}1)]\equal{}6a\plus{}6(4\minus{}a)$
$ f(0)\plus{}f(4)\minus{}6\equal{}24 \Leftrightarrow f(0)\plus{}f(4)\equal{}30$
[/hide]

[quote="mathemagician1729"]Sorry, could you explain how you did that stephen? I used a much more exhaustive method that involved eliminating the variables one by one.[/quote]

the graph of  intersects the graph of $ y\equal{}f(x)$ intersects the graph of $ y\equal{}2x\minus{}1$ at $ x\equal{}1,2,3$

$ \Leftrightarrow$ $ 1,2,3$ are roots of equation $ f(x)\equal{}2x\minus{}1$.

As $ f(x)$ is of degree $ 4$ with leading coefficient $ 1$,

we know $ f(x)\minus{}(2x\minus{}1)\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}a)$ , where $ a$ is a constant.

Could you find $ f(0)$ and $ f(4)$ separately and then add them - or do you need a single equation?  Also how would you do this kind of problem if the leading coefficient wasn't one?


floor function
function
combinatorics proposed
combinatorics

$ n$ points (no three being collinear) are given in a plane. Some points are connected and they form $ k$ segments. If no three of these segments form triangle ( equiv. there are no three points, such that each two of them are connected) prove that $ k \leq \left \lfloor \frac {n^{2}}{4}\right\rfloor$

Let $ a\minus{}b$ mean $ a$ is connected to $ b$. Let the "points" be associated with the numbers from 1 to n. Call a permutation $ f$ of $ [n]$ a "relabeling" if $ a\minus{}b\iff f(a)\minus{}f(b)$
For given n consider a "connecting" of $ [n]$ with maximal number k connections, without a triangle.
Thus, there exists a relabeling $ \pi_1$ of $ [n]$ so that $ \pi_1(1)\minus{}\pi_1(2),\pi_1(2)\minus{}\pi_1(3),...$
Clearly if this weren't the case, $ k$ wouldn't be maximal (we would add in extra connections).
Then there exists a relabeling $ \pi_2$ so that $ \pi_2(p_1(i))\minus{}pi_2(p_1(i\plus{}3))$ whenever $ i,i\plus{}3$ are in $ [n]$.
We continue in this way until we reach a function $ p_1\circ p_2\circ \cdots p_k$ that relabels our connections into:
1-2,2-3,3-4,...
1-4,2-5,3-6,...
1-6,2-7,3-8,...
...
We verify that k has at most:
k-1 + (k-3) + ... + 3 + 1 = (k/2)^2, if k is even
k-1 + ... + 4 + 2 = (k^2-1)/4, if k is odd
elements, solving the problem

The following estimate is not rigorous but it help us solve this one in almost all cases.  :|
$ {k \choose 2} \leq 2 \vdots {n \choose 3}$

[quote="pbornsztein"]This is Turan's theorem...

Pierre.[/quote]
well,actually this is "Mantel's theorem" a special case of "Turan's theorem",Turan's theorem says that:

[b]if a simple graph on $ n$ vertices has more than $ M(n,p)$ edges,then it contains a $ K_p$ as a subgraph,where:
$ M(n,p) \equal{} \frac {p \minus{} 2}{2(p \minus{} 1)}n^2 \minus{} \frac {r(p \minus{} 1 \minus{} r)}{2(p \minus{} 1)}$
in which $ n \equal{} t(p \minus{} 1) \plus{} r$ , $ 1\leq r\leq p \minus{} 1$[/b]

also another proof for Mantel's theorem is the following one:
let $ G$ have $ n$ vertices,numbered from $ 1$ to $ n$,and no triangles.we give vertex $ i$ a weight $ z_i\geq 0$ such that $ \sum z_i\equal{}1$ and we wish to maximize $ S\equal{}\sum z_iz_j$,where the sum is taken over all edges $ \{i,j\}$.suppose that vertex $ k$ and vertex $ l$ are not joined.let the neighbors of $ k$ have total weight $ x$,and those of $ l$ total weight $ y$,where $ x\geq y$.since $ (z_k\plus{}\epsilon)x\plus{}(z_l\minus{}\epsilon)\geq z_kx\plus{}z_ly$,we do not decrease the value of $ S$ if we shift some of the weight of vertex $ l$ to the vertex $ k$.it follows that $ S$ is maximal if all of the weight is concentrated on some complete subgraph of $ G$,i.e. on one edge.therefore $ S\leq\frac 14$.on the other hand,taking all $ z_i$ equal to $ \frac 1n$ would yield a value of $ \frac{|E|}{n^2}$ for $ S$.therefore $ |E|\leq\frac 14 n^2$.


AwesomeMath
summer program
Mafia
geometry
search
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number theory

OK. I know there's like billion feeds about the life at AMSP and stuff like that, but they're getting old, so.. Here's a new one?

Some questions for all of you guys who went to AMSP before, because this is going to be my first year @ AMSP.

1. How much freedom do we get at camp? (Like, I have a math circle @ UTD with the director of this camp actually, Dr. Andreescu... and I know the campus itself pretty well, but after classes and stuff, can you just walk out of where you live and just go around the campus whenever you feel like it as long as it's not class time and not after curfews?)

2. I might be underestimating my mathematical abilities? Can you tell me like how hard the intermediate and maybe a little about the entry-level.. (coz I suck in geo and might be taking just geo for entry-level.. hm..??) I mean, most of people are in intermediate I guess.. So just, generally how hard is it? Because I really have no idea...

3. Do you get to do a lot of stuff other than math at camp? (Things like little sports, recreational activities, and yea.. Oh, and are there practice rooms in the music building or anything  :wallbash: becoz I HAVE to practice clarinet!!! Just because I have to.. OR ELSE I am going to be last chair for the first semester when I get back at school, and then maybe have a chance at 2nd to last chair.. ha..)

4. I just want to know what you guys' overall experiences were, what you did, which profs are good, stuff like that. And oh, yes, btw, I'm going to UTD, so. yea.

5. P.S. And.. the food. Can you eat out order,, stuff like that.. And how is the cafeteria dining hall .. food?

1. i think i answered this question in Awesomemath Camp and Life and Process.
btw, the answer to ur question is yes.

2. best advice: challenge yourself. intermediate is pretty good. you might be a bit overwhelmed but you'll catch on. just keep trying.
lol. personally i thought comp. geo. was prettyyyy hard. but i was a slacker. ://

3. yes . POOLLLL [swimming & the ball kind]; MAFIAAAAA; six flags; hanging out; card games; practice music [if you want]

4. my experience cant be described really, well it could but id just go on & on bout how great it was. and i think ive done that quite a few times already so just one word: amazing.

5. food was not very good. you'll live tho. there is a 'pub' restaurant. and chick-a-fil.

Heh. This will be interesting.

I wonder if I'm allowed to bring my drum set? xD

errrrrr..... ud probably have to pay extra for plane carry on stuff.
but ummm.... lol ur roomies may not like you so mucfh if ur drumming into the night :P

Maybe if I can convince my mom to get me a small, compact electric drum set, I could fit it in the dorm. AND I could play with headphones. I live like a half-hour drive away, so airplane issues are non-applicable.

Plus I'm still questioning what people do there...

I mean, duh, math! What am I thinking.. And besides that, how much free time do we get. Is there like "homework" (haha...) and stuff?

I would really like to play a lot of tennis, because I am a competitive player and not playing in any tournaments in July hurts my record bad enough, and never mind not practicing..

Oh Also... Should I bring my clarinet.. I mean (of course STEALING or anything is not an issue? Because it costs about 2500 and if I lost it/ got it stolen.. MY PARENTS WILL KILL ME!!!) ... I just don't know if I have time for math (of course), and tennis and practicing clarinet in addition to that... hm.

Any clues?

Wow, a clarinet costs more than a drum set? Whoa.

Anyway, it should be okay to bring it. Just bring a lockable case to put the clarinet in, and then some way to lock the case to something in your room (like a bed). This way, it's impossible to steal the clarinet, as to steal it, you would also have to steal the bed, making you look rather conspicuous.

I THINK there are tennis courts there...the AwesomeMath website is not very clear about things like this.

there is ALOT of free time. basically ~2 hrs for lunch a little less. then after your 2nd class [past 4 or 5? idr] ur free to do whatever. if my memory serves me right.
erm. there's norm. no "homework" depends on teh class u take. plus its optional. mod had hwk. i think. i wasnt in it though.

yes. there are tennis courts. right by the basket ball courts.
but its hot. -____-;;

yeah. u can. no stealing. however. some guys' violin got crushed last year cuz they were wrestling and fell on it. so keep it somewhere safe.

Although Santa Cruz may be different than TX.

oh. oops. -__-;;
so idk bout the tennis courts but i think everything else is the same. in terms of schedule nd stuff.

Well, I also looked into COSMOS at UC Santa Cruz, and they allow access to tennis courts (along with a lot of other things). So there's definitely tennis courts at UCSC, it's just a question of whether they'll let you use them.

At COSMOS, does "among other things" include a tap-dancing floor?

I think that's what "dance floor" would be.

at UTD do they allow us to leave campus? I heard we leave campus to go to Six Flags but are we allowed to go to like the mall to pick up stuff or something

You won't be leaving campus unless some AMSP-sponsored activity is going on. I don't know of any malls around Richardson anyway, and you'd need a car if you wanted to get anywhere. Last year, two kids left campus and walked to Chili's. They got busted. Not sure what happened.

If you need anything [like shampoo, pillow... hair dye...] you can fill out a sheet and someone will go buy it for you [at walmart] [you have to pay tho duh.]

If you have ever gone to CTY, then here's a comparison:

You have much more freedom at AMSP than CTY. (At CTY we just had a quad, two dorms, a few classrooms, and a cafeteria. At AMSP we had a whole CAMPUS.)
You do NOT have to attend any of the non-educational activities (I was overstimulated at CTY).
You get to MAKE activities at AMSP (for example, ultimate, CTF, mafia).
Curfew gives you five whole ours out of the last class and then one hour in your room, then lights out at AMSP. At CTY it was one hour for each.
No CTY campus will have Naoki Sato, Titu Andreescu, and Zuming Feng teach on it the same year. But it happened at Texas AMSP last year.

AMSP>CTY.

Lol. Well put. From my experience, CTY is pretty horrible in a lot of ways. I am looking forward to AMP.

[quote="sooozyy"]okay so if you made the decision to attend amsp this summer, GOOD JOB!

this camp will be very very very fun :]   (well i hope it'll be fun for you!)

you don't really get homework but it's all up to you pretty much. i guess if you want to cout your practice problems as your "HOMEWORK" then i guess that can be your homework.

you will have to work on team contest problems though, so keep that in mind.

as for the instrument issue, the security isn't the issue. i mean... i had my violin in my room the whole time [i wont say the cost....but its less than my car.. =P and i have a lexus so i guess you can kinda estimate the cost i suppose] and it wasnt an issue. though i only practiced like twice cuz i got really caught up with hanging out with other people.

just dont do too much math. have some social life :][/quote]

it'll be one of the best three weeks ever ;]

well actually i think it depends on the class you take. counting strats & comp geo didnt have any hwk. but i think mod arth. did.

you dont need to work on team problems. well it depends on how "intense" your team is.

make sure your instrument is in a safe place.... esp if ppl r roughhousing. having a bunch of ppl falling on ur violin then having all the strings break is not cool [happened last year i think]

have fun. im sure you all will XD

haha so essentially a certain two people ... have been dominating the responses to the questions here. I'd just like to comment that there is a fair amount of variation as to what people like to do. For example, when I was a student at AMSP, I spent say 1/2 of my free time doing math problems. The rest I spent doing various activities and hanging out with people. You can pretty much choose how to spend your time whatever way you like. For some people, socializing is best, for others it is more enjoyable to do math. You'll find groups of people in all ranges.

As for homework, I'm not sure what the policy is going to be. I think that I'll assign homework but not force people to turn it in. Or possibly have problems for people to present at the board the next day and if nobody does it, then I'll just do it. But I'm not sure how much freedom I'll get to decide these things.

1. Freedom: you pretty much can do whatever you want on campus during your free time. We don't like the younger kids to go roaming around but that isn't really a problem for you. And yes there is a curfew and I generally did not have a problem with this last year but a lot of kids tend to stay up too late and then they are sleepy during classes.

3. Activities: there are plenty of things to do. The assistants hold activities. Last year, we did assorted sports, swimming, mafia, board games, etc. Other people think of things to do on their own. Some days there are special things. I wouldn't worry here, there is always stuff to do. I never practiced music, but some others did.

4. Well I think that I'll be holding stuff that will be helpful (for geometry) but they may be more in the olympiad range.

5. I had no complaints about the food. I mean like maybe some people have high standards or something but the food was generally good (and I stuffed myself each time haha). But I did order out once or twice a week for pizza of Chinese food.

I might as well post this in this thread.

I got accepted, and my mother wants to know if there is anyone from around the Philadelphia area that is also going, to possibly take the same flight that I will be on, so that I "won't get bored". And that I won't feel too alone/scared on the plane, apparently.

(send a PM if you can)

I think I'll respond too.

1) Yes, as Altheman said, it's really up to you. I mean, you can't go off campus or anything, but how you spend your time outside of the six hours of classes every day is basically your own choice; I think I was more like Michelle than Altheman, but I almost never saw some of my friends outside of class because they presumably were being studious and taking advantage of the supplementary math lectures instead. There is a curfew, but some of the mentors don't enforce it as well as others (*coughcough*)

2) Don't worry: you can switch classes in the first few days if you feel you're not in the right class. You might feel overwhelmed, though, if you take something in a subject you're not very familiar with (like geometry for you). Don't let that bother you; I think you should probably switch out if you have absolutely no clue what is going on, but otherwise... if you can follow the class and you think its curriculum will help you, even if you don't know ANYTHING they're teaching (hey, that's actually a good thing, right?), then you might as well stay.

3) Oh, yes! Definitely. As I said earlier, how you spend your time outside of the six hours of classes is basically your own choice, so you can practice music if you want to... basically any time, so long as you're not TOO loud and it's not past curfew or something.

4) I very much liked it, and it inspired me to go to another math camp this summer as well (although not AMSP, unfortunately!) To be honest, I'd say it was one of the most fun experiences of my life so far. As far as professors go, my Modular Arithmetic teacher, Dorin Andrica, taught very very well. On the other hand, my Computational Geometry class seemed to shuffle teachers around every few days or so; Mr. Sato, who directs the WOOT course that AoPS offers, taught for about a week, and he was an excellent teacher.

5) Food was... decent to good; some days it didn't look very appetizing, but usually it was pretty good. Once or twice we got some Chinese takeout as our dinner, which was probably the worst meal I had there. But if you absolutely detest the food for the day, you can go to the cafe nearby, which was consistently very good (at least if you go to UTD). During lunch you can also go to the cafeteria and buy pizza, Chik-Fil-A, sushi, etc. A lot of my friends ordered pizza out pretty often, also.

hahaha.
Altheman, way to forget #2 :P

completely agreed with leon's #4.

although many of you may be very engrossed in getting done as many math problems as possible in 3 weeks, chillax a little and have fun. and err.... vice versa as well? learn something ;P [which you will -hopefully- even if you dont do math outside of class]

haha well I didn't think that I could offer a useful opinion for that question...

By the way I always read "Playing Mafia" but what is Mafia?

[url=http://lmgtfy.com/?q=play+mafia]nobody uses google anymore :([/url]

Question: Do you think they'll mind if I practice piano for like, several hours a day? Also if there is some sort of piano available can I get a non-broken one, preferably a grand somehow?

I think my time will probably be used like, SC (if there is some way to get a computer with internet access), piano, and geometry :O

[quote="hasan4444"]Sorry "andersonw"  :D I don't poor search engines you know I have my own but they are hidden  :P
Are you coming to TX of CA?[/quote]

I'm going to TX.

Hmmm maybe I should start learning how to play SC...

[quote="serialk11r"]Question: Do you think they'll mind if I practice piano for like, several hours a day? Also if there is some sort of piano available can I get a non-broken one, preferably a grand somehow?[/quote]

i dont think anyone really cares what you do with ur free time [provided ur not breaking any laws]. go ahead.
um. there is a grand piano. but sometimes there are college students around. the upright ones in the practice rooms r almost always open. but r pretty crappy. but playable.

y is everone obsessed with starcraft??theres no where you can play it there, unless you want to bring ur PC...

What Starcraft is an awesome game, duh. When the "netbook" thingies begin to have processors with core 2 architecture I'll buy one so I can play Starcraft wherever I want :D (jk, but those things are pretty nifty)

Hey, were you supposed to email in your selections for classes or tell them when you got there?


Consider the points $ O(0, 0, 0)$, $ A$ $ (\sqrt{2}, 0, 0)$ and $ B$ $ (0,\sqrt{2}, 0)$ and a sphere of radius $ \sqrt{6}$, tangent to $ OA$, $ OB$ and $ AB$. Given that the centre of the sphere lies above the z-axis, what are its co-ordinates?

another to-be-done-under-2-mins problem.... but there is no restriction on math to be used....

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=255886[/url]


AMC
AIME
geometry
USA(J)MO
USAMO
inequalities
geometric transformation

I'll open this thread to provide advice and to invite other participants to share advice. Perhaps in a while a moderator can stickify this thread so that it becomes a FAQ file for participants on this forum.

How to Do Better on [url=http://www.unl.edu/amc/e-exams/e4-amc08/amc8.html]AMC 8[/url]

The AMC 8 25-question, multiple-choice exam tests only STRICTLY prealgebra mathematics, so it ought to be easy for eighth graders, but in many states of the United States NO ONE aces it, year after year. Take this test, which is always offered in November, the first school year you take a prealgebra course, or earlier if you have read a lot about mathematics. Your eighth-grade year is the last year you can take the AMC 8 test.

To prepare well for the AMC 8, do as many [url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#2]old AMC 8 exams[/url] as possible under exam conditions: the exact same time limits, same limits on tools and helps, and same time of day as the actual test. A reasonable schedule is to do one AMC 8 practice exam every two weeks for the three months preceding each year's November test date. You can do more old tests over a longer span of months if you have them, and you can go up to twice a week if you want to get lots of practice. But doing the whole tests in the actual exam time limit is MUCH better than doing a problem or two every once in a while. DEFINITELY refer to the official AMC solutions for the old tests--you can learn a lot of math from those.

Keep a record of which questions you miss on which tests--you could eventually build a "killer AMC" test out of all the problems you missed from the practice tests you have done, as a thorough review of your topics that most need improvement. As on all AMC tests, usually the earlier questions are easier than the later questions on each test, so learn how to allocate your time and make sure all the early questions are easy for YOU because you practiced enough.

You can get a thorough review of prealgebra math by looking at good discussions of that level of math. The [url=http://www.artofproblemsolving.com/Books/AoPS_B_V1.php]Art of Problem Solving Volume 1[/url] has some good review of math up to an even higher curriculum level, and has lots of good contest problems. Other book suggestions for REALLY learning prealgebra math can be found in an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=33140&highlight=#33140]old thread advising parents of young math learners[/url] about how to get started in math study.

How to Do Better on [url=http://www.unl.edu/amc/e-exams/e5-amc10/amc10.html]AMC 10[/url]

There is a limited supply of [url=http://www.unl.edu/amc/d-publication/pubform.html]old AMC 10 tests[/url], and you should obtain all of them. Take the AMC 10 seriously--it can get you into AIME, which can get you into USAMO. Take all the old tests and review the official AMC solutions. You should start taking this test, which is offered twice each February, the school you first take a math course after your beginning algebra course.

The AMC 10 tests curricular math up to geometry, with an emphasis on THOROUGH, practiced understanding. So make sure you have learned and relearned secondary geometry. Much of the content of The [url=http://www.artofproblemsolving.com/Books/AoPS_B_V2.php]Art of Problem Solving Volume 2[/url] and nearly all of the content of the preceding volume is useful for taking the AMC 10. Find "college algebra" books or books about geometry you haven't used before for additional review materials.

Take the test shrewdly: the penalty for guessing is HUGE on the AMC 10 and AMC 12, so know how many questions you must get correct to get a qualifying score for the AIME, and then don't answer too many questions. Answer only the questions you are really sure you know the answers for, and CHECK your work. It's important to stay organized in your scratch work to [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]stop making stupid mistakes[/url].

How to Do Better on [url=http://www.unl.edu/amc/e-exams/e6-amc12/amc12.html]AMC 12[/url]

The AMC 12 (formerly the AHSME) has been around for a long time, so there is a HUGE supply of [url=http://www.unl.edu/amc/d-publication/d6-books/exprobbooks.html#A]former tests[/url] with which to practice, including quite a few with the current 25-question, multiple-choice-with-penalty-for-guessing, calculators-permitted format. It is a test of strictly precalculus math, but it is not easy, and [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2004-12a/04perfect1012AB.html]perfect scorers[/url] are quite rare. Practice on former exams under exam conditions for a few months before the official test dates in February.

Review all of the usual precalculus math topics, especially geometry, thoroughly before taking the AMC 12. Volume 2 of The Art of Problem Solving is essential preparation, as is looking at hard problems from precalculus math books other than the one used in your school.

Again, the AMC 12 has a severe penalty for guessing, so answer enough questions, not too many, if you are a first-time test-taker.

How to Do Better on [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url]

To do well on the AIME, practice with the abundant supply of [url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#14]old AIME tests[/url], and take this 15-question, integer-answer test under test conditions--which means WITHOUT a calculator. Make sure to study the official AMC solutions thoroughly; they are much better than the [url=http://www.kalva.demon.co.uk/aime.html]Brand X[/url] partial solutions available for free on the Web. A first-time qualifier on the AIME who hasn't completed a precalculus course is probably best off practicing the first five questions from several old AIMEs, to get used to the question format, and making sure to qualify for AIME again next year.

There is no formal penalty for wrong answers on the AIME, but because it is not a multiple-choice test, it is easy to reach a wrong answer and nearly impossible to get a right answer by guessing. Work carefully, check your assumptions and reread the question before you start writing out steps, and definitely write out steps. The more you think about how to put a reality check on your answers, the better your answers should be. There is a [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimeanswerforms.html]sample AIME answer sheet[/url] online to help you practice the answer format. Note [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimestats.html]how low the average score on the AIME[/url] is, and remember that everyone who takes the AIME takes it by invitation, and plan accordingly for a TOUGH test.

How to Do Better on [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]USAMO[/url]

The USAMO test is the last step in qualifying for [url=http://www.unl.edu/amc/a-activities/a6-mosp/mosp.html]MOP[/url], and accordingly it is a very challenging test, intended to identify the best of the best among math contest participants in the United States. It is a two-day, six-question, essay-proof examination. You must be able to write mathematical proofs well to have a chance of scoring well on USAMO. [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamoscoring.html]Low scores are common[/url] even among the select group who are invited to take the USAMO, so prepare for a challenging test. As usual, lots of previous USAMO tests are [url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#16]available from AMC[/url] or [url=https://www.maa.org/ecomtpro/timssnet/books/PBS.cfm]from MAA[/url] and you should definitely become familiar with genuine previous problems and official solutions.

Find books that guide university students on how to write proofs, for example Scheinerman's Mathematics: A Discrete Introduction or the like to learn how to write proofs.

More editing to follow, and I hope others will join in with advice here.

Good luck.

For all pre-USAMO contests, harder problems are at the end.  Don't let the first few problems throw your timing off.

btw, USAMO is now 9 hours with 4.5 hours each day.

The AMC 8 and AMC 10 are offered at different times, so see about taking both if possible.

[quote="BHorseMath"]okay....I am very new to these tests and have never taken them before but I am ok at math.  I am looking at trying them next year.  Which test do u think I should look to taking the AMC 8 or 10?  I will be an 8th grader next year taking algebra.  I am acing pre-algebra right now and have a straight 100 in the class.[/quote]

this year i spent a lot of time preparing for AIME.  I brought the old AIME booklets and studied them.  My advice is that, focusing on the easy problems.  this year I only tried the first 9 problems and got 7 of them right.  For a Sophomore or below a 7 on the AIME can qualify you for the USAMO.  I know a kid who is also very good at math.  He tried 12 problems but only got 1 of them right because of stupid mistakes.

My school doesn't offer the AMC 8 tests, but I'm trying to get them to.  I'll be in 7th grade next year and that's when I'm going to take it.  I read AoPS volumes 1 and 2 and got Mathematical Circles (it's supposed to be a good math book).  Hopefully this will be all I need to do well.

Looks like you're all set.  I would actually suggest that you take the AMC 10 or 12 (in addition to AMC 8).  Without having read either AoPS book, I nearly qualified for the AIME via the AMC 10 in 6th grade (missed by 1 point :-( ).  That goes to show that your odds are looking pretty good next year.

I did a previous AMC 10 test a while ago.  I did twenty questions (had to stop for dinner and never got back to it) and got about 19 correct.  Is that good?  I doubt I could do well enough on it in 7th grade to qualify for the AIME (only the top 1% get to go - that's probably not me).  I may take it in eighth grade, I don't know yet.

[quote="nr1337"]I did a previous AMC 10 test a while ago.  I did twenty questions (had to stop for dinner and never got back to it) and got about 19 correct.  Is that good?[/quote]

Getting 19 correct on the AMC 10 and leaving the other 6 blank will readily qualify you for AIME. My son's strategy this year was 18 answered (all correctly) and 7 left blank, and he not only qualified for AIME with points to spare as a sixth-grader, but was the state winner for ALL grades on the AMC 10B. Most people who take the AMC 10/12A/B don't think carefully enough about how the tests are scored (the penalty for a wrong guess is SEVERE) and end up answering too many questions with wrong answers.

Good luck next year.

19 correct and 5 unanswered gives you a score of 126.5!  This more than qualifies; 120 is the fixed threshold and the top 1% threshold is usually slightly lower.  Given this, you should definitely be able to make AIME with a little more practice.  Besides, the experience is always good, no matter how unlikely it is that you do well.  If that weren't true, a lot of people wouldn't be taking the USAMO ;-).  Plus it's fun :-).

P.S. the formula for AMC score is 6*correct + 2.5*unanswered.

Nineteen was a typo that should be 18, but that's still a score of 120.5.  Next time I see my principal, I'll have to ask him these things.  Anyway, I have all summer long to prepare.  I don't think getting to take the AMC 10 should be a problem, because my high school uses the tests.

[quote="BHorseMath"]Which test do u think I should look to taking the AMC 8 or 10?[/quote]

Thanks for raising that question, which belongs in the FAQ. In general, one should take the AMC 8 as soon as one begins taking prealgebra, and take the AMC 10 as soon as one begins taking the earlier of geometry or algebra 2. Or one should take the AMC 10 as soon as one has worked many of the problems in AoPS volume 1, and AMC 12 as soon as one has begun working on AoPS volume 2, in all cases taking each test on the EARLIEST possible occasion, the better to get practice before losing eligibility.

[Aside to moderator:] Could you kindly split off from this thread the interesting discussion of the Cauchy-Shwarz inequality, which appears to be confined to messages that deal only with that topic?

[i]Moderator note: Done.[/i]

I say take all the AMCs as early as possible regardless of how much math you know.  In fact, if I had a child, I would encourage him/her to take the AMC 8, 10, and 12 as soon as he/she was beyond arithmetic.  It never hurts to try, anyway.

By the way... School math doesn't reflect how you will be doing on the AMC so don't be discouraged / encouraged by your grades... One of my classmates doesn't have really good grades in his math class, but he beats me on a lot of tests.

Yea well my school doesn't offer the tests anyhow they don't even know what they are.  Which is kinda sad...  I am going to try to get them to do the AMC tests though but if they arent' going to which I doubt that they will then I will see if I can take them at another school.  Would that be a good thing to do takeing it at another school?

[quote="BHorseMath"]Would that be a good thing to do takeing it at another school?[/quote]

It really doesn't make a difference, unless you're caught up on being a "school winner" or whatever.  Just make sure you get to take it :)

Yes, it would be good to take it at another school if you can't take it at your own.  Although, of course, it would be better if you could take it at your own school.

As I said my school has never heard of them and I don't even know of a school closer than 2 hours away that does them.

Try to encourage your school to administer it - if they absolutely refuse, do a little research; there must be [i]somewhere[/i] not too far away that you could take it.

my school is not a math school.  We do a math contest called tmsca thats really easy compared to others and thats it.

I've never took AMC but I think I'll do okay in AMC10 next yr
since I'll by that time would be pretty much covering all stuffs in geometry and trig that time and AoPS volume one...
I don't do well in AMC10 test now but I think i just need bit  practice
because my first year mathcounts rank in chapter was somewhere in 40th while this yr was 4th..

Hmm. I tried one of the old AIME sets and I answered 7 - got them all right. I need to study up on my trig. and algebra 2 though. The logs and cosines really through me off because I have trouble making sense out of them. I could get our school to offer them this year in 8th, but they will start to in 9th, when I'll be competing w/ my best friends - Who managed to take them this year in high school (they are in 9th grade) and got 103's. They passed for AIME. How? Isn't the cutoff 110?

[quote="white_horse_king88"] I need to study up on my trig. and algebra 2 though. [/quote]

My school doesn't offer algebra 2 so I took it with high school students