AMC USA(J)MO USAMO ratio Princeton college Yale

Recently, the new Newsweek Ranking of the "Top Public High Schools in America" came out. The rankings are solely based off of one formula:
(total number of AP/IB tests taken) / (number of graduating seniors).
The idea of this ranking is to give the average amount of AP tests that each student graduates with. What are your opinions on this ranking? Do you believe it is a fair way to rank schools?
Facts:
1. Magnet schools that require a standardized test for admittance for more than 50% of their student population are excluded.
2. The important number is the number of tests taken, not the score on the test.
3. It is possible to take an AP test without taking an AP course.
4. It is also possible to take an AP course without taking the corresponding AP test.
5. Some states pay for the AP fee (~$80) with state funds. Other states do not.

Even without the above facts, should standardized tests play such a large role in saying what school is "better" than another? Can you think of any other ways to rank school performance?

No way. This is a stupid way to rank schools. # of AP/IB taken is way off - a much more accurate way would be AP score/points per graduating senior (though they would need to wait until after summer, but that's ok). But just basing off AP/IB is not perfect, either, though I don't know what would be better.

PS. We got 13th last year and 41th this year... that's just because our administration is insane about rankings like this and they push kids who shouldn't be taking AP's to take them.

[quote="stupidkid"]Yea I read that also and thought the methodology is flawed.
How is Thomas Jefferson not in the top 100? considering they have the most USAMO qulifyers. Same with Exeter.
They should take into consideration the colleges that graduates from each school goes to and other standarized test scores such as SAT.[/quote]

it says magnet school are excluded.

O sorry I didn't know what "magnet schools" meant.
Yea just go to newsweek.com

most of the schools i know that are listed top 100 (troy21, interlake24, newport32, bellevue41) and even top 1000 have a history of pressuring students too much.

[quote="aznphatso"]I think this is flawed in the sense that many schools don't allow kids to take certain APs while others have the same AP calss but they're pushovers. Also, not all the schools in the nation offer the same APs. This index that Newsweek uses also doesn't address the average GRADE on all those AP exams.

The title of this list shouldn't be "Top X Schools" but rather exactly what it is, "Schools with the highest ratio"[/quote]

Exactly my point. The grade is far more important than the number of exams. Nice title ;)

lol how englightening to know my school isn't in the top 1000. and btw, i agree that the system for ranking is not exactly the most accurate way to measure a school's performance. but then again, it would be hard to develop a good, thorough ranking system anyway.

The index that is used, called the challenge index, is absolutely absurd.

I won't go into the problems of only basing the index on AP tests because even if it were right to base rankings only on AP tests, the challenge index: number of AP tests given/number of seniors is completely flawed.

My school, Scarsdale High School, is only around #70 on the list because we only allow people with a good chance of getting a 3 or higher to take AP tests.  For that reason, 97% of AP tests given at SHS are given a grade of 3 or higher.
I'd also mention that the challenge index does not count juniors or below who take AP tests.

On the other hand, many schools who give many more AP tests have a much, much lower passing rate.  The top schools on the list only pass slightly more than a majority of the AP tests given.  If SHS doubled the number of AP tests given, and students failed most of the extra ones, we would have the same passing rate as the other schools and would be in the top 10.

It should be based on the number of passed AP tests, not the number of given AP tests.  Theoretically, a small school with 100 graduating seniors could give all of them 10 AP tests.  Even if they failed all of them, the challenge index would be 1000/100, or 10.0, making that school the highest in the nation.

my school didn't make the list either. might have something to do with living in south carolina. and also that everyone is affraid to take ap classes. i'm guessing our score using that formula would be about 0.5

so is most everyone in agreement that a better calculation would be (total number of points scored on APs)/ (total number of graduating seniors) ? can you think of anything else that should, perhaps, be taken into account, such as:
1. SAT/ACT scores
2. number of internships taken
3. extracurriculars given
4. things like number of theatre productions, dance shows, recitals, etc.
5. number of competitions won (in all aspects, including math team, etc)
6. i think something about the number of disadvantaged kids should be included. a great school, i believe, should not only serve those who are gifted AND talented, but also those who are talented, but have not been given the luck to be born into a well-to-do family.

can anyone think of any other factors? what makes a "great" school?

My school ranks #636.  We have a very high ratio of AP Exams Taken per Subsidized Lunch (140.3)   :D

I agree with you, Maria, but you have a few small discrepancies (sp?):
1. You can "force" a student to take a class by offering no alternatives, but you cannot "force" them to take the tests. The ranking was based on the number of tests taken.
2. If the fee is waivered, it does not have to be paid.
3. In some states, like Texas, the government pays for the AP tests. Thus, this eliminates a small amount of "rich people bias" IN THOSE STATES.

[quote="bleumoose"]
3. In some states, like Texas, the government pays for the AP tests. Thus, this eliminates a small amount of "rich people bias" IN THOSE STATES.[/quote]

Seriously? I'm moving there! :D

[quote="bleumoose"]3. In some states, like Texas, the government pays for the AP tests. Thus, this eliminates a small amount of "rich people bias" IN THOSE STATES.[/quote]Ummm....I live in Texas, and, at least at our school, we have to pay for the AP tests ourselves.

[quote="eryaman"]Ummm....I live in Texas, and, at least at our school, we have to pay for the AP tests ourselves.[/quote]

Yea. 82 per test is no small deal, imo. It adds up for people taking 6 or 7...

my apologies. i may have gotten the state wrong.  :blush:

just putting in a word edgewise, but look at the [url=http://www.msnbc.msn.com/id/7760504/site/newsweek/]newsweek faq[/url]

6. Why don’t I see famous public high schools like Stuyvesant in New York City or Thomas Jefferson in Fairfax County, Va., or the Illinois Mathematics and Science Academy in Aurora, Ill. or Lowell High in San Francisco on the NEWSWEEK list?
I do not include any high school that accepts more than half of its students into the school based on highly competitive academic criteria like grades and test scores. All of those schools you name are terrific places with some of the highest average test scores in the country, but it would be deceptive for me to put them on this list. The Challenge Index is designed to honor schools that have done the best job in getting average students into college level courses. It does not work with schools that have no, or almost no, average students. I want a list that measures how good the schools are, not just how good their students are.

There are some magnet schools on the NEWSWEEK list, but only those that admit students through lottery-driven, first-come-first-serve, ethnic balance or other formulas that do not draw just the students with the highest grades and scores.

7. But aren’t all the schools on the list doing very well with AP or IB? So why rank them and make some feel badly that they are on the lower end of the scale?
That is exactly right. Every school on the list is in the top four percent of American high schools based on this measure. They have all shown exceptional AP and IB strength. I am mildly ashamed of my reason for ranking, but I do it anyway. I want people to pay attention to this issue, because I think it is vitally important for the improvement of American high schools. Like most journalists, I learned long ago that we are tribal primates with a deep commitment to pecking orders. We cannot resist looking at ranked lists. It doesn’t matter what it is—SUVs, ice cream stores, football teams, fertilizer dispensers. We want to see who is on top and who is not. So I rank to get attention, nothing more, in hopes people will then argue about the list and in the process think about the issues it raises.

[quote="bleumoose"]\
1. You can "force" a student to take a class by offering no alternatives, but you cannot "force" them to take the tests. The ranking was based on the number of tests taken.[/quote]
woops, well my school forces people to take the test once they are in the class (you get a 55 for the class if you dont' take the test). and i just thought it's the same everywhere

[quote="dareonion"]So I rank to get attention, nothing more, in hopes people will then argue about the list and in the process think about the issues it raises.[/quote]
nice argument?

so a little on rankings and then ideas about education

obviously newsweek's system is flwed in some way.  i propose a more complex system that no one will use because it requires too much research :)

catalogue schools by retention/graduation within standard time frame.  track students individually from entrance into a school: students that transfer to another school would be removed from consideration; students who graduate within one additional semester or less of the standard rate (4 years for a 9-12, 3 for a 10-12) would be considered a stage 1 success; students who cannot graduate in that time frame or drop out would be considered a stage 1 failure.   subsequently measure stage 1 successes: students who chose a college and graduate in their program within a similar standard timeframe would be considered stage 2 successes; students who chose a vocational school and graduate and maintain employment after graduation would also be successes under stage 2;  students who go straight into the workforce and maintain employment would also be successes;  students who were unable to keep a job for a signifigant ammount of time and students who were unable to find a job despite true effort would be considered failures as well as those who could not graduate fom their post secondary schools in a standard ammount of time. this leaves ambiguous the posistion of students who upon graduation choose neither additional schooling nor employment but instead chose to rely on someone else for support, but i personally would consider those students failures.

as for education reform i would switch the focus of high school education to creative thinking and problem solving as illustrated through the core curriculum, such that students would be gaining skills instead of knowledge.  this would allow colleges to focus on knowledge and assume the students knew how to learn before they got there. this would also make employees with only a diploma more valuable because they understand how to fix the supply problem insead of the capital of hungary !!

thats just my ideas take them however you want.

just commenting, in Florida all AP tests are paid for students attending a public school as long as they are enrolled in the course and acheive a passing grade. (and show up for the test). Students who take the test without taking the course (even if it's not offered, like at my high school) have to pay. It's still a good deal for Florida students, though the education system in Florida sucks, so I'm not really sure it's a tradeoff.

[quote="stupidkid"]
Yea just go to newsweek.com[/quote]

That's a novel idea... if today was still May 10th. Why didn't anyone bother to post a link?

the list is available at [url=http://www.msnbc.msn.com/id/7723397/site/newsweek/]www.msnbc.msn.com/id/7723397/site/newsweek/[/url]

and by fates_killer's standards, my school (public though it is) would pretty much head that list. as 99.9% of our graduates go on to attend and graduate from college (ok, i may be exaggerating... it may be 99%). a very large amount go on to grad school. and also, to JBL, i have an answer to your question, but you probably aren't going to like it, because it is (pretty much) immeasurable: passion. how many of the students at X High School enjoy going there? how many of them get up every morning and go in, eager to learn? how many of them stay late for school sponsored activities? that is truly the mark of a good school but it is, by far, immeasurable, as i said. you cannot measure the intangible, so one must resort to what can be measured.

[quote="bleumoose"]ahow many of the students at X High School enjoy going there?[/quote]

err, 0? :D Probably higher because of the people who just go to school for social reasons and fail every (or almost every) class... which I doubt shows achievement or passion. I just don't think there is a good way to rank schools; besides, all it does is make schools try to get the best ranking under the used formula and not really teach kids - take my school for example: the AP Calculus AB section has grown from 2 classes to 4 classes in two years. Why is this? Are kids getting smarter? Perhaps, but not that many. The administration is pushing kids to take higher-level courses (basically AP) so that our ranking in Newsweek gets better. Somehow that doesn't seem like the best way to teach...

well, at my school, there is a "citizehship" requirement: being in at least one club every term you are in school. even though many people join clubs just becasue they like what those clubs stand for, some just join to get that required credit. as a result, the time spent in school, bleumoose, is directly based on how much people want to be there, sometimes they just have to!

Okay, you're right, I don't like it :-)

Two reasons why:
(1) "passion" is a rather difficult thing to quantify.  Frankly, though, measurability wasn't a criterion I wanted in an answer to my question and it's not *completely* unquantifiable, just not easily quantifiable.
(2) More importantly, though, I think it still doesn't meet my condition.  I think it will be correllated with family wealth.  Most importantly the removal of pressures that comes with wealth, and additionally the appreciation for learning that is more prominent among better-to-do Americans, will cause that measure to correlate with family wealth.

In fact, though, I don't think my question was entirely fair.  After all, without considering any particular metric it is evident that the best high schools in the country overwhelmingly serve the upper classes, so finding a meaningful factor that correllates with "goodness" but not wealth would be exceedinly difficult (if not a mathematical impossibility).  I suppose your suggested "passion measure" is useful because it's not directly caused by the wealth of an area, but I think schools would certainly be hurt only by the lack of wealth, such as in a lack of available afterschool activities and thus fewer things to attract the passion of students.  On the other hand, though, many high schoolers like their schools (or are even passionate about them) for reasons unrelated to academic measures -- just check out the obession with high school football in many places.


Let $ {a_{1},...,a_{r}}$ denote a sequence of integers. For each non-empty subsequence A of S, we defne $ p(A)$ to be the product of all integers in $ A$. Let $ m(S)$ be the artihmetic average of $ p(A)$ over all non-empty subsets $ A$ of $ S$. If $ m(S) = 13$ and $ m(S\cup{a_{r+1}}) = 49$ for some positive integer $ a_{r+1}$, determine the
values of $ a_{1},...a_{r}$ and $ a_{r+1}$.

Canada MO 1988, Problem 5.

Given a set of size $ r$, we know that the number of non-empty subsets is $ 2^{r}-1$.  Hence we have that $  \sum_{A\subseteq S}p(A) =13(2^{r}-1)$ where $ A$ is not the empty set.  We have a similiar equality for the set $ S\cup \{a_{r+1}\}$, and so we can deduce from for all non-empty sets $ A$

$ \sum_{A\subseteq S\cup\{a_{r+1}\}}p(A)=(a_{r+1}+1)\sum_{A\subseteq S}p(A)+a_{r+1}$

that we must have $ 49(2^{r+1}-1)=13(2^{r}-1)(a_{r+1}+1)+a_{r+1}$.  This simplifies to $ \frac{85\cdot 2^{r}-36}{13\cdot 2^{r}-12}=a_{r+1}\in\mathbb{N}$, and this forces $ r=3$ and $ a_{r+1}=7$.  Now, in the original equation, we have

$ a_{1}a_{2}a_{3}+a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3}+a_{1}+a_{2}+a_{3}=13(2^{3}-1)=91$ and so

$ (a_{1}+1)(a_{2}+1)(a_{3}+1)=92=2\cdot 2\cdot 23$ which means that if $ a_{1},a_{2},a_{3}$ are positive integers, the only possibility is that $ a_{1}=1,a_{2}=1,a_{3}=22$ and its permutations.  Thus $ S=\{1,1,22\}$ and $ a_{4}=7$.


What am I supposed to bubble for age on the AIME form? My teacher and I noticed it said less than or equal to, as it did on the AMC form. I am 16 years old but my birthday has already passed (i.e. 16.5 years old). Am I supposed to write 17? I wrote 16 on the AMC form.
I hope this doesn't matter much :D

Don't overthink it, the $ \le$ only applies to the number it is next to (meaning $ \le11$ or whatever the number actually is). It's just meant to prevent them from having to go all the way up to age 8ish since there are so few 8ish-11ish-year-olds who take the AMC.

Bubble 16.  That is your age as commonly used in the US.

Steve Dunbar
MAA Director, American Mathematics Competitions


function
quadratics
algebra
calculus
calculus computations

For positive real numbers $a,\ b,$ prove that there exists the function satsfying the following conditions (1), (2), (3).

(1) $f(x)$ is the continuous real-valued function at $0\leq x\leq a$ such that $f(0)=b,\ f(a)=0.$
(2) For all $x_{1},\ x_{2}$ such that $0\leq x_{1}<x_{2}\leq a,\ f(x_{1})>f(x_{2})$ holds.
(3) Let $V_{x}$ be the volume of the solid produced by the revolution of the part enclosed by the x-axis and y-axis about the x-axis and $V_{y}$ be the one about the y-axis. $V_{x}=V_{y}$ holds.

Trivial when $a=b$, so assume $a>b$. For a positive $p$ let $f_{p}(x)=b(1-x/a)^{p}$. Computations yield $V_{x}=\frac{\pi a b^{2}}{2p+1}$ and $V_{y}=\frac{2\pi a^{2}b}{(p+1)(p+2)}$*. We have $V_{x}<V_{y}$ for $p$ close to $0$ but $V_{x}>V_{y}$ for large $p$. So, $V_{x}=V_{y}$ for some $p$ (you can solve a quadratic equation to find such $p$).

(*) Using cylindrical shells. :roll:


calculus
integration
trigonometry
calculus computations

Evaluate

$ \int_0^{\frac{\pi}{2}} e^x\left\{\cos \ (\sin x)\cos ^2 \frac{x}{2}\plus{}\sin \ (\sin x)\sin ^ 2 \frac{x}{2}\right\}\ dx$.

$ I\equal{}\int_0^{\frac{\pi}{2}} e^x\left\{\cos \ (\sin x)\cos ^2 \frac{x}{2}\plus{}\sin \ (\sin x)\sin ^ 2 \frac{x}{2}\right\}\ dx$

$ \cos 2x \equal{} 2\cos^2 x \minus{} 1 \equal{} 1 \minus{} 2\sin^2 x$

$ \cos^2 \frac{x}{2} \equal{} \frac{1\plus{}\cos x}{2}$
$ \sin^2 \frac{x}{2} \equal{} \frac{1 \minus{}\cos x}{2}$

$ I\equal{}\int_0^{\frac{\pi}{2}} e^x\left\{\cos \ (\sin x)\frac{1\plus{}\cos x}{2}\plus{}\sin \ (\sin x)\frac{1 \minus{}\cos x}{2}\right\}\ dx$

$ \equal{}\frac{1}{2}\int_0^{\frac{\pi}{2}} e^x\left[ (\cos (\sin x) \plus{} \sin (\sin x))\plus{}\cos x(\cos (\sin x) \minus{} \sin (\sin x))\right]dx$

$ \equal{}\frac{1}{2}\int_0^{\frac{\pi}{2}} [e^x(\cos (\sin x) \plus{} \sin (\sin x))]' dx$

$ \equal{}\frac{1}{2}\left[ e^\frac{\pi}{2}(\cos (\sin \frac{\pi}{2}) \plus{} \sin (\sin \frac{\pi}{2})) \minus{} e^0(\cos (\sin 0) \plus{} \sin (\sin 0)) \right]$

$ \equal{}\frac{1}{2} ( e^\frac{\pi}{2}\cos 1\minus{}1 )$

where did sin 1 go? =(

$ \equal{}\frac{1}{2} ( e^\frac{\pi}{2}(\cos 1\plus{}\sin 1)\minus{}1 )$


A small but interesting problem :

[color=blue]Problem[/color]
Let $n$ be an integer prove that there exist an integer like $P$ such that $P-1$ and $P+1$ both have at least $n$ different prime factors.

Take different primes $p_1, p_2,...,p_n$ and $q_1,q_2,...,q_n$.
Then juse Sunzi (or however it's called in China) to find an integer $P>1$ with $P\equiv 1 \mod p_i$ and $P \equiv -1 \mod q_i$.
This integer will work.

Yes thats right :)  ,well if we wana  have some thing else do to ,find a Prime number like $P$ which satisfeis the conditions

Isn't it "Sun Ze"? :)

Anyways, I prefer "the Chinese remainder theorem" ;)

That prime number exists by Dirichlet's theorem.
But I wonder if there is a more elementary way  :?

Sun Zi? The man who wrote [u]The art of war[/u]? :)

Yes, CRT's definitely better ;)


Prove that $\frac{1}{x}+\frac{1}{cosx} \geq \frac{2}{\sqrt{sinx}}$. Can equality hold?

I'm going to assume you mean on $(0, \frac{\pi}{2})$.  Note that because $\tan{x}\geq{x}$ on this interval (well-known), we have $\sin{x}\geq{x*\cos{x}}$, which implies $\frac{1}{\sqrt{x*\cos{x}}}\geq{\frac{1}{\sqrt{\sin{x}}}}$.  Now, by AM-GM we have $\frac{1}{x}+\frac{1}{\cos{x}}\geq{\frac{2}{\sqrt{x*\cos{x}}}}$, and the inequality follows.

equality cannot hold because $\tan x>x$ for $x\in(0,\frac{\pi}2)$


hello, we get $ \sum_{i\equal{}1}^{100}i^5\equiv 0 \mod 4$.
Sonnhard.

Thanks, Sonnhard. I had $ (2n)^{5}$ as $ 0$ mod $ 4$ and then $ (4k\plus{}1)^{5}$ and $ (4k\minus{}1)^{5}$ cancelling. (Sorry, my CTRL key has gone again.)

@Sonnhard : I think AndrewTom might want to ask how you got that direct answer so here goes....
We use here the fact that \[ n^m \equiv (n \pm xk)^m (mod \ k)\] for any integer x.
Therefore ,We have \[ (4k\plus{}1)^5 \equiv 1^5 (mod \ 4)\]
\[ (4k\plus{}2)^5 \equiv 2^5 (mod \ 4)\]
\[ (4k\plus{}3)^5 \equiv 3^5 (mod \ 4)\]
\[ (4k)^5 \equiv 0^5 (mod \ 4)\]
Therefore, everything depends on $ 1^5 \plus{}2^5 \plus{}3^5 \plus{}4^5$
\[ 1^5 \equiv 1 (mod \ 4)\]
\[ 2^5 \equiv 0 (mod \ 4)\]
\[ 3^5 \equiv \minus{}1^5 \equiv \minus{}1 (mod \ 4)\]
\[ 4^5 \equiv 0 (mod \  4)\]
Adding up we get,
\[ Ans\ : \ \boxed{0}\]


(Note: because I cannot LATEX, I will use = instead of the three bars sign)

In aops Volume 1, in page 44, they say, "If a = b (mod m) and p = q (mod m), then for all positive integers c, we have:

...

5. (a+b) = (p+q) (mod m)
6. ab = pq (mod m)

You may need to chew on those last two a bit, though they are among the most useful. They just restate the fact that we can mod out before of after we add or multiply."

Let a be 3, b be 8, c be 4, and d be 9 and let the mod m be 5.

By "property " 5, we get 11 = 13 (mod 5), an obviously false relation.
By "property" 6, we get 24 = 36 (mod 5), an obviously false relation.

(or choose any values you want for a, b, p, q, and m)

Could aops have meant instead:

5. a + p = b + q (mod m)
6. ap = bq (mod m)

or did I do something wrong?????????

Any help or insight into this question is greatly appreciated. I thank you in advance.

What you proposed with the $ b,p$ switched is what they meant.

Er..in my book, that is not how it is written..

5. $ (a \plus{} b) \pmod{m} \equiv a \pmod{m} \plus{} b \pmod{m}$
6. $ ab \pmod{m} \equiv (a \pmod{m} )( b \pmod{m})$

Let $ a \equal{} 5$ and $ b \equal{} 10$ and $ m \equal{} 10$

5. $ 5 \pmod{10} \equiv 5 \pmod{10} \plus{} 10 \pmod{10}$, so $ 5 \equiv 5 \pmod{10}$
6. $ 150 \pmod{10} \equiv (5 \pmod{10})(b\pmod{10})$, so $ 0 \equiv (5)(0) \equal{} 0 \pmod{10}$

EDIT: i don't know of that is the same..

[quote="theax"]007 math, are you looking at aops volume 1: the basics, 7th edition, page 44?

Does the chart of page 44 start like this (again I'm going to use = instead of the three bars sign because I cannot LATEX)
"Let's summarize what we can do with congruences. If a = b (mod m) and p = q (mod m), then for all positive integers c, we have:

1. a + c = b + c (mod m)
2. a - c = b - c (mod m)
3. ac = bc (mod m)
4. a^c = b^c (mod m)"

If you are, hmm....
that's weird.
Anyone else?[/quote]

my book does not have that p = q (mod m) thing.  ur 1-4 and my 1-4 are the same...

though we are talkin about 5 & 6...

look at this, page 6:
http://www.artofproblemsolving.com/Books/Vol1/exc1.pdf

There were a few changes made to these pages at some point in the last few years; it's possible that you have an older book that had some typos here.


Hi everyone!

my problem:

[quote]If $ f: X\minus{}>R$ is a function that for each $ \epsilon > 0$ exists such simple function g satisfying $ |f(x)\minus{}g(x)| <\equal{} \epsilon$ for each $ x\in X$, then f is measurable and bounded.[/quote]

since every simple function is bounded, we at once know, that either is our function f, cause:
$ \minus{} \epsilon \plus{} g(x) <\equal{} f(x) <\equal{} \epsilon \plus{} g(x)$, so that's obviously not the problem here. this whole measure stuff doesn't get into my intuition and I don't have any idea how to try to solve this task. if i knew, that for each $ \epsilon$ i could get a measurable function g, it would be obvious, that f is measurable too [wouldn't it?], but can i really always have a measurable function g?

i would be very grateful for any hints. hope my english isnt terrible enough to disturb the sense of this post.

rahl.

Sorry, I'm having a little trouble understanding the question.  You might be missing the assumption that $ g$ is measurable since if $ N$ is a nonmeasurable set, $ f\equal{}\chi_{N}$ is a non-measurable simple function and we can always approximate $ f$ with simple functions-itself.


Duke college HMMT email FTW ARML geometry

Hi folks,

The Walton MathFest is on Saturday, April 18th. We've got the standard tests. (Allen, Claus, and Henry have been working hard on these.) A ciphering round similar to the Duke ciphering or HMMT GUTS round. Our standard scavenger hunt.
And of course the math talks. We would like to invite everyone to come. If you are interested and haven't responded yet, please send me an email and let me know that you are coming. We need to get a good count so that we can prepare!

Thanks,
Tom
TFulton42@gmail.com <--- new email old one still works

Yay Mathfest ftw ^__^

Question, has there been a ciphering round in the past? I don't remember ciphering rounds...o.o

But i DO remember the scavenger hunt!! -clapclapclap- :D

[quote="gtf"]Nope, the ciphering round is new. It will be a mixed round with folks from different teams![/quote]

As in people from different schools can form a team?

That's intense. :O

[quote="timwu"]So we finally got enough people to come to the competition tomorrow. I just e-mailed the walton math fest to register. Is this ok?

Also, I can't find the schedule for this competition? Does anyone know when it starts?[/quote]

We will be irritated at you, but it's probable. Don't know for sure unless Fulton/Annie OKs it. You might not get lunch, but it should be fine besides that.

Registration starts at 8:30. Tests start at 9:00.

[quote="Arvind_sn"]gtf wrote:
Nope, the ciphering round is new. It will be a mixed round with folks from different teams!

As in people from different schools can form a team?

That's intense. :O[/quote]

Oh yes it is.

Or to clarify Alan's response. You will be on a team with folks from different schools. We [b]DO[/b] expect the ARML folks to help the non-ARML folks. Think of it as developing for the future.

And Tim, we order some extra pizza just for you! Well, not really, but we will hopefully have plenty.

Tom

Thanks, Mr. Fulton, though I actually did not eat pizza but chicken hamburger (hopefully that's its correct name).

Congrats to the Super Varsity winners:
1. Robert
2. Miles
3. Mohan Prasad Akkinei (Northview)

BTW, was chaos ciphering only awarded to top two? I left that a bit earlier to a lecture (after problem 50) so I didn't get to see our team's final score.

Great tournament!

I enjoyed myself every bit of it, and the lectures were great. Thanks to Dr. (Aah I forgot his name but he did the first lecture), Mr. Fulton, Dr. Koch, Miles, and anyone else I might have forgotten.

The tests were well made, as was the scavenger hunt.

Congrats to all the winners :D

Chaos ciphering was originally supposed to be awarded to the first 6-person team, but since we actually had 4 person teams and the first place only had 3 people, we just gave the fruit snacks to the top two.

Aside from that, how did you guys like the problems? Henry wrote the Junior Varsity Test, I wrote most of the Varsity Test, and Claus wrote most of the Super Varsity Test. I wrote most of the Chaos round and the Scavenger Hunt. Were the problems good (aside from the occasional typo)? And how did you guys like the setup and organization? I'm (probably) mostly going to be in charge next year, so it won't be as good at all as when Annie organized it this year... Please comment!

I would strongly encourage a different tiebreaker question.  How could we write an elegant proof when we were struggling through the questions?

This next part is nitpicking.

1.  For the Pascal pyramid problem, I thought that the second row would have 6 1's instead of just three.  That was a perspective fault on my part, but sometimes diagrams can be helpful.

2.  $ f^n(x)$ can be interpreted as composition of functions or as the nth derivative.  I guessed the answer right, but it may be good to clarify the point next time, especially since there was calculus on this test.

[quote="azn6021023"] $ f^n(x)$ can be interpreted as composition of functions or as the nth derivative.  I guessed the answer right, but it may be good to clarify the point next time, especially since there was calculus on this test.[/quote]
Actually, $ f^n$ usually means $ f$ composed with itself $ n$ times pretty unambiguously.  Derivatives are usually written as $ f^{(n)}$ in my experience.  The first four places I'm looking online all use this, as well, with no mention of $ f^n$ :  [url=http://en.wikipedia.org/wiki/Notation_for_differentiation#Lagrange.27s_notation]Wikipedia[/url], [url=http://www.maths.manchester.ac.uk/~cds/articles/derivative.pdf]this thing[/url], [url=http://web.mit.edu/wwmath/calculus/differentiation/notation.html]this thing[/url], [url=http://mathworld.wolfram.com/Derivative.html]Mathworld[/url].

The only possible confusion of which I'm aware is that sometimes people unfortunately take $ f^n(x)$ to mean $ \bigl[ f(x) \bigr]^n$, i.e., $ f(x)$ multiplied with itself $ n$ times.  Also, some people simply forget the notation, so it's still probably a good idea to define such notation.

The setup and organization were excellent.

I thought the super varsity questions were generally engaging, but sometimes unpolished---for example, there were sometimes terms that were not well defined.  Also, there were some small TeX errors :  quotes should be written like ``this'' or ``this", not like "this", so that they come out like $ \text{``this''}$, not like $ \text{"this"}$.  And \cdot is better for multiplication than *, as $ a \cdot b \cdot c$ looks a lot better than $ a*b*c$.

Overall, though, great job!  I really enjoyed everything.

My apologies for angering the great Boy Soprano.  I guess I never really paid attention to where the parentheses were.

Sorry, I wasn't angry; I'd just never seen that used for a derivative, but I didn't know whether it was used, so I wanted to check.  I guess tone is hard to communicate in print. :oops:  I am rarely angry, and I find it hard to imagine getting angry over mathematical notation (notwithstanding certain "notation warz" that occurred at Hampshire over notation for dihedral groups), although I can become exercised.  But that's because I care about math, not because I'm angry.

The Walton MathFest tests and solutions are up at [url]http://www.waltonmathteam.org[/url]. For some reason, clicking on the Home, the logo, or the "Previous" links directs you to the registration page and there's no way for you to get back to the front page, so you'll have to go to [url]http://www.waltonmathteam.org[/url] again. Also, where it says "Super Varsity" the second time, it should say "Scavenger Hunt." I think Henry did this late at night...

We are aware that there are some LaTeX errors and vague notation, but we (Henry and I) are too lazy to go back and fix them. I think all of the math mistakes have been fixed though.

Have fun and feel free to ask about anything.


geometry
AwesomeMath
summer program
3D geometry
pyramid
AMC
AIME

Here's the link to this really addicting Flash Game.

[url]http://babarageo.com/flash/ginormo/[/url]

I've literally spent hours on it.

Tell me how you like it!

I've beat the False God, so I can give you guys hints.

I'm having some trouble against Poseidon. I took him down to about 1/3 once, but he always kills me. (it takes forever to damage him :mad: )

What's your strength?

What's your sword level?

If possible, enchant your sword with rubies, although I know you can't have many of those yet.

If you go back to the boss that's the big cloud and can only be damaged by magic, he'll occasionally drop rubies.  Also, death clouds.

once you learn magic and increase your vitality, you can earn lots of money by taking the dark mist down over and over again.

I am currently devising a strategy for taking down the sand worm. I have strategies for all the others too.

well not so much for the minotaur.

Minotaur is easy.  just stay to the top left of him.  For the fire minotaur, just keep +water enchentments.  I'm on the last 3 stages.   :D

@alex: So are you stuck at the Gate Dragon?

[hide="a sort of cheat"]once you get past the mutant cyclops, you will have access to the vitality shrine. If you can, upgrade vitality and go defeat the dark mist a few times. Vitality increases the damage made by magic. Once you get a high enough vitality, you can defeat the dark mist in a couple blows, and you can increase on strength. Do that, since that increases the damage done by sword. Also, once you go to the magic university, only the wide shot is worth enough to buy. You can buy all the others just to be awesome, but I advise against it.[/hide]

[hide="a time saver"]After you get past dark mist, there will be a merchant that sells all sorts of things. If you can, get the speed-up boots. That is a huge time saver, even though it has no use in battle.[/hide]

Vitality helps magic?

I thought intelligence did...

I advise the Cross Napalm - the cluster shots do more damage than the Circler and Wide Shot.

For those of you who are at around the 4-6th area, what I did was just kill the minotaur over and over again to upgrade my sword.

I think sword size affects damage, and it's cheaper than Str, so once you have a big sword, you can kill loads of stuff a lot more easily.

I was surprised that vitality helps magic, but I keep increasing my vitality and I really am doing more damage with my magic attacks...or at least it looks like that to me...

Cross Napalm is the best!

The Dark Mist is way better for easy gold.

I think sword size just lets you hit more monsters at a time...

actually, sword length increases damage crazy.

I never farmed Dark Mist, so I wouldn't know - even now I've only killed it about twice.

I never knew that about Vitality.

Actually, I think that is sword LEVEL that increases damage.

Maybe. Increasing width and length do make enemy's easier to hit, so maybe levels increase damage.

I am stuck on the witch on level 8, any tips?

[quote="andrewjjiang97"]I can't beat the pixie. Whenever I equip a helper, s/he always dies right before the pixie does. Help?[/quote]

[hide="Spoiler"] You aren't supposed to kill the pixie. Just dodge her attacks (or pump your armor/endurance up enough so that it doesn't matter) for a minute, and she'll just leave and give you 100 gold[/hide]

[quote="Iggy Iguana"]How do you beat the lizard assasin. He keeps duplicating himself[/quote]

[hide="Spoiler"] The Lizard Assassin will change positions every time you swing your sword, so don't swing it randomly. Use a fire based sword, and if that doesn't work, pump it up with more rubies (you can get them easily from Efreet). Make your sword very large, enough so that it can almost take up the whole screen, and if you can afford chaser spell by now (100mil), you should probably try to get it[/hide]

Just use Indigo=Roger, dodge the lizard's attacks, and wait  :lol: It may take some time this way, but you can kill it without ever swinging your sword.

[hide="Spoiler"]In the 8th world, there is a place which seems to have nothing inside. Keep going in and out and eventually Chaser Spell will be available. Make sure you have 100 mil though, before you try.[/hide]

Well, there are internet walkthroughs... I know everything but I haven't beaten the game yet, for example.

[quote="piegraph"]@Bogtro:  Have you beat the game since you like know everything?[/quote]

As of 45 minutes ago, I finally defeated the Avatar :D

How do you defeat the demons on 15th level.
Where do you get Hyper Armor?

Hyper Armor is obtained by defeating the Golden Armor, which is a 4th tier Living Armor. Be advised that it takes extremely long to kill (I took 40 minutes to kill it [don't worry I was listening to Simpsons episodes while I was doing it {wearing winter gloves helps for comfort}]), so equip the ring that doesn't let you escape a battle if you are trying to find it.

[hide="Golden Knight info"]Additionally, it is considerably weak to emeralds. I beat it in roughly 15 minutes with 500 emeralds (lol, it doesn't actually take too long to get them). My advice: bring a book.[/hide]

[hide="Demons"]Unfortunately, there does not seem to be an elemental weakness (that I noticed at least). Basically, just either practice dodging or pump your stats. Don't come back to them until much later in the game when you've pumped up your stats a LOT[/hide]

[quote="Iggy Iguana"]Where's the best place to get emeralds?[/quote]

[hide="Spoiler"]
Perhaps surprisingly, on area 2 with the Orcs. You should be able to kill them in 1 hit by now, and they drop emeralds very often. If you don't seem to be getting emeralds, try increasing your luck, on area 12[/hide]

actually, im lucky and get emeralds from the 1st stage easily


I have posted it before but it had a mistake, but I can't edit because it closed, so I post it again, of course, without mistake  :P
Let a,b are integer and a>b>2. Prove that
$a^b<b^a$

And it has a generation, ex. $3^8 > 4^6 > 6^4 > 8^3$
Let's prove it!! Very easy!!


trigonometry logarithms function AMC AIME

[u][i][b]Potpurri of Problems:[/b][/i][/u]

1.What is the coefficient of the term $ x^3 \cdot y^5 \cdot z^2$ in the expansion of $ (x \plus{} y \plus{} z)^{10}$?

[hide="Answer Key"]The answer key says 2520[/hide]

2.How many solutions are there to the equation?
$ \cos (\ln x) \equal{} 0$ such that $ 0 < x < 1$

[hide="Answer Key"]I keep getting 0, but the answer key says there is [b]1[/b] solution.[/hide]

3.Let $ n$ be a positive integer. The function $ F(n)$ is the number of positive divisors of $ n$. For
example, $ F(6) \equal{} 4$. The number of $ n$ between $ 1$ and $ 10,000$, inclusive, in which $ F(n)$ is even
is what?

[hide="Answer Key"]The answer key says 9900. That means there are 100 numbers that satisfy n, but I got something different. $ 1^2, 2^2, 3^2,.....100^2$ those are 100 numbers right there, and we still have to try other powers....so I dont get how the answer is 9900.[/hide]

4.A certain cookie store sells six different types of cookies. How many ways are there to fill a
box of cookies if the box holds ten cookies and cookies may be placed in any order?

[hide="Answer Key"]The answer key says the answer is $ 3003$, but my answer is way bigger than that.[/hide]

Definitely does NOT belong in Interm. Topics.

[hide="1"]
$ \frac{10!}{3!5!2!}\equal{}2520$
[/hide]

[hide="3"]
It is obvious that only perfect squares have an odd number of factors...just use the # of factors formula.  Then the answer will just be 10000-100=9900.
[/hide]

Dude are you from Michigan? Because these are from the MMPC!
[hide="2"]The only value that works in this range is cos(-90), so there is one solution.[/hide]

[hide="4"]Set up an equation:
$ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \equal{} 10$. By balls and urns, this has $ \binom{15}{5} \equal{} 3003$ ways.[/hide]

And AIME gave the solutions to the other 2.

[quote="AIME15"]Definitely does NOT belong in Interm. Topics.

[hide="1"]
$ \frac {10!}{3!5!2!} \equal{} 2520$
[/hide]

[hide="3"]
It is obvious that only perfect squares have an odd number of factors...just use the # of factors formula.  Then the answer will just be 10000-100=9900.
[/hide][/quote]Why is it not Intermediate? Since the solution to 3 is 'obvious', as you claim, and the question is simple... There are much harder questions in Intermediate and they weren't moved to Pre-Olympiad for some odd reason.  :roll:

[quote="Poincare"]I think he means that the problems are too easy for intermediate (maybe)...[/quote]Actually, these problems are appropriate for Intermediate. It's a shame that higher level forum users would berate on students posting a bit easier problems; thus, this is the result of users 'scared' of posting easier problems on Pre-Olympiad or Olympiad Section.


function
induction
algebra unsolved
algebra

Find all continuous functions $ f: R^\plus{}\to R^\plus{}$ satisfy:
1) $ f(2x)\equal{}2f(x)$;
2)$ f((f(x))^2)\equal{}xf(x)$
3)$ f(x)\epsilon N^*$  for $ x\epsilon N^*$

2) implies $ f$ is injective.
From 1), we have $ 2a,\frac a2\in Im(f)$ when $ a\in Im(f)$. Then by continuity, $ f$ is surjective.
bijective and continuity imply $ f$ is strictly increasing.
Let $ f(a)\equal{}1$. Then $ f(1)\equal{}a$ by 2). $ a\ge1$ from 3). Then by monotonicity, $ f(1)\equal{}1$.

We can use induction to show $ f(n)\equal{}n$ for natural $ n$. Assume $ f(m)\equal{}m$ when $ m< n$. Then if $ 2|n$, $ f(n)\equal{}2f(\frac n2)\equal{}n$. If $ 2\not| n$, $ n\minus{}1\equal{}f(n\minus{}1)<f(n)<f(n\plus{}1)\equal{}2f(\frac{n\plus{}1}2)\equal{}n\plus{}1\Rightarrow f(n)\equal{}n$.

Now we see $ f(x)\equal{}x$ if $ x$ has finite binary expansion, because there exists natural $ n$ such that $ 2^nx$ is natural.

We can approach any $ x$ with a series of number which have finite binary expansion, we see
$ f(x)\equal{}x,\forall x$ by continuity.


inequalities
LaTeX
inequalities proposed

Given positive real numbers a, b, and c such that a + b + c = 1, show that (a^a)(b^b)(c^c) + (a^b)(b^c)(c^a) + (a^c)(b^a)(c^b) <=1

please guys, reply with solution as soon as possible

One can notice that (a,b,c) majorizes (a+b+c,0,0), as a+b+c>a, a+b+c>a+b+0 and a+b+c=a+b+c+0+0. That means, we can apply the Muirhead inequality to get:
(a^a)(b^b)(c^c)+(a^b)(b^c)(c^a)+(a^c)(b^a)(c^b)<=(a^(a+b+c))(a^0)(b^0)+(b^(a+b+c))(a^0)(c^0)+(c^(a+b+c))(b^0)(a^0)=a+b+c=1.
We get equality only for a=b=c=1/3.
Excuse me, but I haven't learned LateX yet.
[/code]

We could use AM-GM in an onesentence-proof with a, b and c as weights since 1=a+b+c;
$ {a}^{b}{b}^{c}{c}^{a} \leq (ab\plus{}bc \plus{}ac)$ and so on for each product..
 $ \Rightarrow  (a^a)(b^b)(c^c) \plus{} (a^b)(b^c)(c^a) \plus{} (a^c)(b^a)(c^b) \leq  {(a\plus{}b\plus{}c)}^{2}\equal{}1$

In fact, AM-GM here is a special case of the inequality of weights:
xy+yz+zx>=x^y*y^z*z^x


If 64 is divided into three parts proportional to 2, 4, and 6, the smallest part is:

$\textbf{(A)}\ 5\dfrac{1}{3} \qquad
\textbf{(B)}\ 11 \qquad
\textbf{(C)}\ 10\dfrac{2}{3} \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ \text{None of these answers}$

I assume you mean ratios are $ 2: 4: 6$, or $ 1: 2: 3$

The smallest is $ \dfrac {64}{1\plus{}2\plus{}3}\equal{}\dfrac{32}{3}$, so C.

[hide="Solution"]

Let the constant be $ x$ so that the three numbers are $ 2x, 4x, 6x$. Then:

$ 2x\plus{}4x\plus{}6x \equal{} 64$
$ 12x \equal{} 64$

We are looking for $ 2x$ so dividing the above with 6, we get $ 2x \equal{} \frac{64}{6} \equal{} \frac{32}{3}$, which can be expressed as C. [/hide]

[hide]
2x+4x+6x=64
therefore x=16/3
and smallest will be 32/3 i.e C
[/hide]

I have to post a solution.

So $2x+4x+6x=12x=64\implies x=\frac{16}{3}$, so $2x=\frac{32}{3}\implies \boxed{\textbf{(C)}}$.


ARML geometry incenter perimeter trapezoid trigonometry LaTeX

I created a set of 8 geometry problems to practice for ARML.
(Some of the problems were taken from other contests.)
You have 10 minutes for each pair of problems, just like the actual contest.

This is not a competition, so you can post solutions in this thread, but [u][i]please hide your solutions[/i][/u].

Have fun!

1. Given square $ABCD$, $E$ is the midpoint of $CD$. $F$ is on $BC$, $G$ is on $AB$ and $H$ is on $AD$ such that $AE\perp EF$, $EF\perp FG$, $FG\perp GH$. Compute $\frac{AH}{HD}$.

2. In $\triangle ABC$, $AB=3$, $BC=7$, $CA=5$. The angle bisector of $\angle A$ meets $BC$ at $D$. If $I$ is the incenter of $\triangle ABC$, compute $DI$.

3. Hexagon $ABCDEF$ is inscribed in a circle. If $AB=BC=CD=5$ and $DE=EF=FA=3$, compute the length of the radius of the circle.

4. $\triangle ABC$ is a right triangle with $AB=9$, $BC=12$ and $AC=15$. $D$ is on $BC$ and $E$ is on $AC$ such that $ED\parallel AB$. If the perimeter of trapezoid $ABDE$ and the perimeter of $\triangle EDC$ are equal, compute the area of trapezoid $ABDE$.

5. In quadrilateral $ABCD$, $AB=12$, $CD=10$, $\angle ABC=80^\circ$ and $\angle BCD=70^\circ$. $P$, $Q$, $R$, $S$ are the midpoints of $AD$, $BD$, $BC$, $AC$, respectively. Compute the area of quadrilateral $PQRS$.

6. Two lines $l$ and $m$ are not parallel. Points $A$, $B$, $C$, and $D$ are on line $l$ in this order, and points $H$, $G$, $F$, and $H$ are on line $m$ in this order. $AE$, $BF$, $CG$, and $FH$ intersect at $P$. If $AB=4$, $BC=2$, $CD=6$, $EF=3$ and $FG=1$, compute $GH$.

7. Quadrilateral $ABCD$ is inscribed in a circle with side $AD$, a diameter of length $4$. If $AB=BC=1$, compute $CD$.

8. Square $ABCD$ is on the xy plane. $P (\frac{25}{8}, 0)$, $Q (3,4)$, $R(0,\frac{25}{6})$ and $O (0,0)$ are on $AB$, $BC$, $CD$ and $DA$, respectively. Compute the side length of square $ABCD$.

Edit: Added #8.

[hide]
Problem 1:
Without loss of generality, let the square's side length be 1.
In triangle $ADE$, $AD: DE = 2: 1$, since $DE = 1/2$.
Because of angle equivalences, triangle $ADE$ is similar to triangles $ECF$, $FBG$, and $GAH$.
So $EC: CF = FB: BG = GA: AH = 2: 1$.
Now $EC = 1 - 1/2 = 1/2$ and $CF = 1/4$ and $BF = 1 - 1/4 = 3/4$ and $BG = 3/8$ and $GA = 1 - 3/8 = 5/8$ and $AH = 5/16$ and$HD = 1 - 11/16 = 11/16$.
Then $AH\HD = 5/11$.

Problem 2:
Let X be the point of tangency between ABC's incircle and side a.
Let $K$ be the triangle's area, and $s$ be the triangle's semiperimeter or $\frac{15}{2}$.
Since $K = rs$ and $K = \sqrt{s(s-a)(s-b)(s-c)}$ we have $r = \frac{\sqrt{3}}{2}$.
Consider right triangle IDX.
By the Angle Bisector Theorem, $BD = \frac{3*7}{3+5} = \frac{21}{8}$.
We also know that $BX = s - b = \frac{5}{2}$.
So $DX = \frac{21}{8}-\frac{5}{2} = \frac{1}{8}$.
Of course, $IX = r = \frac{\sqrt{3}}{2}$.
By the Pythagorean Theorem, $ID = \sqrt{DX^2+IX^2} = \sqrt{\frac{1}{64}+\frac{3}{4}} = \sqrt{\frac{49}{64}} = \frac{7}{8}$.[/hide]

These are my solutions for Problem 1 and 2.

thanks for these problems!
ill just do 5 because it is nice and i know how to do it  :lol:
[hide="number 5"]
since $P$ is the midpoint of $AD$ and $Q$ is the midpoint of $BD$, by similar triangles, $\triangle ABD \sim \triangle PQD$, and $PQ=\frac{1}{2}AB=6$
similarly,
$\triangle ABC \sim \triangle SRC$, and $RS=\frac{1}{2}AB=6$
$\triangle DAC \sim \triangle PAS$, and $PS=\frac{1}{2}CD=5$
$\triangle DBC \sim \triangle QBR$, and $RQ=\frac{1}{2}CD=5$
Therefore $PQRS$ is a parallelogram. Since $\angle BRQ=\angle BCD=70$ and $\angle CRS=\angle CBA=80$, $\angle QRS=\angle SPQ=180-80-70=30$
$[PQRS]=2\left(\frac{1}{2}\right)(5)(6)\sin{30}=\boxed{15}$
[/hide]

[hide="Number 7"]OH MAN, PTOLEMY IS LIKE SO COOL, I CAN'T RESIST BUT TO DO NUMBER 7 EVEN THOUGH I HATE THIS LATEX NOW

$\triangle{ABD}$ and $\triangle{ACD}$ are right triangles. So if we let $CD=x$, and apply Ptolemy, we have $4+x=\sqrt{15(16-x^2)} \Longrightarrow x=\boxed{\frac{7}{2}}$[/hide]

Yeah, I made #4 a bit easy because #3 is relatively hard.
(Remember, you have 10 minutes to work on #3 and #4)

Does anyone want to try #3, 6 or 8? (These are harder) :)

[hide="Problem 3 sir"]I once heard of a theorem that one can rearrange the sides of an inscribed polygon. Such overheard snippets often come in handy.

Let's do that, alternating 3s and 5s. If we connect 3 adjacent pairs of sides, we will form an equilateral triangle inscribed in the circle. Let's call it Frank.

Frank combined with the triangle formed by a side of the Frank and two sides of the hexagon makes a cyclic quadrilateral.   Frank's angles are 60, so the opposite angle (between the sides of 3 and 5) is 120 since cyclic quadrilaterals have supplementary opposite angles. With this info, law of cosines gives the side of Frank to be 7.

Radius of the inscribed circle is 2/3 the height of Frank (or use extended law of sines), given by s sqrt3/2 = 7sqrt3/2, so that leaves 7sqrt3/3.
[/hide]

[hide="3 clarification"]Thinking about it a little, it seems pretty intuitive that if you consider the polynomial to be defined as points on the circle with certain spacing between them, then you can rearrange those points without changing the properties of the circle. QED[/hide]

Alternate solution to Problem 3 (just because I didn't get the whole 'Frank' thing):
[hide]
As Cicatriz, rearrange the sides so they alternate between 2 and 3. Since the hexagon is now symmetric, it's not hard to show that all its angles are the same, and of course you only have to realize that it's true in a competition such as the ARML. So all angles in the hexagon are $120^\circ$.

If we let $O$ be the center of the circle, and we call our new vertices $A'$, $B'$, etc. in clockwise order around the edge, look at triangles $A'B'C'$ and $A'OC'$. We know $A'B' = 5$ (without loss of generality) and $B'C' = 3$, and angle $A'B'C'$ is $120^\circ$, so we can find by the Law of Cosines that $A'C'$ is 7.

On triangle $A'OC'$, on the other hand, we have that $A'O = OC' = r$ for $r$ the radius of the circle, and (again by symmetry) angle $A'OC'$ is $120^\circ$, and $A'C'$ is 7. Again using the law of cosines:
$49 = x^2+x^2-2x^2\cos{120^\circ}$
$49 = 2x^2 (1 - \cos{120}) = 2x^2 (\frac{3}{2}) = 3x^2$
$\frac{49}{3} = x^2$
$x = \frac{7\sqrt{3}}{3}$.[/hide]

I have alternates for 3 and 7. (Unlike what was posted, i rearranged on #7 but not #3)
[hide="3"]
$\angle EBC =\angle BCF =\angle BEF =\angle EFC=60^\circ$ because each inscribed angle intercepts 1/3 of the circle. Drawing in $BE$ and $CF$ creates equilateral triangles which reveal $BE=CF=8$. Using the Law of Cosines, $BF=\sqrt{8^2+3^2-8\cdot 3}=7$.

Since three points uniquely determine a circle, the circumcircles of $ABCDEF$ and $BEF$ are the same. By the Extended Law of Sines, this is equal to $\frac{7}{2\sin 60^\circ}=\boxed{\frac{7\sqrt{3}}{3}}$.
[/hide]
[hide="7"]
Rearrange the sides into quadrilateral $PQRS$ with $QR=CD$, $PQ=RS=1$, and $PS=4$ and is still a diameter. Then $PR=QS=\sqrt{15}$ by Pythagorean Theorem. Ptolemy gives $4\cdot CD+1=15\Rightarrow CD=\boxed{\frac 72 }$.
[/hide]

[hide="Answers to number 6 and number 8"]
6. $2$

8. $5$
[/hide]

I can't solve #8 (although I can confirm the answer). How do you do it?

EDIT: Same thing for #6.

Ok, here are the solutions.

[hide="6"]
Draw line $l'$ that goes through $E$ and is parallel to line $l$. $\overrightarrow{BF}$, $\overrightarrow{CG}$, $\overrightarrow{DH}$ meet $l'$ at $B'$, $C'$ and $D'$, respectively.

Since $l$ and $l'$ are parallel, $EB': B'C': C'D' = AB: BC: CD=2: 1: 3$.
Applying mass points or Menelaus Theorem to $\triangle EC'P$, we get $C'G: GP=1: 2$.

Note that $PC'$ is the median from $P$ to $ED'$ in $\triangle PED'$.
Since $G$ divides the median into $1: 2$, $G$ is the centroid of $\triangle PED'$.

Hence $EG: GH=2: 1$, so $GH=\boxed{2}$

[/hide]

[hide="8"]
Let $S$ be the point on $BC$ such that $SO\perp PR$.

Since the slope of $PR$ is $\frac{-\frac{25}{6}}{\frac{25}{8}}=-\frac{4}{3}$, the equation of $\overleftrightarrow{SO}$ is $y=\frac{3}{4}x$.

Note that $SO=PR$. (For proof, consider a pair of congruent right triangles containing $SO$ and $PR$)
Then $SO=PR=\sqrt{\left(\frac{25}{8}\right)^2+\left(\frac{25}{6}\right)^2} =\frac{125}{24}$.

Letting $S(4a, 3a)$, we have $5a=\frac{125}{24}\Longleftrightarrow a=\frac{25}{24}$, so $S\left(\frac{25}{6},\frac{25}{8}\right)$

Then the equation of $BC$ (= the equation of $QS$) is $y-4=\frac{\frac{25}{8}-4}{\frac{25}{6}-3}(x-3) \Longleftrightarrow y=-\frac{3}{4}x+\frac{25}{4}$

Since the distance between $BC$ and $O$ is equal to the side length of the square, the answer is $D=\frac{\frac{25}{4}}{\sqrt{1^2+\left(-\frac{3}{4}\right)^2}} =\boxed{5}$

[/hide]

What is mass points? What is Menelaus's Theorem?

Menelaus' Theorem:
http://mathworld.wolfram.com/MenelausTheorem.html

For mass point geometry, search the forum :)


[hide]
The scale factor is uniform in all dimensions, so the volume of the liquid is $\left(\frac{3}{4}\right)^{3}$ of the container.

The remaining section of the volume is $\frac{1-\left(\frac{3}{4}\right)^{3}}{1}$ of the volume, and therefore $\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height when it is consolidated at the tip.

So, the liquid occupies $\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height, or $12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)$

$\boxed{m+n+p=052}$
[/hide]

[quote=4everwise]A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the liquid is $m-n\sqrt[3]{p},$ where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m+n+p.$[/quote]

Very late revive, but the fourth sentence is incomplete (I think). It doesn't make sense, although TZF's solution does answer the question.


inequalities
number theory proposed
number theory
Kazakhstan

Find all integer solutions $ (a_1,a_2,\dots,a_{2008})$ of the following equation:

$ (2008\minus{}a_1)^2\plus{}(a_1\minus{}a_2)^2\plus{}\dots\plus{}(a_{2007}\minus{}a_{2008})^2\plus{}a_{2008}^2\equal{}2008$

We use from Cauchy-Swartz inequality -->$a_1=a_2=...=a_(2008)=1$

We use from Cauchy-Swartz inequality -- > Answer: $a_1=2007, a_2=2006 ... a_{2008}=0$


When you fold a piece of paper, can you naturally assume that it will form two symmetric shapes divided by the fold, hence corresponding parts are congruent? (as in a proof)

Or, do you have to prove the above statement geometrically?

I would say that the answer to your question depends on context.  Certainly, in most applications, if the problem said "fold a sheet of paper in half", I would assume that means folding a standard rectangular sheet so that you get two smaller congruent rectangles (probably folding the long edges, but possibly not).  However, there are many problems which will provide more information that makes it clear this is not happening.  Sometimes they fold triangular pieces of paper so that a vertex ends up on the opposite side.  Sometimes they fold a rectangular piece of paper so that opposite vertices coincide.

If you have a particular problem you are wondering about, we may be able to be of more help.

This seems to be related to the USAMTS...

I am discussing a general concept rather than a specific problem.

You would have to prove it.  You risk losing points if you don't.

Thanks. I will do so.

The problem took me about 5 minutes to find the solution, yet an hour+ to type up the work. :evil:


quadratics Vieta algebra polynomial SFFT articles videos

[quote="lkryptonicl"]3. The real numbers x and y satisfy the equations $y(2x-3y)=19$ and $x(3x-4y)=94$. Evaluate $|x-y|$.[/quote][hide]$3x^{2}-4xy=94$, $2xy-3y^{2}=19$.. subtract to get $3x^{2}-6xy+3y^{2}=75$, $(x-y)^{2}=25\Longrightarrow |x-y|=5$[/hide]

[hide="2"]$\frac{10-\sqrt{100-12m}}{2m}=\frac23\cdot\frac{10+\sqrt{100-12m}}{2m}$

$30-6\sqrt{25-3m}=20+4\sqrt{25-3m}$

$10=10\sqrt{25-3m}$

$m=8$

The sum of the roots is $-\frac{-10}8=\boxed{\frac54}$.[/hide]
[hide="3"]$2xy-3y^{2}=19;\,3x^{2}-4xy=94$

Subtract the first equation from the first to get $3x^{2}-6xy+3y^{2}=75$.

Then, divide by 3 and factor the LHS to get $(x-y)^{2}=25$.

Finally, take the square root of both sides to get $|x-y|=5$.[/hide]

[quote="lkryptonicl"]1. The equation $2x^{4}-3x^{3}-14x^{2}-22x-8=0$ has two real and two complex solutions. What is the product of the two complex solutions?

2. One root of $mx^{2}-10x+3=0$ is $\frac{2}{3}$ of the other root. What is the sum of the roots?

3. The real numbers x and y satisfy the equations $y(2x-3y)=19$ and $x(3x-4y)=94$. Evaluate $|x-y|$.[/quote]
[hide="1"]
Using the RRT and guessing and checking, we see that 4 and -.5 are roots.
Dividing, we get the quadratic $x^{2}+2x+2=0$
Using vieta's, we see that 2 is the answer.
[/hide]

[quote="diophantient"]
Using the RRT and guessing and checking, we see that 4 and -.5 are roots.
Dividing, we get the quadratic $x^{2}+2x+2=0$
Using vieta's, we see that 2 is the answer.
[/hide][/quote]

Is there an easier way to do it rather than using RRT?

[hide="2"]
Let the two roots be $a$ and $\frac{2}{3}a$.

Using Vieta's, $a+\frac{2}{3}a=\frac{5}{3}a=\frac{10}{m}$ and $a\cdot \frac{2}{3}a=\frac{2}{3}a^{2}=\frac{3}{m}$.

So $\frac{5}{3}a=\frac{10}{3}\cdot\frac{2}{3}a^{2}$, which gives $a=\frac{3}{4}$, and thus the sum of the roots is $\frac{5}{3}a=\boxed{\frac{5}{4}}$.[/hide]

[quote="diophantient"][quote="7h3.D3m0n.117"][quote="t0rajir0u"]The [b]Rational Root Theorem[/b] states that if a polynomial $P(x) = a_{n}x^{n}+...+a_{0}$ with integer coefficients has a rational root, that rational root must be of the form $\frac{p}{q}, (p, q) = 1$ where $p | a_{0}, q | a_{n}$.[/quote]

Oh I thought it was another trick like SFFT.  :lol:[/quote]
What's SFFT?[/quote]

Simon's Favorite Factoring Trick. AoPS Wiki it and at the bottom of the article Mr. R explains it in a video!

[quote="7h3.D3m0n.117"][/quote][quote="diophantient"][quote="7h3.D3m0n.117"][quote="t0rajir0u"]The [b]Rational Root Theorem[/b] states that if a polynomial $P(x) = a_{n}x^{n}+...+a_{0}$ with integer coefficients has a rational root, that rational root must be of the form $\frac{p}{q}, (p, q) = 1$ where $p | a_{0}, q | a_{n}$.[/quote]

Oh I thought it was another trick like SFFT.  :lol:[/quote]
What's SFFT?[/quote][quote="7h3.D3m0n.117"]

Simon's Favorite Factoring Trick. AoPS Wiki it and at the bottom of the article Mr. R explains it in a video![/quote]
uhh, isn't it just foil?

Uh, what are you talking about? Simon's is just foil in reverse.

(x-a)(y-b)=xy-ay-bx+ab
(x+a)(y+b)=xy+ay+bx+ab

then put in whatever you want for a and b
the main crux of a problem involving sfft is adding something to both sides so it will work

So does anyone have a nice solution for #1?

Yeah, RRT seems like the way to go. It isn't that cumbersome.

Without RRT for #1, you can group the real and complex factors like \[(x^{2}+px+q)(x^{2}+p'x+q')=2x^{4}-3x^{3}-14x^{2}-22x-8=0\]
and go from there (no I'm not making a garbage post, I've seen a method before for doing this...)

$2x^{4}-3x^{3}-14x^{2}-22x-8=(2x+1)(x^{3}-2x^{2}-6x-8)=(2x+1)(x-4)(x^{2}+2x+2)$.
We can use the Horner diagram to get this.
Then, use Viète theorem we get the product of the two complex roots is 2. :D


[b]The Question:[/b]

Do we say a function attains its maxima/minima even if it is discontinuous at that point.My teacher said yes, but in a question in my book, there is a question in which the function is discontinuous at a point(minima point for both sub-functions) and the LHL is lesser than the RHL at that point but the answer says the minima is indeterminate. :S

Check out section 6 of this link: http://www.geocities.com/pkving4math2tor3/3_app_of_the_der_part_1/3_02_crit_pt_and_extrema.htm

It all depends on the value the function takes at the points of discontinuity. The fractional part function $ \begin{cases}f(x)\equal{}x\minus{}k& k\le x<k\plus{}1\end{cases}$ takes on its minimum value of zero at each integer, and has no maximum. The least upper bound of its values is $ 1$, but this is never achieved.


advanced fields
advanced fields unsolved

In 1996, Nesterenko showed that $ \pi$ et $ e^{\pi}$ are algebraically independant. The proof is quite complicated (and shows a more general result).
Do you know a (more direct/simpler) proof?

I see no reason why such a result should have a direct or simple proof. As far as I know, it is still an open problem whether $ \pi$ and $ e$ are algebraically independent over $ \mathbb Q$.


Find the limit: $ \lim_{n\rightarrow \infty}{\left(\dfrac{\sqrt{1\plus{}3n^2}}{7n^2\minus{}2}\right)^{\frac{\sqrt[3]{n^2\plus{}2n}}{3n\minus{}5}}}$

hello, with the rules of De L'Hospital i got $ 1$ as the searched limit.
Sonnhard.

To get an idea for what this does, think in very rough terms. The expression inside parentheses is like $ c/n$, while the power is like $ d/n^{1/3}$, where $ c, d$ are constants. Actually, forget about $ c, d$ for the moment. What does $ (1/n)^{1/n^{1/3}}$ do as $ n\rightarrow\infty$?


blogs AMC AIME USA(J)MO USAMO videos MATHCOUNTS

Hapy New Year to everyone.

What are your goals for this year?

err why don't we get off the road to spamworld?

my goals:
I just wanna get to MOP.....other than that, maybe getting 1000 songs on my iPod? dunno

edit:I forgot
Happy new years everybody!

Yes Happy New Year to all.

My Goals: Don't be stupid as often, make fewer careless errors, and continue being happy with life.

dibbs, i was the first one to post after 12:00am aops time in a blog or post! yay me!
-jorian

[quote="Ignite168"][quote="PenguinIntegral"][quote="Ignite168"]goals are gay[/quote]
you are gay

EDIT: Hinder is gay also[/quote]

you are gay with metal bands like iron madien[/quote]

I really don't think you should say things like that, especially not on a public forum.

Anyway.  Yes my goals for next year are, uh, to get good grades and to
[img]http://www.flatrock.org.nz/topics/working/assets/get_rich_quick.jpg[/img]

I swear I was not paid for this.

Here in Central Time (UTC-6), we still have ~45 min before 2007.

Goals:

(1) Increase AIME score to 10 and make USAMO

(2) Win a gold medal at Science Olympiad (http://www.soinc.org)

(3) Win my state piano competition

(4) Make USABO semifinals (it's my first year)

(5) Post more...

[quote]EDIT: Hinder is gay also[/quote]

Didn't we already have this discussion?

And homophobia bad.

Goals; keep up my GPA, take 2 or 3 AP's this year (probably not), be more time efficient, and be smarter. Cut video games (though my post christmas attempt to do so only led to a very distracted brain).

Make RSI, MOP...well those are goals, not resolutions, though I hope one of them comes true...
I have lots of goals in different things...
Practice violin a good amount even though it's track season
Play Go and pass 10k by the end of the year (or try to get within a few stones of my friend who introduced me to the game).  Come to think about it, that should come pretty quickly, so maybe 7k is a good goal.

oh, I didn't see this topic...

goals:

Get at least a 6 in AIME.
Make Mathcounts Nationals.
Get at least 5 in NYSML Individuals.
Get to Spelling Bee States.
Don't flunk HMMT.
Be able to do at least 1 USAMO problem.

Anyway, Happy New Year to all who are not in a timezone that is still in 2006.

Happy new year.

I don't do MC. Our school and most of our middle schools don't do it so yeah.. never heard of it till this year.

Goals:

Make AIME and get an AIME score of 4+

Place in the top 10 at all my in-state tournaments even though they aren't written well at all.

Top 3 on Alabama Statewide Contest.

umm.... yeah.... and help our team place at local competitions!!! we haven't yet because other people goof off and aren't good  :lol:

[quote="Ignite168"][quote="PenguinIntegral"][quote="Ignite168"]goals are gay[/quote]
you are gay

EDIT: Hinder is gay also[/quote]

you are gay with metal bands like iron madien[/quote]
you are gay with rush limbaugh and hal turner.

AT THE SAME TIME

i would also like to note that approximately one third of your immediate family is also gay

[quote="PenguinIntegral"][/quote][quote="Ignite168"][quote="PenguinIntegral"][quote="Ignite168"]goals are gay[/quote]
you are gay

EDIT: Hinder is gay also[/quote]

you are gay with metal bands like iron madien[/quote][quote="PenguinIntegral"]
you are gay with rush limbaugh and hal turner.

AT THE SAME TIME

i would also like to note that approximately one third of your immediate family is also gay[/quote]

Wow nice insult.

HAPPY NEW YEAR!!!

List also what you plan to do that will change your life forever in 2007.

[quote="PenguinIntegral"][/quote][quote="Ignite168"][quote="PenguinIntegral"][quote="Ignite168"]goals are gay[/quote]
you are gay

EDIT: Hinder is gay also[/quote]

you are gay with metal bands like iron madien[/quote][quote="PenguinIntegral"]
you are gay with rush limbaugh and hal turner.

AT THE SAME TIME

i would also like to note that approximately one third of your immediate family is also gay[/quote]

you are gay with your father, al gore, and a monkey who is beating his meat

AT THE SAME TIME IN TWO REALITIES

i would also like to note that your mother does not love you and you should go watch bukkake

[quote="Ignite168"][/quote][quote="PenguinIntegral"][/quote][quote="Ignite168"][quote="PenguinIntegral"][quote="Ignite168"]goals are gay[/quote]
you are gay

EDIT: Hinder is gay also[/quote]

you are gay with metal bands like iron madien[/quote][quote="PenguinIntegral"]
you are gay with rush limbaugh and hal turner.

AT THE SAME TIME

i would also like to note that approximately one third of your immediate family is also gay[/quote][quote="Ignite168"]

you are gay with your father, al gore, and a monkey who is beating his meat

AT THE SAME TIME IN TWO REALITIES

i would also like to note that your mother does not love you and you should go watch bukkake[/quote]

you are gay with your father, al gore, and a monkey who is beating his meat, hal turner, rush limbaugh, your other father,  the ugly half of new york city,  three moose,  Hinder, a Mormon pastor, and a  piano

AT THE SAME TIME IN FOrTY-FOUR DIMENSIONAL SPACE In INFINITE REALITIES

I also add that all three of your gay fathers do not love you and the best you will be in life is a buffalo neuterer who will get third place in the world buffalo neutering competition and then commit suicide with a horse anus

You also have poor musical taste and are ugly

Hinder is also gay

You two may find enjoyment in insulting each other, but I personally would prefer if you did it somewhere else and without using the word "gay" in a derogatory manner.  I'm sure many others feel the same way about this.

Happy New Year.

chill out with the gay stuff. Besides no one cares if you're gay or not.

Those insults were funny, although not appropriate.

[quote="junggi"][quote="Piggypi314"]Ah yes well, I'm not actually.  Really, just because I don't believe in using certain words to insult other people does not make me gay.  I was trying to point something out that other people might not have seen, but oh well.  Sorry I brought it up, I'm sure.[/quote]
uh i thought mdk was talking to penguinintegral and ignite?[/quote]

yeah he was talking to them.


function
real analysis
real analysis unsolved

I need to come up with an example of a sequence of functions that is not continuous at any point, but which converges uniformly to a continuous function.  Any suggestions?

for every natural number $ k$, let $ f_k$ be defined by $ f_k(x)\equal{}\frac{1}{k}$ for $ x\in \mathbb{Q}$, $ f_k(x)\equal{}0$ for $ x\not\in \mathbb{Q}$. $ f_k$ is not continuous anywhere because for every real number there exist both rational and irrational sequences converging to it. the $ f_k$ however converge uniformly to $ f(x)\equiv 0$.


[color=darkblue]Se considera o coarda fixa $[AB]$ intr-un cerc dat $w=C(O,R)$
astfel incat $AB<2R$ si un punct fix $C\in (AB)$.
Notam $CA=a$, $CB=b$ si arcul mic $m$ definit de coarda data $[AB]$.
1. In functie de $a$, $b$, $R$ sa se determine $\min_{X\in m}CX$.
2. Sa se precizeze si sa se construiasca pozitia punctului de minim.[/color]

[hide="pozitie"]Pp. ca $a\leq b$. Atunci consideram cercul $C'(C,a)$, care intersecteaza a doua oara cercul $C$ in $Q$. Perpendiculara din $C$ pe $AQ$ intersecteaza cercul $C$ in $X$, astfel realizandu-se minimul.[/hide]
[hide="lungimea segmentului"]Se demonstreaza usor ca punctele $O$, $C$ si $X$ sunt coliniare. Ducem perpendiculara $OR$ pe $AB$.
$CX=OX-OC$
$CX=R-\sqrt{(2R-a-b)R}$[/hide]
sper sa nu fi gresit la vreun calcul.

Din inegalitatea triunghiului avem $OC+CX \geq OX$.
$OC$ si $OX$ sunt fixe, iar egalitatea se atinge cand $O,C,X$ sunt coliniare.
Oricum, partea grea deja o facuse deja mychrom.

[quote="Virgil Nicula"][quote="mychrom"]Pp. ca $a\leq b$. Atunci consideram cercul $C'(C,a)$, care intersecteaza a doua oara cercul $C$ in $Q$. Perpendiculara din $C$ pe $AQ$ intersecteaza cercul $C$ in $X$, [u][b]astfel realizandu-se minimul[/b][/u]. Se demonstreaza usor ca punctele $O$, $C$ si $X$ sunt coliniare. Ducem perpendiculara $OR$ pe $AB$. Deci $CX=OX-OC$ $\Longrightarrow$ $CX=R-\sqrt{(2R-a-b)R}$. Sper sa nu fi gresit la vreun calcul.[/quote][color=darkblue]O.K., insa cred ca ai gresit la calcule. Imi permit [u][b]o mica corectie [/b][/u]: puterea punctului $C$ fata de cercul dat este $p(C)=-CA\cdot CB=OC^{2}-R^{2}$ de unde rezulta $OC=\sqrt{R^{2}-ab}$. Asadar, $\boxed{CX=R-\sqrt{R^{2}-ab}}$. Asa este, insa poti totusi dovedi de ce [u][b]astfel se realizeaza punctul de minim [/b][/u]?[/color][/quote]

scuze :oops: , aveti dreptate, am gresit putin la calcule.
$CX$ intersecteaza cercul $C'$ in $M$, si se demonstreaza usor, folosind faptul ca $M$ este mijlocul arcului $AQ$ din $C'$.

[quote="perfect_radio"]
Oricum, partea grea deja o facuse mychrom.[/quote]
[color=darkblue]Asa este, Perfect_radio ! Problema nu era grea, insa cu numeroase implicatii in inegalitati geometrice. Daca notam $X_{m}\in w\cap (OC$, atunci $OC+CX_{m}=$ $OX_{m}=$ $OX\le OC+CX$ $\Longrightarrow$ $CX_{m}\le CX$, adica $X_{m}$ este punctul de minim.

[b]Remarca.[/b] Vezi problema de la http://www.mathlinks.ro/Forum/viewtopic.php?t=137067  care, cu ajutorul acestei probleme de extrem, se poate generaliza (in alta reformulare) de la centrul de greutate $G$ al unui triunghi $ABC$ la orice punct interior acestuia.[/color]


If $ \overline{AD}\equal{}\overline{DB},\;\overline{BE}\equal{}\overline{EC},\;\overline{AF}\equal{}\overline{FG}\equal{}\overline{GC}.$ Calculate $ x$.

Greetings  :D

As well, I'd like to see the procedure strokes.

Greetings.   :D

Please if anyone can get a solution is the welcome.

Greetings  :)

Using dot product EG.EA we have:
[0,a/2]-[c,0]=[-c,a/2]
[c/3,2a/3]-[0,a/2]=[c/3,a/6]
[-c,a/2].[c/3,a/6]=0 because EGandEA are perpendiculars.So a=2c and angleBEA=pi/4 and angleCEG.Uing analitycal geometry we find that EG interection with DF =P[c,a] so BAPC is a retangle. Angle EPC=pi/4 and angle DPA=atan(c/2a)=atan(1/4).
So angle X=pi/2-pi/4-atan(1/4)=pi/4-atan(1/4)

I hope you understand the solution.


Find $ t$ such that $ (t,5)$ lies on the line through $ (0,3)$ and $ (\minus{}8,0)$.

[quote="GameBot"]Find $ t$ such that $ (t,5)$ lies on the line through $ (0,3)$ and $ ( \minus{} 8,0)$.[/quote]
You are given 2 points, and 1 of them is a y-intercept. Therefore, it is quite simple to realize that finding the slope and taking the intercept creates the basic slope-intercept form, $ y \equal{} mx$. So:
$ \frac {y_2 \minus{} y_1}{x_2 \minus{} x_1} \equal{} \frac {0 \minus{} 3}{ \minus{} 8 \minus{} 0} \equal{} \frac {3}{8}$

So, we have:
$ y \equal{} \frac {3}{8}x \plus{} 3$

Now, input 5 for y:
$ 5 \equal{} \frac {3}{8}x \plus{} 3$
$ 2 \equal{} \frac {3}{8}x$
$ \frac {16}{3} \equal{} x$.

Therefore, $ \boxed{\frac {16}{3}}$.


probability
limit
logarithms
function
combinatorics unsolved
combinatorics

Can anyone help me solve this problem?

Taking a random graph G=(V,E) of n vertices such that V is any set of size n. V={1,2,...,n}.
For any two distinct vertices: v1,v2 belong to G, the edge (v1,v2) belongs to E with probability p and does not belong to it with probability q=1-p where 0<p<1 is arbitrary fixed. Show that the probability of G being connected converges to 1 as n->oo.

Thanks.

In the following a $ C_k$ denotes a connected component of order $ k$.

For a given subset $ S$ of $ V(G)$ with $ |S| \equal{} k$, the probability that $ S$ is a $ C_k$ is $ P(S\equal{}C_k) \leq (1\minus{}p)^{k(n\minus{}k)}$   (1)
because we have to ensure that none of the vertices in $ S$ is connected to a vertex which is not in $ S$, and we omit the conditions that $ S$ has to be connected.

Now, assume that the connected component of $ G$ with minimal order has order $ k > \frac n 2$. Then, since any two connected components are disjoints, it follows that there can be no more than one connected component, whicn means that $ G$ is connected.
Therefore $ G$ is disconnected if and only if there exist a $ C_k$ with $ 1 \leq k \leq \frac n 2$.

Let $ X_k$ be the number of connected components with order $ k$.
It follows from above that the probability that $ G$ is disconnected is
$ P_n \equal{} P(\sum_{k\equal{}1}^{\frac n 2} X_k \geq 1) \leq \sum_{k\equal{}1}^{\frac n 2} P(X_k \geq 1).$  (2)

Now, for each $ k$, we have
$ P(X_k \geq 1) \leq \sum_{|S|\equal{}k} P(S\equal{}C_k) \leq \binom n k (1\minus{}p)^{n(n\minus{}k)}$ from (1).

It follows that
$ P_n \leq \sum_{k\equal{}1}^{\frac n 2} \binom n k (1\minus{}p)^{n(n\minus{}k)}$
$ \leq \sum_{k\equal{}1}^{\frac n 2} (\frac {en}{k})^k (1\minus{}p)^{n(n\minus{}k)}$  from classical evaluation
$ \leq \sum_{k\equal{}1}^{\frac n 2} (\frac {en}{k} e^{\minus{}p(n\minus{}k)})^k$ since $ 1\minus{}x \leq e^{\minus{}x}$ for $ x>0$
$ \leq \sum_{k\equal{}1}^{\frac n 2} (ene^{\minus{}\frac {pn}{2}})^k$  since $ 1 \leq k \leq \frac n 2$
$ \leq \sum_{k\equal{}1}^{\frac n 2} ene^{\minus{}\frac {pn}{2}})^k$  for sufficientely large $ n$
$ \leq \frac {en^2}{2}e^{\minus{}\frac {pn}{2}}$.

From this last bound, it is clear that $ \lim_{n\rightarrow n} P_n \equal{} 0$ which means that the probability that $ G$ is connected has limit $ 1$, as desired.

I hope for not too many mistakes.

Pierre.

The argument given above goes a good deal further than was asked by the problem (down to $ p \equal{} \frac {c \ln n}{n}$, for $ c > 4$).  Actually, I think you can get all the way down to $ c > 1$ by making a few small adjustments to it (exercise: prove that for $ c < 1$ the graph is almost surely NOT connected)

When $ p$ is fixed, you can simplify the argument by making the following observation:

To show $ G$ is connected, it suffices to show that every pair of vertices has a common neighbor.

As far as I remember, the initial result from Erdös and Rényi states that for any pertinent function $ w(n)$ which diverges to $ \plus{}\infty$, almost every graph with $ p \equal{} \frac {\log (n) \plus{} w(n)}{n}$ is connected (which means that the probabilty that it is connected converges to $ 1$), and almost every graph with $ p \equal{} \frac {\log (n) \minus{} w(n)}{n}$ is disconnected.

Pierre.


The equation  $x^5+ax^4+bx^3+cx^2+dx+e=0$  has only real roots, namely $x_1<x_2<x_3<x_4<x_5 .\;$  Suppose that $z_1<z_2<z_3$ are the roots of $10x^3+6ax^2+3bx+c =0 .$   [b] Prove that [/b]
\[ \frac{x_5-x_1}{z_3-z_1}\le \sqrt{\frac{10}{3}}\; .  \]

With Vieta I get a new inequality, but it seems aimless to prove...
 we just need a connection for $f(x)$ and$f'(x)$ max -min of their roots


For every positive integer $ n$, let $ z(n)$ be the number of zeros at the end of $ n!$.  As $ n$ approaches infinity, what value does $ \frac{z(n)}{n}$ approach?

As usual, the number of 0s is the same as the power of 5 in the factorization, which is
\[ z(n)\equal{}\sum_{i\equal{}1}^k \left\lfloor\frac{n}{5^i}\right\rfloor\]
where $ k$ is the largest integer with $ 5^k\leq n$
Writing $ y\minus{}1\leq \lfloor y \rfloor \leq y$ and dividing by $ n$:
\[ \left( \sum_{i\equal{}1}^k \frac{1}{5^i}\right)\minus{}\frac kn \leq z(n)/n \leq \sum_{i\equal{}1}^k \frac{1}{5^i}\]
As $ n$ goes to infinity, the $ k/n$ term goes to 0, and the sums on each side become infinite geometric series so the answer is the sum of that series, $ 1/4$

Something I thought was somewhat interesting is that in binary, the answer 1.

[hide]$ z(n)\equal{}\left\lfloor \frac{n}{5} \right\rfloor \plus{} \left\lfloor \frac{n}{25} \right\rfloor \plus{} \left\lfloor \frac{n}{125} \right\rfloor \plus{}\cdots$, so asymptotically, we have $ z(n)\sim \frac{n}{5}\plus{}\frac{n}{25}\plus{}\frac{n}{125}\plus{}\cdots \equal{} \frac{\frac{n}{5}}{1\minus{}\frac{1}{5}}\equal{}\frac{n}{4}$, and thus, $ \frac{z(n)}{n}\sim \frac{1}{4}$.[/hide]

[quote="caffeineboy"]Something I thought was somewhat interesting is that in binary, the answer 1.[/quote]

That is exactly why the $ p$-adic exponential $ \exp(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {x^n}{n!}$ converges $ p$-adically for all $ x$ divisible by $ p$ if and only if $ p \neq 2$. (if $ p \equal{} 2$ you need $ 4 | x$)

[quote="JBL"]And what about base 12?   :)[/quote]

[hide="an informal argument"]If $ 3^k$ is the highest power of 3 dividing $ n!$ and $ 4^j$ is the highest such power of 4, then the number of zeros at the end of $ n!$ is $ \min(k,j)$, since a zero occurs if and only if a 3 is paired up with a 4 among the factors of $ n!$.

The highest power of 3 asymptotically approaches $ \frac {n}{3} \plus{} \frac {n}{3^2} \plus{} .... \equal{} \frac {n}{2}$, while the highest power of 4 approaches $ \frac {1}{2}(\frac {n}{2} \plus{} \frac {n}{2^2} \plus{} ... )\equal{} \frac {n}{2}$ (i.e. half the maximum power of 2). These two are equal, so the minimum is obviously $ \frac {n}{2}$.

Thus, $ \lim_{n\rightarrow\infty}\frac {z(n)}{n} \equal{} \frac {n}{2}$ [/hide]


geometry
circumcircle
incenter
geometry unsolved

Daer Mathlinkers,
let ABC be a triangle, 1 the circumcircle, I the incenter, 2 the circle passing through A, I and C,
B' the center of 2,
3 the circle passing through A, I and B,
C' the center of 3,
U a point on 2, V a point on 3 so that IV is orthogonal to IU.
Prove that B'U and C'V meet on 0.
Sincerely
Jean-Louis

Dear Mathlinkers,
sorry for the typos: read
Prove that B'U and C'V meet on 1.
Sincerely
Jean-Louis

[hide="Solution"]Denote E be the intersection of C'V abd B'U.
$ \angle AC'V\plus{}\angle AB'U\equal{}2\angle AIV\plus{}2\angle AIU\equal{}2\angle VIU\equal{}180^o$
$ \Rightarrow\angle AC'E\plus{}\angle AB'E\equal{}180^o$, consequently, A,B',C',E lie on a circle.
Since B',C' lie on the circumcircle 0 of triangle ABC, hence result.[/hide]


search
algebra
polynomial
abstract algebra
group theory
superior algebra
superior algebra unsolved

show that $ F_p$ is cyclic ($ F_p$=$ Z/pZ$ with p prime)

[quote="ZetaX"]Additive group: trivial.
Multiplicative group: use the search function.[/quote]

A little hint for the multiplicative group.
It's a group of order $ p \minus{} 1$.
First fact : if $ d$ is a divisor of this order, there are at most $ d$ elements $ x$ of the group such that $ x^{d} \equal{} 1$ (roots of a polynomial of degree $ d$).
Second fact : a finite abelian group is a direct product of cyclic subgroups, each of order a power of a prime. (Classical result.)
If $ q$ is a prime divisor of $ p \minus{} 1$, the first fact leads to a proof that there is only one cyclic factor (in the decomposition of our group) whose order is a power of $ q$.
Thus, the multiplicative group modulo $ p$ is a direct product of cyclic subgroups whose orders are mutually coprime, thus it is cyclic.

The second fact is a bit difficult to prove. Perhaps there exists a proof which doesn't rest on it. (I didn't use the search function.)

Friendly,
P.

P.S. Now I have used  the search function. Look at :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=root+primitive&t=27050


Let $a,b,c$ be positive numbers such that $abc=1$. Show that
\[ \frac{1}{4\sqrt{ab}-a}+\frac{1}{4\sqrt{bc}-b}+\frac{1}{4\sqrt{ca}-c}\geq 1  \]

(whenever the denominators are not zero)

Edit: [b][size=150]Ignore this problem![/size][/b]

[quote="phamduyhiep"]Let  $\frac{m}{n}$ = $\sqrt{a}$,$\frac{n}{p}$ = $\sqrt{c}$,$\frac{p}{m}$ = $\sqrt{b}$ with m,n,p>0.
We have to prove that
$\frac{m^2}{4mn-p^2}$+$\frac{n^2}{4np-m^2}$+$\frac{p^2}{4pm-n^2}$ $\geq$ 1
From the Cauchy-Schwarz inequality ,we have
($\frac{m^2}{4mn-p^2}$+$\frac{n^2}{4np-m^2}$+$\frac{p^2}{4pm-n^2}$)($4mn$+$4np$+$4pm$-$m^2$-$n^2$-$p^2$) $\geq$ $(m+n+p)^2$
So it is  enough to prove that $(m+n+p)^2$ $\geq$  $4mn$+$4np$+$4pm$-$m^2$-$n^2$-$p^2$, which is true .
We have equality iff a=b=c=1.
(The same idea:
 $\frac{1}{k\sqrt{ab}-qa}$ +$\frac{1}{k\sqrt{bc}-qb}$+$\frac{1}{k\sqrt{ca}-qc}$$\geq$ $\frac{3}{k-q}$ with k>q>0)[/quote]
$4mn-p^2$ may be negetive.........
so I think the original problem is false...

OK,you are right,zhaobin.But I think the original problem is true.I will try again.

[quote="zhaobin"]I don't think so..
let $b \to (\frac 1 {16})^-,a=1$[/quote]

You are right zhaobin  :blush:


Find all continuous functions from R to R such that :$ 9f(8x)\minus{}9f(4x)\plus{}2f(2x)\equal{}100x$.
have fun :)

Let $ g(x)\equal{}f(x)\minus{}\frac{5}{2}x$, then $ g$ continuous and $ 9g(8x)\minus{}9g(4x)\plus{}2g(2x)\equal{}0$. Plug in $ x\equal{}0$ to get $ g(0)\equal{}0$.

Now take any arbitrary real $ r$, and let $ a_n\equal{}g\left( \frac{r}{2^n}\right)$. Then $ a_0\equal{}g(r),a_1\equal{}g(r/2)$, and the equation and continuity imply
\[ 0\equal{}9a_n\minus{}9a_{n\plus{}1}\plus{}2a_{n\plus{}2},\ \lim_{n\rightarrow \infty} a_n\equal{}0\]
The recurrence equation implies
\[ a_{n\plus{}2}\minus{}(3/2)a_{n\plus{}1}\equal{}3(a_{n\plus{}1}\minus{}(3/2)a_n)\Rightarrow a_{n\plus{}1}\minus{}(3/2)a_n\equal{}3^n(a_1\minus{}(3/2)a_0)\]
\[ a_{n\plus{}2}\minus{}3a_{n\plus{}1}\equal{}(3/2)(a_{n\plus{}1}\minus{}3a_n)\Rightarrow a_{n\plus{}1}\minus{}3a_n\equal{}(3/2)^n(a_1\minus{}3a_0)\]
Isolating $ a_0$, we get
\[ a_0\equal{}\frac{2}{3}\left( \frac{a_{n\plus{}1}\minus{}(3/2)a_n}{3^n}\minus{}\frac{a_{n\plus{}1}\minus{}3a_n}{(3/2)^n}\right)\]
for all nonnegative integers $ n$. For arbitrarily large $ n$, the right hand side tends to 0, so $ g(r)\equal{}a_0\equal{}0$. Since this is true for all $ r$, we have $ g\equiv 0,\boxed{f(x)\equal{}\frac{5}{2}x}$.


Hello  :lol:
f(x,y) means a function with 2 variables

While searching the maximum value of f(x,y) can we put y=a?
Thanks  :wink:

As long as you search every value of a, yes.... I don't quite get what you're asking.

Suppose I take $f(x,y)=-(x-1)^{2}-(y-1)^{2}$. It's maximum is zero, and this is achieved when x=y=1. Had we fixed y=5, we would never have found this maximum.

OOOK, THIS IS THE HOLE EXERCISE:

[b]Exercice [/b]: Find the maximum of   [img]http://www.gnux.be/latex/data/e554d343fc343679a4e1491bdf0b039b.png[/img] with  [img]http://www.gnux.be/latex/data/f7f3e0b3522f7fc908b8b368a21aaed8.png[/img]

[b]Solution found in a book[/b]:
 [img]http://www.gnux.be/latex/data/517d91eaf0cf591a6ffd682039ead509.png[/img]
.........

And then they puted [img]http://www.gnux.be/latex/data/906a2168101e86d5465fab7bb79ee68e.png[/img] and they searched the maximum of :  [img]http://www.gnux.be/latex/data/858084a62f387c474c9f490fefdae8a4.png[/img] WITH k FIXED

See what I mean??
And thanks  :)

is there an other method to define the max of a function with two variables
cause lagrange multiplier is a litle bit complex

thanx

[quote="t0rajir0u"]$k$ may be fixed, but it is also [i]indeterminate.[/i]

Anyway, the general method is known as the method of [url=http://mathworld.wolfram.com/LagrangeMultiplier.html]Lagrange multipliers[/url].[/quote]
K, thnx :wink:


abstract algebra
induction
superior algebra
superior algebra unsolved

Let $ A$ be a Noetherian local ring, and $ I$ a proper ideal of $ A$; assume that $ \textrm{proj dim } I < \infty$. Prove that $ I$ is generated by an $ A$-sequence iff $ I/I^2$ is a free module over $ A/I$.

I assume that by an $ A$-sequence a regular sequence $ A$ considered as module over itself is meant.

[hide="Hint"]Refine the statement that a basis of $ I/I^2$ serves as regular sequence and vice versa.

Proof by induction on $ r\equal{}proj dim I$ where the ring $ A$ may change in the induction step as long as it is Noetherian and local. ;)

Use for $ r\equal{}0$ that a finitely generated projective $ A$-module is free.

The induction step is dominated by the mantra that $ proj dim(I/x_1I)\equal{}proj dim (I)\minus{}1$ for every regular element $ x_1\in I$.[/hide]


There is a simble question...can a sudoku be a magic square?

Does a magic square have main diagonals having all different numbers?
if not, the answer is obvious.

[quote="thatin"]Lets ignore the fact that usually the numbers must be different... :roll:[/quote]

Yes, ignoring that fact, it can definitely be a magic square.

now... the obvious question is... how many magic Su doku there are... :P


On their website, it says:

Applicant Informed of Admission Decision: Feb 18, 2009.

Are they going to inform us the decission through email or post mail ?

My mom told me that somebody from AMSP called today and said that I didn't submit the admission test, but I'm pretty sure I did (by checking my sent mails).  Though I resent my solutions anyway.  I hope that's not a (wry) euphemism for saying I'm rejected.

still no email indicating admission or the opposite for me. I hope I put down my email address correctly...

also, I submitted the check, personal information form, reference letters, and problems separately. is that what I was supposed to do?

Hmm...I got accepted, but the email mentioned the Santa Cruz site, while I was thinking of going to the Texas one...I don't remember putting any choice as to which one I would like to go to on the application though. Should this concern me?

[quote="hello123"]I was unaware of the due date extension and my solutions contained an error but I was accepted anyway. Does anyone know how to do number 9 on the admission test? It seems difficult to find a clean solution. Also is there a nice solution that gets you a simple answer instead of an ugly expression for 10 if you use Lagrange?[/quote]

If i remember which one was 9 correctly, the sum of squares is an invariant.

For number 10, was it really supposed to be 2009x^2009? If the polynomial had 2009 roots there would not be enough information, and also 2009, 8,...1 doesn't imply any obvious sequence...if it was 9, the problem would be trivialized though.

accepted :) How do you do number 8 and 10?

accepted as well.  :) my solutions were pretty sketchy, though; I was highly nonrigorous on some and couldn't solve #7 or #9 (I think those were the two) at all.


We, the attendees of Jenks Mathcounts do well in states. Almost every year we do well in the state of Oklahoma. We owe it all to our coach danb80. He gives us all the info and practice we need in order to make our Mathcounts skills stronger, memorable, and better. He is the source of our strong based team and for our reputation on the state. All of you guys, DonkeyKong, jonathanchou711, carphead0, mr.k_74, and more. Yeha.

BOOYAH!! GO MR. BEATY!!   U PROVIDE US WITH A LOT OF PRACTICES!!! W00T!!! GO MR. BEATY!!!!!    :gleam:  :first:  :thumbup:

raised eyebrow
 :huuh:  :read:
are you joking?
[size=200]seriously... [/size]

Nope! It's awesome. I like it. But it's only for OK students.

i assume by ok
you mean oklahoma
but this
has

[size=200]a
b
s
o
l
u
t
e
l
y

n
o

r
e
l
e
v
a
n
c
e
.
i
t

i
s

r
a
n
d
o
m

l
i
k
e

t
h
i
s

s
e
n
t
e
n
c
e[/size]

if this post is about
mr.b80
saying  [i]it's[/i] awesome is probably referring to mathcounts
 not this random "club"


This question does not make sense unless you define what you mean by a face of a graph.  Typically, "face" is taken to only have meaning for planar graphs, which $K_{5}$ is not.  (And whatever your definition is, it better not depend on the particular way that I draw $K_{5}$.)

oops.... :oops:

OK, here's the real question:

Find the number of faces of $K_{4}$.

Every three vertices defines a face, so its at least 4C3 = 4, and simply drawing it shows it is indeed 4.

[quote="1=2"]I can find more...

[hide="hint"]What would Euler say?[/hide][/quote]
Euler also says 4. V-E+F=2. V=4,E=6, so F=4.

oohh... :oops:

[hide="what I thought..."]I thought that you could join faces to make a new face...

stupid arithmetic error...[/hide]


trigonometry
geometry
algebra proposed
algebra

Find ALL solutions to the following equation:

$\frac{1}{2 sin10} = \frac{sinx}{sin(40-x)}$, where $0\le x \le 40$ in degrees.

Obviously $x=30$ is a solution.Because $f(x)=\frac{sinx}{sin(40-x)}$ is increasing we get that this is the only solution.

Thanks!

somehow.. i couldn't see that f(x) was increasing... :D

I forgot to mention.. This result came from CMO 1998, and it trivializes the following problem:

The triangle ABC has ∠A = 40 and ∠B = 60. X is a point inside the triangle such that ∠XBA = 20 and ∠XCA = 10. Show that AX is perpendicular to BC.

I know it's geometry, but it's related to algebra.. Right?  :D  :P


I have a BUNCH of friends on my User Control Panel.  Not enough to constitute the entire AoPS population, but a LOT of friends.  Where did these "friends" come from?  Some are users that I've never corresponded with.  How did my friends get chosen?  Since not ALL AoPSers are on my friends list, but many random ones are, it seems weird...

What's the point in friends and foes..is foe someone you hate in this case (I already know the definition)? Is friends someone you like?

I didn't know it was that hard to read
[quote]Friends enable you quick access to members you communicate with frequently. If the template has relevant support any posts made by a friend may be highlighted.[/quote]
and
[quote]Foes are users which will be ignored by default. Posts by these users will not be fully visible and personal messages will not be permitted. Please note that you cannot ignore moderators or administrators.[/quote]

For now though these features are not yet enabled, but they will be in the future.

Can you see other people's friends and foes, like you can on slashdot? (I hope people can't see other people's foes, unless you can only see the foes of the people who have made you their friend. But I do think having public friend lists is good.)

No, your friends and foes list is personal. Others cannot see it (or know that you have them on your list, either as a friend or as a foe).

I'm sorry.
I don't know where you got that quote from or anything. Friends and Foes--- I don't know

*PS. Are you teaching any classes?


quadratics algebra polynomial conics parabola

A quadratic polynomial satisfies $ f(x)>\equal{}0$ for all $ x$. $ f(1)\equal{}0$, and $ f(3)\equal{}3$. What is $ f(5)$?

[hide="Solution"]

Because $ f(1) = 0$, this means that $ (x-1)$ is a factor of the quadratic polynomial ($ ax^2 + bx + c$).

Therefore, we have

$ (x-1)(x-z)$ where $ z$ denotes the other root of the polynomial. If $ f(3)$ leaves a remainder of $ 3$, we can set up the following:

\begin{align*}
(3-1)(3-z) &= 3 \\
6-2z  &= 3 \\
z &= \frac{3}{2}
\end{align*}

Our polynomial takes the form of $ (x-1)(x-\frac{3}{2})$ Plugging in the value of $ 5$ for $ x$, we arrive at our answer of $ \boxed{14}$.

[/hide]

[quote="bigman"]A quadratic polynomial satisfies $ f(x) > = 0$ for all $ x$. $ f(1) = 0$, and $ f(3) = 3$. What is $ f(5)$?[/quote]

[quote="n0vad3m0n"][hide="Solution"]

Because $ f(1) = 0$, this means that $ (x - 1)$ is a factor of the quadratic polynomial ($ ax^2 + bx + c$).

Therefore, we have

$ (x - 1)(x - z)$ where $ z$ denotes the other root of the polynomial. If $ f(3)$ leaves a remainder of $ 3$, we can set up the following:
\begin{align*} (3 - 1)(3 - z) & = 3 \\
6 - 2z & = 3 \\
z & = \frac {3}{2} \end{align*}
Our polynomial takes the form of $ (x - 1)(x - \frac {3}{2})$ Plugging in the value of $ 5$ for $ x$, we arrive at our answer of $ \boxed{14}$.

[/hide][/quote]

There are a few mistakes found in your solution. First is how do you know $ z$ is the other root when the condition on the quadratic is $ f(x)\ge0$? If that's the case, a graph of this quadratic is a parabola crossing the x-axis twice. But if $ f(x)\ge0$, the parabola lies on or above the x-axis, which clearly indicates either $ f(x)$ has a double root or no real roots. :ninja:

Second is what do you mean by "$ f(3)$ leaves a remainder of $ 3$"? $ f(3)=3$, it's a value, there are no remainders from a certain quotient.

[hide="Hints"]Since $ f(1)=0$ and $ f(x)\ge0$ for all $ x$, let $ f(x)=a(x-1)^2$ for some constant $ a$. Use $ f(3)=3$ and voila, after a few algebraic manipulation you should get the value of $ f(5)$.
[hide="My answer is"]12[/hide][/hide]

Just to add to 10000th hint, you have $ y \equal{} a(x\minus{}1)^2$ because of the restriction that all outputs of $ x$ are either 0 or positive. So, if we had something like:

$ y \equal{} a(x\minus{}1)(x\minus{}2)$

This would not satisfy this requirement because if we put in $ x \equal{} 1.5$, it would give us negative number. So, it has to be $ (x\minus{}1)^2$.  :)


calculus
integration
LaTeX
parameterization
function
calculus computations

hey, im a first time poster, thought id try u guys out :)

i have these 2 integrals and must prove them. plz give me a hand

here is the first one

integral  :inf: to 0  x^(2a) * e^(-px^2) dx  where a and p are const

i must show that this equals (((2a-1)...5.3.1) / (2^(a+1).p^a)) .  (:pi: /p)^(1/2)

the second one is similair

integral  :inf: to 0  x^(2a+1) * e^(-px^2) dx  where a and p are const

i must show that this equals (n!)/(2a^(n+1))

i apologise for the keyboard characters, i suggest u guys add a helping hand to teach people how to use the html for math symbols, cos i got no idea.  but yeah any suggests or solutions to these proofs are more than welcome!!!

thx

Such problems belong here, in "Calculus Computations and Tutorials."

Also: learning LaTeX would be nice, but barring that, you should at least put enough parentheses in what you say to make it unambiguous. Here's what I think you said:

Show that

$\int_0^{\infty}x^{2n}e^{-px^2}dx = \frac{(2n-1)\cdots5\cdot3\cdot1}{2^{n+1}  p^a}\cdot\frac{\sqrt{\pi}}p.$

(I used $n$ instead of $a$ because the context makes it clear that this parameter is a natural number.)

In order to do this, you must use an integration by parts to create a reduction formula. You can either find the appropriate integration by parts directly in this problem, or you can use a substitution to turn it all into the Gamma function, where you can use the recursion for that function.

Let $u=px^2,$ which we rewrite as $x=\frac{\sqrt{u}}{\sqrt{p}}.$ Then $dx=\frac{du}{2\sqrt{p}\sqrt{u}}.$

$\int_0^{\infty}x^{2n}e^{-px^2}dx=\int_0^{\infty}\frac{u^n}{p^n}e^{-u} \frac{du}{2\sqrt{p}\sqrt{u}}$

$=\frac1{2p^n\sqrt{p}}\int_0^{\infty}u^{n-\frac12}e^{-u}du = \frac1{2p^n\sqrt{p}}\,\Gamma(n+\frac12).$

From there use the fact that ${\Gamma(\frac12})=\sqrt{\pi}$ and the recursion $\Gamma(z+1)=z\Gamma(z).$

And essentially the same substitution for the second problem, leading to Gamma of an integer rather than a half integer. $\Gamma(n+1)=n!.$

As for the symbols: it's LaTeX, not html. Look on this site for both a LaTeX tutorial and a forum for discussion LaTeX.

cheers buddy, that is very helpful

yes, i think i will take the time out to check out the Latex symbolling, it makes it a lot nicer to read.

just out of curiosity, where does the integration by part method lead to? i am a little unfamiliar with the gamma function. im checkin out this gamma function now.

Because i initally tryed using this technique, yet i found the integral that i started with was worse than the initial one.  ;)

In doing the integration by parts directly, here's what you want to do:

In $\int u\,dv = uv = \int v\,du$, you want to let $dv = x\,e^{-px^2}dx$ so that $v = \frac1{2p}e^{-px^2}.$ Then $u$ can be the remaining powers of $x.$

Try it. You'll find that it knocks the power of $x$ down by 2 for each step. The integral you get at the bottom depends on whether you have a chain of even powers or a chain of odd powers.

If it's the chain of even powers, you need to know that

$\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}2.$


depends on your state. i see you live in florida. last year, in pennsylvannia, a 40 could get you first place written. theres some states where 40 would get you something like 10th written

[quote="27r"]What score do people usually get on the state test to get in the top 4 and go to Nationals?[/quote]

Usually it depends on the results of the other people participating. You'll have to do be within say 5%-10% of the top scorers.

40 is first written? I got 38 last year in Texas and I didn't make nationals.

But on 40 tenth...I think that's only for states like California.

well in my state two years ago, 37 was first, 33 made it to nats.

[quote="abacadaea"]texas is a hard state, what place did you get?[/quote]

6th, but 5th written.

[quote="saintmichaelmathcounts"]minnesota is a fairly easy state, last year, 36 won, and a 34 got you into nationals, but watch out, we had to of the top ten at nationals, and we are relentless.[/quote]

In NY, 37 won, and fourth place tied with fifth with 34.

why is california such a tough state? i mean, we don't even do well at nationals. what was the winning score in 07. iv been trying to find that out. i know some years back, someone was saying that they got a 43 and got 6th.

[quote="unimpossible"]i would say unless you're in California, a 40 is a guarantee to get to Nationals.[/quote]

Texas also (possibly).

In MA, last year, a 37 got you last spot for Nats.

Same for Michigan.

I think a 36 at least should do it for Ohio (last year was a 34 cutoof).

I think in NJ, 39 is the borderline, you might get in, you might not... but 40 should just about gurantee you 4th.

Does it really matter? You should just be preparing and trying to do your best instead of asking what you need.

27r lol =).  I know where you live. and who you are. and etc.  But... florida -- generally in the middle.  It really depends on the test and the people taking the test xD.  Just do your best.  But aim for around 40 during practice.

Are the state rounds harder than the sect rounds? I'm new to this
[/i]

its a pretty sizable leap from chapters ("sects") to state

yes, I know you like to stalk people, mathcrazed. ._.

[quote="sumguy727"]Are the state rounds harder than the sect rounds? I'm new to this
[/quote]

They are supposed to get harder as they go along.
However, if chapter had lots of probability questions, for example, and you're not very good at them, but then state had more geometry questions, which you might be better at, you would probably think state was easier.

Texas is ultrahard...
you have to get like 45 to get first written

California is pretty hard too....and that's where I am =/

California is VERY competitive.  Last year the top 6 or so tied with 41, so they actually had to use tiebreaker procedures.

ra, you're reviving a lot of old topics  :roll:


Prove that if $a,b,c \geq 0$ then:
$ \sum \sqrt{a^4+a^2b^2+b^4} \geq \sum a \sqrt{2a^2+bc}$

Hojoo Lee

It is trivial to prove that sqrt(x^2+xy+y^2)>=sqrt(3)/2*(x+y) for x,y>=0. THus the LHS is at least sqrt(3)/2 sum (a^2+b^2)=sqrt(3)*sum a^2. From Cauchy, the square of the RHS is at most (sum a^2)(sum a^2+2bc)<=3(sum a^2)^2.