MATHCOUNTS AMC USA(J)MO USAMO AIME Putnam articles

Hey all, since 7th grade, I have participated in 4 different math contests. I have spent time studying for them, but I have never been announced a winner at any math contest, since I would do bad on the actual test. Since the end of 8th grade, I have dedicated an entire summer to doing math problems. In ninth grade, I was never even invited to a single math contest since there were more experienced people on the math club. However, I did finally see a faint light in the tunnel when I qualifyed for the AIME's in 2004. It showed that studying could pay off a little. Now, I am starting 10th grade, after another summer of preparing for math contest, (I have participated on AoPS, mathjams, read the AoPS Vol. 2 and vol. 1, and have done ALL of the AMC's (10 and 12 for every year), I would like to do well on the tests this year. However, I don't know how much effort it takes to qualify at a regional final. I really want to do well at State math contests, but since 8th grade I have lost my confidence in them. Could people please post stories of how they improved over the years at these math contests? If after so much studying, this year I still do not win any contests, I would lose all my hope in winning, since I have really tried. So what does it really take to win?

Well..keta..this makes the two of us, but i must say that I haven't put as much work into trying to do well in math contests like u have....but i really also would like to know..and get some inspiration..I do know one thing though...my math teacher (although i really don't like her that much) says the difference between an ordinary person and a hero is just 5 minutes becuase when an average person quits a hero will last for 5 more minutes and this quote gave me hope and inspiration to keep on going even though things don't seem to go the way you want them to. I have also wondered about whether this was all worth it towards the end...but I find that it is becuase its not just winning math contests or the getting into the highest math class...but the satisfaction of what you learn in math and even though i don't win, I still love math and knowing that I will make a great mathematican when I grow up or eventually i'll become one is what I'm going for....winning is just a short term thing...but doing what you love and succeeding in it in life stays longer. So I guess what I'm saying is don't give up..I know that I won't

there are lots of ppl in the country, and you have to admit it, some of them are smarter than you.  there are only so many that could get perfect papers in USAMO/IMO at grade nine/ten.  back on topic, do what you find what you do enjoyable...  it is wrong to participate in a contest for the sole purpose of wanting to win, because that will make you fraustrated and even depressed, which is not good.  It is not the end of everything if you dont win a contest.  Besides they dont mean much anyways.  just relax and have fun.

bugzpodder, I do agree that there are some who are smarter than others, and I think makes the different is whether you really like to do math during your free time or not. For me, I simply like to do math for fun, however, I feel that trying my hand at contests is not a bad thing. At our school, many people (not just a few) have won math contests, yet many of them still view me as simply obsessed with math. I feel that with enough hard work, only chance can influence one's performance. If so many people at our school can win, then so can I.  :cool: I feel that no matter how small the probabilities, if one works hard and attends enough math contests, they are bound to succeed at least once.

Generally, if you like to do math, you're going to do it a lot, making you better. :)  But I think that there isn't really a point where only chance can affect your performance on a math contest. You can always get faster at doing calculations, for one thing.

Keta, I am a twelfth grader with a story similar to yours. I participated in Mathcounts in grade school and never did exceptionally well, but the contests really sparked my interest in mathematics. When I entered high school and heard about the USAMO and IMO, I made it my goal to become a member of the USA IMO team. Freshman year I qualified for the AIME, but I only scored a 2. Sophomore year, I barely qualified for the USAMO, and I scored a zero. Junior year I got the highest score in Indiana on the AMC 12, a 10 on the AIME, and an 8 on the USAMO even though I had a bad first day when taking the USAMO. I also now always get top places at regional and statewide competitions. I know that I will almost certainly never be a member of the USA IMO team, but setting that high goal has led me to experience several lesser successes and to appreciate and love mathematics even more. I hope you will experience many similar successes in your own high school career, but most importantly, just keep doing what you love.

In my opinion, qualifying for the AIME is enough of an achievement. You've succeeded already, in my eyes at least. Keep competing, it can't hurt. Keep my sig in mind, too.

[quote="keta"]Hey all, since 7th grade, I have participated in 4 different math contests. I have spent time studying for them, but I have never been announced a winner at any math contest, since I would do bad on the actual test.[/quote]

Yeah, what the other participants said. The one point I would focus on is staying "loose" (relaxed) during the actual test, by taking everyone else's advice and remembering that the win/lose aspect of taking math competition tests is secondary to the personal growth aspect. Tell yourself, "I'm giving myself this challenge to grow as a person," and simply DON'T THINK ABOUT what the score might be while you are taking the test. You will do better, and you will enjoy it more. And the cool thing is, you'll get the personal growth, which is the best thing to get from competition participation.

Throughout most of school I would get pretty jacked/stressed over contests.  Until senior year, when I chilled out.  Didn't cram the week before events like AIME/USAMO (didn't touch a book the couple days before).  By far, those were my best results.  It's hard to get in that place where you're doing the problems for the joy of finding the answer rather than the competition, but most of the top students are there.  You find a lot of smart, hard-working stressed-out people coming in 10th-15th out of 50 who would be way better off finding a way to chill out.  The hard work should come weeks, months, or even years before the big dates.  The big dates should be fun.

It's kind of weird because on the actual competition, I do better than I do in practices.  And also, remember that your biggest rival is [color=red]yourself.[/color]

And unfortunately, I am just bad at REAL math competitions where you have limited time

Also, you've got to remember that a ton of it's luck. Luck luck luck luck luck. If you have bad luck on the day of the contest, you definitely won't do as well. Don't worry, you're luck will catch you someday and you'll get to your expectations.

[quote="Fiachra"]Okay, I admit I have no experience of these actual competitions. But in my experience it is possible to do far better on the day of the exam than you would ever do in practice, simply because you can focus harder on this once off basis with a definite goal. One reliable test of success in a math contest is when, afterwards, you find yourself able to do something you couldn't before. If so, then in my opinion it's worth it, regardless of position or score.[/quote]

I'm not sure about the first part. Many people do worse in competition than they do in practice, because of more pressure or whatnot. But, some people thrive under pressure too.

But the second part's definitely true. Learning something new from it is always worth it.

Really, when you look at the evidence, math contests don't really determine who will be the best at math.  Many great mathematicians (John Nash on the Putnam Exam, for example) have been beaten on contests by others who later turned out to be much worse at math than them.  Contest don't REALLY measure ability to do real-world, abstract, difficult mathematical research problems which no one has solved before.  Seriously, the only real purpose of contests is to have better credentials for being accepted to college or graduate school and/or to win a scholarship.

Each math contest detemines, to some extent, a certian skill(s) in math.

But, the ability, or the willingness to do serious mathematical research is not measured by the contest.

[quote="goodyfresh741"] Seriously, the only real purpose of contests is to have better credentials for being accepted to college or graduate school and/or to win a scholarship.[/quote]
I wouldn't say that. Contests are supposed to be for fun. Like some people find collecting belly lint fun. In the end, nobody [i]really[/i] cares who won and who lost. They're not something to sweat blood over.

[quote="goodyfresh741"]Really, when you look at the evidence, math contests don't really determine who will be the best at math.[/quote]

I'm not so sure I can agree, based on a [url=http://groups.google.com/groups?&threadm=47e1ef3e.0307242052.673d3777%40posting.google.com]sci.math newsgroup post[/url] I read once in which it was said that many of the best mathematicians on that newsgroup are Putnam winners, as was one Nobel Prize winner in physics (Richard Feynman) and a few Fields medalists. There have also been dead-tree articles about the productivity of the mathematics research careers of IMO winners as contrasted with unselected Ph.D. mathematicians, and the IMO winners do pretty well by the usual criteria of papers published or papers cited by other mathematicians. What "evidence" did you have in mind?

In any event, I sure don't think it is actively harmful to participate in math competitions. Not everyone does as well in competition math as in post-doc research math, and vice versa, but generally competition math participants learn a good bit of math, and a good bit of personal characteristics like not giving up easily, that are beneficial for a future research career.

[quote]Really, when you look at the evidence, math contests don't really determine who will be the best at math. Many great mathematicians (John Nash on the Putnam Exam, for example[/quote]

To add to token adult here are my thoughts:
Sure, Nash was not a Putnam Fellow, But Feynman, TT Wu, G. Wilson, R. Mills just  to name a few,  were.

And at least two (Wilson and Feynman)  won Noble prize. . But for every Putnam fellow,    there are thousand(s) of "non-Putnam" fellows.  So unless you can count a FEW THOUSAND(s) (or millions - depends on  how you define  "non winners)  for-example-non-Putnam-fellows-Nobel- prize winners"  (instead of just John Nash )  your "example"    is not  even close to   an "evidence". ..
(Consider some claiming no discriminate based on sex  - and the  presenting evidence:
[quote] " Among our 50,000 employees we did hire  Ms. Jane Doe  If you look at the evidence  we really dont have any Bias) [/quote]

IOW,  if one looks at the evidence like you are looking,  you will see that, say  the likelihood of a  Putnam fellow (or IMO participant)  to win a noble prize,  is MILLION TIMES more than an ordinary college student. .. Statistically speaking the evidence is better than most of the other things we take for granted. (For example, just because   George Burns  did not get cancer,  is no evidence that smoking is not bad for you)

This is NOT to say that math contests are an end to itself, or they are every thing. NOT at all.  Like  many things they do measure some thing, and there is   a strong  correlation that some contests may  make you a better mathematician. Of course, this does NOT mean   a contest winner is always better than a "non-contest winner  ... but  [b] no one claims  that anyway[/b]. Primary reason to enter those, contests, IMHO, is to enjoy... and in that process, if you meet nice people, increase your knowledge, better your chances of getting into a   good college ... why not.

But you are right, one has to keep the perspective.  Math contests are not every thing, but for those who enjoy  - Have fun.

Just my thoughts.

I mostly agree with Gyan - I think contests are supposed to allow meeting people with same passion, to exchange informations. It's like a spark, somehow, like a huge kick which move you forward, no matter what place do you get.  Low score in competition isn't bad because it's motivating to harder training [at least in my case, but, I guess, each of us, has something like an ambition].

Does anyone (esp USAMO qualifiers) have study habits that they can share?

In 10th grade, i scored a 490 on the PSAT math.  After studying for a couple of months, i could easily get an 800 on the practices.  i was informed about these contests, which was just two months ago.  And at first i looked at the questions for amc/aime/usamo and i was humbled really. i could answer 5-7 questions of the amc 12 back in  august.  But now, after studying i took the test again, and i came out answering almost 20 of the questions.  Studying pays off! Same thing happened on the aime, i couldnt answer a single question in august, but now im getting like 7 or 8.  Point is, studying pays off, it really does!

Confidence is really important on these tests.  You have to have a one on battle with these questions, killing as many questions as you can.   Treat it like a game, have fun with it.  Dont be intimidated or scared by any question.

winning is does not mean win, losing does not mean lose.  going to competition and winning it to assure oneself as good in math is a false speculation that one is good at math.   if one is good at math, one does not need to study math.

[quote="maximocapcom"]if one is good at math, one does not need to study math.[/quote]

Say what?!

Don't you study math? I think every professional mathematician on the face of the earth would disagree with the denotation of your statement. What connotation do you have in mind? How does one know who is good at math?


Consider a particle in a 1-D box with impenetrable walls at x=0 and x=L. Then the eigen function for this problem is of the form,
Psi(x) = sqrt{2/L}sin(n(pi)x/L)
This is always of odd parity. The question which is troubling me is that, in this case though the eigen function seems to be of odd parity mathematically, when see their eigen states, they are alternately symmetric and anti-symmetric i.e odd and even parity.
On the contrary, if the origin is shifted to the centre of the box in which case the particle is confined between -L/2 to +L/2, then there are 2 sets of eigen functions. One set of functions is of odd parity and other set of functions is of even parity. Since both are of same width, how can there exist such a difference? Could someone point out the flaw in my understanding?

The parity is defined with respect to some reference point...
keep this in mind, and figure out the reference points for the different configurations.
you will find that in each case the parity of the wavefunction will be the same, but with different reference points.


geometry
incenter
trigonometry
inequalities proposed
inequalities

Let $ I$ be the incenter of triangle $ ABC$ and let $ m_a,m_b,m_c$ be the lengths of the medians from vertices $ A,B$ and $ C$, respectively. Prove that

$ \frac {IA^2}{m_a^2} \plus{} \frac {IB^2}{m_b^2} \plus{} \frac {IC^2}{m_c^2}\leq\frac {3}{4}$

I think the correct statement is this : $\boxed{\ \frac {IA^2}{m_a^2}+\frac {IB^2}{m_b^2}+\frac {IC^2}{m_c^2}\ \le\ \frac 43\ }$ .

Since : $\{\begin{array}{cccc}
IA=\frac {bc}s\cdot\cos\frac A2 & \implies & IA^2=\frac {bc(s-a)}s  \\ \\
\begin{array}{cccc}
m_a^2=\frac {2(b^2+c^2)-a^2}4 \\ \\
2(b^2+c^2)\ge (b+c)^2\ \end{array}\| & \implies & m_a^2\ge s(s-a) \end{array}\|\implies$

$\implies\sum\ \frac {IA^2}{m_a^2}\ \le\ \sum\ \frac {bc(s-a)}{s^2\cdot (s-a)}=\frac {ab+bc+ca}{s^2}\le\frac {(a+b+c)^2}{3s^2}=\frac 43$ .


calculus
derivative
function
calculus computations

[u]X      f        f'[/u]
1    2     3
2    1     4

Given the table above, where f is differentiable, find the derivative of f(f(x)) at x=2.

I have the answer.  The answer is 12, but that doesn't help me, because I don't know how to get there.  Can anyone help me?[/u]

By the chain rule, the derivative of $f(f(x))$ is $f'(f(x))\cdot f'(x).$

Thus what you are looking for is $f'(f(2))\cdot f'(2).$

Since $f(2)=1,\ f'(f(2))=f'(1)=3,$ and $f'(2)=4.$

Thus the derivative is $3\cdot 4=12.$

This was an exercise in understanding composition of functions and the chain rule.

Thank you.  I get it, now.  I guess I need to brush up on the rules of derivatives.


Source: ACoPS
Show that $\frac{1}{\sqrt{4n}}\leq \left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\cdot\cdot\cdot\left(\frac{2n-1}{2n}\right)<\frac{1}{\sqrt{2n}}$.

[quote="ThienA2PctSchool"]This is from Canada Mo. But what's ACoPS?[/quote]
I found this pbm. in "Art and Craft of Problem Solving" book, and it didn't cite the source of the problem...


conics
ellipse
analytic geometry
geometry
rectangle
cyclic quadrilateral

[quote="dengmi"]Can a non-cyclic quadrilateral be inscribed in a non-circular elipse?[/quote]

Yeah, consider plotting it on the coordinate plane. You have four points on the quadrilateral and four unknowns in the conic section for an ellipse (and this is just the non-tilted kind), so you should be able to solve for the equation of the ellipse.

the quadrilateral has to be convex though

I think a non-square rectangle can be inscribed in an elipse.  First, inscribe the rectangle in a circle.  Then, shrink the axis that corresponds to the smaller side.  You now have an ellipse with an inscribed rectangle in it.

[quote="Hokkage"]A similar question: Can a cyclic quadrilateral be inscribed in a non-circular ellipse?
Or, Can [b]any[/b] cyclic quadrilateral be inscribed in a non-circular ellipse?[/quote]

Yeah, it just has to be a tilted ellipse, because then you'll have a bunch of variables (by the conic section formula) and you can just make it not a circle.

And Altheman is right, it has to be convex (obvious if you imagine it).


[quote="sen"]let $a$, $b$, $c$, $d$ be real numbers. Prove that the numbers $a - b^2$, $b - c^2$, $c - d^2$, $d - a^2$ cannot all be larger than $1/4$[/quote]

[hide="yeah"]
well obviously, a, b, c, and d must be between 0 and 1, otherwise it's a trivial case.

case 1: a = b = c = d

a - a^2's max value is 1/4, so if a = b = c = d, none of the following can be larger than 1/4 (they can all equal 1/4, however)

-=-=-=-=-

Case 2: a, b, c, d not equal, a > d

If a, b, c, and d are not equal, then one must be greater than another: assume that a > d. If a > d, then d - a^2 will be less than 1/4.

-=-=-=-=-

Case 3: a, b, c, d not equal, d > a

If a, b, c, and d are not equal, then one must be greater than another: assume that d > a.

If d > a, then a > b in order for a - b^2 to be greater than 1/4.
If a > b, then b > c in order for b - c^2 to be greater than 1/4.
If b > c, then c > d in order for c - d^2 to be greater than 1/4.
If c > d, then d > a in order for d - a^2 to be greater than 1/4.

Contradiction.
[/hide]

No Casework!

[hide]
The sum: (a-b^2)+(b-c^2)+(c-d^2)+(d-a^2)=a(1-a)+b(1-b)+c(1-c)+d(1-d)

The maximum value of the parabola a-a^2 is 1/4, and the same goes with the other terms.

So (a-b^2)+(b-c^2)+(c-d^2)+(d-a^2)<=4(1/4)=1

Since the sum is <= 1, then the terms cannot all be larger than 1/4
[/hide]

Instead of using AM GM (which is only defined for positive reals), you could just say the maximum value of the parabola $a-a^2$ is 1/4 etc.

[hide]
Assume that all the given expressions are greater than $\frac{1}{4}$.
Add all of them to obtain $a+b+c+d-a^2-b^2-c^2-d^2>1$
Taking everything to one side, we obtain
$\\1-a-b-c-d+a^2+b^2+c^2+d^2<0\\ a^2-a+\frac{1}{4}+b^2-b+\frac{1}{4}+c^2-c+\frac{1}{4}+d^2-d+\frac{1}{4}<0\\ \left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2+\left(d-\frac{1}{2}\right)^2<0$
This is not true, thus the conclusion follows.[/hide]
This problem seems familiar.
EDIT:Forgot to hide.
Masoud Zargar

It's one of the problems in [i]Mathematical Olympiad Challenges[/i].  Maybe that's why it is familiar  :?:

Yup. I've seen this there.
What's the source of this problem?

The source is supposedly.. from [b][u](andreescu)[/u][/b]?

but yeh boxedexe.. thats the most efficient and quick method to solve it..  :lol:

inequalitiess.. simplest things go a long way

$x^2 \geq  0$


function
induction
algebra proposed
algebra

Let $ f: \mathbb {R} \to \mathbb{R}$ be a function such that for all real numbers $x$ and $y$ with $x \neq y$, we have
\[ f \left(\frac{x\plus{}y}{x\minus{}y} \right)\equal{}\frac{f(x)\plus{}f(y)}{f(x)\minus{}f(y)}.\]
Prove that $ f(x)\equal{}x$ for all $x \in \mathbb{R}$.

$ y \equal{} 0$  $ \implies$  $ f(1) \equal{} \frac {f(x) \plus{} f(0)}{f(x) \minus{} f(0)}$

$ \implies$  $ f(x)f(1) \minus{} f(1)f(0) \equal{} f(x) \plus{} f(0)$  $ \implies$  $ f(x)[f(1) \minus{} 1] \equal{} f(0)[f(1) \plus{} 1]$

if $ f(0)\not \equal{} 0,f(1)\not \equal{} 1$  $ \implies$  $ f(x) \equal{} c$ , $ c\in\mathbb{R}$

if $ f$ is not constant function and $ f(0) \equal{} 0$ $ \iff$   $ f(1) \equal{} 1$

$ f(\frac {x \plus{} y}{x \minus{} y}) \equal{} \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)}$

Let $ x \equal{} yk(k\not \equal{} 1)$  $ \implies$  $ f(\frac {k \plus{} 1}{k \minus{} 1}) \equal{} \frac {f(yk) \plus{} f(y)}{f(yk) \minus{} f(y)} \equal{} \frac {f(k) \plus{} 1}{f(k) \minus{} 1}$

$ \implies$  $ (f(k) \plus{} 1)(f(yk) \minus{} f(y)) \equal{} (f(k) \minus{} 1)(f(yk) \plus{} f(y))$

$ \implies$  $ f(yk) \equal{} f(k)f(y)$

$ \implies$  $ [f(x \plus{} y) \plus{} f(x \minus{} y)]f(y) \equal{} [f(x \plus{} y) \minus{} f(x \minus{} y)]f(x)$

and $ f(1) \equal{} 1$  $ \implies$  by the induction

$ f(x \plus{} 1) \plus{} x \minus{} 1 \equal{} [f(x \plus{} 1) \minus{} x \plus{} 1]x$  $ \implies$  $ f(x) \equal{} x$  for $ x\in \mathbb{N}$

$ f(\frac {x \plus{} y}{x \minus{} y}) \equal{} \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)}$

Let $ x \equal{} \minus{} y$  $ \implies$  $ f(x) \equal{} \minus{} f( \minus{} x)$  $ \implies$ $ f(x) \equal{} x$  for $ x\in\mathbb{Z}$

and $ f(xy ) \equal{} f(x)f(y )$  $ \implies$  $ f( x ) \equal{} x$  for $ x\in\mathbb{Q}$

$ f(x)$ is an increasing function

$ \exists$ $ u\in\mathbb{R}$  $ f(u) > u$

We can $ u < r \equal{} f(r) < f(u)$ and $ r\in\mathbb{Q}$

Contradiction!  Similarly $ f(u) < u$

$ \implies$  $ f(x) \equal{} x$

[quote="Dimitris X"]Let $ f: \mathbb {R} \to \mathbb{R}$ is a function such that $ f \left(\frac {x \plus{} y}{x \minus{} y} \right) \equal{} \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)},\forall x \not \equal{} y$.Prove that $ f(x) \equal{} x,\forall x \in \mathbb{R}$.[/quote]
Let $ P(x,y)$ be the assertion $ f\left(\frac {x \plus{} y}{x \minus{} y}\right) \equal{} \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)}$

$ P(0,x)$ $ \implies$ $ f( \minus{} 1) \equal{} \frac {f(0) \plus{} f(x)}{f(0) \minus{} f(x)}$ and so $ f(x)(1 \plus{} f( \minus{} 1)) \equal{} f(0)(f( \minus{} 1) \minus{} 1)$. So, if $ f( \minus{} 1)\ne \minus{} 1$, $ f(x)$ is constant, which is impossible.

So $ f( \minus{} 1) \equal{} \minus{} 1$ and $ f(0) \equal{} 0$

Then $ P(x, \minus{} x)$ $ \implies$ $ f( \minus{} x) \equal{} \minus{} f(x)$ and so $ f(1) \equal{} 1$.

Then, comparing $ P(x,1)$ and $ P(xy,y)$, we get $ \frac {f(x) \plus{} 1}{f(x) \minus{} 1} \equal{} \frac {f(xy) \plus{} f(y)}{f(xy) \minus{} f(y)}$ and so $ f(xy) \equal{} f(x)f(y)$

From this, it's easy to conclude :
1) $ f(x) > 0$ $ \forall x > 0$ and $ f(x) < 0$ $ \forall x < 0$
2) Using then $ P(x,y)$ : $ x > y > 0$ $ \implies$ $ f(x) > f(y) > 0$ and so $ f(x)$ is an increasing function.
3) $ f(x) \equal{} x^a$ $ \forall x > 0$ such that $ \ln(x)\in\mathbb Q$

From 3) and 2), and since $ \{x > 0$ such that $ \ln(x)\in\mathbb Q\}$ is dense in $ \mathbb R^ \plus{}$, we get that $ f(x) \equal{} x^a$ $ \forall x\in\mathbb R^ \plus{}$

Then $ P(2,1)$ $ \implies$ $ 3^a \equal{} \frac {2^a \plus{} 1}{2^a \minus{} 1}$ whose unique solution is $ a \equal{} 1$ and so $ f(x) \equal{} x$ $ \forall x > 0$

And so $ f(x) \equal{} x$ $ \forall x$

[u][b]For cnyd :[/b][/u]
[quote="cnyd"]...$ \implies$  $ [f(x \plus{} y) \plus{} f(x \minus{} y)]f(y) \equal{} [f(x \plus{} y) \minus{} f(x \minus{} y)]f(x)$

and $ f(1) \equal{} 1$  $ \implies$  by the induction

$ f(x \plus{} 1) \plus{} x \minus{} 1 \equal{} [f(x \plus{} 1) \minus{} x \plus{} 1]x$  $ \implies$  $ f(x) \equal{} x$  for $ x\in \mathbb{N}$[/quote]

You cant use the induction simply here since you need at least $ 2$ initial values different and you just have one $ f(1)\equal{}1$. You should show first that $ f(2)\equal{}2$

[quote="cnyd"]...   $ f( x ) \equal{} x$  for $ x\in\mathbb{Q}$[/quote]

The request is about all $ \mathbb R$

$ f(2) \equal{} 2$ is really difficult to be proved i think!!!!

But $ \forall x>0$  $ \implies$  $ f(x)>0$

[quote="cnyd"]But $ \forall x > 0$  $ \implies$  $ f(x) > 0$[/quote]

You're right. So you got it.
I just wanted to say that some steps were missing in your demo. :)


The other angle of the triangle with 100 degrees in it is 180-100-40=40. angle A is 180-110-40=30 degrees.


inequalities
function
algebra
polynomial
superior algebra
superior algebra theorems

Hi

I'm probably far from the first one who thinks of this, but what inequalities can be proven in any ordered field

for instance : AM-GM can be written in a purely algebraic way :

edit :
$a_1^2\cdots a_n^2\leq (\frac{a_1^2+\cdots a_n^2}{n})^n$

(I realised that I still had to asume positivity, which can be dodged by taking all squares)

the Holder inequality in some cases can also be described in a purely algebraic way

Tricks like using homogenity or convex functions won't work here anymore.

If necessary : my definition of a field K that is ordered :

there is a subset $A\subseteq K\backslash\{0\}$ and every nonzero element in the field is either in A, or $-a$ is in A, but never both
A is closed under sum and multiplication
I say $a\geq b$ wen $a-b$ is zero or in A

Artin and Schreier proved that every correct polynomial inequality can be proved by writting it as sum of squares (of rational functions).
This should give that the power-mean, Holder, Rearrangement ... inequalities will still work.

do you have an url or some more specific information?
that's quite a powerful theorem!

ZetaX, forgive me my rudeness, but I cannot find anything on that topic?
I have found an Artin Schreier theorem concerning extensions that are algebraically closed, but nothing about inequalities.  Wikipedia doesn't help either.

The Artin-Schreier theorem about closures isn't that much appart.
http://www.hindawi.com/books/977594502X/B977594502X00028X.pdf (page 190) sketches the idea of the prove for the sum of squares (and names further results).
The idea of real closed fields applies for all ordered fields ( http://en.wikipedia.org/wiki/Real_closed_field ), thus the theorem should also work there.

Uhm I took a quick look at that pdf, essentially, is it true that you are claiming :

every polynomial over any ordered field, that is zero or more everywhere, in a finite number of variables, is always a sum of squares?

But even then, how does this help me proving correctness of something like AM GM in any ordered fields? :?

Couldn't you write AM-GM as sum of squares $s_i(x)^2$ ($x=(x_k)_k$) in the real closure of $\mathbb{Q}$ then, giving that the sum of the $s_i(x)^2$ as rational functions are positive in any real closed field (they must contain the real closure of $\mathbb{Q}$, I think, since all ordered fields have characteristic $0$)¿ And then using that every ordering of the real closure continues the ordering of the first field¿


Suppose $x_1,x_2,...,x_n$ are positive reals such that $\sum_{i=1}^n x_i=\sum_{i=1}^n \frac{1}{x_i}$.Prove that $\sum_{i=1}^n \frac{1}{n-1+x_i} \le 1$.

I think it have been posted.And it is an inequality of VASC.
the mothed is to prove the equvaite inequality:
If:$\sum_{i=1}^n \frac{1}{n-1+x_i} =1$
we have:$\sum_{i=1}^n x_i \le \sum_{i=1}^n \frac{1}{x_i}$
use submittion and try. :)


if $ a,b$ are the roots of $ x^2\minus{}2cx\minus{}5d\equal{}0$ and $ c,d$ are the roots of $ x^2\minus{}2ax\minus{}5b\equal{}0$.then prove that$ a\plus{}b\plus{}c\plus{}d\equal{}30$

[hide="Hint"]$ a^{2}\minus{}2ac\minus{}5d\equal{}0$ and $ c^{2}\minus{}2ac\minus{}5b\equal{}0$ adding both these and using $ (a\plus{}c)\equal{}(b\plus{}d)$,$ ac\equal{}25$ we get $ (a\plus{}c)^{2}\minus{}150\minus{}5(a\plus{}c)\equal{}0$.
Solving for $ a\plus{}c\equal{}15$ gives $ b\plus{}d\equal{}15$ and the desired result.[/hide]

[quote="aadil"]if $ a,b$ are the roots of $ x^2 \minus{} 2cx \minus{} 5d \equal{} 0$ and $ c,d$ are the roots of $ x^2 \minus{} 2ax \minus{} 5b \equal{} 0$.then prove that$ a \plus{} b \plus{} c \plus{} d \equal{} 30$[/quote]

It's not necessarily true.  It did not state that a and b are to be different from c and d.

For instance, $ \minus{} 5, \minus{} 5$ are the roots of $ x^2 \plus{} 10x \plus{} 25 \equal{} 0$ and $ \minus{} 5, \minus{} 5$ are the roots of $ x^2 \plus{} 10x \plus{} 25 \equal{} 0$, but $ a \plus{} b \plus{} c \plus{} d \equal{} \minus{} 20$.

Four equations in four unknowns:
[hide]

$ (x \minus{} a)(x \minus{} b) \equal{} x^2 \minus{} 2cx \minus{} 5d$

$ x^2 \minus{} (a \plus{} b)x \plus{} ab \equal{} x^2 \minus{} 2cx \minus{} 5d$

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

$ (x \minus{} c)(x \minus{} d) \equal{} x^2 \minus{} 2ax \minus{} 5b$

$ x^2 \minus{} (c \plus{} d)x \plus{} cd \equal{} x^2 \minus{} 2ax \minus{} 5b$

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

a + b = 2c
ab = -5d
c + d = 2a
cd = -5b

There are sets of four numbers that do add up to 30, and they can be seen at this website:

http://www.quickmath.com --> Equations, Solving, Advanced.  I typed in "a,b,c,d" vertically under "Variables."

[/hide]

Actually this a very similar type of question that came in IIT-JEE(2006) and it was stated that $ a,b,c,d$ are distinct numbers.Thats why i assumed  them to be distinct.

There are four solutions; two satisfy the desired condition, one is mentioned by Arrange your tan, and the last is the not-very-interesting $ a \equal{} b \equal{} c \equal{}d \equal{} 0$.


Solve for $ w$ and express as a common fraction: $ \frac{1\frac16}w\equal{}\frac{42}3$.

When you have proportions, you have to cross-multiply to solve. This gives us:

$ 3 \cdot \frac{4}{3} \equal{} 42w \implies \frac{12}{3} \equal{} 42w \implies \frac{2}{21} \equal{} w$.

I got a different answer.  We can simplify $ \frac{42}{3}$ and change the mixed number into a fraction so that we have
\[ \frac {\frac {7}{6}}{w} \equal{} 14\]
Multiplying each side by $ \frac {w}{14}$:
\[ w \equal{} \frac {\frac {7}{6}}{14} \equal{} \frac {7}{6\cdot14} \equal{} \boxed{\frac {1}{12}}\]


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

Given a quadrilateral $ABCE$, where $AB \perp BC$, $CE$ bisects $\angle{ABC}$, $BC$ extends to point $Z$, with $C$ between $B$ and $Z$, $EC$ bisects $\angle{ACZ}$ on line $BC$, and $\angle{CAB}=30$, what is the value of $\angle{BEA}$?

Ok, so my friend says to use basic trig to solve the problem...I favored Law of Sines...I think his work looks like he worked backwards from the answer, but I guess I'm not the one judging... :maybe: I'll give you notice about our solutions for each step...

(1) Create a square around $ABCE$, with $AB$ and $BC$ on adjacent sides.  Label the two new corners $D$ (close to $A$ and $E$) and $F$ (close to $C$ and $E$) and the intersection of $AC$ and $BE$ a point $G$. (me and him)

(2) If $BE$ bisects $\angle{ABC}$, then $\angle{ABE}=\angle{CBE}=45$.  If $EC$ bisects its respective angle, then $\angle{ACE}=\angle{ECF}=\frac{180-(90-30)}{2}=60$, and $\angle{CEF=30}.$ (me and him)

(3) Let $EF = \sqrt{3}$.  Then $CF=1, BC=\sqrt{3}-1, AB=3-\sqrt{3}, AD=2\sqrt{3}-3$, and $DE=\sqrt{3}$. (me and him)

-SPLIT-

(4) $tan \angle{DEA}=\frac{AD}{DE}=2-\sqrt{3}$.  Then, by taking $tan (3\angle{DEA})$ (and this is the step in question...where did he pull that rabbit out of? :ninja: ), it equals $1$ and thus $\angle{DEA}=15, \angle{BEA}=45-\angle{DEA}=30$ (him)

(4) Solve for $AC$ and $AE$ and apply the Law of Sines to get the same answer. (me)

Is one way better than the other :?:

[i am pretty sure you mean BE bisects angle ABC]

$E$ is the excenter of $\triangle ABC$, so $\angle EAC=\frac{180-30}{2}=75$, $\angle EBA=45$, so $\angle BEA=180-45-75-30=30$


[u][b]Problem:[/b][/u]k is a non-negative integer and the integers a1, a2, ... , an give at least 2k different remainders on division by n+k. Prove that among the ai there are some whose sum is divisible by n+k.

You may find a solution in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2734]this[/url] thread.


Sequence $a_{1}, a_{2}.....$ is defined recursively by $a_{1}=0, a_{2}=100, a_{n}=2a_{n-1}-a_{n-2}-3$.  Find the greatest term of the sequence.

[hide="Solution"]We have

\begin{eqnarray*}a_{n}-2a_{n-1}+a_{n-2}&=&-3\\ a_{n-1}-2a_{n-2}+a_{n-3}&=&-3\\ &\vdots&\\ a_{3}-2a_{2}+a_{1}&=&-3\end{eqnarray*}

Summing up all the equalities, and denoting $S=a_{1}+\dots+a_{n}$, we get

$S-a_{2}-a_{1}-2(S-a_{n}-a_{1})+S-a_{n}-a_{n-1}=-3(n-2)$

$-a_{2}-a_{1}+2a_{n}+2a_{1}-a_{n}-a_{n-1}=-3(n-2)$

$a_{n}-a_{n-1}=-3n+6-a_{1}+a_{2}=106-3n$

If $106-3n>0$, we'll have $a_{n}>a_{n-1}$, hence the sequence will increase, and vice versa, and $106-3n>0\implies n\leqslant 35$. Therefore the greatest element is $a_{35}$.
[/hide]
Is it necessary to find its value? Or just its index?

Can we find a closed form expression for a term $a_{n}$? I'm new to recursive sequences, and would like to find an explicit way to calculate an Nth term. Can you guys show me how to do this?

You have it third row from the bottom in my post titled "Continuation".

Gotcha. Is there a general method I could use to derive a close form?

Say you are given the expression $a_{n}= 3a_{n-1}+2a_{n-2}-9a_{n-3}$ IF you are given the first three values, say $a_{1}= 12$, $a_{2}= 30$, and $a_{3}= 33$

Something on the subject you can find here: [url]http://en.wikipedia.org/wiki/Recurrence_relation[/url]


If there is a circle inscribed in an isosceles trapezoid with bases 8 and 18, what is the area of the trapezoid?
All sides of the trapezoid are tangent to the circle (of course...)

let me try

[hide]
the leg is 4 + 9 = 13

draw height, which forms a right triangle with one leg as hypotenuse

sqrt(13^2 - 5^2) = 12

area is 13*12 = 156
[/hide]

The circle is inscribed, so it's tangent to the trapezoid.
Two tangents drawn from a point outside a circle are always congruent.  :)

Point of Tangency! :P

Should've known :blush:

 :D  :D  :D :alien:

[b]Also, if a trapezoid $ABCD$ HAS an inscribed circle.
Then we have $AB+CD=AD+BC$.[/b]


Fie $a,b,c \: \geq  0$ cu $abc = 2$ , demonstratzi inegalitatea:

$a^{3}+b^{3}+c^{3}\geq  a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$


can some one post some number theory problems?

thanks

There are plenty of problems in NT Olympiad Section, in the Resources also... Or just search  :wink:


Functional Analysis
real analysis
real analysis unsolved

Let $H$ be a finite dimensional Hilbert space, $n>1$ and $a_{1},a_{2},...,a_{n},b_{1},...,b_{n}$ in $H$ such that $\parallel a_{i}-a_{j}\parallel \geq \parallel b_{i}-b_{j}\parallel $. If the intersection of the closed balls $B(a_{i},r_{i})$ is non void, so is the intersection of the closed balls $B(b_{i},r_{i})$. What is happening in infinitely dimensional Hilbert spaces?

Harazi, is your proof for the finite-dimensional case short? I remember trying to prove it and coming up with this ugly thing which I didn't really check thoroughly.

As for the second question, it follows from the finite dimensional case and the weak compactness of a ball in an arbitrary Hilbert space that the intersection is still non-empty. I wrote this down in more detail in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=416746&sid=1553dfc3ee27d07c00c51bb00725e0ca#416746]another topic[/url] (se the last post).

P.S.

This Kirszbraun thingie is becoming increasingly popular:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=lipschitz&t=114701[/url]

No, the proof for the finite dimensional case is not short and I found it quite ugly, this is why I posted the problem. I am sorry, I didn't notice it was already posted.


linear algebra
matrix
combinatorics unsolved
combinatorics

(APMC 2001)  Find the largest possible number of subsets of {1, 2, ... , 2n} each with n elements such that the intersection of any three distinct subsets has at most one element.

I have some trouble with this problem. Please help!

Can you give me your ideas about this problem?
I don't have any ideas.

Here's a simple idea that leaves only a few special cases to consider.

The condition can be rephrased as saying that every pair of elements in $ \{1, 2, \ldots, 2n\}$ may belong to at most $ 2$ subsets in our collection. Now, the overall number of pairs is $ \binom{2n}{2}$ and every $ n$-element subset contains $ \binom{n}{2}$ such pairs, so we must have

$ k \binom{n}{2} \leq 2\binom{2n}{2}$

where $ k$ is the number of subsets satisfying the condition. This simplifies to

$ k \leq \frac {8n \minus{} 4}{n \minus{} 1}$.

This already tells us that the number of sets we can hope to achieve is a constant independent of $ n$; in fact, for $ n \geq 6$ the upper bound is just $ 8$.

To improve this upper bound, it is convenient to use a matrix representation (we won't use any linear algebra, it's just convenient to look at columns as well as rows): Given a collection of $ k$ subsets of $ \{1, 2, \ldots, 2n\}$ we can write down a matrix with $ k$ rows and $ 2n$ columns where the element in row $ i$, column $ j$ is $ 1$ iff the $ i$th set in the collection contains $ j$, and otherwise it is $ 0$. We call the number of $ 1$'s in column $ j$ the [i]weight[/i] of this column. The conditions of the problem state:

1) Each row has exactly $ n$ 1's. In particular the overall number of $ 1$'s is $ kn$, and the average weight of the columns is $ k/2$.

2) A pair of columns may not contain three (or more) $ 1$'s in the same location.

Note that columns of weight $ 3$ and up must be distinct, by condition 2.

Suppose now that we can find $ 5$ subsets satisfying our condition, namely $ k \equal{} 5$. Thus on average, each column must have weight $ 2.5$.

* Is there a column of weight $ 5$? If there is, then we can't have any columns of weight $ 3$ or $ 4$ since that would violate condition 2. Thus all other columns must have weight $ 2$ at most, and as soon as $ n \geq 4$ this forces the average below $ 2.5$. So $ n$ must be very small in this case.

* Is there a column of weight $ 4$? If so, there can only be $ 6$ columns of weight $ 3$ (those where the $ 1$'s are not all chosen at the same locations as those in the $ 4$-column). Even if we're able to put all $ 6$, all the other columns have weight $ 2$ or less which puts the average below $ 2.5$ once $ n > 8$.

* If there are no columns of weight $ 4,5$, we can have at most 10 distinct $ 3$-columns. The other columns have weight at most $ 2$ we can sustain the required average up to $ n \equal{} 10$.

[b]Conclusion[/b]: If $ n > 10$ then finding $ 5$ sets meeting the requirements is impossible.

Finding $ 4$ such sets, however, is trivial for any $ n$: just use $ A, B, \bar{A}, \bar{B}$ where $ A$ and $ B$ are arbitrary subsets of size $ n$ which are distinct and not the complement of each other. any three sets chosen from this small collection must include two complementary ones, so the intersection is empty.

We now have tight lower and upper bounds for $ n > 10$. For small $ n$ we have the following:

* When $ n \equal{} 3$ the upper bound is $ 10$, and it can be achieved.

* When $ n \equal{} 4$ the upper bound is $ 9$, I don't think it can be achieved.

* When $ n \equal{} 5$ (EDIT: had a mistake here earlier) I think I can show that $ 7$ sets cannot be achieved, so the upper bound is likely $ 6$ at this point and forward.

* When $ n \equal{} 10$ it's easy for find $ 6$ sets: just use all possible columns of weight $ 3$. I'm sure this is tight.

So, I still don't have a slick argument for either the lower or upper bound for $ n \leq 10$, but the margins are pretty narrow. I'll think about this some more.

Thank you very much, randomgraph. I will read it carefully and think about finishing the solution.

OK, one small correction: you can actually take $ 8$ sets for $ n \equal{} 5$. Here's the matrix, where I've marked off the numbers by rows to make things a bit clearer. It's easier to look down the columns in order to understand what's going on. I suspect this is a well-known error-correcting code or at least some block design.

$ \begin{pmatrix} 1 & 1 & & & & 1 & 1 & 1 & & \\
2 & & 2 & 2 & & 2 & & & 2 & \\
3 & & 3 & & 3 & & 3 & & & 3 \\
4 & & & 4 & 4 & & & 4 & 4 & \\
& 5 & 5 & 5 & & 5 & & & & 5 \\
& 6 & 6 & & 6 & & 6 & & 6 & \\
& 7 & & 7 & 7 & & & 7 & & 7 \\
& & & & & 8 & 8 & 8 & 8 & 8 \end{pmatrix}$

Right now I have precise values for $ n \equal{} 3, 5$ and $ n \geq 10$. Still missing a good argument to cover them all. BTW what does "APMC" mean? If this is taken from a competition, I must really be missing something fundamental since those ad-hoc constructions for small values of $ n$ are really not likely to be asked about in a live event.

APMC = Austrian-Polish Mathematical Contest.
Yes, the problem was given there... :(
EDIT: Wait, it was given with the additional assumption that $ n > 10$. Kalva probably dropped this as too simplifying :D

  darij

Aha!! I feel much better now. Thanks  :)

For $ n > 10$ the argument is really pretty short and simple. The special cases for $ n \leq 10$ can be handled by somewhat tedious case bashing - nothing too complex but definitely not suitable for competition. I guess kalva was having in mind people solving this "off line" with time on their hands, which makes this much more reasonable.

BTW you can take the matrix from my previous message and remove two complementary columns (say $ 2,3,5,6$ and $ 1,4,7,8$) to get $ 8$ sets for $ n\equal{}4$ as well. It's not hard to show that $ 9$ sets is impossible with $ n\equal{}4$, using the same reasoning I used above with a bit more case consideration. So $ n\equal{}3,4,5$ are done.

Yes, in the original APMC 2001 paper I remember that there was the condition n > 10. But in kalva this condition is dropped.

Nice discussion. Now, we look for the rest cases.


LaTeX inequalities calculus integration MATHCOUNTS logarithms

I found this in a very old math book, and I need help with it.

[quote]Without the use of a calculator, find the integer whose square root is nearest to (sqrt 2)^[(sqrt 3)^(sqrt 5)].[/quote]

I really don't know how to use Latex to create exponents with radicals or exponents with exponents. If anyone knows how, please tell me! With an entire page of algebraic manipulation, I was able to limit the expression between the square root of 8 and four. Therefore, the integer, which will be called N, is in the inequality 8 < N < 16. It was after this I became stuck. Does anyone know how to finish the problem?

so the question is find n such that $ \sqrt {n} = \sqrt {2}^{(\sqrt {2}^\sqrt {3})$ Is this a calculus book by any chance?

[quote="mathemonster"]so the question is find n such that $ \sqrt {n} = \sqrt {2}^{(\sqrt {2}^\sqrt {3})}$ Is this a calculus book by any chance?[/quote]

Actaully it would be $ \sqrt {2}^{\sqrt {3}^{\sqrt {5}}}$

Using a calculator I found the answer was
[hide]11[/hide]but I don't have the faintest idea how to do it without a calculator...

We could get to $ n =\sqrt2^{2\sqrt {3}^{\sqrt {5}}}$

You have it right math monster, but the square root of 2 is raised to the power of the square root of three, which in turn is raised to the power of the square root of five. Mewto55555 has it written correctly, although we are looking for hte integer that is closest to the above expression. The book in question is math book with difficult questions, but it has some calculus in it since it has sections on integration, Taylor series, and other calculus topics.

Ok then if it's so easy, please evaluate $ 1.73^{2.23}$

[quote]Well, if you were fresh out of 8th grade and had to memorize the mathcounts toolbook things, then this problem would be a piece of cake. (they make u memorize, the square r00ts of numbers)[/quote]

No, not really. Aside from the fact that I know how to [url=http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html]calculate square roots without a calculator[/url], there exists the problem of raising numbers to rational approximations.

I don't fully remember the title as I do not own the book, although I'm sure that it had something to do with "Mathematics Olymipics". There were no solutions in it due to it being a workbook.

A possible approach:

Let $ \sqrt {n} \equal{} x \equal{} \sqrt {2}^{\sqrt {3}^{\sqrt {5}}}$. Take the log of both sides to get $ \log x \equal{} \sqrt {3}^{\sqrt {5}} \log{\sqrt {2}}$. Let $ k \equal{} \sqrt {3}^{\sqrt {5}}$. Take the log of both sides: $ \log k \equal{} \sqrt {5}\log \sqrt {3}\equal{} (\sqrt{5}/2)\log{3}$. We know that $ \sqrt {5}\approx 2.236$, so use a log table to approximate the value of $ \log k$, and then find $ k$ in the table. Multiply $ k$ by $ (1/2)\log 2$ to find $ \log x$. From this, $ x$ can be found; square it to get $ n$.

But I guess this would give a bad approximation, and log tables (assuming you're allowed to use them) might not have the value you want...

It seems that the difficulty of the question relies on whether or not one can use a calculator...


Ring Theory
superior algebra
superior algebra unsolved

A ring $ R$ is said to be a regular ring if for every element $ a$ in $ R$, there exist at least one element (also),b in $ R$, so that $ aba\equal{}a$. Define $ Z(R)$ is a subring,so that every element in $ Z(R)$ commutes with all element in $ R$. Prove or disprove, if $ R$ is a regular ring, then $ Z(R)$ is also regular. Thank you for your attention.

See page $ 14$ of [url]http://www.ma.utexas.edu/mp_arc/c/03/03-181.pdf[/url]

Thank you, but can you explain more briefly (step by step) because I don't understand about the explanation in the link you have given to me :D

Prove that :
Each element in a group $ G$ can be expressed as a square of the other element in $ G$ (In other words, each $ b$ in $ G$, there exist $ a$ in $ G$ so that $ b\equal{}a^2$) if and only if the order of the group is an odd number.


Let M and N are subspaces of a vectorspace and lets assume that

dim(M+N)=dim(M intersection N) + 1

I need to prove that

either M C N and then dim(N)=dim(M) + 1
or N C M and then dim(M)=dim(N) + 1

Anyone has any idea how to solve the problem???

first of all you should post these problems in the [url=http://www.mathlinks.ro/Forum/index.php?f=346]matrix-linear algebra forum[/url]

for this question: suppose that $M\not\subset N$ then
$dim(M+N)>dim(N)$ so $dim(M+N)\geq dim(N)+1$ and obviously
$dim(N)+1\geq dim(N\cap M)+1$
so $dim(N)=dim(N\cap M)$ and $N\subset M$ ... :roll:

Sorry for posting here...I am kinda new and did not look carefully I guess. I will keep in mind for further postings.

on the solution:
I don't understand why
$dim(N)+1\geq dim(N\cap M)+1$
when it is rather the opposite as it is given that dim(M+N)=dim(MnN)+1 and when you substitute you get
$dim(N\cap M)+1\geq dim(N)+1$

[quote="terebide"]I don't understand why
$dim(N)+1\geq dim(N\cap M)+1$[/quote]

It's because $N\cap M\subset N$, so $\dim(N\cap M)\leq \dim(N)$.


Δίνεται τρίγωνο ABC και σημείο P του επιπέδου του .
Έστω ότι οι εφαπτομένες του περιγεγραμμένου κύκλου του τριγώνου ΑΒC sta B,C τέμνονται στο D και έστω ότι η CP τέμνει την AB στο Μ και η BP τη CA στο N .Δείξτε ότι τα σημεία D,M,N είναι συνευθειακά αν και μόνο αν το P είναι σημείο του περιγεγραμμένου κύλου του ABC .

Αν το σημείο $D,$ είναι το σημείο τομής των εφαπτομένων στις κορυφές $B,$ $C,$ του $%Error. "bigtriangle" is a bad command.
ABC$ όπως νομίζω, τότε τα πράγματα δεν είναι τόσο δύσκολα όσο φαίνονται (εφαρμογή διάσημου γνωστού θεωρήματος, που δεν φαίνεται με την πρώτη ματιά).

Κώστας Βήττας.

Μία παρόμοια πρόταση είναι η εξής:

Δίδεται τρίγωνο $\bigtriangleup ABC,$ εγγεγραμμένο σε κύκλο $(O)$ και έστω $BE,$ $CF,$ δύο τυχούσες σεβιανές του, οι οποίες τέμνουν τον $(O),$ τα σημεία $E',$ $F',$ αντιστοίχως. Αποδείξτε ότι οι ευθείες $EF,$ $E'F'$ και η εφαπτομένη του $(O)$ στο σημείο $A,$ τέμνονται στο ίδιο σημείο.

Κώστας Βήττας.

Ναι οι εφαπτόμενες είναι στα B,C .Συγγνωμη για την παράλειψη. :(

Να δώσω hint ?Είναι πολύ ωραία ως άσκηση συμφωνείτε ?(Και έχει και μια πάρα πολύ ωραία λύση .... :wink:


geometry
parallelogram
similar triangles

Let $ABCD$ be a parallelogram.  Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$  Given that $PQ=735$ and $QR=112,$ find $RC.$

[hide]

Considering $\triangle RBC \sim \triangle RDP$ and $\triangle RQB \sim \triangle RCD$,
by similar triangles, $\frac{BR}{RD}=\frac{QR}{RC}$ and $\frac{BR}{RD}=\frac{CR}{RP}$
$\frac{CR}{RP}=\frac{QR}{RC}$
\[ \begin{eqnarray*} RC^2 &=& QP \cdot RP\\ RC &=& \sqrt{112(735+112)}\\ &=& 308 \end{eqnarrya*} \]

[/hide]

I think it's easier with a picture (from Kalva)
[hide]We know $\triangle{PCD}$ is similar to $\triangle{CBQ}$, along with a lot others..., so calling $RC$ $x$, we have $\frac{x}{112}=\frac{847}{x}$ and $x=\boxed{308}$[/hide]

[img]http://www.kalva.demon.co.uk/aime/soln/diag986.jpg[/img]

[quote="Iversonfan2005"][img]http://www.kalva.demon.co.uk/aime/soln/diag986.jpg[/img][/quote]How come the image isn't showing up?? :huh:
Alternatively we could consider $ RQA\sim CQB$,$ PRD\sim CRB$

$ RQA\sim CQB\implies 735/(112 \plus{} RC) \equal{} PA/BC$

$ PRD\sim CRB\implies 847/RC \equal{} (PA \plus{} AD)/BC \equal{} (PA \plus{} BC)/BC \equal{} PA/BC \plus{} 1 \equal{} 735/(112 \plus{} RC)\plus{}1$

solving $ 837/RC \equal{} 735/(112 \plus{} RC)\plus{}1$ gives 308.

[hide="projective geometry overkill"]Let \(RC = x\). First, note that \(\triangle PAQ \sim \triangle PDC\). By projecting \(\overline{PC}\) onto \(\overline{PD}\), we have
  \[(Q,P;R,C)\overset{B}{=}(A,P;D,P_\infty).\]
  Computing cross ratios gives us the following.
  \begin{align*}
    \frac{QR}{PR}\cdot\frac{PC}{QC} &= \frac{AD}{PD} \\
    &= \frac{PD-PA}{PD} \\
    &= 1 - \frac{PA}{PD} \\
    &= 1- \frac{PQ}{PC} \quad (\text{since } \triangle PAQ \sim \triangle PDC).
  \end{align*}
  Substituting the values given in the problem and letting \(u = 112+x\) gives us
  \begin{align*}
    \frac{112}{735+112}\cdot\frac{735+112+x}{112+x} &= 1 - \frac{735}{735+112+x} \\
    \frac{112}{735+112}\cdot\frac{735+u}{u} &= \frac{735+u-735}{735+u} \\
    \Rightarrow u^2-2\cdot112u - 112\cdot735 &= 0
  \end{align*}
  Applying the quadratic formula then gives us
  \[u = \frac{2\cdot112\pm \sqrt{2^2\cdot112^2 + 4\cdot 112\cdot735}}{2} = 112 \pm \sqrt{112(112+735)}.\]
  Since \(u = x+112,\) we have that
  \[x = \sqrt{112\cdot847} = \fbox{308}.\][/hide]


Two regular polygons of the same number of sides have sides 48m and 55m in length respectively.
What is the length of the side of the another regular polygon of the same number of sides, if its area
is equal to the sum of the two ? Please explain how you got the answer. Thanks in advance.

[hide="most likely incorrect"]The area of the regular $ n$-gon is directly proportional to the square of its side length. Thus (where $ c$ is some constant), the area of two polygons are $ 48^2c$ and $ 55^2c$, respectively. The sum of their areas is $ (48^2 \plus{} 55^2)c$, and the $ n$-gon with this area has a side length of $ \sqrt{48^2 \plus{} 55^2} \approx
\boxed{71.13}$.[/hide]


Is it known whether $a^m = b^n + 1, a, b \in \mathbb{Z}, m, n \in \mathbb{Z}^+$ necessarily means that either $(m-1)(n-1) = 0$ or $(a, b, m, n) = (2, 3, 3, 2)$? In other words, are there any nontrivial consecutive powers besides $8$ and $9$?

This is not stupid question, this is a Catalan conjecture, which had been prooved only recently  ;)  .


limit
integration
calculus
real analysis
real analysis unsolved

For a positive value of $ c$, the limit

\[ L \equal{} \lim_{x\rightarrow\plus{}\infty}x^{c}e^{\minus{}2x}\int_{0}^{x}e^{2t}\sqrt{et^{2}\plus{}1}dt\]

exists, is finite and nonzero. Find this value of $ c$, and the limit.

Inside the square root, is it really $ et^2$ or should it be $ e^{t^2}$?

I typed it out exactly as in my source (1982 APICS contest, problem 3)... but it could still be a typo if replacing $ et^2$ with $ e^{t^{2}}$ makes the solution much more straightforward.

this problem should go in calculus computations and tutorials, and it should be $ et^2$ and not $ e^{t^2}$.

write the integrand of $ L$ as
\[ \frac {\int^x_0 e^{2t}\sqrt {et^2 \plus{} 1}dt}{e^{2x}x^{ \minus{} c}}
\]
and use L'hopital to reduce the behavior of $ L$ to the behavior at infinity of
\[ \frac {x^{c \plus{} 1}\sqrt {ex^2 \plus{} 1}}{2x \minus{} c}
\]
counting degrees, we get that $ c \equal{} \minus{} 1$, and all that's left to do is compute
\[ \lim_{x\to\infty} \frac {\sqrt {ex^2 \plus{} 1}}{2x \plus{} 1} \equal{} \frac {\sqrt {e}}{2} \equal{} L
\]

yes.
also we can do this by computing asymptotical behavior of integral when $ x \to \infty$
We can write exact asymptotic , but in this case it's not important for as. Hence, in this casae it's simple to do.
\[ \int_{0}^{x}e^{2t}\sqrt{et^{2}\plus{}1}dt \approx \sqrt{e}\int_{0}^{x}e^{2t}tdt \approx \sqrt{e}x\int_{0}^{x}e^{2t}dt \approx \frac{\sqrt{e}}{2}xe^{2x}\]

hence limit will be "good" if $ c\equal{}\minus{}1$


limit
inequalities
combinatorics proposed
combinatorics

Let $ H(n)$ be the number of simply connected subsets with $ n$ hexagons in an infinite hexagonal network. Also let $ P(n)$ be the number of paths starting from a fixed vertex (that do not connect itself) with lentgh $ n$ in this hexagonal network.
a) Prove that the limits \[ \alpha: \equal{}\lim_{n\rightarrow\infty}H(n)^{\frac1n}, \beta: \equal{}\lim_{n\rightarrow\infty}P(n)^{\frac1n}\]exist.

b) Prove the following inequalities:
$ \sqrt2\leq\beta\leq2$
$ \alpha\leq 12.5$
$ \alpha\geq3.5$
$ \alpha\leq\beta^4$


Could anyone tell me if there are "NEW IDEA" related to
nash-non-cooperative games ? I believe the mathematician from
Princeton University had really contributed an idea 'perfectly'
applicable in today's changing condition.

                                                 Respectfully,
                                               A.Eddington

Hi Arthur,
could you please rephrase that, would like to try help


calculus
derivative
limit
function
trigonometry
integration
real analysis

Suppose $f(t)$ is defined on $[a,b)$ and continuously differentiable in $(a,b)$ and that the right derivative exists at $a$. Is it true that $f'(a+) = f'_{\text{right}}(a)$ ?

I think it is well-know :)
Here is the solution(I define $f'_{+}(a)=f'_{right}(a)$):
Because:
$f'_{+}(a)=\lim_{x\to 0^{+}}\frac{f(x+a)-f(a)}{x}$
By Largrange's theorem:
There is a $\xi \in (a,a+x)$ such that:
$f(x+a)-f(a)=f'(\xi)x$
Thus:$f'_{+}(a)=\lim_{x\to 0^{+}}f'(\xi)$
When $x \to a^{+}$,we have$\xi \to a^{+}$
Thus we complete the proof. :)

[quote="Kalle"]Yes, I thought of that too. But how do you know that the limit $f'(a+)$ actually exists?[/quote]
Oh,Sorry.I mistook the problem.I think it is wrong.Because $f'(a+)$ actually may not exist as you said.
Consider the function
$f(x)=0,x=0;f(x)=x^{2}\sin \frac{1}{x},x \neq 0$ :)

Ok. Then I don't understand my text :(. Here is what it says:

"Let $\gamma$ be a piecewise differentiable curve in $\mathbb{C}$. This means a complex-valued, continuous function defined on a compact real interval which is continuously differentiable except at a finite number of points, where at least the left and right hand derivatives exist. Thus it is described by an equation $z = z(t)$ where $a \le t \le b$ for some real numbers $a$ and $b$ and $z'$ is continuous except at a finite number of jump discontinuities".

How does he know that they are jump discontinuities? Is he wrong? At the very least, we need to know that $z'$ is bounded in $[a,b]$ (to define a complex integral). I don't understand. Do you?

:( I think I may mistake something again.
I think that you post is a conclusion,it must have the reason write in the book in the above passage.Because it says "this mean" what is this does the book properly refer to?

I edited the text. It is now the complete passage. I think your example is a counterexample to what he is saying. What we want is $\int_{a}^{b}f(z(t)) z'(t) dt$ to exist for continuous $f$.


Hello fellow members and hello Valentin!

       On the old board/forum Valentin, you had a specific part of the board for math-links.  I think it will be a good idea to reinstitute a specific part of the board for math-links once again.  I have already taken initiative in bringing to the forum math-links I have collected from the net and from other people like Mathfanatic.  Our archive of links can be found in the beginner's corner under the subforum algebra.

Sincerely,

Santa's little Elf

P.S.  Attention all Members!  Please feel free to private message me your links to great math/science related webpages!

I also think it will be useful that we can get access on beta version to the
LINKS, there was some material for olympiad training, interesting math-web page etc...

the links section will be back, more better than before. please have a little bit of patience ... :) :D remember this is BETA (you folks know what beta means, don't you? :D) so not everything is back yet. I am making milestones up to importance, you can see the progress in the Beta release announcement :)


this is a pretty easy problem:

how many pairs of prime numbers such as $(p,q)$ exist where:
$\frac{1}{p}+\frac{1}{q}+\frac{1}{pq}=\frac{1}{n}$, where $n\in\mathbb{N}$?

One obvious solution is if p = -q.

But assuming you mean positive distinct primes...
[hide="small hint"]
1/p + 1/q + 1/pq is the same as $ \frac{p+q+1}{pq} $.
[/hide]


I have just found out about the AMC test, and really want to take it. I tried to solve some problems from previous years, and I was able to solve a lot of them. what do I have to do to take the test? do I take it at my school or something? I am in 9th grade.

i am new to this forum

Frequently Asked Question #4
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


Prove that
$ x^4 \plus{} y^4 \plus{} z^4 \minus{} t^4 \equal{} 1991$
has no integral solutions

[hide="Solution"]
$ x^4\plus{}y^4\plus{}z^4\minus{}t^4\equiv 1991\equiv 7\bmod 16$

Fourth powers are $ 0$ or $ 1$ $ \bmod 16$, so this is impossible.
[/hide]


Ok, this is probably the easiest question you'll find, but I need to know everything you all know on a+bi...has to do something with imaginary numbers. I know the whole i squared = radical -1 thing, but I would like a detailed explanation for this.
P.S. sorry of this is to easy for the section I put it under. This is not really a theorem... :blush:

[quote="eryaman"]Basically its the standard form of a complex number, where a is real.[/quote] true, but b is real also...

think of this.  Every number can be written as a+bi.  Only in some numbers like '4' for example, a would be 4 and b would be 0.  So this yeilds a new classification of numbers.  Complex numbers.  This encompasses all real numbers and all imaginary numbers.  Note every number is complex (I'm not sure about infinity, but for all intents and purposes don't worry about it :] )

The conjugate of z=a+bi is a-bi, and to denote it you put a bar over z.
|z|=sqrt(a^2+b^2)

In the complex coordinate system, a determines the x coordinate and b determines the y coordinate.


logarithms
inequalities
search
geometry
similar triangles

I think there are some "simple" geometric methods. I can prove the two variable case :D

EDIT:
Wikipedia has a pretty "simple" (still relative) proof that relies only on basic algebraic methods. Although, for one to think of this is pretty advanced. It can be found [url=http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof]here.[/url]

Wait, was this meant to be a challenge or something? Or were you curious of the way to prove AM-GM?

[quote="surge"]I think there are some "simple" geometric methods. I can prove the two variable case :D[/quote]
[hide="One of the geometric proofs"]
Consider a semicircle with diameter $AB$ and center $O$.
Let $P_1$ be the point on the semicircle so that $P_1O\perp AB$.
Let $C$ be an arbitary point on $AB$, and let $P_2$ be the point on the semicircle such that $P_2C\perp AB$.

The arithmetic mean of $AC$ and $BC$ is $\frac{AC+BC}{2}=P_1O$
The geometric mean of $AC$ and $BC$ is $\sqrt{AC\cdot BC} = P_2 C$ (Consider similar triangles $\triangle ACP_2$ and $\triangle P_2CB$)

Since $P_2C\le P_1O$ for any arbitary point $P$, we have $\frac{AC+BC}{2}\ge \sqrt{AC\cdot BC}$.
[/hide]

You can do it with Lagrange Multipliers.  Also, it's really easy to prove simple cases like $n=2$:

$\left(a-b\right)^{2}\geq 0\implies a^{2}+b^{2}\geq 2ab\implies \left(a+b\right)^{2}\geq 4ab\implies\frac{a+b}{2}\geq \sqrt{ab}$

I meant it as a challenge. see how many different ones I can find.

Here is one classic easy way

(Assuming all are non-negatives)
Start with 2 variables:

$(x_1-x_2)^2 \ge 0 \implies x_1^2 + x_2 ^2 \ge 2 x_1 x_2$   which  proves for two variables. (Let $a_1 = x_1^2$  and $a_2=x_2^2$  etc )

($\frac{a_1+a_2} 2 \ge \sqrt{a_1 a_2}$

and we can  also write
$\frac{a_3+a_4} 2 \ge \sqrt{a_3 a_4}$
Cobining above two we get: (and applying  AM-GM for two variables again for
$\frac{a_1+a_2} 2 + \frac{a_3+a_4} 2 \ge \sqrt{a_1 a_2}+ \sqrt{a_3 a_4}$
But  $\frac{\sqrt{a_1 a_2}+ \sqrt{a_3 a_4} } 2 \ge \sqrt[4] {a_1a_2 a_3 a_4}$

So we get $\frac{a_1 +a_2+a_3+a_4} 4 \ge \sqrt[4] {a_1a_2a_3a_4}$

We can thus prove it for 4 and then 8 etc... in fact for any n of the form $2^k$  we have
$\frac{a_1+a_2+a_3+\dots + a_n} n \ge \sqrt[n]{a_1a_2a_3\dots a_n}$

Now to prove it for general $m$, all you have to do is to assume $a_{m+1} = a_{m+2} = \dots = x = \frac{a_1+a_2+\dots+a_m} m$

and we get $\frac{m x + (n-m) x} n \ge \sqrt [n]{a_1a_2 \dots a_m x^{n-m}}$
of $x \ge \sqrt[m] {a_1a_2a_3\dots a_m}$
QED

Doesn't seem too hard...

By Jensen's,

\[ \frac{\log{a_1}+\log{a_2}+...+\log{a_n}}{n} \le \log{\frac{a_1+a_2+...+a_n}{n}} \]
because the curve $\log{x}$ is concave down.  Then, the left side becomes $\log{\sqrt[n]{a_1a_2...a_n}}$, and from there we take out the logs on both sides and we're done.

It's concave up.  The problem with that approach, of course, is proving Jensen's inequality, or at least proving that $f''(x)\geq 0\implies f$ is convex (as the other methods are more or less self contained)

[quote="Centy"]Killing a fly with a sledgehammer:

Muirhead: $(1, 0, \ldots, 0)$ and $\textstyle (\frac{1}{n}, \ldots, \frac{1}{n})$.[/quote]

Circular reasoning anyone?

[quote]By Jensen's..[/quote]
Yes that is nice, but  perhaps not that elementary..
One trick, one can use  to make it  more elementary, IMO,  is

A)  For    all non-negative $x$,  prove  $e^x \ge ex$

B)  Let  AM be $A = \frac{a_1+a_2+a_3+\dots+ a_n} n$ and

Use:
$e^{(a_1/A)} \ge e (a_1/A)$
   $e^{(a_2/A} \ge e (a_2/A)$
 $e^{(a_3/A)} \ge e (a_3 /A)$
  $\dots$
 $e^{(a_n/A)} \ge e (a_n/A)$
Multiplying together,  You get  $e^n \ge e^n \frac{a_1a_2a_3 \dots a_n} {A^n}$

 divide by $e^n$ and take nth root and that will    give you the result you want.
(This is also an old classic mehtod)

I personally like the proofs that use an algorith, such as one with weights on a balance. so the center or balancing point is the average of the #s. so half the wight is above or below that point or all the weight is at that point. then we take one weight and move it closer to the center so we must move a weight on the other side and also move it toward the center. and the center stays the same but the product increases.

another approach uses the fact that if $a+b=k$ for some fixed value $k$ then the value of $ab$ increases as $a$ and $b$ become closer together (namely $\frac{k}{2}$). and you just do this with a set number of values.

[quote="blahblahblah"][quote="Kalle"]Is this really an intermediate topic? I have seen two different proofs, and neither are high school level.[/quote]

There are proofs at a high school level, yes.

It has also been done a million times on the forum, including a couple of times by myself, and I won't do it again here.[/quote]

I agree with you blahblahblah. I'm going to lock this one because if you really want to find more about the different proofs about Arithmetic Mean-Geometric Mean, please use search engine. It's much quicker.

http://www.google.com/search?hl=en&lr=&q=Arithmetic+Mean-Geometric+Mean+Inequality


1) How many ways can I make a collection of 40 bottles from 8 different types?
This has already been answered. Thank you.

2) Repeat the same problem with the condition: there must be between 4 and 7 bottles of EACH type in the collection.

The problem is equivalent to the following one (I'm referring to the second problem):

Find the number of nonnegative ordered solutions to $ a_1 \plus{} a_2 \plus{} \dots \plus{} a_8 \equal{} 40$ such that $ 4\le a_i\le 7$ for each $ i \equal{} 1,2,\dots, 8$.

By solution(s) we will mean nonnegative integer solution(s). Rewrite the equation as
\[ (a_1 \minus{} 4) \plus{} (a_2 \minus{} 4) \plus{} \dots \plus{} (a_8 \minus{} 4) \equal{} 8
\]
It has $ \binom{8 \plus{} 8 \minus{} 1}{8 \minus{} 1}$ solutions.

Assume that $ a_i > 7$ for some $ i$. Then rewrite the equation as
\[ (a_1 \minus{} 4) \plus{} (a_2 \minus{} 4) \plus{} \dots \plus{} (a_i \minus{} 4 \minus{} 4) \plus{} \dots \plus{} (a_8 \minus{} 4) \equal{} 4
\]
which has $ \binom{4 \plus{} 8 \minus{} 1}{8 \minus{} 1}$ solutions. Thus when one of the $ a_i$'s exceed $ 7$ there are $ \binom{8}{1}\binom{4 \plus{} 8 \minus{} 1}{8 \minus{} 1}$ solutions.

Note that two of the $ a_i$'s cannot exceed $ 7$ simultaneously.

Therefore by the inclusion-exclusion principle the number of desired solutions is
\[ \binom{8 \plus{} 8 \minus{} 1}{8 \minus{} 1} \minus{} \binom{8}{1}\binom{4 \plus{} 8 \minus{} 1}{8 \minus{} 1}
\]


trigonometry
inequalities unsolved
inequalities

Prove that   $\frac{2sin \frac{\pi}{12}cos \frac{\pi}{24}}{2cos \frac{\pi}{24}-1} < 0,53$

I want to know the right side is 0.53 or 0,53?

I think both are right !
 0.53 or 0,53 is mean $\frac{53}{100}$  :P


Two non-zero real numbers, $ a$ and $ b,$ satisfy $ ab \equal{} a \minus{} b$. Which of the following is a possible value of $ \frac {a}{b} \plus{} \frac {b}{a} \minus{} ab$?

$ \textbf{(A)}\minus{}\!2 \qquad \textbf{(B)}\minus{}\!\frac {1}{2} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ 2$

[hide]$\mathbb{D}$[/hide]

yes i know its frustrating to just to see a letter blablah but if its right i'll post the solution.I'm even sick right now.

[quote="xxxyyyy"][hide]$2$

$y = (a/b) + (b/a) - (ab)$
or, $y = ({a^2 + b^2 - (ab)^2})/{ab}$

since $ab = a-b$, we get...
$y = 2{ab}/{ab}$,
or $y = 2$[/hide]

where did I made mistake?[/quote]

no, this is correct and the intended solution

[hide]$=\frac{a^2+b^2}{ab}-ab$
$=\frac{(a-b)^2+2ab}{ab}-ab$
Let $x=ab=a-b$, so we rewrite as $\frac{x^2+2x}{x}-x=x+2-x=2$.

$\boxed{E}$.[/hide]

[quote="Silverfalcon"][quote="shadysaysurspammed"]Ok,bad calculation mistake.[/quote]

That's why it is necessary to put full solution. If you feel sick and do not feel like that you want to post a full solution, [b]just not post anything at all[/b].[/quote]

Sure thing

[hide]I used simplification and substitution to get the answer of 2.

$2$[/hide]

[hide]
$ab = a - b b = a/(a+1)$
so then substituting everything for b..
you end up with eventually
$(a + 1) + 1/(a+1) - a^2/(a+1)$ combining, we get $2a+2/a+1$
2[/hide]

Or we could use SFFT to solve for the possible pairs (a,b). There is only one pair that works because $a,b$ cannot be 0.

@above $a,b$ are positive reals, and SFFT only works for integers.


Any quick and easy solutions?

1. The number of pairs of integers $ (a, b)$ with $ \minus{}500 < a < 500$ and $ \minus{}500 < b < 500$ that satisfy
the equation $ 3a \plus{} b \equal{} \frac {10ab}{2a \plus{} b}$ is?

2. What is the number of pairs of integers that satisfy the following two equations:
$ \cos(xy) \equal{} x$
$ \tan(xy) \equal{} y$

[hide="Answers"]
1.832
2.1
[/hide]

1. [hide] Equivalently, $ (3a\plus{}b)(2a\plus{}b)\equal{}10ab$
                              $ 6a^2\plus{}5ab\plus{}b^2\equal{}10ab$
                              $ 6a^2\minus{}5ab\plus{}b^2\equal{}0$
                              $ (3a\minus{}b)(2a\minus{}b)\equal{}0$

So, either $ b$ is $ 3$ times $ a$, or $ b$ is $ 2$ times $ a$.

Case 1: $ b$ is $ 3$ times  $ a$
            There are 166 positive and 166 negative cases (since $ 166\times{3}\equal{}498$ and the same with negs).
                  $ 166\plus{}166\equal{}332$
Note: (0,0) is not counted since it would make the RHS of the original problem undefined.

Case 2: $ b$ is $ 2$ times $ a$
           There seems to be a slight problem since I count 249 positive and 249 negative cases.
                  $ 249\plus{}249\equal{}498$

                   $ 332\plus{}498\equal{}\boxed{830}$? Ah well, the method is what's important, right? :D [/hide]

[hide="#1"]Rearranging it we have

$ 6a^2 \minus{} 5ab \plus{} b^2 \equal{} 0$

$ (3a \minus{} b)(2a \minus{} b) \equal{} 0$

$ b \equal{} 3a$ or $ b \equal{} 2a$

So, now we just count the numbers divisible by 3 or divisible by 2. (between -500 and 500)

Smallest multiple of 3: - 498
Greatest     "       "  " : 498
Total #: 332

Smallest multiple of 2: -498
Greatest    "        "  " : 498
Total #: 498

So the number of solutions is $ 332 \plus{} 498 \equal{} \boxed{830}$[/hide]

[hide="#2"]$ \sin (xy) \equal{} xy$

Now let's think what [b]integer[/b] values are in the output of the cosine function.. Just $ \minus{} 1, 0 , 1$. Great! Easy, now we just have 3 cases:

Case 1 : x = -1

We have that $ \cos ( \minus{} y) \equal{} \minus{} 1$, which is true if $ y \equal{} \minus{} 180$. So lets check it. $ \sin (180) \equal{} 180$...Not true..
Case 2: x = 0

We have that $ \cos (0) \equal{} 0$....not true..

Case 3: x = 1

We have $ \cos (y) \equal{} 1$ which yields $ y \equal{} 180$. $ \sin (180) \equal{} 180$...not true..

So in total, there are $ \boxed{0}$ solutions.[/hide]

What overlap? If I'm understanding you correctly, then every $ b$ which is a multiple of $ 6$ is counted twice and so should be subtracted from the overall count. But the question asks for ordered pairs. So while the $ b$s may be equivalent, the $ a$s will be different. Ex: (4,12) and (6,12).

You are correct. I didn't think about that lol.

[quote="KING1"]
[hide=Answers]
1.832
2.1
[/hide ][/quote]

This is what the answer key shows. You guys got different answers. For number one your answer was just 2 away, so I think you guys missed one number from each multiple. For number two...I think you totally forgot about the second equation.

[quote="KING1"][quote="KING1"]
[hide="Answers"]
1.832
2.1
[/hide ][/quote]

This is what the answer key shows. You guys got different answers. For number one your answer was just 2 away, so I think you guys missed one number from each multiple. For number two...I think you totally forgot about the second equation.[/quote]
[hide]For #2, $ tan(xy) \equal{} \frac {sin(xy)}{cos(xy)}$, and that implies that $ sin(xy) \equal{} xy$. Substitute $ xy \equal{} a$ and the only value that $ sin(a) \equal{} a$ is when $ a \equal{} 0$. Therefore this implies that $ cos(a) \equal{} x$, so $ cos(0) \equal{} 1$. So we have that the only solution that would work is $ x \equal{} 1$ and $ y \equal{} 0$. [/hide][/hide]

Oh ok, I had an error in my solution (specifically case #3).


logarithms
real analysis
real analysis solved

a<b, $f,g:[a,b]\rightarrow R_{+}^*$, $C^{1}$ such that

$f'>1,g'>0$

$g',\ln (f')$ are strictly increasing

Prove that there exist $c \in [a,b]$ such that

$\left(\frac{f(b)-f(a)}{b-a}\right )^{\frac{g(b)-g(a)}{b-a}}=f'(c)^{g'(c)}$

:D  :D  :D
Another great Moubinool's joke?

[quote="blahblahblah"]Anyone want to post a solution?[/quote]
Why?

Call the ugly quantity with all the secant lines (I'm in a hurry here) $Q$

Since $f,g$ are convex and increasing, $\exists p,q$ such that

$1<f'(p)^{g'(p)}<Q<f'(q)^{g'(q)}$

Now $f'(x)^{g'(x)}$ is increasing, so by the IVT there exists $c\in [a,b]$ such that $Q=f'(c)^{g'(c)}$

Am I missing something?

No, you're not!!

[quote="blahblahblah"]
Since $f,g$ are convex and increasing[/quote]

 But it is a too strong hypothesis. Can you replace it by something weaker, Moubinool?? :)


So if we use a computer program as our proof, do we have to paste in the code, or can we just say what the program did and prove that it worked?

See item #9 of the [url=http://www.usamts.org/Rules/U_RulesComplete.php]rules[/url].

Well, I meant did we need to paste the code into our proof our summarize it in English?

Or c) All of the above?

Not the code. If you used a solver, for example, just plug the numbers in and show that they satisfy the equation. If you used an enumeration program, describe what it does. For example, "my program first places the vowels of FATHER in the even places, and then places the consonants in the remaining places." I still have yet to hear what you do when you use a program to evaluate a rather large expression (involving summations, for example).


ratio
geometry
perimeter
symmetry
congruent triangles

[quote="GameBot"]What is the ratio of the measure of a regular hexagon's longest diagonal to the measure of its perimeter? Express your answer as a common fraction.[/quote]

let the side length be x. $ \frac{2x}{6x}\equal{}\frac{1}{3}$.

We can observe that the longest diagonal is twice the value the side length of a regular hexagon by dividing the hexagon into 6 equilateral congruent triangles by using the lines of symmetry in the hexagon that cut through the vertices.


[quote="t0rajir0u"]Show that $(-1) \times (-1) = 1$.
:D
[hide="Hint"] Define $1$ (what's special about it?).  Now define $-1$ in terms of $1$. [/hide][/quote]
I did a proof for a question like this before, we'll start of with,
$0=0\times 0$
$0=(1-1)(1-1)$
$0=(1)(1)+2(1)(-1)+(-1)(-1)$<---distributive law
add 1 to both sides,
$1=1-2+1+(-1)(-1)$
$1=(-1)(-1)$. quod erat demonstrandum :)


superior algebra
superior algebra unsolved

I need to prove an intersection of 3 groups is nonempty. In special one of the form $B\,\cap\, b_{1}g^{-1}C g \cap b_{1}g D g^{-1}$, with $b_{i}\in B$, $B\le C$, $B\le D$, $C\cap D=B$, $g\in G$, a group containing $B,C,D$.

I have one lemma: that for $X,Y,Z\le G$ and $y,z\in G$ such that $X(Y\cap Z) = XY \cap XZ$ then we have that $X\cap yY \cap zZ$ is nonempty iff each of the pairwise intersections is nonempty.

Does anyone know some alike lemma's?

The statement looks strange: you mention $b_{i}$ but use only $b_{1}$. The element $b_{1}$ obviously belongs to the intersection.

If you meant $B\,\cap\, b_{1}g^{-1}C g \cap b_{2}g D g^{-1}$, then the intersection may be empty. Let $G=S_{6}$ and let $B=C=D$ be generated by the transposition $(12)$. Let $g$ be the shift by $2$ elements: $g=(135)(246)$. Then $g^{-1}Cg$ is generated by $(56)$, while $gDg^{-1}$ is generated by $(34)$. If $b_{1}\ne b_{2}$, then $b_{1}g^{-1}C g \cap b_{2}g D g^{-1}=\varnothing$.

Yes, $b_{1}$ and $b_{2}$. But in my case C and D were given groups (but with a too long history of properties). Of course it's not true in general, but I hoped someone knew conditions under which it would be true (like $X(Y\cap Z)=XY\cap XZ$ -- which was not fulfilled, too bad).

But never mind, my original problem has been solved, at last.


inequalities
triangle inequality
absolute value
inequalities open

Hey guys,

Can anyone give me the proof for the Triangle inequality for absolute value ? i.e

|a-b|<=|a-c|+|c-b|

I got stuck and cant do anything with it :(


geometry
geometric transformation
rotation
probability
number theory
least common multiple
greatest common divisor

After SOMEONE finally sent in their answers... cough lol the guy at camp cough.

Here are the results.

In 1st place, with a score of 34: stevenmeow :winner_first:
In 2nd place, with a score of 29: 27r  :winner_second:
In 3rd place, with a score of 28: RisingFlame  :winner_third:

Good job to all 3 of you. Your prize is the satisfaction of knowing you beat... some other people. =D

The rest of the results are in the attached excel file. Since none of you PMed me that you cared about your results being shown...

Here you go... Also, the answer key is there... Feel free to challenge my answers because err... no one *actually* proofread the target. Yongyi just was like. the wording is wrong. here. =D

Wow... some interesting information:

No one got #22, #29, or #30 right on the sprint. The only question I would've expected NO ONE to get right was #30. It was surprising that no one got #22 or #29 right either. :(

None but one person got #5 right on the target. :o
I actually proofread that question and found that mathcrazed's original answer was incorrect as well!

No one got #6 or #8. Although I would've expected no one to get #8. :P

Wow, my score on this beat my score at Nats :P
That shows how badly I messed up there.

Ya.. err on #5 lol. I had like. 9.123214%... which was because window's calculator SUCKS.
You can start posting solutions also

[hide="Problem 29"]
Because I'm too lazy to look at the problem, I'll just say the colors are red, green, and blue.

Fix the rotation by putting a color on top. Let's say it's red. It doesn't matter because all we're doing is fixing the rotation. Now there are 3 DISTINCT choices for the bottom face.

If it is red, then we have two distinct colors left to place on the side faces. Now this is just like seating two Bobs and two Alices around a table. There are only two ways to do this.

If it is either blue or green (two possibilities), then we have three distinct colors BUT we only have four colors left. For now, assume the bottom face is blue. Then we have the colors BRGG to place on the side faces and there are 3 ways to do this. HOWEVER, putting red opposite the blue is an extra rotation because in our original configuration, the red WAS opposite the blue on the top and bottom. So there are only really $ 2 \cdot 2 \equal{} 4$ possibilities in this case.

Add up: $ 2 \plus{} 4 \equal{} \boxed 6$. Yay! :D [/hide]

Oh ya. Forgot to thank all the proofreaders who actually did something... Perfect628, Dynamo729 (kinda =D), and Yongyi781. Thanks to Yongyi for being faster than the rest of you lol... and Cognos599, who I almost forgot xD. #h34n1 kinda. lol . Lot's of lols. HAHA LOL.

OK. I'll start some solutions.  Also added the tests so anyone who wants to post solutions can.
[hide="1"]
We just do $ .8*.7\equal{}.56$ So the amount subtracted was $ 44$%[/hide]
[hide="2"]
Well-known problem. Since the worst case scenario is getting one of each color twice, we require one more to get 3 of one color. So the answer is 7.[/hide]

[hide="3"]Draw a radius from the center to a vertex of the triangle. Then draw the perpendicular to one of the sides containing that point. Since it's an equilateral triangle, we have a 30-60-90 triangle. The length of 2 is opposite the 60 degree angle, so the hypotenuse has length $ \frac{4}{\sqrt{3}}\equal{}\frac{4\sqrt{3}}{3}$[/hide]

[hide="4"]Since $ 25^{x\plus{}2}\equal{}5^{2x\plus{}4}\equal{}3125$
$ 2x\plus{}4\equal{}5$
$ x\equal{}\frac{1}{2}$
$ 16^x\equal{}4$
[/hide]
[hide="5"]
The ones that Christine will win on: $ 1, 3, 5, 6, 9, 12, 15, 18, 24, 25, 30, 36$ and the number of outcomes:
$ 1\plus{}2\plus{}2\plus{}4\plus{}1\plus{}4\plus{}2\plus{}2\plus{}2\plus{}1\plus{}2\plus{}1\equal{}24$

So the probability she loses is $ \frac{1}{3}$
[/hide]

[hide="6"]Simple formula: LCM*GCF=Product of two numbers. So the LCM is obviously 60.[/hide]
[hide="7"]
The two differ by 76. This means the two square roots must differ by an even number.

Since both are some number squared, we can say that $ (b\minus{}a)(b\plus{}a)\equal{}76$.  However, we know both of these factors must be of the same parity.  Since 76 is even, both must be even. The only factorization that satisfies this is $ 2*38$. Solving gives us 20 and 18 as solutions.  So $ n\equal{}324\minus{}125\equal{}199$[/hide]

[hide="8"]Draw the two circles. Connect the intersection points to both centers. A 120 degree arc means 1/3.[/hide]

[hide="9"]$ 5^{20}*2^{24}\equal{}10^{20}*16$ which has 22 digits. $ 10^{20}$ has 21 digits and 16 adds one more.

[/hide]

#10 asked for it in feet, which is why some of you missed it.

[hide="11"]There are three red marbles. Only one of them has a blue marble in the other box. $ \frac{1}{3}$[/hide]
[hide="12"]
COUNT ... I'm too lazy to[/hide]
[hide="13"]

Either use magic... cough euler's totient function. or notice that $ 7^2\equal{}49\equiv 1\mod16$.

So, the answer is either $ 7$ or $ 1$. 109 is odd. $ 7$
[/hide]

[hide="14"]Two cases, A goes horizontally/vertically or it goes diagonally.

First case: $ 1*4*3$
Second case: $ 1*4*5$
Add them up to get $ 32$[/hide]

How would you go about solving number 13 using Euler's Totient Function?

It is well known that $ a^{\phi(b)}\equiv 1\mod b$ for $ GCD (a,b)\equal{}1$.

Since $ \phi(16)\equal{}8$, we have that $ 7^{108}\equiv 1\mod 16$. So $ 7^{109}\equiv 7 \mod 16$

how did you arrive at 7^108 being equivalent to 1 mod 16?

... -_- whoops lol. $ 108$ isn't divisible by $ 8$ is it xP. Sorry. $ 7^{104}\equiv 1 \mod 16$
Then you check $ 7^5$ to get $ 7$. XD

Noting $ 7^2\equiv 1\mod 16$ is a better way though, but sometimes (with large numbers), it's really hard.

Is there going to be another mock test
:jump:
it would be cool

There will be... BTW. PEOPLE PLEASE POST SOLUTIONS?!

Note: I didn't do this test, so I may be wrong. Deathly wrong.  :P
For #15, I'm not too sure, since it seems like $ AD$ can be as big as you want, which would make the area infinitely large. But if $ CD$ also has to have integer length, then...
[hide="15?"]Draw a figure, looking like so:
[asy]pair A=(3,4), B=(9,4), C=(9,0), D=(0,0), X=(3,0);
draw(A--B--C--D--cycle); draw(A--C, dashed); draw(A--X, dashed);
label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW);
label("$4$", (3,2), W); label("$2\sqrt{13}$", (6,2), NE); label("$4$", (9,2), E);[/asy]
From the Pythagorean Theorem, $ AB\equal{}6$. Now for both $ AD$ and $ CD$ to be integer values, the triangle part on the left will have to be a 3-4-5 triangle, so $ AD\equal{}5$, and $ CD\equal{}AD\plus{}3\equal{}9$. Then the area is $ .5\cdot4\cdot(6\plus{}9)\equal{}30$.[/hide]
[hide="16"]If you have $ l,w,h$ as the length, width, and height, respectively, then the amount of cubes that remain unpainted is $ (l\minus{}2)(w\minus{}2)(h\minus{}2)$. It's somewhat obvious that you should try to make these values as equal as possible, giving something like $ l\equal{}w\equal{}5, h\equal{}6$, and $ (5\minus{}2)(5\minus{}2)(6\minus{}2)\equal{}\boxed{36}$. (You can always check all the cases to make sure that this is correct.)[/hide]
[hide="17"]For the sake of simplicity, let $ ac\equal{}x$. Then $ x\plus{}b\equal{}1$ and $ bx\equal{}6$. Substitute $ b\equal{}1\minus{}x$ into the second equation to get $ x(1\minus{}x)\equal{}\minus{}6$, which when expanded out has solutions $ x\equal{}\minus{}2,3$. The final answer is $ \minus{}2\plus{}3\equal{}\boxed{1}$.[/hide]
[hide="18"]When only one number remains, every integer from 1 to 2008 will have been added to the total. There will also have been 2007 operations, so that will subtract 2007 from the total. The final answer is $ \frac{2008\cdot2009}{2}\minus{}2007\equal{}2015029$.[/hide]
[hide="19"]$ k_1\equal{}1$, $ k_2\equal{}k_3\equal{}k_4\equal{}2$, $ k_5\equal{}k_6\equal{}\cdots\equal{}k_9\equal{}3$, $ k_{10}\equal{}k_{11}\equal{}\cdots\equal{}k_{16}\equal{}4$, $ k_{17}\equal{}k_{18}\equal{}\cdots\equal{}k_{25}\equal{}5$, and $ k_{26}\equal{}k_{27}\equal{}\cdots\equal{}k_{29}\equal{}6$. There is 1 value of 1, 3 values of 2, 5 values of 3, 7 values of 4, 9 values of 5, and 4 values of 6. Add up $ 1\cdot1\plus{}3\cdot2\plus{}5\cdot3\plus{}7\cdot4\plus{}9\cdot5\plus{}4\cdot6\equal{}\boxed{119}$[/hide]

Hm they look good to me (I didn't actually look at them but don't tell anyone...) but yeah I misread 10 questions. Which is a record for me. But whatever, I fail anyway.

[quote="Brut3Forc3"]
[hide="19"]$ k_1 \equal{} 1$, $ k_2 \equal{} k_3 \equal{} k_4 \equal{} 2$, $ k_5 \equal{} k_6 \equal{} \cdots \equal{} k_9 \equal{} 3$, $ k_{10} \equal{} k_{11} \equal{} \cdots \equal{} k_{16} \equal{} 4$, $ k_{17} \equal{} k_{18} \equal{} \cdots \equal{} k_{25} \equal{} 5$, and $ k_{26} \equal{} k_{27} \equal{} \cdots \equal{} k_{29} \equal{} 6$. There is 1 value of 1, 3 values of 2, 5 values of 3, 7 values of 4, 9 values of 5, and 4 values of 6. Add up $ 1\cdot1 \plus{} 3\cdot2 \plus{} 5\cdot3 \plus{} 7\cdot4 \plus{} 9\cdot5 \plus{} 4\cdot6 \equal{} \boxed{119}$[/hide][/quote]

Note the sum involves $ n(2n\minus{}1)$ also.

If I gave more terms, it would be hard to calculate like that.


probability
geometry
3D geometry
prism
rectangle
inequalities
trapezoid

Lets see if you can solve these 2 questions:

A bag contains 10 marbles. Each marble is either blue or green, and there are at least two marbles of each color. When three marbles are drawn without replacement the probability that all of them are green is the same as the probability that exactly one of them is blue. How many of the 10 marbles are blue???

A particular right prism has a rectangular octagonal base. Its base edges are 4 cm each and its height is 3 cm. Its volume can be expressed in the form (a(b + radical c)) cubic cm and c is not divisible by the square of a prime. What is that value of a+b+c???

Using the hypergeometric distribution because this is without replacement. If we start with the possibility that there are 2 blue and 8 green we compute:

$ {blue \ total \choose blue \ picked} { green \ total \choose \ green \ picked}$

This is for all 3 green:
$ { 2 \choose 0 } { 8 \choose 3} \equal{} 56 \ ways$

This is for exactly 1 blue and 2 green picked:
$ { 2 \choose 1 } { 8 \choose 2} \equal{} 56 \ ways$

So there are 2 blue marbles in the bag

How is it that there are 56 ways, can you explain, then i will let you know if you are correct.
can you solve this one?

A particular right prism has a rectangular octagonal base. Its base edges are 4 cm each and its height is 3 cm. Its volume can be expressed in the form (a(b + radical c)) cubic cm and c is not divisible by the square of a prime. What is that value of a+b+c???

Fine, you are correct. Solve the second one

Great, can you answer any of these problems??

A 20 gallon container is filled halfway filled with a mixture that is 90% vinegar and 10% water. How many gallons of water must be added for the mixture to become 60% vinegar and 40% water?

The points (x,y) represented in this table lie on a strait line. When the equation of this line is written in the form y=Ax + B, what is the value of A + B?

        [u] [b] X_____Y[/b][/u]
          2_____7
         t-2____v
          t_____v+6

A 20 gallon container is filled halfway filled with a mixture that is 90% vinegar and 10% water. How many gallons of water must be added for the mixture to become 60% vinegar and 40% water?
[hide="solution"] The container has $ \frac{1}{10}*20\equal{}2$ gallons of water.  Since we want 40% water, we have:

$ \frac{2\plus{}x}{20\plus{}x}\equal{}\frac{2}{5}$

$ 5(x\plus{}2)\equal{}2(x\plus{}20)$

Solving this gives us x=[b]10[/b].[/hide]

Ok, can anyone solve the second one??

The points (x,y) represented in this table lie on a strait line. When the equation of this line is written in the form y=Ax + B, what is the value of A + B?
[u][b]
X_____Y [/b][/u]
2_____7
t-2____v
t_____v+6

Solution:
[hide]$ A \equal{} \frac{(v\plus{}6)\minus{}(v)}{t\minus{}(t\minus{}2)} \equal{} \frac{6}{2} \equal{} 3$,
so $ B \equal{} 7 \minus{} 2A \equal{} 7 \minus{} 2\cdot 3 \equal{} 1$. So $ A\plus{}B \equal{} 3 \plus{} 1 \equal{} 4$.[/hide]

Can you solve any of the 2 problems below?

In a right circular cone shaped cup (Vertice pointing downward) with a height of 20cm and a radius of 6cm, how many centimeters deep would the water be when exactly 3/5 of the cup's volume contains water?? Express your answer as a decimal rounded to the nearest tenth.

Using only the letters from the set {B,C,D,F,G,H,J}, all of the possible four-letter "words"  are written in alphabetical order starting with BBBB. What is the 2010th word in the list???

Hi;

[quote]
Using only the letters from the set {B,C,D,F,G,H,J}, all of the possible four-letter "words" are written in alphabetical order starting with BBBB. What is the 2010th word in the list???[/quote]

If BBBB is numbered 1 then the 2010 word is HJBB.

Ok, thanks. Can you answer this problem?

For a given arithmetic series, the sum of the first 50 terms is 200 and the sum of the next 50 terms is 2700. What is the first term in the series?? Express your answer as a decimal to the nearest tenth.

nightmare1251 are you publishig a book? :rotfl:  :rotfl:  :rotfl:

For a given arithmetic series, the sum of the first 50 terms is 200 and the sum of the next 50 terms is 2700. What is the first term in the series?? Express your answer as a decimal to the nearest tenth.

Solution:
Let $ a$ be the first term of this arithmetic progression and $ d$ the difference of the progression.The sum of the first 50 terms is $ 200\equal{}50a\plus{}1225d (1)$ and the sum of the next 50 terms is $ 2700\equal{}50a\plus{}3725d (2)$. So if we make $ (2)\minus{}(1)$ we get that $ 2500d\equal{}2500$, so $ d\equal{}1$. And we get that $ 50a\equal{}200\minus{}1225$ so $ a\equal{}\minus{}20.5$

Can you please explain how you got 200 = 50a + 1225d(1) and how you got 2700 = 50a + 3725d(2).
Explain mostly on how you got the 1225d and the 3725d and how you got 2500d  = 2500 and determined that a was equal to -20.5? I was just wondering because i solved it a different way.

I think those problems are too hard, here are some others:

A book shelve contain seven identical math books, three identical science books, and two identical history books. If the books are randomly arranged on a bookshelf, what is the probability that all the math books are together, all of the science books are together, and all of the history books are together? Express your answer as a common fraction.

There are $ \frac{12!}{7!3!2!}\equal{}7920$ ways to arrange the books. Of these, there are only $ 3!\equal{}6$ ways to arrange them under the conditions. Thus, $ \frac{6}{7920}\equal{}\boxed{\frac{1}{1320}}$.

Good job, can you answer this:

A bag contains four times as many red marbles as white marbles. Two marbles are drawn without replacement and the probability that they are different colors is 1/3. How many marbles are there in the bag?

I think that is too hard. Answer any of these problems within 12 hours if you can:

Trapezoid ABCD has bases AB and CD of lengths 27 and 41 units, respectively, and its legs are 13 and 15 units. Diagonals AC and BD intersect the trapezoid's median at X and Y. What is the length of segment XY?

In rectangle ABCD, side BC is twice as long as side AB. Point E is on side CD with DE:EC = 1:2. Segments BE and AC intersect at point F. The area if quadrilateral AFED is what fraction of the area of the rectangle ABCD? Express your answer as a common fraction.

A bag contains four times as many red marbles as white marbles. Two marbles are drawn without replacement and the probability that they are different colors is 1/3. How many marbles are there in the bag?

Do you think you can answer the third question in the link below:

http://mathcounts.org/Page.aspx?pid=355

[quote="nigh0848"]Can somebody please explain to me how the answer to #11 was found?[/quote]

Can u restate the question?

I believe he means the answer posted in post #11 of this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1744317#1744317

Yes, exactly! Post #11. I'm very curious on how to solve it

If you're referring to bobbym's post, then I can try to show you it.

[quote] Using only the letters from the set {B,C,D,F,G,H,J}, all of the possible four-letter "words" are written in alphabetical order starting with BBBB. What is the 2010th word in the list??? [/quote]

Ok, let's start with B. If "B" is the first letter, then for the 3 remaining places, we can fit 7 different letters into each place. Therefore, the number of ways to make a word in the form "$ Babc$" is just $ 7*7*7\equal{}343$

Since we know that $ 2010/343\equal{}5....R 295$, we can figure out that the 2010th word begins with the letter H. this is since the 5 in the quotient represents that the 5 letters {B,C,D,F,G} have already been used. If we assume it's in the form "$ HBab$", we can see there are 49 possibilities for "$ ab$".

Since $ 295/49\equal{}6......R 1$, we know it's not in the form "$ HBab$", but actually in the form "$ HJab$". The 6 in the quotient represents that we have already used the 6 letters {B,C,D,F,G,H} as the second letter. Since there is only 1 in the remainder, we can tell it is the very first word that begins with "$ HJab$", which means it's $ \boxed{HJBB}$

Hope this helped.

Thank you so much for your time and help!

Thanks for the post too! I was also wondering how to finger out post #11

[quote="AmericanPi"]octagon base's area

if octagon is ABCDEFGH

connect AD, HE, BG, and CF

the area of the four triangles formed is 4 ea, so together they are 16

the four outer rectangles are each 4 x 2sqrt2, so together they are 32sqrt2

the inner square is 4 x 4 = 16.

base area is therefore 32 plus 32sqrt2

volume

then u multiply by height and divide three

final ans is 32+32\sqrt{2}[/quote]

Wouldn't it be nicer to write (or type) $64\sqrt2/2$ instead 32+32\sqrt{2}[/quote] of and answer instead of ans???

Woah. Dude. Major revival.

Read Stickies maybe?

Why is everyone doing this one kid's homework?


Prove that, for all positive real numbers $a,b$ and $c$,
\[\frac{a(b+c)}{a^{2}+bc}+\frac{b(c+a)}{b^{2}+ca}+\frac{c(a+b)}{c^{2}+ab}\le\frac{a+b+c}{\sqrt[3]{abc}}\le \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\]


I have a problem for my introductory aerodynamics class that gives you a weight (force) of a piston which is 1000N and they tell you that it's cross sectional area is 0.05 square meters. it asks for the gauge pressure. I know that P = F/A, and i also know that the total pressure = atmospheric pressure + gauge pressure. So, is F/A = the total pressure or is it equal to the gauge pressure? Sorry if this is a dumb question but i need to know it for an upcoming final exam in about 3 weeks.

My guess is that they want gauge pressure; even if not, you can always specify that in your answer.

[quote="ROLEX4life"]well they give you atmospheric pressure but they ask for gauge and absolute pressure. My question is what type of pressure does computing F/A give you? I am assuming it's absolute pressure but i just want to be sure.[/quote]
F/A is the definition of pressure as a concept. Atmospheric pressure, in simplest terms, is pressure where F equals the weight of the atmosphere. Absolute pressure has that weight included, gauge pressure does not.

Usually what you compute is gauge pressure, and then you add atmospheric pressure to that to get absolute pressure.


integration
inequalities
real analysis
real analysis unsolved

Let $ \minus{}\infty \le a < b \le \infty$, $ 1 \le p < \infty$, and $ 0 < \epsilon_0 < \infty$. Prove that there exists a finite constant $ C \equal{} C(\epsilon_0,p,b\minus{}a)$ such that for every $ \epsilon\in(0,\epsilon_0]$, and for every $ f \in C^2((a,b))$
\[ \int_a^b |f'(x)|^p dx \le C \left( \epsilon \int_a^b |f''(x)|^p dx \plus{} \frac 1\epsilon \int_a^b |f(x)|^p dx \right).
\]

In 1-D, we have
\[ \left( {\int_a^b {\left| {u'\left( x \right)} \right|^p dx} } \right)^2  \leqslant c_1^2 \left( {\int_a^b {\left| {u''\left( x \right)} \right|^p dx} } \right)\left( {\int_a^b {\left| {u\left( x \right)} \right|^p dx} } \right)
\]
which gives
\[ \int_a^b {\left| {u'\left( x \right)} \right|^p dx}  \leqslant c_1 \left( {\varepsilon \int_a^b {\left| {u''\left( x \right)} \right|^p dx}  \plus{} \frac{1}
{\varepsilon }\int_a^b {\left| {u\left( x \right)} \right|^p dx} } \right)
\]
for all $ 0<\varepsilon$.

[quote="volodja"]How do you prove the first inequality?[/quote]

This comes from
\[ \left\| {Du} \right\|_{L^q }^2 \leqslant C\left\| u \right\|_{W^{2,r} } \left\| u \right\|_{L^p }
\]where $ u \in L^p  \cap W^{2,r} ,1 \leqslant p,r \leqslant \infty ,\frac{1}
{q} \equal{} \frac{1}
{2}\left( {\frac{1}
{p} \plus{} \frac{1}
{r}} \right)$ known as Galiardo-Nirenberg's interpolation inequality.


calculus
integration
trigonometry
calculus computations

Write $ \tan^{2}{x}\equal{}\sec^{2}{x}\minus{}1$.  The hard part is integrating $ x\sec^{2}{x}$, but this is just $ \frac{x}{\cos^{2}{x}}\equal{}\frac{1}{1\minus{}\sin^{2}{x}}$.  Can you take it from here?

[hide]$ \int x \tan^2x \,dx \equal{} \minus{} \int x \sin x (\minus{} \sin x)(\cos x)^{\minus{}2}\,dx$

$ \equal{} \minus{} \left[\minus{}(\cos x)^{\minus{}1} \cdot x \sin x \plus{} \int (\cos x)^{\minus{}1}(\sin x \plus{} x \cos x)\,dx\right]$

$ \equal{} x \tan x \minus{} \int \tan x\,dx \minus{} \int x\,dx$

$ \equal{} x \tan x \plus{} \ln|\cos x| \minus{} \frac{x^2}{2} \plus{} C$.[/hide]

$ \int x\tan^{2}x\ dx$
$ \equal{}\int x\sec^{2}{x}\minus{}\int x dx$
Integration by parts, let $ u\equal{}x, du\equal{}dx, dv\equal{}\sec^2 x, v\equal{}\tan x$
$ \equal{}x\tan x \minus{}\int \tan x\minus{}\int x dx$
$ \equal{}x\tan x\plus{}\ln|\cos x|\minus{}\frac{x^2}{2}\plus{}C$


Is there a easy way of taking a formula such as co2 and making it like carbon dioxide? I know how to do the easy ones just now the harder ones. If you need more information on my question, just ask!

There are several laws governing different types of compounds. Not really just an easy way. After practicing for a while, you get used to which ones use which rules, so I suggest using a textbook or http://www.google.com/webhp?hl=en#hl=en&source=hp&q=chemistry+nomenclature+rules
In this case and with many covalently bonded compounds, it's just using the subscripts and turning them into prefixes, like NO2 is nitrogen dioxide.


I am not sure if you're looking for answer, or just post the question. In
any case, 221=13*17

so $ 221^{221}$ = $ 13^{221}$*$ 17^{221}$

Now each factor can be placed in 222 different way (0-221), so
there should be $ 222^{2}$ different factor including 1 and itself.

Jeff
[quote="Lazarus"]How many factors does $ 221^{221}$ have?[/quote]

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I agree with Indianmath and gjshao. The answer is $ 49284$.

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