Do you have suggestions for a Team format (3-4 middle school students per team; 12-20 teams) that is perhaps more interactive and/or spectator friendly than "Solve 10 questions in 20 minutes".  20-40 minutes total.

Thanks

Perhaps a sort of countdown format?

Number the students on each team 1234.  1 has a problem against 1.  The loser goes out, and the next team member comes.  If a team loses all its players, it loses.  Repeat with other teams (obviously you'll need a bracket).

Of course, the disadvantage is a) there could be a very strong team member, so members 2-3-4 don't get to play, and b) it's not very interesting for students not so good at countdown, as they might only get one problem.

I think AIME15's idea is good.

It is easy to do this in FTW.
Each team arrange the order of their players by FTW rating.
The highest rating will play the last.

We can use AIME15's knock out system.
With such a system, it is better we have less team (say 5 teams) but more players on each team (say 10 players on a team).
In such a system every player will have a chance to play.
Particularly weak players.
The drawback of such a system: Whichever team has the most strongest player (e.g. AIME15 on FTW) will win.
The result is kind of predictable.

Playing math is like playing chess. Better player will win most of the time.

[quote="SuperNerd123"]But maybe not everyone plays FTW a lot[/quote]

We can play in any other format that is different from FTW.
Maybe a new format other than FTW will have more surprises in end results.

I can donate an Internet server so we can run any math game on it.
We just need someone(s) to write a nice math tournament program that we can play on.

Maybe you could have lifelines where you ask the audience

yeah, that's really gonna work tennis :)

@math champion: why not? maybe there are teachers or parents in the audience who are good at math

Have you seen "Are you smarter than a 5th grader" The audience isn't exactly rock-solid in the elementary subjects. :)

Uhh, you realize the show specifically picks stupid people. I think the idea is a good one.


To write a book 1890 digits were utilised. How many pages does the book have?

[hide="EDITTED EDIT Miscounted"]
Possibly miscounted.
Number of digits per the first 100, by counting, $ 189$
$ 1890-189=1701$

Then $ 300$ for every hundred.
With $ \left\lfloor \frac{1701}{300}\right\rfloor = 5$
Leaving $ 201$ with three digits per page after $ \frac{201}{3}=67$
With $ 66$ being the last two digits since the $ 67$ includes the $ 00$ as the last two digits

Making $ \boxed{100+500+66 = 666 \ \text{pages}}$.
[/hide]


How can one prove that $\sqrt{2}^{\sqrt{2}}$ is irrational?

Well, it follows from Gelfond's theorem that it's even transcendental, because it's of the form $a^b$, where $a\ne 0,1$ is algebraic and $b$ is algebraic and irrational.


Let $x,y,z>0$ such that $xyz(x+y+z)=1$.Prove the inequality:

              $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\leq\frac{9}{16}\cdot(x+y)(y+z)(z+x)$ :)

Apply $9(x+y)(y+z)(z+x) \ge 8(xy+yz+zx)(x+y+z)$, then we have to prove $(xy+yz+zx)(x+y+z) \ge 2 \sum \frac{1}{x+y}$
or equivalently, $(xy+yz+zx) \ge 2xyz \sum \frac{1}{x+y}$
Now, R.H.S. $\le 2xyz \sum \frac{1}{2 \sqrt{xy}} = \sqrt{xyz} (\sum \sqrt{x}) \le \sqrt{xyz} \sqrt{3(x+y+z)} = 1$
L.H.S. $\ge \sqrt{3xyz(x+y+z)} = 1$. Proof finished.

Very nice! That's exactly my solution. ;)  :P


Find Ceiling[Sqrt[1+c]-1], where c=2999*Log_2[10]. Without using a calculator.

Did you do all that without a calculator? I think "without a calculator" means to do it some nice way without so much arithmetic..

Also, you just did a whole bunch more steps than you ever needed to -- you could have just divided 2999 by .3010 and then added 1, and you would have arrived at the same thing without most of the intermediate steps.

By the way, there is a nice algorithm similar to long division for doing square roots by hand -- anyone interested?

i did do all of that without a calculator...

yeah, i sort of realized this morning that there were quite a few extraneous steps. (my excuse? it was late)

i have been playing around with several methods that wouldn't require knowing that log_2(10)=.3010, but haven't made significant progress..

i mean...if we set x = sqrt(1+2999log_2(10)), we see that

x^2-1 = 2999*log_2(10), and we are trying to find x-1...i was hoping to find some nice factoring method or something, but was not able to..

[quote="JBL"]By the way, there is a nice algorithm similar to long division for doing square roots by hand -- anyone interested?[/quote]

Certainly. I once had a teacher who said that such an algorithm existed. When questioned, though, she couldn't show it to us.  :(

Advance apologies for difficulties formatting.

Okay, so this is pretty cool.  Say you have some number, like, I dunno, 56447, and you want to find its square root to the tenths-place.  The basic rule is you need 2 digits to get 1 digit (this makes sense, no?), so on either side of the decimal point we group digits in pairs.

:sqrt:05 64 47 . 00

Now, we start with the left-most pair, and we find the largest number whose square is less than that digit.  In this case, that's 2.  So we write a 2 above that pair of digits, square the 2 and subtract it from the 05, and bring down the next pair of digits.

__2
:sqrt:05_64_47_._00
_04
_01_64

Now, this is the tricky bit.  We take the string of digits we have already found (which at this point is just a 2) and multiply it by 20 (in this case giving us 40).  Then we find the largest value of x such that (40 + x)*x :le: 164.  Now, the way this is written that looks nicest when you set it up as a multiplication problem 4x * x = 164.  Anyhow, however you write it, you get that the next digit is 3 (since 44*4 = 176).  43 * 3 = 129.  So we write that and subtract and bring down the next pair:

__2__3
:sqrt:05_64_47_._00
_04
_01_64
_01_29
____35_47

Now we double, bring down, and solve (as closely as possible) the multiplication problem 46x * x = 3547.  8 is too big, so 7 it is.  467*7 = 3269.  Then we get

__2__3__7 .
:sqrt:05 64 47 . 00
_04
_01_64
_01_29
____35_47
____32_69
____02_78_00

Then we solve 474x * x = 27800, and we have 4746*6 = 28476 which is too big while 4745 * 5 = 23725.  So, to one decimal place, :sqrt:56447 = 237.5.  A quick calculator check gives 237.586 et cettera.

If you've read that, there are 2 questions I have for you:

(1) Did it make sense?
(2) If the answer to number 1 is "yes," prove the algorithm works :).

OK I didn't read all through it cos I think I've seen it before  . I tried proving it then, don't have a "formal" proof but basically you could see that.. [hide]its sort of like repeatedly completing the square. You start off with a square plus something else, and then you take part of that something else and put it inside the square (thats sort of where the "doubling" or multiplying by 20 or whatever comes in, the 2xy term is (x+y)^2.. but anyway yeah this probably doesn't make sense..[/hide]

[quote="JBL"]If you've read that, there are 2 questions I have for you:

(1) Did it make sense?
[/quote]
It made sense after I read it three times. Thanks

While visiting my cousins then living in Trinidad, I was surprised to see this square-root algorithm being taught as part of the school curriculum for I think 6th graders.  It seemed pointless to require it for all students.


A teacher gave a test to a class in which $ 10\%$ of the students are juniors and $ 90\%$ are seniors.  The average score on the test was $ 84$.  The juniors all received the same score, and the average score of the seniors was $ 83$.  What score did each of the juniors receive on the test?

$ \textbf{(A)}\ 85 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 93 \qquad \textbf{(D)}\ 94 \qquad \textbf{(E)}\ 98$

[hide]
Assume there are $100$ students
$10$ of them are juniors and $90$ of them are seniors
Average score on the test=$84$
Sum of everyone's score=$84*100$
Sum of the scores of seniors=$83*90$
$8400$(total score)$-7470$(senior's score)$=930$(sum of the juniors)

We assumed there were $10$ juniors so the average would be $\frac{930}{10}=93$
[/hide]

For fun, assume there are $10$ students.  :)

[hide]Okay, we assume that there are 10 students.  Nine are juniors with an average of 83, so the one junior must have had a score capable of raising the average the 84.  Similar to before (except with smaller numbers),

$(83*9+x)/(10) = 84$

Simplifying,

$747+x=840$

So $x=93$, answer choice $C$[/hide]

You don't even need to multiply all of that out:
[hide] Say there are 100 points on the test.  Thus, a senior gets 83 points.  The average is 84 points, so the seniors are all 1 point from the average.  Now, there are 9 times as many seniors than juniors.  So for every one point 1 senior misses, 1 junior will have to account for (1)(9)=9 extra points.  So, each junior must get the average+9=93, which is our answer.  [/hide]


geometry
perimeter
geometric transformation
reflection

is it true that the orthic triangle of a triangle has the lowest perimeter? i seem to be getting stuck with my refllection approach..

Yes. This problem is called Fagnano's problem. You can read more about it [url=http://www.cut-the-knot.org/triangle/Fagnano.shtml]here[/url].


If $ R(R^2\minus{}A)<\minus{}8$ what is the lowest possible value for A if A is positive?

I'm not sure if this is in the right problem section because I don't know the solution.

I am attempting to find what value of R minimizes A. So That makes them constant if I am not mistaken.

A and R are both variables. Here's a restatement of the problem:

[i]Find the minimum positive value of $ A$ such that there exists a positive* number $ R$ for which $ R(R^2\minus{}A)<8$[/i]

[hide]$ R(R^2\minus{}A) < \minus{}8 \implies R^3 \minus{} RA < \minus{}8 \implies RA > R^3\plus{}8 \implies A > R^2 \plus{} \frac{8}{R}$

Observe that $ R^2\plus{}\frac{8}{R} \equal{} 3 \cdot \frac{R^2 \plus{} \frac{4}{R} \plus{} \frac{4}{R}}{3} \geq 3 \sqrt[3]{R^2 \cdot \frac{4}{R} \cdot \frac{4}{R}} \equal{} 3 \sqrt[3]{16} \equal{} 6 \sqrt[3]{3}$

Where the inequality step in the above is AM-GM.

Therefore, $ A > 6 \sqrt[3]{3}$.[/hide]

*- I wrote this in even though the OP didn't because the answer is pretty dull without it (the answer would otherwise be that it can be arbitrarily close to 0).

Your answer can't be right because it says A is approximately 8.6 and if $ R \equal{} 1.6$ then $ A \equal{} 7.56$

I did mean to specify that R is positive.

Good call. The error was in the very last step: $ 3 \sqrt [3]{16} \equal{} \color{red}{6 \sqrt [3]{3}}$.

I meant to say, $ 3 \sqrt [3]{16} \equal{} \boxed{6 \sqrt [3]{2}}$.


function trigonometry quadratics calculus derivative limit algebra

A function $ f(x)$ is [i]periodic[/i] if there is a positive number $ p$ such that $ f(x\plus{}p) \equal{} f(x)$ for all $ x$. For example, $ \sin x$ is periodic with period $ 2 \pi$. Is the function $ \sin(x^2)$ periodic? Prove your assertion.

I think this is right..

[hide="Solution"]

The answer is yes.

First of all, use $ x \plus{} p$ as the problem did to get:

$ \sin((x \plus{} p)^2) \equal{} \sin(x^2 \plus{} 2px \plus{} p^2)$

Now, our emphasis is with $ \sin(x^2)$ so expand it like this;

$ \equal{} \sin(x^2) \cos(p^2 \plus{} 2xp) \plus{} \cos(x^2) \sin(p^2 \plus{} 2xp)$

This'll equal to $ \sin(x^2)$ if $ \cos(p^2 \plus{} 2xp) \equal{} 1$ and $ \sin(p^2 \plus{} 2xp) \equal{} 0$. Now, this can become true in infinite ways, and one being $ p^2 \plus{} 2xp \equal{} 2 \pi$. Thus, solving for $ p$ using the Quadratic Formula:

$ p \equal{} \minus{} x \plus{} \sqrt {x^2 \plus{} 2 \pi}$

One way we can do is by letting $ x \equal{} 1$ and getting $ p \equal{} \minus{} 1 \plus{} \sqrt {2 \pi \plus{} 1}$. [/hide]

Hmm, my idea seemed to work until I checked my graphing calculator and it doesn't match...  :| I mean, I know that I could've set $ p^2 \plus{} 2xp \equal{} 2 n \pi$ where $ n$ is an integer but that's it...

Using the quadratic formula, you get $ p\equal{}\minus{}x\pm\sqrt{x^2\plus{}2n\pi}$. This must hold true for all $ x$, not just $ x\equal{}1$.

[quote="Xantos C. Guin"]Using the quadratic formula, you get $ p \equal{} \minus{} x\pm\sqrt {x^2 \plus{} 2n\pi}$. This must hold true for all $ x$, not just $ x \equal{} 1$.[/quote]

Yeah I acknowledged that fact at the end (although I didn't explicitly write it). I wasn't sure if it was right though because calculator said something else but then I could've been wrong on the calculation. Thanks for the confirmation!  :)

Actually, I was saying that $ \sin(x^2)$ is not periodic.

The zeros of $ \sin(x^2)$ are at $ \sqrt{k\pi}$ for $ k \in \mathbb{Z}$. So, $ p\equal{}\sqrt{m\pi}\minus{}\sqrt{n\pi}$ for some $ m,n \in \mathbb{Z}$.

I'm not sure where to go from here, but I'm fairly certain that $ \sin(x^2)$ is not periodic.

[quote="Xantos C. Guin"]Actually, I was saying that $ \sin(x^2)$ is not periodic.

The zeros of $ \sin(x^2)$ are at $ \sqrt {k\pi}$ for $ k \in \mathbb{Z}$. So, $ p \equal{} \sqrt {m\pi} \minus{} \sqrt {n\pi}$ for some $ m,n \in \mathbb{Z}$.

I'm not sure where to go from here, but I'm fairly certain that $ \sin(x^2)$ is not periodic.[/quote]$ \sin(0^2)\equal{}0$
It looks too simple so I must be overlooking something[hide]If a period $ p$ exists, then for $ x\equal{}p$
$ \sin(p^2)\equal{}0$, which means
$ p^2\equal{}n\pi$ for some integer $ n$, and $ p\equal{}\sqrt {n\pi}$.
Then for  $ x\equal{}mp ($for any integer $ m)$
$ x^2\equal{}m^2n\pi$, and $ \sin(m^2n\pi)\equal{}0$ again.
However, between $ x\equal{}0$ and $ x\equal{}p$ there were $ n\minus{}1$ zeroes,while between $ x\equal{}p$ and $ x\equal{}2p$ there were $ 3n\minus{}1$ zeroes, so $ p\equal{}\sqrt {n\pi}$ cannot be the period.[/hide]Or how about[hide]$ \sin(x^2)$ is positive on both sides of $ x\equal{}0$, but there is no other number $ p$ where $ \sin(p^2)\equal{}0$, and $ \sin(x^2)$ is positive on both sides of $ x\equal{}p$[/hide]

there are a lot of ways to do this.  assume that it is periodic with period $ p>0$

as xantos notices, the k-th nonnegative zero $ z_k$ is at $ \sqrt {k\pi}$, so the difference between consecutive zeroes is $ \Delta_k \equal{} \sqrt {k\pi} \minus{} \sqrt {(k \minus{} 1)\pi}$.  on one hand, $ \Delta_k$ is strictly decreasing.

on the other hand, let $ p$ be a period.  then $ p$ and $ \sqrt {\pi} \plus{} p$ are $ z_j$ and $ z_{j \plus{} 1}$ for some $ j > 1$, so $ \Delta_j \equal{} \Delta_1$, which is a contradiction.

What's essentially going on is that each "quasi-period" is getting shorter because $ x^2$ increases at an increasing rate.  A glance at the graph should be enough to convince you that this function is not periodic.

[hide="Rigorously"] The derivative of a periodic function is periodic, but $ 2x \cos (x^2)$ is clearly not periodic - its amplitudes increase without bound. [/hide]

Thank you for all the responses. My calculator didn't look like it was periodic but ha, I didn't really understand this much anyway.  :D

Now t0rajir0u, how does having $ 2x \cos(x^2)$ not periodic helps about $ \sin(x^2)$ not being periodic though? Like... if the derivative of function is not periodic, that means the function is not periodic? I didn't really understand this part.

If a function is periodic, its derivative must also be periodic:
[hide="hidden for those with severe calculus allergies"]
If $ f(x)$ is periodic with period $ p$, then $ f(x \plus{} p) \equal{} f(x)$ and $ f(x \plus{} p \plus{} \Delta x) \equal{} f(x \plus{} \Delta x)$.
So, $ f'(x \plus{} p) \equal{} \lim_{\Delta x \to 0}\dfrac{f(x \plus{} p \plus{} \Delta x) \minus{} f(x \plus{} p)}{\Delta x} \equal{} \lim_{\Delta x \to 0}\dfrac{f(x \plus{} \Delta x) \minus{} f(x)}{\Delta x} \equal{} f'(x)$
Thus, $ f'(x)$ is also periodic.
[/hide]
The contrapositive is also true: if the derivative of a function is not periodic, then the function is also not periodic.
Since $ \frac {d}{dx}\sin(x^2) \equal{} 2x\cos(x^2)$ is not periodic, $ \sin(x^2)$ is also not periodic.

[quote="t0rajir0u"]What's essentially going on is that each "quasi-period" is getting shorter because $ x^2$ increases at an increasing rate.  A glance at the graph should be enough to convince you that this function is not periodic.
 The derivative of a periodic function is periodic, but $ 2x \cos (x^2)$ is clearly not periodic - its amplitudes increase without bound. [/quote]

I don't really understand why the derivative is periodic

EDIT:
Never mind..

Answer: Not periodic.
Proof: Assume $\sin((x+p)^2) = \sin(x^2)$ for all $x$. Then
$$ 0 = \sin(x^2+2px+p^2) - \sin(x^2) = \cos\left(x^2+px+\frac{p^2}{2}\right)\sin\left(px+\frac{p^2}{2}\right)$$
for all $x$. However, $\cos\left(x^2+px+\frac{p^2}{2}\right)=0$ for only countably many $x$ and $\sin\left(px+\frac{p^2}{2}\right)=0$ for only countably many $x$ too, so $\cos\left(x^2+px+\frac{p^2}{2}\right)\sin\left(px+\frac{p^2}{2}\right)$ cannot be identically zero. QED


[b]Show that :

if $ x,y \in R^{ \plus{} }$  and $ xy \equal{} 1$ we have :

$ \displaystyle{\frac {1}{\sqrt {1 \plus{} 3x}} \plus{} \frac{1}{\sqrt {1 \plus{} 3y}}} \geq 1$

 :roll: [/b]

Try substituting 1 by xy a few times, square it, and then rewrite the inequality in the form:
$ \frac{y}{y\plus{}3}\plus{}\frac{x}{x\plus{}3}\plus{}\frac{2}{\sqrt{(y\plus{}3)(x\plus{}3)}}\geq 1$
Now multiply by $ (y\plus{}3)(x\plus{}3)$
What will be to prove is:
$ 2\sqrt{(y\plus{}3)(x\plus{}3)}\geq 8$ Which folows by AM-GM :wink:

[quote="polskimisiek"]
$ 2\sqrt {(y \plus{} 3)(x \plus{} 3)}\geq 8$ Which folows by AM-GM :wink:[/quote]

Thanks  :wink:  but what do you mean by AM-GM  :?:

That a strange term for me.

Arithmetical Mean, Geometrical Mean :wink:
It follows like that:
$ 2\sqrt {(y \plus{} 3)(x \plus{} 3)}\geq 8 \iff \sqrt{(y\plus{}3)(x\plus{}3)}\geq 4 \iff xy\plus{}3(x\plus{}y)\plus{}9\geq 16\iff 3(x\plus{}y)\geq 6$
Which by AM-GM:
$ 3(x\plus{}y)\geq 3(2\sqrt{xy})\equal{}6$ :wink:

I got it .

Thanks a lot

but this is the 1st time for me I see AM-GM :

Thanks again  :wink:


If the ecuation $2x^4+x^3+mx^2+x+2=0$ has all the roots real, what is the interval for $m$ ?
And in general, what is the condition for an ecuation to have all the roots or $k$ roots real ?

And for the second question ?
"In general, what is the condition for an ecuation to have all the roots or  $k$ roots real ? "

There isn't one- at least, there isn't anything that can be simply written down. The idea of a discriminant can be generalized, but it can't tell the difference between no real roots and four. This sort of problem always has to be attacked with a patchwork of methods.

[quote="Diogene"]You divide by $x^2$ , put $y=x + \frac 1x$  and the equation becomes : $2y^2 + y + m-4 = 0$
Now it's easy : the equation (second degree)  with $y$ has to have real roots $y_1,y_2$ , and the equation (still second degree) $x+\frac 1x = y_k$ has to have real roots
 :cool:[/quote]

How'd you think to do that?  I can see it from the $2x^4$, $2$... $1x^3$, $1x$... but is there something else that led you to that?


The number $ 10!$ ends with how many zeroes?

$ 10!$ contains $ 5$ factors of $ 2$ and $ 2$ factors of $ 5$. Therefore, it ends with $ \boxed{2}$ zeroes.


At the end of a professional bowling tournament, the top $ 5$ bowlers have a playoff.  First, #5 bowls #4.  The loser receives 5th prize and the winner bowls #3 in another game.  The loser of this game receives 4th prize and the winner bowls #2.  The loser of this game receives 3rd prize and the winner bowls #1.  The winner of this game gets 1st prize and the loser gets 2nd prize.  In how many orders can bowlers #1 through #5 receive the prizes?

$ \textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ \text{none of these}$

I am probably wrong, but I would say:

(E) None of these

as I got solution to be 14 :S

Ihatepie is correct.  There are $ 4$ games in every playoff, and each game has $ 2$ possible outcomes.  For each sequence of $ 4$ outcomes the prizes are awarded differently.  Thus, the answer is $ 2^4\equal{}16 \rightarrow \boxed{\textbf{(B)}}$.

[quote="AIME15"]Ihatepie is correct.  There are $ 4$ games in every playoff, and each game has $ 2$ possible outcomes.  For each sequence of $ 4$ outcomes the prizes are awarded differently.  Thus, the answer is $ 2^4 \equal{} 16 \rightarrow \boxed{\textbf{(B)}}$.[/quote]

I have found my mistake... I have done it with $ 2^4$ but for a stupid reason I made it $ 2^4\minus{}1\minus{}1$

Another way is to just take smaller cases, you find them all to be powers of 2.


What is the largest number of distinct positive integers such that no four of them have sum divisible by 4?

generalize to n

The [url=http://www.mathlinks.ro/viewtopic.php?t=271583]EGZ[/url] (Erdős, Ginzburg, Ziv) Theorem states that any set of $ 2n\minus{}1$ numbers has $ n$ whose sum is divisible by $ n$. However, for $ 2n\minus{}2$ numbers, we can take $ \{\underbrace{0,\dots,0}_{n\minus{}1},\underbrace{1,\dots,1}_{n\minus{}1}\}$.

Therefore, the answer to your specific question is $ 6$ and the answer to the general question is $ 2n\minus{}2$.

[quote="Fraenkel"]What is the largest number of distinct [b]positive integers [/b]such that no four of them have sum divisible by 4?

generalize to n[/quote]

So we cannot have 0 in our set.

Ah, well, modulo $ n$, so we can take $ \{\underbrace{n,\dots,n}_{n\minus{}1},\underbrace{1,\dots,1}_{n\minus{}1}\}$.  :blush:

Also "distinct," but both of these conditions are obviously irrelevant since we can add a bunch of multiples of $ n$ to all of any such set of numbers to make them both positive and distinct.


This question might seem trivial, but could you please help me understand it?

A 1955 AHSME question is reproduced below: "A train traveling form Aytown to Beetown meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles more before the accident, it would have bben just one hour late. What is the usual rate of the train?"

We all know that the answer is easily obtained by ...
[hide="Not very well written solution"] setting the times equal to each other, but of course we would have to add the one hour; a simpler way, as stated in the solutions, is to realize that the only difference is the rates at which the 80 miles are covered, and the difference obtained is equal to the difference of the times, which turns out to be precisely one hour[/hide]

The solution above implied that the "if 80 miles part" was at the [i]same[/i] rate as the usual rate - the rate from Aytown to Beetown before the accident - that is, you are traveling the 80 miles after one hour. And you would be right.

But wouldn't it also be remotely possible to interpret the hypothesis - "If the train had covered 80 miles more before the accident..." as the train covering the 80 miles [i]during[/i] the one hour; that is, the rate would now be x + 80.

I mean, although this might not be the interpretation that you get when you read the problem (although this was what I first interpreted it as - I justified myself with the accident occuring "after 1 hour". So therefore, the rate must have been increased by 80), isn't is even remotely possible to interpret the question  this way (which would, obviously lead to much different answers, and therefore you would get the problem wrong)?

I think that the part that says "...after which it proceeds at four-fifths of its usual rate..." sort of implies that the problem is referring to the speed before the accident as the "usual rate". But it is very possible to interpret the problem your way.

wait - so the question could have had another answer?????????

I'm just saying that the question is not 100% clear, and that your interpretation is not "wrong" by any means.

Thanks!
But can anyone confirm math154's conclusion????????

Hmm...
theax, if no one is answering the post, it might just be a sign that this should be moved into the intermediate forum...
Unless however, someone decides to give a response...

$ 1\plus{}\frac{1}{2}\plus{}\frac{d\minus{}R}{\frac{4}{5}R}\equal{}1\plus{}1\plus{}\frac{80}{4}\equal{}\frac{1}{2}\equal{}\frac{d\minus{}R\minus{}80}{\frac{4}{5}R}$, so $ R\equal{}20 \text{ mph}$.

Thanks AIME15, but my real question is... well, refer to my first post, which is currently unanswered yet - with the exception of math154, whose work has yet to be verified -, please

i know that there are a lot of smart people out there....

What are you asking?  Yes, I suppose it's possible for the problem to be misinterpreted in the way you described.  But that is an incorrect interpretation, and you will get a wrong answer.

But the question is: how do you know that the interpretation is incorrect? Sure you know that the answer is incorrect if you check the answer key. Why would it be incorrect? What's wrong with it?

Any help is greatly appreciated. Thanks in advance!

I guess the best explanation that I can give is this:  "80 miles" is a [i]distance[/i], not a [i]rate[/i].  So simply traveling "80 miles more" does not affect the rate at which you are traveling.


geometry
3D geometry
prism
calculus
integration
ratio
rectangle

A right rectangular prism $P$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$  A plane parallel to one of the faces of $P$ cuts $P$ into two prisms, one of which is similar to $P,$ and both of which have nonzero volume.  Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist?

sorry for bringing this back up but no1 ever posted a solution heh

[hide]

we let the plane cut $c$ into a length $R$ (it is not hard to check that the plane can only cut $c$ and not $a$ and $b$)

so we have cases where

$b\ge a \ge R$

$R \ge b \ge a$

$b \ge R \ge a$

then since they must be similar, we have the ratios:

$\frac{c}{b} = \frac{b}{a}$

$\frac{c}{R} = \frac{b}{b} = 1$

$\frac{c}{b} = \frac{b}{R}$

in which case we see that the 2nd case is not possible, giving us the original prism, and that the first and third cases are the same

hence we just have:

$\frac{c}{1995} = \frac{1995}{a} \implies ac = 1995^2$

since $c \ge a$, our answer is just going to be half the number of divisors of $1995^2$ which is:

$(3^4 - 1) / 2 = \boxed{40}$

[/hide]

[quote="gotztahbeazn"]sorry for bringing this back up but no1 ever posted a solution heh

[hide]

we let the plane cut $ c$ into a length $ R$ (it is not hard to check that the plane can only cut $ c$ and not $ a$ and $ b$)

so we have cases where

$ b\ge a \ge R$

$ R \ge b \ge a$

$ b \ge R \ge a$

then since they must be similar, we have the ratios:

$ \frac {c}{b} \equal{} \frac {b}{a}$

$ \frac {c}{R} \equal{} \frac {b}{b} \equal{} 1$

$ \frac {c}{b} \equal{} \frac {b}{R}$

in which case we see that the 2nd case is not possible, giving us the original prism, and that the first and third cases are the same

hence we just have:

$ \frac {c}{1995} \equal{} \frac {1995}{a} \implies ac \equal{} 1995^2$

since $ c \ge a$, our answer is just going to be half the number of divisors of $ 1995^2$ which is:

$ (3^4 \minus{} 1) / 2 \equal{} \boxed{40}$

[/hide][/quote]

But why don't we consider the third side of the rectangle prism?  How do we know that once two sides of the rectangle prism satisfies the problem then it will be a solution?

Apologies for reviving an old thread, but I was just doing this problem for practice and I believe that the answer is $ 40$.  We must ignore the case where $ a\equal{}b\equal{}c$ because the similar prism which this produces is the original prism and the problem states that both prisms have nonzero volume.


Is there any way of determining whether a is a quadratic residue in mod m or not.

Thanx :)

Usually you will just have to find $1^{2},2^{2},3^{2}\dots (\frac{m}{2})^{2}$
but if $m$ is prime, you can use the [url=http://www.artofproblemsolving.com/Wiki/index.php/Quadratic_residues]Legendre Symbol[/url] or quadratic reciproity.

And if it is not prime, you could factor $m$ and test if the number is a quadratic residue $\mod$ all prime powers dividing $m$.


topology
vector
real analysis
real analysis unsolved

Let $ X$ be Banach space. Let $ x^{**} \in X^{**}$. Is it true that $ x^{**}$ is weak* continuous as a map $ X^* \to \mathbb{C}$ (so $ X^*$ is given the weak* topology) [b]only[/b] if $ x^{**}$ is in $ X$ (i.e. in the range of the canonical imbedding)?

Sure. Consider continuity at $ 0$ and conclude that there exist finitely many $ x_j\in X$ such that if all $ x_j(x^*)\equal{}0$, then $ x^{**}(x^*)\equal{}0$. Now forget topology and everything else and just observe that if $ E$ is a linear space and $ f,f_i$ are finitely many linear functionals on $ E$ such that $ f(y)\equal{}0$ whenever all $ f_i(y)\equal{}0$, then $ f$ is a linear combination of $ f_i$.

Hmm, that would be very neat. I am so bad with this stuff so far. So as I understand it, continuity at 0 means this:

For each $ \delta > 0$ there exist $ x_1,..,x_n$ in $ X$ and an $ \epsilon$ such that $ x^{**}\left( \{x^* \in X^* \, : \, |x_j(x^*)| < \epsilon \, \forall j \}\right) \subset \{z\in \mathbb{C} \, : \, |z| < \delta\}$

How do I pass to the limiting case?

Fix any $ \delta>0$ you wish. Take a vector $ x^*$ such that $ x_j(x^*)\equal{}0$ for all $ j$ and multiply it by any number $ c$. You should still have $ |x^{**}(cx^*)|\equal{}|c|\cdot|x^{**}(x^*)|<\delta$ whence $ x^{**}(x^*)\equal{}0$.

I think the proof may also be finished by applying separation of convex sets:

Let $ \delta \equal{} 1/2$, and suppose we are in the real case (otherwise take real parts). Let $ M \equal{} \{(x^{**}(x^*), x_1(x^*),..,x_n(x^*))\, : \, x^* \in X^*\}$, a convex subset of $ R^n$ whose closure does not contain $ (1,0,\ldots, 0)$. By some form of separation of convex sets, we can find constants such that

$ x^{**}(x^*) \plus{} c_1x_1(x^*) \plus{} ... \plus{} c_nx_n(x^*) \leq 1$ for all $ x^*$

which implies that the LHS is 0

edit: I didn't see your reply when I began writing this. It took some time to put my thoughts together  :P

Yes. This is also a valid approach. The key point is to reduce everything to a finite-dimensional question. After that you have many ways.


[quote="Beat"]Let $p>0, q>0$ and $n\ge 0$ are integer numbers. How many integer numbers can be represented in the form $ip + jq$, where $i$ and $j$ are non-negative integers, and $i + j \leq n$.[/quote]
If  gcd(p,q)>1 we can change p1=p/gcd(p,q),q2=q/gcd(p,q) (Number of solution for (p,q) and (p1,q1) equal). Let gcd(p,q)=1, and p<q. We  have ip+jq=(i-q)p+(j+p)q, and i-q+j+p<i+j. Therefore for all solution (i1,j1) we have only one solution (i,j) with i<=q-1. If i1*p+j1*q  is not equal i2*p+j2*q if 0<=i1<=q-1,0<=i2<=q-1. Therefore if n>=q-1 we have (n+1)+n+(n-1)+...(n+2-q)=q*(2n+3-q)/2 different means. If n<q-1 we have 1+2+...+(n+1)=(n+1)(n+2)/2 different means.


AMC AIME geometry trigonometry function perimeter inradius

Well, I was actually able to figure out this problem, except that the solution required the use of trig functions applied to random numbers, so a calculator, which is not allowed on the AIME, was needed.

The inscribed circle of triangle ABC is tangent to AB at point P, and its radius is 21. Given that AP = 23, PB = 27, find the perimeter of the triangle.

*sigh* its also so annoying when you manage to find a relationship between some geometrical objects using, say, stewart's theorem and the pythagorean theorem, just to use them alongside one another in equations and later find out that one follows directly from the other, rendering it useless.

Fierytycoon

OK. Just to label some other points, let the inscribed circle be tangent to BC at Q and CA at R.
We know AP=AR=23 and BP=BQ=27, we want to find CR(=CQ).

Let angles AOP=x,  BOQ=y,  COR=z. We know that x+y+z=180 and that tan x=23/21, tan y = 27/21, tan z = CR/21.

tan z = tan (180-x-y) = -tan(x+y) = -((tan x + tan y)/(1 - tan x tan y)) = -(50/21)/(-180/441) = 5 and 5/6 or 35/6.
So CR = 21 x 35/6 = 122.5.

In total the perimeter is 345.

This doesn't look quite right, but it's what I got.

We express the area of the triangle in 2 ways. First we use K=sr (semiperimeter*inradius), and then we use 1/2*b*c*sin(A). We call the point of intersection on BC Q and the incenter I. Then we let a=CQ. The semiperimeter is then 23+27+a, so the area is 21*(50+a). Also, we calculate sin(PAI)=21/sqrt(23^2+21^2)=21/sqrt(970). PAI=IAQ, so sin(BAC)=2*21/sqrt(970)*23/sqrt(970)=2*21*23/970 (which I guess need not be simplified for the time being). Now we equate the two expressions for the area: 21*(50+a)=25(23+a)*2*21*23/970, and then we solve for a. If I didn't make an arithmetical error (big if), then a=245/2, so the semiperimeter is 245/2*50=345/2, so the perimeter is 345. That's an appropriate AIME answer, but it looks a bit large. But it should be right if I didn't add or multiply something wrong.

Thanks for the solutions, I understand how to solve it now.

Fierytycoon

An even shorter way is to use Heron's Formula and the Area-Inradius Formula.

K =  :sqrt:(s(s-a)(s-b)(s-c)) = s*r
So (s-a)(s-b)(s-c) = s*r^2

Using simlar notation as Zeta, then
(23)(27)(x) = (x + 50)*21^2
621x = 441x + 22,050
180x = 22,050
x = 245/2

And P = 2s = 2x + 100 = 345


geometry
trigonometry
trig identities
Law of Cosines
power of a point
Pythagorean Theorem

Points $ A$, $ B$, $ C$, $ D$ are on a circle such that $ AB\equal{}10$ and $ CD\equal{}7$. If $ AB$ and $ CD$ are extended past $ B$ and $ C$, respectively, they meet at $ P$ outside the circle. Given that $ BP\equal{}8$ and $ \angle APD\equal{}60^\circ$, find the area of the circle.

[hide="Ugly solution"]

First of all, we apply Power of a Point on $ P$ to get $ (18)(8)\equal{}(PC\plus{}7)(PC)$. From this, $ PC\equal{}9$. Now, we apply Law of Cosines on $ \triangle{PAD}$ to get \[ AD\equal{}\sqrt{256\plus{}324\minus{}2(18)(16)\cos 60^\circ}\equal{}2\sqrt{73}\] Again, applying Law of Cosines, this time on $ \triangle{PBD}$, we have $ BD\equal{}\sqrt{64\plus{}256\minus{}2(16)(8)\cos 60^\circ}\equal{}8\sqrt{3}$. Now, notice that $ BD^2\plus{}AB^2\equal{}AD^2$, so $ \angle{ABD}\equal{}90^\circ$ and $ AD$ is the diameter. Hence, the area of the circle is $ (\sqrt{73})^2\pi\equal{}73\pi$

[/hide]

Using Power of a Point, we can find that $ CP\equal{}9$. Since $ AP\equal{}18$ and $ m\angle CPB\equal{}60^\circ$, we know that $ \triangle ACP$ is a $ 30\minus{}60\minus{}90$ triangle, so that $ AC$ is perpendicular to $ DP$. Then $ \triangle ACD$ is right, so $ AD$ is a diameter of the circle. Simple Pythagorean Theorem gives that $ AD\equal{}2\sqrt{73}$, so the radius is $ \sqrt{73}$ and the area is $ \boxed{73\pi}$.


Let $ABCD$ be a convex quadrilateral. $O$ is a point of intersection of its diagonals. Prove that $ABCD$ is circumscribed iff $\frac{1}{r_{AOB}}+\frac{1}{r_{COD}}=\frac{1}{r_{BOC}}+\frac{1}{r_{DOA}}$

The problem is correct; it says circumscribed, not inscribed.  It was also posted here, but no solution was ever presented:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64321]www.artofproblemsolving.com/Forum/viewtopic.php?t=64321[/url]


Zach has chosen five numbers from the set {1, 2, 3, 4, 5, 6, 7}. If he told Claudia what the product of the chosen numbers was, that would not be enough information for Claudia to figure out whether the sum of the chosen numbers was even or odd. What is the product of the chosen numbers?

[hide="A way to find the answer but not to show that it is unique"]Notice that $ 2\cdot 6 \equal{} 3\cdot 4$ and the sums of the factors on the LHS and RHS have different parities. So $ 1\cdot 2\cdot 5\cdot 6\cdot 7 \equal{} 1\cdot 3\cdot 4\cdot 5\cdot 7 \equal{}\boxed{420}$ works.[/hide]


function
advanced fields
advanced fields unsolved

prove that a smooth function $ f: \mathbb{R}^2\rightarrow \mathbb{R}$ cannot be one  to one.

You don't need smoothness, just continuity. Choose $ t\in\mathbb R$ so that the sets $ \{(x,y)\colon f(x,y)<t\}$ and $ \{(x,y)\colon f(x,y)>t\}$ are both nonempty. The union of these sets is not connected. But $ \mathbb R^2$ minus a point is connected.


Euler
trigonometry
geometry
circumcircle
analytic geometry
conics
ellipse

If a triangle ABC has the property that its Euler Line is parallel to its side BC, then prove $\tan B \tan C=3$.

I'd be really intrigued to see this solution. I worked on the problem for maybe 10 minutes and came to what I thought could work as a strategy but I fell short of being able to really start the problem.  :blush: What level contest would this problem show up?? I mean, I'd have a LOT of trouble with a problem of this difficulty. Here's what I came up with if anybody cares to tell me if I was even remotely on the right track:

[hide]
It's only necessary to use two of the 4 characteristic points of Euler's line. I chose to use the orthocenter and the centroid with no specific idea in mind. For Euler's line to be parallel to BC, perpendiculars dropped from the orthocenter and the centroid to BC must be equal in length. I'm guessing that when this is true the triangle will have some characteristics that would allow you to prove the trigonometry part of the problem... but that's really where I fall short.
[/hide]

I had problem solving this problem as well. :blush:
I think you are on the right track. But I haven't completed this problem yet. I will try again and post again. As for difficulty, maybe a harder COMC problem?? :P

Nice solution, but I can't see the picture. I've always had this problem though. :(

[quote="SM4RT"]That's a nice solution... I had a problem with seeing the attachments too but it turns out i had some bugs and major problems on my comp. it got formatted and now it seems to work..maybe you have the same problem (like a bug or something)[/quote]
lol, Thanks. I'll check it out.

this question is sort of easy...i soved it mentally...

[quote="lightrhee"]Let $\alpha$, $\beta$, $\gamma$ be $\frac{B+C-A}2$, $\frac{C+A-B}2$, $\frac{A+B-C}2$, respectively. Then,
$\tan B \tan C = 3$
$\leftrightarrow \tan A+\tan B+\tan C = 3\tan A$
$\leftrightarrow \tan B+\tan C = 2\tan A$
$\leftrightarrow \tan (\gamma+\alpha)+\tan (\alpha+\beta) = 2\tan (\beta+\gamma)$
$\leftrightarrow \cot \beta+\cot \gamma = 2\cot \alpha$ (because $\tan(\frac{\pi}2-\theta) = \cot \theta$)
$\leftrightarrow \frac{\cos \beta}{\sin \beta}+\frac{\cos \gamma}{\sin \gamma}= 2\cdot \frac{\cos \alpha}{\sin \alpha}$
$\leftrightarrow \frac{\cos \beta \sin \gamma+\cos \gamma \sin \beta}{\sin \beta \sin \gamma}= 2\cdot \frac{\cos \alpha}{\sin \alpha}$
$\leftrightarrow \frac{\sin (\beta+\gamma)}{\sin \beta \sin \gamma}= 2\cdot \frac{\cos \alpha}{\sin \alpha}$
$\leftrightarrow \frac{2 \cos \alpha}{\cos (\beta-\gamma)-\cos (\beta+\gamma)}= 2\cdot \frac{\cos \alpha}{\sin \alpha}$
$\leftrightarrow \cos (\beta-\gamma)-\cos (\beta+\gamma) = \sin \alpha$
$\leftrightarrow \cos (\beta-\gamma) = 2 \cdot \sin \alpha$

Now, look at the picture. Red, yellow, blue angles are $\alpha$, $\beta$, $\gamma$, respectively. It's pretty easy to show that this is the case.
$\cos (\beta-\gamma)$
$= \cos \angle OAH$
$=\frac{AH}{OA}$
$=\frac{2OM}{OC}$
$=2\sin \angle OCM$
$=2\sin \alpha$
and we are done.[/quote]
this solution is a bit too long and too complicated...just draw the circumcircle of ABC...and things will be a lot clearer...there's no need to use trig... :|  :|  :P  :|

[quote="max_tm"]uuuh maybe I'm just being thick but I really don't know where you're going with this. If you do have a solution that doesn't use trig then that's incredible let's see it but lightrhee's solution is well done.[/quote]

I'd love to see a solution without trig too. Please post it.

lol, you can use coordinates, if you want.
Set the coordinate of B: $(0,0)$, A:$(a,b)$, C:$(c,0)$
and you can get the vertical coordinate of O and H
$\frac{a^{2}+b^{2}-ac}{2b}and \frac{a(c-a)}{b}$
Since they are equal, you can get an equation and simplify it in one step
$3a^{2}-3ac=b^{2}$ which is $tanBtanC=3$

BTW, it's obvious that the locus of A is a ellipse

i'd love to post the solution...but wut software do u use???i'm pretty new...so i don't know...sry...

i can explain a little though...if u could just follow along (if my explanation is not too vague for u...)

let O be the circumcenter of ABC, if u draw the perpendicular OM to BC, it should be parallel to the altitude AD, and it should be 1/3 of AD because the euler line is parallel to BC (the orthocenter...1:3)...now if u draw the circumcircle of ABC, and extend AD and meet the circle at E, then there would be AD*DE=BD*CD
and tanBtanC=(AD/BD)*(AD/CD)=(AD*AD)/(BD*CD)=(AD*AD)/(AD*DE)=AD/DE
now if u could just draw the perpendicular ON to AD, then there would be AN=EN, and OMDN would be a rectangle, so OM=DN, and because AD=3OM=3DN, AN=AD-DN=3DN-DN=2DN=2AD=EN=DE+DN, which means OM=DE...and i guess i dont have to point out wut that means...

could someone plzzzzz tell me wut sofware u use...cuz i believe my explanation is very very vague...i'm really new to math...

yes I see where you are going with the solution but I am a little confused at the 3rd line...??? I donno maybe I'm just missing something in my logic from one line to the next? Also, on the bus today I was thinking of a solution with the incircle and using BC as tangent. Will post a solution if I have the time and insight to complete it!  :P

[quote="lightrhee"][quote="bryant124"][quote="Phelpedo"]No. The Latex is built into this system, but you need Geometer's Sketchpad or something for diagrams. Check out the latex tab on the left side.[/quote]
yeah...i downloaded it...and then i put my solution there...dont have the image though...plz dont laugh at me if i have any typos or misuse of symbols...just correct me...thx...[/quote]

That's really nice and simple solution. The third line is a typo though. That line implies that D=P=M.[/quote]
o...thx for pointing it out...AD is supposed to be the altitude on BC...


Let $ABC$ be an acute-angled triangle and $D$ the midpoint of $BC$. Let $E$ be a point  on segment $AD$ and $M$ its projection on $BC$. If $N$ and $P$ are the projections of $M$ on $AB$ and $AC$ then the interior angule bisectors of $\angle NMP$ and $\angle NEP$ are parallel.

you should not post problems into the "solved" section (it is a kind of archive for problems from the unsolved section which have been solved). if you know the solution of a problem and want to present the problem to others, then post it in the "proposed&own" section.

as for this one, it's not important however, since it already has a thread:
http://www.mathlinks.ro/Forum/viewtopic.php?t=25947 , so let's continue discussing it there.

  dg

Let ES and EK be perpendicular to AB and AC we will prove ∠SEK and ∠NEP have common angle bisector. we have to prove ∠ENM = ∠EPM.
Let TR be a line parallel to BC passing through E so by simple angle chasing TEMN and REMP are cyclic. ∠ENM = ∠ETM and ∠EPM = ∠ERM.
ME is perpendicular bisector of TR so ∠MTR = ∠MRT so ∠ENM = ∠EPM as wanted.
we're Done.

As $\angle NMB = 90^{\circ} -\beta$, $\angle NME = \beta$, $\angle EMP = \gamma$, and $\angle CMP = 90^{\circ}$, the angle bisector of $\angle NMP$ is parallel to that of $\angle BAC$. Now if $B_{1}, C_{1}\in BC$ are such that $\overline{EB_{1}}\parallel \overline{AB}$, and $\overline{EC_{1}}\parallel \overline{AC}$, we require the angle bisectors of $\angle B_{1}EC_{1}$ and $\angle NEP$ to coincide. This, however is equivalent to $\angle B_{1}EN = \angle C_{1}EP\iff \angle ANE =\angle APE$. Denoting the first by $x$ and the second by $y$, from sine law applied on $\triangle ANE, \triangle APE, \triangle NEM, \triangle PEM, \triangle ABD, \triangle ACD$, we compute:
\[\frac{\sin x}{AE}\cdot\frac{EM}{\cos x} = \frac{\sin\angle EAN}{NE}\cdot\frac{NE}{\sin\beta}=\frac{\sin\angle EAN}{\sin\beta} = \frac{\sin\angle EAP}{\sin\gamma} =\frac{\sin\angle EAP}{PE}\cdot \frac{PE}{\sin\gamma} =\frac{\sin y}{AE}\cdot\frac{EM}{\cos y} \]
Therefore $\tan x = \tan y\Longrightarrow x=y$, and we're done.


Rem is right, induction is probably the easiest way.  The only non-trivial part of the induction is the fact that

$2<\left (1+\frac{1}{n}\right )^{n}<e<3$ for all $n\ge 6$.

[hide="Okay fine here is the induction proof: (Base cases left to reader :D )"]
[b]1) $n! < \frac{n^{n}}{2^{n}}$[/b]
Assume $(n-1)!< \left( \frac{n-1}{2}\right)^{n-1}\implies n! < n \left( \frac{n-1}{2}\right)^{n-1}$
We only need to show that \[n \left( \frac{n-1}{2}\right)^{n-1}< \frac{n^{n}}{2^{n}}=\left( \frac{n}{2}\right)^{n}\\ \leftrightarrow 2n(n-1)^{n-1}< n^{n}\leftrightarrow 2<\left( \frac{n}{n-1}\right)^{n-1}= (1+\frac{1}{n-1})^{n-1}\] [/hide]
It remains to prove that $n>6 \implies 2<(1+\frac{1}{n-1})^{n-1}$, or simply prove that $f(n)=(1+\frac{1}{n-1})^{n-1}$ is increasing (since $f(7)>2$. To prove its inreasing there are many methods :wink:. Now prove its limit is less than 3 :wink:  and you're done with the problem using similar equations...[/hide]

[quote="zgorkster"]Rem is right, induction is probably the easiest way.  The only non-trivial part of the induction is the fact that

$(1+\frac{1}{n})^{n}<e$[/quote]
Use $lim(1+\frac{1}{n})^{n}=e$

Yes, everyone knwos that... but what's the nicest proof you know of it??

You can't PROVE that $\lim_{n\to\infty}\left(1+{1\over n}\right)^{n}=e$. You can only prove that the limit exists and then decide to denote that limit by $e$.

Alternatively, you must define $e$ by some other limit. Then you can PROVE the above statement.


Find all solutions $ (x,y)\in \mathbb{R}\times\mathbb R$ of the following system: $ \begin{cases}x^3 \plus{} 3xy^2 \equal{} 49, \\
x^2 \plus{} 8xy \plus{} y^2 \equal{} 8y \plus{} 17x.\end{cases}$

Here is my solution to this problem .
Clearly $ x\neq 0$ so from the first equation we get $ y^2 \equal{} \frac {49}{3x} \minus{} \frac {x^2}{3}$

Subtituting in the second equation we get $ x^2 \plus{} 8xy\plus{}\frac {49}{3x} \minus{} \frac {x^2}{3} \minus{} 17x \minus{} 8y \equal{} 0$ or equivalently

$ (x \minus{} 1)(2x^2 \plus{} 24xy \minus{} 49x \minus{} 49) \equal{} 0$  . Now if  $ x \equal{} 1$ we get $ y \equal{} 4$ or $ y \equal{} \minus{} 4$

If $ 2x^2 \plus{} 24xy \minus{} 49x \minus{} 49 \equal{} 0$ we get $ 2x^2 \plus{} 24xy \equal{} 49 \plus{} x^3 \plus{} 3xy^2$  or

$ 2x \plus{} 24y \equal{} 49 \plus{} x^2 \plus{} 3y^2$ or $ (x \minus{} 1)^2 \plus{} 3(y \minus{} 4)^2 \equal{} 0$ so we get again the same solution $ x \equal{} 1$ and $ y \equal{} 4$

Finally the solutions are $ (x,y) \equal{} \left\{(1,4),(1, \minus{} 4)\right\}$


A positive integer M divided by six gives a remainder of three. Another positive integer N divided by six gives a remainder of five. What is the remainder when the sum $ M\plus{}N$ is divided by six?

$ M\plus{}N\equiv3\plus{}5\equal{}8\equiv\boxed2 \pmod 6$

Or, you could use examples.

$ M$ could be $ 9$. ($ 9 \div 6 \equal{} \text{remainder of } 3$)

$ N$ could be $ 11$. ($ 11 \div 6 \equal{} \text{remainder of } 5$)

$ M \plus{} N \equal{} 20$

When you divide $ 20$ by $ 6$ you could a remainder of $ 2$. So our answer is $ \boxed{2}$


basically i can't figure out this question.. it's review for grade 11 math..

Express 3* 2^x + 2^x as a power of 2.

please please help!!  :|

that doesn't make sense to me...

the answer in the back said 2^(x + 2)

but i don't get how you got 3+1

i get the part at (4)2^x.. but how did you get (3+1)??? i know i must sound stupid

oh wow.. thanks so much!! i feel stupid but i get it now!  :lol:


How many permutations of the word SEATTLE have the 'A' between the two 'T's?

[hide=" click to reveal hidden content"]

If we treat the arrangement $ TAT$ as a single letter, then we have $ 5$ letters, two of which (the two $ E$'s) are the same. This gives $ \frac{5!}{2!} \equal{} 60$ permutations.

[/hide]

But what about if there are other letters in between the Ts excluding the A? Or do those not count.

I think those count too.

If they do:

[hide="Solution"]We can use complementary counting to count the number of ways where the A is not between the Ts. We proceed with casework.

Case 1: A is the first letter. Then there are six spots after that, which the T's can allocate to in $ \binom{6}{2}$ ways. The The remaining 4 letters, S,E,E,L, can rearrange in 12 ways.

Similarly, we have when A is in the second spot, the T's must be after the A, in $ \binom{5}{2}$ ways, and the remaining letters can rearrange in 12 ways.

If A is in the third place, the T's can be both before the A, or both after the A. If they are both after, we have $ \binom{4}{2}$ ways, and then 12 ways for the rearrangement. If they are before, we have $ \binom{3}{2}$ ways and then 12 ways.

If A is in the fourth, then the T's can both be before or after the A. If they are both after, there are $ \binom{3}{2}$ ways, and then 12 ways for the remaining letters, if they are before, by symmetry there are $ \binom{3}{2}$ ways and then 12.

The rest of the cases follow by symmetry. (So we just double the count of the first three cases.) [/hide]

@limac. I'm sure you're correct: looking at the question again, I see that I misread it. Thanks.


Does anybody know how to solve problems similar to this;

 28x^6 - 3x^3 + 5 = 0   ?

There was somebody I know who couldn't solve it and I realized that I couldn't either.

hello, by substituting $ x^3\equal{}t$ we get a quadratic equation in $ t$
$ 28t^2\minus{}3t\plus{}5\equal{}0$.
Sonnhard.

Oh, thanks. I knew that it was something like that, but I kept on factoring x^3, instead of actually replacing it with a variable. Thank you very much  :lol: .


function
trigonometry
algebra proposed
algebra

Find all functions $ f$ defined on the set of all real numbers with real values, such that $ f(x^2 \plus{} f(y)) \equal{} y \plus{} f(x)^2$ for all $ x, y$.

Let the first term of $f(x)$ is $ax^{n}$ such that $n>2$
Taking $x=y$ yields $f(x^{2}+f(x))=x+f(x)^{2}$
lets look for the left side.
İts first term is  $a^{n+1}x^{n^{2}}$ and right side's first term is $ a^{2}x^{2n}$ so as you see
Degrees of the x's must be equal.thus we obtain $n=2$ which is wrong because of $n>2$ rule.

Lets take $n=2$ and look for first term.if you calculate you can easily see there is no solution.for $n=1$ we get
$f(x)=x$ so solution is $f(x)=x$

İ looked for socrates limk and i guarantee that my solution is the simplier one.

[quote="Mathematicalx"]Let the first term of $f(x)$ is $ax^{n}$ [/quote]
What would be the first term $ax^n$ of $f(x)=\sin x$ ?

Yes , i looked for not functions,i looked for polynoms which's coefficients are all integers

So you got a simpler solution to a simpler problem. :lol:
Congrats.


Find all ordered triples of primes $(p,q,r)$ such that:
$p$ divides $q^r+1$; $q$ divides $r^p+1$; $r$ divides $p^q+1$.

note that the three congruences :
i) q divides r+1 ii)p divides q+1 iii)r divides p+1
cannot hold together so at least one of them is false.
wlog let gcd(q,r+1)=1
we have that r^p +1 = 0 mod q
hence r^2p =1 mod q
hence  q-1 is divisible by p.
since q^r+1 =0 mod p hence 2=0 mod p hence p=2.
either r divides p+1 or not. if r does not divide p+1 then q=2. hence r=5.
the other case is that p=2,r=3, q=5

the other solutions can be obtained in similar manner.


Two circles $ P$ and $ Q$ that have equivalent radii pass through the other's center and intersect each other at points $ A$ and $ B$. Find the area of the circle that is tangent to the quadrilateral $ PAQB$ and whose center lies on the midpoint of $ AB$ in terms of $ r$. (If extra variables are necessary, let me know. I haven't done this problem yet.)

$ PAQB$ is a rhombus with side length $ r$. $ PQ \equal{} r$.  Let the radius of the inscribed circle be $ R$, and its center be $ C$.  We draw the altitude from $ C$ to $ AQ$.  We see that $ \triangle APQ$ is equilateral since all of its sides have length $ r$.  Therefore, $ \angle AQP \equal{} 60^{\circ}$.  Thus, we have in our right triangle that $ \frac {2R}{\sqrt {3}} \equal{} r/2$, so [u]$ R \equal{} \frac {r\sqrt {3}}{4}$[/u].

[hide]Immediately, we see that $ PAQB$ is a rhombus with side length $ r$. Also, the length of $ PQ$ is $ r$. Then $ PAQB$ is composed of two equilateral triangles, so we can draw the radii from the center of the inscribed circle to the sides of $ PAQB$ and use trigonometry:
\[ R\equal{}\frac{r}{2} \sin \frac{\pi}{3}\equal{}\boxed{\frac{\sqrt{3}}{4}r}.\][/hide]


[quote="Cezar Lupu"]Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:

       \[ ab(a^{2}+b^{2})+bc(b^{2}+c^{2})+ca(c^{2}+a^{2})\leq\frac{2}{27}(a+b+c)^{4}.  \][/quote]

Let $a=2, b=1, c=\frac{1}{2}$
$LHS=\frac{119}{8}$
$RHS=\frac{7^4*2}{2^4*27}=\frac{2401}{216}$

So $LHS>RHS$

[quote="Cezar Lupu"]\[ ab(a^{2}+b^{2}) + bc(b^{2}+c^{2}) + ca(c^{2}+a^{2}) \leq \frac{2}{27}(a+b+c)^{4}.  \][/quote]

Could you check it, and then post the right form? As it is now, it has a very small chance of being true.

Thanks.

[Edit] I posted in the same time with eLeM.

The general problem was posted here by me.  :P


In the square $ ABCD$ ,$ E \in DC$ so that $ D \in (EC)$    and   $ \frac{ED}{EC}\equal{}\frac{1}{4}$.If  $ F \in (BC)$ so that $ BF\equal{}DE$ and $ EF \cap AD\equal{}\{M\}$ then find the ratio $ \frac{AM}{MD}$.

WLOG, let $ AB \equal{} BC \equal{} CD \equal{} AD \equal{} 4$, then $ ED \equal{} BF \equal{} 1$, and $ FC \equal{} 3$. Notice that $ \triangle EFC \sim  \triangle EMD$ by AA similarity. So,

$ \frac{FC}{MD} \equal{} \frac{EC}{ED}$

$ \frac{3}{MD} \equal{} \frac{5}{1}$

$ MD \equal{} \frac{3}{5} \implies AM \equal{} \frac{17}{5}$

$ \therefore \frac{AM}{MD} \equal{} \frac{\frac{17}{5}}{\frac{3}{5}} \equal{} \boxed{\frac{17}{3}}$


calculus
integration
geometry
trigonometry
function
logarithms
calculus computations

Find the area bounded by $ 0\leq y\leq \frac{x|\sin x|}{1\plus{}|\cos x|},\ 0\leq x\leq n\pi \ (n\equal{}1,\ 2,\ \cdots).$

Our area is $ S\equal{}\int_{0}^{\pi n}{x\cdot \frac{|\sin x|}{1\plus{}|\cos x|}dx}$.
Function $ g(x)\equal{}\frac{|\sin x|}{1\plus{}|\cos x|}$ is pereodic with period $ T\equal{}\pi$. That is why we can write:
$ S\equal{}\frac{1}{\pi n}\cdot \int_{0}^{\pi n}{\frac{|\sin x|}{1\plus{}|\cos x|}dx} \cdot \int_{0}^{\pi n}{x dx}\equal{}$
$ \equal{}\frac{1}{\pi n}\cdot 2n\ln 2\cdot \frac{\pi ^2 n^2}{2}\equal{}\pi n^2\ln 2$

[size=117][color=brown][b]Thank you, [u]Kunny[/u] ! Nice and easy problem ![/b][/color][/size]

[quote="kunny"][color=darkred]Find the area bounded by $ 0\leq y\leq \frac {x|\sin x|}{1 \plus{} |\cos x|},\ 0\leq x\leq n\pi \ (n \equal{} 1,\ 2,\ \cdots)$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ I\equiv \int_0^{n\pi}\frac {x|\sin x|}{1 \plus{} |\cos x|}\ \mathrm {dx} \equal{}$  $ \int_0^{n\pi}\frac {(n\pi \minus{} x)|\sin (n\pi \minus{} x)|}{1 \plus{} |\cos (n\pi \minus{} x)|}\ \mathrm {dx}$  $ \implies$ $ \boxed {I \equal{} \frac {n\pi}{2}\cdot J}$ ,

where $ J \equal{} \int_0^{n\pi}\frac {\sin x}{1 \plus{} |\cos x|}\ \mathrm {dx}$ $ \implies$ $ \boxed {J \equal{} n\cdot K}$ , where

$ K \equal{} \int_0^{\pi}\frac {\sin x}{1 \plus{} |\cos x|}\ \mathrm {dx} \equal{}$ $ \int_{ \minus{} \frac {\pi}{2}}^{\frac {\pi}{2}}\frac {\cos x}{1 \plus{} |\sin x|}\ \mathrm {dx}$ $ \implies$ $ \boxed {K \equal{} 2\cdot L}$ , where

$ L \equal{} \int_0^{\frac {\pi}{2}}\frac {\cos x}{1 \plus{} \sin x}\ \mathrm {dx} \equal{} \int_0^{\frac {\pi}{2}}\frac {(1 \plus{} \sin x)'}{1 \plus{} \sin x}\ \mathrm {dx}$ $ \implies$ $ \boxed {L \equal{} \ln 2}$ .

In conclusion, $ I \equal{} \frac {n\pi}{2}\cdot J \equal{} \frac {n^2\pi}{2}\cdot K \equal{} n^2\pi\cdot L \equal{} n^2\pi\ln 2$ $ \implies$ $ \boxed {\ I \equal{} n^2\pi\ln 2\ }$ .[/color]

[quote="Virgil Nicula"][size=117][color=brown][b]Thank you, [u]Kunny[/u] ! Nice and easy problem ![/b][/color][/size]

[quote="kunny"][color=darkred]Find the area bounded by $ 0\leq y\leq \frac {x|\sin x|}{1 \plus{} |\cos x|},\ 0\leq x\leq n\pi \ (n \equal{} 1,\ 2,\ \cdots)$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ I\equiv \int_0^{n\pi}\frac {x|\sin x|}{1 \plus{} |\cos x|}\ \mathrm {dx}$ $ \implies$ $ \boxed {I \equal{} n\cdot J}$ [/color][/quote]

Why do you think so?

Check integral for $ n \equal{} 2$. I think it will be $ 4\pi \ln 2$

$ \int_{0}^{\pi}{\frac {x|\sin x|}{1 \plus{} |\cos x|}dx} \equal{} \pi \ln 2$
$ \int_{\pi}^{2\pi}{\frac {x|\sin x|}{1 \plus{} |\cos x|}dx} \equal{} 3\pi \ln 2$

and

$ \int_{0}^{2\pi}{\frac {x|\sin x|}{1 \plus{} |\cos x|}dx}\ne 2\cdot \int_{0}^{\pi}{\frac {x|\sin x|}{1 \plus{} |\cos x|}dx}$

i first splitted the integral as n*the integral between 0 to pi
then i split the integrals from 0 to pi/2 and pi/2 to pi

then evaluating the modulus of the 2 separate integrals and evaluating them i am getting the answer as n*pi/2

$ S \equal{} \int_{0}^{\pi n}{x\cdot \frac {|\sin x|}{1 \plus{} |\cos x|}dx} \equal{} n\int_{0}^{\pi }{x\cdot \frac {|\sin x|}{1 \plus{} |\cos x|}dx} \plus{} \frac {n(n \minus{} 1)\pi}{2}\int_{0}^{\pi }{\frac {|\sin x|}{1 \plus{} |\cos x|}dx}$

Sorry ! Yes, yes, yes ! I corrected above. Thank you !

can anybody please tell me where i go wrong??

$ \int_{0}^{\pi n}{\frac {x |\sin x|}{1 \plus{} |\cos x|}dx} \equal{} \int_{0}^{\pi n}{\frac {(\pi n \minus{} x) |\sin (\pi n \minus{} x)|}{1 \plus{} |\cos (\pi n \minus{} x|)}dx} \equal{}$
$ \equal{} \int_{0}^{\pi n}{\frac {(\pi n \minus{} x) |\sin x|}{1 \plus{} |\cos x|}dx} \equal{} \pi n \int_{0}^{\pi n}{\frac {|\sin x|}{1 \plus{} |\cos x|}dx} \minus{} \int_{0}^{\pi n}{\frac {x |\sin x|}{1 \plus{} |\cos x|}dx} \equal{}$
$ \equal{} \pi n^2 \int_{0}^{\pi}{\frac {|\sin x|}{1 \plus{} |\cos x|}dx} \minus{} \int_{0}^{\pi n}{\frac {x |\sin x|}{1 \plus{} |\cos x|}dx}$
So:
$ \int_{0}^{\pi n}{\frac {x |\sin x|}{1 \plus{} |\cos x|}dx} \equal{} \frac {\pi n^2}{2} \int_{0}^{\pi}{\frac {|\sin x|}{1 \plus{} |\cos x|}dx} \equal{} \frac {\pi n^2}{2} \cdot 2\ln 2 \equal{} \pi n^2 \ln 2$

Here is a simultaneous to write ! I have not time to correct because at mine  [b]preview[/b] does not go, I use directly [b]submit[/b]. Sorry !

Sorry for that I don't know the right answer, but my answer is same as Yuriy Solovyov's one.

Here is my solution.

$ I_n \equal{} \int_0^{n\pi} \frac {x|\sin x|}{1 \plus{} |\cos x|}\ dx$

$ \equal{} \sum_{k \equal{} 1}^{n} \int_{(k \minus{} 1)\pi}^{k\pi} \frac {x|\sin x|}{1 \plus{} |\cos x|}\ dx$ for $ k\geq 1,$

$ \equal{} \sum_{k \equal{} 1}^n \int_0^{\pi} \frac {(x \plus{} (k \minus{} 1)\pi)|\sin \ (x \plus{} (k \minus{} 1)\pi)|}{1 \plus{} |\cos \ (x \plus{} (k \minus{} 1)\pi )|}\ dx$

$ \equal{} \sum_{k \equal{} 1}^n \left[\int_0^{\pi}\frac {x|\sin x|}{1 \plus{} |\cos x|}\ dx \plus{} (k \minus{} 1)\pi\int_0^{\pi} \frac {|\sin x|}{1 \plus{} |\cos x|}\ dx\right]$

Now let $ I \equal{} \int_0^{\pi} \frac {x|\sin x|}{1 \plus{} |\cos x|}\ dx$ Let $ \pi \minus{} x \equal{} y$

$ \equal{} \int_{\pi}^0 \frac {(\pi \minus{} y)|\sin y|}{1 \plus{} |\cos y|}\ dy \minus{} \int_0^{\pi} \frac {|\sin y|}{1 \plus{} |\cos y|}\ dy$

$ \equal{} \pi \int_0^{\pi} \frac {|\sin y|}{1 \plus{} |\cos y|}\ dy \minus{} I\Longleftrightarrow I \equal{} \frac {\pi}{2}\int_0^{\pi} \frac {|\sin x|}{1 \plus{} |\cos x|}\ dx$

$ \therefore I_n \equal{} \sum_{k \equal{} 1}^n \left(\frac {\pi}{2}\int_0^{\pi} \frac {|\sin x|}{1 \plus{} |\cos x|}\ dx \plus{} (k \minus{} 1)\pi \int_0^{\pi} \frac {|\sin x|}{1 \plus{} |\cos x|}\ dx\right)$

$ \equal{} \sum_{k \equal{} 1}^n \left(k \minus{} \frac {1}{2}\right)\pi \int_0^{\pi} \frac {|\sin x|}{1 \plus{} |\cos x|}\ dx$

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

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

$ \equal{} \frac {1}{2}n^2 \pi \left(\left[ \minus{} \ln (1 \plus{} \cos x)\right]_0^{\frac {\pi}{2}} \plus{} \left[\ln (1 \minus{} \cos x)\right]_{\frac {\pi}{2}}^{\pi}\right)$

$ \equal{} \frac {1}{2}n^2\pi \cdot 2\ln 2 \equal{} \boxed{\pi n^2\ln 2}.$

What Long Proves. I will try to make one shorty.

$ f$ can be rewritten:

$ f(t) \equal{} \frac {|\sin (t)|}{|\cos (t)| \plus{} 1} \equal{} \begin{cases} \tan \left(\frac {t}{2}\right) & 0 < t < \frac {\pi }{2} \\
\cot \left(\frac {t}{2}\right) & \frac {\pi }{2} < t < \pi \end{cases}$

It's easy to prove that the period is $ T \equal{} \pi$:
Another step is to prove that the function is symmetric about the line $ x \equal{} \frac {\pi}{2}$. For this step we will simultaniously proove that and evalute the integral.

$ \int_{\frac {\pi }{2}}^{\pi } \cot \left(\frac {t}{2}\right) \, dt \equal{} \underbrace{\int_0^{\frac {\pi }{2}} \tan \left(\frac {z}{2}\right) \, dz}_{t \equal{} z \plus{} \pi } \equal{} 2 \underbrace{\int_0^{\frac {\pi }{4}} \tan (z) \, dz}_{\int \tan (x) \, dx \equal{} \minus{} \log (|\cos (x)|)} \equal{} \log (2)$

$ S \equal{} \underbrace{\int_0^{n T} t f(t) \, dt}_{\underbrace{t \equal{} n T \minus{} z}_{f(t) \equal{} f(n T \minus{} t)}} \equal{} \frac {1}{2} n^2 T \int_0^T f(z) \, dz$


Let $ a;b;c$ be positive real numbers satisfying $ abc \geq 1$
Prove that:
$ \frac{3a^3\plus{}2b^2\plus{}c}{\sqrt{4a^3\plus{}b^2\plus{}c}}\plus{}\frac{3b^3\plus{}2c^2\plus{}a}{\sqrt{4b^3\plus{}c^2\plus{}a}}\plus{}\frac{3c^3\plus{}2a^2\plus{}b}{\sqrt{4c^3\plus{}a^2\plus{}b}}\geq 3\sqrt{6}$

ChienThan: an inequality, not a inequality

[quote="pardesi"]can u please exlain ur first step Shishkin[/quote]
because $ x^2\minus{}2xy\plus{}y^2  \ge 0$ so $ \frac{x}{{\sqrt y }} \ge \sqrt {2x \minus{} y}$

It's very nice solution


Suppose that $ n$ is expressed in the decimal representation : $ n \equal{} \overline{a_ma_{m \minus{} 1}\dots a_1}$,where $ a_m\neq 0$.

Find all values of $ n$ such that $ n \equal{} (a_m \plus{} 1)(a_{m \minus{} 1} \plus{} 1)\cdot \cdots (a_1 \plus{} 1)$.

[quote="kunny"]Suppose that $ n$ is expressed in the decimal representation.

Find all values of $ n$ such that $ n \equal{} (a_m \plus{} 1)(a_{m \minus{} 1} \plus{} 1)\cdot \cdots (a_1 \plus{} 1)$.[/quote]
Do you mean that $ n \equal{} \overline{a_ma_{m \minus{} 1}\dots a_1}$,where $ a_m\neq 0$.If so then the problem is quite easy:
Solution:Suppose $ m\geq 3$.
If at least one of the numbers $ a_{i}\leq 8,1\leq i\leq m \minus{} 1$:
$ 10^{m \minus{} 1}a_{m}\leq n \equal{} (a_m \plus{} 1)(a_{m \minus{} 1} \plus{} 1)\dots(a_1 \plus{} 1)\leq 9(10^{m \minus{} 2}a_m \plus{} a_m)$,so after dividing to $ a_{m}\neq 0$:
$ 10^{m \minus{} 1}\leq 9\cdot10^{m \minus{} 2} \plus{} 9$,or
$ 10^{m \minus{} 2}\leq 9$,which is not true.
So $ n \equal{} \overline{a_m99\dots 9}$.
But then
$ n \equal{} 10^{m \minus{} 1}a_m \plus{} 10^{m \minus{} 1} \minus{} 1 \equal{} (a_m \plus{} 1)10^{m \minus{} 1}$,which is not possible,hence
$ m\leq 2$.
If $ m \equal{} 2$,then
$ 10a_2 \plus{} a_1 \equal{} (a_1 \plus{} 1)(a_2 \plus{} 1)$,from where $ a_1$ is an even number.
So if $ a_1 \equal{} 0,2,4,6$,then the equation has no solutions,hence $ \boxed{a_1 \equal{} 8,a_2 \equal{} 1}$.
if $ m \equal{} 1$,then
$ a_1 \equal{} a_1 \plus{} 1$,but it is impossible.
Answer $ n \equal{} 18$.

Actually, the $ a_m\equal{}0$ thing doesn't matter since it neither affects the product or the value given to the decimal number


[quote="mathVNpro"]Prove the following inequality:[/quote]
Let $ a,b$ and $ c$ be three positive real numbers . Prove that\[4\left |\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}+\frac{a^3-b^3}{a+b} \right|\leq(b-c)^2+(c-a)^2+(a-b)^2.\]

[quote="sqing"][quote="mathVNpro"]Prove the following inequality:[/quote]
Let $ a,b$ and $ c$ be three positive real numbers . Prove that\[4\left |\frac{b^3-c^3}{b+c}+\frac{c^3-a^3}{c+a}+\frac{a^3-b^3}{a+b} \right|\leq(b-c)^2+(c-a)^2+(a-b)^2\][/quote]
$\sum_{cyc}\frac{a^3-b^3}{a+b}=\frac{(a-b)(b-c)(c-a)(ab+ac+bc)}{(a+b)(a+c)(b+c)}$.
Let $a\geq b\geq c$.
Hence, we need to prove that
$(a^2b+a^2c+b^2c-ab-ac-bc)(a+b)(a+c)(b+c)\geq2(a-b)(a-c)(b-c)(ab+ac+bc)$.
We see that $(a-b)(a-c)(ab+ac+bc)\leq a^2b^2\Leftrightarrow a^2+ab+b^2\geq ac+bc$, which is obvious.
Hence, it remains to prove our  inequality for $c=0$ and $a\geq b$, which gives $(a^2-ab+b^2)(a+b)ab\geq2a^2b^2(a-b)$, which is obvious.


Hi,
suppose f: A→B is conformal, B is bounded, c is a closed curve in A.
Is f(c) then always a closed curve in B?

let $ C: [0,1]\rightarrow A$ be a closed curve , i.e $ C(0) \equal{} C(1)$
since $ f$ is conformal then $ f$ is one-to -one
this implies $ foC : [0,1]\rightarrow B$ and  $ f(c(0)) \equal{} f(c(1))$ , i.e, $ foC$ is closed curve

In fact, we really dont need the fact that B is boudned and f is conformal, we just need f is 1-1 analytic


Let a,b,c be the sides of a triangle. Prove that

  3 \leq (a 2 +b 2 )/(ab+c 2 ) + (b 2 +c 2 )/(bc+a 2) + (c 2 +a 2 )/(ca+b 2)<4

For the first part it is sufficient to apply the AM-GM inequality and then \prod (a 2 +b 2 ) \geq \prod (c 2 +ab), which can be easily proved by Cauchy-Schwartz. For the second part is suffices to prove that (a 2 +b 2  )/(ab+c 2  ) <2(a 2 +b 2  )/(a 2 +b 2 +c 2 ). But this is equivalent with c 2 >(a-b) 2 , which is true in all triangles.

I think I should say that this inequality was published by Marius Mainea in Arhimede mathematical Magazine and was not given in mathematical olympiads.


Among all possible representations of the positive integer $ n$ as $ n\equal{}\sum_{i\equal{}1}^k a_i$ with positive integers $ k, a_1 < a_2 < ...<a_k$, when will the product $ \prod_{i\equal{}1}^k a_i$ be maximum?

If we have two numbers $ a$ and $ a\plus{}k$ ($ k>1$), the product is $ a^2\plus{}ak$. If we reduce the difference between these numbers and mantain the sum, so that they become $ a\plus{}1$ and $ a\plus{}k\minus{}1$, then the product becomes $ a^2\plus{}ak\plus{}k\minus{}1>a^2\plus{}ak$. Hence, the product will be maximum when the value of $ \max_{i\equal{}1}^k\{a_i\}\minus{}\min_{i\equal{}1}^k\{a_i\}$ in minimum.

That doesn't tell you how to choose the best possible number of factors.  (As it happens, one should choose as many as possible such that all are larger than 1, but this needs to be proved.)  The resulting product is sequence [url=http://www.research.att.com/~njas/sequences/A034893]A034893[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] -- the entry and accompanying [url=http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Doslic/doslic15.pdf]article (.pdf)[/url] do not note that this appeared on the Miklos Schweitzer competition.


Let be $ a,b,c>0$ such that $ abc\equal{}1$ and two of the numbers are smaller or equal than $ \frac{1}{2}$ . Prove that :

$ 6(a\plus{}b\plus{}c)\minus{}(a^2\plus{}b^2\plus{}c^2)\le 15$

[hide]Completing the squares:

\begin{eqnarray*}
0&\le& 15+\sum_{cyc}a^2-6\sum_{cyc}a\\
0&\le& 15+\sum_{cyc}(a^2-6a)\\
0&\le& 15+\sum_{cyc}(a-3)^2-27\\
12&\le& \sum_{cyc}(a-3)^2
\end{eqnarray*}

But if WLOG, $ a,b\le\frac{1}{2}$ and $ c\ge 4$, then $ |a-3|,|b-3|\ge 2.5$, $ |c-3|\ge 1$, so the RHS is greater/equal to $ (2)(2.5^2)+1^2=13.5>12$, as desired. [/hide]

[hide="solution"]WLOG let $ a,b\leq \frac{1}{2}$. Therefore $ c\geq 4$. Our inequality is congruent to
\[ (6a\minus{}a^2)\plus{}(6b\minus{}b^2)\plus{}(6c\minus{}c^2)\leq 15\]
The maximum $ 6a\minus{}a^2$ can be is when $ a$ is maximized, which is when $ a\equal{}\frac{1}{2}$. Then the value is $ 3\minus{}\frac{1}{4}\equal{}2.75$. This goes for $ 6b\minus{}b^2$ as well. Since $ c\geq 4$, $ 6c\minus{}c^2\equal{}c(6\minus{}c)$ is maximized when $ c$ is as close as possible to $ \frac{0\plus{}6}{2}\equal{}3$ (the average of the zeroes of $ 6c\minus{}c^2$), or when $ c\equal{}4$. Then we would have $ 6c\minus{}c^2\equal{}24\minus{}16\equal{}8$.

Therefore the maximum value of $ (6a\minus{}a^2)\plus{}(6b\minus{}b^2)\plus{}(6c\minus{}c^2)$ is $ 2.75\plus{}2.75\plus{}8\equal{}13.5\leq 15$. Therefore

\[ 6(a\plus{}b\plus{}c)\minus{}(a^2\plus{}b^2\plus{}c^2)\leq 15\][/hide]


Let $x_{1}, x_{2}, \ldots , x_{n}$ be a sequence of integers such that
(i) $-1\le x_{i}\le 2$, for $i=1,2,3,\ldots , n$;
(ii) $x_{1}+x_{2}+\ldots+x_{n}=19$; and
(iii) $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=99$.
Let $m$ and $M$ be the minimal and maximal possible values of $x_{1}^{3}+x_{2}^{3}+\ldots+x_{n}^{3}$, respectively. Then $M/m=$

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

[hide="not rigorous"] The minimum is choosing as few $2$'s as possible, namely $0$. Obviously choosing $0$ has no effect, so we only pick $1$'s and $-1$. With $59$ $1$'s and $40$ $-1$'s, we get a $m = 19$.

The maximum is choosing as many $2$'s as possible. Say we pick $k$ of them. The sum of the squares is at most $4k+a$ where $a$ is the number of $\pm 1$'s and the sum is at least $2k-a$. We must have

$4k+a = 99$ and $2k-a \le 19$

which gives us $k \le 19$. Maximizing, we choose $k = 19$, $a = 23$. The resulting set of numbers is $19$ $2$'s, $21$ $-1$'s, and $2$ $1$'s, giving us

$M = 19 \cdot 8-21+2 = 19 \cdot 7$.

So $M/m = 7$.[/hide]


geometry
geometric transformation
inequalities theorems
inequalities

It ' s too old and I think nothing is from yourself


limit
ratio
integration
logarithms
real analysis
real analysis unsolved

I will give the obvious response and say Stirling works:

$ \lim_{n\rightarrow \infty}\frac{n}{\sqrt[n] n!}= \lim_{n \rightarrow \infty}\frac{n}{\sqrt{2 \pi n}^{\frac{1}{n}}\left( \frac{n}{e}\right)}= \lim_{n\rightarrow \infty}\frac{e}{\sqrt{2 \pi n}^{\frac{1}{n}}}=e$

[b]Lemma (well-known - [u]criterion of the radical[/u]).[/b] $ b_{n}>0\ ,\ n\in N^{*}\ ,\ \lim_{n\to\infty }\frac{b_{n+1}}{b_{n}}=L\in \overline{\mathcal R}$ $ \implies$  $ \lim_{n\to\infty }\sqrt [n]b_{n}=L$.

In the our case $ a_{n}=\frac{n}{\sqrt [n]n!}=\sqrt [n]b_{n}$, where $ b_{n}=\frac{n^{n}}{n!}$ and $ \frac{b_{n+1}}{b_{n}}=\left (1+\frac{1}{n}\right)^{n}\rightarrow e$. In conclusion, $ a_{n}\rightarrow e$.

If $ s_{n}= \frac{n!}{n^{n}}$ the [b]ratio test[/b] gives $ \frac{s_{n+1}}{s_{n}}= \left( \frac{n}{n+1}\right)^{n}\to \frac{1}{e}$. So the [b]root[/b] test gives the same limit but $ |s_{n}|^{1/n}= \frac{(n!)^{1/n}}{n}\to \frac{1}{e}$. Take reciprocals to get the answer.

EDIT: Virgil Nicula, I didnt see, I wasnt stealing. I hope its okay.  :blush:

[quote="misan"]in how many [i]different [/i]ways can you show  :oops:

$ \lim_{n\to\infty}\;\frac{n}{\sqrt[n]{\;n!\;}}\;=\;\boxed{e}$[/quote]

$ log \frac{\sqrt[n]{n!}}{n}= \frac{1}{n}\sum_{k=1}^{n}log(\frac{k}{n}) \to \int_{0}^{1}ln(x)dx =-1$, when $ n \to \infty$ so
$ \lim_{n\to\infty}\;\frac{n}{\sqrt[n]{\;n!\;}}\;=\;\boxed{e}$

[quote="didilica"]In view of these calculations it is natural to ask the following question:

What is the value of the following limit:

$ \lim_{n\rightarrow \infty}n\left(\frac{n}{\sqrt[n]{n!}}-e\right)=?$.[/quote]

[color=darkred][b]Why $ \lim_{n\rightarrow \infty}n\left(\frac{n}{\sqrt[n]{n!}}-e\right)$ ?

Exactly,[/b] $ \lim_{n\rightarrow \infty}\frac{n}{\ln n}\cdot \left(e-\frac{n}{\sqrt[n]{n!}}\right)=\frac{e}{2}$.[/color]

Thank you Virgil.

The reason I was asking that question is that I was interested in the asymptotic behavior of the sequence $ \frac{n}{\sqrt[n]{n!}}-e$.

I checked the calculations, you are right, I was not thinking of multiplying the sequence $ \frac{n}{\sqrt[n]{n!}}-e$ by a less powerful infinite.

Thank you for the hint.

[size=117][color=purple][b]It is O.K., [u]Didi[/u] ! I understood from the outset what you say.

Here is a proof for this interesting asymptotic behavior of the sequence $ e-\frac{n}{\sqrt[n]{n!}}$.[/b][/color][/size]

[color=darkblue][b]Lemma.[/b] $ \boxed{\lim_{x\to\infty}x[1-\ln(1+\frac{1}{x})^{x}]=\frac{1}{2}}$. Indeed, $ f(x)\equiv x[1-\ln(1+\frac{1}{x})^{x}]=$ $ x[1-x\ln(1+\frac{1}{x})]=$ $ x^{2}[\frac{1}{x}-\ln (1+\frac{1}{x} ) ]$ $ \implies$

$ f(x)=\frac{u(x)}{v(x)}$, where $ \{\begin{array}{c}u(x)=\frac{1}{x}-\ln(1+\frac{1}{x})\\\\ v(x)=\frac{1}{x^{2}}\end{array}$. Apply the [b]l'Hospital's theorem[/b] : $ \{\begin{array}{c}u'(x)=-\frac{1}{x^{2}(x+1)}\\\\ v'(x)=-\frac{2}{x^{3}}\end{array}\implies$ $ \lim_{x\to\infty}\frac{u'(x)}{v'(x)}=\frac{1}{2}$ $ \implies$ $ \lim_{x\to\infty}f(x)=\frac{1}{2}$.

[b]Particular case.[/b] $ \boxed{\lim_{n\to\infty}n[1-\ln(1+\frac{1}{n})^{n}]=\frac{1}{2}}$.

[b]Exercise.[/b] $ \boxed{a_{n}=\frac{n}{\ln n}(e-\frac{n}{\sqrt [n]{n!}})\rightarrow\frac{e}{2}}$.

[b]Proof.[/b] $ a_{n}=-e\cdot (\frac{n}{e\sqrt [n]{n!}}-1)\cdot\frac{n}{\ln n}=$ $ e\cdot \frac{\frac{n}{e\sqrt [n]{n!}}-1}{\ln\frac{n}{e\sqrt [n]{n!}}}\cdot u_{n}$, where $ \alpha_{n}=\frac{n}{e\sqrt [n]{n!}}\rightarrow 1$, $ \lim_{n\to\infty}\frac{\alpha_{n}-1}{\ln\alpha_{n}}=1$

and $ u_{n}=\frac{-\ln\frac{n^{n}}{e^{n}n!}}{\ln n}=$ $ \frac{x_{n}}{y_{n}}$, where $ \{\begin{array}{c}x_{n}=-\ln\frac{n^{n}}{e^{n}n!}\\\\ y_{n}=\ln n\end{array}$. Apply the [b]Stolz's theorem[/b] :

$ x_{n+1}-x_{n}=$ $ n+1+\sum_{k=2}^{n+1}\ln k-(n+1)\ln (n+1)+n\ln n-n-\sum_{k=2}^{n}\ln k=$ $ 1-\ln (1+\frac{1}{n})^{n}$

and $ y_{n+1}-y_{n}=\ln (n+1)-\ln n=$ $ \ln(1+\frac{1}{n})=$ $ \frac{1}{n}\cdot\ln(1+\frac{1}{n})^{n}$. Therefore,

$ \frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=\frac{n[1-\ln(1+\frac{1}{n})^{n}]}{\ln(1+\frac{1}{n})^{n}}\rightarrow\frac{1}{2}$ ([b]using the particular case of the above lemma[/b]). In conclusion, $ u_{n}\rightarrow\frac{1}{2}$ and $ a_{n}\rightarrow\frac{e}{2}$.[/color]

very nice, virgil nicula !  :)  :showoff:


Pour some soda water or soft drink to a glass and drop some raisins into it. What happens, and why?

Raisins are denser than the liquid, so initially they sink to the bottom of the glass
Then, there's a release of carbon dioxide bubbles. The bubbles stick to the rough surface of a raisin, the raisin is lifted because of the increase in buoyancy. But, when the raisin reaches the surface, the bubbles pop, and the carbon dioxide gas escapes into the air. This causes the raisin to lose buoyancy and sink. As long there are bubbles, the raisins will continue to rise and sink.


analytic geometry
trigonometry
limit
function
real analysis
real analysis unsolved

This is just for people's amusement - i won't post a solution for some time as it spoils the fun. Anyone who knows a solution from Bollobas' book please don't post it!
We work in the open unit disc in the plane. Point and Line play the following game - Point chooses a point $ x_{1}$ of the open unit disc $ \mathbb{D}$   then Line draws a line $ l_{1}$ through $ x_{1}$. Point chooses a point $ x_{2}$ on $ l_{1}$ contained in $ \mathbb{D}$ and so on. Line wins if the sequence $ (x_{n})_{n \equal{} 1}^{\infty}$ converges, otherwise Point wins. Show that Line can force a win.

Edit - I guess the disc should be closed...

I think maybe you left out extra conditions because a toddler could figure out the following..

Fix a point (a,b) somewhere at the top of the circle to give you some space below and fix some small $ \varepsilon$. Draw a line of length $ \varepsilon$ straight down from x. At the kth step draw a line from (a,b) to (a,b+$ \frac {\varepsilon }{{2^{k \minus{} 1} }}$). Sequence of points converges by Cauchy criterion.

I wonder how it looks like in n dimensions if Line is allowed to choose hyperplanes of dimension n-1 in a closed unit ball (and, of course, we have some tougher conditions).

[quote="§outh§tar"]Draw a line of length $ \varepsilon$ straight down from x.[/quote]  Either you haven't understood the problem or I don't understand your solution at all: Line must draw the entire line (or, if you like, chord of the circle) through the point Point has chosen -- she doesn't get to choose the length of the segment she draws (except to the extent that it's determined by the direction she chooses to draw the chord in).

Whether the disk is closed or open doesn't make a difference.  (If Point could escape in the open disk, he can just choose to restrict his moves to the open disk of radius $ \frac{1}{2}$ and play his winning strategy there.)  The "natural" strategy is for Line to draw the chord perpendicular to the radius through the point that Point has chosen, but I wasn't able to quickly come up with a smart estimate to show that this actually prevents Point from escaping.

If the disk was closed then surely the line can "weave" its way to the boundary. Make the line go orthogonal to the vertical where the vertical is pointing upwards to the furthest point on the circle from the current point.

That way, the point can only go closer and closer to the boundary, unless it decides to be stubborn and not move, in which case it has converged to its current point.

Obviously this technique should work for an open disk too and also I would think for higher dimensions as §outh§tar said.

Forcing Point closer to the boundary does not ensure that Line wins without further argument.  (What is to keep Point from spiraling around the origin in a smaller and smaller annulus?)

Represent the points in polar coordinates as $ (r_j,\theta_j)$ and employ the "natural" strategy. We have that the sequence $ (r_j)$ is nondecreasing, and WLOG we can have $ (\theta_j)$ also nondecreasing.

The key equation is that $ r_{j \plus{} 1} \equal{} r_j\sec(\theta_{j \plus{} 1} \minus{} \theta_j),$ or $ \cos(\theta_{j \plus{} 1} \minus{} \theta_j) \equal{} \frac {r_j}{r_{j \plus{} 1}}.$

I haven't worked out all the details, but it looks to me that you can cause trouble by letting $ \theta_{j \plus{} 1} \minus{} \theta_j$ behave asymptotically like $ \frac {C}{j}$ for some constant. Then $ \theta_j$ would increase without bound (at a logarithmic rate) but with $ \cos(\theta_j) \equal{} 1 \minus{} O(j^{ \minus{} 2}),$ the infinite product for the $ r_j$ can still converge.

Edit: OK, details. Use the closed disk, because I'm going to let $ r_j\to 1.$

Let $ r_j\equal{}\frac{j}{j\plus{}1}.$ Then,

$ \frac{r_j}{r_{j\plus{}1}}\equal{}\frac{j(j\plus{}2)}{(j\plus{}1)^2}\equal{}1\minus{}\frac1{(j\plus{}1)^2}\equal{}\cos(\theta_{j\plus{}1}\minus{}\theta_j).$

Then $ \theta_{j\plus{}1}\minus{}\theta_j\equal{}\cos^{\minus{}1}\left(1\minus{}\frac1{(j\plus{}1)^2}\right)\equal{}\frac{\sqrt{2}}{j\plus{}1}\plus{}o(j^{\minus{}1}).$

Then $ \lim_{j\to\infty}\theta_j\equal{}\infty$ by comparison to the harmonic series.

Point escapes, continually circling the ring and refusing to be cornered.

I'm guessing for this to not work, you will get to a point where letting \theta_{j+1} = \theta_j +\frac{C}{j} will not be permissible, i.e. will take you outside of the circle. That is to say if you just "reset your axis" after each turn so that \theta_j = 0 , the distance from the boundary prevents you from choosing such a value of \theta_j for your next turn.

Perhaps the relation \theta_j+1-\theta_j < inverse_cos(r_j) and r_j converges to one shows that the sum of values of theta converge.

Will have a go at this when I get some time :)

§outh§tar - Line can't choose the length of the lines - the problem is trivial otherwise.

Kent - I haven't taken the time to read your argument but I have a feeling there's a mistake somewhere in there as it is actually Line who has the winning strategy.

JBL- I think whether or not the disc is open does make a difference, if not point just chooses points getting closer and closer to the boundary and wins.

But EBz89, if the point can always beat the line on an open disk, surely it can just "pretend" that when playing on the closed disk, it is on an open one and by the same token, win that game?

[quote="EBz89"]JBL- I think whether or not the disc is open does make a difference, if not point just chooses points getting closer and closer to the boundary and wins.[/quote]  It's now been asserted by one person in this thread that being forced to approach the boundary results in a win for Line, and by you that being forced to the boundary results in a win for Point.  The former, as I noted, is wrong without further argumentation; the latter is just wrong.  (What's wrong with converging to a point on the boundary, after all?  Indeed, this is the only place Point can be "trapped.")

Kent's computation seems right to me ... which is unfortunate, since it means the "natural" strategy fails.  So, perhaps the problem is in error, or perhaps there's some other strategy (but I would be more likely to bet on the former).

[quote="JBL"]... or perhaps there's some other strategy (but this seems rather doubtful to me).[/quote]
Here's a vague notion, with no supporting details.

Under the natural (perpendicular) strategy, I had Point always play to the left (counterclockwise) so that $ \theta_j$ would increase.

Perhaps Line could try tilting the chord (by a carefully chosen amount) so that if Point plays to the left without increasing $ r_j$ by too much, then $ \theta_j$ won't increase by enough. By playing to the right, Point could actually get $ r_j$ to decrease by a little bit, but at the price of backing $ \theta_j$ up into the territory of angles already used.

... that's going to take a lot more work to flesh out. We need a strategy for Line that uses not just the most recent point but also the point before that.

I've found that \theta_j+1 < arccos(r/cos(\theta_j)cos(\theta_j-1)........cos(\theta_1))

(The thetas here are what we add on each time, I "reset the axis" so we just have a value of r at each turn)

I suppose if you could show this to have a sum which converges, you would be done, although I feel I've gone down the wrong road, sorry!

JBL - for some reason I had it in my head that point wins if the sequence doesn't converge to a point of the disc $ \mathbb{D}$ however it now occurs to me that this not stated in the problem and as such presumably isn't one of the rules. My mistake! (And yes, I agree that with the correct rules the open/closedeness of the disc is irrelevant.) Anyway, nice to see the problem is resulting in so much discussion.

Can you post the solution to this infuriating problem now?

Here's the hint that Bollobas gives (in his problem book [i]The Art of Mathematics[/i]): Line can force Point to stay outside larger and larger compact sets, thereby forcing convergence.  The solution given is 2.5 pages of nontrivial work.  The problem was posed and solved by Maly and Zeleny.

The hint seems fairly useless but, please, give me a week (I haven't read the book, so I assume I am allowed to try). I just cannot believe that I overlooked this nice thread completely.  :o

I can give a more substantive hint, but it looks like you don't want it yet.  Let me know if you change your mind.

Apparently, Maly and Zeleny used this Point-Line result to give an alternative proof of Buczolich's result that there is a differentiable function $ f \colon \mathbb{R}^2 \to \mathbb{R}$ such that $ \nabla f(0) \equal{} 0$ and $ | \nabla f | \ge 1$ almost everywhere.  In turn, Buczolich's result solved Weil's gradient problem (in the negative), which had been open for nearly 30 years.


I made this topic because I was too lazy to wait until kyyuanmathcounts' opened.
You can't say scores or difficulty, only whether you made states or not.
Also, say what state youre from if you did make states

btw, I didn't have chapter cuz delaware's cool like that :coool:

Yes, going from Santa Clara Valley chapter to Northern California competition.

California's so populated we have a north and south... Last year, I was 4th in the north, but some guy from southern got the last national team spot, and I'm like gaaaaaaaaaaaaah

3 teams out of 5 could go from my chapter. We had a very strong team this year, finishing with a high score (but exact info unrevealable) and we made states easily.

Yes we're going to states (yay!). Our team got 5th, and the top 5 teams out of 20 something teams go to state, so yay!  (Eastern Chapter, KS).

Our chapter had.. uh.. I can't remember, but a lot of teams.
Top 6 from our chapter would make it. We got 5th! yay

Top 4 teams were guaranteed, 5-6 hada chance at making it (and they did). My team was #3.

Top 12 individuals went to state and I got first individually.

congrats to all state qualifiers! (wow i never realized that AoPS was that smart)

I don't have a team, but I made it to the states individually. The other teams that passed the chapter competition were pretty strong.

But I don't need a team. :P

Yes. WE GOT 1ST IN CHAPTER!!!!
[i][b]Most[/b][/i] of the other teams were bad.

[color=orange][size=75][solafidefarms reminds you that you are not to make ANY, ANY, remarks on ANY question. [/size][size=150]!!![/size][size=75]][/color][/size]

1st? Wow! Good job.

I was in 14th. I missed the countdown round by 2 places.  :furious:

Good job to everyone!!!

I was 11th. I missed the countdown round by 1 place  :furious:

My school, LTMS, won the Bluegrass Chapters.
Our Team placed 1st
I placed 8th (by winning a TON OF TIEBREAKERs).
I was on the team. (so it didn't matter if I placed or not)
I only beat one person in countdown. :bomb:
There were 12 people who did countdown.

I was 9th overall last year and missed Countdown by one.

This year I did well at chapter but it doesn't matter much -- state is what counts, and in Indiana... eek.

In Northern Virginia, the top team plus two individuals made states. We swept all six.

I just realized that chapters are over tommorow... wow I am stupid

In Pennslyvania there are to many chapters. Only top team and top 4 individual make it individuals on team count. But I made it from coutdown

Nice
I made it last week with what it says in my signature

Darn, I got second in written, if the competition thingy was over I could announce how many points the other guy beat me by. :bomb:
apusey do you live in orange county?

I live in the Lake Charles Chapter of Louisiana. I got second on written, first in countdown, and our team got first. Also, the individual who got first was on our team.

So yeah, we beat the other teams pretty badly.

Aight, competition over!     [img]http://www.vtk.ugent.be/forums/images/smiles/yeah.gif[/img]

I got 1st! :winner_first:

My team made it too, 4th i.e.

Now can we discuss the chapter competition?

Okay this is what happened

I am from Oakland Chapter one of the most toughest chapters in the entire state of Michigan.

I got fourth place before countdown, and then after countdown I was first place.

Our team was second place (Detroit Country Day School)

In our chapter Countdown counts towards your final ranking, so I technically was the first place individual and countdown.

Also, as with Biffandoc, are we allowed to discuss Chapter Round Questions

[quote="mathgeniuse^ln(x)"]Biffandoc, are we allowed to discuss Chapter Round Questions[/quote]
We can say scores and about the competition.
I advise that you don't talk about specific questions until MATHCOUNTS posts them because there may be people still taking it (delays)

I got to state. My team lost by a quarter of a point  :mad:  again, just like last year at state  :mad: .

did you guys get 65.75? because that would suck soooo bad

no. my team (individuals) got 46 (me), 41, 40, and 34. And we got a 10 on the team; the other school just had better individuals.  If my brother (6th grade) had been on the team instead of the 34, we would have won :mad: because he got a 37.

i made it in oklahoma.
my team got 3rd, and i got 1st individually. yippee! :lol:
my team was going to state anyway, top 4 go to state.
this year actually was a bad year for my city, we usually get first or second :(

[quote="xboteb13"]Yep, I'm going to State! Only the top 4 teams and top 4 individuals go to Florida's State Competition. Fortunately, two of my team's members (including me) made it in the top four in Countdown, and the team placed 2nd overall with a score of 9 out of 10 (IF THE OTHERS HAD LISTENED TO ME, WE WOULD HAVE HAD 10 OUT OF 10!!!!!!!!!!!!!!!!!!!!!!!!!!!)[/quote]

In my chapter in florida only 2 teams go to state. Our team came in third(it stunned the whole room) after we had two consectutive years of winning state. I had very bad luck during countdown....... of the 2 questions i got both werre the only 2 i didn't know how to do i got 6th!!!!! and i had made state last year............

How does the Countdown work when official? Does it alone determine the ranking, or is it worked into the written test scores?

In my chapter it determined the official ranking

[quote="tz"]I got moved 1 place back in countdown.
The 6 countdown trophies were all claimed by the 4 members of my team (including me) and the 2 others were claimed by 2 people from the 2nd place team.[/quote]

6 trophies???
we only got 2
our team missed first place by half a point
but were going to state anyways

Alone? As in the written test was purely for getting into the Countdown Round?

[color=red]Moderator's note:  Now, you may talk about the Chapter competition.[/color]

my team go first in our chapter and i got third individually (but was beaten by a teammate and an individual from our school, so it's not so bad)
I'm going to state (this saturday.)
deja vu... didn't I already post this?

The homeschool team I'm on is going to state (we got 6th). I got 4th individually (same as last year) and 4th countdown (countdown is unofficial for us). I can't believe that states are tommorow.


Try to solve this problem by producing linear equations or by any other method>

On family night, tickets to a ball game cost $4.50 for the first person and$2.50 for each accompanying member of the same family. In one section 480 ticket holders paid $1,550. How many$4.50 tickets were sold for that section?

Equation 1: 4.5x + 2.5y = 1,550
Equation 2: ????????????????????????

Please HELP! Thanks!

It should be at preolympiad or idermidiate


What do you get if you do well on these contests?

Are there prizes? certificates? anything?

... and if so, how well does one have to do?

I don't think that anyone actually cares about this contest because it's so easy and informal.

Well, I think you'll get a certificate if you get a decent mark. But yeah, it is really informal.

It's actually nice to have some contests that do not need to be prepared for...^^

I've heard that if you get a perfect score on all of the contests in a particular year, you get a very nice plaque - apparently the nicest of all the high school math contest plaques, some sort of blue shiny thing.  But I haven't got a perfect score anymore... so I don't care anymore.


ratio
AMC
AIME
LaTeX
Support
arithmetic sequence

a BIG [size=200]LOL[/size] to Treething.

If anybody has a problem with my questions, please look away.

I post questions here that I have not been able to solve myself.

Otherwise, why post questions, right?

I have better things to do than to post questions here like a maniac.

I am preparing for a June 2007 state exam, which I must pass to get my license.

This is only the millionth time I've had to say this on AOPS.

To Treething---->LOOK AWAY!

Looking away is simple, right?

:rotfl:

[hide]I tried solving it, but the answer ends up being irrational...[/hide]

[quote="Interval"]If anybody has a problem with my questions, please look away.

I post questions here that I have not been able to solve myself.

Otherwise, why post questions, right?

[b]I have better things to do than to post questions here like a maniac.[/b]

I am preparing for a June 2007 state exam, which I must pass to get my license.

This is only the millionth time I've had to say this on AOPS.

To Treething---->LOOK AWAY!

Looking away is simple, right?[/quote]

*cough* Does anyone else notice the irony here?

interval, 4 the millionth time, post in [size=200]CLASSROOM MATH!!!!![/size]

[/hide]

[quote="Treething"]Over a period of four days, Interval posts 139 annoyingly basic math questions. On the first day, he posts 10 questions.

On the first three days of his posting spree, the amount of questions in each day form an arithmetic sequence.

On the last three days of his posting spree, the amount of questions in each day form a geometric sequence.

How many questions does he post on the last day?[/quote]
OMG! I solved it!!!!
[hide]So the first day he solves 10 problems. The secind day he posts 10+r problems. So on the third day he posts 10+2r problems. So on the fourth day he posts $\frac{10+2r}{10+r}(10+2r)$. So adding all the days and setting it equal to 139 you should get the following equation $\frac{7r^{2}+100r+400}{r+10}=139$. So simplyfing this you should now get $7r^{2}-39r-990=0$. Now if you have a CASIO calculator, you can let it do the solving. When you do you should get that r=15. So plugging it back in you should get this
1st day           2nd day         3rd day         4th day
10 posts          25 posts         40 posts       64 posts
So the arithmetic sequence is adding 15 and the geometric ration is 1.6 So one the last day he posted 64 problems[/hide].
This was very annoying problem but at the same time very hilarious.

Oh, I must've made a mistake somewhere...

[quote="mathnerd314"]Treething, :rotfl: Is the question even possible?

Interval, if you are looking for a math tutoring website (as you have expressed in your thread, "AoPS Math Tutors"), run a google search. This is not a math tutoring website.[/quote]
Treething is a master at writing questions (though this is incredibly similar to an old AIME question)
[hide]10, 10+r, 10+2r, $\frac{(10+2r)^{2}}{10+r}$
$20+3r+\frac{(10+2r)^{2}}{10+r}=139$
$30r+3r^{2}+100+40r+4r^{2}=1190+10r$
Now one just needs to solve this quadratic.[/hide]

cmon guys, just because we all know that interval's posts are really annoying, doesnt mean we have to make fun of his posting.  just dont answer them (although it is a little irritating when half of all the topics on the first page are by interval)

I agree with jli that if you find that someone is posting problems that are too easy for you, then don't read their posts.

Most of Interval's questions do indeed fall under 'High School Basics'; they are part of the standard high school curriculum.  Most are too hard for 'Classroom Math in Middle School,' so this is the appropriate place to post them.  I do suggest Interval space out the posts a little more in time, so as to 'mix' the posts in more with posts from other people.  However, if people are really bothered by the incidence of Interval's posts, they can solve this problem by posting other appropriate problems of their own.

While this is not a 'math tutoring' website, it is a place people should feel free to come and ask for help with problems without being abused.  It isn't a place someone should come and have the Community routinely do their homework for them.  However, I don't think that's what Interval is doing.

As a side note - while I (again) suggest spacing out the frequency of questions, it is nice to see a potential teacher taking such an interest in mathematics as to spend a great deal of time on the site.

^^^^^^^^^^^^^^^^^^^^^^^^

Give a beer to rrusczyk!

Anyway I have no problem helping Interval out except it is wierd that he comes to us before he consults a mathbook.

[quote="Interval"]I never post questions WITHOUT consulting my mathbooks (more than 1 book) and various net sites.  The questions posted here are the ones I played with on paper and was not able to solve.

interval[/quote]

wow then your textbooks suck, no offense

mine is extremely simple it splits the homework problems so that if you have trouble, you can refer to an example with a similar problem

to avoid this kind of problem again here are two things i recommend....

1. LEARN LATEX!!!! you have near 700 posts AND you seem to grasp certain latex concepts but im guessing u r too lazy to add dollar signs...

2. Post what work you have so far on problems you don't understand....

and i remember 1 problem where there were 8 #s on a spinner, 1-8, and what was the prob. of spinning an odd #? could you really not get that 1?

Um... this is starting to be not about math.

I agree. Either have a mod close this matter once and for all or either move it to the support forum?

Yes, this is definitely off topic.

Locked.


Let $a$,$b$ and $c$ be real numbers such that $abc=1$. Prove that a least one of the numbers $2a-\frac{1}{b}$,$2b-\frac{1}{c}$,$2c-\frac{1}{a}$ does not exceed $1$.


See the rules [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205339]here[/url].

Reminder PM's will be sent at the start.

You are NOT ALLOWED to edit posts!  Except if you're me.

Also, I will post vote counts.  DO NOT post vote counts!

NO PM alliances, unlike in Elect the Executioner I.  If you want to make an alliance, it must be publicly stated on this thread.

SIGNUPS:
1. pacman2812
2. cf249
3. mewto55555
4. budi713
5. Brut3Forc3
6. isabella2296
7. crazychinesemaniac
8. Bugi
9. chenhsi
10. mathblitz

[quote=]i will join[/quote]
Yes, I will.

/in definitely

EtE is awesome!

I modded/started 4 games at MafiaScum.

im in but you will have to pm me frequently because school work may get in my way

I've decided to bump it back down to nine...using random.org, I will pick one of you to leave.  You are numbered in the order that you signed up.  I pick ten different numbers from 1-10, and whoever's number is picked last dies.

[hide="pacman2812"]
[color=green]YOU ARE SAFE[/color]

[hide="cf249"]
[color=green]YOU ARE SAFE[/color]

[hide="mewto55555"]
[color=red]SORRY, YOU DIED.[/color]
[/hide][/hide][/hide]
[hide]
[quote="random.org"]
Here is your sequence:

    * 2
    * 4
    * 9
    * 10
    * 1
    * 5
    * 7
    * 6
    * 8
    * 3

Timestamp: 2008-10-27 00:47:05 UTC[/quote][/hide]

[b]Elect the Executioner Round I begins now.[/b]

[b]POINTS TO GIVE[/b]
pacman2812: 3
cf249: 3
budi713: 3
Brut3Forc3: 3
isabella2296: 3
crazychinesemaniac: 3
Bugi: 3
chenhsi: 3
mathblitz: 3

[b]POINTS RECEIVED[/b]
pacman2812: 0
cf249: 0
budi713: 0
Brut3Forc3: 0
isabella2296: 0
crazychinesemaniac: 0
Bugi: 0
chenhsi: 0
mathblitz: 0

ONCE POINTS ARE GIVEN, THEY MAY NOT BE TAKEN BACK.  YOUR POINTS MUST BE IN BOLD, OR ELSE THEY WILL NOT BE COUNTED.

[b]THE DEADLINE IS HALLOWEEN NIGHT, 11:59 PM[/b]

[b]1 to Brut3Forc3, 1 to chenhsi.[/b]
Go scummers!

[hide="CLICK FOR POINT COUNT"]
[b]POINTS TO GIVE[/b]
pacman2812: 1
cf249: 3
budi713: 3
Brut3Forc3: 3
isabella2296: 3
crazychinesemaniac: 3
Bugi: 3
chenhsi: 3
mathblitz: 3

[b]POINTS RECEIVED[/b]
pacman2812: 0
cf249: 0
budi713: 0
Brut3Forc3: 1
isabella2296: 0
crazychinesemaniac: 0
Bugi: 0
chenhsi: 1
mathblitz: 0

ONCE POINTS ARE GIVEN, THEY MAY NOT BE TAKEN BACK.  YOUR POINTS MUST BE IN BOLD, OR ELSE THEY WILL NOT BE COUNTED.

[b]THE DEADLINE IS HALLOWEEN NIGHT, 11:59 PM[/b][/hide]

hmm, well obviously I will vote [b]pacman2812[/b]

shouldnt have killed me pacman....

[b]1 vote to james4l[/b]

[b]Vote: Pacman2812[/b] GG both of you two

GG all.

I do notice that the votes have been alternating...

[b][u]Who will be the Executioner?[/b][/u]
[b]james4l (3):[/b]cf249, mathblitz, pacman2812
[b]pacman2812 (3):[/b]james4l, budi713, crazychinesemaniac
[b]Not voting (3):[/b]isabella2296, Bugi, Dojo

With [b]9[/b] players, it's [b]5[/b] to win.

[b][u]Who will be the Executioner?[/b][/u]
[b]james4l (5):[/b]cf249, mathblitz, pacman2812, Dojo, Bugi
[b]pacman2812 (3):[/b]james4l, budi713, crazychinesemaniac
[b]Not voting (1):[/b]isabella2296
With [b]9[/b] players, it's [b]5[/b] to win.

[b]james4l[/b] is the Executioner...if he wants to sacrifice himself for pacman2812 is his decision.

Um just a question...
Why would he want to do that? :maybe:

well it is an obvious decision... [b]pacman2812[/b]

Good game, once again, it was a well-executed one. (no pun intended)