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USA(J)MO

We're taking the AMC 10 A

tomorrow?

Haha, wow.

And after my performance on the AIME (I)

I don't even want to take it.

(And I'm really sorry if this somehow goes against the ban of the discussion of the AIME II).

At least people at your school don't mistake the urinal for a water fountain.

At least your school lets you take the AMC

At least your school didn't ban rock paper scissors because people were betting money on it.

At least your school doesn't respond with blank silence when you ask it what 2 and 2 is.

Seriously, it's like talking to a wall. Except it's four walls and a ceiling, with classrooms inside.

At least people at your school don't think you are a nerd becasue you are always reading a P. G. Wodehouse.

[quote="zeb"]Seriously, it's like talking to a wall. Except it's four walls and a ceiling, with classrooms inside.[/quote]

And what would happen if you painted that wall? In particular, what if the walls were 12 feet long, 10 feet wide, and 8 feet high in each room, there were 3 rooms, and doorways and windows occupied 60 square feet in each room?

I messed up on that question because I transposed the last two digits :wallbash:

Then I would get an 8 on the AIME.

It's always more interesting when you're right on the cutoff line. I think it makes life exciting. (generally, I put things off until the last minute as well. The adrenaline rush sometimes makes up for the mind-numbing stupidity of the assignment. Of course, when I say the last minute, I mean the LAST minute - as in midnight, a week after it's due, submitted by e-mail with a late pass. I generally do this when the assignment is worth a major part of my grade. Also, which story is better: "I filed my taxes six weeks ahead of time," or, "My house was surrounded by choppers and guys with machine guns started pouring out to rescue the hostage I took to buy me an extra five minutes to finish claiming dependants. I electronically submitted the tax return, and we made it out of the building 3 and a half seconds before it exploded - and that's how me and your mother met for the first time."?)

[quote="b-flat"]My school got a new web security system.  Of course, it blocks almost every single site, including all .org sites  :?:   Yes, Wikipedia & my school's website & one of my teacher's websites included.  Also, they block most pictures from showing up on a google image search and they block AoPS and Mathlinks.

However, some people seemed to be on MySpace...  :roll:[/quote]

Same. They block AoPS, Myspace, Google Images (so its impossible to do a project at school now), and most times random (yet helpful) info sites. However, they close the filters at 2:15 PM so it's possible to access some sites (like AoPS) at that time but some sites are blocked 24/7 (Myspace, Google Images, etc.)

i can get on AoPS from school, but then again, since nobody else [color=blue]in my school (and only one other in the district)[/color] knows about AoPS, there's no point in blocking it

[quote="pianoforte"]i can get on AoPS from school, but then again, since nobody else knows about AoPS, there's no point in blocking it[/quote]
Hey, you're that person from XP Math, aren't you?

AoPS actually is well known. My school doesn't block it.

[quote="zeb"][quote="Temperal"]AoPS actually is well known.[/quote]

Stop dreaming. You remind me of the good old days when I convinced myself that all of my friends new what Pi Day was, and they were just pretending not to. :o

All the shattered dreams... the memories, they hurt! :noo:[/quote]
It 's the most well-known math website on the internet. You can't argue with that.
It's like saying "Peter the Great is not well known because less than half of people know who he is."

Oh sh-t, I completely left out "nobody else [b]in my school[/b]". that's what i meant  :oops:

Oh, I'm the only one in my school who knws the triangle inequality... :huh:

oh my teacher denies that "i^i" has infinite values....

I'm in 7th grade and i have to make the keys for answers to practice contests (like matheletics). Of course, I'm not uncredibaly good at math (you think 21 in mathcounts states is good?)

Atleast your school isn't my school!!
In my school(hell),
a)The teachers didn't know that there is an IMO until I told them so.
b) People think I'm a nerd coz I read Arthur Clarke, Wodehouse and Asimov between classes and I get an aggregate of about 92%(YUKK).
c)One of the teachers was caught reading "Alice in Wonderland" in class.
d) The students(or atleast majority of them) think that 3 is an irrational number and that you CAN divide anything by zero.
e)The vice-principal likes to eat cotton candy during morning prayers.
f)The cafetaria lady uses the drinking water fountain to wash her feet.
g)The hindi teacher speaks tamil, the english teacher speaks hindi and the maths teacher speaks rubbish.
h)95% of the students cannot survive without plastering themselves with make up atleast twice an hour.
i)The worst-ALL the students...THINK IT'S COOL TO FLUNK IN MATH!!!
Oh dear me, what is this world coming to... :)

[quote="Padmavathi"][quote="junggi"][quote="Padmavathi"]
c)One of the teachers was caught reading "Alice in Wonderland" in class.[/quote]
hey what's wrong with that[/quote]
[b]LOL she was supposed to be taking class[/b] but since she was reading the book(Children's Classics) for the first time, :D  she found it too interesting to put down  :lol: and so she didn't take class. Ten minutes later, the principal walked in!!  :rotfl:[/quote]
lol
but i heard it's a good book(from a buddy of mine in high school)

[quote="Padmavathi"]Atleast your school isn't my school!!
In my school(-),
a)The teachers didn't know that there is an IMO until I told them so.
b) People think I'm a nerd coz I read Arthur Clarke, Wodehouse and Asimov between classes and I get an aggregate of about 92%(YUKK).
c)One of the teachers was caught reading "Alice in Wonderland" in class.
d) The students(or atleast majority of them) think that 3 is an irrational number and that you CAN divide anything by zero.
e)The vice-principal likes to eat cotton candy during morning prayers.
f)The cafetaria lady uses the drinking water fountain to wash her feet.
g)The hindi teacher speaks tamil, the english teacher speaks hindi and the maths teacher speaks rubbish.
h)95% of the students cannot survive without plastering themselves with make up atleast twice an hour.
i)The worst-ALL the students...THINK IT'S COOL TO FLUNK IN MATH!!!
Oh dear me, what is this world coming to... :)[/quote]

Wow, India sounds like a craphole.

But I thought 3 WAS an irrational number!  :roll:

My school's perfectly fine.

[quote="MissSmartyPants"]does your school
a)close early because the teachers want to watch the World Cup?
b)Think that a fashion show is the most important event in the calendar?
c)Wonder why you are so enthusiastic about taking part in the GOOGLE seminar??
d)Have english teachers who say -"Shakespeare is boring..lets all close our books and chat"
No I'm not sayin my skool is awful and that I am fully justified in b*tching about them...but it feels good!![/quote]

i wish my school was like yours.

because other than part b (it's my birthday lmao) i'm going to agree

[quote="Padmavathi"][quote="Ignite168"]
Wow, India sounds like a craphole.[/quote]
Of course it isn't!!! :mad:  :mad:  :mad:
Well,sorry...but i HATE it if anyone insults India.
You may think I'm doing it too, but my school is in a minority.I'm just bringing out the funny side of it and it is NOT any basis to judge a country on.
The only reason my school is so "good" is because it is trying to keep up the [b]British[/b] reputation of 140 years it has and it's being controlled by a power hungry Bishop and a money hungry Church.. :lol: [Which isn't exactly how a true Indian would have liked it to be!!]
Just because all the students of a school get above 90% in the ICSE exams and just because it's won the loads of awards doesn't mean that the students are all brilliant and life is great. All I'm doing is trying to tell people that. If people are going to start commenting on a country with a billion-plus population based on what I've just said then I really think that's dumb.
Even though you may find it hard to believe, Ignite,mine isn't the only school in India and u really shouldn't judge our country on the basis of it...
 :)[/quote]

you tell him

Whoa, and I thought I had complaints when I had to get my head of math department to get off his lazy ass to give me the AIME I this year, because he forgot.

my friend got suspended cause he brought a water gun to school

and the day he came back (today), he wears a box, baggies for socks, shoe boxes for shoes, soda cans for earings, and a buncha random other stuff for french class for a fashion show

wow. real. smart.

some kids in my school are so smart they can approximate pi to about 2.
 :rotfl:
jorian

[quote="7h3.D3m0n.117"][quote="Temperal"]Uh, sorry if I sound like your teachers, but what's the IMO? :blush:[/quote]

International Math Olympiad.

First to get there:

AMC
AIME
USAMO
IMO

if you do well in the first 3 of course.[/quote]
plus for IMO, you need to do well on the TST given (to black moppers only?) at MOP. I think the team is decided by USAMO score+TST score or something like that

[quote="randomdragoon"]At least your school doesn't have a poster up comparing the principal to some gangster rapper guy type person.[/quote]

Are you SERIOUS?! :rotfl:

My school recently had a fundraiser or something, and it was supposed to be a competition between eighth and ninth grade. Over the announcements they claimed that the eighth grade was winning, but due to the increased competition from last week, it would be extended another week.

[hide="the kicker"]Last week was spring break.[/hide]

[quote="7h3.D3m0n.117"][quote="Temperal"]Uh, sorry if I sound like your teachers, but what's the IMO? :blush:[/quote]

International Math Olympiad.

First to get there:

AMC
AIME
USAMO
IMO

if you do well in the first 3 of course.[/quote]

You racist, the US is not the only country in the world.


function
geometric series
combinatorics solved
combinatorics

Let O(n) signify the number of ways we can express n as a sum of ODD integers (positive).
Let D(n) signify the number of ways we can express n as a sum of DIFFERENT integers (positive).
Find a relation between O(n) and D(n).
TIP : O(n)=D(n)
:)
:)
 :D

Hey this can be done with a computer programm :) :D.. hmm.. mathematical.. dunno :?

cheers!

let u_n denote the no. of partitions into unequal parts and v_n deote the partition into odd parts
U(x) = (1+x)(1+x^2)(1+x^3)(1+x^4)..... is the generating function for u_n
after some thought
V(x) = (1+x+x^2+x^3+x^4....)(1+x^3+x^6+x^9...)(1+x^5+x^10+...)
and all powers of x  containing multiples of odd numbers.
now all that remains to be shown U(x) = V(x)
sum up the geometric series and watch wot unfolds!!!they are equal :D  :D
best regards


linear algebra
matrix
algebra proposed
algebra

Prove that always for given 2 positive real numbers $ a$ and $ b$\ we can find positive reals $ x$ $ y$ and $ z$ such that:

$ a\equal{}\frac{y\plus{}z}{2x}$ and $ b\equal{}\frac{x\plus{}z}{2y}$

It doesn't hold. We get $ z \equal{} 2ax \minus{} y \equal{} 2by \minus{} x\implies y \equal{} {2a \plus{} 1\over 2b \plus{} 1}x$, and now $ z \equal{} 2ax \minus{} {2a \plus{} 1\over 2b \plus{} 1}x \equal{} {4ab \minus{} 1\over 2b \plus{} 1}x$. If $ 4ab < 1$, then $ z$ is negative (assuming $ x,y$ are positive).

$ 2ax \minus{} y \equal{} z \equal{} 2by \minus{} x$
Let $ y \equal{} 1$
$ 2ax \minus{} 1 \equal{} 2b \minus{} x$
$ x \equal{} \frac {2b \plus{} 1}{2a \plus{} 1}$
So our numbers may be: $ z \equal{} t, y \equal{} 1, x \equal{} \frac {2b \plus{} 1}{2a \plus{} 1}$
As we may see for every $ a, b$ exists infinitely many triples.

It is OK I think Also condition $ 4ab > 1$ should be given.

I posted this problem just to check is it true that:

If $ a$, $ b$ and $ c$ are positive reals such that $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1$ then there always exist positive reals $ x$, $ y$ and $ z$ such that:

$ a \equal{} \frac {y \plus{} z}{2x}$, $ b \equal{} \frac {x \plus{} z}{2y}$ and $ c \equal{} \frac {x \plus{} y}{2z}$

The last problem comes down to the system

$ \begin{matrix} \minus{}2a & 1 & 1 & \vline & 0\\ 1 &\minus{}2b & 1 & \vline & 0\\ 1 & 1 & \minus{}2c & \vline & 0\end{matrix}$

whose determinant is $ D\equal{}2(a\plus{}b\plus{}c\plus{}1\minus{}4abc)\equal{}0$, hence it has infinitely many solutions, being homogenous. Now it's easy to find $ y\equal{}{2a\plus{}1\over 2b\plus{}1}x, z\equal{}{a\plus{}b\plus{}1\over c(2b\plus{}1)}x$


conics
ellipse
quadratics
linear algebra
matrix
hyperbola
calculus

Prove that $ 2 x^2\plus{}5 y^2\plus{}3 x y\plus{}4 y\plus{}5 y\minus{}10 \equal{} 0$  is an equation of ellipse. Further find the equation of directrix and the foci.

Actually I have a question about this type of problems. I know that it is enough to find Canonical form (i.e. $ x^2/a^2\plus{}y^2/b^2\equal{}1$). But I don't know how to find the correct transformation. Is there any elegant method for doing this? or I just need to guess it (blind guess, or guess using a graphing calculator).
Please solve this problem in details. Thanks in advance.

[quote="Moonmathpi496"]But I don't know how to find the correct transformation. Is there any elegant method for doing this?[/quote]
You can take the fact that I just moved this into the linear algebra section as a hint.

Key words:
Quadratic form.
Positive definite.
Symmetric matrix.
Eigenvalues, eigenvectors, diagonalization.

for an eq.
$ ax^2\plus{}2hxy\plus{}by^2\plus{}2gx\plus{}2fy\plus{}c\equal{}0$
for this to be an ellipse
$ h^2\minus{}ab<0$ and
determinant of
$ \begin{bmatrix}a & h & g \\
h & b & f \\
g & f & c \\
\end{bmatrix}$ is not equal to 0 .
for an hyperbola
$ h^2\minus{}ab>0$ and determinant is again not equal to 0(If its rectangular one additional condition is $ a\plus{}b\equal{}0$)

The center is found in this way:partial derivative on x and y and solve the system .


function
calculus
derivative
limit
trigonometry
real analysis
real analysis unsolved

Prove that a Differentable function in $ [ a,b] $ has a derivative which is continous in $ [ a,b] $.

Can somebody please also post related results!!!

There's nothing to prove.

It follows directly from the definitions of differentiability and continuity,

[b]Def. [/b]If $\lim_{x\to a} f(x)=f(a)$ then $f$ is continous at $a$.

[b]Def. [/b]If $\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ exists $f$ is differentiable at $a$ and the limit is the derivate.

since it is clear that $\lim_{x\to a}  f'(x) =\lim_{x\to a}  \Big( \lim_{y\to x} \frac{f(y)-f(x)}{y-x} \Big)=\lim_{y\to a} \frac{f(y)-f(a)}{y-a}=f'(a)$

Er..It may not be continuous.

Consider the function

$f(x) = \begin{cases}
x^2\sin \frac{1}{x}, x\neq 0\\
0, x = 0
\end{cases}$

For $x\neq 0$, $f'(x) = 2x\sin \frac{1}{x} - \cos\frac{1}{x}$
And $f'(0) = \lim_{x\to 0} \frac{f(x)-f(0)}{x} = \lim_{x\to 0} x\sin \frac{1}{x} = 0$.

Thus $f$ is differentiable everywhere. But $\lim_{x\to 0} f'(x)$ does not exist.

You're absolutely right liyi!

It's a classic example and it's a shame I didn't recall it.

But I think my false proof looks pretty convincing anyway...
...can you tell me more specificly what's wrong?
Is it the of the limits?

And about the question, was it a trick question?
It obviously tricked me... :P

Generally, you cannot interchange taking limits, which is the flaw in your proof.

Does anyone ahve another proof or counter-proof.
Actually this question came in context of the qquestion that if $f$ is differentiable on $ [a,b] $ , then $ f'$ takes on every valur between $ f'(a) $ and $ f'(b) $.
I concluded that this is equivalent to proving that if $f$ is differentable , $ f'$ is continous. Probably I concluded  wrong..??

Yes, you concluded wrong.  But that result is in Rudin, for example (or any other analysis text), so I won't bother posting the proof.

[quote="Rushil"]Does anyone ahve another proof or counter-proof.
Actually this question came in context of the qquestion that if $f$ is differentiable on $ [a,b] $ , then $ f'$ takes on every valur between $ f'(a) $ and $ f'(b) $.
[/quote]
Yes. $f'$ has intermediate value property. But it might be discontinuous. From this we also know that a function which has intermediate value property is not necessary to be continuous. (It is a little amazing, isn't it?)

It is sufficient to show that $f'(a)<0$ and $f'(b)>0$ then there exists $\xi\in(a,b)$ such that $f'(\xi)=0$. (Here we assume that $a<b$).

$f'(a)<0$ implies that $f(a)$ is not the max value;
$f'(b)<0$ implies that $f(b)$ is not the max value;
thus $f(x)$ attains its max value, say, at $x=\xi$ in $(a,b)$.

The Fermat's Lemma tells us that $f'(\xi) = 0$.

and how does that conclude my question??? can you plz explain properly!!

Okay.

Suppose $f$ is differentiable on $[a,b]$. $c$ is a real number between $f(a)$ and $f(b)$, then there exists $\xi\in(a,b)$ such that $f(\xi)=c$.

Proof. It is trivial when $c=f(a)$ or $c=f(b)$.
Without loss of generality, assume that $f(a)<c<f(b)$. Let $g(x)=f(x)-cx$, then $g$ is also differentiable, $g'(x) = f'(x) - c$.
$g'(a) = f(a) - c < 0, g'(b) = f(b) - c > 0$.
Then from the argument above, there exists $\xi$ such that $g'(\xi) = 0$, i.e., $f'(\xi) - c = 0$.

[quote="Peter VDD"]I should have read this topic before this year's IMC :([/quote]
hehe, but isn't it an exercise in your analysis textbook?

No. :( I don't have a good textbook either. Any recommendations?


I/ Supply the correct form of the word in parentheses:

1. What lovely...........children! (MANNER)
2. I'll never forget the............I felt in that situation. (HUMILIATE)
3. Despite going to German classes twice a week, I don't fell I'm making much..........with the language.(HEAD)
4. She's so...........that she won't let anything stand in the way of her ambition. (MIND)
5. That was a very..............thing to say. (HURT)
6. Children can be very............(SPITE)
7. The police were............... Why should anyone want to steal a wastepaper basket? (MYSTIFY)
8. It's...........of him to lose his temper like that- he's usually very calm.(CHARACTER)
9. He stood at the door to make sure that no one............the party. (GATE)
10. The athletes take part in the Olympic Games in the true spirit of...........(SPORT).

II/ Rewrite the sentences:

1. If nothing unfortunate happens, I'll see you next week.
All being............................................
2. Fortunately, the boat hadn't left.
By a stroke.........................................
3. I need an early night.
I could..............................................
4. I 've considered the advantages and disadvantages and I 've decided not to go.
I 've weighed.....................................
5. He pretends that his latest business failure wasn't important.
He laughed........................................

I get it from a English contest in my country :( I think it is very hard so Who can help me :D

1) Mannered
2) Humiliation
5) Hurtful
6) Spiteful
7) Mystified

3 and 4 seem to require different words

Try to figure out the rest.

3. headway

4 determined

8 uncharacteristic

9 gatecrashed

10 sportsmanship

1. All being well, I'll see you next week.

2. By a stroke of luck, the boat hadn't left.

3. I could do with an early night.

4. I weighed up the pros and cons and I've decided not to go.

5. He laughed off his failure.


Solve for positive integers $ x, y$ and $ z$,

(i) $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} 1$

(ii) $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \ge 1$.

[hide="1"] Clearly, none of x, y, or z can be 1, otherwise the sum is greater than 1.

Firstly, let $ x\le{y}\le{z}$. We now go through some casework.

[hide="Case 1, x=2"] This means $ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {1}{2}$. We can see that y>2. If y=3, we find z=6. If y=4, z=4. If $ y\ge{5}$, $ \frac {1}{y} \plus{} \frac {1}{z}\le\frac {2}{5}$, so there are no more solutions.

Our solutions for this case are $ (x, y, z) \equal{} (2, 3, 6), (2, 4, 4)$.[/hide]

[hide="Case 2, x=3"] This means $ \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {2}{3}$. However, $ 3\le{y}\le{z}$ so $ \frac {1}{y} \plus{} \frac {1}{z}\le\frac {1}{3} \plus{} \frac {1}{3} \equal{} \frac {2}{3}$, so we must have $ (x, y, z) \equal{} (3, 3, 3)$.[/hide]

[hide="Case 3, x is greater than or equal to 4"] Then $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\le\frac {1}{4} \plus{} \frac {1}{4} \plus{} \frac {1}{4} \equal{} \frac {3}{4}$, contradiction. Thus there are no solutions here.[/hide]

Our solutions are $ (x, y, z) \equal{} (2, 3, 6), (2, 4, 4),$ and $ (3, 3, 3)$, and all permutations.[/hide]

Similar logic for 2.


calculus
integration
trigonometry
calculus computations

Solve $\int \frac{tan^{2}\theta}{cos\theta}d\theta$

I already have a solution for this and will post it after a couple of days but I would like to see how you all would solve that.

Anyway, my solution uses integration by parts

write $\tan^{2}(\theta)$ as $\sec^{2}(\theta)-1$
so you now have
$\int\sec^{3}(\theta)-\sec(\theta)d\theta$
which is much easier to work with. just use integrations by parts and you should know how to solve $\int\sec(\theta)d\theta$

$\int \frac{\tan^{2}\theta}{\cos \theta}d\theta =\int \frac{\sin^{2}\theta}{\cos^{3}\theta}d\theta =\int \frac{\sin^{2}\theta}{\cos^{4}\theta}\cos \theta d\theta =\int \frac{t^{2}}{(1-t^{2})^{2}}dt$ where $t=\sin \theta$

So here is the way I solved it

Solve $\int \frac{tan^{2}\theta}{cos\theta}d\theta$

Write integral like this $\int tan\theta\cdot\frac{tan\theta}{cos\theta}d\theta$

Integrate by parts

Let  the 2 choices be:

$(tan\theta)\prime=\frac{1}{cos^{2}\theta}$

and

$\int \frac{tan\theta}{cos\theta}d\theta = \frac{1}{cos\theta}$

So  $\int tan\theta\cdot\frac{tan\theta}{cos\theta}d\theta=tan\theta\cdot\frac{1}{cos\theta}-\int\frac{1}{cos^{2}\theta}\cdot\frac{1}{cos\theta}d\theta$

That integral on the right is yucky. Now I can't explain the following. ARE YOU WATCHING ME NOW!

$\int tan\theta\cdot\frac{tan\theta}{cos\theta}d\theta=\frac{1}{2}(tan\theta\cdot\frac{1}{cos\theta})-\frac{1}{2}\int\frac{1}{cos\theta}d\theta$

The final solution is $\frac{1}{2}(\frac{tan\theta}{cos\theta}-ln(tan\theta+\frac{1}{cos\theta}))$

comments?

They factored out the $\frac{1}{\cos^{2}\theta}$

I just edited my previous post and erased a couple of $d\theta$ symbols(seems I had sprinkled too liberally)

Ragingbullrocky

That is obvious, but how did that $\frac{1}{cos^{2}\theta}$ dissapear? Can you explain that manipulation in more detail?

Seems nobody can explain where the $\frac{1}{cos^{2}\theta}$ went

I will come back in two days with the answer. Hint: Kunny actually came close to the solution!

 :rotfl:

I'm back!! and you no longer have to scratch your head about this

Solve $\int\frac{tan^{2}\theta}{cos\theta}d\theta$

Write integral like this

$\int sin\theta\cdot\frac{sin\theta}{cos^{3}\theta}d\theta$  (THANKS KUNNY!!)

Integrate by parts and let the 2 choices be:

$sin\theta\prime = cos\theta$

and

$\int\frac{sin\theta}{cos^{3}\theta}d\theta = \frac{1}{2cos^{2}\theta}$

so $\int sin\theta\cdot\frac{sin\theta}{cos^{3}\theta}d\theta = sin \theta\cdot \frac{1}{2cos^{2}\theta}-\int cos\theta \cdot \frac{1}{2cos^{2}\theta}d\theta$

$=\frac{1}{2}\cdot\frac{tan\theta}{cos\theta}-\frac{1}{2}\int \frac{1}{cos\theta}d\theta = \frac{1}{2}(\frac{tan\theta}{cos\theta}-ln(tan\theta+\frac{1}{cos\theta}))$


I have a table. The code is shown below. When I put the code in a separate new document, it works okay. But when i incorporate the code into my file, the backslash line in the table disappear. how to solve it...

But the way, how to make the numbers and text in the table smaller? the table looks very busy.

\documentclass{report}
\usepackage{array}
\usepackage{slashbox}
\setlength{\arrayrulewidth}{1pt}
\setlength{\doublerulesep}{\arrayrulewidth}
\begin{document}

\begin{table}
\begin{minipage}[h]{5in}
\caption{my caption}
\begin{tabular}{|c|c|c|c|c|c|}\hline
\backslashbox{PI\protect \footnote{PI = Price Increase}}{TL\protect \footnote{TL = Time Length}}&1&2&3&4&5\\ \hline
100\% & \multicolumn{1}{!{\vrule width 0.5pt}c|}{603.32}&215.28&-89.96&-330.07&-518.95\\ \hline \cline{2-3}
200\% &1512.22&930.24 &\multicolumn{1}{!{\vrule width 0.5pt}c|}{472.45}&112.33&-170.94\\ \hline \cline{4-4}
300\% &2421.11 &1645.21 & 1034.86 & \multicolumn{1}{!{\vrule width 0.5pt}c|}{554.74}&177.07\\ \hline \cline{5-5}
400\% & 3330.01 & 2360.17 & 1597.27 & 997.15 & \multicolumn{1}{!{\vrule width 0.5pt}c|}{525.08} \\ \hline \cline{6-6}
500\% & 4238.91 & 3075.14 & 2159.68 & 1439.56 & 873.09 \\ \hline
600\% &5147.81 & 3790.10 & 2722.09 &1881.97 &1221.10 \\ \hline
\end{tabular}
\label{siteindex45}
\end{minipage}
\end{table}
\end{document}

Not sure how you expect us to know what else in an unknown file is causing the backslash to disappear. Take out parts of the file until you work out what is causing the problem. Then, if necessary, post the smallest file that causes the problem.

To make the table text smaller put \small before \begin{tabular}.

These are the all I set before \begin{document}
I tried but can't find anything that cause the disappearance of the backslash line

\documentclass[12pt, fleqn]{report}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{float}
\usepackage{eepic}
\usepackage[sort]{natbib}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{color}
\usepackage{array}
\usepackage{slashbox}
\setlength{\arrayrulewidth}{1pt}
\setlength{\doublerulesep}{\arrayrulewidth}
 \setlength{\textwidth}{6.5in}
\setlength{\textheight}{8.5in}
\setlength{\evensidemargin}{0in}
\setlength{\oddsidemargin}{0in}

\setlength{\topmargin}{0in}

\setlength{\parindent}{0pt}
\setlength{\parskip}{0.1in}
\setlength{\abovecaptionskip}{0cm}
\setcitestyle{round}
\renewcommand{\thesection}{\arabic{section}}

it is solved. .I add one more package \usepackage{pict2e}

This is how you analyze problems such as this.
1) Use a document with only the table in it and cut out irrelevant parts of the table.
2) Take out around half of the lines of the preamble, this would tell you which half causes the problem.
3) Repeat 2) with whatever half is causing the problem.
4) Keep repeating 2) until you find the cause: the eepic package.
5) This gives you a minimal example such as:
[code]\documentclass{report}
\usepackage{eepic}
\usepackage{slashbox}
\begin{document}
\begin{tabular}{|c|c|}\hline
\backslashbox{first}{second}&1\\ \hline
a& b \\ \hline
\end{tabular}
\end{document}[/code]Removing the \usepackage{eepic} line puts back the backslash.
6) Read the documentation for eepic.
7) This tells you it works with \specials. These are macros only used when you compile to DVI. Indeed if you go straight to PDF the log will tell you
[i]Non-PDF special ignored![/i]
8) Solution: Either don't use eepic or else compile using LaTeX=>PS=>PDF so \specials aren't ignored


how many nonzero angles are formed in a fan of rays with 17
 rays?

I can't seem to understand the question... :huh: Could you elaborate?

I think his question is pretty clear: If you have 17 rays starting at a given point, how many nonzero angles between these rays do you get?

[i]Solution.[/i] An angle is formed by a pair of different rays, and 17 rays yield $\binom{17}{2}=136$ pairs of different rays. Hence, one gets 136 angles.

Of course, this question is completely misplaced in the Geometry Unsolved section. I'm moving it into Intermediate.

  Darij


[quote="AYMANE"]Thanks Kunny :)
But thus what is the minimum (I need a value)[/quote]

$ -1-(a-1)(b-1)(c-1)=ab+bc+ca-abc$
$ MIN=3\min_{a+b+c=3 \\ a,b,c>0}ba+bc+ca-abc$
$ =-3-3\max_{a+b+c=3 \\ a,b,c>0}(a-1)(b-1)(c-1)$

May be i am not clear
But i need a value
example:Min=1
Thanks

The minimum is $ \epsilon$, where $ \epsilon\to 0$. Take $ a\to 3$ and $ b\to 0,c\to 0$

Easy to see that M=a^2(b+c)+b^2(a+c)+c^2(a+b)=(a^2+b^2+c^2)-(a^3+b^3+c^3) (1).
By cauchy-shwarz inequality we have: (a^3+b^3+c^3)(a+b+c)>=(a^2+b^2+c^2)^2, or
(a^3+b^3+c^3)>=1/3(a^2+b^2+c^2). So, from (1), M<=1/3(a^2+b^2+c^2)(9-(a^2+b^2+c^2).
N=a^2+b^2+c^2. Then, M<=(9N-N^2)*1/3.  As average square is more then average arithmetical,
N>=3. So, M<=1/3*3*6=6, because 1/3(9N-N^2) is strictly increasing function. So, M<=6. The equality achieves, when a=b=c=1.

The inequality is equivalent to:
\[ (a-1)^{2}(b+c-2a)+(b-1)^{2}(c+a-2b)+(c-1)^{2}(a+b-2c)\ge0\]
It is correct if we using Vornicu Schur's inequality

[quote="zaizai-hoang"]The inequality is equivalent to:
\[ (a-1)^{2}(b+c-2a)+(b-1)^{2}(c+a-2b)+(c-1)^{2}(a+b-2c)\ge0 \]
It is correct if we using Vornicu Schur's inequality[/quote]
what ineq are u talking 'bout ; zaizai
i've expanded it but im not really clear 'bout the minimum

Clearly, the maximum is 27/4, minimum is zero(if take a,b,c>=0).


Steve Olson was a guest today on the radio program The Connection. Joining him on the program were Melanie Wood and Gabriel Carroll. If you missed the show you can listen to it here: http://www.theconnection.org/shows/2004/04/20040423_b_main.asp

Thanks for the link.  I listened to the show and enjoyed it, although I came away with some very strong feelings about it.

I'd be interested in hearing other AoPSers' opinions on the show - did you like it?  Did it do justice to [i]Count Down[/i]?  Did it do justice to math competitions in general?  Were the things discussed during the show interesting?

I enjoyed it a lot.  I haven't read Count Down so I can't say if it did it justice.  They did a good job on portraying math competitions in my opinion.  They seemed to address the major points of competitions.  Like the lack of girls.  They also explained to the public that the olympiad contests are more than just doing math.

The confusion behind the meaning of ``olympiad'' in the show bothered me. I know that there is an olympiad for elementary school students (in which you don't have to prove anything; I don't know why it's called an olympiad) and the olympiads like the USAMO and IMO because I have participated in both types. But why should someone else who hasn't participated in either one know the difference? I think that might have been incredibly confusing to many listeners.

[quote="Ravi B"]Yes, my daughter's school participates in the "Mathematical Olympiads for Elementary and Middle Schools" ([url]http://www.moems.org[/url]).  For young kids (4th to 8th grade), it has pretty good, short-answer problems that you have to solve in about 5 minutes each.  But its name is very confusing; it has nothing to do with olympiad-level problems for high school students.[/quote]

Really easy questions that 1/6 of the sixth graders aced or missed one whose level isn't that high at all probably 50 on AMCs.


algebra function domain inequalities logarithms

How many integers are in the domain of the function

\[ f(x)\equal{}2\log(2x\minus{}1)\minus{}3\log(8\minus{}x)\]

[hide="Solution"]$ f$ is defined when $ 2x\minus{}1>0\implies x>1/2$, and when $ 8\minus{}x>0\implies 8>x$.
The number of integers in the range $ 7\geq x\geq 1$ is $ \boxed{7}$.[/hide]

[quote="DaReaper"]How many integers are in the domain of the function
\[ f(x) \equal{} 2\log(2x \minus{} 1) \minus{} 3\log(8 \minus{} x)
\]
[/quote]

[hide]
Because you can only take the log of positive numbers, the following inequalities must hold
1.)
$ 2x \minus{} 1 > 0 \\
2x > 1 \\
x > \frac {1}{2}$

2.)
$ 8 \minus{} x > 0 \\
8 > x$

combining these answers gives you
$ 8 > x > \frac {1}{2}$
x can be any integer from 1 to 7, so the answer is 7.[/hide]

The domain of the logarithmic function is $ [0,\infty)$. Therefore,

$ 2x\minus{}1>0\rightarrow x>1/2$

$ 8\minus{}x<0\rightarrow x<8$

$ D: (1/2,8)$. There are 7 integers on this interval.

The logarithm function only takes real positive numbers.

No, the domain of $ \log$ is $ (0,\infty)$.

$ \log0$ is undefined.


Triangle ABC is given.

From a point P distinct from A,B and C draw lines through B and C

Let L,M be the feet of the perpendiculars from A to these lines.

Find the position of P when the distance from L to M is greatest.

Locus of the points L is a circle (C') with diameter AB, locus of the points M is a circle (B') with diameter AC, where the circle centers C', B' are midpoints of the sides AB, AC. The 2 most distant points $L_0 \in (C'),\ M_0 \in (B')$ are intersections of these 2 circles with their center line C'B', such that $L_0$ is in the exterior of the circle (B') and $M_0$ is in the exterior of the circle (C'). The desired point P is the intersection of the lines $BL_0,\ CM_0.$


calculus
integration
inequalities
function
LaTeX
real analysis
real analysis unsolved

$ f$ real continous functions on $ [0;1]\minus{}\minus{}\minus{}\minus{}> R\plus{}$

prove that $ (\int_{0}^{1}f(x)dx)(\int_{0}^{1}x^3f(x)dx)\geq (\int_{0}^{1}xf(x)dx)(\int_{0}^{1}x^2f(x)dx)$

Cezar Lupu & Mihai Piticari

Define a measure $ \mu(A) \equal{} \int_Af(x)dx$.  By Holder's inequality we have
\[ \int_0^1{x^2d\mu}\leq\left(\int_0^1{x^3d\mu}\right)^{\frac {1}{2}}\left(\int_0^1{xd\mu}\right)^{\frac {1}{2}}
\]
and similarly,
\[ \int_0^1{xd\mu}\leq\left(\int_0^1{x^2d\mu}\right)^{\frac {1}{2}}\left(\int_0^1{d\mu}\right)^{\frac {1}{2}}
\]
Multiplying the two inequalities and squaring yields the result.

Multiply $ \int_0^1 xf(x) dx$ on both sides, and happy cauchying.

I believe the result is still true if we replace $ (x, x^2, x^3, [0, 1])$ with $ (g(x), h(x), g(x)h(x), [a, b])$ for any interval $ [a, b]$ on which functions $ g, h$ are increasing.

Moubinool: you can use the LaTeX command \to to make a nicer-looking arrow: $ \to$.

An other one  :lol: : $ \left (\int_{0}^{1}f(x)dx\right )\left (\int_{0}^{1}g^{a\plus{}b}_{(x)}f(x)dx \right  )\geq \left (\int_{0}^{1}g^a_{(x)}f(x)dx \right )\left (\int_{0}^{1}g^b_{(x)}f(x)dx \right  )$ where $ f,g : [0,1]\rightarrow R^\plus{}$ are integrable and $ a ,b \geq 0$   :cool:


algebra
polynomial
real analysis
real analysis solved

Find all polynomials f with real coefficicients such that $\frac{1}{f} $ has it's primitive a rational function.

Since in $\mathbb{C}$ $f=(x-x_1)^{\alpha_1}\cdots(x-x_n)^{\alpha_n}$ with $x_1,\ldots,x_n$ distints,  then $\frac{f}{(x-x_i)^{\alpha_i}}$ are coprimes
so exists polinomilas $a_i,\ldots a_n$ such that
\[1=a_1\frac{f}{(x-x_1)^{\alpha_1}}+\ldots+a_n\frac{f}{(x-x_n)^{\alpha_n}}\]
\[\frac{1}{f}=\sum_{i=1}^n \frac{a_{i}}{(x-x_i)^{\alpha_i}}\]
Now $a_i(x)=b_i(x-x_i)$ so the condition that we want is that the coef. of the lineal term of the $b_i$ must be nules.
Well those are call the residues of $1/f$ and they can be obtain integrating  $1/f$ arround closed path that  surround each $x_i$... it can be found in any complex theory book.


How would you change bases from one to another directly instead of through base 10. For instance .221(base 6) to base 9?

good question

i believe it has to do with divison and remainders.  lets see:

(((original base/final base)-whole part)*final base)=ones place

(((whole part/final base)-whole part of new)*final base)=next place

if decimal

(((original base/final base)-whole part)*final base)=tenths place

(((whole part/final base)-whole part of new)*final base)=next place right

Thanks, its a little confusing but i think i get the idea, but in the first part where you write first base/final base minus whole part then multiplied by final base wont u get a decimal sometimes? THen what would you do?

multiplying by final base will never give a decimal unless the original number is a decimal.  In that case you would convert it to a full whole number by shifting the decimal place.  After calculations you would shift it back

It's usually easier to convert to decimal and then to another base, except when you're converting from bases in which one is a power of another (i.e. binary and octal, base 3 and base 9, etc.).

i agree but the reason i did it this way was because he specifically asked for it not to go through decimal

If we are converting between base $p$ and base $q$, and if there exist positive integers $s$ and $t$ such that $p^s=q^t$, then groups of $s$ digits in base $p$ can be exactly converted to groups of $t$ digits in base $q$.

For the specific case given, there are no such positive integers $s$ and $t$, so we cannot use this shortcut.  So let's go back to basics.  What does $.221_6$ mean?  $0.221_6=\frac{2}{6}+\frac{2}{6^2}+\frac{1}{6^3}=\frac{2\times6^2}{6^3}+\frac{2\times6}{6^3}+\frac{1}{6^3}=\frac{85}{6^3}$

Now, let's convert this fraction to base 9:
$\frac{85}{6^3}=\frac{85}{2\times 2\times 2\times 3\times 3\times 3}=\frac{85}{9\times 24}=\frac{3\times 24+13}{9\times 24}=\frac{3}{9}+\frac{13}{9\times 24}$

The first digit after the radix point is 3.

Continuing...
$\frac{13}{9\times 24}=\frac{13}{9\times 3\times 8}=\frac{13\times 3}{9^2\times 8}=\frac{39}{9^2\times 8}=\frac{4\times 8+7}{9^2\times 8}=\frac{4}{9^2}+\frac{7}{9^2\times 8}$

The second digit after the radix point is 4.  Continuing...
$\frac{7}{9^2\times 8}=\frac{7\times 9}{9^3\times 8}=\frac{63}{9^3\times 8}=\frac{7\times 8+7}{9^3\times 8}=\frac{7}{9^3}+\frac{7}{9^3\times 8}$

The third digit after the radix point is 7.  Continuing, we will find that
$0.221_6=\frac{3}{9}+\frac{4}{9^2}+\frac{7}{9^3}+\frac{7}{9^4}+\cdots=0.34\overline{7} _9$

Is it possible to easily convert a number from one base to another without going through base 10?

i believ rvc showed a nice example of such a task.  Nice job

How about if you have a non-terminating decimal that you are trying to move to a different base?

Getting stuck on a problem and can't work my way out of it.

Thanks.


abstract algebra
vector
superior algebra
superior algebra unsolved

Let R be a PID. Let K,L,M be R-modules such that
K+M isomorphic to L+M (direct sums). Prove that K is isomorphic to L.

this is not true, consider $R = \mathbb{Z}, K = M = \mathbb{R}, L = 0$.

did you forget some restriction? ($End_{R}(M)$ semilocal? ;-))

yes, E.G. Evans proved this in 1973 - I think it's very difficult ...

[quote="-oo-"]this is not true, consider $R = \mathbb{Z}, K = M = \mathbb{R}, L = 0$.
 [/quote]
So you mean $\mathbb{R}\oplus\mathbb{R}\cong0\oplus\mathbb{R}$ as groups?

yes, this comes from the structure as Q-vector space.


[quote="tochong84"]Hi, is there anyone here who knows how to solve this problem? Please help me with it

Characterize the primes p for which congruence
    [n^2 congruent -1 (mod p)]
has a solution.[/quote]

We need that the Legendre symbol,
$\left(-1/p\right)=1$

By one of the properties of the Legendre symbol (consequence of Euler's criterion),
$(-1/p)\equiv (-1)^{(p-1)/2}(\mod p)$
That means we want,
$1\equiv (-1)^{(p-1)/2}(\mod p)$

So if $p=4k+3$ we got that,
$1\equiv-1$ a contradiction.

And if $p=4k+1$ we got that,
$1\equiv 1$ which is true.
So all primes satisfy $p\equiv 1 (\mod 4)$

But, this proof only works for odd primes (because all these theorem I used relied on that fact).
I leave it to to to confirm if $p=2$ works or doesnt.

Thank you for the solution!
I did try solving the problem using regular properties of congruence but none worked. I didnt know Legendre symbol and that theorem of Euler. Well but it makes sense to me now.
Thank you very much


I don't understand  ur  problem.  Maybe it's wrong
This is my proof.

From conditions,  we have: $ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2) \equal{}4(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$
Or $ a^3\plus{}b^3\plus{}c^3\equal{} 3\sum a^2b \plus{}12abc$

The other hand, the inequality is equavilent to $ a^3\plus{}b^3\plus{}c^3 \leq 3\sum a^2b$

Or $ 12abc \leq 0$

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4(ab \plus{} bc \plus{} ca)$.

Prove that: $ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{(a \plus{} b \plus{} c)^3} \leq\ \frac {1}{2}$
 :)[/quote]
Are you sure for this inequality my friend???
We have to prove that:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 6abc \le a^3 \plus{} b^3 \plus{} c^3 \plus{} 3 \sum_{sym}a^2b \plus{} 6abc \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \le 3 \sum_{sym}a^2b$
But from the condition we have:
$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \equal{} 4(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} \sum_{sym}a^2b \equal{} 12abc \plus{} 4\sum_{sym}a^2b \Leftrightarrow a^3 \plus{} b^3 \plus{} c^3 \equal{} 3\sum_{sym}a^2b \plus{} 12abc$
Putting this result to the first we have to prove that
$ 3\sum_{sym}a^2b \plus{} 12abc \le 3 \sum_{sym}a^2b \Leftrightarrow 12abc \le 0$
which is a contradiction....Am i right???
Ooops second....

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 4(ab \plus{} bc \plus{} ca)$.

Prove that: $ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{(a \plus{} b \plus{} c)^3} \leq\ \frac {1}{2}$
 :)[/quote]
$ \sum a^{3}\equal{}3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}3abc$
$ (a\plus{}b\plus{}c)^{3}\equal{}6(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
=> ineq isn't true !  :(

Sorry my frieds. Indeed, I had a mistake when created it. I just edited by my another result.  :oops:
[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$

Prove that: $ \frac {25}{3\sqrt [3]{4}\sqrt [3]{ab \plus{} bc \plus{} ca}} \geq\ \frac {a \plus{} 1}{b \plus{} 1} \plus{} \frac {b \plus{} 1}{c \plus{} 1} \plus{} \frac {c \plus{} 1}{a \plus{} 1}$

When does equality occurs?
 :)[/quote]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$

Prove that: $ \frac {25}{3\sqrt [3]{4}\sqrt [3]{ab \plus{} bc \plus{} ca}} \geq\ \frac {a \plus{} 1}{b \plus{} 1} \plus{} \frac {b \plus{} 1}{c \plus{} 1} \plus{} \frac {c \plus{} 1}{a \plus{} 1}$

When does equality occurs?
 :)[/quote]
Equality occurs when $ (a,b,c)\equal{}(0,1,2)$  :)


geometry
MATHCOUNTS
3D geometry
AMC
AMC 10
AMC 10 A
ratio

Hi! I had great difficulty with geometry so I created a Marathon for Geometry.
Feel free to post any Geometry (Difficulty lvl. Mathcounts - AMC 10/12) problems.
Hopefully there are a lot of problems. If a problem is already solved please post a new problem.
So let's start off!!!
Problem:
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder. (Source: [i]AMC 10)[/i]

Let $ r$ be the radius of the cylinder. Then the height is $ 2r$. The slant height of the cone is $ \sqrt{12^2\plus{}5^2}\equal{}13$.

Using similar triangles with the cross sections of the cone and cylinder, we have

$ \frac{5\minus{}r}{5}\equal{}\frac{2r}{12}$
$ \frac{5\minus{}r}{5}\equal{}\frac{r}{6}$

Cross multiplying, we have

$ 5r\equal{}30\minus{}6r$
$ 11r\equal{}30$
$ r\equal{}\boxed{\frac{30}{11}}$

Marathon means post a new problem...

From this year's AMC10A:

Four congruent circles are inscribed in a larger outer circle, with each inner circle externally tangent to the two adjacent small circles and internally tangent to the larger outer circle. Find the exact ratio of the sum of the four smaller circles' areas to the area of the larger outer circle. (Express in simplest radical form.)

[hide=" Solution"]So, when you draw 4 externally tangent congruent circles, internally tangent to the larger, outer circle, you are going to have "open spaces" (spaces not covered by any of the triangles). Thus, you will  have to find the ratio of the radius of the larger circle to the radius of each of the smaller circles. To do so, we must use an alternate shape (preferably a [u]regular [/u] quadrilateral) in the center. [I am sorry I do not know how to plot shapes on $ {LaTeX}$]

Out of symmetry, the quadrilateral in the center must be a square. Because of this symmetry the square's side(s) will span from one radius of a circle to the other, tangent circle, making the side $ {2r}$. Therefore its diagonal is $ {2r\sqrt {2}}$. We can now compute the length of the vertical diameter of the large circle as $ {2r \plus{} 2r\sqrt {2}}$. Hence $ {2R \equal{} 2r \plus{} 2r\sqrt {2}}$, and thus $ {R \equal{} r \plus{} r\sqrt {2} \equal{} r(1 \plus{} \sqrt {2})}$.

Then the area of the large circle is $ {L \equal{} \pi R^2 \equal{} \pi r^2 (1 \plus{} \sqrt 2)^2 \equal{} \pi r^2 (3 \plus{} 2\sqrt 2)}$. The area of four small circles is $ {S \equal{} 4\pi r^2}$.

Now, let's substitute in for our main variable, $ {r}$. For ease of calculation, let's use $ {5}$ ; substituting $ {5}$ for $ {r}$ into the equation $ {4\pi r^2}$ (the sum of the smaller circle) we get $ {100\pi}$ . Now substitute $ {5}$ into $ {\pi r^2 (3 \plus{} 2\sqrt 2)}$ , that gives us $ {25\pi(3 \plus{} 2\sqrt2)}$ ,which is the larger circle's area.

Our last step is to now find the ratio of $ {100\pi}$ and $ {25\pi(3 \plus{} 2\sqrt2)}$ . I find it easy to distribute the $ {25\pi}$ to $ {(3 \plus{} 2\sqrt2)}$ first. By doing so, we get $ {75\pi \plus{} (50\sqrt2)\pi}$ set this to $ {100\pi}$  do some basic arithmetic (I used a calculator  :blush: ) and we arrive at our answer: $ {12 \minus{} 8\sqrt2}$ which simplifies to $ \boxed{4(3 \minus{} 2\sqrt2)}$ .

Just for reference, an alternative way to conduct the last step is to make the equation $ {25\pi(3 \plus{} 2\sqrt2)}$ = $ {(100pi)x}$ and solve for $ {x}$ , and get $ \boxed{4(3 \minus{} 2\sqrt2)}$ .[/hide]

If I made a mistake, please notify me ASAP.  :)

[hide=" New Problem"]An object moves $ {8}$ cm in a straight line from $ {A}$ to $ {B}$, turns at an angle $ {\alpha}$, measured in radians and chosen at random from the interval $ {(0,\pi)}$, and moves $ {5}$ cm in a straight line to $ {C}$. What is the probability that $ {AC < 7}$? (2003 AMC 12B) [/hide]


What is Simon's Favorite Factoring Trick? :maybe:

I'm pretty sure this was asked before(probably a lot of times)
you could do a search on the forums, or find it on AoPSwiki

Indeed. The most recent one is from 5 september: http://www.mathlinks.ro/Forum/viewtopic.php?p=622171#622171

It's actually just FOIL in reverse.

$(a+b)(c+d) = ac+bd+ad+bc$

We rewrite $x = a, y = c, n = b, m = d$ and we get the expression

$(x+n)(y+m) = xy+ny+mx+nm$

Which is the familiar form (we regard $x, y$ as variables and $m, n$ as constants so the expression makes a little more sense).  For example,

$xy+3x+4y+12 = (x+4)(y+3)$


algebra
polynomial
number theory proposed
number theory

let P be a real polynomial with integer coefficients such that there is an infinite subsequence {$P(k)$} with the property that the subsequence has only finitly many prime divisors prove that P is the form $P(x)=(ax+b)^n$

Nobody managed to solve this problem ( I wrote that in Crux) and thus it remains an open question. So, do you have a solution? Then send it to Crux!!!!

I dont have sol but I heard from my friend that had difficult sol btw it is realy nice who can solve it without brute force?

I really don't know how you can solve with brute force such a problem and that is why I beg you to post your friend's solution. And tell him to send it to Crux.


Let's be given $x_{1}\ge x_{2}\ge \cdots >0$ sastisfying $\sum^{\infty}_{n=1}\frac{x_{n^{2}}}{n}\le 1$
Prove that : $\sum_{n=1}^\infty \frac{x_{n}}{n}\le 3$

[u][b]Important[/b][/u] it's $x_{n^{2}}$ not $x_{n}^{2}$

This one is easy:

$\sum_{n=1}^{\infty}\frac{x_{n}}n = \sum_{k=1}^{\infty}\Big( \frac{x_{k^{2}}}{k^{2}}+\ldots+\frac{x_{k^{2}+2k}}{k^{2}+2k}\Big) \le \sum_{k=1}^{\infty}x_{k^{2}}\Big( \frac{1}{k^{2}}+\ldots+\frac{1}{k^{2}+2k}\Big)$,

so it suffices to show that
$\frac{1}{k^{2}}+\ldots+\frac{1}{k^{2}+2k}\le \frac{2k+1}{k^{2}}\le \frac{3}{k}$ for all $k$.

But this is trivial. :)


A chord PQ with midpoint R is drawn in a circle with diameter AB.  Perpendiculars PS and QT are dropped onto AB.

a) Prove that $RST$ is isosceles.

b) Prove that $RST$ is equilateral if and only if $2PQ=AB$.

I got part (a), but couldn't get (b).

Just put in (a) just in case.

Thanks

[hide]
Let $O$ be the center of the circle. Then $\overline{OR}\perp \overline{PQ}$. The pairs of right angles make quadrilaterals $SORP$ and $TORQ$ cyclic. The cyclic quadrilaterals give \[\angle RST=\angle RSO=\angle RPO=\angle QPO\]
\[\angle RTS=\angle RTO=\angle RQO=\angle PQO\]
Since $OP=OQ$, we have $\angle PQO=\angle QPO$, so $\angle RST=\angle RTS$ making $\triangle RST$ isosceles. This proves (a).

For (b), note that $\triangle RST$ is equilateral $\iff \angle RST=60^\circ \iff \angle PQO=60^\circ\iff \triangle PQO$ is equilateral. But this happens whenever $PQ=OP$; since $2OP=AB$, $\triangle RST$ is equilateral if and only if $2PQ=AB$.
[/hide]

Wow thank you very much.

also, could someone delete the post I accidentally submitted.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi there
Normal subgroups of S_n:
We say that if n>4 then all normal subgroups of S_n are (e),A_n,S_n (see eg Topics in group theory by Jamali),but  the case n=4 seems to be somehow difficult .How can Ihandle this one.(Herstein Permutaitions section)
i'd be thankful for your help

There is an easy way: the only normal subgroups when $n\geq 5$ are $\{e\} , A_n , S_n$.
Now only find them for the four remaining cases (only $n=4$ is a little bit of work)

Let $ N $ be a normal subgroup of $ S_4 $. Since a normal subgroup must be self-conjugate,  every conjugacy class of $ S_4 $ must either be disjoint from $ N $, or contained entirely in $ N $. It is easy to show that two elements of $ S_n $ are conjugates if and only if they have the same cycle structure. For, if we have two permutations $ \tau $ and $ \sigma $ with the same cycle structure, and $ f: [1,n] \mapsto [1,n] $ a bijection (permutation) which takes the numbers in the cycle notation of $ \tau $ into the corresponding numbers and cycles in $ \sigma $, then we have $ \tau = f \sigma f^{-1} $. The converse is similar.

This means that we can count the size of each conjugacy class by looking at all possible cycle structures. So they have sizes 1, 6, 3, 8, 6, corresponding to cycles structures 1,1,1,1; 1,1,2; 2,2; 1,3 and 4 respectively. A subgroup must contain the identity element, which forms the conjugacy class of size 1, and it's order must divide 4! = 24, so checking possible sums, they can only be 1, 1 + 3 = 4, 1 + 3 + 8 = 12, or the whole group itself. The first, third and fourth correspond to $ \{e\}, A_4, S_4 $. The second is the Klein group, which turns out to be normal in $ S_4 $ as well.

[quote="Origin"]The second is the Klein group, which turns out to be normal in $S_4$  as well.[/quote]
This needs some clarification: the group in question has elements $e$ and the three products of two disjoint transpositions. There are other ways to embed an isomorphic subgroup of $S_4$, but those are not normal.


solve in $ \mathbb{R}$

$ a\plus{}b\plus{}c\plus{}d\equal{}12$ and $ abcd\equal{}27\plus{}ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd$

im not sure about your problem but if you meant
$ a,b,c,d \in \mathbb R^\plus{}$ and $ abcd\equal{}27\plus{}ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd$
Then [hide="solution"] $ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd \geq 6\sqrt{abcd}(AM\minus{}GM)$

$ abcd \geq 27\plus{}6 \sqrt{abcd}$

$ (\sqrt{abcd}\minus{}9)(\sqrt{abcd}\plus{}3) \geq 0$

$ \sqrt{abcd}\geq 9$ $ \implies$ $ \sqrt[4]{abcd}\geq 3\equal{} \frac{a\plus{}b\plus{}c\plus{}d}{4}$

and we also know that $ \frac{a\plus{}b\plus{}c\plus{}d}{4} \geq \sqrt[4]{abcd}$ by AM-GM

so $ \frac{a\plus{}b\plus{}c\plus{}d}{4}\equal{}\sqrt[4]{abcd}$ $ \implies$ $ a\equal{}b\equal{}c\equal{}d\equal{}3$
[/hide] is $ a\equal{}b\equal{}c\equal{}d\equal{}3$


The distance AB is 12. The circle center A radius 8 and the circle center B radius 6 meet at P (and another point). A line through P meets the circles again at Q and R (with Q on the larger circle), so that QP = PR. Find QP^2.

been done before http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1983&p=76385


logarithms
combinatorics proposed
combinatorics

About Hanoi I Just remember the Hanoi's bank because the [b]A-TEAM[/b](my favorite TV programme) were blamed, unfairly, about a burglary in this bank

[quote="ZetaX"]We have a stack of $n \geq 1$ cards, numbered from top to bottom as $1,2,...,n$. We have a number $k$ of spaces available to form auxiliary stacks. The task is to rebuild the original stack on one of the spaces, according to the following rules:
- in each step, only one card can be moved
- card number $m$ can only be put on an empty space or on top of card number $m+1$

What is the smallest number $k$ of spaces such that the task can be performed¿[/quote]

I re-state the problem, a little

There are:
- one source stack (Src)
- one destination stack (Dest)
- [b]k[/b] auxiliary stacks (Aux)

Restrictions:
- in each step, only one card can be moved
- card number $m$ can only be put on an empty space or on top of card number $m+1$
- when the [u]source stack[/u] is empty, we [u]cannot[/u] put any card on it

Let $f(n)$ denote the minimum number of the auxiliary stacks that they are needed.

[color=olive][size=150][hide="Some Examples"][/size]
If $n=1$ of course we don't need any auxiliary stack, hence $f(1)=0$

If $n=2$, we move the card 1 to the auxiliary stack, then we move card 2 to the destination stack, and finally put 1 on top of 2

$\begin{array}{|c|c|c|}\hline Src&&Dest\\ \hline 2,1&&\\ \hline 2&1&\\ \hline &1&2\\ \hline &&2,1\\ \hline \end{array}$

If $n=3:$

$\begin{array}{|l|l|l|l|}\hline Src&&&Dest\\ \hline 3,2,1&&&\\ \hline 3,2&1&&\\ \hline 3&1&2&\\ \hline 3&&2,1&\\ \hline &&2,1&3\\ \hline &1&2&3\\ \hline &1&&3,2\\ \hline &&&3,2,1\\ \hline \end{array}$

If $n=4:$

$\begin{array}{|c|c|c|c|}\hline Src&&&Dest\\ \hline 4,3,2,1&&&\\ \hline 4,3,2&1&&\\ \hline 4,3&1&2&\\ \hline 4,3&&2,1&\\ \hline 4&3&2,1&\\ \hline &3&2,1&4\\ \hline &&2,1&4,3\\ \hline &1&2&4,3\\ \hline &1&&4,3,2\\ \hline &&&4,3,2,1\\ \hline \end{array}$

From the above tables we can see that $f(3)=f(4)=2$

The procedure in the case $n=4$ could be explained like this:

[b]M1:[/b] Move $(2,1)$ to some auxiliary place.
[b]M2:[/b] Move $(4,3)$ to Dst.
[b]M3:[/b] Move $(2,1)$ to Dst.

If we continue this procedure, we will find:

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\ \hline f(n)&0&1&2&2&3&3&3&3&4&4&4&4&4&4&4&4\\ \hline \end{array}$
[/color]
[/hide]$\hline$

[color=brown][size=150]Theorem[/size] $f(n) = [\log_{2}(n-1)]$, where $[x]$ means the integer part of $x$

Or else, if $2^{k-1}<n\leq 2^{k}$ then $f(n)=k$[/color]

[color=blue][size=150][hide="Proof"][/size]

Suppose that the theorem holds for any number $\leq n-1$. We will prove that it holds also for $n$

Let $1\leq a\leq b$ natural numbers such that $n=a+b$. We divide the move to three parts:

In the first move ([b]M1[/b]) we move the top $a$ cards from Src to some Aux
In the second move ([b]M2[/b]) we move the remaining $b$ cards from Src to Dest
In the third move ([b]M3[/b]) we move the $a$ cards to Dst

In $M1$ we need $f(a)$ auxiliary places and 1 more to hold the $a$ cards.
In $M2$ we need $f(b)$ Aux places. But notice that 1 more Aux is engaged.
In $M3$ we need $f(a)$ Aux places.

In total, we will need $\max\{f(a),f(b)\}+1$ Aux places. But $a\leq b$, so according to the theorem $f(a)\leq f(b)$
So we'll need $f(b)+1$ places

We want to find $a,b$ such that $f(b)+1$ is minimum.

Let $m$ be the smallest natural number such that $m\geq \frac{n}{2}$
$m\leq b\Rightarrow f(m)\leq f(b)$

So the min value of $f(b)+1$ occurs when $b=m$ and then $f(n)=f(m)+1$

Suppose that $2^{k}<n\leq 2^{k+1}$

If $n$ is even, then it can be written in the form $n=2^{k}+2\cdot q$, where $1\leq q\leq 2^{k-1}$
If $n$ is odd, then it can be written in the form $n=2^{k}+2\cdot q-1$, where $1\leq q\leq 2^{k-1}$

At any case the smallest number $m$ such that $m\geq \frac{n}{2}$ is $m=2^{k-1}+q$

But then $2^{k-1}<m\leq 2^{k}\Rightarrow f(m)=k\Rightarrow$

$f(n)=f(m)+1 = k+1$

We proved that $2^{k}<n\leq 2^{k+1}\Rightarrow f(n)=k+1$, so the theorem holds also for $n$[/color][/hide]


function
limit
logarithms
real analysis
real analysis unsolved

Let $f: \mathbb{R}\to\mathbb{R}$ a derivable function on $\mathbb{R\1}$ such that there is $M>0$ such that

                                             \[ |xf'(x) ln x-f(x)|\leq Mx( ln x)^{2}, \]
 for any $x\in\mathbb{R\1}.$ Prove that $\lim_{x\to 1} f(x)=0.$ ;)

I assume that $f: (0,\infty)\mapsto\mathbb{R}.$ We only need $f$ to be defined in a neigborhood of $1,$ anyway.

$x\ln x\,f'(x)-f(x)=x(\ln x)^2\left(\frac1{\ln x}f'(x)-\frac1{x(\ln x)^2}f(x)\right)$

$=x(\ln x)^2\left(\frac1{\ln x}f(x)\right)'.$

Thus our condition becomes:

$\left|\left(\frac1{\ln x}f(x)\right)'\right|\le M.$

So if $g(x)=\frac1{\ln x}f(x),$ then $g$ satifies a Lipschitz condition and is thus uniformly continuous on $(1-\delta,1)$ and $(1,1+\delta).$ A uniformly continuous function can be continuously extended, so $\lim_{x\to1^-}g(x)$ and $\lim_{x\to1^+}g(x)$ both exist. Hence, $g$ is bounded on a (punctured) neighborhood of $1.$

Thus $\lim_{x\to1}f(x)=\lim_{x\to1}(\ln x)g(x)=0.$

Note: the choice of what to multiply by to find a function whose deriviative we can estimate was not a lucky random guess; it was motivated and directed by the technique of multiplying factors for first order linear differential equations.


modular arithmetic
algorithm
number theory
prime numbers
group theory

For an odd $ n > 1$, prove that n does not divide $ 3^n \minus{} 1$.

[hide="Solution"]
We know that for all positive integer values n, 3^n is odd.
Hence, 3^n-1 must be even.
Since n is restricted to being odd, therefore n does not divide into 3^n-1, as required.
[/hide]

Thanks for the speedy response!

umm....
counterexample:

6 is even and 3 is odd. However $ 3|6$

[hide]
If $ 3|n$, then clearly $ 3\not|3^n \minus{} 1$ so $ n\not| 3^n \minus{} 1$.
If $ n$ is a prime other than $ 3$, then $ \gcd(3,n) \equal{} 1$, and so by Fermat's Little Theorem, $ 3^n\equiv 3\pmod n$, so $ n\not|3^n \minus{} 1$.
If $ n$ is composite, then consider a prime $ p^k$ dividing $ n$, and do some stuff that I'm too lazy to do since Catalyst gave a solution anyway :P[/hide]

Edit: Fixed serious oversight. :ninja:

Sorry about that, had a brain cramp  :blush:

[hide="solution"]
Consider this, all numbers greater than are either prime numbers, or can be written as a product of prime numbers.
Therefore, for a number to be divisible by a composite number, it must also be divisible by its prime factors.

So, we consider only the prime numbers.
Aside from 2, all prime numbers are odd and all non-prime odd numbers are products of odd prime numbers.

By Fermat's Theorem,
$ 3^{n\minus{}1}\equiv 1 \pmod{n}$ where 3 is not divisible by $ n$

The only possibility that $ n\mid 3$ is when $ n \equal{} 3$
It's easy to see that $ 3^n\minus{}1$ is not divisible by 3.

Then by congruences, multiplying both sides by 3 give
$ 3^n\equiv 3 \pmod{n}$
Subtracting both sides by 1,
$ 3^n\minus{}1\equiv 2 \pmod{n}$
Hence, $ n\nmid 3^n\minus{}1$ for all odd $ n$.[/hide]

Hopefully, I didnt make any dumb errors again  :lol:

Yongy781: Check the statement of Fermat's Little Theorem again.

yfl20673: Your proof seems to assert that $ n|3^{n} \minus{} 1$ implies $ p|3^{p} \minus{} 1$ for all $ p|n$, which isn't necessarily "true" (of course we can't talk about this being true or not, because the problem statement tells us that $ n|3^{n} \minus{} 1$ is impossible.

Anyway, this problem is a bit trickier than it looks:
[hide]Let $ p$ be the smallest prime dividing $ n$, and assume that $ p\ge5$ ($ p\neq2$ since $ n$ is odd and we can take care of the case in which $ p \equal{} 3$ easily. Then, $ p|3^{n} \minus{} 1$, so $ 3^{n}\equiv1\pmod{p}$. If $ p\neq 3$, we also have $ 3^{p \minus{} 1}\equiv1\pmod{p}$. But note that $ (n,p \minus{} 1) \equal{} 1$, since $ p \minus{} 1$ is divisible by no primes greater than or equal $ p$, and $ n$ is divisible by no primes less than $ p$.

From here, we can let $ c$ be the order of 3 modulo $ p$ (explanation below), to get that $ c|p \minus{} 1, c|n$, so $ c \equal{} 1$ and $ 3\equiv1\pmod{p}\Leftrightarrow p|2$, contradiction.

Anyway, the [b]order[/b] of $ a$ modulo $ p$ is defined to be smallest positive $ c$ such that $ a^{c}\equiv1\pmod{p}$. The fact we used above is that if $ a^{n}\equiv1\pmod{p}$, we in fact have $ c|n$. The proof is as follows: Write $ n \equal{} xc \plus{} y$ using the division algorithm, so that $ x,y$ are integers and $ 0\le y\le c \minus{} 1$. Then, $ a^{n}\equiv a^{xc \plus{} y}\equiv (a^{c})^{x}a^{y}\equiv a^{y}\pmod{p}$. Assuming this is equivalent to 1, if $ y > 0$, we have a smaller positive integer $ y$ such that $ a^{y}\equiv1\pmod{p}$, a contradiction, so $ c|n$.

In group theory 1337, this is just the order of $ a$ in $ (\mathbb{Z}/p\mathbb{Z})^{\times}$.[/hide]

[quote="Yongyi781"][hide]
If $ 3|n$, then clearly $ 3\not|3^n \minus{} 1$ so $ n\not| 3^n \minus{} 1$.
If $ 3\not|n$, then $ \gcd(3,n) \equal{} 1$, and so by Fermat's Little Theorem, $ 3^n\equiv 3\pmod n$, so $ n\not|3^n \minus{} 1$.[/hide][/quote]

I think Yongyi781's second argument can be fixed it one considers the canonical prime decomposition of n, I think.

No, that's essentially what yfl20673 tried to do, which, as I mentioned in my eariler post, unfortunately doesn't work.


inequalities
geometry
inradius
trigonometry
inequalities unsolved

Is it true that in all acute angled triangles following inequality holds:

$ \tan(A) \plus{} \tan(B) \plus{}\tan(C) \geq \frac{s}{r}$, where $ s$ and $ r$ are semiperimeter and inradius of triangle $ \triangle ABC$, respectively  :?:

This is a problem from the current edition of Mathematical Reflections. So I think we are not yet allowed to discuss it over the internet.

[hide="@ delegat"]Hi how are you? I met you in Madrid remember? :) [/hide]

[quote]This is a problem from the current edition of Mathematical Reflections. So I think we are not yet allowed to discuss it over the internet. [/quote]

Exactly But I am just asking whether is it true or not ? I am afraid that they made mistake (that happens almost in each issue) :roll:

[quote]yes, it's true...[/quote]

OK Thank you very much!

Here is my solution:

Let $ T$ denote the area of triangle $ ABC$. Verify that
\[ \sum\tan A = \sum\frac {bc\sin A}{bc\cos A} = \sum\frac {2T}{bc\cos A}
\]
Hence it remains to show that
\begin{align*} & \sum\frac {2T}{bc\cos A}\ge\frac sr = \frac {T}{r^2} \\
\Leftrightarrow & \sum\frac {1}{bc\cos A}\ge\frac {1}{2r^2}\end{align*}
Let $ a = x + y,b = y + z,c = z + x$. Then we have $ bc\cos A = \frac {b^2 + c^2 - a^2}{2} = z^2 - xy + yz + zx$. Combining this with $ r^2 = \frac {xyz}{x + y + z}$ it remains to show that
\[ \sum\frac {1}{z^2 - xy + yz + zx}\ge\frac {x + y + z}{2xyz}
\]
From Cauchy-Schwarz inequality,
\begin{align*}\sum\frac {1}{z^2 - xy + yz + zx} & = \sum\frac {1}{(y + z)(z + x) - 2xy} \\
& = \sum\frac {(x + y)^2}{(x + y)^2(y + z)(z + x) - 2xy(x + y)^2} \\
& \ge\frac {4(x + y + z)^2}{2(x + y)(y + z)(z + x)(x + y + z) - 2\sum xy(x + y)^2} \\
& = \frac {2(x + y + z)^2}{(x + y)(y + z)(z + x)(x + y + z) - \sum xy(x + y)^2} \\
& = \frac {x + y + z}{2xyz}\end{align*}
which was what we wanted.


Suppose $ P$ is the point $ (5,3)$ and $ Q$ is the point $ (-3,6)$. What is the midpoint of $ \overline{PQ}$? Express your x and y coordinates as fractions as needed.

Midpoint: (1,9/2) by averaging the coordinates.

There's something wrong with this. On my computer, it says that the answer is 19/2, which is completely wrong. Help?!  :wallbash_red:  :wallbash: :wallbash_red:

Next time this happens, just press report error.


calculus
integration
real analysis
real analysis unsolved

let $ f: [0,1]\to\mathbb R$ continuous and let $ g_n(x)\equal{}f(x^n)$ for $ x\in[0,1]$ for $ n\in\mathbb N.$ Prove that $ \int_0^1 g_n(x)\,dx$ converges to $ f(0).$

i solved this by using the DCT, but it's quite easy, i'm looking for another way by not using that powerful tool.

Fix $ \delta>0.$ We claim that $ g_n(x)$ converges to $ f(0)$ uniformly on $ [0,1\minus{}\delta].$ This, because $ x^n\to 0$ uniformly on that interval. So,

$ \int_0^1g_n(x)\,dx\minus{}f(0)\equal{}\int_0^{1\minus{}\delta}(g_n(x)\minus{}f(0))\,dx\plus{}\int_{1\minus{}\delta}^1(g_n(x)\minus{}f(0))\,dx$

$ \left|\int_0^1g_n(x)\,dx\minus{}f(0)\right|\le \int_0^{1\minus{}\delta}|g_n(x)\minus{}f(0)|\,dx\plus{}\int_{1\minus{}\delta}^12M\,dx$

where $ M$ is the maximum value of $ |f(x)|$ on $ [0,1].$

The first integral on the right tends to zero as $ n\to\infty$ because of the uniform convergence. The second integral is bounded by $ 2M\delta.$

Hence we can make $ \left|\int_0^1g_n(x)\,dx\minus{}f(0)\right|$ as small as we please.

ahhh, thanks a lot but i still don't study uniform convergence.

Interesting: uniform convergence is a Riemann integral tool and is usually dealt with in a course earlier than one would encounter the DCT, which is a Lebesgue integral tool.

Suppose $ 0\le x\le 1\minus{}\delta.$ Then $ |x^n\minus{}0|<(1\minus{}\delta)^n.$ Suppose for any $ \epsilon>0$ there exists $ \eta$ such that if $ |y\minus{}0|<\eta,$ then $ |f(y)\minus{}f(0)|<\epsilon.$ (That's the definition of $ f$ being continuous at $ 0.$) Choose $ N$ so that if $ n>N,$ then $ (1\minus{}\delta)^n<\eta.$ Then for $ n>N,$

$ \int_0^{1\minus{}\delta}|f(x^n)\minus{}f(0)|\,dx\le \int_0^{\delta}\epsilon\,dx\equal{}(1\minus{}\delta)\epsilon<\epsilon.$

Of course, that was basically the uniform convergence argument - just without using those words.

[quote="Kent Merryfield"]Interesting: uniform convergence is a Riemann integral tool and is usually dealt with in a course earlier than one would encounter the DCT, which is a Lebesgue integral tool.
[/quote]
i'm teaching myself some of stuff! haha, thanks for your second solution!


How many variantions of six digit are there it the six digit has 4 different numbers.

I'm sorry, is this supposed to make any sense?


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector

Let ABC be a triangle. On side BC we have point A'. Midperpendicular of side A'B has common point with side AB - M. Midperpendicular  of side A'C has common point with side AC - N. Straight line k goes through point N and M. Let point A'' be an image of point A' in axial symetry (line k). Prove that A'' lies on circle circumscribed about a traingle ABC.

Hope you understand my translation..

[color=blue][b]Problem.[/b] Let ABC be a triangle. Let A' be a point on the line BC. The perpendicular bisector of the segment A'B intersects the line AB at a point M. The perpendicular bisector of the segment A'C intersects the line AC at a point N. Let k be the line MN. Let A'' be the reflection of the point A' in the line k. Prove that the point A'' lies on the circumcircle of triangle ABC.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°.

Let the line MN intersect the line BC at a point Q.

Since the point M lies on the perpendicular bisector of the segment A'B, we have MA' = MB. Thus, the triangle A'MB is isosceles, and it follows that < MBA' = < BA'M. In other words, < MBQ = - < MA'Q. But since the point A" is the reflection of the point A' in the line k, we have < MA"Q = - < MA'Q. Thus, < MBQ = < MA"Q, and it follows that the points M, Q, B and A" lie on one circle. In other words, the point A" lies on the circumcircle of triangle BMQ. Similarly, the point A" lies on the circumcircle of triangle CNQ. But the Miquel theorem, applied to the complete quadrilateral formed by the lines AB, CA, BC and MN, yields that the circumcircles of triangles BMQ, CNQ, ABC and AMN have a common point. The point A", lying on the circumcircles of triangles BMQ and CNQ and not coinciding with the point Q (in general), must be this common point; hence, this point A" lies on the circumcircles of triangles ABC and AMN. This already proves more than the problem asked for.

  Darij


If $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
\[ \sum_{cyc} \frac {1}{\sqrt {a \plus{} \frac 1c \plus{} \frac 12}} \ge \sqrt 2
\]

See here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265723 :wink:


Will equilibrium and elasticity or fluids be on this year's prelim (f=ma) exam, i.e. has it been on previous exams (a lot or a little) ?

Equilibrium of what? Chemistry?

There has been a question on buoyancy on the 2008 and 2007, I think, but only minor stuff.

I don't think there's been anything major on elasticity either, although it would be perfectly fair if they had something having to do with oscillations.

That buoyancy question in the 2007 F=ma only had to do with forces, not fluid dynamics, so rest assured.

If you mean dynamic equilibrium, the answer is yes. Quite a few questions requires balancing forces and torques in the presence of springs and whatnot.


Two of the 12 winners of this year's USAMO are from NYC.  Congratulations to Tao Ran Chen (Bayside) and Krishanu Sankar (Horace Mann)!  They will be competing for spots on the IMO team.  The last student from NYC to make the IMO team was Daniel Stronger (1997).

Another student from NYC qualified for Red MOP, the olympiad summer program for 9th graders.  Congratulations to Chong Gu (Cardozo)!

Update: Krishanu made the IMO team.  Congratulations!  Here is a link to the US team:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209822[/url].


By support files, I mean the aux, log, and other files that are created when the tex file is compiled. I was wondering if there was a way to configure Kile to output those supporting files to a separate directory so the directory containing the tex file doesn't get cluttered up.. maybe there is a button/shortcut way to sort of purge the support files? Manually deleting them is getting a bit tedious ;)

That would be a silly thing to do. Quite often you need to run latex more than once and the second and third time it needs to read the content of the aux, bbl files etc which is why they are created. They are also needed if you make changes to the tex file and recompile.
You could move copies of the tex and pdf files to another directory by writing a bash file that
1. runs pdflatex or whatever with the correct options;
2. copies the tex, pdf files to another directory.
Then tell Kile to run the bash file instead of pdflatex.


Find the domain of $ f(x) = \frac{1} { [|x-1|] + [|7-x|] - 6 } $ where $[] $ is the gretaest integer fn..


ceiling function
floor function
combinatorics unsolved
combinatorics

We call $A_1, A_2, \ldots, A_n$ an $n$-division of $A$ if
(i) $A_1 \cap A_2 \cap \cdots \cap A_n = A$,
(ii) $A_i \cap A_j \neq \emptyset$.
Find the smallest positive integer $m$ such that for any $14$-division $A_1, A_2, \ldots, A_{14}$ of $A = \{1, 2, \ldots, m\}$, there exists a set $A_i$ ($1 \leq i \leq 14$) such that there are two elements $a, b$ of $A_i$ such that $b < a \leq \frac{4}{3}b$.

I think there's a mistake in your statement, since given condition (i) implies given condition (ii) for nonempty $A$...

In fact the condition (i) is : \[A_{1}\cup A_{2}\cup ... \cup A_{n}=A;\]

If $m \geq 56$, then some $A_{i}$ must contain at least two of $\{42, 43, \ldots, 56\}$, and any two such elements satsify the condition.

If $m < 56$, then define a family of sequences $a_{i,j}$ by $a_{i,0}= i$, and $a_{i,j+1}= \lceil \frac{3 a_{i,j}}{4}\rceil-1$.  Define $B_{i}=\{a_{i,j}> 0\}$ and define $A_{m}= B_{m}$, $A_{i}= B_{i}\setminus \bigcup_{j = i+1}^{m}B_{j}$ for $\lceil \frac{3m}{4}\rceil \leq i \leq m$.  Then for any $i \in \{1,2, \cdots,m\}$, define $b_{j}$ by $b_{0}= i$, $b_{i+1}= \lfloor \frac{4b_{i}}{3}\rfloor+1$.  Some $b_{j}$ will be eventually lie between $\lceil \frac{3m}{4}\rceil$ and $m$, and then $i$ will be contained in $A_{b_{j}}$.  So the $A_{i}$ are disjoint, their union is $\{1,2,\cdots,m\}$, and they satisfy the above condition.


Let $ABCD$ be a cyclic quadrilateral. I note: the intersection $I\in AC\cap BD$, the midpoints $M$, $N$, $P$, $R$ of the sides $[AB]$, $[BC]$, $[CD]$, $[DA]$ respectively, the projections $X$, $Y$, $Z$, $T$ of the point $I$ on the lines $AB$, $BC$, $CD$, $DA$ respectively and the intersections $G\in MP\cap NR$, $W\in XZ\cap YT$, $U\in MP\cap XZ$, $V\in NR\cap YT$. prove that $GW\perp UV$.

i have one very ugly solution with complex geometry.

i have no ideas to solve it with elementary ways.

According to the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55836]Quadrilateral[/url] in the high school section, $MP \perp YT, NR \perp XZ$, because the quadrilaterals MYPT and NZRX are both kites and the above pairs of perpendicular lines are their diagonals. Hence, UW, VW are 2 altitudes of the triangle $\triangle UVG$ to the sides VG, UG, respectively, W is its orthocenter and GW is the 3rd altitude to the side UV. :P


integration
logarithms
calculus
calculus computations

Solve the differential equation $ xx''\minus{}(x')^2\plus{}(x')^3\equal{}0$.

The main idea is this....Rearranging, we have

$ \frac{(x^{\prime })^{2}\minus{}xx^{\prime \prime }}{(x^{\prime })^{2}}\minus{}x^{\prime }\equal{}0$

$ \frac{x}{x^{\prime }}\minus{}x\equal{}A$

$ \int dt\equal{}\int \frac{A\plus{}x}{x}dx$

Hence $ t\equal{}A\ln |x|\plus{}x\plus{}B$


algebra
polynomial
floor function
irrational number
algebra open

Is it true that if $f$ is any non-constant polynomial with integer coefficients and a is irrational, then the sequence $ [na]$ cannot contain all the numbers $f(n)$ with n natural?

We need additional conditions. Otherwise take an arbitrary $a\in(0,1/2)$.

Upps, of course $a>1$. Are there still contraexemples?

:?  Isn't it an immediate corollary of Weyl's theorem that if $a>1$ is irrational then $\{\lfloor na \rfloor \mid n \in \mathbb{N} \}$ cannot contain all values $f(n)$ for any non-constant polynomial $f$? For if $f \in \mathbb{C}[x]$ is any such polynomial, then $f(\mathbb{N}) \subseteq \mathbb{N}_0$ implies that $f \in \mathbb{Q}[x]$ has rational coefficients, so by Weyl's theorem the set $\big\{f(n)/a \big \}$ is dense in $(0,1)$. In particular, as $a>1$, there is an $n$ such that $\big\{ f(n)/a \big\} < 1 - \frac{1}{a}$, which means that the interval $\Big( \frac{f(n)}{a}, \frac{f(n) + 1}{a} \Big)$ does not contain integers. Hence, $f(n) \neq \lfloor ka \rfloor$ for all $k$.

I don't know is it good or not to have super strong theorems for all cases. :?

At least we know that the assertion is true :). Anyway, this question asks for a proof that if $f \in \mathbb
{Q}[x]$ is non-constant and $b \in (0,1)$ is any irrational number then there is an $n$ such that $\{bf(n)\} < 1-b$; I'd like very much to see an elementary proof of this ...


Have you heard of the Bernouilli numbers?

[quote="thanhnam2902"]Hi every One!

I have a question for you:

Find formula of this sum $ S \equal{} 1^k \plus{} 2^k \plus{} 3^k \plus{} .... \plus{} n^k$

How many way to done this problem?

If it's hard, you can aswer this question "How many way to done this problem at $ k \equal{} 4,k \equal{} 5,k \equal{} 6?$".

Thank you very much.

Thanhnam2802 :P .[/quote]
What do you mean? Should we find the number of ways to solve the problem? That's just about impossible - there are probably infinitly many ways to do it..

[quote="Mathias_DK"]What do you mean? Should we find the number of ways to solve the problem? That's just about impossible - there are probably infinitly many ways to do it..[/quote]
 :rotfl:  :P  :rotfl:

BTW. This has been posted on the forums a lot of times. People like Dariz have replied. Try searching.
It's not too difficult.

Thank every One, I will seaching, I known is not difficult.


Given that
\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& \text{y is the number formed by reversing the digits of x; and}\\ &(3)& z=|x-y|. \end{eqnarray*}How many distinct values of $z$ are possible?

Sorry for bringing this topic up.. But if you do the problem in the way how kalva translated.. it can be $10$ lol....
Because it doesn't say that $n'$ is between $100$ and $999$?
so $910$ becomes $019$ lol
This is how kalva translated: $n$ is an integer between $100$ and $999$ inclusive, and so is $n'$ the integer formed by reversing its digits. How many possible values are there for $|n-n'|$?

but i know.. in the form how AoPS translated.. it's $9$.
 :D   :rotfl:

[quote=bpms][quote="4everwise"]Given that
\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& \text{y is the number formed by reversing the digits of x; and}\\ &(3)& z=|x-y|. \end{eqnarray*}
How many distinct values of $z$ are possible?[/quote]
[hide]$|100x+10y+z-100z-10y-x|=$
$|99x-99z|=$
$99|x-z|$
So all multiples of 99 work, there are 9 of them.[/hide][/quote]

Sorry to revive this post, $99|x-z|$ but what happens if x=z.

[quote=cardinals08]Please, can anyone answer?[/quote]

Ok, I just realize |x-z| can only be 0-8 inclusive not include 9.

[quote=icematrix2]this seems way too easy for an AIME problem[/quote]

Was there a point to bump this thread? Do you have a legitimate question to ask?

Combo AIME grind: Let $x=\overline{abc}$, so $x=100a+10b+c$ and $y=100c+10b+a$. Subtracting gives $|99(a-c)|=99|a-c|$. $a$ can range from $1$ to $9$ and $c$ can range from $1$ to $9$, so there are $\boxed{8}$ possibilities.


number theory
greatest common divisor
algebra solved
algebra
Iran
1997
Perfect Squares

If $m$ and $n$ are positive integers such that
\[3n^2+n=4m^2+m\]
prove that  $n-m = k^2$  where $k$ is an integer.

[quote=Arne]Rearranging we get $(n - m)(3n + 3m + 1) = m^2$.

It suffices to prove that $\text{gcd}(n - m, 3n + 3m + 1) = 1$.
Then both $n - m$  and $3n + 3m + 1$ are squares.

Now suppose a prime $p$ divides $n - m$.
Then it divides the LHS; hence it divides the RHS, so $p|m^2$.
Hence $p|m$. Because we assumed $p|n$, we get $p|3(n + m)$.
So $p$ does not divide $3n + 3m + 1$.
This suffices to establish that $n - m$ and $3n + 3m + 1$ are coprime.
Hence the result.[/quote]

Why do we have to prove that $\gcd(n-m,3n+3m+1)=1$?

[quote=grupyorum]Because the product of pairwise coprime numbers is perfect $n^{th}$ power iff each of the numbers is a perfect $n^{th}$ power.

As an alternative, observe first $n>m$, and set $d={\rm gcd}(m,n)$, with $m=dm_1,n=dn_1$. This yields, $3dn_1^2+n_1= 4dm_1^2+m_1$, namely, $n_1-m_1=d(4m_1^2-3n_1^2)$. Now, if $p\mid n_1-m_1$ and $p\mid 4m_1^2-3m_1^2$ a prime, then $n_1\equiv m_1\pmod{p}$, yielding that $p\mid m_1^2$, and thus, $p\mid m_1$, which together with $p\mid n_1-m_1$ brings $p\mid n_1$. Thus, $p\mid (m_1,n_1)=1$, a contradiction. Thus, $4m_1^2-3n_1^2$ is coprime to $n_1-m_1$, and therefore must be one (observe, it is positive).

This yields, $n_1-m_1=d={\rm gcd}(m,n)$, yielding $n-m=d^2$, as claimed.[/quote]

How $P|4m_1^2-3n_1^2$ imply $p|m_1^2$

[quote=amar_04][quote=Arne]Rearranging we get $(n - m)(3n + 3m + 1) = m^2$.

It suffices to prove that $\text{gcd}(n - m, 3n + 3m + 1) = 1$.
Then both $n - m$  and $3n + 3m + 1$ are squares.

Now suppose a prime $p$ divides $n - m$.
Then it divides the LHS; hence it divides the RHS, so $p|m^2$.
Hence $p|m$. Because we assumed $p|n$, we get $p|3(n + m)$.
So $p$ does not divide $3n + 3m + 1$.
This suffices to establish that $n - m$ and $3n + 3m + 1$ are coprime.
Hence the result.[/quote]

Another way to show why $\gcd(n-m,3m+3n+1)=1$.

First of all let $\gcd(n-m,3m+3n+1)=k\implies k|n-m$ and $k|3m+3n+1$. Now, $k|3n-3m\implies k|6m+1$. Also $k|6n-6m\implies k|6m+6n+1$ and $k|6m+6n+2\implies k|1$. So, $k=1$.[/quote]

How $k\mid 6m+6n+1$ $?$
Thanks :)


Prove that the principle of mathematical induction is equivalent to the minimum principle.

$\mathbb{N}_{0}=\{0,1,2,\dots\}$
1. The principle of math. induction implies the minimum principle.
Assume that $A$ is a nonempty subset of $\mathbb{N}_{0}$, and $A$ hasn't got the smallest element.
Consider $\phi(n): =(A\cap\{0,1,\dots,n\}=\varnothing)$.
By the principle of math. induction, $\phi(n)$ holds for all $n\in\mathbb{N}_{0}$ : $\forall_{n\in\mathbb{N}_{0}}\phi(n)$, that is $\forall_{n\in\mathbb{N}_{0}}A\cap\{0,1,\dots,n\}=\varnothing$.
Hence $A=\varnothing$ - contradiction.

2. The minimum principle implies the principle of math. induction.
Easy indirect proof.


geometry
3D geometry
algebra
function
domain
vector
linear algebra

Let $g$:$M(n,C)\rightarrow M(n,C)$ a fixed map, and $A_{0}\in M(n,C)$ fixed
$j\neq 1$ complex number fixed such that $j^{3}=1$

Find all $f$:$M(n,C)\rightarrow M(n,C)$ such that for all

$X\in M(n,C)$
$f(X)+f(jX+A_{0})=g(X)$

I'll use $\omega$ rather than $j$ for the cube root of unity.

$f(X)=\frac12\cdot\left(g(X+\frac1{1-\omega}A_0)+g(\omega^2X+\frac1{1-\omega}A_0) -g(\omega X+\frac1{1-\omega}A_0)\right)$

The nature of the domain is completely irrelevant, as long as it is a vector space over $\mathbb{C}$.


Hey,

A company has moved to new premises and the management have reason to believe that the number of internal telephne calls will increase. At the old premises the daily number of calls could be described by a normal distribution with a mean of 110. For 5% of the days, the number of calls was either less than 80 or above 140.

Question 1: Calculate the standard deviation.

How do I do that?

$ 5\%$ of the outcomes in a normal distribution occur outside of two standard deviations from the mean in either direction, so letting $ \sigma$ be the standard deviation, we have:

$ 140\equal{}\mu \plus{}2\sigma$
$ 80\equal{}\mu \minus{}2\sigma$

Since $ \mu \equal{}110$, the answer to both of these equations is $ \sigma \equal{}15$. I believe that this is the solution to this problem.


solve
$ \sin 3x\plus{}3\cos x\equal{}2(\sin x\plus{}\cos x)\sin 2x$

[quote="aktyw19"]solve
$ \sin 3x \plus{} 3\cos x \equal{} 2(\sin x \plus{} \cos x)\sin 2x$[/quote]

The equation may be written $ \sin(x)\cos(2x)\plus{}\sin(2x)\cos(x)\plus{}3\cos(x)$ $ \equal{}2\sin(x)\sin(2x)\plus{}2\cos(x)\sin(2x)$

$ \iff$ $ 3\cos(x)\equal{}2\sin(x)\sin(2x)\plus{}\cos(x)\sin(2x)\minus{}\sin(x)\cos(2x)$

$ \iff$ $ 3\cos(x)\equal{}2\sin(x)\sin(2x)\plus{}\sin(x)$ $ \equal{}4\sin(x)^2\cos(x)\plus{}\sin(x)$

$ \iff$ $ \cos(x)(3\minus{}4\sin(x)^2)\equal{}\sin(x)$

Squaring and using $ u\equal{}\sin(x)^2$, we get $ (1\minus{}u)(3\minus{}4u)^2\equal{}u$ $ \iff$ $ 16u^3\minus{}40u^2\plus{}34u\minus{}9\equal{}0$ $ \iff$ $ (2u\minus{}1)(8u^2\minus{}16u\plus{}9)\equal{}0$

And so $ u\equal{}\sin(x)^2\equal{}\frac 12$ and so $ x\equal{}\frac{\pi}4\plus{}k\frac{\pi}2$

Checking back in the original equation, we get the solutions $ \boxed{x\equal{}\frac{\pi}4\plus{}k\pi}$

hello, your equation can be written in the form
$ \sqrt{2}\sin(\frac{\pi}{4}\minus{}x)(2\plus{}\sqrt{2}\sin(\frac{\pi}{4}\plus{}2x))\equal{}0$
from here we get
$ x\equal{}\minus{}\frac{3}{4}\pi\plus{}2k\pi$
$ x\equal{}\frac{\pi}{4}\plus{}2k\pi$
Sonnhard.


AMC USA(J)MO USAMO geometry inradius trigonometry quadratics

We should do some more math on this forum. This especially goes for the junior high kids.

[USAMO 00/2] Let $ S$ be the set of all triangles $ ABC$ for which

$ 5\left(\dfrac{1}{AP} \plus{} \dfrac{1}{BQ} \plus{} \dfrac{1}{CR}\right) \minus{} \dfrac{3}{\min \{AP, BQ, CR\}} \equal{} \dfrac{6}{r},$

where $ r$ is the inradius and $ P, Q, R$ are the points of tangency of the incircle with sides $ AB, BC, CA$, respectively. Prove that all triangles in $ S$ are isosceles and similar to one another.

[hide="my solution"]

Notice that the condition,

$ 5\left(\dfrac{1}{AP} \plus{} \dfrac{1}{BQ} \plus{} \dfrac{1}{CR}\right) \minus{} \dfrac{3}{\min \{AP, BQ, CR\}} \equal{} \dfrac{6}{r}$

is equivalent to

$ 5\left(\dfrac{r}{AP} \plus{} \dfrac{r}{BQ} \plus{} \dfrac{r}{CR}\right) \minus{} \dfrac{3r}{\min \{AP, BQ, CR\}} \equal{} 6$

$ \Rightarrow 5\left(\tan\dfrac{A}{2} \plus{} \tan\dfrac{B}{2} \plus{} \tan\dfrac{C}{2}\right) \minus{} 3\tan (\max\{\dfrac{A}{2}, \dfrac{B}{2}, \dfrac{C}{2}\}) \equal{} 6$

Without loss of generality, let us assume that $ C$ is the maximum angle in the triangle.

$ \Rightarrow 5\left(\tan\dfrac{A}{2} \plus{} \tan\dfrac{B}{2}\right) \plus{} 2\tan\dfrac{C}{2} \equal{} 6$

$ \Rightarrow 5\left(\tan\dfrac{A}{2} \plus{} \tan\dfrac{B}{2}\right) \plus{} 2\tan(90 \minus{} \dfrac{A\plus{}B}{2}) \equal{} 6$

$ \Rightarrow 5\left(\tan\dfrac{A}{2} \plus{} \tan\dfrac{B}{2}\right) \plus{} \dfrac{2}{\tan\frac{A\plus{}B}{2}} \equal{} 6$

$ \Rightarrow 5\left(\tan\dfrac{A}{2} \plus{} \tan\dfrac{B}{2}\right) \plus{} \dfrac{2 \minus{} 2\tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2} \plus{} \tan\frac{B}{2}} \equal{} 6$

$ \Rightarrow \dfrac{5(\tan\frac{A}{2} \plus{} \tan\frac{B}{2})^2 \plus{} 2 \minus{} 2\tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2} \plus{} \tan\frac{B}{2}} \equal{} 6$

$ \Rightarrow 5\tan^2\dfrac{A}{2} \plus{} 5\tan^2\dfrac{B}{2} \plus{} 8\tan\dfrac{A}{2}\tan\dfrac{B}{2} \plus{} 2 \equal{} 6\tan\dfrac{A}{2} \plus{} 6\tan\dfrac{B}{2}$

Let us substitute $ x : \equal{} \tan\dfrac{A}{2}, x : \equal{} \tan\dfrac{B}{2}$.

$ 5x^2 \plus{} (8y\minus{}6)x \plus{} (5y^2\minus{}6y\plus{}2) \equal{} 0$

By the Quadratic Formula,

$ x \equal{} \dfrac{6\minus{}8y\pm\sqrt{64y^2\plus{}36\minus{}96y\minus{}20(5y^2\minus{}6y\plus{}2)}}{10}$

$ \equal{} \dfrac{6\minus{}8y \pm 2\sqrt{\minus{}9y^2\plus{}6y\minus{}1}}{10} \equal{} \dfrac{6\minus{}8y\pm2\sqrt{\minus{}(3y\minus{}1)^2}}{10}$

But since $ x : \equal{} \tan\dfrac{A}{2}$, it must be a real number, as $ A$ is a real number. Therefore,

$ 3y\minus{}1 \equal{} 0 \Rightarrow y \equal{} \dfrac{1}{3} \Rightarrow \tan\dfrac{B}{2} \equal{} \dfrac{1}{3}$

We can plug this value in to the formula for $ x$:

$ x \equal{} \dfrac{6 \minus{} 8(\frac{1}{3})}{10} \equal{} \dfrac{1}{3} \Rightarrow \tan\dfrac{A}{2} \equal{} \dfrac{1}{3}$

Therefore, since $ \tan \dfrac{B}{2} \equal{} \dfrac{1}{3} \equal{} \tan \dfrac{A}{2}$, $ A \equal{} B$, so the triangles in $ S$ are isosceles.

Also, since $ A \equal{} B \equal{} 2 \arctan \dfrac{1}{3}$, by Angle-Angle Similarity, all triangles in $ S$ are similar.

[/hide]

why do you say this especially goes for jhs kids?

besides that is usamo and i suck at it :(

[quote="Aryth"]because you guys keep talking about predictions for mathcounts when they're most likely horribly inaccurate, and they won't make YOU any better.  :wink:
you might suck now, but that's why you should try to get better[/quote]
Brutal

hey, well life is brutal sometimes. so be brutal back.
except, please don't brute force  :roll:

Maybe we could start an Illinois Math Marathon

Sweet, some math. Too bad I'm too lazy to work on it right now....maybe later :P

Maybe i can make a mathcounts mock test!!!!!!!!!!!!!!!!!!!!!!! :D  :)

is that sarcastic? is that a joke? i sure hope so.

[quote="Aryth"]We should do some more math on this forum[/quote]
In 5 days you have posted 1 problem!

mock test should be ready tomorrow
and stop triple posting!!!!!!! :mad:  :mad:  :mad:

Here's the Target Round: (no signups needed, you can do this for fun If you wanna do it please [b]PM[/b] me the answers and I'll score it)

1. Callie bought six ounces of coffee cream and 10 ounces of coffee. She first mixes both the cream and the coffee in one large pitcher, then pours 3/5 of the coffee mixture into her cup. What fraction of the coffee mixture in Callie's coffee cup is cream? Express your answer as a common fraction.

2. How many integers between 1 and 100 inclusive are divisible by 4 but not divisible by 5?

3. Bobby has a square based toolshed measuring 15 feet on each side. He tethers his dog to one vertex of the shed with a 10-foot leash. What is the area of the region where his dog can roam freely? Express your answer in terms of pi.

4. On GINORMO Sword, Lawrence kills 5 Golems in 2 minutes. If he played for 1.75 hours, how many Golems can he murder?

5. Paul decides to make a 4 feet by 6 feet montage of smilies, either with Starwars or Spam emoticons. Starwars emoticons cover 2 square feet of space and Spam emotes cover 1 square foot of space. What is the maximum number of smilies Paul can fit on the montage?

6. Xiaoyu is collecting a pyramid structure of red and green apples. On the top level, she puts 1 red apple. On the second level, she puts 4 green apples, and on the third level she puts 9 red apples. If she follows this pattern and makes the pyramid 20 rows high, how many green apples would be in her pyramid?

7. While staying at the Denver Marriott Tech Center for the 2008 Lockheed Martin MATHCOUNTS Competition, Darryl enters an elevator that travels through 20 floors. He enters the elevator on the second floor. Darryl tosses a fair, six-sided die. If the number appearing is prime, Darryl moves down one floor, otherwise, he goes up one floor. What is the probability that after 5 die rolls, Darryl would be on the 5th floor or higher? Express your answer as a common fraction?

8. Kevin leaves his pet monkey on the computer to spam in gauss1181's blog. The monkey is ordered to use a 26 letter keyboard to type "HEY GAUSS WANNA SPAM WITH ME SPAM SPAM SPAM SPAM SPAM". If the monkey does not know how to type in English letters and instead just strikes random keys on the keyboard, what is the probability that the monkey would spell GAUSS correctly AND spell SPAM correctly AT LEAST ONCE? Express your answer as a common fraction.

apparently it was not a joke =(

poor kevin all of your efforts are for naught

oh wow lol when i first looked at this problem i thought that $ P$ was opposite $ A$ in the triangle and that would make this problem quite difficult...but it is just like the little tangent segment so its actually pretty easy...

i'm pretty sure my pet monkey does know how to type in English letters.  :huh:

it is very offended  :mad:

ambiguity adds to the offensive nature of that problem...

kevin got excited because he thought that someone thought he was important, but alas he goes back to knowing that he is worthless and unloved =(.

MOCK MC TARGET. In the true spirit of MathCounts.  :roll:

[hide="Problems 1 and 2"]

1. If 1 apple is worth 13 oranges, 7 oranges are worth 113 peaches, 43 peaches are worth 1337 watermelons, and 666 watermelons cost 3451213 dollars, then 707 apples cost ____ dollars.

2. Find the area of the pentagon ABCDE with sides AB, BC, CD, DE, EA having lengths of 7879; 7883; 7901; 7907; 7919, given that CDEA is cyclic.

[/hide]

[hide="Problems 3 and 4"]

3. Approximate the fifth root of the 30th prime number to the nearest $ 10^{ \minus{} 15}$.

4. Approximate the thirtieth root of the 5th prime number to the nearest $ 10^{ \minus{} 15}$.

[/hide]

[hide="Problems 5 and 6"]

5. Find the largest prime number less than 34790!

6. How many domino tilings are there of an $ 18 \times 18$ checkerboard?

[/hide]

[hide="Problems 7 and 8"]

7. Explicitly list all solutions to the system:

$ x_1 \equal{} x_2$

$ x_2 \equal{} x_3$

$ x_3 \equal{} x_4$

...

$ x_{2008} \equal{} x_{2009}$

$ x_{2009} \equal{} x_1$

8. Calculate $ 1234567890987654321G$ where $ G$ is Graham's Number

[/hide]

[quote="Aryth"]darn, steven, you got me there. -sniff-  :([/quote]

it is stephen, now i feel unloved because you can't spell my name...WAHHHHH =(

woah that is so far beyond mathcounts level...

and stupidityismygam are you even from IL???? if you are are you stephen s. from roosevelt M.S.?

it was a joke.
and he is stePHen w. from texas.

hey, I think #7 is particularly beautiful though. doncha agree?

I think #2 looks fun... brahmagupta a bunch of times? or hmm... yeah maybe and then PIE to count areas blah

darn you actually need to THINK for #2  :huh: ? i failed at writing this then.

[hide="REPLACEMENT #2"]

2.  Sergei performs a card trick. Out of an audience of 65 people, he chooses a volunteer. He asks the volunteer to pick a card from a randomly-shuffled deck of cards. Then, the card is replaced. Sergei takes the deck of cards, shuffles it randomly, then throws them into the air, and catches one of the cards. He repeats this trick 69 times. What is the probability that he chooses Zumbot as his volunteer, and catches the correct card every time? Calculate to 150 significant digits.

[/hide]

oh well. you still have to write all the 0's  :roll:

[quote="not_trig"][quote="CatalystOfNostalgia"]darn this is quite tricky but i am pretty sure that #2 is trivial. (65*52)^69<<100^150 so i'm quite sure it's 0.00000...0[/quote]

umm wouldn't you need it to be >> not <<[/quote]

no, he's taking 1 over both sides, so sign swap.

but it should sign swap to >> ?

because like you want 1/x < 1/y   $ \implies$ x>y?....

This seems like a pretty fruitful discussion... hmmm so which way should it be...

no, dude I am pretty sure I am right

like... ok

the probability is 1/(stuff) > 1/10^150

so... I think I am right? :-/ (haha NECK [just for alex])


pls post ur scores  of inpho 09
is ne body applying re evaluation ?

wow nice da.
so you're through to the bio camp then ? tittu will be so proud of you .


Do you happen to know what Millon's Test, Hopkins-Cole Test and Ninhydrin Test's
chemical equations, principle, structure, compound responsible for the visible result
when used with a protein (egg white)? haha :p sorry for bothering you...
                                                                  -MintChocolate

i hope this is what you want
http://www.chemistry.mcmaster.ca/~chem2o6/labmanual/expt11/2o6exp11.html
http://www.cerlabs.com/experiments/10875404480.pdf


quick question:

[asy]import graph;
unitsize(1inch);
picture pic;
draw(pic, unitcircle);
draw(unitcircle);
pair O;
O=(0,0);
label("I",O+.85dir(60));
label("II",O+.85dir(30));
label("III",O+.85dir(0));
label("IV",O+.85dir(330));
label("V",O+.85dir(300));
label("VI",O+.85dir(270));
label("VII",O+.85dir(240));
label("VIII",O+.85dir(210));
label("IX",O+.85dir(180));
label("X",O+.85dir(150));
label("XI",O+.85dir(120));
label("XII",O+.85dir(90));
filldraw(Circle((0,0),.05),black);

pair H,M;
H = (O + .6dir(-((180)/2) + 90));
M = (O + .75dir(-(6(180)) + 90));
draw(O--M,linewidth(1.5), EndArrow);
draw(O--H, linewidth(2), EndArrow);[/asy]

How do I make the arrow size smaller?

The code was

[code]
import graph;
unitsize(1inch);
picture pic;
draw(pic, unitcircle);
draw(unitcircle);
pair O;
O=(0,0);
label("I",O+.85dir(60));
label("II",O+.85dir(30));
label("III",O+.85dir(0));
label("IV",O+.85dir(330));
label("V",O+.85dir(300));
label("VI",O+.85dir(270));
label("VII",O+.85dir(240));
label("VIII",O+.85dir(210));
label("IX",O+.85dir(180));
label("X",O+.85dir(150));
label("XI",O+.85dir(120));
label("XII",O+.85dir(90));
filldraw(Circle((0,0),.05),black);

pair H,M;
H = (O + .6dir(-((180)/2) + 90));
M = (O + .75dir(-(6(180)) + 90));
draw(O--M,linewidth(1.5), EndArrow);
draw(O--H, linewidth(2), EndArrow);

[/code]

Like this?
[asy]import graph;
unitsize(1inch);
picture pic;
draw(pic, unitcircle);
draw(unitcircle);
pair O;
O=(0,0);
label("I",O+.85dir(60));
label("II",O+.85dir(30));
label("III",O+.85dir(0));
label("IV",O+.85dir(330));
label("V",O+.85dir(300));
label("VI",O+.85dir(270));
label("VII",O+.85dir(240));
label("VIII",O+.85dir(210));
label("IX",O+.85dir(180));
label("X",O+.85dir(150));
label("XI",O+.85dir(120));
label("XII",O+.85dir(90));
filldraw(Circle((0,0),.05),black);

pair H,M;
H = (O + .6dir(-((180)/2) + 90));
M = (O + .75dir(-(6(180)) + 90));
draw(O--M,linewidth(1.5), EndArrow(6));
draw(O--H, linewidth(2), EndArrow(6));[/asy]
By the way, there is no need to post the full code separately if the picture works: we can see it by clicking on the picture :).

Thanks, I should have thought of trying that!


geometry
incenter
circumcircle
angle bisector
similar triangles
cyclic quadrilateral

9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$.
([i]A. Badzyan[/i])

Let $ID\perp MN$, where $D\in MN$. Let also E the midpoint of arc AC. Then I,D,N,B are concyclic, $\angle INB=\angle IDB$. Therefore we just have to show that $IM//BD\Leftrightarrow\angle EIM=\angle IBD=\angle INM\Leftrightarrow IE^2=EM\cdot EN\Leftrightarrow AE^2=EM\cdot EN$.
This is true since $\triangle AEM\sim\triangle NEA$.

Congrats! Very nice solution!
The official solution uses $AP=IP$ (easy angle calculation) to obtain $\triangle IMP\sim \triangle INP$ via $\triangle AEM\sim\triangle NEA$. And it immediately follows the conclusion :)

[quote="Myth"]9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$.[/quote]

Short remark about language: "$M$ is [b]a[/b] midpoint of $AC$" should be "$M$ is [b]the[/b] midpoint of $AC$" instead (the segment AC has only one midpoint, so the definite article should be used). The same error appears in your next sentence. Just wanted to point this out (no offence - just a little bagatelle mistake you do from time to time).

Cool problem from Badzyan, by the way. Here is a slightly different solution:

Let $I_c$ and $I_a$ be the excenters of triangle ABC opposite to the vertices C and A. Then, it is well-known that the midpoint N of the arc ABC on the circumcircle of triangle ABC is the midpoint of the segment $I_cI_a$. Now, since the internal and the external angle bisector of an angle are perpendicular to each other, we have $AI_c\perp AI_a$ and $CI_c\perp CI_a$, so that $\measuredangle I_cAI_a=90^{\circ}$ and $\measuredangle I_cCI_a=90^{\circ}$; thus, the points A and C lie on the circle with diameter $I_cI_a$. Consequently, $\measuredangle I_aAC=\measuredangle I_aI_cC$. In other words, $\measuredangle IAC=\measuredangle II_cI_a$. Similarly, $\measuredangle ICA=\measuredangle II_aI_c$. Hence, the triangles IAC and $II_cI_a$ are similar. Now, in similar triangles, corresponding points form equal angles. Since the points M and N are the midpoints of the respective sides AC and $I_cI_a$ of the similar triangles IAC and $II_cI_a$ (and thus, corresponding points in these triangles), we thus have $\measuredangle IMA=\measuredangle INI_c$. In other words, < IMA = < INB. And we're done.

  darij

[quote="Valiowk"]Actually, I think that the official solution is even nicer. :)

Just to point out, in Myth's post, $P = E$.[/quote]
Of course the official solution is nicer ;)
and if you have noticed, the point D in my solution is useless actually, because $\angle IMA=\angle INB\Leftrightarrow
\angle EIM=\angle INM$ can also be shown by angle chasing, and after all, it is just the same as the official solution.

Sorry for my typo. Indeed, $P=E$ ($P$ comes from the official solution).

[quote="mecrazywong"]Of course the official solution is nicer ;)
and if you have noticed, the point D in my solution is useless actually, because $\angle IMA=\angle INB\Leftrightarrow
\angle EIM=\angle INM$ can also be shown by angle chasing, and after all, it is just the same as the official solution.[/quote]
Yes, I realised that your solution is about the same as the official solution.  It's just that yesterday, when I was reading your solution, I somehow didn't really like the part with the cyclic quadrilateral and parallel lines :oops: , because like you say, it can be done even more easily by angle chasing. What I meant was that if Myth feels that yours is a "very nice solution", then the official solution should be "nicer". :P ;)

Okay, I'm just being picky. :lol:


For $ x>0$ and $ m\in\mathbb{N}$ Prove that:

$ \frac{x(x+1)\cdots(x+m-1)}{m!}\geq x^{1+\frac{1}{2}+\cdots+\frac{1}{m}}$

Wow that was a coOl solution for this CoOl iNequality :D