for $a,b,c>0$ prove:
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\frac{1}{3\sqrt[3]{abc}}\geq \frac{1}{abc+ab+a}+\frac{1}{abc+bc+b}+\frac{1}{abc+ac+c}$

I took it from another thread, but now it's after deadline  :D

or see here http://www.mathlinks.ro/Forum/viewtopic.php?t=117113,but the solution I gave is totally same as robpal's :D


Let A and B be two subsets of the real numbers. It is known that A  B+Z and B  A+Z
for any positive real number . (Note that X +Z is the set consisting of all numbers x+n,
where x 2 X and n 2 Z.)
(a) Is it necessarily true that A = B?
(b) If B is bounded is it necessarily true that A = B?

Bomb

please use latex, it is a bit hard to understand

Why $\LaTeX$ shouldn't work for you¿ Read Valentins introduction ( http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 ) and try !


function
trigonometry
real analysis
real analysis solved

Use Mean-value Theorem for
$f(x,y)=\sin\pi x+\cos\pi y$
to prove that there exist $\theta\in (0,1)$ such that
$\frac{2}{\pi} = \cos\frac{\pi\theta}{2}+\sin\frac{\pi(1-\theta)}{2}$

Take $x=\frac{\theta}{2}$ and $y=\frac{1-\theta}{2}$
$\theta=0\Rightarrow f(x,y)=2$
$\theta=1\Rightarrow f(x,y)=0$
$f$ is continuous ...


quadratics
algebra
polynomial
quadratic formula

[quote="gracexcor3"]how do you factor these two?

5x^2-22x-40

AND

-2x^2+15x-9

thanks![/quote]
There had been many numerous posts on how to factor quadratics. You should simply refer to those. But here's a hint: use the quadratic formula, if you know what it is.

[quote="gracexcor3"]how do you factor these two?

5x^2-22x-40

AND

-2x^2+15x-9

thanks![/quote]
They are quadratics.
So they can be factored into the form $ (x\minus{}y)(x\minus{}z)$
Refer to other threads if you need help.

Although you can only do that with monic polynomials, Peijin.

Taking a look at the discriminant of the first equation, which is $ 22^{2}\minus{}4(5)(\minus{}40)\equal{}1284$, the expression can't be factored into a nice form since 1284 isn't a perfect square.

It is factorable, but you have to use irrationals.

The second one has a discriminant of $ 225\minus{}72\equal{}153$, so again it's not factorable unless you consider irrational numbers

you can aalways do the cross method

5 times x to the 2nd power minus 22x-40
1                                                   1
5                                                   40
1                                                   2
5                                                   20
1                                                   4
5                                                   10
and so on

these numbers have to equal the middle term(without the variable)

lol, just try factoring it

and to do that stuff, you do
[code]
$5x^{2}-22x-40$
[/code]
Which produces
$ 5x^{2}\minus{}22x\minus{}40$

It's the form of factoring a quadratic in a way to find the solutions.

[quote="Ryuk"]Yes, using the quadratic equation would be the easiest way to solve this problem.

However, if you want to actually factor it, you need to use completing the square.

First, reduce (a) to 1 (The coefficient of x^2)

Then, move (c) to the right side of the equation. (Number that isn't multiplied by x)

Divide (b) by 2 and then square it. (Don't really do this in equation, it's off to the side)

Add the number you received previously to both sides.

Now, you should be able to factor the left side.

Then find the square root of each side

The rest from here should be self explanatory.[/quote]

You do not [b]need[/b] to use the complete the square method.  It may be helpful, but certainly not here.

Remember, Quadratic Formula<-->Completing the Square.


floor function
number theory
relatively prime

How many numbers less than $ 529$ are relatively prime to $ 462$?

Uh...not too sure about this solution...

[hide="?"]
So, finding $ \phi(462) \equal{} 120$. So, there are 120 numbers less than $ 462$ that are relatively prime to $ 462$.

Now, we have to look at the numbers from $ 462\minus{}528$, inclusive.

Suppose $ a<462$.  If $ a$ and $ 462$ are coprime, or $ (462,a)\equal{}1$ then $ a\plus{}462$ should be coprime too, by the Euclidean Algorithm.

$ (a\plus{}462, 462) \minus{}> (462,a) \minus{}> 1$

So, we just need to find the numbers from $ 1\minus{}66$ (462 is not going to be relatively prime to $ 462$, so I'll just disregard that case) that are relatively prime to $ 462$, which factors to $ 2,3,7,11$.

To find this, we'll count how many numbers are NOT relatively prime.  This is just $ \lfloor 66/2 \rfloor \plus{} \lfloor 66/3 \rfloor \plus{} \lfloor 66/7 \rfloor \plus{} \lfloor 66/11 \rfloor \minus{} \lfloor 66/(2\times 3) \rfloor \minus{} \lfloor 66/(2\times 7) \rfloor \minus{} \lfloor 66/(2\times 11) \rfloor \minus{} \lfloor 66/(3\times7) \rfloor \minus{} \lfloor 66/(3\times 11) \rfloor \minus{} \lfloor 66/(7\times 11)  \rfloor \plus{} \lfloor 66/(2\times 3\times 7) \rfloor  \plus{} \lfloor 66/(2 \times 3 \times 11) \rfloor  \plus{} \lfloor 66/(2 \times 7 \times 11) \rfloor \plus{} \lfloor 66/(3\times 7 \times 11) \rfloor  \minus{} \lfloor 66/(2\times 3\times 7\times 11) \rfloor$

Which equals to $ 33\plus{}22\plus{}9\plus{}6\minus{}11\minus{}4\minus{}3\minus{}3\minus{}2\minus{}0\plus{}1\plus{}1\plus{}0\plus{}0\minus{}0 \equal{} 33\plus{}16 \equal{} 49$.

So, there are $ 49$ numbers from $ 1\minus{}66$ that are NOT relatively prime to $ 462$.

That means that there are $ 17$ numbers that are coprime.

So, the answer is $ 120\plus{}17\equal{}\boxed{137}$
[/hide]

[quote="CatalystOfNostalgia"]You have not answered the question.[/quote]

wow im stupid. Just like countdown yesterday :|
i agree with zellex
 :lol:


If b is a positive integer greater than 1, the base b expansion of a positive integer is denoted $ (d_kd_{k\minus{}1}...d_1d_0)_b$ where the $ d_i$ are base b digits.  Thus, the base 5 expansion of $ (54)_{10}$ is $ (204)_5$ since in base 10, $ 54\equal{}2(5^2)\plus{}0(5^1)\plus{}4(5^0)$.  How many of the integers between one and one thousand, inclusive, have a base 5 expansion that contains at least one zero?

Notice that $ 1000\equal{}13000_5$. It is easier to count how many numbers have no zeros in them. 1 digit: $ 4$, 2 digit: $ 4^2$, 3 digit: $ 4^3$, 4 digit: $ 4^4$. For 5 digit numbers with no zeros, notice the first digit must be 1 and the second digit must be 1 or 2, for a total of $ (1)(2)(4)^3$ possibilities. The total amount without zeros is $ 4\plus{}16\plus{}64\plus{}256\plus{}128\equal{}468$, so the amount of numbers that contain a zero is $ 1000\minus{}468\equal{}532$


Pentru orice numar natural $\ n$ definim expresia $\ E(n)=n(n+1)(2n+1)(3n+1)...(10n+1)$.Determinati cel mai mare divizor comun al numerelor $\ E(1),E(2),\dots$.

Cine poate da o solutie la problema asta??

In esenta se gaseste cmdc al E(1) si E(2) si apoi se arata ca divide orice E(n.)

Pai asa m-am gandit si eu, Bianca.Dar poti da o solutie completa?As dori sa vad si alta rezolvare, diferita de a mea.


Let $ 2$ triangle $ ABC,A'B'C'$ satisfy $ A\in B'C',B\in C'A',C\in A'B'$ and $ \widehat{ABC}\equal{}\widehat{A'B'C'},\widehat{BCA}\equal{}\widehat{B'C'A'},\widehat{CAB}\equal{}\widehat{C'A'B'}$.
Prove that: distance from $ 2$ orthocentre of $ 2$ this triangle to circumcentre of triangle $ ABC$ are equal.

Already posted. See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143090[/url].


(a) Find the maximum value of the expression $ x^{2}y-y^{2}x$ when $ 0\le x \le 1$ and $ 0 \le y \le1$.

(b) Find the maximum value of the expression $ x^{2}y+y^{2}z+z^{2}x-x^{2}z-y^{2}x-z^{2}y$

when $ 0 \lex\le 1, 0\le y\le 1, 0 \le z \le1$.

[quote="AndrewTom"](a) Find the maximum value of the expression $ x^{2}y - y^{2}x$ when $ 0\le x \le 1$ and $ 0 \le y \le1$.

(b) Find the maximum value of the expression $ x^{2}y + y^{2}z + z^{2}x - x^{2}z - y^{2}x - z^{2}y$

when $ 0 \lex\le 1, 0\le y\le 1, 0 \le z \le1$.[/quote]

(a) By $ x^2\le x$, we get $ x^{2}y - y^{2}x\le x(y-y^2)\le y-y^2=\frac 14-(y-\frac 12)^2\le \frac 14$.

(b) $ x^{2}y + y^{2}z + z^{2}x - x^{2}z - y^{2}x - z^{2}y=(x-y)(x-z)(y-z)$. As the expression is cyclic, we can assume $ x=max\{x, y,z\}$. For a maximum of it, we must then have $ y>z$, hence $ x=1$ and $ z=0$, which leaves us with $ y(1-y)$ as in (a). So the max is again $ \frac 14$. :)


Let $ T: F^{\infty}\rightarrow F^{\infty}$, $ T(x_{1}, x_{2},\ldots)=(x_{2},x_{3},\ldots)$. Find a basis for $ \textrm{Ker }T$ and what is $ N(T^{k})$?

I hope I am not missing something here, because this seems straightforward.

$ T$ is just a shift operator.  So $ \ker T=(k,0,0,\cdots)$ where $ k\in F$, so a basis can be given by $ k=1$.  Similarly for $ T^{k}$ (which just shifts $ k$ times), $ \ker T^{k}= (c_{1},c_{2},\cdots,c_{k},0,0,\cdots)$ where the $ c_{i}\in F$.


There's a $ \frac{1}{129}$ chance that he picked the heads-heads coin, and that coin has a $ \frac{1}{1}$ chance of being heads on the next flip. The probability that he picked the heads-heads coin and the next flip will heads is $ \frac{1}{129}\times\frac{1}{1} \equal{} \frac{1}{129}$
There's a $ \frac{128}{129}$ chance that he picked a heads-tails coin, and those coins have a $ \frac{1}{2}$ chance of being heads on the next flip. The probability that he picked a heads-tails coin and the next flip will be heads is $ \frac{128}{129}\times\frac{1}{2} \equal{} \frac{64}{129}$
Since these two sequences of events cannot both occur, (ie., he cannot pick both the heads-heads coin and a heads-tails coin), we simply add the probabilities of each to obtain the answer. Thus, the answer is $ \frac{1}{129} \plus{} \frac{64}{129} \equal{} \frac{65}{129}$

[quote="aidan"]There is a pile of 129 coins on a table, all unbiased except for one which has heads on both sides. Bob chooses a coin at random and tosses it eight times. The coin comes up heads every time. What is the probability that it will come up heads the ninth time as well?[/quote]

this is one of those tricky weighted probability questions.

so well, let's look at the possibility that it came up heads eight times in a row. it seems pretty unlikely.

$ \frac{1}{129}\cdot1 \plus{} \frac{128}{129}\cdot\frac{1}{256} \equal{} \frac32\cdot\frac{1}{129}$

and now the probability of the coin picked being the biased one is $ \frac{\frac{1}{129}}{\frac32\cdot\frac{1}{129}}\equal{}\frac23$, so the probability of the coin being the unbiased one is $ \frac13$.

so our answer is $ \frac23\cdot1 \plus{} \frac13\cdot\frac12 \equal{} \boxed{\frac56}$

Isn't the probability of the coin coming up heads the same each time? So is the condition that it comes up heads 8 times in the first 8 throws important?

That's kinda what I was thinking...but I'm not completely sure.

Suppose we have the same setup (128 regular coins and 1 two-headed coin) and I choose one coin at random, flip it 10,000 times, and get heads every time.  Given this information, is the probability that I chose the irregular coin still 1/129, or do you think it has changed?

Suppose instead we have the same setup (128 irregular coins and 1 two-headed coin) and I choose one coin at random, flip it 8 times, and get heads 7 times and tails once.  Given this information, is the probability that I chose the irregular coin still 1/129, or do you think it has changed?

Lazarus' solution-idea is correct (haven't checked the details), if not explained as thoroughly as could be.  This is just a standard application of Bayes' Theorem on conditional probabilities, but "standard" applications of Bayes' Theorem tend to be very tricky if you aren't used to it.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221979 for another recent example from the fora.


limit
logarithms
algebra
binomial theorem
calculus
calculus computations

Proove the following limit: \[\lim_{x\to 0+}\frac{1}{x^{x}}=1\]

how do you prove \[\lim_{x\to 0^{+}}e^{x\ln x}= e^{\lim_{x\to 0^{+}}x\ln x}\]

[quote="mathematica"]how do you proove \[\lim_{x\to 0^+} e^{x\ln x} = e^{\lim_{x\to 0^+} x\ln x}\][/quote]

In short,

$f$ continuous, $\lim_{x\to x_0}g(x)=b$

We want to show that  $\lim_{x\to x_0}f(g(x))=f(\lim_{x\to x_0}g(x))$

Take $any$ sequence ${x_n}$ that $x_n\to x_0$. We have $sequence$  ${g(x_n)}$, and $\lim_{n\to \infty}g(x_n)=b$

So $\lim_{n\to \infty}f(g(x_n))=f(b))$ because f is continuous at b.

could you please use the epsilon-delta notation?

If you're just learning analysis (which seems likely), it would be a good exercise for you to prove some of the statements used here.  I know of a proof that uses only the binomial theorem, but I don't have time to post it now (it's in Walter Rudin's Principles of Mathematical Analysis)


One thing I'm having a bit of a hard time getting into my cranium is why I keep running across the statement that halogens are just missing one electron to fill their shell.  I do understand valence electrons but since all elements in this group have seven valence electrons and beyond the second shell, each "shell" is large enough to accomodate much more than just one more electron beyond the seven valence electrons.

Anyone have a simple way of explaining the "just missing one electron to fill their shell" statement?    :D

the one missing electron does not mean that the one electron is all the atom needs to complete its shell!

but rather, when they recieve the extra electron, they became more stable in Noble conficgration (because of the electron config.)

yea this also affects it's ionization energy and electron affinity...if you add one more electron, you'll have a stabler atom..


graph theory
combinatorics proposed
combinatorics

Let $m = 30030$ and let $M$ be the set of its positive divisors which have exactly $2$ prime factors. Determine the smallest positive integer $n$ with the following property: for any choice of $n$ numbers from $M$, there exist 3 numbers $a$, $b$, $c$ among them satisfying $abc=m$.

I think it's $11$. $30030=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13$. We regard the six prime divisors as the vertices of a graph, and turn the problem into the following, more general one:

Find the minimal $e=e(t)$ with the property that every graph on $2t$ vertices with $e$ edges has a perfect matching (in our case, $2t=6$).

The answer should be $\frac{(2t-1)(2t-2)}2+1$ (which gives $11$ when $2t=6$), I believe. We can prove this as follows:

$\frac{(2t-1)(2t-2)}2$ certainly isn't enough, since we can take a complete graph on $2t-1$ vertices, and an isolated vertex. This graph obviously has no perfect matching. On the other hand, $\frac{(2t-1)(2t-2)}2+1$ is enough, since it has been shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hamiltonian&t=45004]here[/url] that any graph on $u$ vertices with $\frac{(u-1)(u-2)}2+2$ edges is hamiltonian. We apply his to $u=2t$ to conclude that our graph, which has $2t$ vertices but only $\frac{(2t-1)(2t-2)}2+1$ edges, must have a hamiltonian path $x_1,x_2,\ldots,x_{2t}$. Our perfect matching will then consist of the pairs of vertices $(x_i,x_{i+1}),\ i\in\{1,3,\ldots,2t-1\}$.

I'm sorry to ask, Grobber, but how is relation between two vertices defind?

Two vertices=prime numbers are connected iff their product occurs in the set of that $n$ numbers.


geometric series
calculus
calculus computations

If $ r>1$, then prove that $ \sum_{n\equal{}1}^{\infty} \frac{1\plus{}r\plus{}r^2\plus{}\cdots \plus{}r^{n}}{(1\plus{}r)^n}$ converges and find the sum $ S$ in terms of $ r$.

[quote="kunny"]If $ r > 1$, then prove that $ \sum_{n = 1}^{\infty} \frac {1 + r + r^2 + \cdots + r^{n}}{(1 + r)^n}$ converges and find the sum $ S$ in terms of $ r$.[/quote]

The solution is very easy. We have
\[ S(r): = \sum_{n = 1}^{\infty}\sum_{i = 0}^{n}\frac {r^i}{(1 + r)^n} = \sum_{n = 1}^{\infty}\frac {r^{n + 1} - 1}{(r - 1)\cdot (1 + r)^n}.\]
Hence, if $ r > 1$, the sum indeed converges. Next, we have for $ r > 1$
\begin{align*} S(r) & = \frac {r}{r - 1}\sum_{n = 1}^{\infty}\left (\frac {r}{1 + r}\right )^n - \frac {1}{r - 1}\sum_{n = 1}^{\infty}\frac {1}{(1 + r)^n} = \\
& \\
& = \frac {r}{r - 1}\cdot\frac {r}{1 + r}\cdot\frac {1}{1 - \frac {r}{1 + r}} - \frac {1}{r - 1}\cdot\frac {1}{1 + r}\cdot\frac {1}{1 - \frac {1}{1 + r}} = \\
& \\
& = r + 1 + \frac {1}{r}. \end{align*}

[quote="M@ri@n"][quote="kunny"]If $ r > 1$, then prove that $ \sum_{n \equal{} 1}^{\infty} \frac {1 \plus{} r \plus{} r^2 \plus{} \cdots \plus{} r^{n}}{(1 \plus{} r)^n}$ converges and find the sum $ S$ in terms of $ r$.[/quote]

The solution is very easy. We have
\[ S(r): \equal{} \sum_{n \equal{} 1}^{\infty}\sum_{i \equal{} 0}^{n}\frac {r^i}{(1 \plus{} r)^n} \equal{} \sum_{n \equal{} 1}^{\infty}\frac {r^{n \plus{} 1} \minus{} 1}{(r \minus{} 1)\cdot (1 \plus{} r)^n}.\]
Hence, if $ r > 1$, the sum indeed converges. [/quote]

Why does the sum converge?

That's new to me  :roll: , but I examined it by Google.:

[quote="kunny"]
Why does the sum converge?[/quote]
Let $ S(r): \equal{} \sum_{n \equal{} 1}^{\infty}\sum_{i \equal{} 0}^{n}\frac {r^{i}}{(1 \plus{} r)^{n}} \equal{} \sum_{n \equal{} 1}^{\infty}\frac {r^{n \plus{} 1} \minus{} 1}{(r \minus{} 1)\cdot (1 \plus{} r)^{n}}$.

Let $ A(r): \equal{} \sum_{n \equal{} 1}^\infty\left(\frac {r}{r \plus{} 1}\right)^n$, and $ B(r): \equal{} \sum_{n \equal{} 1}^\infty\left(\frac {1}{r \plus{} 1}\right)^n$. Becouse $ r > 1$ both $ A(r)$ and $ B(r)$ converge as geometric series, and becouse $ S(r) \equal{} \frac {r}{r \minus{} 1}A(r) \minus{} \frac {1}{r \minus{} 1}B(r)$, $ S(r)$ converges.

The reply what I expected is $ 0<\frac{r}{1\plus{}r}<1,\ 0<\frac{1}{1\plus{}r}<1$ for $ r>1$, although you might say the fact is self-evident.

Anyway M@ri@n7s answer to $ S$ was correct.


How did you do at the Chapter MATHCOUNTS competition?  Feel free to list your test and countdown results as well as your team results.  How many of you will be competing at the state level?

Here's our school's results for the chapter:

School: CMS (its been abbreviated)
Chapter: Seacoast
State: New Hampshire
----------
Team Results:
Our Team got indiv. places: 1,2,3,4
Our team average: 38.25 +20 team round points
We got 1st in chapter by 20 points
----------
My Results:
I got a 37 and placed 3rd
----------
Individual Results:
Our Individuals got places: 5,7,10,-- (I think)
----------
Counddown Results (Unofficial in our chapter)
Our qualifiers got: 1,3,3,...
----------
Sheer ownage of our chapter

School: Smith Middle School
Town: Glastonbury, Connecticut
Chapter: Greater Hartford

School Results:
Our Team placed 2nd :( :
Placings were 4, 12, 19, 123; Team Round=8
Another individual placed 6th with a 37

My Results:
1st Countdown (Unofficial with 16 competitors)
4th Written (39 {8 Target, 23 :(  :(  :(  :(  :( Sprint})

Three of our people made countdown, but two failed to advance past the first round

I hope I do better at states... I was pretty upset with my performance at Chapters...

We had like a 10 person tie, but they said i missed a problem earlier or something.

Ya... ours came down to a bunch of tiebreakers... that's how it always is in CT... one problem drops you like 2-10 spots at best, depending on where you are in the rankings....
Go Braves... I feel bad for you... must have a tough tough chapter...

Dallas chapter:

i got 29/7 (43)

second place

team got 8 and second place

Clayton/Henry/Fayette county chapter
Cherish Christain Home Educators
Georgia
Had 1 individual in top 6, he got #1 spot
And he took that team to the top!

If you didn't figure it out, that's me.

I don't know the exact score yet

Billy

School: BCS

Chapter: Oakland

State: Michigan

My Score: 44 (28/8), 2nd place

School Score: 55.25 (35.25/10), 5th place

Countdown (10 person Ladder): 3rd  :(

Made it to states

[quote="MCrawford"][quote="GoBraves"]14th individual (State- NC)
Individual Score-39 (8 target, 23 sprint)
School score- Dont even ask.....

Did I make it to states- No :(

Hockey you are lucky........[/quote]

That sounds like an extremely tough chapter.[/quote]

In my chapter, the 7th place written was 42.  2nd place team had no people in countdown, but an average written of 39.75.

Hm, Connecticut's state competition is going to be pretty interesting...

I want to get a 25/7 as my baseline for states... but that's realy baseline.. but I kno that I can get a 28 to thirty on a state sprint (was 2003 considered easy? I got a 30 on it...) and I'm hoping that I don't make too many stupid mistakes.. it just involves having 10-15 minutes to check....

I thought 2003 states was easy, but there were some weird scores on it. I took it as practice last year and got a 46, but people I knew who are really smart only got mid-30s on it. So look for mixed results. Top 4 in Connecticut 2003- 36,35,34,33

Edit- oops, never mind. that's 2002. CT's 2003 had 43, 42, 40, 39

In 2003, I got a 43 (29/7).  The mistakes I made were, to say the least, retarded.  And CA had 2 perfects, so a 43 was 6th (in our Northern CA competition)

[quote="uberl33tmage"]I did a lot better this year. i made it in the 4th seat. Left_unattended, I'm surprised. BTW nice name.. WHAT HAPPENED TO YOU!!!? I was expecting a #1 or 2. I'm sorry you had to miss nationals.. :( .[/quote]
Who is she? (left_unattended)

The SC Nats team has long been a joke, coming in 53rd last year.  Watch Coach John Rushman turn it around with a great SC team ths year.  Watch Phillip Cross in the Countdown Round.  Remember, I called it!  SC in the top 20!

Something cool: The Crayton Middle School team placed 1st regionally, and Hand Middle was 3rd.  AC Flora High School feeds from Crayton; Dreher High feeds from Hand Middle.  So the 3rd place Hand team went to a high school competition and came in 1st.  That's 3rd in middle school, but 1st in high school.  That shows how advanced MathCounts is!!!

[quote="Press alt f4"]I'm from the Greater Houston Chapter which is probaboly (sp?) one of the hardest Chapters. I was over all disapointed with my preformance, the two individuals from our school did better than me. Oh well it was only a point, and I'm still going to state. Well here are the results.
Team -
1st place team score (63.75) (yes it really is that high, they had two 46's)
2nd place team score (61) :(
My Team #3 score (60.75)  :(  ONE QUESTION!!
Individuals -
#1 - 2 (46)
#3 (45)
#4 - 6 (44)
#7 - 10 (42)
#11 ME! (41)  :([/quote]

Whos PressAltF4?
Are you Drew?
Im Dennis and Im also from that chapter.
I got a perfect and was on the winning team.

Btw, 2nd place team (Doerre) got a 60.75 and you guys (First Colony) got a 60.50.

Too bad this is Jenks' last year on top of both Tulsa and Oklahoma...

Got scores. 36,5 for team. Individuals got 40,22,14,14.

7th in chapter with 36 :( , Second in Countdown round :lol: Did not make it to state, countdown round unofficial :( 7th grade .beat  :first: all 7th graders in chapter.  I'll be coming next year, but Calafornia State looks pretty hard!

I got 40 points and my team got first:

[color=darkblue][size=75][Moderator rcv's note:  Off-topic discussion of the Northwest Chinese School moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31669]this thread[/url]][/size][/color]

I got 22 on the Sprint Round and all eight on the Target Round, giving me a score of 38 points.  That placed me second overrall before the countdown round.  I pulled ahead to first place in the countdown round :D and went on to state.

I'm pretty sure I got a 28/7 for 42 which made me 1st place, and I also won countdown. 2nd place was a 40 from the same team... third place was a 30-something from the same team.. I'm pretty sure we got really really close to a 60 for overal team score though. Our chapter is so easy!

Mine:

1st in Countdown
2nd Indivisual (Score - I think I got a 16 on target but I don't know what my sprint score was)

Team:
2nd place we got team members 2nd, 3rd 4th 5th, but some random school comes up and beats us in the team round, yep that's how bad we did as a team
Also another indivisual not in the team got 6th and another 12th

This is the first year that we didn't win in our chapter :(

In our chapter competition, we got 4th place team (Out of 5 teams!)

um...
Houston Chapter--
My scores were:
25/8 (41)
11th place....

Team results:
individuals got 11th, 37th, 64, 68 (out of like 300 students)
individ scores were 41, 38, 29, 28
Team score of 7
Our team got 6th in chapter, advanced to state

Hmm...I actually dunno my score, but my coach does, he just hasn't told me (grr!)

Anyway, I'm in the Greater Cleveland Chapter, I guess you could call it, which is about average difficulty I'd say with 15 teams and over 100 people. Our team placed first written and I took first individual and a teammate took third; an individual not on the team finished somewhere in the top 16 advancing him to states with the team. This is weird: our chapter does a team countdown round that is unoffical; we go down there as a team against 3 others at a time and face off; the winners of each bracket then go to the finals. We swept down there, getting 3 points instantly both rounds we were up. It was great!


SFFT
videos
Vieta
special factorizations

Solve

$ x\plus{}y\equal{}7$

$ \frac{x}{y} \plus{} \frac{y}{x} \equal{} \frac{25}{12}$

hello, hint plugg $ y\equal{}7\minus{}x$ in the second equation.
The solutions are $ (4,3)$ or $ (3,4)$.
Sonnhard.

[hide] The second equation can be rewritten as $ \frac{x^2\plus{}y^2}{xy}\equal{}\frac{25}{12}$. Squaring the first equation we find that $ x^2\plus{}2xy\plus{}y^2\equal{}49 \implies x^2\plus{}y^2\equal{}49\minus{}2xy$. Substituting into our second equation, we have

$ \frac{49\minus{}2xy}{xy}\equal{}\frac{25}{12}$

$ \implies \frac{49}{xy}\minus{}2\equal{}\frac{25}{12}$

$ \implies \frac{49}{xy}\equal{}\frac{49}{12} \implies xy\equal{}12$.

Now, we can use SFFT to show that $ (x\plus{}1)(y\plus{}1)\equal{}20$, which when combined with the first equation, yields the solutions as $ (4,3)$ and $ (3,4)$.[/hide]

Simon's Favorite  Factoring Trick (with American spelling). If you google this, you'll find two videos on it: they are excellent.

SFFT is very useful.

From the second equation, $ 12x^2\minus{}25xy\plus{}12y^2\equal{}0 \implies 12(x\plus{}y)^2\equal{}49xy \implies xy\equal{}12$

By Vieta's, it is then just the solutions to $ a^2\minus{}7a\plus{}12\equal{}0$ , i.e. $ (3,4) or (4,3)$


geometry
algorithm
number theory
greatest common divisor
analytic geometry
graphing lines
slope

I used to have a "great" prime factorization program that a bit bruteforce...

it gives the factors, ordered from smallest to largest and in a list :)
The problem is that its still bruteforce :(
what's your factorization algorithm?

Basically I prompted N, let M=1, and if the gcd of M and N is greater than 1, then we divide N by M, store to N, and we post M, let M=M+1 if gcd(M,N)=1, and so on, until $ M > \sqrt {N}$, at which we post N. So for 4759, you would wait until M is about 70, and then it posts 4759. Or for 10023, we have 3, 13, and 257.

It's the algorithm for testing for primes.

EDIT: It uses 93 units of space. If anyone has a smaller program, I'd like to see it.

I find distance, slope, and midpoint formulas helpful.

Here is my code (TI-84):

ClrHome
Disp "COORDINATES","(X,Y),(A,B)"
Prompt X,Y,A,B
ClrHome
B-Y->C
A-X->D
B+Y->E
A+X->F
(E+F)/2->H
sqrt(C^2+D^2)->I
Disp "DISTANCE:",I
Disp "DISTANCE^2:",I^2
Pause
ClrHome
F/2->J
E/2->K
Disp "MIDPOINT:","X VALUE",J,"Y VALUE",K
Pause
ClrHome
If D=0
Then
Disp "NO SLOPE"
Else
Disp "SLOPE:",C/D->Frac,C/D->Dec

Uses 234 units of space.  Very efficient.

Oh, and if you don't wanna work for your programs, go [url=http://www.ticalc.org/pub]here.[/url]

For a factors program (which I find to be more useful), I just use a while statement, testing whether $ n$ is divisible by any integers 1 to $ \sqrt {n}$.  Simply do if n mod i = 0 display it, then divide n by i and display that.

For prime factorization use the factor() function on TI-89  :) .

...anyone here program in TI-assembly code?
I made Tetris on the 84 with it...  :)

1=2, could you post the exact code for your factoring program? Thanks.

I do have a "last digit" program that chess64 gave me.

[hide="code"]Prompt N
1->M
Lbl 0
If M>sqrt(N)
Then
Goto 1
Else
If gcd(M,N)=1
Then
M+1->M:Goto 0
Else
Disp M:Pause
N/M->N
If N=1
Then
Goto 1
Else
Goto 0
Lbl 1
Disp N[/hide]

What's a "last digit" program?

My math teacher made this program (I think).

Input B
(0)->A
Lbl 1
(A+1)->A
(B/A)->C
int(C)->D
If (C-D) $ \neq$ 0 or (D-C) $ \neq$ 0
Goto 1
(B/C)->E
int(E)->F
If (E-F) $ \neq$ 0 or (F-E) $ \neq$ 0
Goto 1
Disp C
Disp E
Pause
If A<C or A<E
Goto 1

[hide="Quadratic Solver"]a+bi
Disp "AX^2+BX+C=0"
Prompt A,B,C
If A=0:Then
If B $ \neq$ 0
Disp "X=",-C/B->Frac
If B=0
Disp "BAD PROBLEM"
Stop:End
B^2-4AC->D
Disp "X=",-B/(2A)+sqrt(D)/2A->Frac
Disp -B/(2A)-sqrt{D}/2A->Frac[/hide]

Here is a not-very-useful program that can tell you the coefficient of a term in the expansion of (ax+by)^c.

ClrHome
Disp "(AX+BY)^Z"
Prompt A,B,Z
ClrHome
Input "X TERM?",T <---See Note, Below.
Input "POWER?",P
If P>Z
Then
Disp "ERROR!"
Else
A^P->D
D * Z nCr P-> E
Z-P->F
ClrHome
If T$ \neq$0
Disp E,"X=",P,"Y=",F
If T=0
Disp I,"X^",F,"Y=",P

Note: If T=0, then you know the Y term power.  If T$ \neq$0, then you know the X term power.

[quote="Brut3Forc3"]Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them.  :wink:
Oh, and ernie, your If-Then-Else needs an End somewhere...[/quote]

I know.

It was my teacher's program.

And I almost never put End's.

It still works. :P

I have a program for graphing the Mandelbrot set. And I did not make this, one of my friends did. Pretty useless though.

[hide]

[code]

-2->Xmin
1->Xmax
1->Ymax
-1->Ymin
ClrDraw
AxesOff
a+bi
(Xmax-Xmin)/94->H
(Ymax-Ymin)/62->V
For(I,0,93)
For(J,0,61)
Xmin+IH->X
Ymin+JV->Y
X+Yi->C
0->Z
For(N,1,<insert number>) *The larger the number, the more accurate the program is
If abs(Z)<=2
C->Z
End
If abs(Z)<=2
Pt-On(real(C),imag(C))
DispGraph
End
End
StorePic 7

[/code]

Note the last step is optional.
[/hide]

Also, Shoelace algorithm

[hide]

[code]

ClrHome
Prompt N
For(C,1,N)
Prompt X
Prompt Y
ClrHome
X->L1(C)
Y->L2(C)
End
L1(1)L2(N)->S
L1(N)L2(1)->T
For(D,2,N)
S+L1(D)L2(D-1)->S
T+L1(D-1)L2(D)->T
End
ClrHome
abs(S-T)/2->A
Disp A

[/code]

N is the number of vertices of the polygon and X and Y are the coordinates

[/hide]

I've also made MONOPOLY on my calc! It's like 3 KB though and doesn't handle chance cards...

will release code soon

[quote="james4l"]basically for my program of factoring (it got mem-deleted)
[hide]
[code]
Prompt A
{1}->L1
While fPart(A/2)=0
A/2->A
augment(L1,{2})->L1
End
1->B
While B <= sqrt(A)
B+2->B
While fPart(A/B)=0
A/B->A
augment(L1,{B})->L1
End
End
A->L1(1)
L1->L1
[/code]

-> means store variable and <= means less than or equal to
[/hide]

This code cuts time in about half by skipping even numbers larger than 2  :)[/quote]

There you go, 1=2.

The reason why Labels and Gotos are bad are that they require massive amounts of time searching for the label, so While and For loops are often MUCH faster.

My factoring program factors typical 7-digit numbers in about 3-4 seconds, and for each digit added on, the time increases about 1.5 fold

ah.

nice program. I can actually scroll through too! :D

It cut the time in a fourth for finding that 4759 is prime, and has the memory capacity for $ 2^{2^5}\plus{}1$.

Yeah, but if TI-84 doesn't SUPPORT Java/C then what's the point?

Read the title. Not computers, calculators...  :roll:

[quote="1=2"]What's a "last digit" program?[/quote]

Like, find the units digit of $ 3^{1337}$. I'll post later when I get home.

[quote="1=2"][quote="Brut3Forc3"]Just a general comment: Lbl and Goto's are bad. Like.... bad. Try not to use them.  :wink: [/quote]

1) How bad is bad?
2) I don't know how to eliminate them in my factoring program. Can you tell me how?[/quote]
[url]http://xkcd.com/292/[/url]
That's how bad.

I've also made some programs -a snake game, a linear diophantine equation solver, a work generator for simplifying fractions in the form of $ \frac {a \plus{} bi}{c \plus{} di}$ (I got lazy in class), and an epic animation that composed of for loops to generate lag, and outputing long strings that I had calculated myself... it had a bomb, snake, pong, and everything... too bad I broke my calculator... :ewpu:  :bomb:  :furious:
EDIT: Oh, I also made a fake menu that let you fake deleting memory, so you could keep your programs on a test. But that was on assembly... BASIC is way too laggy to actually fake anything.

Tinspire has crazily limited programming capabilities.

[hide="Pascal's Triangle"]Input "ROW?", A
If A < 0
Then
ClrHome
Disp "ROWS CANNOT BE","NEGATIVE."
Pause
prgmPSCLTRI
End
If A >= 0
Then
-1 -> B
Lbl C
B + 1 -> B
If B <= A
Then
Disp A nCr B
Pause
Goto C
End
If B > A
Then
Stop:End
[/hide]

[hide="Binomial Expander"]ClrHome
Disp "(AX+BY)^C"
Prompt A,B,C
-1 -> D
Lbl E
D + 1 -> D
If D <= C
Then
ClrHome
Output (3, 1, C nCr D * A ^ (C - D) * B ^ D)
Output (4, 1, "*X^")
Output (4, 4, C - D)
Output (5, 1, "*Y^")
Output (5, 4, D)
Pause
Goto E
End
If D > C
Then
Stop:End[/hide]

[hide="Factor"]Input A
0 -> B
Lbl C
B + 1 -> B
If B <= sqrt A
Then
If A/B = int (A/B)
Then
Disp B, A/B
Pause
Goto C
End
End
If B > sqrt A
Then
Stop:End[/hide]

WARNING: All three use Lbl's and Goto's. :P

[hide="Pascals Triangle"]Input "ROW?", A
If A < 0
   Then
   ClrHome
   Disp "ROWS CANNOT BE","NEGATIVE."
   Pause
   prgmPSCLTRI
End
If A >= 0
   Then
   -1 -> B
   While B <= A
      B + 1 -> B
      Disp A nCr B
      Pause
   End
End
[/hide]
I think this works better. (Quote me to see the structure better.)

ok... all this hype about goto is ridiculous
all practical programs dont change significantly
oh well here is an uber pwn program

note that => is the store function... too lazy to make a LaTeX code for it
:Delvar z
:ClrHome
:For(A,1,16)
:randInt(1,4,A)=>$ L_1$
:For(B,1,A)
:Output(1,B,$ L_1$(B))
:End
:For(B,1,100A)
:End
:ClrHome
:For(B,1,A)
:Delvar C
:While c>4 or c<1
:getKey=>C
:C-91+(C=95)+13(C=82)=>C
:End
:If C$ \neq L_1$(B)
:Goto 1
:End
:Z+1=>Z
:End
:Lbl 1
:ClrHome
:Disp "SCORE:"
:Output(1,7,Z)

It is NOT ridiculous, most bruteforce-ish programs that encounter too many Gotos and Lbls and Menus slow down considerably. Consider my RPG game, in around 500 moves, and 120 battles (battles used Labels back then) the program slowed down about 60%. In about 1200 moves the game is unplayably slow [which was why I had savegames].

[quote="stevenmeow"]but that game isnt practical; most practical programs should be like under 300 bytes
anyway who has tried my program?
at least it has no goto's except for 1 and that is used only once :D
the programs pretty beast[/quote]

Not always. Many practical programs like spreadsheets, probability simulators, and "calculator in calculator" programs can take upward of 10kb.

And consider this (admittedly impractical) program:

0->A
Lbl A
A+1->A
Disp A
Goto A

This should be obvious that it is a counter that displays 1,2,3,4,5,6,7,...

But at higher numbers [I think it is at 6000-10000, not so sure] it displays ERR:MEMORY.

Of Course we can substitute this particular program with:

For(A,1,999999)
Disp A
End

Yes, One-Time Gotos/Lbls are generally O.K. But if the Gotos and Lbls are being used like loops that is intolerable to the calculator. Personally I don't use Gotos or Lbls, I use subroutines.

whats a subroutine? and you dont have to close your parentheses and quotes
EDIT:
For(Q,1,9999999
GarbageCollect
End
ill put some thought into making an actual program later but now im too lazy

[quote="ternary0210"]Easy:
Input screen:
NO!
NO!
NO!
NO!
NO!
NO!
NO!
Code:
Lbl 1
Disp "NO!(factorial)"
Goto 1[/quote]

A better way (the program must be called "NO"):

[code]Disp "NO!"
prgmNO[/code]

did i mistype something? because as far as im concerned, all stevenmeow's program does is see if you're patient enough to wait before you type in a number
Oh Wait, i missed a clrhome

I made this really beast program that cuts time in chemistry homework basically in HALF. Since its an epic pain to keep looking up atomic masses and then typing them in, it would work so much better if I could just have the masses stored in variables on my calculator. This would work great for single-letter elements, but then it fails for two letter ones. So, using omnicalc to type the lower case letters, I made a program, that, when turned on, will parse my calculations and replace any single uppercase letter of an uppercase letter followed by a lowercase letter by the corresponding atomic of that element.

This works much better than all the programs people at my school tried to make a program that just accept a string and then prints out the corresponding mass of one element at a time :P

oh, woohoo. my quartic solver failed so i just made a minigame
[hide]im sure this can be shortened and improved massively but w/e
delvar c
clrhome
for(a,1,4
randint(10^(A-1),10^A-1,8)-L1
for(B,1,8
output(b,1,b
output(b,2,")
output(b,3,L1(b))
End
for(B,1,250+50A
delvar d
getkey->d
if D<=94 and D>=72:Then
D-91+13(D<85)+13(D<75)->d
for(E,1,8
If L1(d)>L1(E)
goto 1
end
output(C,1,"*16 spaces*
10^(A)->L1(D)
End
End
for(E,1,8
If L1(Q) =/= 10^(A)
goto 1
end
C+1->C
end
lbl 1
disp "score:
output(1,7,c
[/hide]

Ok, if you have TI-84+ (SE or not), then you have a polynomial root finder and simontaneous equation solver.

ono my previous program had a bunch of errors. ill try again
this one is a bit lengthy(350 bytes or so)
->=sto
[hide]clrhome
disp "score:
delvar a
randint(0,2,5)->L1
randint(0,0,5)->L2
getkey
For(b,1,100
output(1,7,"*4spaces*
output(1,7,a
for(C,1,5
if L1(c)=0
output(8,3C-1,"*space*
if L1(C)=1 or L1(C)=2
output(8,3C-1,L1(C)-1
end
getkey->c
if int(c/10)=1:then
C-10->C
if L1(c)=2
A+1->A
if L1(C)=1
A-1->A
if L1(C) =/= 0
0->L2(C)
0->L1(C)
output(8,3C-1,"*space*
end
for(d,1,5
L2(D)+1->L2(D)
If L1(D)=/=0 and L2(d)=4:then
output(8,3D-1,"*space*
0->L1(d)
0->L2(D)
end
if L1(d)=0 and L2(D)=2>:then
randint(1,2)->L1(D)
0->L2(D)
end:end:end
disp "game finished
pause:clrhome[/hide]

I made a program for simulating a risk battle (basically no max on number of soldiers),
and one for calculating chances of each outcomes (max 40 soldiers on defender and offender).

I also made a program that displays a word and moves it in a direction in the screen...and bounce off edges...

My factoring program(uses a Goto) uses less memory than james4l's one, which doesn't use a Goto. The difference is only 18 bytes, though, and I can't scroll up on mine. :(

@lifeisacircle: Your shoelace algorithm is very practical! Only 140 bytes of space!

@ernie: What does your factor program exactly do? When I input 75, it displayed 1, 75, then it ended. When I put in 144, it displayed 1,144, 2,72, 3,48, and 4,36. Then it ended.

Oh, that's I think because I typoed.  :blush:

[hide="Factor Program"]Input A
0->B
Lbl C
B+1->B
If B <= sqrt A
Then
If A/B = int(A/B)
Then
Disp B,A/B
Pause
End
Goto C
End
If B > sqrt A
Then
Stop:End[/hide]

That should work.

[hide="Modular Arithmetic Program"]ClrHome
Disp "A^B MOD N"
Prompt A,B,N
If A<0 or B<0 or N<0 or  int(A) $ \neq$ A or int(B) $ \neq$ B or int(N) $ \neq$ N
Then
ClrHome
Disp "A,B, AND N MUST","BE WHOLE NUMBERS"
Pause
prgmMODARITH
Else
Lbl C
(B/N-fPart(B/N))+(fpart(B/N)+N)->B
If B>N
Then
Goto C
Else
fPart(A^B/N) * N->D
If D>0.9
Then
1->D
End
If D<0.1
Then
0->D
End
Output(5,1,"A^B MOD N=")
Output(5,11,D)[/hide]

Is there anyway to display a radical symbol in an answer? (Such as the solution to the quadratic formula but preferablly using radicals instead of decimals). I believe this is possible now, thanks to James' factoring program, but I'm too lazy to think of how to make it.

[quote="1=2"]I thought that it would be cool to make a topic about programming calculators, like the TI-84, or the TI-89(If it is possible, the TI-$ n$spire).

I currently have three programs for the TI-84. One on Binet's formula, one for the area of a regular polygon, and one for the prime factorization of a number(can anyone teach me to scroll? :oops: ), if anyone needs it.[/quote]
what's the code for the prime factorization progam?  :)

heh sry for reviving the topic  :oops:

but I just got a TI-$ n$spire and I want to know what programs I can put into it. I can program it with the TI-84+ keypad, but not with the TI-$ n$spire keypad. Does anyone know any good programs?

Thanks in advance :lol:


Do the competition sets have the countdown round or solutions? Here's the link:

[url]http://secure.sportsawardsonline.com/applications/default/store/index.asp?product_category_id=1000009[/url]

yuppers! i know because i bought 1999 meself!

My coach has those books (at least the covers look the same) and they DO have countdown rounds.


[quote="Virgil Nicula"][color=darkred][b]Let $ ABC$ be a triangle. Prove that[/b]  $ \{\begin{array}{cc}1\blacktriangleright &\boxed{\ \sum\sin\frac{A}{2}\ \le\ \sqrt{\frac{4R+r}{2R}}\ \le\ \frac{3}{2}\ }\\ \\ 2\blacktriangleright &\boxed{\ \sum\cos\frac{A}{2}\ \le\ \sqrt{\frac{3(4R+r)}{2R}}\ \le\ \frac{3\sqrt 3}{2}\ }\end{array}$[/color][/quote]

[color=darkblue][b]Consequence.[/b] $ \boxed{\ (\sum\sin\frac{A}{2})^{2}\ +\ (\sum\cos\frac{A}{2})^{2}\le\ 8\ +\ \frac{2r}{R}\ \le\ 9\ }\ \ (*)$

Apply the inequality $ (*)$ to the orthic triangle of an acute triangle $ ABC$ and obtain :

$ \{\begin{array}{cc}1\blacktriangleright &\boxed{\ (\sum\sin A)^{2}+(\sum\cos A)^{2}\le 8(1+\cos A\cos B\cos C)\ }\\ \\ 2\blacktriangleright &\boxed{\ \frac{p^{2}+r^{2}+2Rr-7R^{2}}{8R^{2}}\le\cos A\cos B\cos C\le\frac{1}{8}\ }\\ \\ 3\blacktriangleright &\boxed{\ ab+bc+ca\le 2R(4R+r)\ }\end{array}$[/color]

[size=150]Very nice, Virgil Nicula![/size]

$ 1.\blacktriangleright$ $ \tan\frac{A}{2}\cdot\sin B\plus{}\tan\frac{B}{2}\cdot\sin A\ge 4\cdot\sin\frac{A}{2}\sin\frac{B}{2}$
$ \implies\sum\tan\frac{A}{2}\cdot\left(\sin B\plus{}\sin C\right)\ge 4\cdot\sum\sin\frac{B}{2}\sin\frac{C}{2}$
On the other hand, $ \tan\frac{A}{2}\cdot\left(\sin B\plus{}\sin C\right)\equal{}\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\cdot 2\cdot\sin\frac{B\plus{}C}{2}\cos\frac{B\minus{}C}{2}\equal{}\cos B\plus{}\cos C$
It implies that $ \sum\cos A\ge 2\cdot\sum\sin\frac{B}{2}\sin\frac{C}{2}$
$ \implies\sum\cos^{2}\frac{A}{2}\ge\left(\sum\sin\frac{A}{2}\right)^{2}$
$ \implies\frac{3}{2}\plus{}\frac{1}{2}\cdot\sum\cos A\ge\left(\sum\sin\frac{A}{2}\right)^{2}$
$ \implies 2\plus{}\frac{r}{2\cdot R}\ge\left(\sum\sin\frac{A}{2}\right)^{2}$
$ \implies\sqrt{\frac{4R\plus{}r}{2R}}\ge\sum\sin\frac{A}{2}$

$ 2.\blacktriangleright$ $ \sum\cos\frac{A}{2}\le\sqrt{3\cdot\left(\sum\cos^{2}\frac{A}{2}\right)}\equal{}\sqrt{\frac{3\left(4R\plus{}r\right)}{2R}}$

Note that the last line reduces into:

$ 4Rr\plus{}r^{2}\plus{}s^{2}\le 2rR\plus{}8R^{2}\iff s^{2}\le 8R^{2}\minus{}2rR\minus{}r^{2}$

Note this bound is slightly weaker than the bound given by blundon's inequality (the weaker version without all the squareroots, etc):

$ 4R^{2}\plus{}4rR\plus{}3r^{2}\le 8R^{2}\minus{}2rR\minus{}r^{2}\iff$
$ 4R^{2}\minus{}6rR\minus{}4r^{2}\equal{}2(R\minus{}2r)(2R\plus{}r)\ge 0$


In a school,each teacher must choose $ 10$ students to join a math contest,a group student is call "acceptable" with teacher $ A$ if in this group has at least one student who is choosen by teacher $ A$.With any 6 teacher,exits a group is "acceptable" with them(all 6).Prove that we can choose $ 10$ students join the contest such that this group is "acceptable" with all teacher.

[quote="mylove"]With any 6 teacher,exits a group is "acceptable" with them(all 6).[/quote]

Isn't this always the case? (Just pick one distinct student from each teacher's group of ten... and then pick four more students however you wish.)

With any 6 teacher,always exits a group is "acceptable" with them.


function
geometry
geometric transformation
real analysis
real analysis unsolved

Here is a problem I just cannot seem to crack:

 Suppose f is continuous on [0,1] and that f(0) = f(1). Prove that for any positive integer n, there exists some x in the interval [0,1-(1/n)] such that
 f(x) = f(x+(1/n)).

 [Comment:If one assumes that f > =0 (or f<= 0) the problem is straightforward enough. For example let f >= 0. Without loss of generality assume f(0) =f(1) =0  and let g(w) = f(w) - f(w+ (1/n)) for w in [0,1-1/n].  Then g(0) = -f(1/n) < 0 and g(1-(1/n)) = f(1-(1/n))-f(1) = f(1-(1/n))>0 . So g changes sign between 0 and 1-(1/n) and so by the Intermediate Value Theorem, there is an x: g(x) = 0 . That is, there is an x such that f(x)=f(x+(1/n)). The problem seems much harder to do for the case where f changes sign at least once in [0,1]. ]

Actually, if $ f\geq 0$ in $ [0,1]$, you can't assume $ f(0)\equal{}0$ (the function might be something like $ (2x\minus{}1)^2$ which is nonnegative but drops below $ f(0)$).

If you let $ g(x)\equal{}f(x)\minus{}f\left( x\plus{}\frac{1}{n}\right)$, look at
\[ g(0),g\left( \frac{1}{n}\right),g\left( \frac{2}{n}\right),\cdots ,g\left( 1\minus{}\frac{1}{n}\right)\]
They can't all be strictly positive or strictly negative.

Thankyou for your reply.

If f is continuous on [0,1], then f - k is continuous on [0,1] where k is a constant. So if f(0) = f(1) = k then F(0) = (f-k)(0) = 0 etc.
So a translation disposes of the problem provide the resulting function >=0 or <=0. Is this reasoning faulty ??

I can see that, making no assumptions about f, that g(0)+g(1/n)+g(2/n)+ . . . +g(1-1/n) = 0 and so not all terms in this sum are of the same sign. But how does one proceed from here ? How is the proof completed  - ie., how does one now set out the argument from this point ?

Now you have a continuous function that either has the value 0 at all those points (in which case ...) or that has a positive value at some point and a negative value at some other point (in which case ...).  Fill in the blanks  :wink:

[i]Now you have a continuous function that either has the value 0 at all those points (in which case ...) or that has a positive value at some point and a negative value at some other point (in which case ...). Fill in the blanks [/i]

Thankyou !![/i]


Let two circles $ \Gamma_1$ and $ \Gamma_2$ intersect at $ A$ and $ B$. Points $ P$ and $ Q$ lie on $ \Gamma_1$ and $ \Gamma_2$ respectively such that line $ PQ$ is the common tangent of $ \Gamma_1$ and $ \Gamma_2$. Line $ AB$ and $ PQ$ intersect at $ E$. Point $ F$ is the foot of perpendicular from $ A$ to line $ PQ$. Point $ G$ is such that $ EG$ is the diameter of the circumcircle of $ EBF$. Prove that $ P,Q,B,G$ lie on a circle.

Let $ AT_1 \cap \Gamma_1\equal{} A_1$ and similarly define $ A_2$.$ T_1,T_2$ are the centers.
Let $ G' \equal{} (PQB) \cap A_1A_2$.
we must prove that $ G'A \perp PQ$.
we have $ QA \perp PG'$ because we have $ \angle PQA \plus{} \angle G'PQ \equal{} \angle ABQ \plus{} \angle QBA_2 \equal{} \angle ABA_2 \equal{} 90^\circ$.
similarly we have the perpendicularity $ PA \perp G'Q$.
so $ A$ is the orthocenter of  $ PQG'$ and we have $ G'A \perp PQ$.

nice proof shoki. :D I made this problem by inversion from an IMO problem.


If $ g(x) \equal{} x^2$ and $ f(x) \equal{} 2x \minus{} 1$, what is the value of $ f(g(2))$?

you have g(x)=x^2, f(x)=2x-1 and f(g(2)). Plug in 2 for the function.

g(2)=2^2=4

Now you know g(2) is 4. so, f(g(2))=f(4). Now plug in 4 for the other function.

f(4)=2(4)-1=[b]7[/b]


Let $ x\in\mathbb{R}$. Find a closed formula for the infinite sum $ \sum_{i\equal{}0}^{\infty}x^{2^{\minus{}i}}\minus{}1$


topology
calculus
geometry
set theory
Galois Theory
superior algebra
superior algebra unsolved

[quote="Interval"]I am not going to ask a question in this category for it is too advanced for me.  However, I am curious about what ACTUALLY is the toughest math course in school.

Some say topology, others set theory and still others say differential calculus plus differential geometry.

Honestly, what is the toughest math course out there?[/quote]

Since a mathematician is only interested in the things he does not understand, anything in math is hard. But Algebra & Topology is fun yeah

I'd say everyone is naturally good at different things. Some people naturally get algebra, but think analysis is really hard. Others vice versa. I just took a class in Galois theory which is famed to be quite challenging, but I'd take a test in that class anyday over a test on probability/combinatorics which some think is below college level. So I'd say it all depends on what areas you take a liking to.


My daughter just asked me why the orthocenter of a triangle is usually called $H$.  I confessed that I didn't know why.  Does anyone know?

H - the point of intersection of [b]H[/b]eights?

I always assumed it was the next letter after G, the centroid.
I guess I though a,b,c are vertices, d,e,f are medial triangle points (?), g centroid, h orthocenter, i incenter, etc. now O would be circumcenter because it is a commonly used circle (?) and the circumcenter is also like the origin of the triangle in terms of vectors usually...

just some crazy thoughts of mine

Don't know if this is anything helpfull, but in german the orthocenter is called "[b]H[/b]öhenschnittpunkt", meaning that this point is the intersection of the altitudes (german: "Höhen").

Indeed... H stands for Horthocentre :D. At least that's what I like to think.

The German thing may be right, but what about this...

Once you've labeled the vertices a, b, c, and the endpoints of the cevians d, e, f, the leftovers are g (centroid), h (orthocenter), i (incenter). The circumcenter, which is generally more useful and intuitive (and probably well-known long before the other three), gets the default name for a circle's center.

Just a guess, anyway.

The explanation with [b]h[/b]eight seems to be definitely most convincing for me.

the point of intersection of Heights=Höhenschnittpunkt?

Makes it more convincing, though the other argument (abcd...gh...) is good too


limit
logarithms
floor function
integration
real analysis
real analysis unsolved

Let $ a_n $ be the maximum integer such that : $ 5^{a_n} < 1^1.2^2.3^3...n^n $
 Find $ \lim_{n\to \infty}\frac{a_n}{n^2} $ ?

$\lim_{n\to\infty}\frac{a_n}{n^2}=\infty.$

More precisely, $\lim_{n\to\infty}\frac{a_n}{n^2\ln n}=\frac1{2\ln5}.$

Are you sure , Kent ? My teacher said the answer is $ \lim_{n\to \infty}\frac{a_n}{n^2}=\frac{1}{8} $
  Can you write down your solution ?

$a_n = \lfloor \ln_5 2^2 \cdot 3^3 \cdot \ldots \cdot n^n \rfloor = \lfloor \sum^n_{j = 2} j \cdot \ln_5 j \rfloor \geqslant \lfloor \sum^n_{j = 2} j\cdot \ln_5 2 \rfloor \geqslant 2 \sum_{j = 2}^n j = n^2 + n - 2$, so $\lim_{n\rightarrow \infty} \frac{a_n}{n^2} \geqslant \lim_{n \rightarrow \infty} \frac{n^2 + n - 2}{n^2} = 1$. Thus youre teacher is wrong.

In fact $a_n = \frac{1}{\ln 5} \sum_{j = 1}^n j \cdot \ln_{} j + O ( 1 )$, and $\sum_{j = 1}^n j \ln j = \int_1^n x \ln x %Error. "tmop" is a bad command.
{dx} + O ( n \ln n ) =
\frac{n^2 \ln n}{2} - \frac{n^2}{4} + O ( n \ln n )$. Thus $a_n = \frac{n^2 \ln n}{2 \ln_{} 5} - \frac{n^2}{4 \ln 5} + O ( n \ln n )$ and $\lim_{n
\rightarrow \infty} \frac{a_n}{n^2} = + \infty$. Also from this asymptotic result we get, just as pointed Kent, $\lim_{n \rightarrow \infty} \frac{a_n}{n^2 \ln n} = \frac{1}{2 \ln 5}$.


Let $S=(x_1, x_2, ..... , x_n)$ be a set of reals which are at least 1. We count the number $f(S,I)$ of the sums $\sum {a_i.x_i}$, where $a_i$ is either 0 or 1, that lies on an interval $I$ with length 1. Find the maximum of $f(S,I)$ for any interval $I$ and any set $S$.

I need a precisation:
Is the interval closed? Also can the numbers be $1$?
If both questions are affirmative, my answer is smth like $C_n^{[\frac {n-1}2]}+C_n^{[\frac n 2]}$.
Otherwise it's $C_n^{[\frac n2]}$

As far as I remember, the right statement corresponds to your second answer. It easily follows from Sperner's theorem on antichain.

Pierre;

The interval here is closed such as $(a, a+1)$.
  Pbornztein, would you mind giving me a clear solution to this problem which I 'm very interested in. Thank you.

I don't quite understand: I always thought that a closed interval is $[a;a+1]$ (so $a,a+1 \in I$)

I checked my archives and the right statement deals with an open interval of length 1.
The desired maximum is therefore $C_n^{[ \frac n 2 ]}$ which follows by Sperner's theorem.
Indeed, for each $A \subset S$, let $s(A)$ be the sum of the elements of $A$ (with $s( \emptyset ) = 0$).

Let $I$ be an open interval with length 1.

Lemma.
If $A,B$ are two distinct subsets from $S$ such that $A \subset B$ then at least one of the numbers $s(A)$ and $s(B)$ is not in $I$.

Proof of the lemma.
Let $x \in B \setminus A$. Then $x \geq 1$ and $s(B) \geq s(A) + x \geq s(A) + 1$. Thus, $s(A)$ and $s(B)$ cannot both belong to a same open interval with length 1.

Now, since each number of the form $\sum_{i=1}^n a_ix_i$ is of the form $s(A)$ for some subset $A$ from $S$, we deduce that the maximum number of such numbers which belong simultanously in an open interval of length 1 is not greater than the maximum length of an antichain formed by subsets from $S$. According to Sperner's theorem, this proves that the desired maximum is not greater than $C_n^{[ \frac n 2 ]}$.

Now, it is straightforward to verify that if $x_i = 1 + \frac 1 {2^{p+i}}$ where $[ \frac n 2 ] \leq p $ and $I = ] [ \frac n 2 ]  , 1 + [ \frac n 2 ]  [$ then, for each subset $A$ from $S$ with $|A| = [ \frac n 2 ] $ then $s(A) \in I$.

Pierre.

This result has been generalised by Kleitman to higher dimensions :
Let $u_1, \cdots, u_n$ be vectors in $\mathbb {R}^d$ with $||u_i|| \geq 1$ for all $i$.
Among the $2^n$ linear combinations of the form $\sum_{i=1}^n a_iu_i$, where $a_i \in \{-1;1 \}$, the maximum number which lie in a given open ball with radius 1 is $C_n^{[ \frac n 2 ]}.$

References :
[1] D. Kleitman, 'On a lemma of Littlewood and Offord on the distribution of linear combinations of vectors', Advances Math., 5 (1970), p.155-157.
[2] M. Aigner, G.M. Ziegler, 'Proofs from The Book', Springer.

Pierre.


geometry
topology
advanced fields
advanced fields unsolved

Let L/K be a field extension, and V a K-variety in L^n.
A point g on V is called a "generic point of V over K" if the Zarisky closure of {g} is V.

Prove the following:

-If L is algebraically closed and the trascendence degree of L over K (that is, the size of a maximal subset of L which is algebraically independent over K) is at least n, then every V has generic points.

-If V has generic points, then it is irreducible (that is, it's not the union of two other proper subvarieties).

-The following are equivalent:
i) g is a generic point of V over K.
ii) Ideal_K({g}) = Ideal_K(V).
iii) K[V] is isomorphical to K[g_1,... g_n]  \leq L, where g = (g_1,... g_n).
iv) for all a \in V, there is a homeomorphism h:K[g_1,... g_n] -> K[a_1,... a_n] such that h(g_i) = a_i for all i.

-If g and g' are generic points of V over K, there is an isomorphism h:K[g_1,... g_n] -> K[g_1',... g_n'] such that h(g_i) = g_i' for all i.

-Let L' be an extension of L, V a K-variety in L^n, CV its Zarisky closure in (L')^n. If g \in V is a generic point of CV, it is of V also.

-Let g be a generic point of V over K, and d:=Trascendence degree_K({g_1,...,g_n}). Then for all a \in V such that Trascendence degree_K({a_1,...,a_n}) = d, a is a generic point of V.

Yep, I know it is long. Can't help it, sorry. And please forgive any grammar or spelling errors (if any); after all, English is not my mother tongue.

Thank you guys. Take care and good luck.

[quote="Fibonacci"]Let $L/K$ be a field extension, and $V$ a $K$-variety in $L^n$.
A point $g$ on $V$ is called a "generic point of $V$ over $K$" if the Zarisky closure of $\{g\}$ is $V$.

Prove the following:

\begin{enumerate}
\item If $L$ is algebraically closed and the trascendence degree of $L$ over $K$ (that is, the size of a maximal subset of $L$ which is algebraically independent over $K$) is at least $n$, then every $V$ has generic points.

\item If $V$ has generic points, then it is irreducible (that is, it's not the union of two other proper subvarieties).

\item The following are equivalent:
\begin{enumerate}\item $g$ is a generic point of $V$ over $K$.
\item $\textrm{Ideal}_K(\{g\}) = \textrm{Ideal}_K(V)$.
\item  $K[V]$ is isomorphic to $K[g_1,... g_n]  \subset L$, where $g = (g_1,... g_n)$.
\item  for all $a \in V$, there is a homeomorphism $h:K[g_1,... g_n] \to K[a_1,... a_n]$, such that $h(g_i) = a_i$ for all $i$.
\end{enumerate}

\item If $g$ and $g'$ are generic points of $V$ over $K$, there is an isomorphism $h:K[g_1,... g_n] \to K[g_1',... g_n']$, such that $h(g_i) = g_i'$ for all $i$.

\item Let $L'$ be an extension of $L$, $V$ a $K$-variety in $L^n$, $\overline V$ its Zarisky closure in $(L')^n$. If $g \in V$ is a generic point of $\overline V$, it is also one of $V$.

\item Let $g$ be a generic point of $V$ over $K$, and $d=\textrm{Trascendence degree}_K(\{g_1,...,g_n\})$. Then for all $a \in V$ such that $\textrm{Trascendence degree}_K(\{a_1,...,a_n\}) = d$, $a$ is a generic point of $V$.
[/quote]

For questions of this sort, the best references are elementary algebraic geometry books. Start with Harris, Joe. Algebraic Geometry, a first course, Springer. (http://www.ajorza.org/papers/harris/)

I will suggest the answer to your second question: If $V$ is reducible, there exist nonempty disjoint open and closed $A,B$ such that $V=A\cup B$. If $g\in A$, then $\overline{\{g\}}\subset\overline A=A$, so $A=V$ which is a contradiction.


This was question #18 on the 2005 AMC 12A. I know the answer is 100 but when I try the problem I keep on getting 101  :wallbash_red:. Anyone have a good solution?

    Call a number "prime-looking" if it is composite but not divisible by $2, 3,$ or $5$. The three smallest prime-looking numbers are $49, 77,$ and $91$. There are $168$ prime numbers less than $1000$. How many prime-looking numbers are there less than $1000$?

You can look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367462#p367462]here[/url].

Did you forget that 1 is neither prime nor composite?

yeah I forgot that 1 isn't composite :oops: .

[quote="joml88"][quote="K81o7"]Use a complement argument.
# of prime looking numbers=1000-(prime numbers)-(multiples of 2, 3, or 5 except 2, 3 and 5)-1, because 1 is not prime nor composite, but isn't prime looking.

Hence, our answer is $1000-168-(-3+500+333+200-166-100-66+16)-1$ which doesn't seems to come out correctly, but that's the jist of how I would do it...[/quote]

I think you messed up on how many numbers <1000 are divisible by 2, 3, and 5 when you counted.  It should be 33, not 16.[/quote]
 :what?: Oh yeah...haha...that's what I would mess up on the AMC...thinking [1000/30]=16 :cursing:  :fool:  :wallbash_red:


A robot is programmed to do the following moves in a coordinate system: start at the origin facing in the positive $ y$ direction and move $ 1$ unit. For each subsequent move, it is to turn $ 90^\circ$ clockwise and go forward $ 1$ unit farther than it went in the previous move. What is the sum of the coordinates of the robot's position after it just moved $ 14$ units in one direction?

This is just the addition of the following:

1-2-3+4+5-6-7+8+9-10-11+12+13-14
Group them in pairs:
-1+1-1+1-1+1-1
Group them in pairs again:
-1. Solved.

The answer is 15.

After the first 2 moves, turning points for the robot lie on either the function $y=-|x|$ or $y=|x-1|$ on the coordinate plane.

On its second turn, the robot is on $(2,1)$. On its sixth turn, it is on $(4,3)$. These two points form one diagonal of a 2x2 square, and after mapping out the other turning points, diagonals of a 2x2 square can be seen between any two consecutive turning points in the same quadrant. The difference between any two consecutive turning points in the same quadrant is also 4 (as in the first quadrant will have the turning point #'s 2, 6, 10, 14, 18, 22, 26, 30,... quadrant 2 will have #'s 5, 9, 13, 17, 21,...differences of 4)

After the walk of 14 units, the robot ends up in the 1st quadrant of the coordinate plane.

Thus, on its tenth turn, the robot must lie on the point $(6,5)$, and on its 14th turn, it will lie on $(8,7)$
$8+7=15$

james4l, you're way does work, and it's much quicker than mine... (Fortunately)

The sequence would be 1+2-3-4+5+6-7-8+9+10-11-12+13+14.

Grouping the 1st and 3rd, 2nd and 4th, 5th and 7th, 6th and 8th, 9th and 11th, and 10th and 12th, you get a bunch of -2's, 6 of them to be exact. The last two numbers add up to 27.

$27+6(-2)=15$ as desired.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

[i find it interesting, not sure if it's already posted]

Let $ f : [0, 2] \to R$ be a function 3 times derivable, with $x_0 = 1$ point of relative extremum and $f'''(x) \geq 0, \forall x \in [0, 2]$.
Prove $\int_1^2 f(x) dx \geq \int_0^1 f(x) dx$.

$f(x) = f(1) + f'(1)(x - 1) + \frac{1}{2}f''(\theta (x)){\left( {x - 1} \right)^2}$
So, we prove that $\begin{array}{l}
\int\limits_1^2 {f''(\theta (x)){{\left( {x - 1} \right)}^2}} dx \ge \int\limits_0^1 {f''(\theta (x)){{\left( {x - 1} \right)}^2}} dx\\
 \Leftrightarrow f''({c_1})\int\limits_1^2 {{{\left( {x - 1} \right)}^2}} dx \ge f''({c_2})\int\limits_0^1 {{{\left( {x - 1} \right)}^2}} dx
\end{array}$

it 's the result


When we type small letters in some Commands like
$\text{mathcal}$
 the letters will change to some NEW symbols.
For example if you type $\{$ abcdefghijklmnopqrstuvwxyz $\}$
in mathcal we will have \[\mathcal{abcdefghijklmnopqrstuvwxyx }\] Does any one know  another command like this , that can Create symbols?

It not create new symbols, this is just because mathcal command is not defined for lowercase characters.
And if you want symbols "already created" :D  see [url=http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]this[/url].

$\mathcal{VERY,,,USE,,,FULL,,,ARTICLE }$

I tried to type lots of them but i get the usual POTENTIOL error .

For istance i typed the command [code] \tent [/code] page 59 in your article but it didnt work.

Anyway , Thanks.

\tent requires the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ifsym]ifsym[/url] package, which provides extra symbol fonts, to be loaded which you can do on your own computer but not here.

Under every table of symbol's command, you will find the package required for this table. So you must put tis in the preamble:[code]\usepackage{packagename}[/code]


algebra
polynomial
superior algebra
superior algebra unsolved

what is the min. polynomial of 3^(1/4)+3^(1/2) over Q(sqrt(3))?

It is fairly easy to make a good first guess.  If $ \alpha \equal{} \sqrt[4]{3}\plus{}\sqrt{3}$, then $ (\alpha\minus{}\sqrt{3})^2\equal{}\sqrt{3}$.  So we guess that $ (x\minus{}\sqrt{3})^2\minus{}\sqrt{3}$ is our minimal polynomial.  The only thing you have to do is show that this polynomial doesn't have any roots in $ \mathbb{Q}(\sqrt{3})$.

Equivalently, show that $ \sqrt[4]{3}$ is not in $ \mathbb{Q}(\sqrt{3})$.  Do you know how to do that?


Is set { p/q | where p, q are prime } dense in $ [0, \infty )$ ?

Yes, you can easily reduce it to this generalisation of Bertrand's theorem: for any $ c>1$ there exist an $ N$ such that there is a prime in $ [n,cn]$ for any $ n>N$.

I am afraid that i does not follow this way. But if youe have simple proof just show it :)

It suffices to show: for any $ s > 0$ and any $ \epsilon >0$, there exists primes $ p,q$ such that $ s \leq \frac pq \leq s\plus{}\epsilon$.

Take $ c\equal{} 1 \plus{} \frac{\epsilon}{s}>1$ and then $ N$ as in the cited lemma above (that lemma is a consequence of the prime number theorem). Take a prime $ q$ such that $ qs>N$ and a prime $ p \in [sq,csq]$ which exists by the lemma. Then $ \frac pq \geq \frac{sq}{q} \equal{} s$ and $ \frac pq \leq \frac{csq}{q} \equal{} cs \equal{} s \plus{} \epsilon$, as claimed.

Thank you very much! I have wrongly understood that theorem - just did not noticed "for any c>1"  :blush:.
Could you please give some references to that version of Bertrands theorem?

To be honest - I do not see how that version of Bertrand's theorem follows from prime number theorem. Could you (or someone) please provide more info about it ?

See my post in http://www.mathlinks.ro/viewtopic.php?t=180277 .


group theory
abstract algebra
superior algebra
superior algebra solved

If p is a prime.  How can it be shown that a group with order 2p must have a normal subgroup of order p?

It has an element $x$ of order $p$. This element generates a subgroup of index $2$ and therefore is normal.


During a certain election campaign, $p$ different kinds of promises are made by the different political parties ($p>0$). While several political parties may make the same promise, any two parties have at least one promise in common; no two parties have exactly the same set of promises. Prove that there are no more than $2^{p-1}$ parties.

Let $A$ be some subset of promises.
Considering $(A,A^C)$, you cannot pick both of them because that means that those parties doesn't share a promise.
Since there are $2^{p-1}$ of those, there are at most $2^{p-1}$ parties.
The inequality is tight, by choosing one promise, have all parties share it, and then taking every subset of the other $p-1$ promises.


8 regular, identical, six-sided dice are arranged in a 2x2x2 cube.Two touching dice are said to "match" if the side they share has the same number on both dice. If the 8 dice are arranged randomly, what is the probability that all 12 pairs of touching dice "match." Note: it is known that on regular dice the opposite sides always add up to seven.

Here's a hint. The answer can be expressed in the form [(1/2^a]*[(1/3^b]. Solve for a and b.


There is an island with a certain number of inhabitants, 10 of whom have a red face. For some reasons, a red face problem is a topic that is never-to-talk on this island. No mirrors or anything on the island, and nobody is allowed to say to the person who has a red face that the person has a red face. And law of island says that if a person learns that he has a red face, he have to commit a suicide at 11:00pm that day. Since the topic is never-to-talk, there are no ways for red faced people to realize that he has a red face, so everyone lives happily until an alien visits the island. Right before leaving the island, an alien tells everyone: “There are some people on this island who have a red face! You should all start thinking about it now.” Ten days later (that is, on the the tenth 11:00pm after this declaration), all ten red-faced inhabitants commit suicide.

Question: How did they realize that they had a red face?
What did an alien say to the islanders that they didn’t know already?
What was an alien's contribution?

Anyone have an answer?

There have been DOZENS of questions already, almost exactly like this one:

One example: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117110

EDIT: Another. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89127

I kind of get the main part of the question, but I still don't get

What did an alien say to the islanders that they didn’t know already?

What was an alien's contribution?

parts... Can you help me with this? I mean they all knew that there are at least 9 people before alien came..

Your question

[quote]What was the alien's contribution?[/quote]

is very strange. Isn't the alien's contribution what he told the red-faced people, and therefore the same as what he told them that they didn't know?

Anyways, in case you don't have the main question completely figured out, here's the answer:

[quote] At the beginning, they all know that there is at least 1 zero.

If 1 midnight passes without anyone leaving, everyone will know that there are at least 2 zeroes because if there were 1 zero, that person would know his number.
If 2 midnight pass without anyone leaving, everyone will know that there are at least 3 zeroes because if there were 2 zero, those people would know their numbers.

AND SO ON UNTIL

If 99 midnight pass without anyone leaving, everyone will know that there are at least 100 zeroes because if there were 99 zero, those person would know thier numbers.

No one leaves before the 99th night because the information is not enough.
At the 100th midnight, the people with 0 will know their number because they know the information that there are at least 100 zeroes, and they see only 99 zeroes.
[/quote]

Except that was for the problem with 100 people, so you just shift the numbers a bit. Plus, they commit suicide instead of "leaving", as that answer says. Plus, replace "know they have a 0" with "know they have a red face"

Anyways,
As for what he told them... this is pretty complicated to think about, I'll try to come back to answer later.


trigonometry
calculus
integration
number theory
relatively prime
complex analysis
complex analysis unsolved

Let $ k,n \in \mathbb{N}$, with $ n\geq 1$. Prove that $ z\equal{}\cos{\frac{2k\pi}{n}}\plus{}i\sin\frac{2k\pi}{n}$ is a primitive $ n$th root of unity if and only if $ k$ is relatively prime to $ n$.

Let $ (k,n)\equal{}d$, and write $ k\equal{}da, n\equal{}db$. then $ 1\equal{}(\text{cis} \frac{2k\pi}{n})^r \equal{} \text{cis} \frac{2ar\pi}{b}\Leftrightarrow \frac{2ar\pi}{b}$ is an integral multiple of $ 2 \pi \Leftrightarrow \frac{ar}{b} \in \mathbb{Z} \Leftrightarrow b|r$. so the smallest possible value of $ r$ is $ b$.

If $ d\equal{}1$, then $ b\equal{}n$, so the smallest possible value of $ r$ is $ n$, i.e. $ \text{cis} \frac{2k\pi}{n}$ is a primitive $ n$-th root.

If $ d>1$, $ b<n$, so the smallest possible value of $ r$ is $ <n$, i.e. $ \text{cis} \frac{2k\pi}{n}$ is not a primitive $ n$-th root.


geometry
area of a triangle
Heron\u0027s formula

Three circles of radii 20, 36 and 70 are tangent to each other such that the center of each circle is outside the two other circles. Find the number of square units in the area of the triangle whose vertices are the centers of the three circles?

[quote="darktreb"]Hero's Formula cleans this up nicely. All the primes in the radical match up to form squares.

(By the way the sides of the triangle are 56 106 and 90)[/quote]

Do $56^2+90^2$ and take the square root. what can be said about this triangle?

[hide="took me closer to twenty seconds"]
28-45-53 triangle x2

28 x 90 = (27 + 1) x 9 x 10 = (243 + 9) x 10 = [b]2520[/b]

Hardest part was realizing it was a 28-45-53...

If you added them, it wouldn't take much longer, as 28 squared should be naturally 784, and 45 squared equals 2025 (4 times 5, add a 25), which adds up to 2809, which is 53 squared (50-something squared equals 25 plus that something, with that something squared tacked on the end).
[/hide]

With a calculator, it should take about 10 seconds to get it using Heron's Formula, but probably about the same as you took if you're working Heron's out on paper. ;)


Let $ x,y,z$ be positive real numbers such that $ xyz\equal{}1$. Prove that \[ \frac{x}{y^{2}\plus{}2}\plus{}\frac{y}{z^{2}\plus{}2}\plus{}\frac{z}{x^{2}\plus{}2}\ge 1\]

first inequality for me in a long time : )
[hide="solution"]
after multyplying its equivalent to proving
$ \sum_{cyc}x^{3}z^{2}\plus{}2\sum_{cyc}x^{3}\plus{}2\sum_{cyc}xz^{2}\plus{}4\sum_{cyc}x\geq 9\plus{}2\sum_{cyc}x^{2}z^{2}\plus{}4\sum_{cyc}x^{2}$
since we have $ x^{3}\plus{}x\geq 2x$ and $ x^{3}z^{2}\plus{}xz^{2}\geq 2x^{2}z^{2}$ we have that
$ \sum_{cyc}x^{3}z^{2}\plus{}\sum_{cyc}xz^{2}\geq 2\sum_{cyc}x^{2}z^{2}$
$ 2\sum_{cyc}x^{3}\plus{}2\sum_{cyc}x\geq 4\sum_{cyc}x^{2}$ and we know that
$ \sum_{cyc}xz^{2}\plus{}2\sum_{cyc}x\geq 9(xyz)^{\frac{4}{9}}\equal{} 9$
suming these three inequalities we have the desired one.
[/hide]

Thank you. Nice solution. :) If anyone has a solution without expanding, please be sure to post it. :wink:

The following inequalities hold for $ xyz \equal{} 1$ too:

$ 1.$
\[ \frac{x}{2y^{2}\plus{}1}\plus{}\frac{y}{2z^{2}\plus{}1}\plus{}\frac{z}{2x^{2}\plus{}1}\ge 1\]
$ 2.$
\[ \frac{2x}{y^{2}\plus{}5}\plus{}\frac{2y}{z^{2}\plus{}5}\plus{}\frac{2z}{x^{2}\plus{}5}\ge 1\]
Have fun! :wink:

I think for each of these the substitution $ x\equal{}\frac{a}{b}, y\equal{}\frac{c}{a}, z\equal{}\frac{b}{c}$ is really powerful because you get $ a^{3}$ on the top of fractions, but for now I cannot find any nicer way than to mass expand.
So here is my solution to 2:
After the substitution, the inequality becomes:
$ \sum_{cyc}\frac{a^{3}}{2c^{2}b\plus{}a^{2}b}\geq 1$
Expand to get:
$ \sum_{cyc}a^{3}(2b^{2}a\plus{}c^{2}a)(2a^{2}c\plus{}b^{2}c)\geq (2c^{2}b\plus{}a^{2}b)(2b^{2}a\plus{}c^{2}a)(2a^{2}c\plus{}b^{2}c)$
Simplify to get:
$ 4\sum a^{6}b^{2}c\plus{}2\sum a^{6}c^{3}\plus{}2\sum a^{4}b^{4}c\plus{}\sum a^{4}c^{3}b^{2}\geq  4\sum a^{5}b^{3}c\plus{}2\sum a^{5}c^{3}b\plus{}9a^{3}b^{3}c^{3}$
where $ \sum$ means cyclic sum.
Then we get:
$ LS\equal{}\sum 2(a^{6}b^{2}c\plus{}b^{4}a^{4}c)\plus{}\sum (a^{6}c^{3}\plus{}a^{4}c^{3}b^{2})\plus{}\sum (2a^{6}b^{2}c\plus{}a^{4}c^{3}b^{2})\geq 4\sum a^{5}b^{3}c\plus{}2\sum a^{5}c^{3}b\plus{}9abc$ by AM-GM and the result follows.

[quote="rem"]I think for each of these the substitution $ x \equal{}\frac{a}{b}, y \equal{}\frac{c}{a}, z \equal{}\frac{b}{c}$ is really powerful because you get $ a^{3}$ on the top of fractions, but for now I cannot find any nicer way than to mass expand.
So here is my solution to 2:
After the substitution, the inequality becomes:
$ \sum_{cyc}\frac{a^{3}}{2c^{2}b\plus{}a^{2}b}\geq 1$
[/quote]

Without expanding we have $ \sum\frac{a^{4}}{2c^{2}ba\plus{}a^{3}b}\geq\frac{\left(\sum a^{2}\right)^{2}}{2\sum c^{2}ba\plus{}\sum a^{3}b }\geq 1$
$ \Leftrightarrow$
$ \sum a^{4}\plus{}2\sum c^{2}ba\geq 2\sum c^{2}ba\plus{}\sum a^{3}b$
$ \Leftrightarrow$
$ \sum a^{4}\geq\sum a^{3}b$,which is just rearrangements :D.

Another pretty nice way without expanding is to use Cauchy-Shwartz. Continuing from the substitution from my first post we have:
$ (\sum_{cyc}\frac{a^{3}}{2c^{2}b\plus{}a^{2}b})(\sum 2c^{2}b\plus{}a^{2}b)\geq (\sum a^{\frac{3}{2}})^{2}$
So in order to prove the original inequality we need to prove:
$ \frac{(\sum a^{\frac{3}{2}})^{2}}{\sum 2c^{2}b\plus{}a^{2}b}\geq 1$
$ \Leftrightarrow (\sum a^{\frac{3}{2}})^{2}\geq\sum 2c^{2}b\plus{}a^{2}b$
$ \Leftrightarrow\sum a^{3}\plus{}\sum 2\sqrt{a^{3}b^{3}}\geq\sum 2c^{2}b\plus{}a^{2}b$   (*)
But we have for (*):
$ LS\equal{}\sum(\frac{2}{3}c^{3}\plus{}\frac{2}{3}\sqrt{c^{3}b^{3}}\plus{}\frac{2}{3}\sqrt{c^{3}b^{3}})\plus{}(\frac{1}{3}a^{3}\plus{}\frac{1}{3}\sqrt{a^{3}b^{3}}\plus{}\frac{1}{3}\sqrt{a^{3}b^{3}})\geq\sum2c^{2}b\plus{}a^{2}b$
by AM-GM so (*) is true, and inequality 2 follows.

The same technique can be applied for number 3. After the substitution the inequality becomes:
$ \sum\frac{2a^{3}}{c^{2}b\plus{}5a^{2}b}\geq 1$
We again apply Cauchy:
$ (\sum\frac{2a^{3}}{c^{2}b\plus{}5a^{2}b})(\sum c^{2}b\plus{}5a^{2}b)\geq (\sum\sqrt{2}a^{\frac{3}{2}})^{2}$
So in order to prove the given inequality we need to prove:
$ (\sum\sqrt{2}a^{\frac{3}{2}})^{2}\geq\sum c^{2}b\plus{}5a^{2}b$
$ \Leftrightarrow\sum 2a^{3}\plus{}\sum 4\sqrt{a^{3}b^{3}}\geq\sum c^{2}b\plus{}5a^{2}b$ (**)
But we have for (**):
$ LS\equal{}\sum(\frac{1}{3}c^{3}\plus{}\frac{1}{3}\sqrt{c^{3}b^{3}}\plus{}\frac{1}{3}\sqrt{c^{3}b^{3}})\plus{}(\frac{5}{3}a^{3}\plus{}\frac{5}{3}\sqrt{a^{3}b^{3}}\plus{}\frac{5}{3}\sqrt{a^{3}b^{3}})\geq\sum5a^{2}b\plus{}c^{2}b$
by AM-GM, therefore (**) is true, and inequality 3 follows.

Very nice solutions guys, and my solution was sort of like them too. :wink:  Anyway I was just going to post the following inequality:

Let $ a,b,c>0$. Show that \[ \frac{a^{3}}{(8a^{2}\plus{}b^{2})c}\plus{}\frac{b^{3}}{(8b^{2}\plus{}c^{2})a}\plus{}\frac{c^{3}}{(8c^{2}\plus{}a^{2})b}\ge\frac{1}{3}\]
But your methods really kill it! :(  :D Still if someone has other solutions please be kind to post it. :)


calculus
number theory proposed
number theory

Let the divisors of $ n$ be $ 1\equal{}d_1<d_2<....<d_k\equal{}n$.Determine all positive integers $ n$ such that $ n\plus{}1\equal{}d_6^2\plus{}d_7^2$

[quote="SUPERMAN2"]Let the divisors of $ n$ be $ 1 \equal{} d_1 < d_2 < .... < d_k \equal{} n$.Determine all positive integers $ n$ such that $ n \plus{} 1 \equal{} d_6^2 \plus{} d_7^2$[/quote]

Rather difficult.  For the moment, the only things I found are : $ d_6$ and $ d_7$ are coprime and $ n\equal{}pd_6d_7$ with some prime $ 2\leq p\leq d_6$

If $ p\equal{}2$, a rather ugly calculus gave me only two solutions :

$ n\equal{}144$ with $ d_1\equal{}1$, $ d_2\equal{}2$, $ d_3\equal{}3$, $ d_4\equal{}4$, $ d_5\equal{}6$, $ d_6\equal{}8$, $ d_7\equal{}9$ and $ 8^2\plus{}9^2\equal{}144\plus{}1$

$ n\equal{}1984$ with $ d_1\equal{}1$, $ d_2\equal{}2$, $ d_3\equal{}4$, $ d_4\equal{}8$, $ d_5\equal{}16$, $ d_6\equal{}31$, $ d_7\equal{}32$ and $ 31^2\plus{}32^2\equal{}1984\plus{}1$

But it remains the case $ p>2$  :(


[quote="Yimin Ge"]Let $M(n)=\{-1,-2,\ldots,-n\}$. For every non-empty subset $T$ of $M(n)$, let $P(T)$ be the product of all elements of $T$. Find the sum of all $P(T)$.[/quote]

If $h(n)$ is the requested sum, it's easy to see that $h(n)$ is the sum with all subsets without $-n$ (hence $h(n-1)$), plus the sum of all subsets containing $-n$ + something else (hence $-nh(n-1)$), plus the sum of subset $\{-n\}$). $\Rightarrow $ $h(n)=(1-n)h(n-1)-n$

And , since $h(1)=-1$, $h(n)=-1$ $\forall n\in\mathbb{N}$

--
Patrick

for any subset T of M(n) which does not contain -1, there is a unique T' which contain all the elements of T and -1. Thus the sum of every pair of T and T' (there are 2^(n-1)-1 pairs), giving the sum of P(T) equal to -1.


Αν η $ f$ είναι δυο φορές παραγωγίσιμη στο $ R$ και υπάρχει $ c$ πραγματικός : $ \frac{f(x)\minus{}f(y)}{x\minus{}y}\ne f'(c)$ για κάθε $ x \ne y$ τότε να δειχτεί ότι $ f''(c)\equal{}0$

[quote="r_boris"]Αν η $ f$ είναι δυο φορές παραγωγίσιμη στο $ R$ και υπάρχει $ c$ πραγματικός : $ \frac {f(x) \minus{} f(y)}{x \minus{} y}\ne f'(c)$ για κάθε $ x \ne y$ τότε να δειχτεί ότι $ f''(c) \equal{} 0$[/quote]

[b]ΛΥΣΗ[/b]
Έστω ότι $ f''(c)\neq0$. Ας υποθέσουμε χωρίς βλάβη της γενικότητας ότι $ f''(c)>0$. Θα αποδείξουμε ότι υπάρχει $ \epsilon >0$ τέτοιο ώστε $ f''(x)>0, \forall x\in [c\minus{}\epsilon,c\plus{}\epsilon]$.

[i]Απόδειξη:[/i] Έστω ότι αυτό δεν ισχύει, οπότε για οσοδήπτε μικρό $ \epsilon>0$ υπάρχει $ \xi\in [c\minus{}\epsilon, c\plus{}\epsilon]$ τέτοιο ώστε $ f''(\xi)\leq 0$. Τότε για $ \epsilon\to 0$, λαμβάνουμε $ f''(c)\leq0$, που είναι άτοπο από την αρχική μας υπόθεση. Άρα πράγματι η $ f$ είναι κυρτή σε μια περιοχή του $ c$. $ \square$

Τότε ως γνωστόν,η γραφική παράσταση της $ f$ θα βρίσκεται πάνω από την εφαπτομένη της στο $ c$ στο διάστημα που αυτή είναι κυρτή, δηλαδή $ f(x)\geq g(x)\equal{}f'(c)(x\minus{}c)\plus{}f(c)$ για κάθε $ x\in [c\minus{}\epsilon, c\plus{}\epsilon]$. Θεωρούμε τώρα μια μετατόπιση της $ g(x)$ προς τα πάνω και συγκεκριμένα την ευθεία $ h(x)\equal{}g(x)\plus{}\delta$ με $ 0<\delta<min\{f(c\minus{}\epsilon)\minus{}g(c\minus{}\epsilon),f(c\plus{}\epsilon)\minus{}g(c\plus{}\epsilon)\}$ η οποία τέμνει τη γραφική παράσταση της $ f$ σε δύο σημεία $ (a,f(a))$ και $ (b,f(b))$ που βρίσκονται εκτέρωθεν του $ c$ και ως εκ τούτου είναι διαφορετικά μεταξύ τους. Ο συντελεστής διεύθυνσης της $ h$ είναι ίσος με $ f'(c)$, επομένως $ \lambda_h\equal{}\frac{f(b)\minus{}f(a)}{b\minus{}a}\equal{}f'(c)$. Η τελευταία σχέση όμως οδηγεί σε άτοπο λόγω της αρχικής υπόθεσης του προβλήματος και έτσι η απόδειξη ολοκληρώνεται. $ \blacksquare$

[quote="Nick Rapanos"][quote="r_boris"]Αν η $ f$ είναι δυο φορές παραγωγίσιμη στο $ R$ και υπάρχει $ c$ πραγματικός : $ \frac {f(x) \minus{} f(y)}{x \minus{} y}\ne f'(c)$ για κάθε $ x \ne y$ τότε να δειχτεί ότι $ f''(c) \equal{} 0$[/quote]

[b]ΛΥΣΗ[/b]
Έστω ότι $ f''(c)\neq0$. Ας υποθέσουμε χωρίς βλάβη της γενικότητας ότι $ f''(c) > 0$. Θα αποδείξουμε ότι υπάρχει $ \epsilon > 0$ τέτοιο ώστε $ f''(x) > 0, \forall x\in [c \minus{} \epsilon,c \plus{} \epsilon]$.
[quote]
[i]Απόδειξη:[/i] Έστω ότι αυτό δεν ισχύει, οπότε για οσοδήπτε μικρό $ \epsilon > 0$ υπάρχει $ \xi\in [c \minus{} \epsilon, c \plus{} \epsilon]$ τέτοιο ώστε $ f''(\xi)\leq 0$. Τότε για $ \epsilon\to 0$, λαμβάνουμε $ f''(c)\leq0$,[/quote] που είναι άτοπο από την αρχική μας υπόθεση. Άρα πράγματι η $ f$ είναι κυρτή σε μια περιοχή του $ c$. $ \square$

Τότε ως γνωστόν,η γραφική παράσταση της $ f$ θα βρίσκεται πάνω από την εφαπτομένη της στο $ c$ στο διάστημα που αυτή είναι κυρτή, δηλαδή $ f(x)\geq g(x) \equal{} f'(c)(x \minus{} c) \plus{} f(c)$ για κάθε $ x\in [c \minus{} \epsilon, c \plus{} \epsilon]$. Θεωρούμε τώρα μια μετατόπιση της $ g(x)$ προς τα πάνω και συγκεκριμένα την ευθεία $ h(x) \equal{} g(x) \plus{} \delta$ με $ 0 < \delta < min\{f(c \minus{} \epsilon) \minus{} g(c \minus{} \epsilon),f(c \plus{} \epsilon) \minus{} g(c \plus{} \epsilon)\}$ η οποία τέμνει τη γραφική παράσταση της $ f$ σε δύο σημεία $ (a,f(a))$ και $ (b,f(b))$ που βρίσκονται εκτέρωθεν του $ c$ και ως εκ τούτου είναι διαφορετικά μεταξύ τους. Ο συντελεστής διεύθυνσης της $ h$ είναι ίσος με $ f'(c)$, επομένως $ \lambda_h \equal{} \frac {f(b) \minus{} f(a)}{b \minus{} a} \equal{} f'(c)$. Η τελευταία σχέση όμως οδηγεί σε άτοπο λόγω της αρχικής υπόθεσης του προβλήματος και έτσι η απόδειξη ολοκληρώνεται. $ \blacksquare$[/quote]

Νίκο δεν είναι σαν να λες ότι $ f''$ συνεχής ? ή μου διαφεύγει κάτι?
Επειδή και στην δικιά μου λύση έχω ανάλογο πρόβλημα περιμένω να δω αν υπάρχουν άλλες λύσεις και μετά θα ποστάρω την λύση μου

[quote="cmad"][quote="Nick Rapanos"]
Έστω ότι $ f''(c)\neq0$. Ας υποθέσουμε χωρίς βλάβη της γενικότητας ότι $ f''(c) > 0$. Θα αποδείξουμε ότι υπάρχει $ \epsilon > 0$ τέτοιο ώστε $ f''(x) > 0, \forall x\in [c \minus{} \epsilon,c \plus{} \epsilon]$.
[/quote]

Δε χρησιμοποιείς πουθενά στην απόδειξη του λήμματος τη δεδομένη σχέση $ f(x) \minus{} f(y) \neq f'(c)(x \minus{} y)$. Όμως η υπόθεσή σου δεν ισχύει για τυχαίες συναρτήσεις. (για τον ίδιο λόγο που δεν ισχύει το: $ f'(x_0) > 0 \Rightarrow f$ αύξουσα κοντά στο $ x_0$ για $ f$ παραγωγίσιμη.) Διόρθωσέ με αν έχω λάθος.

In any case, here goes:

Θεωρούμε την $ g(t) : \equal{} f(t) \minus{} f(c) \minus{} f'(c)(t \minus{} c)$, $ t \in \mathbb{R}$. Από τη δεδομένη ανίσωση, η $ g$ βγαίνει $ 1 \minus{} 1$. Ως $ 1 \minus{} 1$ (και συνεχής), η $ g$ θα είναι μονότονη. Επομένως, η $ g'$ δεν αλλάζει ποτέ πρόσημο. Άρα, η $ g'$ έχει ακρότατο στο $ c$ (αφού $ g'(c) \equal{} 0$), και άρα $ g''(c) \equal{} 0$.[/quote]
Βασικά το ίδιο λέμε
Η λύση μου στο attachment

[quote="cmad"]By the way, νομίζω ότι η άσκηση αυτή πρέπει να μεταφερθεί στις σχολικές. Δεν έχει τίποτα μη-σχολικό η λύση της.[/quote]
Έχεις δίκιο απλώς δεν είναι πολύ εύκολη :wink: το έκανα για θέμα του 2007


function
calculus
integration
logarithms
inequalities
number theory proposed
number theory

Let ${\mathbb N }$ be the set of positive integers.

 Consider a function $ F: {\mathbb N}\to {\mathbb N} $  which satisfies properties :
\[   1)\; \;   F(1)=0  , F(2)=1\]
\[   2) \; \;  F(xy)=F(x) +F(y)   ,\;  \forall (x,y) \in {\mathbb N}\times {\mathbb N}  \]
and $ 3) \; \;  F(x)= F(x-1) $  when  $ x $  is a prime number from ${\mathbb N} $ .

[b]Questions:[/b]
 i)  The function $ F:{\mathbb N}\to {\mathbb N} $ is  well-defined ?

 ii) Let [.] denotes integral part , and $a(x):=\frac{\ln{x}}{\ln{3}}\; ,\;
A(x):=\frac{\ln{x}}{\ln{2}} . $

[i] It's true that for  $ x=3,4,... $ the inequalities
\[
   [a(x)] \le F(x)\le  [A(x)]
\]
are verified ? [/i]

Hi,

In your problem statement, you say the function $F : N \rightarrow N$ where $N$ is the set of positive integers. Then you say that $F(1) = 0$, but $0$ is not a positive integer.

Moreover, you say $F(p) = F(p-1)$ if $p$ is prime. It then follows that $F(2) = F(1) = 0$. But you say that $F(2) = 1$.

Could you please explain this?

Thanks,
Sambit

I think you should simply understand N as it is: the set of naturals {0,1,2,...}

As for the other one, that's quite strange indeed. I think the problem will require x>2 or x>3 somehow... or the question on part 1 is just no :D

i) let $A_n$ be the subset of $N$ of those number which are only divisible by the first n primes. if $F$ is well defined for the first n primes, then for $A_n$ by 2). but by 3), it is then well-defined for the n+1-th prime since $p_{n+1}-1$ can only be divisible by smaller primes(let $p_n$ be the n-th prime). we inductively see that $F$ is well-defined.
ii) (the inequalities even hold for 1 and 2) the first observation is, that if the inequalities hold for all primes, then they hold everywhere by 2). the second inequality holds for 2, thus for $A_1$, thus for 3 by propery 3), thus for $A_2$, thus for 5 and so on.
the second inequality holds for 2 and 3($F(3)=1$). we will show that equality holds iff $n$ is a power of 3. this is true in $A_2$. now, if want to go from $A_n$ to $A_{n+1}$, we have to show that $F(p_{n+1})>\frac{ln(p_{n+1})}{ln(3)}$. but $F(p_{n+1})=F(p_{n+1}-1)=F(2)+F(\frac{p_{n+1}-1}{2})\geq 1+\frac{ln(p_{n+1}-1)-ln(2)}{ln(3)}>\frac{ln(p_{n+1})}{ln(3)}$ since $ln(x+1)-ln(x)<ln(3)-ln(2)$.

Peter

[quote="Peter VDD"][quote="flip2004"]Thank you !
Indeed there is a mistake . Please consider tha N={1,2,...}= the set of positive integers, N_0={0,1,...}= the set of non-negative integers, and F:N---> N_0 .
Moreover, instead of 3) please consider
3')   F(p)=F(p-1) when p is prime > 2 .
Regards,proposer.[/quote]

No... N_0 is by definition {1,2,...} and N is by definition {0,1,2,...} that are just the international standards all around the world ;)
So it's a F: N_0 --> N

greetz, Peter[/quote]

I have always seen N_0 as {0,1,2,3,...} and N as {1,2,3,4,...}

so I guess that just means people are going to have to specify in their problem statements, thats all.


inequalities
Cauchy Inequality
inequalities proposed

Prove that, for any positive real numbers $a,b$ and $c$,
\[ \sum_{cycl}\frac{a^2+bc}{a^2+(b+c)^2}\le\frac{18}{5}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2} \]

(My friend & I solved it)
WLOG,assume that $a+b+c=3$.We have to prove that $\frac{2}{5}(a^2+b^2+c^2)\geq\sum\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}$
This is an easy problem because:
$\frac{2}{5}a^2-\frac{4a^2+(3-a)^2}{4(a^2+(3-a)^2)}\geq\frac{11}{25}a-\frac{11}{25}\Leftrightarrow(a-1)^2(80a^2-168a+171)\geq 0\ldots$ :lol:

Nice proof! Hiep.

I found this inequality the same time as this one
[url]http://www.artofproblemsolving.com/Forum/topic-79611.html[/url]

[quote="probability1.01"]A similar inequality that I just thought of:

Prove that $\sum \frac{a^{2}+bc}{2a^{2}+(b+c)^{2}}\le \frac{7}{4}-\frac{1}{4}\cdot \frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$[/quote]
We can prove it by the same method. :)


integration
trigonometry
calculus
calculus computations

[quote="bythecliff"]Can anyone solve the following?

$ sin 3x dx\plus{}2y (cos 3x)^{3}dy \equal{} 0$?

I'm having trouble integratig $ \frac{sin 3x}{(cos 3x)^{3}}$.[/quote]

[hide]$ \int\frac{\sin3x}{\cos^{3}3x}dx \equal{}\minus{}\frac{1}{3}\int\frac{d(\cos3x)}{\cos^{3}3x}dx$[/hide]

pluricomplex:

$ \int\frac{\sin{3x}}{\cos^{3}{3x}}\,dx \equal{}\minus{}\frac{1}{3}\int\frac{d(\cos{3x})}{\cos^{3}{3x}}$

Solution

$ \sin{3x}\,dx\plus{}2y\cos^{3}{3x}\,dy \equal{} 0$

For now, assume $ \cos^{3}{3x}\neq 0$

$ \Rightarrow\frac{sin{3x}}{\cos^{3}{3x}}\,dx\plus{}2y\,dy \equal{} 0$

$ \Rightarrow\int\frac{sin{3x}}{\cos^{3}{3x}}\,dx\plus{}\int 2y\,dy \equal{} c$

Let $ u \equal{}\cos{3x}$

$ \Rightarrow du \equal{}\minus{}3\sin{3x}\,dx$

$ \Rightarrow\minus{}\frac{1}{3}du \equal{}\sin{3x}\,dx$

$ \Rightarrow\minus{}\frac{1}{3}\int\frac{1}{u^{3}}\,du\plus{}\int 2y\,dy \equal{} c$

$ \Rightarrow\minus{}\frac{1}{3}\left(\minus{}\frac{1}{2u^{2}}\right)\plus{}y^{2}\equal{} c$

$ \Rightarrow\frac{1}{6 u^{2}}\plus{}y^{2}\equal{} c$

$ \Rightarrow y^{2}\equal{} c\minus{}\frac{1}{6\cos^{2}{3x}}$

This is not the complete solution, as we have ignored the cases for $ \cos^{3}{3x}\equal{} 0$.

Solving this:

$ \cos{3x}\equal{} 0$

$ \Rightarrow 3x \equal{} (2n\plus{}1)\frac{\pi}{2}$

$ \Rightarrow x \equal{} (2n\plus{}1)\frac{\pi}{6}$

These values of $ x$ are all solutions to the differential equation too (for any $ y$). This is because if you put these values into the differential equation $ dx \equal{} 0$ is obtained.


trigonometry
geometry
AMC
AIME
ratio
function
Pythagorean Theorem

If $ \tan{x} \plus{} \tan{y} \equal{}25^{\circ}$, and $ \cot{x} \plus{} \cot{y} \equal{}30^{\circ}$, what is $ \tan{x\plus{}y}$?

Another one:

Compute $ \displaystyle \sin{\theta}$

http://www.math.ku.edu/mathawareness/98competion/56.pdf

(Its problem 2 In the 10-12th grade division)

$ 1/\tan x \plus{} 1/\tan y \equal{} (\tan x \plus{} \tan y)/(\tan x \tan y) \equal{} 25/(\tan x \tan y) \equal{} 30$

$ \tan x \tan y \equal{} 5/6$

$ \tan(x \plus{} y) \equal{} (\tan x \plus{} \tan y)/(1 \minus{} \tan x \tan y) \equal{} 25/(1 \minus{} 5/6) \equal{} 25*6 \equal{} 150$

$ \sin \theta \equal{} \sin(\pi/2 \minus{} 2\arcsin(1/\sqrt {5})) \equal{} \cos(2\arcsin(1/\sqrt{5}) \equal{} 1 \minus{} 2/5 \equal{} 3/5$

Um the last line you wrote seemed like a formula because I don't see how you got  it from manipulation, can you please derive $ \tan{\left(x \plus{} y\right)} \equal{} \frac {\tan{x} \plus{} \tan{y}}{1 \minus{} \tan{x}\tan{y}}$?

Well, just take $ \frac{\sin{(\alpha \plus{} \beta)}}{cos{(\alpha \plus{} \beta)}}$, expand it, and divide the numerator and denominator by $ \cos{\alpha}\cos{\beta}$.

sorry, but I don't quite understand what you're saying. How do you "expand" these terms? (Sorry I haven't done a lot of trig)

[url]http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities[/url]. Volume 2 has a good proof using the area of a triangle.

If I read through vol 2, will I have enough knowledge of trig and complex numbers up till the AIME level?

[quote="mathemagician1729"]If $ \tan{x} \plus{} \tan{y} \equal{} 25^{\circ}$...[/quote]

I don't understand how the tangent of something can equal an angle. Tangent is a ratio of sides. Are you saying that it equals tangent of 25 degrees?

For your second problem, you could also solve it like this

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Call the square $ ABCD$ with midpoints $ M$ and $ N$ on $ AD$ and $ CD$, respectively. By Pythagorean Theorem, we have $ BN\equal{}BM\equal{}\frac{\sqrt{5}}{2}$ and $ MN\equal{}\frac{\sqrt2}{2}$. Hence, by Law of Cosines on triangle $ BMN$, \[ \frac{5}{4}\plus{}\frac{5}{4}\minus{}2(\frac{5}{4})\cos\theta\equal{}\frac{1}{2} \] Solving gives $ \cos\theta \equal{}\frac{4}{5}$. Since $ \sin^2\theta \plus{} \cos^2\theta \equal{}1$, we have $ \sin\theta \equal{} \frac{3}{5}$

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[quote="grn_trtle"]It doesn't equal an angle. It equals a degree measurement.
$ 25^\circ$ is a real number.

If it helps, converting to radians gives $ 0.436332$.[/quote]

I still don't understand what the question is asking for. Tangent is a function that gives you a fraction as an answer, such as 5/12 or 2. You can take the tangent of a measurement of an angle, but tangent will give you a number, and I don't understand what $ 25^\circ$ stands for. Perhaps an example will help me.

What does the $ \Tan \frac {\pi}{4}$ equal in degrees?

$ \frac{\pi}{4}\equiv 45^\circ$.

Now, realize that $ \tan: \mathbb{R}\to\mathbb{R}$ is a function that takes a real number input and gives a real number output. $ 25^\circ$ is a numerical output (an angle measurement). It's just in some unit.

Now, it might be confusing if you had to draw a line that's $ 25^\circ$ long, but in this case it's all algebraic.

Note: $ \tan$ maps a real number to a real number, [b]not necessarily[/b] a rational number. (Simple terms: it doesn't always give you a fraction).

Um, that's not even what he's asking.
Anyway, those are supposed to be 25 and 30, not 25 and 30 degrees.

I was correcting his statement "Tangent is a function that gives you a fraction as an answer"

Anyway, the problem statement itself was wrong, so whatever :D

[quote="Brut3Forc3"]Oh, I see now. But I think his use of the word fraction was meant to mean opposite/adjacent, a ratio of side lengths, as opposed to a measure of angle (in degrees).[/quote]
That is what I meant, but I also didn't want to use the word ratio, because that does not denote a value.

My question was supposed to read $ \tan{\frac{\pi}{4}}$.

While degrees is written as a value, the value represents a degree measurement. Tangent on the other hand is a ratio, which has a real answer. $ \tan {\frac{\pi}{4}}\equal{}1$ for example. I don't see a way to say what 1 equals in degrees. Without a definition of how to change a real number into a degree measurement, the equation made no sense. I am still trying to find out what the definition is, if such a definition exists. Seeing as the problem was incorrectly written, it does not seem to be a problem, but if there is a definition, I would like to know it.


Maybe it's well-known, but I didn't manage to find it on the forum. Let $\varphi(\cdot)$ be the totient function. Prove that for all positive integers $n$ we have $\sum_{k=1}^{n}\frac{\varphi(2k)}{k}\le n \le \sum_{k=1}^{n}\frac{\varphi(2k-1)}{k}$.

As $\phi (2k)$ is not relativly prime with $2,4,...,2k$ so $\phi(2k)\leq k$. Obvious $\sum_{k=1}^{n}{\frac{\phi(2k)}{k}}\leq n$. For right side it is $\phi(2k-1)\geq k$ because $2k-1$ is relativly prime with $2,4,...,2k-2$ and $(2k-1,2k-3)=1$ (this doesn't work for $k=1$, but we have $\phi(1)\geq 1$). So $\sum_{i=1}^{n}{\frac{\phi(2k-1)}{k}}\geq n$.

It's not true that $2k-1$ is coprime to $2,4,6,...,2k-2$, e.g. take $k=5$.