probability
combinatorics unsolved
combinatorics

Let $ F$ be the family of the subsets of the set $ A\equal{}\{1,2,...,3n\}$ which have the sum of their elements divisible by $ 3$
For each element of $ F$,compute the square of sum of its element.What is the value of the sum of all the obtained numbers?

We will prove that every element of $ A$ belongs to exactly
\[ f(n) \equal{} \frac {2^n (2^{2n \minus{} 1} \plus{} 1)}{3}
\]
members of $ F,$ after which computation is immediate and the total sum becomes
\[ S \equal{} \frac {2^n (2^{2n \minus{} 1} \plus{} 1) n (3n \plus{} 1)(6n \plus{} 1)}{6}
\]
The number of members of $ F$ that contain $ i \in A$ is the same for all $ i \equiv_3 j,$ so we denote it by $ f(n, i \bmod 3).$ Then we can assume $ i \in \{1, 2, 3\}$. If it is $ 3,$ then it can be or not be in a member of $ F$ with equal probability, so $ f(n, 3) \equal{} \tfrac{|F|}2.$ Now denote by $ \{b_1, \ldots, b_{3n}\}$ a set in $ F$, with $ b_i \equal{} 1$ if $ i \in \text{the set}$ and $ b_i \equal{} 0$ otherwise. Then $ f(n, 1) \equal{} f(n, 2) \equal{} \tfrac{|F|}{2},$ because the map $ F \mapsto F$ that sends
\[ \{b_1, b_2, b_3, b_4, \ldots, b_{3n}\} \mapsto \{1 \minus{} b_1, 1 \minus{} b_2, b_3, b_4, \ldots, b_{3n}\}
\]
bijectively maps members of $ F$ to which $ 1$ belongs to members to which $ 1$ doesn't (and similarly for $ 2$). We have reduced our assertion, then, to finding $ |F|.$ We know that
\[ |F| \equal{} 2^n \times \text{\# of } \subset \{1,2,4,5,\ldots,3n \minus{} 2,3n \minus{} 1\} \text{ whose sum of elements } \equiv_3 0
\]
It is enough to prove that this number above equals $ g(n) \equal{} (2^{2n} \plus{} 2) / 3.$ The subsets of $ \{1,2,4,5 ,\ldots, 3n \minus{} 2, 3n \minus{} 1\}$ whose sum of elements is divisible by $ 3$ can be split in two classes, with one consisting of supersets of $ \{3n \minus{} 2,3n \minus{} 1\}.$ This class has $ g(n \minus{} 1)$ members, while the other has $ 2^{2n \minus{} 2}$ elements (choose among the first $ 2n \minus{} 2$ elements arbitrarily, and then you have a single choice for the last one.) The relation $ g(n) \equal{} g(n \minus{} 1) \plus{} 2^{2n \minus{} 2}$ and $ g(1) \equal{} 2$ ($ F$ contains the empty set) gives $ g(n) \equal{} 2 \plus{} 2^2 \plus{} 2^4 \plus{} \cdots \plus{} 2^{2n \minus{} 2} \equal{} (4^n \plus{} 2) / 3$ and we're done.

bodan, it asks for the sum of the squares of the sums, not the sum of the squares of the elements ;)

So then we also need to find, for a set $ X \equal{} \{a, b\} \subset A,$ how many members of $ F$ contain $ X.$

This time it's $ \tfrac{|F|}{4},$ because $ a$ and $ b$ can be taken to be in $ a \in \{1,2,3\}$ and $ b \in \{4,5,6\},$ (this works only for $ n > 1,$ we have to check $ n \equal{} 1$ separately) and again the map
\[ \{b_1, b_2, b_3, b_4, \ldots, b_{3n}\} \mapsto \{1 \minus{} b_1, 1 \minus{} b_2, b_3, b_4, \ldots, b_{3n}\}
\]
takes bijectively members of $ F$ that contain $ a$ and contain $ b$ to members that do not contain $ a$ but contain $ b,$ and since the total number of members that contain $ b$ was proven to be $ \tfrac{|F|}{2},$ the result follows.

So, for $ n \geq 2,$
\[ \sum_{3 \,|\, a \plus{} b \plus{} \cdots} (a \plus{} b \plus{} \cdots)^2 \equal{} \sum_{3 \,|\, a \plus{} b \plus{} \cdots} \sum_{a,b} ab \equal{} \frac {|F|}{2} \sum_{1 \leq i \leq j \leq 3n} ij
\]
and since we know $ |F|,$ we can simplify this.

We can generalize this as follows: denote, for a subset $ \{a_1, \ldots, a_k, a_{k\plus{}1}, \ldots\} \equal{} X \subset A$ with $ k \leq n,$ the $ k\minus{}$th symmetric sum of $ X$ by \[ \sigma_k (X) \equal{} a_1 \cdots a_k \plus{} a_1 \cdots a_{k\minus{}1}a_{k\plus{}1} \plus{} \cdots\] Let $ s(X)$ be the sum of elements of $ X,$ and put \[ S_k (A) \equal{} \sum_{3 \,|\, s(X)} \sigma_k (X)\] Then each $ k\minus{}$element subset of $ A$ appears in the sum $ S_k$ exactly $ \tfrac{|F|}{2^{k}}$ times (because $ k \leq n,$ we can put the elements of this $ k\minus{}$element subset each in $ \{3i \plus{} 1, 3i\plus{}2,3i\plus{}3\},$ for $ 0 \leq i < k;$ then partition the members of $ F$ in $ 2^k$ classes of the same size, only one of which contains the entire $ k\minus{}$element subset.) So
\[ S_k (A) \equal{} \frac{2^{n\minus{}k\plus{}1}(2^{2n\minus{}1} \plus{} 1)}{3} \sum_{1 \leq i_1 < \cdots < i_k \leq mn} i_1 \cdots i_k\]

See also http://www.mathlinks.ro/viewtopic.php?t=290343 :)


geometry
analytic geometry
trigonometry
function
inequalities
HCSSiM
probability

[quote="stupidkid"]How can the area of the quadrilateral in problem 3 be $8i$?[/quote]
I think on the Math Jam he realized he made a mistake on that.  Here are my solutions.  Feel free to criticize, especially problem 2 (I used a solution that, while workable, isn't particularly elegant):

Yeah, that's wrong, as I found out in the Math Jam.  I thought that coordinates => area

I'm trying to figure out how we could have gotten the same answers yet my solutions are 2-3 times as long.  :huh:

I wasn't at the math jam, but from what I've gathered on the forums, my solutions for 2, 4, and 5 are definitely correct.  I'm fairly confident that my solutions for 1 and 3 are also correct, and these are all brief, with pictures too!  The only question is... were they too brief... we'll see...

Edit:  Bah, #3 wrong, I simplified root(4r^6) into 4r^3.  Hopefully I'll only lose a point for that.  Everything else is correct.

[quote="sonar"]Post away  :lol:[/quote]
I loved your solution to #2 - so simple and elegant (especially the way you generalized it).

I can't post mine because I handwrote it. Oh well.  :(

My solutions sounded really good. What they didn't accomplish in equations, they made up for in language, hehe. More points for me!  :D

here are mine.  i think i was kind of too undetailed for some stuff.

1/lnx, your solutions weren't that bad.  :)

All the trig on number 5 made me dizzy for a while though.  :D

I'll follow suit and post my solutions. If you feel like it, you can read or skim over my solution to #1, which was probably way overdone. I defined a function [i]f[/i] which is basically equivalent to modulus, except I wasn't so sure about modular arithmetic so I didn't want to risk making a mistake.

I like the way I did #5. I haven't seen any other people do it that way. It was kinda long, though.

My solution to #2 was also pretty good. It was actually just a page, making it my shortest one.

[quote="1/(ln x)"]i wasn't in the math jam and yes i knew that i got a lot of them wrong, but here are my solutions. Criticize away :)[/quote]
I think your problem with #4 is you assumed it was sufficient for $a + b$ to be expressable in more than one way as a sum.  In fact, you find that $a + b$ cannot be expressable as a sum of squares.  You also say at one point that $a \neq a'_1$ and $b \neq b'_1$, and then reverse that inequality in the next paragraph - so, I'm not sure why you ended up with the solution you did.

As a general note, did anybody else provide a motivation for their factorization in #3?  I tried to, because just showing the factorization seemed a bit arbitrary.

[quote="zanttrang"]Feel free to criticize, especially problem 2 (I used a solution that, while workable, isn't particularly elegant):[/quote]

But your solution for 3 makes up for it  :)

[quote="zanttrang"]I loved your solution to #2 - so simple and elegant (especially the way you generalized it).[/quote]

Thanks, HCSSiM Probability Maxi represent  :P

[quote="stupidkid"]Wow, your solutions are very detailed...I kinda said, "Similarly it can be shown...." quite a lot. Haha, probably will get some points deducted for that.[/quote]

Not necessarily.  Dr. Schram noted in last year's posts that if the method is exactly the same, saying "similarly ..." is acceptable.  I'll try and find the exact post to give a bit of context.

Here they are: I had to hunt up my tex compiler again to recompile. I don't really mind you guys getting my ID number - that's public information - and I trust you guys not to use my name. ;)

I think your sum notation is wrong on #2, and also the answer is wrong... but so is mine :( Nice job on #1 and #5 though! :)

[quote="jb05"]All right, I don't have PDF writers, and all my solutions are handwritten, so I'm just going to choose my favorite (and one of my shortest), #3:

[hide]Let $w=rz$ and $q=r^4$. Then the expression reduces to $w^4+(10q-2)w^2-16qw+(q^2+10q+1)=0$ or $9q^2+(10w^2-16+10)q+(w^4-2w^2+1)=0$. We use the quadratic formula on the above (in my proof I go through it, but here I will just give the result) and get $q=-(9w^2-18w+9)/9$ or $w=-(w^2+2w+1)/9$. Therefore, $-q=(w-1)^2$ or $-9q=(w+1)^2$. So $w=1\pm i\sqrt q$ or $w=-1\pm 3i\sqrt q$. Since w=r*z and $q=r^4$, $z=1/r\pm ir$ or $z=-1/r\pm 3ir$ (again, sorry but I'm leaving out intermediate steps to save space). Since this figure has 2 sets of 2 points each having identical x-coordinates, it is a trapezoid. The average of the bases is (6r+2r)/2=4r and the height is 2/r. Multiplying, the area is always 8.[/hide][/quote]
Nice job -- that's a really clever substitution. :)

[quote="solafidefarms"]Yes, I had the wrong sum. :( I couldn't figure out how to include an image, so I didn't have a diagram included with the last one, and I forgot to *prove* that the segments connecting the midpoints of 2 sides of a triangle has half the length of the third side and also that if a quadrilateral has an inscribed circle tangent to all 4 sides the sum of any 2 opposite sides is equal to any other sum. Hopefully I won't lose a point for that.[/quote]
You don't have to prove that the midsegment of a triangle is half the length of the corresponding side. That's already a well-known fact. ;)

[quote="chess64"]Some of you guys should learn how solutions are properly written. ;) I thought that a lot of them were too informal. Look through the published solutions for year 16, for example. :)

Well here are my solutions. I didn't get #5 and #2 is wrong :stink: but I think the rest are good.

Feel free to laugh though! :D[/quote]
Nice diagram on #3. :)

I somehow kept thinking that the $10r^4$ term was $4r^4$ for some reason and so I missed that factoring thing and didn't get the answer.  :wallbash_red:  :wallbash_red:  :wallbash_red:

I'll post the pdf version of my solutions.  Maybe I shouldn't have sent in a dvi...

EDIT: I posted something completely different, something from my math club.  My bad.  :D

Here's the zip I sent in, anyways...

EDIT2: Sorry i had to delete it...was taking up space...but ask me if you want it...

[quote="K81o7"]I'll post the pdf version of my solutions.  Maybe I shouldn't have sent in a dvi...[/quote]

[quote="USAMTS"]We will only accept the following file types: DVI, PDF, RTF, TEX, TXT, TIF, GIF, BMP, JPG, EPS. [/quote]
You should be fine.

p.s. What problem is your solution for? Doesn't look like something from the USAMTS. :huh:
Either something's messed up or I'm just crazy . . .

[quote="ZennyK"]
p.s. What problem is your solution for? Doesn't look like something from the USAMTS. :huh:
Either something's messed up or I'm just crazy . . .[/quote]
No, you're not crazy - I'm not sure what his solution is for either, but it doesn't look like USAMTS stuff.

:o Whoops....  :D
This is the solution i wrote for a different problem, but i converted at the same time.  I'll put my USAMTS solutions in when I get home...
I'm such an idiot...
Just pretend you didn't see it.  :?

[quote="solafidefarms"]Wayyyyyy too informal, but it *might* do the job.[/quote]

Whose thing are you talking about?

Well here's the pdf wtih my solutions...and the diagram i drew for #5...


trigonometry
function
real analysis
calculus
integration
derivative
calculus computations

Inverse Laplase transformation of 1 / (s^2 + 1)^3? That s^2 is what baffles me. Someone help please? Much thanks.

Are you sure this isn't an inverse Lapla[b]c[/b]e transform problem? Is this possibly what the problem really said?

Find $f(t)$ such that $\mathcal{L}(f)(s)=\frac1{(s^{2}+1)^{3}}.$

One possible approach (and a messy one):

Let $y(t)=\sin t.$ Then $\mathcal{L}(y)(s)=\frac1{s^{2}+1}.$

The function we want is $f(t)=y*y*y(t).$

Admittedly, computing those convolution integrals isn't nice.

The transform of $tf(t)$ is $-F'(s)$. So the transform of $t^{2}\sin t$ should be $\frac{2(3s^{2}-1)}{(s^{2}+1)^{3}}$. Well, this is not quite what you want, but it may give you an idea.

Try [url=http://www.eecircle.com/applets/007/ILaplace.html]this applet[/url], but don't trust it blindly.

And where the sine is, so also should be the cosine. The Laplace transform of $\cos t$ is $\frac{s}{s^{2}+1}.$

Using the derivative relationship mlok gave, try making yourself a little table with the Laplace transforms of $\sin t,\ t\sin t,\ t^{2}\sin t,\ \cos t,\ t\cos t,$ and $t^{2}\cos t.$

One last perspective:

The function we want is the solution of this differential equation:

$y^{[6]}+3y^{[4]}+3y''+y=0;\ y(0)=y'(0)=y''(0)=y'''(0)=y^{[4]}(0),\ y^{[5]}(0)=1$

This solution has the form

$y(t)=(At^{2}+Bt+C)\cos t+(Dt^{2}+Et+F)\sin t$

There's an implied $6\times 6$ system of linear equations to solve for the coefficients of that linear combination; you might want to seek electronic help with that linear system.


Hmmm? I can do 70 wpm without punctuation. (Real sentences, reasonably common words; I read off a sentence and then type it from memory, so if it's not grammatically complete or spelled correctly :D I am slooow.)

70-100+, depending on what I'm typing (the words, whether there are numbers/symbols, capitalization, etc.)

According to the typing program at school I can get up to 720 :D  :D
Really I am usually at about 100-150 wpm

around 75-85

on micro type pro, the fastest i got was 117, but it was easy words like "fast hide good lukc to you he she it" :D

Approximately 110...
But yeah, that's bad grammar and stuff.

I can consistently type around 130. [for extended periods of time]

Reach peak speeds of 150.

If you don't believe me you haven't played trivia against me under the old AOPSTI format.

If I type with complete accuracy, and am going pretty quickly, maybe 70 wpm.  I generally don't just speed type making grammar mistakes, but just now I did a sample, typing the last paragraph of my english assignment sheet, with a few words mispelled at the end, and got around 90-100 wpm.

I was like 35 GWAM..and I might have gotten like 40 a couple times

Somewhere over 100, with good grammar and punctuation, etc.

Who said over 200? I don't believe you. :P

I want to see a video of typing that fast.

[quote="biffanddoc"][quote="la mafioso"]biff world record was 360 so stop lying  ;)

I think you meant words by the way biff, he meant gwam (5 characters a word), not numbers ;)[/quote]
I can type six words in one second...
but they're all "a"[/quote]

i want see u do that ;)

On typing programs, i get 160.  Consistantly, i get about 70 if im concentrating (usually not).

I type between 100-150 and I use proper grammar and stuff. It's really not that hard unless you don't type very often. By the way, I believe Lucky707. He killed us in Trivia.  :D

Depends on the Language. In English I can type around 120-150 wpm. In Romanian I can easily do 150-180 wpm (without special characters, ie replacing â with a, î with i and so on). With diacritics, it goes way down to 50-80 wpm.

I usually get 60-70 wpm, and the highest I got was like 110 or something.

WOW you know what....all of you people are like KRAZY and must have 11 brains one on each finger+ one in ur head. I only have one. I average 50 and 60 when i am feeling good 80 when I am typing really really fast

I did a 2 minute online typing test just now, with gross speed of 82 wpm and a net speed of 73 wpm.  That's about where I figured I was at.

am i just slow???? my words per minute is at 45 right now...

Around 45-50 WPM with numbers, punctuation and high accuracy. I go ALOT faster with no numbers and symbols.

[quote="probability1.01"]According to an online test, I type at around 80 wpm at 98+ accuracy (an insertion/deletion throws everything off, and you can't backspace). But the passage I typed was pretty complex, and it lasted 3 minutes. When I'm typing off the top of my head instead of reading, I'm guessing my speed goes over 100 wpm.[/quote]

I usually type faster when I'm reading something, because otherwise I don't always know what I want to say. However, I generally get lower accuracy when I do that, because I'm typing so fast that I don't always think about what I'm doing, and of course it doesn't let you backspace. >_<

I'm at 52 words per minutes, but it's probably higher on a normal day, cause I'm really really tired right now.

INSANELY FAST. if I really really have to. usually though i'd guess... 100-150 rpm? lol, wpm, not rpm.

Err.. well the last time i measured was 3 years ago and then i did 55. But now, on average of just typing casually (no rush) i do 70. When im in a rush i can do over 100 easily. And its cool to type while your reading something (ie copying it onto computer) - i duno its just a cool feeling.


Let $ a,\ b,\ c$ be real numbers.

Prove: $ (a^2 \minus{} ab \plus{} b^2)|a \minus{} b| \plus{} (b^2 \minus{} bc \plus{} c^2)|b \minus{} c| \plus{} (c^2 \minus{} ca \plus{} a^2)|c \minus{} a|$

$ \plus{}2(\sqrt{1\minus{}a^2}\plus{}\sqrt{1\minus{}b^2}\plus{}\sqrt{1\minus{}c^2})\leq 12.$

[color=red]Last edited[/color]

[quote="kunny"]Let $ a,\ b,\ c$ be real numbers.

Prove: $ (a^2 \minus{} ab \plus{} b^2)|a \minus{} b| \plus{} (b^2 \minus{} bc \plus{} c^2)|b \minus{} c| \plus{} (c^2 \minus{} ca \plus{} a^2)|c \minus{} a|$

$ \geq \frac {4}{27}(a \plus{} b \plus{} c)^4 \plus{} \frac 13 (ab \plus{} bc \plus{} ca)^2$[/quote]

Was the inequality supposed to be reversed from the above? Because $ a\equal{}b\equal{}c\equal{}1$ immediately yields a contradiction.

Cheers,

Rofler

Thank you for pointing out it, but I have changed the problem a little bit.

[quote="kunny"]Let $ a,\ b,\ c$ be real numbers.

Prove: $ (a^2 \minus{} ab \plus{} b^2)|a \minus{} b| \plus{} (b^2 \minus{} bc \plus{} c^2)|b \minus{} c| \plus{} (c^2 \minus{} ca \plus{} a^2)|c \minus{} a|$

$ \plus{} 2(\sqrt {1 \minus{} a^2} \plus{} \sqrt {1 \minus{} b^2} \plus{} \sqrt {1 \minus{} c^2})\leq 12.$

[color=red]Last edited[/color][/quote]
How about if $ a>1,\, b>1\,\,$ or $ \,\,c>1$ ?

We are to suppose that $ a,\ b,\ c\in{[\minus{}1,\ 1]}$, which is common sense, isn't it? :wink:

[quote="kunny"]We are to suppose that $ a,\ b,\ c\in{[ \minus{} 1,\ 1]}$, which is common sense, isn't it? :wink:[/quote]
Common sense? I do not think so.

Why? Can you say $ \sqrt{1\minus{}a^2}<0?$ :P


when is the IMO going to be?

GO USA!!!! :winner_first:    :yoda:  :thumbup:

The contest itself is on July 25 and 26. See [url=http://www.imo2007.edu.vn/]the official site[/url] for more details.


geometry
calculus
probability
number theory

My math team participate in the City of Chicago Math League and its format is as follows:

Each contest consists of a 20 question exam which must be completed in 50 minutes. Contests are administered at three sites; schools are assigned to sites based on their locations. Competition is divided into four grade level/topic areas: Algebra 1 (9th grade), Geometry (10th grade), Algebra 2 (11th grade), and Pre-Calculus (12th grade). An advanced student who has already completed the course for her grade level (for example, a sophomore who has already completed geometry) can take an "Accelerated" version of that contest, or can take a higher-level contest. As a result, there are eight levels of competition.

Rather than having people practice independently at practice (which is really boring for them and for the group), I tried to do different things with the old practice tests. In the past couple of months I did:

1) Relays (of another contest)
-> Our team captain did a Ti-89 workshop.
2) Speed rounds (one on one and category vs. category)
3) Counting and Probability Seminar (I go through all the old tests and extract all the problems related to the topic. I explained the basics, then each team solve their own probability problems then present the hard ones).
4) Musical chairs (where the ones left standing up had to write their memorized topic formulas- that is given on an index card- on the board)

I ran out of ideas so fast that I am freaking out now. What are other FUN and productive ways to practice for such contests on this format? It is really hard because the team is divided among FOUR topic areas (algebra-precalc) and FOUR subs (in-progress and accelerated).

I am a co-captain and I really want the math team practices to be "better than math class" .

Any ideas for me? Thanks!

Well, since you had a C/P seminar, you might as well have a basic number theory seminar  :D
sorry, that wasnt very creative

actually, i would say occasionally take a break and do some challenge problems. Although it isn't from the contest, it would indirectly help them by improving their problem solving skills. they like challenge problems, right?  :maybe:


algorithm
number theory open
number theory

Hi , i'm not god on mathematics sow please help me.I'm working on software who must calculate an algorithm :
    i have 16 rows with 3 values and i have to calculate the maximum of options for this algo.
  ex:   1.   a1  a2  a3
          2.   b1  b2  b3
          3.   c1  c2   c3
          ....................
          ....................
        16.   q1  q2  q3
 the algo is like the game for socker .you must choice 1 value for 1 string : a1
 or a2 or a3 for first string so on until the end.So , my problem is what algo to use ?Please help me bigs..!!!


Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be such that

$ f(\frac{x\plus{}y}{2})\plus{}f(\frac{2xy}{x\plus{}y})\equal{}f(x)\plus{}f(y)$ for all positive $ x,y$. Prove that

$ 2f(\sqrt{xy})\equal{}f(x)\plus{}f(y)$ for all positive $ x,y$.

Not sure if the following link gives the answer.
http://www.mathlinks.ro/viewtopic.php?t=241518


is it true or not:if  k=/=1 there is n>1 such that 2^n =k(modn) are there infinitely many such n?

I think that this problem is an open problem, i dont think anybody has found a counterexample, but i dont think the proof is known.


[quote="dondigo"]Is is true that: ${{ktp^{l}}\choose{k}}\equiv{0}\pmod{p}$ where $p$ is a prime and the rest are positive integers :?:[/quote]

Is it true. It is also true that $\binom{ap^{b}}{c}\equiv 0 \pmod{p}$ for $p | \frac{ap^{b}}{c}$, because $\binom{ap^{b}}{c}= \frac{ap^{b}}{c}\times \prod_{i=1}^{b-1}{\frac{ap^{b}-i}{i}}$, and since $p^{x}| i \Rightarrow p^{x}| (ap^{b}-i)$, thus if $p | \frac{ap^{b}}{c}$ then $p | \binom{ap^{b}}{c}$. The original is a trivial consequence of this.


Solve in the equation.

$ [\frac{3x^2}{x^2\plus{}x\plus{}1}]\equal{}x$,$ x\in R$ :blush:

__________________________________
$ \textbf{Azerbaijan Land of Fire }$

Let $ f(x)\equal{}3x^2/(x^2\plus{}x\plus{}1)$. First note that $ 3x^2\ge0$ and $ x^2\plus{}x\plus{}1\equal{}(x\plus{}1/2)^2\plus{}3/4>0$, so $ \min{f(x)}\equal{}0$ when $ x\equal{}0$. Furthermore,
\[ f(x)\le4\Longleftrightarrow(x\plus{}2)^2\ge0,\]so $ \max{f(x)}\equal{}4$ when $ x\equal{}\minus{}2$. Thus $ \lfloor{f(x)}\rfloor\equal{}x\equal{}\{0,1,2,3,4\}$.

[list][*]$ x\equal{}0\implies\lfloor{f(x)}\rfloor\equal{}0$, which works.
[*]$ x\equal{}1\implies\lfloor{f(x)}\rfloor\equal{}1$, which works.
[*]$ x\equal{}2\implies\lfloor{f(x)}\rfloor\equal{}1$, which does not work.
[*]$ x\equal{}3\implies\lfloor{f(x)}\rfloor\equal{}2$, which does not work.
[*]$ x\equal{}4\implies\lfloor{f(x)}\rfloor\equal{}2$, which does not work.[/list]
Hence $ x\equal{}\boxed{0,1}$.


geometry
calculus
integration
rectangle
analytic geometry
conics
parabola

Is shoelace theorem the only way to solve this problem other than graphing and counting squares:

What is the area of a triangle with vertices at the points (-8,5) (3,1) and (5,-4)

There are various other ways, but they're definitely more complicated for this kind of problem.

oh okay ty, i guess i will just learn the shoelace theorem  :(

Heron's formula simplifies fairly nicely in almost all cases with integral coordinates.

Shoelace gives nice answers because these polygons can be inscribed in a rectangle with integral sidelengths, and you only need to chip off rectangles with integral areas and triangles with area ending in .0 or .5.

Basically, the area of a polygon with only integral coordinates can be expressed in the form $\frac{C}{2}$, where C is an integer.

hmm... I know herons. which is the sq rt of semiperimeter times semi minus first side, semi minus 2nd, and semi minus 3rd.

It is kinda hard sometimes though cuz of ugly numbers yeah...

Shoelace? Is it hard to put in your head? Because everytime I see it I can't draw it myself with those diagonals. IF it gives nice numbers, maybe it's worth memorizing? im thinking about it  :roll:

[quote="bpms"]The other ways involve using the distance formula and heron's to get really ugly numbers which give a clean answer. I'm surprised that the shoelace never gives ugly numbers i.e. radicals.[/quote]

And you'll also be pretty surprised when you find that the area between two parabolas will usually turn out nice, like 17/3 or something.

Shoelace is worth learning, graphing and counting points would work if you wanted to use Pick's, but that takes a long time.

Shoelace is easiest unless you're dealing with three points such that 2 of them have either the same x or y coordinate.

oh okay. I will learn it this weekend and get it stuck in my head.

ALSO, WHEN given 4 coordinates for a tetrahedron, does shoelace work? or does a similar formula work?

I doubt it, how are you suggesting you go about finding the volume?

It works for all polygons, though, as long as all the coordinates are in the correct order.

When would you ever have the coordinates for the vertices of a tetrahedron and be asked to find the surface area?

@bpms:  You shouldn't be surprised that it almost never gives radicals (unless you start with radical coefficients) because if the coordinates are all integers, then the worst thing that could happen would be a fraction  :o

You can also use matrices to find the area of a triangle.

$I+\frac{B}{2}-1$ is equal to the area of any polygon in the coordinate system, where $I$ is the amount of integer coordinates contained inside the polygon and $B$ is the amount of integer coordinates contained on the perimeter of the polygon.

I think you can prove the Shoelace's theorem by thinking the sides of polygon as vectors, calculate ther crossproducts, and summing up.

[quote="Totally Zealous"]Is there a proof for the shoelace theorem?[/quote]

Well obviously, as it's a theorem.  :wink:

Google it, you'll find results.

hmm i've never heard of this theorem before. What is it?

[quote="13375P34K43V312"]Google it, you'll find results.[/quote]

Hey just wondering is it also called shoelace [i]algorithm[/i]?
The person that taught it to me said it was the shoelace algorithm.

Lots of questions, here's some answers:

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

[quote="now a ranger"]on my fact sheet i got from school it says you can find the volume of a tetrahedron with vertices x1 y1 z1, x2 y2 z3, and x4 y4 and z 4

is volume equals 1/6 times the absolute value of x1 y1 z1 1
                                                                 x2 y2 z2 1
                                                                 x3 y3 z3 1
                                                                 x4 y4 z4 1[/quote]

Have you learned vectors?  This is derived directly from the vector product of the three defining vectors of a tetrahedron.

[quote="13375P34K43V312"]Call it whatever you want.

For some odd reason bpms calls the Fermat-Euler theorem Euler's totient. You can call things whatever you want, just know how to use it and (hopefully) why it works.[/quote]
That is how it was referred to when I first saw it. I have seen it called Fermat-Euler and Euler's generalization.

i know a little bit about vectors, just vaguely

yeah, just like for example in algebra. MY teacher last year called the point slope form the "equation writer"

Or like slope intercept the slopin but that's just kinda the south i guess

[quote="now a ranger"]

yeah, just like for example in algebra. MY teacher last year called the point slope form the "equation writer"

Or like slope intercept the slopin but that's just kinda the south i guess[/quote]

LOL.

[quote="goldendomer"][quote="bpms"]Shoelace is much more powerful than pick's[/quote]

I agree.[/quote]

I disagree.  Pick's Theorem can be used to prove a surprising number of very powerful results in some very unexpected areas.  For example, it makes for some very elegant proofs in the theory of continued fractions.

Here's an example of what I'm talking about:

Given positive integers $a, b, c, d$ such that $ad-bc = 1$, show that there does not exist a rational $\frac{b}{a}< r < \frac{d}{c}$ with a denominator (in lowest terms) less than $a+c$.

Duke number I8 was easily done with an application of Shoelace followed up by Pick's:

The hexagon defined by the 6 points $(i, 2^{i}-1)$, where i takes the integer values 0 - 5 has how many lattice points in its interior?

Can you post the solution with Picks & shoelace?  That's not the way I did it...

[hide="A sketch"] Find the area by Shoelace.  The only possible non-vertex border points are on the line between $(0, 0)$ and $(5, 31)$ (there aren't any), so by Pick's we can calculate the number of interior lattice points. [/hide]

Yeah, what he said.  Coordinates will give you area by shoelace.  Then use the area (K) in Pick's:

K = I +B/2 - 1

(I = interior points, B=boundary points)

Obviously there are only 6 boundary points, the given ones:  Since the line between two consecutive points only goes over by 1, while it goes up a lot, it won't hit any lattice points until it goes fully over 1 (to the next given point).

pick's theorem is pretty cool, but I messed up counting the dots under time control once.  :(

can someone please post with latex the shoelace theorem. I have to make sure I don't have it wrong.

It really isn't LaTeX.

You go around the polygon and you write all the coordinates down in order, in two neat columns, starting and ending at the same point.

Then you multiply diagonally. You take the first x-coordinate and you multiply it by the second y-coordinate; the second x by the third y; and so on.

Add all these numbers up.

Do the same thing the other way: the first y-coordinate with the second x-coordinate, etc.

Add all these numbers up too.

(If you've been drawing lines between the numbers you're trying to multiply, you should see a row of X shapes that look like shoelaces.)

Subtract the two totals and divide by two.

yeah i knew that part lol

that's why its calle shoelace, and u multiply by 1/2 when ur done adding and subtracting.

[quote="13375P34K43V312"]And you'll also be pretty surprised when you find that the area between two parabolas will usually turn out nice, like 17/3 or something.[/quote]

Yes, and the same goes pretty much for all polynomials of even degree (and odd degree for that matter, but then you have to define horizontal boundaries...), as long as the exponents are all integers.  But that's all Calculus anyway.

Btw, I personally think that the Shoelace Theorem has more applications, but I can see where Pick's would be useful too...


geometry
circumcircle
incenter
geometric transformation
reflection
power of a point
radical axis

Let $ (I_a), (I_b), (I_c)$ be the excircles of triangle $ ABC. (I_a)$ touches $ AB, AC$ at $ C'_a, B'_a$, similar to $ B'_c, A'_c, C'_b, A'_b$. Let $ O_a, O_b, O_c$ be circumcenters of triangle $ AB'_cC'_b, BC'_aA'_c, CA'_bB'_a$. $ l_a$ be a line which passes through $ I_a$ and perpendicular to $ O_bO_c$. Simiar to $ l_b, l_c$. Prove that $ l_a, l_b, l_c$ are concurrent.

Let $ (O)$ to the circumcircle and $ (I)$ incircle of the $ \triangle ABC.$ Incircle $ (I)$ touches $ BC, CA, AB$ at $ D, E, F.$ Excircles $ (I_a), (I_b), (I_c)$ touch $ BC, CA, AB$ at $ D_a, F_b, F_c.$ Internal angle bisectors $ AI, BI, CI$ cut $ (O)$ again at $ X, Y,Z.$ Perpendicular bisectors $ OX, OY, OZ$ of $ BC, CA, AB$ cut these at  their midpoints $ A', B', C'$ and the circumcircle $ (O)$ again at $ X', Y', Z'.$

[color=red]Lemma 1:[/color] Perpendicular bisectors of $ AD, BE, CF$ cut segments $ A'X', B'Y', C'Z'$ at their midpoints $ P_a, P_b, P_c.$ Similarly, perpendicular bisectors of $ AB_c', BC_a', CA_b'$ cut sgments $ A'X, B'Y, C'Z$ at their midpoints $ O_a, O_b, O_c.$

Parallel of $ BC$ through $ A$ cuts $ (O)$ again at $ A''.$ By Ptolemy for $ AA''BC,$ $ \overline {AA''} \equal{} \frac {c^2 \minus{} b^2}{a}.$ Let $ ID$ cut $ AA''$ at $ D''$ and tangent of $ (O)$ at $ X'$ at $ X''.$ $ \overline {X'X''} \equal{} \overline {A'D} \equal{} \frac {_1}{^2} \overline {D_aD} \equal{} \frac {_1}{^2}(c \minus{} b).$

$ \minus{} \overline {D''A} \cdot \overline {D''A''} \equal{} \frac {_1}{^4} (\overline {A''A} \minus{} \overline {D_aD}) \cdot (\overline {A''A} \minus{} \overline {D_aD}) \equal{}$
$ \equal{} \frac {_1}{^4}\left(\frac {(c^2 \minus{} b^2)^2}{a^2} \minus{} (c \minus{} b)^2\right) \equal{} \frac {p(p \minus{} a)(c \minus{} b)^2}{a^2}$

$ \minus{} \overline {D''D} \cdot \overline {D''X''} \equal{} \overline {DD''} \cdot (\overline {A'X'} \minus{} \overline {DD''}) \equal{} h_a \cdot \left(\frac {r_b \plus{} r_c}{2} \minus{} h_a\right) \equal{}$
$ \equal{} \frac {2pr}{a} \left(\frac {pr}{2(p \minus{} b)} \plus{} \frac {pr}{2(p \minus{} c)} \minus{} \frac {2pr}{a}\right) \equal{} \frac {p^2r^2}{a^2} \cdot \frac {(c \minus{} b)^2}{(p \minus{} b)(p \minus{} c)}$

$ \Longrightarrow$ $ \overline {D''A} \cdot \overline {D''A''} \equal{} \overline {D''D} \cdot \overline {D''X''}$ $ \Longrightarrow$ $ AX''A''D$ is cyclic and $ D_a \in \odot (AX''A''D)$ by symmetry. Its circumcenter is the common midpoint $ P_a$ of $ D_aX''$ and $ A'X'.$ Therefore, $ P_a$ is circumcenter of $ \triangle ADD_a$ and similarly, $ O_a$ is circumcenter of $ \triangle AB_c'C_b'.$ $ \square$

[color=red]Lemma 2:[/color] Nagel lines $ AD_a, BE_b, CF_c$ of $ \triangle ABC$ are perpendicular to $ P_bP_c, P_cP_a, P_aP_b.$ Similarly, Gergonne lines $ AD, BE, CF$ of $ \triangle ABC$ are perpendicular to $ O_bO_c, O_cO_a, O_aO_b.$

Let $ (P_b)$ be circle with center $ P_b$ passing through $ A, C$ and let $ (P_c)$ be circle with center $ P_c$ passing through $ A, B.$ Let $ (P_b), (P_c)$ cut $ BC$ again at $ D_b, D_c,$ respectively. Powers of $ B, E$ to $ (P_b)$ are both equal to $ \minus{} (p \minus{} a)(p \minus{} c)$ and likewise, powers of $ C, F$ to $ (P_c)$ are both equal to $ \minus{} (p \minus{} a)(p \minus{} b)$ $ \Longrightarrow$

$ \overline{D_aC} \cdot \overline{D_aD_b} \equal{} \overline{D_aC} \cdot (\overline{CB} \plus{} \overline{BD_b} \minus{} \overline{CD_a}) \equal{}$ $ \minus{} (p \minus{} b) \left(a \plus{} \frac {(p \minus{} a)(p \minus{} c)}{a} \minus{} (p \minus{} b)\right) \equal{} \minus{} \frac {(p \minus{} b)(p \minus{} c)}{a}$
$ \overline{D_aB} \cdot \overline{D_aD_c} \equal{} \overline{D_aB} \cdot (\overline{BC} \plus{} \overline{CD_c} \minus{} \overline{BD_a}) \equal{}$ $ \minus{} (p \minus{} c) \left(a \plus{} \frac {(p \minus{} a)(p \minus{} b)}{a} \minus{} (p \minus{} c)\right) \equal{} \minus{} \frac {(p \minus{} b)(p \minus{} c)}{a}$

$ \Longrightarrow$ $ AD_a$ is radical axis of $ (P_b), (P_c)$ $ \Longrightarrow$ $ AD_a \perp P_bP_c.$ $ \square$

[color=red]Lemma 3:[/color] I-symmedians of the in/excentral triangles $ \triangle II_bI_c,\triangle II_bI_c,\triangle II_bI_c$ are parallel to Nagel lines $ AD_a, BE_b, CF_c$ of their common orthic triangle $ \triangle ABC.$ Similarly, $ I_a, I_b, I_c$-symmedians of the excentral triangle $ \triangle I_aI_bI_c$ are parallel to Gergonne lines $ AD, BE, CF$ of its orthic triangle $ \triangle ABC.$

Let $ D'$ be midpoint of $ AD.$ Newton line $ A'D'$ of degenerated tangential quadrilateral $ ABDC$ with diagonals $ AD, BC$ goes through its incenter $ I.$. Therefore $ (A'I \equiv AD') \parallel AD_a.$ is midline of $ \triangle AD_aD.$ $ BC$ is antiparallel of $ I_bI_c$ WRT $ \angle I_cII_b.$ The I-symmedian of $ \triangle II_bI_c$ has to cut this antiparallel in half $ \Longrightarrow$ $ A'I$ is the I-symmedian of this triangle.
________________________________________________________________

Midpoints $ O_a, O_b, O_c$ of $ A'X, B'Y, C'Z$ from lemma 1 are circumcenters of the $ \triangle AB_c'C_b', \triangle BC_a'A_c', \triangle CA_b'B_a',$ respectively. By lemmas 2, 3, perpendiculars $ l_a, l_b, l_c$ to $ O_bO_c, O_cO_a, O_aO_b$ through $ I_a, I_b, I_c$ are symmedians of the excentral triangle $ \triangle I_aI_bI_c,$ concurrent at its symmedian point. $ \square$

Dear Yetti,
Thanks you for your solution, but I think that using radical center can be shorter  :)

Let $Y_a,Y_b,Y_c$ be the tangency points of  $ (I_a),(I_b),(I_c)$ with $ BC,CA,AB.$ Note that circles $ (O_a),(O_b)$ and $(O_c)$ are the reflections of the circles $ \odot(AY_cY_b),$ $\odot(BY_aY_c)$ and $\odot(CY_aY_b)$ about the external bisectors of $ A,B,C.$ Circles $\odot (BY_aY_c), $ $\odot(CY_aY_b)$ obviously pass through the Bevan point $ B_e$ of $\triangle ABC,$ namely the concurrency point of the perpendiculars from $ I_a,I_b,I_c$ to $BC,CA,AB.$ Therefore, $ I_a$ has equal power WRT $(O_b),(O_c)$ $\Longrightarrow$ $ \ell_a$ is radical axis of  $ (O_b),(O_c).$ Mutatis mutandis, $ \ell_b,\ell_c$ are radical axes of the pairs $ (O_a),(O_c)$ and $ (O_a),(O_b)$ respectively $\Longrightarrow$ $ \ell_a,\ell_b,\ell_c$ meet at the radical center of $(O_a),(O_b),(O_c)$ (Mittenpunkt of ABC).


calculus
integration
calculus computations

Evaluate $\int \int_R (x-1) dA$, where $R$ is the region in the first quadrant enclosed between $y = x$ and $y = x^3$.

So I did

[hide]
$\int_0^1 \int_{x^3}^x (x-1) dy dx$ but I got the wrong answer (according to the back of the book, anyway)... is there something wrong with this? [/hide]

The answer better be negative -- $x - 1$ is negative over that entire region.

[quote="Bert"]I can try:

$\int^1_0\int^x_{x^3}(x-1)dA=\int^1_0[(x-1)y]^x_{x^3}dx=\int^1_0((x-1)x)-((x-1)x^3)=$

$\int^1_0x^2-x-x^4+x^3=[\frac{1}{3}x^3-\frac{1}{2}x^2-\frac{1}{5}x^5+\frac{1}{4}x^4]^1_0=$

$\frac{1}{3}-\frac{1}{2}-\frac{1}{5}+\frac{1}{4}=-\frac{7}{60}$

Strainge that I find A negatief answer is this ok? What is the solution given by your book?

Greets.[/quote]

Yes, that's what I got, too. The book says $-\frac{1}{2}$, which I guess is wrong; it confused me very much.


I'm looking for a list with more new and interesting tricks, more specifically on things that can be done with the forum-latex configuration on MathLinks.

Some tricks I recently discovered include [list][*]\to can be used instead of \rightarrow[*]$\text{sgn}(t)=\begin{cases}1&\text{ voor }t>0\\0&\text{ voor }t=0\\-1&\text{ voor }t<0\end{cases}$ using \begin{cases} instead of working with \left\{\begin{array}[*]\begin{matrix} instead of also with array[*]\buildrel & \stackrel commands[*]the use of \limits for better sub&super-scripting.[*]...[/list]
No world-shocking new things, but I keep discovering new things over and over and I wondered: there has to be a finite list of these, isn't it? Where can I find such a list?

Again, only taking the packages from this forum into account, else the list would of course not be finite as an infinity of packages can be made.

Most of things you mention either don't use packages or use the amsmath package which is implemented here (you can read what it can do in its [url=http://tug.ctan.org/tex-archive/macros/latex/required/amslatex/math/amsldoc.pdf]User Guide[/url]). Also I strongly recommend getting hold of a copy of [url=http://www.amazon.co.uk/-Guide-LATEX/dp/0321173856/sr=8-1/qid=1158007648/ref=sr_1_1/202-0137577-3200634?ie=UTF8&s=gateway]Kopka & Daly's A Guide to Latex[/url]. Yes there's far more than can be used on a forum, but it is an excellent guide to typing LaTeX mathematics.

Fun exercise: Without looking at the code, can you see how this is done with just one command: $\genfrac{)}{[}{4pt}{2}{1}{2}$
The answer is in both publications above.


The never beginning story. We start at the end of the story and the next person writes a setence or two that should occur immediately before the last post.

And so, Alice and Bob learned the secret of unbreakable encryption.

The great lord of the universe threw them into Nottingham where Robin Hood was.

... which would enable them to escape. Unfortunately, this scheme did not turn out feasible because shortly after that ...

Carmen tried to help them by trying to break the encryption but died trying to do so, thus distracting the great Lord of the Universe...

(By the way, let's keep a rule like in the other story, no less than 5 words per "sentence" and no more than 25 words)

Things were looking rather bleak for them at the moment, what with the reinforcements cut off and the Great Lord returning victorious.

Iversonfan2007 got owned by Max300 and felt bad so...

but then he started watching the match between max300 and iversonfan 2007
[color=green]
[size=42](edited by mod, your post is supposed to come [i]before[/i] the previous one)[/size][/color]

...Dino, Alice's brother, invested in google two years ago, so

Alice, Bob, and Carmen had invested in Enron, and were now bankrupt and in a cage under the control of the great lord.... However,

...Alice, Bob, and Carmen each wanted to invest in an energy company, so

and now that Alice, Bob and Carmen were brought back to life, they were poor and

...Alice and Bob, along with their daughter Carmen, unwisely decided to bid their 30 million dollars they won in the lottery on a six-pack of Gatorade on EBAY.

...but fortune must have felt sorry for Bob, Alice, and Carmen because shortly after, they won the lottery  number 667! They received 30 million dollars, and It was the oddest coincidence ever, but

A long time ago, there lived the first humans - Alice and Bob - and there daughter Carmen. That was much before Adam and Eve, who were only the first in the universe, which was yet to be formed. One day they sat in their drawing room, discussing that at this rate the universe would never be formed.

so they quickly went to the computer and called dell tech support, but then the universe blew up and everything had to start at the [b]very[/b] beginning again.

alkara1, this is the never [b]beginning[/b] story. Read the first post of this thread and understand the rules before posting...

so we have...

[quote]and then... Carmn was tinkering on the picture machine with Alice and Bob, and the knew that if the saw Chuck, they will have to find the square root of 1681. Then, a picture of chuck norris appeared. They went to his computer to compute the square root of 1681, but The dell computer went bezerk, etc etc etc...[/quote]

Carmen said to herself, "Oh my, I may have to go look for a picture of Chuck Norris!"

SRRY

alright then,

Carmen loved chuck norris. she even had a chuck norris life-size house somewhere up in Canada. Then

Alice found herself seeing Carmen stare at a Chuck Norris picture in her sparetime. She also heard Carmen making long-distance phone calls to Canada, saying "He is a bit taller, make the house a bit taller."

[quote="akalra1"]alice is a stalker
she stalked carmen in her free time
alice was astounded at where carmen went: she once followed carmen all the way to canada, and she didn't know why she was there[/quote]

u do know that alice is carmen's mom, right?

Carmen wanted to know why Alice followed her around all the time. One day, she was browsing Myspaces when she saw, written on Bob's blog:

(you know, posts are supposed to be between 5-25 words  :wink: )


I was just looking at that one over the past couple days.  What a coincidence.  Unfortunately I don't have the answers with me.  I think my teacher has them so I can get them for you tomorrow if you still need them by then.

here they are:

E
E
A
C
E

E
B
C
D
B

B
D
D
C
B

D
D
D
B
E

A
B
C
D
D

C
C
B
B
B

Does anyone know a website that has  1991 ASHME problems?

i bet kalva does but anyways you can search "1991 AHSME problems" and you're bound to find something.

I thought that the easiest stuff that Kalva had was Aime(hence no ASHME/AMC).


limit
logarithms
floor function
real analysis
real analysis unsolved

Is it true that:
\[ \lim_{n \rightarrow \infty} ( \{ H_n \} - \{ \ln n \} ) = \gamma \]
Where $H_n$ is the n-th harmonic number $\gamma$ is Euler's constant and $\{x\}$ denotes the fractional part of a real number $x$.

No, it won't be true. Look at the integer parts:

$\lim_{n\to\infty}\lfloor H_n \rfloor=\lim_{n\to\infty}\lfloor \ln n\rfloor=\infty$

so each sequence crosses over every positive integer. But they won't cross over the integers at the same time.

There are infinitely many $n$ for which $\lfloor H_n \rfloor=\lfloor \ln n\rfloor$ and there are infinitely many $n$ for which $\lfloor H_n \rfloor\ne\lfloor \ln n\rfloor.$

Thus the sequence of the differences of the fractional parts should have two cluster points.

In order to prove that, do you just use the asymptotics for harmonic numbers (in the sense that you prove that {ln n}>1-y>0.4 infinitely often), or do you have a more ingenious proof ?


ARML
probability
geometry
trigonometry
analytic geometry
algebra
polynomial

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

[hide="1"] It can be rewritten into 1*2-1^2+2*3-3^2+..+n(n+1)-n^2, so that just becomeds 1+2+3+4+5..+n=210, where n is the kth term.  So from the sum formula, we get n^2+n=420 (skipped a couple of trivial steps), since this is arml, its more practical to guess and you can easily guess that n=20, so the there are [b]20 [/b]terms.
[/hide]
[hide="2"] Huh, this could be the easiest problem but I was thinking way to hard.  Its the minium distance of the outer square minus the maximum distnace of the inner square.  The minimum disttance of the outer square is just $30/\sqrt2$, which is $15\sqrt2$.  The maximum distance of the inner square is just $10\sqrt2$ (its just all about the diagonals of the two squares-they are 45 degrees).  So $15\sqrt2-10\sqrt2=5\sqrt2$..Initially I thought the squares all had to share the same center so I got confused.[/hide]

[hide="5"]Denote kings=K, hearts=H, nonkings=NK, nonhearts=NH, nonheartsandkings=NHK, so we count by 1-ones that dont belong.  The ones that dont belong, there are (36)(35)-for both cards being NHK, (3)(38)(2)-for one being K and the other being NH (there are onnly 3 kings that work because the king of hearts would satisfy this), (12)(47)(2)-for one being H and the other being NK (again you cant count king of hearts).  You mulitpy two for the previous 2 equations because you can get the king or heart on the first or second draw.  However, if you realize, you have actually overcounted something.  When we have the combination K(obiviously this king is nonheart) and NH and NH and K, you have counted the combination K and K (these kings are also nonheart) twice (it will satisfy both categories), same with when you have N and NK and NK and H, you have overcounted the H and H twice, so we subtract (3)(2) and (12)(11), so its (36)(35)+(3)(38)(2)+(12)(47)(2)-(3)(2)-(12)(11).  Put that over 52*51, and that is 413/442, so 1-413/442=[b]29/442[/b][/hide]
[hide="6"] $21+12\sqrt3$ can be rewritten as $(3+2\sqrt3)^{2}$, so the equation simplifies to $\sqrt(4+2\sqrt3)=\sqrt(a)+\sqrt(b)$, so squrting everything we get $4+2\sqrt3=a+b+2*\sqrt(ab)$, so we have a+b=4 and ab=3, easily (if you dont want to use the qudratic formula) we see that b=3 and a=1 (a is less than b) so [b](1,3) [/b]
[/hide]

maybe ill do rest later...number 8 can just be written as sin(4a)=24/25 where a is an angle, and you can derive the quadruple angle formula from sin(2a+2a) and so on..(I think there is a better way using sigmas to find sin(4a) but I forgot...)

I agree with indianamath lol...r these past arml problems? cuz if they are then i'm not goin this year lol

Uh, once you both get into 9th grade , you will both understand all of this.  The basic concept is that $tan^{-1}x=angle$, and $tan(tan^{-1}x)=x$ when defined on a certain interval, and you can then find cos and sin of that angle using the fact that sin=y coordinate/radius cosine=x coordinate/radius and r=sin^2+cos^2 (this is why SOCOTOA (sp?) works)

And acutally you both are problaby better than me at stuff like amc 10

yeah.. SOHCAHTOA
Sine = Opp/Hyp
Cosine = Adj/Hyp
Tangent = Opp/Adj

though i never use it

for number 8, you use something called potential rational zeros, which are:

for any polynomial $ax^{n}+bx^{n-1}+\cdots+z$

the PRZ's are $\pm \frac{\mbox{factors of z}}{\mbox{factors of a}}$

so after dividing through by 4 to get $6x^{4}+25x^{3}+12x^{2}-25x+6 = 0$

you can get the PRZ's: $\pm \frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1, \frac{3}{2}, 2, 3, 6$

now comes the tedious part - divide the original polynomial by one of these, and any of these, until you factor it completely.

umm after that, voila, you get $(x+2)(x+3)(2x-1)(3x-1) = 0$

and the roots are $-2,-3, \frac{1}{2}, \mbox{and}\frac{1}{3}$

undefined:

I checked all four answers that you got with a caculator and they all work, so I think those are the 4 right answers.
However, even though I got the right equation, I guareentee you I wouldnt have gotten all 4 right because Im horrible at fast computation, so my way is prbolbay not the best way.

[quote="stupidkid"][quote="Art of Owna"]wtf for number 8..I can get to the very last step but I cant find the roots of a quartic polynomial.

So $sin(4tan^{-1}x)=24/25$, we can call $tan^{-1}x$ as an angle a.  So we find sin(4a), tahts is sin (2a+2a), which is 2sin2acos2a, which is $4(sina)(cosa)(2cos^{2}a-1)$, which simplifies to $sin(4a)= 8(sina)(cos^{3}a)-4(sina)(cosa)$. This is not so bad once you find sin a. Since we know tan a=x, from the fact that tan is y coordinate/x coordinate, we know that the y coordinate is x, x coordinate is 1, the radius must be $\sqrt(x^{2}+1)$. we can then find cosine and sine of x (cosine is x/r, sine is y/r). Plug that in and we actually find the square roots disappear and we get the fraction $(4x-4x^{3})/(x^{4}+2x^{2}+1)=24/25$.  Cross multiplying, that simplyfies to $24x^{4}+100x^{3}+48x^{2}-100x+24=0$  Now the roots of that eqution are your answers..good luck finding that.[/quote]
wtf? How can you not know how to find the roots of a quartic polynomial? Geez, how do you not know the quartic formula????!! http://planetmath.org/encyclopedia/QuarticFormula.html

Oh btw, its \sin \cos \tan for latex.[/quote]
lmao yeah and we'll all memorize that :rotfl: nice.

[quote="stupidkid"][quote="undefined117"]Maple $>$ TI-89 $\geq$ Art of Owna

 :D[/quote]

Whats with the $\ge$? Equality holds iff doomsday is upon us.[/quote]
lol

well according to my math teacher, "your calculator is only as smart as you are"

or something along those lines.

[quote="stupidkid"][quote="undefined117"]Maple $>$ TI-89 $\geq$ Art of Owna

 :D[/quote]

Whats with the $\ge$? Equality holds iff doomsday is upon us.

Dude Art of Owna, how come you weren't at the top 25 banquet? There were a lot of freshman n00bs.[/quote]
wtf i got all As and an A- in honors bio (I already posted somewhat like this in the "what about other subjects besides math") because I got a B+ on a grading period and she doesnt do percent so there was no way I could get an A in that class after getting the B+.  So my GPA's screwed (3.974 including this year and some middle school stuff they added), meaning I wont be getting a scholorship to some prestigious college...

But there were only 5 asians i think (in top 25), only 2 male asians (one is just specialized at school stuff so he'll get into like northwestern or michigan.., jimmy's actually pretty good at other stuff-he was 1 point away on amc 12 and 1.5 on amc 10 from passing, and he acutally does sports-lol he was my tennis doubles partner on team..most annying kid ever though)

[quote="generating"][quote="undefined117"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8

if i ever get on the ARML team.. i'm hopeless.[/quote]

These are the problems from the individual round of the 1979 ARML event. I don't know how the results were that year, but I-8 is always meant to be a back-breaker of a problem.

Sometimes they're hard because they're tedious and have a load of boring calculations (like this one, or like the one from last year). Sometimes, though, they require a brilliant insight and things work out beautifully if approached from the proper point of view (look at the 1997 and 1998 tests).[/quote]

but there must be an easier way for number 8?  I mean im pretty sure they wont be that mean and have you use the ratioan root theorm and guess the roots in less than 10 mins.

[quote="Art of Owna"][quote="undefined117"]heh but it's pretty sad how i have NO CLUE as to how you got to where you were on number 8

if i ever get on the ARML team.. i'm hopeless.[/quote]
but you passed AIME..so you must have to be good at math (i dont know a single person that passes AIME and sucks at math)[/quote]
no really. i suck at mathcounts, especially at countdown. i REALLY should start reading up on the AoPS books :blush:

but otherwise, i usually need some time to think about my strategies and stuff.

[quote="stupidkid"][quote="Art of Owna"]wtf i got all As and an A- in honors bio (I already posted somewhat like this in the "what about other subjects besides math") because I got a B+ on a grading period and she doesnt do percent so there was no way I could get an A in that class after getting the B+.  So my GPA's screwed (3.974 including this year and some middle school stuff they added), meaning I wont be getting a scholorship to some prestigious college...

But there were only 5 asians i think (in top 25), only 2 male asians (one is just specialized at school stuff so he'll get into like northwestern or michigan.., jimmy's actually pretty good at other stuff-he was 1 point away on amc 12 and 1.5 on amc 10 from passing, and he acutally does sports-lol he was my tennis doubles partner on team..most annying kid ever though)[/quote]
Dude chill out, GPA means nothing, don't be like those Asians who cry over a non A. I have 2 A- and 1 B+, although I'm ranked 24th, so barely in top 25. Btw, top colleges, e.g. Harvard, MIT, Stanford, etc, [i]do not[/i] give out merit based scholarships, only financial aid. You get scholarships from other organizations, such as Siemens and they don't give about your GPA (well some probably do, but those are worthless scholarships anyways).[/quote]

I was being sarcastic about the scholorship, I dont cry over not getting a non-A, and Im not like all those other Asians in general (sterotype..)...I actually told my mom when I got that GPA that I dont really care about my GPA and Ill just make it up with some award (like qualification for USAMO or like Intel and stuff), but she just said GPA is the easiest thing to get, along with National Finalist.  Yea and like I said, if you just get 4.0s, you'll go to a respectable school (Northwestern is actually a good example), but you wont go to any of the top colleges because those colleges value abilities/talent more than your GPA.  For example, Harvard isnt looking for future students that will become ordinary doctors or lawyers that make 200,000 a year (which is way above average still), but they are looking for students that will become future millionaires and billionarires that will be come famous that will give Harvard perstige (becuase no one is going to hear about a ordinary doctor, but if that person becomes famous, then Harvard will be mentioned even more).  4.0=ordinary hardworking, dedicated doctors, lawyers,etc...not bad)
National Award or special talent=better chance at becoming famous and that is what top colleges are looking for out of their alumni (pres. bush couldnt go to yale with his grades..but his leadership qualities and money made yale want him because of what he can and already was)
I mean harvard knows how to find talent-for better or worse (Its got some of the most famous and infamous alumni..but their still well-known)

If I sounded like I was mad or something or wining..that wasnt really inteneded (I mean to be honest I was disappointed but I really dont care about it that much-definenlty not as much as all those other Asians in my grade-or even americans-who begged my orchestra teacher for a higher participation grade..-hmm*cough* udayan?  atealst you wenret the only one)

Meh, dunno about the billionaires part. Those people are just born with the ability to make money. If you look at the world's richest 20 people (forbes.com), I think half of them never even finished college! o.O One dude didn't even [i]go[/i] to college.

Oh btw, USAMO alone is definitely not enough these days to get you into a top school. The competition is crazy now. Go ask Mohit, he'd know.

Yo, Art of Owna, wanna join the CHS webteam? We need some underclassman who either knows Linux or is willing to learn (so that means you'll need to install it and spend however long it takes you to get fluent with it, it took me probably 50 hrs in 2 weeks), and it's Gentoo Linux, one of the hardest to learn. If no one knows Linux, carmelhighschool.net is screwed after I graduate.

Oh also, (damn its the 4th edit), you can only participate in the Intel Talent Search when you're a Senior, so it's too late for college apps anyways. Just fyi.

[quote="stupidkid"]Meh, dunno about the billionaires part. Those people are just born with the ability to make money. If you look at the world's richest 20 people (forbes.com), I think half of them never even finished college! o.O One dude didn't even [i]go[/i] to college.

Oh btw, USAMO alone is definitely not enough these days to get you into a top school. The competition is crazy now. Go ask Mohit, he'd know.

Yo, Art of Owna, wanna join the CHS webteam? We need some underclassman who either knows Linux or is willing to learn (so that means you'll need to install it and spend however long it takes you to get fluent with it, it took me probably 50 hrs in 2 weeks), and it's Gentoo Linux, one of the hardest to learn. If no one knows Linux, carmelhighschool.net is screwed after I graduate.

Oh also, (- its the 4th edit), you can only participate in the Intel Talent Search when you're a Senior, so it's too late for college apps anyways. Just fyi.[/quote]
Uh..I dont know Linux..you can go ask udayan (that new indian kid in that sunday class)-he apparently thinks he knows alot of computers..uh I can probalby learn it, but Im pretty busy these days...but Im taking comp sci next year, if that has anything to do with it..

[quote="stupidkid"]...

How are you busy as a frosh? Man, I didn't do anything freshmen year, just kinda slacked off and played some starcraft.[/quote]

I need to be more busy now..I havent done my homework at home (SRT plus orchestra=free SRT) for the past 6 weeks..and it almost cost me another A-..
but i mean on weekends and stuff I have lots of things to do.  and now I have to volunteer at the library sometime (apparently you need that for college :o )
but i dont play games any more..havent played my XBOX in like 2 months..


modular arithmetic
quadratics
number theory open
number theory

Lets forget $3$ and write $k$ instead, and lets also forget that we have Mersenne primes and thus let $p$ be any odd prime.
Let $f(x) : = x + \frac{1}{x} \mod p$, then we want the size (lets call it $\eta_p(k)$) of the set $\{ f(k^n) | n \in \mathbb{N}\}$.

First lets find out how often $f(x) \equiv f(y) \mod p$ with $x,y \not\equiv 0 \mod p$ happens:
This means $x + \frac{1}{x} \equiv y + \frac{1}{y} \mod p$
$\iff x^2y+y \equiv xy^2+x \mod p$
$\iff (xy-1)(x-y) \equiv 0 \mod p$.
This means that either $x \equiv y \mod p$, the trivial case, or $xy \equiv 1 \mod p$.
But: when $x \equiv \pm 1 \mod p$, then only the case $x \equiv y \mod p$ can occure.

Look at the set $\mathrm{Pow}(k) : = \{ k^n \mod p | n \in \mathbb{Z}\}$ (we can use $\mathbb{Z}$ instead of $\mathbb{N}$ because of Fermat's Little Theorem).
It has size $|\mathrm{Pow}(k)| = ord_p(k)$. Additionally, we can pair up the elements $k^n \mod p$ and $k^{-n} \mod p$ for each $n$, since they give the same value $f(k^n) \equiv f(k^{-n}) \mod p$, and only those are equal (note that $1, -1 \mod p$ will be left allone, but each noted as "pair" with one element). Since different pairs give different values, we have that $\eta_p(k)= \text{\"number of such pairs\"}$.
Thus when $-1 \in \mathrm{Pow}(k)$ ($1$ is always in the set), there will be $\frac{ord_p(k)-2}{2}+2 = \frac{ord_p(k)+2}{2}$ pairs, thus by the above $\eta_p(k) = \frac{ord_p(k)+2}{2} \iff 2 \eta_p(k) = ord_p(k)+2$.
Similar when $-1$ is not in the set: $2 \eta_p(k) = ord_p(k)+1$

This for example gives $\eta_7(3) = 4$, but you stated $=3$ above  :?

To fnd out if $-1$ is in the set, we need to know if the order of $k \mod p$ is even or odd (this suffices to know: when $ord_p(k)$ would be odd, we couldn't have $2 \eta_p(k) = ord_p(k)+2 \mod 2$, and analogous for the other case).
When $s$ is the biggest integer with $2^s | p-1$, we could calculate $k^{\frac{p-1}{2^s}} \mod p$ (since $\frac{p-1}{2^s}$ is the biggest odd divisor of $p-1$) and look if it is $1 \mod p$ or not (the order is odd iff it is $1 \mod p$).
When $4 \nmid p-1$, we just ask whether $k$ is a quadratic residue $\mod p$ or not, which can be checked by Jacobi symbols.

Special case $k=3, p=2^q-1$:
Then $4 \nmid p-1$, thus we use Legendre symbols (Jacobi is not needed since both numbers are prime) and the law of quadratic reciprocity:
$\left( \frac{3}{2^q-1} \right) = - \left( \frac{2^q-1}{3} \right) = -1$.
This shows that the order of $3 \mod p$ is even.
Thus for Mersenne primes, it should always be $2 \eta_p(3) = ord_p(3)+2$, but the above says something different...

[quote="ZetaX"]Thus for Mersenne primes, it should always be $2 \eta_p(3) = ord_p(3)+2$, but the above says something different...[/quote]Okay. Seems PARI/gp does not say the truth or I mis-use it.
I used the program (given by maxal in GIMPS forum):
[code]{ tt(q,b) = local(M);
M=2^q-1;
length(Set(vector(2^(q-1),n,lift(Mod(b,M)^n+Mod(b,M)^(-n)))))
}[/code]The results given in post#1 have been computed with tt(q,3) for q =3..31 .
But the main question is not this mistake about 1 missing element for q=3,5,7,17,19 .

While computing $\eta_p(3)$ for q= 13, I've found a different result than the one stated by Shallit and Vasiga in their paper:
The DiGraph under $x^2-2$ modulo $M_{13}$ has 9 cycles of length 6.
Out of these 9 cycles, I've found that only 5 of these cycles have their elements generated by $g(n)=(3^n+1/3^n) \pmod{M_q}$. Here is a list of one element of each of these 5 cycles: 288, 101, 532, 920, 2627 . The 4 other 6-cycles starting with : 23, 722, 1812, 2255 have no element generated by $g(n)$, even when 0 <= n <= 8191.
Shallit and Vasiga say that all Cycles are generated by $g(n)$ ; computation shows that this is true for q=3,5,7,17,19 but not for q=13 and 31. Discussing with "maxal" in GIMPS forum, we came to compute $\eta(q)$ and $order(3,M_q)$.

For q=3,5,7,17,19, we have: $order(3,M_q) = M_q-1$ .

For q=13,31, we have: $order(3,M_{13}) = \frac{M_q - 1}{9} \ \ ; \ \ order(3,M_{31}) = \frac{M_q - 1}{3}$.

Why these formulae are different ?

Tony

Well, I don't speak PARI  :blush:

I haven't read the paper till now, but it is somewhat natural that not all are of the type above.
It's an open conjecture by Artin that every nonsquare (here $3$) is a primitive root $\mod$ infinetely many primes $p$.
As order of $3 \mod p=2^q-1$, we can expect some big divisors of $p-1$, but not that they are $=p-1$.
And I would expect that the Denominator of that fraction can get as big as they want (I would expect every odd integer to occure, or at least a multiple of every odd integer), but also get $1$ often.

[quote="ZetaX"]Well, I don't speak PARI  :blush: [/quote]The basic of this tool is easy.
I do not like the syntax (why did they not used a C-like syntax ?!), but it is really useful.

[quote]I haven't read the paper till now, but it is somewhat natural that not all are of the type above.
It's an open conjecture by Artin that every nonsquare (here $3$) is a primitive root $\mod$ infinetely many primes $p$.
As order of $3 \mod p=2^q-1$, we can expect some big divisors of $p-1$, but not that they are $=p-1$.
And I would expect that the Denominator of that fraction can get as big as they want (I would expect every odd integer to occure, or at least a multiple of every odd integer), but also get $1$ often.[/quote]Mersenne numbers are very special simple and unic numbers. I think we can expect to find simple properties.
I've computed $\frac{M_q-1}{order(3,M_q)}$ for more values of $q$ and here are the results, showing that  we have the conjecture:
\[ \frac{M_q-1}{order(3,M_q)}=3^n \ \ \text{with} \ \ n=0,1,2  \]
T.

[code]q  (Mq-1)/order(3,Mq)
-----------------
3    1
5    1
7    1
13   9
17   1
19   1
31   3
61   9
89   1
107  1
127  3
521  1
607  3
1279 3[/code]


On $ x\minus{}y$ plane, find the maximum length of the line segment involved the region (including the boarder) bounded by the curve $ y\equal{}x^2\minus{}4$ and the $ x$ axis.

may i ask, how come you like min and max problems sooooo much ?  :maybe: these days you seem to post questions on those things only...

so fix the point on the x axis at $ x\equal{}\alpha \ge 0$

we can draw a line from any point on the curve..
$ (x,x^2\minus{}4)$ to $ (\alpha,0)$

and maximize distance squared
$ d^2\equal{}(x\minus{}\alpha)^2\plus{}(x^2\minus{}4)^2\equal{}x^2\minus{}2\alpha x \plus{}\alpha^2\plus{}x^4\minus{}8x^2\plus{}16$

$ d_\alpha \equal{} 2(\alpha \minus{}x)\equal{}0$ then $ \alpha_c \equal{}0,x,2$
$ d_x \equal{}2x \minus{}2\alpha_c\plus{}4x^3\minus{}16x\equal{}4x^3\minus{}14x\minus{}2\alpha_c\equal{}2(2x^3\minus{}7x\minus{}\alpha_c)\equal{}0$

so critical points at $ 2x^3\minus{}7x\minus{}\alpha_c\equal{}0$

case 1: $ \alpha_c \equal{}0$

$ \to 2x^3\minus{}7x\equal{}0 \to 2x(x^2\minus{}\frac{7}{2})\equal{}0 \to x_c\equal{}0, \pm \sqrt{\frac{7}{2}}$

$ d^2(x_c)\equal{}x_c^2\plus{}x_c^4\minus{}8x_c^2\plus{}16\equal{}x_c^2(x_c^2\minus{}7)\plus{}16\equal{} 16, \frac{15}{4}$

case 2: $ \alpha_c\equal{}x$

$ \to 2x^3\minus{}8x\equal{}0 \to 2x(x^2\minus{}4)\equal{}0 \to x_c\equal{}0, \pm 2$

$ d^2(x_c)\equal{}(x_c^2\minus{}4)^2\equal{}16,0$

case 3: $ \alpha_c\equal{}2$

$ \to 2x^3\minus{}7x\minus{}2\equal{}0 \to 2(x\minus{}2)(x^2\plus{}2x\plus{}\frac{1}{2} )\equal{}0 \to x_c\equal{} 2, \minus{}1 \pm \frac{1}{\sqrt{2}}$

$ d^2(x_c)\equal{}(x_c\minus{}2)^2\plus{}(x^2\minus{}4)^2\equal{}0,$ or $ (\minus{}3\pm \frac{1}{\sqrt{2}} )^2\plus{}((\minus{}1\pm \frac{1}{\sqrt{2}})^2\minus{}4)^2\equal{}$

$ 9 \mp  3\sqrt 2\plus{} \frac{1}{2}\plus{} (1\mp \sqrt 2 \plus{} \frac{1}{2} \minus{}4)^2\equal{}\frac{19}{2} \mp 3\sqrt 2 \plus{}(\frac{5}{2}\pm \sqrt 2)^2\equal{}$

$ \frac{19}{2} \mp 3\sqrt 2 \plus{}\frac{25}{4} \pm 5\sqrt 2\plus{} 2\equal{}\frac{71}{8} \pm 2 \sqrt 2 < 16$ either way..

so max $ d^2$ is 16, meaning max distance is 4. either on the boundary or at x=0 straight up


Is it possible for the fundamental group to be finite but not trivial?

Sure; for example, the fundamental group of $ \mathbb{R} P^2$ is $ \mathbb{Z}/2\mathbb{Z}$.  In fact we can say much more: [b]every group[/b] is the fundamental group of [url=http://en.wikipedia.org/wiki/Eilenberg%E2%88%92MacLane_space]some space[/url].


geometry
rectangle
combinatorics proposed
combinatorics

"Knight-rook" can make horizontal or vertical move like rook,but after each horizontal move he must do vertical or vice versa.We arrange $ n$ traps on the chessboard $ 7*7$,such that "knight-rook" has only finite numbers of moves."Knight rook" gets in to trap if it stands on it ,or goes through the trap(Starting position of "knight-rook" is arbitrarily).Find the minimum value of $ n$.

Does anybody want to see a solution?I can post it, if someone wants. I know two solutions,one of them is mine, and the official one.

Here is my solution,
Assume that we have put $ k$ traps and the table satisfies the condition. Now turn the table into a graph such that its vertices are the squares which haven't got traps and put edges to the adjacent ones.
Lemma: If the number of edges of a simple graph is greater than or equal to its number of vertices,  then it contains a cycle. It is a well-known fact.
At first we got $ 2.7.6$ edges and we have deleted some of them after the trapping operation. A trap can delete at most $ 4$ edges. Another point is we must put at least one trap to the edges of the table and this can delete at most $ 3$ edges. So we have at least $ 84\minus{}4(k\minus{}1)\minus{}3$ edges and exactly $ 49\minus{}k$ vertices. Due to the condition this graph can't contain a cycle. Hence from the lemma $ 84\minus{}4(k\minus{}1)\minus{}3<49\minus{}k$, then $ k>12$.
If we put traps to $ (2,2), (4,2), (6,2), (3,3), (5,3), (7,3), (2,4), (3,5), (5,5), (7,5), (2,6), (4,6), (6,6)$ the knight-rook can only have finitely many moves.

[quote="Umut Varolgunes"]Here is my solution,
Assume that we have put $ k$ traps and the table satisfies the condition. Now turn the table into a graph such that its vertices are the squares which haven't got traps and put edges to the adjacent ones.
Lemma: If the number of edges of a simple graph is greater than or equal to its number of vertices,  then it contains a cycle. It is a well-known fact.
At first we got $ 2.7.6$ edges and we have deleted some of them after the trapping operation. A trap can delete at most $ 4$ edges. Another point is we must put at least one trap to the edges of the table and this can delete at most $ 3$ edges. So we have at least $ 84 \minus{} 4(k \minus{} 1) \minus{} 3$ edges and exactly $ 49 \minus{} k$ vertices. Due to the condition this graph can't contain a cycle. Hence from the lemma $ 84 \minus{} 4(k \minus{} 1) \minus{} 3 < 49 \minus{} k$, then $ k > 12$.
If we put traps to $ (2,2), (4,2), (6,2), (3,3), (5,3), (7,3), (2,4), (3,5), (5,5), (7,5), (2,6), (4,6), (6,6)$ the knight-rook can only have finitely many moves.[/quote]
Your solution is totally the same as the official one,but nobody at olympiad didn't solve the problem in this way.In fact,there was only one solution,it was mine :) .I'll post it later...

I loved the question and i am wondering the solution of yours, Erken.

The first idea that came to my mind was to use tilings(in fact it is standart idea  :P ).But the question was how to use tilings in that kind of problems and after thinking for an hour,I came to conclusion that I must find some "almost fixed" arrangement of these traps and after that find some "cycle" to obtain a contradiction.
[b]Solution:  [/b]
We will call a cell(path)[i] "safety" [/i]if there is no trap(s) on cell(path).
Suppose that we have some arrangement that satisfies to our conditions and contains less than $ 13$ traps.
We will prove that there must be at least $ 12$ traps.
It is clear that in each square $ 2\times 2$ there must be at least $ 1$ trap and at least $ 2$ traps must be in square $ 3\times 3$.
Also it is not hard to show that there must be at least $ 3$ traps in rectangle $ 4\times 3$.
We will cut our board on a small squares and rectangles as it showed on a picture:
[img]http://img185.imageshack.us/img185/7490/adzb0.png[/img]
Therefore,there must be at least $ 12$ traps,and each tagged squares and rectangles there must be minimal number of traps,i.e there must be exactly $ 4$ traps in tagged $ 4\times 4$, exactly $ 2$ traps in tagged $ 3\times 3$ square and so on...
And after that we will turn our picture four times to $ 90^{\circ}$ around the center of a board.From where we conclude that in each green tagged square $ 3\times 3$ there are exactly $ 2$ traps and in each blue tagged rectangly $ 1\times 3$ there are exactly $ 1$ trap(see a picture below :roll: ):
[img]http://img362.imageshack.us/img362/5819/adyr8.png[/img]
And obviously in the center of each tagged $ 3\times 3$ square there is a trap,otherwise there exist  a [i]"safety"[/i] cycle.
[b]Here is the main idea:[/b]
It is easy to show that if we have some $ 3\times 3$ square with $ 2$ traps(one of the in a center) and two [i]"safety"[/i] cells on it, then there exist a path between them.
So if we place our "knight-rook" one some cell inside square $ 3\times 3$ then we can move in a clockwise direction and come to the border of rectangle $ 1\times 3$,and considering several cases ,we finally obtain that "knight-rook" can move to the next square $ 3\times 3$,and doing it four times we will return to the first square $ 3\times 3$,and as it was noticed,we can return to the initial cell,therefore there is a cycle.
[b]Here is an example[/b] of how can traps be placed:
[img]http://img145.imageshack.us/img145/9243/adbl1.png[/img]
,where red squares are traps...
Q.E.D
P.S:I can explain my solution in details,if someone wants ... and sorry for bad pictures... :blush:


Given a sequence of eight integers $x_{1},x_{2},...,x_{8}$ in a single operation one replaces these numbers with $|x_{1}-x_{2}|,|x_{2}-x_{3}|,...,|x_{8}-x_{1}|$. Find all the eight-term sequences of integers which reduce to a sequence with all the terms equal after finitely many single operations.

It's Ducci Sequence [url]http://en.wikipedia.org/wiki/Ducci_sequence[/url]

Since the wikipedia page on Ducci sequence doesn't seem to give an explicit solution, I will provide one.

A few initial observations:

[list=1]
[*]All entries are always integers and after the first operation, all entries are nonnegative.
[*]From $(1)$, from the second 8-tuple onward, the maximum entry is non-increasing over successive operations.
[*]The $8$-tuples ends up with all the same entries at some point iff it eventually ends up as an $8$-tuple of all $0$s.
[*]If after some number of operations we have a sequence of $(a_1, a_2, \ldots, a_8)$ of the form $(kb_1, kb_2, \ldots, kb_8)$ for integers $k, b_1, b_2, \ldots, b_8,$ then the $8$-tuple can be reduced to $(b_1, b_2, \ldots, b_8)$ and this new $8$-tuple eventually ends up as all $0$s iff the old one does.
[*]When viewed $\pmod 2$, the operation is equivalent to applying $(x_1, x_2, \ldots, x_8) \pmod 2 \to (x_1+x_2, x_2+x_3, \ldots, x_8+x_1) \pmod 2$.
[/list]

From $(3)$ it suffices to show that the entries all eventually end up as $0$. Proceed by induction on the maximum entry $m$ in the $8$-tuple resulting from a single operation. This method is legal due to $(1)$ and $(2)$. As a base case, with $m = 0$ we are already done. Consider $m = k > 0$ and assume the claim follows for all nonnegative integers less than $k$. From $(5)$, we observe the following for successive operations $\pmod 2$:
\begin{align*}
(x_1, x_2, \ldots, x_8) &\to (x_1+x_2, x_2+x_3, \ldots, x_8+x_1)\\
&\to (x_1+x_3, x_2+x_4, \ldots, x_8+x_2)\\
&\to (x_1+x_2+x_3+x_4, x_2+x_3+x_4+x_5, \ldots, x_8+x_1+x_2+x_3)\\
&\to (x_1+x_5, x_2+x_6, \ldots, x_8+x_4)\\
&\to (x_1+x_2+x_5+x_6, x_2+x_3+x_6+x_7, \ldots, x_8+x_1+x_4+x_5)\\
&\to (x_1+x_3+x_5+x_7, x_2+x_4+x_6+x_8, \ldots, x_8+x_2+x_4+x_6)\\
&\to (x_1+x_2+\cdots+x_8, x_1+x_2+\cdots+x_8, \ldots, x_1+x_2+\cdots+x_8)\\
&\to (0, 0, \ldots, 0) \pmod 2
\end{align*}

Thus after a finite number of steps, the $8$-tuple is in the form $(2a_1, 2a_2, \ldots, 2a_8)$ for integers $a_1, a_2, \ldots, a_8$ and from $(4)$, the $8$-tuple can be reduced so that all entries are divided by a factor of $2$. Since the maximum entry is non-increasing and we began with $m = k$, we now have $m \le \lfloor k / 2 \rfloor < k$ (for $k > 0$). The $8$-tuple now reaches the form $(0, 0, \ldots, 0)$ in a finite number of steps by the IH. $\blacksquare$\\

[hide="Remark"]
In the list of properties of the Ducci sequence given on its wikipedia page, it says that for $n$-tuples where $n$ is not a power of $2$, the sequence either reaches $(0, 0, \ldots, 0)$ or enters a cycle where all entries are either $0$ or $M$ for a fixed constant $M$. The fact that it at least enters a cycle is clear from $(1)$ and $(2)$ above since the number of $n$-tuples with entries all nonnegative integers at most some fixed maximum is finite, so repeats must eventually occur.

To see that the $n$-tuples must all eventually contain only $0$ and $M$ for a fixed $M$ (if not eventually reaching $(0, 0, \ldots, 0)$), note that this requires the non-increasing maximum entry to stop decreasing and remain stationary, in this case no more than $M$, and in fact, exactly $M$. Indeed, call the $i$-th tuple $x_i$ and its $j$-th entry $x_i^{(j)}$. If the the maximum persists as $M$ for $n$ $n$-tuples $x_i, x_{i+1}, \ldots, x_{i+n-1}$, we can prove that all entries of $x_i$ must be either $0$ or $M$: let $x_{i+n-1}^{(j)} = M$.

Then $\{x_{i+n-2}^{(j)}, x_{i+n-2}^{(j+1)}\} = \{0, M\}$ (with superscripts taken $\pmod n$), as any other pair of integers in $\{0, 1, \ldots, M\}$ could not have achieved this difference. From similar reasoning, each of $x_{i+n-3}^{(j)}, x_{i+n-3}^{(j+1)}, x_{i+n-3}^{(j+2)} \in \{0, M\}$ with at least one being $M$. Repeating this reasoning down to $x_i$, all of its entries must be in $\{0, M\}$ and it is straightforward that this then must always be the case for subsequent $n$-tuples in the iterations.
[/hide]

@above What is the proof of observation 4?


define domain of function $f$ bescribed by the following equation:
$f(x)-f^{2}(x)+f^{3}(x)-\ldots = \sin x$


algebra
polynomial
superior algebra
superior algebra unsolved

Given a field $K$ and an irreducible polynomial $f(x)\in K[x]$, degree of $f$ is at least 3. Is it true that there always exist irreducible polynomial $g(x)\in K[x]$ such that degree(g) is more than degree(f)?

In other words, if degrees of irreducible polynomials are bounded, must they be bounded by 2?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=30853 where I asked this already :)


AMC
AMC 12
probability
AMC 10
AMC 10 A
AMC 10 B
geometry

So... what's everyone's AMC12 goals and fears?(by fears, I mean things like lots of NT, lots of geo, etc.)

My goal : 132+
fear : lots of probability

well I didn't study enough... at all.

So I can't say I'm expecting that much.  120+ would make me happy.

I'd say my main fear is trig. Or any problems that take a lot of time...

Goal would be like 120-125, but not taking it- AMC10A and AMC10B this year.

My fear is that there'll be a lot of math on it.

Goal: 150.  But realistically, 120.

Fear: lots of geo in problems 20-25 :o

Goal: 150.
Realistically: 141.

Fear: Brain freeze.

Goal: 140+ 150 would be awesome

fear- multiple geometry problems in the higher numbers

goal: 130

fear: lots of geo and trig, pretty much problems that take a long time.

Never taken it before, but I am planning to do lots of practices.
Goal: 125.
Fear: Trigonometry.

Goal: 120.
Fear: A little combinatorics, but mainly geo.

Goal: Finding somewhere to take it (my high school is proving to be very stubborn). Otherwise, I'm hoping for a 145.5.
Fear: Geometry.

Goal: 135
Fears: lots of hard geo and trig, or <110

Goal: 132+
Hope: 139.5+

Fear: Any hard trig questions, or any involving the complex plane.

Goal: 100+
Fear: Geo problems, because they suck up a lot of time.

Goal: 150, obviously.  I haven't gotten a perfect score on any AMC's for real, and haven't taken any practice ones.

Fear: Problems that require a deep insight for which I don't have time.

[quote="math154"]Goal: >135.
Fear: Computation.[/quote]
ditto

Goal: 110
Fear: Stupid mistakes and things I don't know

[quote="1=2"]What's with the fear of geometry?[/quote]

geometry problems can be quite time consuming on the AMCs... Especially if they're in the 20-25 range.

Goal: 130+
Fear: Spending too much time on the easy problems and running out of time.

Goal: 120-125 (just want to make AIME and hopefully USAMO)

Fear: Combinatorics and problems that exceed my problem solving capability

Goal: 132+
Fear: Not being able to achieve it and/or brain cramp/nervousness or just failing in general.

haha, my goal of 135 is very unrealistic at the moment. I guess my 'real'  goal is 120+.

Goal: Make AIME
Fears: Careless mistakes, inability to read properly on the day of.

To beat my last year's score of 136.5, though I probably won't since I haven't really studied this year.

[quote="Smartguy"]"there is nothing to fear but fear itself" - FDR[/quote] This is actually the abridged version. He then said, "and hard questions that happen on the AMC 12."

Goal: 144+ (BUT NOT A 145.5 SINCE THAT WOULD MEAN I IGNORED ALL OF MY PRINCIPLES!)

Fear: Not enough BASHYNESS

[quote="Mewto55555"]Goal: 144+ (BUT NOT A 145.5 SINCE THAT WOULD MEAN I IGNORED ALL OF MY PRINCIPLES!)
[/quote]

huh?

The only way to get a 145.5 is to get 24 right and leave one blank. I think he is implying that his principles are to answer all the questions.

Goal: 130+

Fears: That I get less than a 130, do the test too slowly, get sleep deprived the night before (I have to take the 12A this year, and possibly the 12B as well if I fail the 12A...and also you know kids at my school get sleep deprived all the time...), don't beat certain beastly people (e.g. xpmath, abacadaea, laughinghead505, whitesoxfan), or if there's too much NT and/or combo :huh:

Biggest fear ever: I have too many fears. :( Plus homework is eating away at my studying for math contests :(

Goal: 115+
Fears: Modular arithmetic, trigonometry


The mean of a collection of five positive integers is 6.  What is the greatest possible median of the collection?

sum of 5 integers are 30.

the problem is about finding greatest median, so let's make two integers 1.

so it's now 1, 1, _, _, _.

and other 3 integers could be x, x, and x+1.

so it's total 3+3x=30.

3x=27, and x=9.

the answer is 9.


modular arithmetic
number theory unsolved
number theory

Find all $ n,m\in N$ such that

$ n^5\plus{}n^4\equal{}7^m\minus{}1$

I think I found something ...Nice problem

The condition is equivalent to $ n^5\plus{}n^4\plus{}1\equal{}7^m$ or $ (n^2\plus{}n\plus{}1)(n^3\minus{}n\plus{}1)\equal{}7^m$  so
$ n^2\plus{}n\plus{}1\equal{}7^p$ (1) and $ n^3\minus{}n\plus{}1\equal{}7^q$ (2) with $ p\plus{}q\equal{}m$
With eaasy divisibilities we can find that
$ n\equiv 2(\mod\ 7^p)$  . Subtituting in the relation (1) gives $ 7^p|7$  so p=0 or 1 from which we can find the solutions
(m,n)=(0,0)   or (m,n)=(2,2)

[quote="baz"][quote="silouan"]
With eaasy divisibilities we can find that
$ n\equiv 2(\mod\ 7^p)$.[/quote]
Do you mind explaining this?  Thanks.[/quote]

In fact, Silouan wrote : "$ n^2 \plus{} n \plus{} 1 \equal{} 7^p$ (1) and $ n^3 \minus{} n \plus{} 1 \equal{} 7^q$ (2) with $ p \plus{} q \equal{} m$"

And, since $ n^3\minus{}n\plus{}1\equal{}(n\minus{}1)(n^2\plus{}n\plus{}1)\minus{}(n\minus{}2)$, we have $ 7^q\equal{}(n\minus{}1)7^p\minus{}(n\minus{}2)$ and $ n\equal{}2\pmod{\min(p,q)}$
Then, if $ p\leq q$, $ n\equal{}2\pmod{7^p}$ and, with (1), we have $ 7\equal{}0\pmod{7^p}$ and so $ p\equal{}0$ or $ p\equal{}1$
Else, if $ q\leq p$, $ n\equal{}2\pmod{7^q}$ and, with (2), we have $ 7\equal{}0\pmod{7^q}$ and so $ q\equal{}0$ or $ q\equal{}1$

And hence the solution $ 2^5\plus{}2^4\equal{}7^2\minus{}1$

[quote="pardesi"]cancer How come u r posting all the indan MO Postal problems ??[/quote]

I didn't mean to, pardesi, I don't even know anything about it. I got the problems from here: http://www.imomath.com/othercomp/lirymnbiueb/BlrMO07.pdf

I hope you understand me  :)

So feel free to lock any problems which coincide with your Indian problems  :)

And actually, as a side note, I don't think it doesn't look like a good idea that the problems which students can do at home are copied from such sources...  :wink:

Consider equation
$ x^{3m \plus{} 2} \plus{} x^{3m \plus{} 1} \plus{} 1 \equal{} p^a$
a)Prove that if $ p\equiv 5(\mod 6)$ then the equation has no solution.
b)Prove that the equation has only finite solution (we can find out it for each m).
More $ v_p(3m^2 \plus{} 3m \plus{} 1) \equal{} k$ then
$ x^2 \plus{} x \plus{} 1 \equal{} p^c$ where $ c\leq k$

sorry cancer but it does turn out that the people setting the questions are copying from there  :)

anyways the last date for submitting the solutions was sept 12... :P

no harm done.. :D

actually this problem has been posted before [url=http://www.mathlinks.ro/viewtopic.php?t=164126]here[/url] and those who wanted to cheat have already done so  :D
actually my question to cancer was more about the coincidence that two posts of his had postal problems
[url=http://www.mathlinks.ro/viewtopic.php?t=168857]here[/url] is the second one.


Euler
modular arithmetic
number theory proposed
number theory

Prove that if $ (a,n)\equal{}1$ and $ k$ is the number of odd prime factors of $ n$, then $ a^{\frac{\phi(n)}{2^{k\minus{}1}}}$ is congruent to $ 1$ mod $ n$.

let $ n \equal{} 2^rp_1^{r_1}\cdots p_k^{r_k}$. by Euler's Theorem, $ a^{\phi(p_i^{r_i})\prod_{j\neq i}\phi(p_j^{r_j})/2^{k \minus{} 1}}\equiv 1\pmod{p_i^{r_i}}$ because $ \prod_{j\neq i}\phi(p_j^{r_j})/2^{k \minus{} 1}$ is an integer. also $ a^{\phi(2^r)\prod_{j \equal{} 1}^k\phi(p_j^{r_j})/2^{k \minus{} 1}}\equiv 1\pmod{2^r}$ because $ \prod_{j \equal{} 1}^k\phi(p_j^{r_j})/2^{k \minus{} 1}$ is an integer. the result follows.


Im unsure of where this goes (considering its calculus), but a question of the question's wording came up in regard to well, heh, this question:

If $f(x) = x^2 + 6x$, $x > 0$, find the derivative of the inverse of $f(x)$ in terms of $x$.

The function is injective over $x\in[0,\infty)$, so it has an inverse.

You can solve for $f^{-1}$, by using the identity $f(x)=y \Rightarrow f^{-1}(y)=x$

It's straightforward to find that $f^{-1}(x)=\sqrt {x+9}-3$, it follows immediately that

$f^{-1}^{'}(x)=\frac {1}{2\sqrt{x+9}}$, $x\in[0,\infty)$

Thanks. This was just to settle a dispute that a couple people had in my school.


Each side of a convex quadrilateral $ ABCD$ is tangent to a circle of radius $ r$ whose centre is denoted by $ O$. Show that at least one of the circle,circumscribed about triangles $ OAB$ and $ OCD$, respectively, has a radius greater tham or equal to $ r$.

It is easy to prove that $ \angle DOC\plus{}\angle AOB\equal{}180^{\circ} \Rightarrow$ max$ (\angle AOB,\angle DOC) \ge 90^{\circ}$

$ W.L.O.G.$ suppose that $ \angle AOB \ge 90^{\circ} \Rightarrow R_{\triangle AOB} \ge r$


vector
geometry
circumcircle
inequalities unsolved
inequalities

Let $ ABC$ is a triangle, let $ x, y, z \in R$. Prove that: $ \boxed{x^2 \plus{} y^2 \plus{} z^2\ge \minus{}2xycos2C \minus{} 2yzcos2A \minus{} 2zxcos2B}$

[quote="thanhnam2902"]Let $ ABC$ is a triangle, let $ x, y, z \in R$. Prove that: $ \boxed{x^2 \plus{} y^2 \plus{} z^2\ge 2xycos2C \plus{} 2yzcos2A \plus{} 2zxcos2B}$[/quote]
It is wrong . Must be  :  $ {x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xycos(2C) \plus{} 2yzcos(2A) \plus{} 2zxcos(2B) \ge 0}$ ;
Let $ \overline{a}$  ,  $ \overline{b}$ ,  $ \overline{c}$ - vectors , then : $ (\overline{a}\minus{}\overline{b})^2\plus{}(\overline{b}\minus{}\overline{c})^2\plus{}(\overline{c}\minus{}\overline{a})^2 \le 3(\overline{a}^2\plus{}\overline{b}^2\plus{}\overline{c}^2)$ ;

Let $ (O;R)$ is circumscircle of $ \triangle ABC$.

Show that: $ (x.\overrightarrow{OA} \plus{} y.\overrightarrow{OB} \plus{} z.\overrightarrow{OC})^2 \ge 0$

OK!

[color=darkblue]Let $ (O;R)$ is circumscircle of $ \triangle ABC$.[/color]

[color=darkblue]note: [/color]$  \| \begin {array} {c}OA=OB=OC=R\\\overrightarrow{OA}.\overrightarrow{OB} = OA.OB.cos2C = R^2cos2C \\
\overrightarrow{OB}.\overrightarrow{OC} = OB.OC.cos2A = R^2cos2A \\
\overrightarrow{OC}.\overrightarrow{OA} = OC.OA.cos2B = R^2cos2B \end {array}$
$ \forall x, y, z \in R$ [color=darkblue]we have [/color]$ (x.\overrightarrow{OA} + y.\overrightarrow{OB} + z.\overrightarrow{OC})^2 \ge 0$

$ \Longleftrightarrow x^2R^2 + y^2R^2 + z^2R^2+ 2xyR^2cos2C + 2yzR^2cos2A + 2zxR^2cos2B \ge 0$

$ \Longleftrightarrow \boxed{x^2 + y^2 + z^2\ge - 2xycos2C - 2yzcos2A - 2zxcos2B}$

[color=darkblue]OK![/color]


Can someone please explain this problem:

What is the smallest number of disks radius ½ that can cover a disk radius 1?

Solution:
http://web.archive.org/web/20040814192240/www.kalva.demon.co.uk/eotvos/esoln/sol1947c.html

thanks!

The smaller disks need to cover the boundary of the larger disk.  Consider the amount of the boundary that a single disk of radius $ \frac {1}{2}$ can cover.  ($ P$ and $ Q$ in the solution you've linked are the intersections of the boundaries of the two disks.)

Here are some threads on the forum that deal with this or similar problems:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=2047
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=169465
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=56774
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=254821974&t=56482


logarithms
number theory unsolved
number theory

how many digits does the following number have in the base of $ 10$ ?

$ {3}^{40}$

hmmm...i'm confused...is there Really an elementary way to calculate the digits of a number without using Logarithms or whatever???

( actually,the question has given 5 choices to choose from (Iraninan Level 1 contest is a 5-choice-to-choose-from-like exam) : 1) 17 , 2) 18 , 3) 19 , 4)20 ,  5)21   )


trigonometry
geometry
trig identities
Law of Sines

Solve the following system:

$ 3(x \plus{} \frac {1}{x}) \equal{} 4(y \plus{} \frac {1}{y}) \equal{} 5(z \plus{} \frac {1}{z});\ xy \plus{} yz \plus{} zx \equal{} 1$

This is a well-known trick, we can let x=tan(A/2), y=tan(B/2), z=tan(C/2), where A,B, and C are angles of a triangle
Proof here: [url]http://www.goiit.com/posts/show/174404.htm[/url]
Subbing in with half-angle formulas, we get
$ \frac {sin A}{3} \equal{} \frac {sin B}{4} \equal{} \frac {sin C}{5}$.
By the law of sines they are the angles of a 3-4-5 right triangle.
So $ x \equal{} \frac {1}{3}$, $ y \equal{} \frac {1}{2}$, $ z \equal{} 1$  :D
By the way, does anyone have another proof of the substitution?

[quote="thomaskurian89"]Solve the following system:

$ 3(x + \frac {1}{x}) = 4(y + \frac {1}{y}) = 5(z + \frac {1}{z});\ xy + yz + zx = 1$[/quote]

[color=darkred]Another solution:[/color]
[hide]
Let $ xy = a,yz = b,xz = c,a + b + c = 1$

$ 3(x + 1/x) = 4(y + 1/y) = 5(z + 1/z)$ times $ xyz$ for each term, we get $ 3(ac + b) = 4(ab + c) = 5(bc + a)$

$ a + b + c = 1 \implies b = 1 - a - c ,c = 1 - a - b, a = 1 - b - c$

Then

$ 3(ac + b) = 4(ab + c) = 5(bc + a)$

$ \implies 3(ac + 1 - a - c) = 4(ab + 1 - a - b) = 5(bc + 1 - b - c)$

$ \implies3(a - 1)(c - 1) = 4(a - 1)(b - 1) = 5(b - 1)(c - 1)$

$ \because x\not = 0,y\not = 0,z\not = 0$
$ \therefore a\not = 1,b\not = 1,c\not = 1$

$ 3(a - 1)(c - 1) = 4(a - 1)(b - 1) = 5(b - 1)(c - 1)\implies \frac 3 {b - 1} = \frac 4 {c - 1} = \frac 5 {a - 1}$

Suppose $ \frac {b - 1} 3 = \frac {c - 1} 4 = \frac {a - 1} 5 = t$

$  \{\begin{array}{c}{1} a = 5t + 1 \
b = 3t + 1 \
c = 4t + 1 \end{array}$
and $ a + b + c = 1 \implies (5t + 1) + (3t + 1) + (4t + 1) = 1\implies t = \frac { - 1}{6}$

$ \therefore a = \frac 1 6 ,b = \frac 1 2 , c = \frac 1 3$

$ \implies x =\pm \sqrt {\frac {ac} b} =\pm \frac 1 3 , y = \pm \sqrt {\frac {ab} c} =\pm \frac 1 2,z = \pm \sqrt {\frac {bc} a} =\pm 1$

Solution $ (x,y,z) = (\frac 1 3,\frac 1 2,1)$ or $ (x,y,z) = (-\frac 1 3,-\frac 1 2,-1)$
[/hide]

In general, how would someone figure out that a trigonometric substitution would be in order?

The 3-4-5 reminded me of a 3-4-5 right triangle, and I had seen a somewhat similar problem before.


If the tens digit $ d$ is replaced with one of the digits 0-9, what is the probability that the four-digit positive integer $ 12d4$ is divisible by 12? Express your answer as a common fraction.

If 12d4 is divisible by 12, d4 must be divisible by 12 since 1200 is divisible by 12.
Thus, d can equal 2 or 8.  Since there are 10 possible values, the answer is 2/10=1/5.

In order for a number to be divisible by 12, the number must be divisible by 3 and 4.

3 - Sum of digits needs to be divisible by 3 ( 48: 4+8=12, 12/3=4 YES )

4 - Last two digits form a number divisible by 4 ( 124: 24/4=6 YES )

We have the digits 12d4.

To make the sum of the digits divisible by 3, we have:  1+2+d+4=mult. of 3
                                                                                7+d=mult. of 3
So we have d= 2, 5, and 8

Now we check if 24, 54, and 84 are divisible by 4.
24/4=6 YES, 54/4=13.5 NO, and 84/4=21 YES.

So, the probability of the number being divisible by 12 is: 2/10 = [size=150]1/5[/size]

Please do not revive threads this old when you have no contribution.


algebra
polynomial
calculus
integration
algebra unsolved

find all polynoms in $R$
such that
for all rational $r$ there is a rational $x$ such  $P(x)=r$

[color=red]Edited by Megus: Don't use dollar sign before every word, also you shouldn't name a topic like "?????"[/color]

ehh... What do you mean by "T.O.T" ??  :?

By using the Lagrange interpolation formula, we find that $P$ (our polynomial) must actually have rational coefficients. By scaling the coefficients, we may assume it has integral coefficients, so let $P(x)=a_nx^n+\ldots+a_1x+a_0\in\mathbb Z[X]$.

For any integer $m$, we can find a rational $\frac{p_m}{q_m}$ s.t. $P\left(\frac{p_m}{q_m}\right)=m$. After writing this expression, we see that we have $q_m|a_n,\ \forall m\in\mathbb Z$, so all the rational solutions to equations of the form $P(x)=m,m\in\mathbb Z$ belong to $\frac 1{a_n}\cdot\mathbb Z$.

If we assume the degree of $P$ is $\ge 2$, then, as $m\to\infty$, we have $P^{-1}(m+1)-P^{-1}(m)\to 0$ (if $m$ is large enough, then $P^{-1}$ makes sense, since $P$ is injective on a neighbourhood of infinity), so we will find pairs of rationals belonging to $\frac 1{a_n}\cdot\mathbb Z$ as close to each other as we want, but this is a contradiction (two such points must have distance at least $\frac 1{a_n}$ between them).

thank you very much
but how we assum that $p(x)$ has in tegral coefficient

This can be done since you can substitute $x=c\cdot y$ where $c$ is some common multiple of the denominator of high enough powers of the coefficients


AB=CD=EF in cyclic hexagon ABCDEF.
 prove mdpts of BC,DE,FA form equilatrel triangle.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=45874 .

  darij


calculus
integration
function
calculus computations

[color=darkblue]For any $a\in R$, $\int_{0}^{1}|f(x)-a|\ dx\ge\int_{0}^{1}\left|f(x)-f\left(\frac{1}{2}\right)\right|\ dx$, where the continue function $f: [0,1]\rightarrow R$ is monotonously.

[b]Remark.[/b] See the my message from the Kunny's topic http://www.mathlinks.ro/Forum/viewtopic.php?t=106178 [/color]

Let $t$ such that $f(t)-a=0\ (0<t<1) ,$

$\int_{0}^{1}|f(x)-a|\ dx=\int_{0}^{t}|f(x)-a|\ dx+\int_{t}^{1}|f(x)-a|\ dx$

$\geq-\int_{0}^{t}\{f(x)-a\}\ dx+\int_{t}^{1}\{f(x)-a\}\ dx$

$=(2t-1)a-\int_{0}^{t}f(x)\ dx+\int_{t}^{1}f(x)\ dx.$

Determining the value of $t$ for which this equation will take a constant number in regardless of the value of $a,$ yielding $t=\frac{1}{2}\Longleftrightarrow a=f\left(\frac{1}{2}\right).$


inequalities
vector
triangle inequality
calculus
calculus computations

Suppose $k>2,\; x,y\in\mathbb{R}^{k},\; |x-y|=d>0,$ and $r>0$. Prove
a)If $2r>d$, there are infinitely many $z\in\mathbb{R}^{k}$ such that $|z-x|=|z-y|=r$.
b)If $2r=d$, there is exactly one such $z$.
c)If $2r<d$, there is no such $z$.
How must these statements be modified if $k=1\;\text{or}\; 2$?

I've read a little bit of Rudin's "Principles of Mathematical Analysis" and I think that this exercise appears in the first or second chapter.

By the triangle inequality we have that $2r = |z-x|+|z-y| \geq |x-y| = d$, from where we deduce (c).

(b) Note that equality holds iff those vectors are collinear. Using the condition
\[|z-x| = |z-y| , \]
we discover that only one point works (the midpoint $m$ of $[xy]$).

(a) Consider the hyperplane $\mathcal H$ orthogonal to $[xy]$ in its midpoint. The "good" points $z$ are exactly those that are in $\mathcal H$ and that satisfy $|z-m| = t$, where
\[t = \sqrt{r^{2}-\frac{|x-y|^{2}}4}. \]
$k=2$: (a) $\mathcal H$ has dimension $1$, so the "good" points are just two.

$k=1$: (a) No "good" points.


Can someone prove this?

In any triangle with sides a,b,c; and s = half of the perimeter,
then

Area = sqrt(s*(s-a)*(s-b)*(s-c))

I tried this formula with triangles that aren't triangles, for example:

With sides 4,6, and 10, Area = 0.
With sides 4,6, and 11, Area = (imaginary)

This question has been asked here once before. For example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6748]here[/url]. You can also try a search for Heron or Heron's to find other topics.

The best proof uses the Law of Cosines.  It should be in the appendix of your precalc book.