trigonometry
logarithms
calculus
calculus computations

$i^{i}$ is a real or complex number? Prove it.
PS: $i=\sqrt{-1}$

Use Euler's equation $e^{ix}=\cos x+i\sin x$ to express $i$. Now raise that to the $i$ power.

[quote="Kent Merryfield"]The answer (well known) is that while $i^{i}$ is not well-defined (since it depends on which branch of the logarithm you choose), it is real under any choice.

We can write $i=e^{\frac{i\pi}2+2\pi ki}$ for some integer $k.$ (The "simplest" choice would be $k=0.$) Then

$i^{i}=e^{-\frac{\pi}2-2\pi k},$ which is real and positive.[/quote]

Can you prove that $e^{ix}=cosx+isenx$ please?

you can also use power series to prove $e^{ix}=cos{x}+isin{x}$.

[quote="PTynan89"]Well, let $y = \cos x+i \sin x$.

Then, $\frac{dy}{dx}=-\sin x+i \cos x = i(\cos x)+i(i \sin x) = iy$.

So, we have $\frac{dy}{dx}= iy$.

Then, $\frac{dy}{y}= idx$, so $ln y = ix+k$.
So, $y = e^{k+ix}= e^{k}e^{ix}$.

Let $C = e^{k}$, then $y = C e^{ix}$.

So, $C e^{ix}= \cos x+i \sin x$.

We now determine $C$ by setting $x = 0$.
This gives $C*e^{0}= \cos 0+i \sin 0$.
So, $C*1 = 1+i*0$, and therefore $C = 1$.

So, we conclude that $e^{ix}= \cos x+i \sin x$.[/quote]
nice proof. i have never seen it proven this way. most of the time i see people prove that $e^{ix}=\cos(x)+i\sin(x)$ with power series.


1. The sum of the measures of the two base angles in an isoscelese triangle is 60 degrees. If the base of the triangle has a length of 10, find the length of the altitude to the base. I saw that it was a 30 60 90 triangle, but i couldn't find the height that is opposite the 30 degree side.

2. A regular hexagon has an inscribed circle of area 25 pie. Find the sum of the lengths of the diagonals. I am having trouble with this 60 30 90 things even though i know the ratios. x, x root 3, and 2

[hide="1"]Split the big one in half, and we have a 30-60-90 triangle with one of the sides $5.$ The smallest side is $\frac{5}{\sqrt{3}},$ or $\frac{5\sqrt{3}}{3}.$ This is also the altitude of the big triangle, so it's the answer.[/hide]

[hide="2"]There are two ways to do this.

First method: The circle's radius is 5. If we construct a 30-60-90 triangle, the side lengths are $5,$ $\frac{10}{\sqrt{3}},$ and $\frac{5}{\sqrt{3}}.$ The length of the hypotenuse is $\frac{10}{\sqrt{3}},$ or $\frac{10\sqrt{3}}{3}.$ Multiplying this by two gives us the length of a diagonal of the hexagon: $\frac{20\sqrt{3}}{3}.$

Second method: The circle's radius is 5, which is also the apothem of the hexagon. The area of a regular polygon is $\frac{ap}{2},$ where $a$ is the apothem, and $p$ is the perimeter. We know the apothem, so the area is $\frac{5p}{2}.$ The area of a hexagon can also be found if given a side length, which can be expressed as $\frac{p}{6}.$ The area is then $\frac{3\left(\frac{p}{6}\right)^{2}\sqrt{3}}{2}.$ Solving for $p$ gives us $p=\frac{60}{\sqrt{3}}=20\sqrt{3}.$ Now that we know the perimeter is $20\sqrt{3},$ each side is $\frac{10\sqrt{3}}{3}.$ The side length of a hexagon is equal to half of it's diagonal, so a diagonal would be $\frac{20\sqrt{3}}{3}.$[/hide]


Hey Guys,
For some reason, when I click on the "Lessons" tab, I don't get any lessons! It says there are no lessons currently available. I still have 900 karma. I finshed 62 of the required questions for the testing. The lessons were available before and I watched quite a bit of them (but not all). Please help me with this problem.

When you get too many wrong i think that's what happpens... i'm not sure tho.... no garuntees!!!

Hi akhil,

I have looked at your lessons page and it shows many lessons available. Can you provide a screen shot of what you are receiving? Also, include a screen shot of your report.

Thanks.


geometry
3D geometry
pyramid
geometric series

A pyramid is formed by stacking squares upon squares. In each row horizontally, there is only one square. As we go up the pyramid, the area of the square is halved. In each square, the builders inscribe a circle that is tangent to all four sides of the square; the square circumscribes the circle. This pyramid is lifted high into the sky and beyond into the infiniteness by the power of spiritual beings. If the diagnol of the 10th square is equal to 1/2, what is the total area of all of the circles incribed?

For now, I will not post the answer. I am sure people will get this easily, but if need be, the answer will be posted tomorow.

I'm not 100% sure on how to answer to this problem.

First of all, you can't make a pyramid by stacking squares. Cubes? Do you mean volume?

I'm around 80% sure you basically want us to find the sum of the first ten terms of a geometric series, but this problem is weirdly stated. I'm guessing the answer is $\frac{1023\pi}{32}$, but I'm not quite sure if that's the answer the problem is asking for.

"Pyramid" is just an unhappy term. Think of it as a stripe of squares next to one another (or one on the top of another), ever decreasing in size (by the given rule) and continuing to infinity (due to the divine influence). Hence we're dealing with an infinite sequence.

(Note: In spite of divinity's best efforts, the height of the "pyramid" will remain finite...)

[quote="Farenhajt"]"Pyramid" is just an unhappy term. Think of it as a stripe of squares next to one another (or one on the top of another), ever decreasing in size (by the given rule) and continuing to infinity (due to the divine influence). Hence we're dealing with an infinite sequence.[/quote]

oh, right, yeah, I get it. $32\pi$?

Yea, basically, it follows the 1+1/2+1/4+1/8+... sequence


Determine the maximum value of $m^{2}+n^{2}$, where $m$ and $n$ are integers satisfying $m,n\in \{1,2,...,1981\}$ and $(n^{2}-mn-m^{2})^{2}=1.$

For reference, another solution can be found [url=http://www.kalva.demon.co.uk/imo/isoln/isoln813.html]here[/url].


algebra
polynomial
modular arithmetic
calculus
integration
Congruences

Let $\Gamma$ consist of all polynomials in $x$ with integer coefficients. For $f$ and $g$ in $\Gamma$ and $m$ a positive integer, let $f \equiv g \pmod{m}$ mean that every coefficient of $f-g$ is an integral multiple of $m$. Let $n$ and $p$ be positive integers with $p$ prime. Given that $f,g,h,r$ and $s$ are in $\Gamma$ with $rf+sg\equiv 1 \pmod{p}$ and $fg \equiv h \pmod{p}$, prove that there exist $F$ and $G$ in $\Gamma$ with $F \equiv f \pmod{p}$, $G \equiv g \pmod{p}$, and $FG \equiv h \pmod{p^n}$.

That's a way to state Hensel's lemma for $\mathbb{Q}_{p}$.
Proof of the theorem:

We inductively construct polynomials $F_{n}, G_{n}$ such that $F_{n}G_{n}\equiv h \mod p^{n}$.
For $n=1$, we take $F_{1}=f$ and $G_{1}=g$.
For $n\geq 1$, we try to find polynomials $f_{n},g_{n}$ such that we can set $F_{n+1}=F_{n}+p^{n}f_{n}, G_{n+1}=G_{n}+p^{n}g_{n}$.
The only thing we need is
\[h \equiv F_{n+1}G_{n+1}\equiv F_{n}G_{n}+p^{n}(f_{n}G_{n}+g_{n}F_{n}) \mod p^{n+1},\]
thus
\[h_{n}\equiv f_{n}G_{n}+g_{n}F_{n}\equiv f_{n}g+g_{n}f \mod p,\]
 where $h_{n}= \frac{h-F_{n}G_{n}}{p^{n}}\in \Gamma$.
By $rf+sg \equiv 1 \mod p$ we get $h_{n}=rfh_{n}+sgh_{n}\mod p$, thus setting $f_{n}=rh_{n}, g_{n}=sh_{n}$ gives our result.


vector
Functional Analysis
combinatorial geometry
advanced fields
advanced fields unsolved

Seems easy,but I could not give a quick answer on the following question :( :
If $ K$ is a convex and compact set in Banach space $ B$, is it true, that $ conv K$ also compact set? ($ conv K$ is a closed convex hull of $ K$).

Yes, it is. In fact the same statement holds for every locally convex quasicomplete vector space.

The proof relies on the fact that in a such a space being compact is the same as being totally bounded and complete. Now completeness of $ \overline{conv(K)}$ follows from closedness. That the convex hull and the closure of a totally bounded set are again totally bounded are easy exercises.

I've already found this fact. Thanks for your reply.


[quote]A database query syntax error has occurred. This may indicate a bug in the software. The last attempted database query was:

    (SQL query hidden)

from within function "SearchMySQL4::update". MySQL returned error "126: Incorrect key file for table './dbaps/aw_searchindex.MYI'; try to repair it (localhost)[/quote]

ML, FF2, XP.

Were you trying to search for something?

No, I was trying to make an edit to [[AoPSWiki:Chess/Chess Tournament Info]]


Find all integer solutions of the equation:
$ 4x^2 \plus{} 2(x^2\plus{} y^2)^2 \plus{} xy (x\plus{}y)^2\equal{}132$

[hide]We have
$ 4x^2 \plus{} 2x^4 \plus{} 2y^4 \plus{} 6x^2y^2 \plus{} x^3y \plus{} xy^3 \equal{} 132$
Now even if $ x^3y$ and $ xy^3$ were negative, $ x^4 \plus{} y^4$ added would make the sum positive by Muirhead. So in all cases
$ x^4 \plus{} y^4 \plus{} 6x^2y^2 \le 132$
We find the only possibilities are when x and y are either $ \pm1$ or $ \pm2$
I don't think there are any solutions.[/hide]


Find all $ n\in\mathbb{N}$ and $ p$ prime such that $ 3^p\minus{}np\equal{}n\plus{}p$

$ n = \frac {3^p - p}{p + 1}$
So $ p + 1|3^p - p$
$ \Leftrightarrow p + 1|3^p + 1$
Easy to check $ p > 2$ so $ p > q$
Call q is a prime divisor of $ p + 1$ so
$ q|3^p + 1$ so $ q|9^p - 1$
So that
${ q|\gcd(9^p - 1 - 1,9^{q - 1} - 1}) = 8$
It means that $ q = 2$
So $ p + 1 = 2^a$
$ 2^a|3^{2^a - 1} + 1$
So $ a = 2 \rightarrow p = 3$
Easy to check p=3 is solution .

[quote="TTsphn"]
So that
${ q|\gcd(9^p - 1 - 1,9^{q - 1} - 1}) = 8$
It means that $ q = 2$
[/quote]

Could you explain it, please?
Thanks a lot  :blush:

Yes . We have a lemma .
$ \gcd(x^m\minus{}1,x^n\minus{}1)\equal{}x^{\gcd(m,n)}\minus{}1$
Because $ p$ is a prime and $ p>q\minus{}1$ so $ \gcd(q\minus{}1,p)\equal{}1$
So $ \gcd(9^p\minus{}1,9^{q\minus{}1}\minus{}1)\equal{}9\minus{}1\equal{}8$


What are your favorite choping nocturnes   :)

F sharp minor! cuz it actually is a little bit more challenging

I also like the one in C minor.

you should include impromptus as well. Maybe etudes too.

Heck, favourite chopin piece in general!

I love the Grande Polonaise Brilliante, myself.

i like etudes better. I was just thinking of nocturnes because of the title.

Polonaises aren't as good to me.

Really? I like the nocturnes better. Easier for me to play.  :lol:

I like the Nocturne in op.27-2 most, Etudes and Polonaises are pretty good as well, but they are really difficult to perform... :P

What, no love for the scherzos?   :lol:
The only one I've learned so far is Scherzo No. 2, Op. 31.  It's good.

My favorite is the nocturn in c# minor because it was the first chopin piece I ever played (in 5th grade).

I've played Chopin's Nocturne in E-flat Major and E Minor, which are both beautiful songs, particularly because of the beauty of the music that can be produced from the left hand, if played correctly.

I like the nocturnes in D-flat (27-2) and F-sharp major (15-2), well, because I've played them.  :D
Speaking of Chopin, I have to leave in a few minutes to go and play Ballade no. 2 at the SF Conservatory. Should be fun.

I personally like Fantasie Promptu and Nocturne op.9 #2. And basically every other Chopin piece.

All of the Chopin Nocturnes are elegent, Scherzos and Polonaises are very nice as well, so I like all of them. :)


function
complex numbers
algebra unsolved
algebra

Let $f: \mathbb{C} \to \mathbb{C}$ be an additive function satisfying $f(x^2)=f(x)^2$. Must $f \equiv 0$ or $f \equiv \text{id}$? This is trivially so if $f(\mathbb{R}) \subseteq \mathbb{R}$, but I was wondering whether there are non-trivial examples of complex-valued additive functions with $f(x^2)=f(x)^2$ ... Any thoughts?  :)

From that condition with the squares you easily find $f$ to be multiplicative, so, basicly, you're asking for the field automorphisms of $\mathbb C$. I remember asking that question a very long time ago. It received quite a few answers, and I believe there was someone who established the existence of non-trivial (i.e different from $z\mapsto 0,z\mapsto z,z\mapsto\bar z$) automorphisms in a non-constructive fashion, but I can't find the thread anymore, and I don't really remember what the conclusion of the whole thing was  :blush:.

From a discontinuous automorphism of the complex numbers one can produce a discontinuous additive function on the real numbers (which is also a nonmeasurable function). So in a certain provable sense there is no hope of producing specific examples.


inequalities
trigonometry
geometry proposed
geometry

Let $T$ be the centroid of triangle $ABC$. Prove that \[ \frac 1{\sin \angle TAC} + \frac 1{\sin \angle TBC} \geq 4 \]

We have :

$\frac{1}{\sin{TAC}}=\frac{2\cdot{m_a}}{c\cdot{\sin{A}}}\geq\frac{2(b^2+c^2)}{4R\sin{A}\cdot{c}}=\frac{b^2+c^2}{ac}$, here $m_a$ is the median from $A$ and I used $m_a\geq\frac{b^2+c^2}{4R}$.

Similarly $\frac{1}{\sin{TBC}}\geq\frac{a^2+c^2}{bc}$

Adding this 2 relations an using AM-GM we get the result!

let $AT$ and $BT$ meet $BC$ and $AC$ at $A'$ and $B'$ respectively.
let $AT=2A'T=2a'$, $BT=2B'T=2b'$ and $a''$ and $b''$ be the distances from $T$ to $BC$ and $AC$ respectively

then, the inequality is equivalent to

$\frac{2a'}{b''}+\frac{2b'}{a''}\geq 4$, but from am-gm we have that

$\frac{2a'}{b''}+\frac{2b'}{a''}\geq 4\sqrt{\frac{a'}{a''}\cdot\frac{b'}{b''}}\geq 4,$

since $a'\geq a''$ and $b'\geq b''$


calculus
integration
number theory unsolved
number theory

Hi, there's my problem:

Let p is a given prime. Determine all possible pairs of integers (x,y) such that p|(x^2 - y^3) (in words: x^2 - y^3 is divisible by p).

The first observation can be very easy - x and y = 0 or 1 (mod p) are convenient always, but I'm not able to describe all convenient pairs (x,y) in some "good way".
Because when p = 6k - 1 for integral k, then for each x there exist y such that all conditions are satisfied (but I don't know how to describe these "y" and also how to prove this idea). But for example when p = 13 convenient x satisfy x^2 = 1,7,11 (mod p) (so all posible residues aren't convenient). Could anybody give me some hint. Thanks.

You want all solutions $x^2 \equiv y^3 \mod p$. Assume that $p$ dosn't divide $x$ which gives that it also doesn't divide $y$.
Let be $s$ a primitive root $\mod p$ and let be $x \equiv s^a , y \equiv s^b \mod p $ were $0 \leq a,b \leq p-2$.
Then you want $s^{2a} \equiv s^{3b} \mod p$ or eqivalent to that: $2a \equiv 3b \mod (p-1)$.
Then you get $2|b$ and so let be $b=2c$, giving $a \equiv 3c \mod \frac{p-1}{2}$

1. case: $6|p-1$
Then you get $3|a$ and for every such $a$ there are ($\mod \frac{p-1}{2}$) exactly $3$ different $c_i \equiv a \mod \frac{p-1}{6}$ that fulfil the equation.

2. case: (otherwise)
Then you get $a \cdot 3 ^{-1} \equiv c \mod \frac{p-1}{2}$ leading to exactly one working $c$.


Here is an interesting probability question for beginners.

There are 3 doors $A,B,C$, behind one door is a prize.  You are told to pick a door.  You pick door $A$.  Now, I tell you that there is nothing behind door $B$.  From a mathematical standpoint, should you keep your original choice, or switch to door $C$ and why?

[hide="hint"]Regardless of which door you pick, the game show host (Monty Hall), who knows what is behind the curtains, will reveal one of the curtains that doesn't have the prize behind it. There will always be such a curtain, so does it really change anything when he reveals a prizeless curtain?

This is the intuitive way of looking at it. For a simple mathematical demonstration look at the various cases and the probabilities associated with them.[/hide]

Here's the answer

[hide]If you originally picked the wrong door, then changing your choice will gurantee that you get the prize.  Since there is a $\frac{2}{3}$ chance that you originally picked the wrong door as opposed to a $\frac{1}{3}$ chance that you picked the right door, then by changing your choice, you double your chances of winning.[/hide]

I've seen like five threads on this topic :? ...

I changed (corrected) the spelling of the thread title so that this would be more searchable.

Yes, there have been quite a few threads on this famous problem here on AoPS over the years. I find the most helpful hint I can give to the correct solution of the problem is to ask how one would reason if there were 99 doors, and 97 had already been opened.


Συνοπτικά γιατί είναι και περασμένη η ώρα. Χρησιμοποιώντας την τριγωνομετρική μορφή Ceva:

Συμβολίζω :$ \angle{BAD}\equal{}A_1$, $ \angle{DAC}\equal{}A_2$, $ \angle{MCA}\equal{}C_1$, $ DCM\equal{}C_2$, $ DBM\equal{}B_1$, $ ABM\equal{}B_2$

Εχουμε:

$ \frac{\sin{A_1}*\sin{B_1}*\sin{C_1}}{\sin{A_2}*\sin{B_2}*\sin{C_2}}\equal{}1\iff$

$ \frac{\sin{A_1}}{\sin{A_2}}\equal{}\frac{\sin{C_2}}{\sin{B_1}}\iff$

$ \frac{\sin{C}}{\sin{B}}\equal{}\frac{\sin{C_2}}{\sin{B_1}}\iff$

$ \frac{AB}{AC}\equal{}\frac{BM}{MC}$

Αυτό μαζί με το $ \angle{B_2}\equal{}C_1$ μας δίνουν ότι $ BMA$ όμοιο με $ CMA$. Επειδή όμως έχουν κοινή την $ AM$ έπεται ότι είναι ίσα άρα $ AB\equal{}AC$

Βγαίνει και αποδεικνύωντας το αντίστροφο και μετά ισάτοπο απαγωγή.

[quote="stergiu"]Σε ένα ορθογώνιο τρίγωνο [b]ΑΒΓ[/b] φέρνουμε το ύψος [b]ΑΔ [/b]προς την υποτείνουσα [b]ΒΓ[/b].Στο τμήμα [b]ΑΔ [/b]παίρνουμε σημείο [b]Μ[/b]. Αν οι γωνίες  [b]ΑΒΜ [/b]και [b]ΑΓΜ[/b] είναι ίσες  , να αποδείξετε ότι [b]ΑΒ= ΑΓ [/b].

 Μπάμπης[/quote]

[u][b]ΛΥΣΗ[/b][/u]

Η ισογώνια του ύψους $ AD$ είναι η διάμεσος $ AN$. Αν υποθέσουμε ότι υπάρχουν οι ισογώνιες $ Bx,Cy$ των ευθειών $ BM, CM$ (δηλαδή αν υποθέσουμε ότι οι $ BM,CM$ δεν διχοτομούν τις γωνίες $ B,C$) τότε γνωρίζουμε ότι οι $ Bx, Cy$ θα τέμνονται σε σημείο $ P$ της διαμέσου $ AM$ και θα ισχύει $ \angle CBP\equal{}\angle BCP\Longrightarrow BP\equal{}CP$, ΑΤΟΠΟ.

Συνεπώς οι $ BM,CM$ είναι διχοτόμοι των γωνιών $ B,C$ αντίστοιχα, άρα $ AB\equal{}AC$.

$ q.e.d.$

Νίκος Ράπανος


AMC
AMC 12
geometry
3D geometry
tetrahedron
AMC 10
trigonometry

How did Georgia people do on the AMC?
I got a 132 on the AMC12, which is much better than my scores from previous years. :)
I missed number 18 (mostly because I am intimidated by tetrahedrons), number 24, and number 25.
My goal is 135+ on the 12B.

Yay, we're allowed to discuss!!  I missed the last question on AMC 10 b/c I had like a few minutes left and I wanted to spend that time checking rather than doing a question that I doubted I would get right in the little time that I had.  Luckily, I was right to check, b/c I made two careless mistakes which I later corrected.  Yeah, but unfortunately, I had to guess on 25 and guessed incorrectly.  So, yeah, I got a $ \boxed {144}$ and now will go into a corner and weep. :(

Ah yes - I forgot to mention that weeping is not allowed in this thread, especially if you outscored the highest scoring AMC10 taker at my school by over 30 points.

Indeed, I find a score of 144 to be incredibly commendable. I particularly commend you for not getting a 145.5, as that is a sine of sissification, timidity, and weakness. 150 is mbleh, but a 144 means you had the strength to admit you were not perfect and miss one.

(ok, so anyway, don't complain, but don't get a 145.5 either ;))

I got the same as Eric, 132. Understand Santosh beat me again by 1.5 points. I demand a rematch on the B ;). (That way Santosh can beat me by 15 points :D.)

[quote="solafidefarms"]Indeed, I find a score of 144 to be incredibly commendable. I particularly commend you for not getting a 145.5, as that is a sine of sissification, timidity, and weakness. 150 is mbleh, but a 144 means you had the strength to admit you were not perfect and miss one.[/quote]
:huh: Ummm, okay.  Thanks, Billy, for giving me some self-confidence, I think...

PS: What does mbleh mean, and is sissification a word?  Just curious. :lol:

[quote="solafidefarms"]I particularly commend you for not getting a 145.5, as that is a [b]sine[/b] of sissification, timidity, and weakness.[/quote]
If I recall correctly, 145.5 isn't in the range of the sine function (and I don't think that sissification, timidity and weakness are in the domain). :lol:
Sorry, I couldn't resisit.

Bah! The AMC 10 is for the WEAK! And Sitan, don't let Billy give you too hard a time. I seem to remember a pretty good score from him on the 10 last year!

[url]http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-5-1012-Perfect.shtml[/url]

(sarcasm)  Thanks, Mr. Fulton, that means a lot to me...(\sarcasm)

PS: I'm not weak...  (sob)  jkjk, I just did AMC 10 cuz I wanted the experience of taking it so I could do 12 in high school.  Don't ask me why I chose to do so...

PSS: Do you have the MC Nats 2007 test yet?  Sorry if I sound hasty, but it's just a reminder. :)

Yeah, yeah, yeah. I'll try to get it to you early this week. And I think that I'm ordering a bunch of 10's for some of our guys on the B. I think that the 12 has finally gotten hard enough that the 10 is worth while again.

[quote="solafidefarms"]I particularly commend you for not getting a 145.5, as that is a sine of sissification, timidity, and weakness.

Understand Santosh beat me again by 1.5 points. I demand a rematch on the B ;). (That way Santosh can beat me by 15 points :D.)[/quote]

I'm assuming that a 146.5 on the AMC 10 before the scoring change is also "a sine of sissification, timidity, and weakness". In that case, you got beat by a person who is a sissy, timid, and weak.  :) (I think those are in the right parts of speech.)

Anyway, I got a 133.5, same as last year.
Couldn't figure out #18 until after the test. Guessed $ \sqrt {105}$ since $ 105|5\cdot6\cdot7 \equal{} 210$.
On #19, I somehow thought that $ 1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^{14}$ had $ 14$ terms.
Eliminated #24 to C or D and then didn't guess.
On #25, figured out $ (a_{103},b_{103}) \equal{} ( \minus{} 32,16)$. Guessed $ \frac {1}{2^{98}}$ correctly.

I'm still trying to figure out why I guessed on 2 problems that I had a "gut feeling" on but didn't guess on a problem I had narrowed down to 2 choices.

I've heard that Harrison got less than 130. Is that true? probably all the geometry at the end.
Eric said that 133.5 could be enough for state winner. I'm doubting that.

[quote="Xantos C. Guin"]Couldn't figure out #18 until after the test. Guessed $ \sqrt {105}$ since $ 105|5\cdot6\cdot7 \equal{} 210$.[/quote]I fell for the $ 105|210$ trap as well.

[quote]Eric said that 133.5 could be enough for state winner. I'm doubting that.[/quote]Okay, maybe not, but maybe state winner outside of Lassiter. And anything in the 130's seemed amazing having just seen my school's top three AMC12 scores: 132, 112.5, and 91.5. :roll:
And our lowest AMC10 score: 19.5

[quote="Xantos C. Guin"][quote="solafidefarms"]I particularly commend you for not getting a 145.5, as that is a sine of sissification, timidity, and weakness.

Understand Santosh beat me again by 1.5 points. I demand a rematch on the B ;). (That way Santosh can beat me by 15 points :D.)[/quote]

I'm assuming that a 146.5 on the AMC 10 before the scoring change is also "a sine of sissification, timidity, and weakness". In that case, you got beat by a person who is a sissy, timid, and weak.  :) [/quote]

No, actually, it was better to not guess on the old scoring. Now it's just as well on average to guess as to blank, so if you blank the last one, it means you didn't perform the analysis (joyofperfect*1/5 + joyof144 *4/5) > (joyof145.5), which I believe to be true.

I took the 12 and got a 100.5  :( .
I could have guessed on several (and gotten them right), but I was scared of guessing the wrong thing and not making the AIME.

Plan for the AMC 12B-guess on anything that I have even a remote clue about

(this is my first post, but this is Slok Carpenter, you may know me in real life)

[quote="Goldentide"](this is my first post, but this is Slok Carpenter, you may know me in real life)[/quote]

I may know you in real life, but I would deny it!  :D

(See you Wednesday morning!)

Yes, the AMC 10 is definitely for the weak.  I would be perfectly happy with a 145.5.

Anyhow, don't look for a state winner from Lassiter on the A contest (I think I have the highest score here, a 129).

Well, Slok, you qualified for the AIME, and had you guessed, you probably would not have.  Your actual score will probably be immaterial, since you are young.  So you of all people should know not to guess the next time around!

And what is this fuss about #18, for crying out loud?  You're given a very orderly system of three equations for the lengths of three mutually perpendicular sides in a tetrahedron and then asked for its volume.  Serves you right for guessing.  God job in general, though, guys.

By the way, 145.5 is certainly in the range of the sine function.
[hide="Every complex number is in the range of the sine function."]
Okay, for starters, the equation
\[ x^2 \minus{}1 \equal{} 2izx \]
has nonzero solutions for $ x$ for any fixed value of $ z$, since the product of the roots is nonzero.  Thus the equation
\[ \frac{x \minus{} 1/x}{2i} \equal{} z \]
always has a complex solution $ x$, so, the function $ f: \mathbb{C}^* \to \mathbb{C}$ satisfying $ f(x) \equal{} \frac{x\minus{}1/x}{2i}$ is surjective.  Since the function $ g(x) \equal{} e^{ix}$ mapping $ \mathbb{C} \to \mathbb{C}^*$ is surjective, it follows that the function
\[ \sin x \equal{} \frac{e^{ix} \minus{} e^{\minus{}ix}}{2i} \equal{} f \circ g \]
is a surjective map from $ \mathbb{C}$ to itself.  (You can verify the identity (or definition) for sin using power series.)  $ \blacksquare$[/hide]

Hm, while I'm at it, I think I'll attach my writeup of problem 25, in case you're not on the ARML mailing list.

Yes, both the AIME I and the AIME II were 6 last year...or the II might have been 7, i'm not sure...

Last year's AIME I was particularly strange because of the moving sidewalk and orbiting planets problems, problems early on in the test which many people missed. This, in addition to the expansion to 500 qualifiers, made the cut-off especially low last year. Because there will again be about 500 qualifiers, I would imagine that the cut-off will not climb up to 9 as it has in previous years, but depending on the difficulty of the test, I would think that the cut-off could be as low as 6 or as high as 8. I guess you should just shoot for a 10 because that way, you know you'll qualify. :)

I missed the orbiting planets one, but not the moving sidewalk one. My score last time was a 6. Hopefully I can push it up to an 8 or 9 this time. :)
I almost got the rotated triangle one last year, but ran out of time. I still lament about this one from time to time. :(

So, when IS the AIME? I've been hearing dates as early as yesterday and as late as April, after spring break...

AIME I is the coming Tuesday the 18th, and AIME II is in mid-April, or so I've heard.

[quote="Boy Soprano II"][quote="Math Geek"]Dang...   7 or 8?  That's going to be kinda hard... :([/quote]
Well, no offense, but if you can't do that many, you probably don't stand a chance at the USAMO.  (This is not to say that you should not start working on olympiad problems before you can do seven or eight AIME problems.)  Anyhow, if you just do a bunch of practice AIMEs and don't worry much about the time limit, you'll be able to do that many before too long (by which I mean a month or a few).  At least, you'll be able to do five or six, and then you should start olympiad problems, and once you're into those, the AIME seems easy.  If you're interested in math, I don't think it takes a year of hard work, depending on how good you already have gotten.

By the way, it's usually been more like eight in normal years, but I get the sense that the AMC is in general hiking up the difficulty of its contests, so maybe seven will be enough.  Wasn't it six or something last year?

Hm, and Karthik, you should make USAMO this year.[/quote]

Hopefully, I'll make the USAMO this year... I really suck at the AMC because of my horrendously slow math ability... but I think my AIME score should be much better than last year.

Say you don't get the amount of problems right, but you DO get the Index. Do you still make USAMO? Is this even possible? Why am I asking so many questions? (Answer the first question plz, I know the second one, answer the third if ur bored)

[quote="Arvind_sn"]Say you don't get the amount of problems right, but you DO get the Index. Do you still make USAMO? Is this even possible?[/quote]
Well, if you read the USAMO qualifying process, you should be able to figure this out.  I assume that by "amount of problems right" you mean "AIME floor score".

If you make the senior index, i.e., the index everybody can use, i.e., one of the top ~330 USAMO indices in the country, then it is not possible not to make the floor, because
[quote="the USAMO manual"]The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value.[/quote]

(P.S.  The [url=http://www.unl.edu/amc]AMC website[/url] has answers to many of your questions.  :wink: )

If you took the AMC 12, that is impossible since the floor is determined by the lowest AIME score of the people who qualified by the index.

If you took the AMC 10, its possible. Say about the 300th highest index was 123+8*10=203, then any people that have an AMC12 index higher than 203 would qualify. This would include a 104.5/10 and a 145.5/6 if the 104.5 and 145.5 are AMC 12 scores. However, if no one with a 203+ index got less than 7 on the AIME, then the floor would be a 7 and a (AMC 10) 145.5/6 would not qualify but a (AMC 10) 117/7 would. (Yes, this is kinda depressing if ur good at AMC but not AIME).

Your asking so many questions since you have some thing disturbing you, but don't want to talk about it, so you are talking about this. Jk, actually I have no clue and am probably more disturbed myself.

[quote="Xantos C. Guin"]If you took the AMC 12, that is impossible since the floor is determined by the lowest AIME score of the people who qualified by the index.

If you took the AMC 10, its possible.[/quote]
Oh, well, if you took the AMC 10, you don't really have an index at all. :yoda:

Why? Is it because it's for the weak and unpowerfulsez?

[quote="Arvind_sn"]Why? Is it because it's for the weak?[/quote]
If you are good enough to benefit from the USAMO, you should be taking the AMC 12.

Good luck on the AIME tomorrow, everybody!

I have some EMERGENT issue here, if anyone could help, that would be greatly appreciated.

Our math teacher today announced she could not let us take AIME I because she could not find a substitute tomorrow. I've just sent her a e-mail to persuade her to let us take the test in the school library. However, if it didn't work either, is it permitted to take the test on Wednesday or Thursday this week?

Thanks in advance.

Well, there's another test called the AIME II, which you can take in case you miss the first one. As for your specific question, I don't know. But, you could take the AIME II if you organize stuff quickly.

You cannot take the AIME I officially on any date except the official date.


Find the greatest m and the smallest M such that whenever a+b+c=1 and a,b,c>0 we have:
   m<=(ab+bc+ca)( (a^2+1)/(a+b)+(b^2+1)/(c+b)+(c^2+1)/(c+a))<=M.

Unfortunately, this is also my solution. I posted this one since I really wanted a neat solution. But I think it is really hard to find one.

I forgot to note that when a-> 1, b->0, c->0 we have C.H.S -> 1. M=5/3 is archieved when a=b=c=1/3.


A circle is inscribed in isosceles trapezoid $ ABCD$. Diagonal $ AC$ intersects the circle at $ K$ and $ L$ where $ K$ is between $ A$ and $ L$. What is the value of $ \frac{AL\cdot KC}{AK\cdot LC}$?

Is there a solution without too much computation?

[quote="not_trig"]sorry, but is it always possible to [i]inscribe[/i] a circle in an isosceles trapezoid? like if the trapezoid is really long, it wouldn't work[/quote]

To answer your question, no, it's not always possible; but the problem assumes that a circle [b]is[/b] inscribed in $ ABCD$


Pictures of people which may include Ignite168 or may not

oh wow... This definitely looks like a person... And it definitely could be Ignite168...

 :rotfl:


Does anybody know a site that has the answer key to the 2004 Sprint round? I google searched it, but..... somehow I couldn't get access to it

[url=http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm]Here is a good link[/url] - Ad, popup, popunder, and other nastiness warning.


Heres what was said on the AMC 10B/12B Booklet

[quote]Tell [the students] that they may use scratch paper, graph paper,
ruler, compass, protractor and four-function, scientific, or graphing
calculators. No problems on the contest require the use of a calculator.

Therefore, [b]no calculator use will be allowed on this contest[/b]. Remind
students to bring a supply of #2 pencils.[/quote]

Here's the link: [url]http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/CU2008/2008-C+U,Booklet.pdf[/url]
It says above that we can use a four-function, scientific, or graphing calculator, but then it tells us not to use it. Can we?

Unfortunately, it appears that that particular document did not get updated properly.

You definitely [b]cannot[/b] use a calculator on any of the AMC contests this year (AMC 10/12, AIME, or USAMO).


Let $ ABC$ be a triangle, $ M$ is the midpoint of $ AB$, $ N$ is the midpoint of $ AC$, $ BN$ meets $ CM$ at $ O$. IF $ \widehat{BOC}\equal{}90^o$, find the minimum value of $ cotB\plus{}cotC$

hello, let $ \overline{CM}=s_a$ and $ \overline{BN}=s_b$ then we get
$ s_b^2=\frac{a^2+c^2}{2}-\frac{b^2}{4}$
$ s_c^2=\frac{a^2+b^2}{2}-\frac{c^2}{4}$
and from the condition of the problem we get
$ b^2+c^2=5a^2$
No we derive ${{ \cot(\beta)+\cot(\gamma)=\frac{\frac{a^2+c^2-b^2}{2ac}}{\frac{2A}{ac}}}+\frac{\frac{a^2+b^2-c^2}{2ab}}{\frac{2A}{ab}}}=\frac{a^2}{2\sqrt{s(s-a)(s-b)(s-c)}}$.
Inserting $ a^2=\frac{1}{5}\left(b^2+c^2\right)$ and minimize the objective function we get the searched minimum as $ \frac{2}{3}$.
Sonnhard.


Does anyone know where I can get the solutions?

Solutions would be appreciated...I still don't know my score!

Answers are here: http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05AMC12AB-answers.html

Both answers are solutions can be found in the solution packet, which was mailed to your school along with the test booklet.

Well, I think it would be illegal to post the actual solutions verbatim, but I think if you posted the numbers of the questions you don't understand, people (including me) would help you.

Seeing as how ppl are posting solutions....give me a problem and i'll give u the solution

[quote="Sunny"][quote="RC-7th"]
Both answers are solutions can be found in the solution packet, which was mailed to your school along with the test booklet.[/quote]

In my case, they actually sent the wrong test, the AMC 10 instead of 12, so they had to e-mail it to my school for me.  As a result, I don't have a solutions packet[/quote]

Hehe, that's pretty funny. I wonder what the facial expression of the person coordinating the test was when he or she got them.

[quote="yif man12"]
Hehe, that's pretty funny. I wonder what the facial expression of the person coordinating the test was when he or she got them.[/quote]

It was amusing, cuz he didn't open it till I showed up to write it.

[quote="wcs7251"]The answers are nice, but I want to know how to solve them.
Were the results from the MathJams posted?[/quote]Yes.  Click on "Math Jams" at the top of the page, then click on "Transcripts".


geometry
circumcircle
power of a point
radical axis
geometry proposed

ABC is a triangle. E is a point on the median from C. A circle through E touches AB at A and meets AC again at M. Another circle through E touches AB at B and meets BC again at N. Show that the circumcircle of CMN touches the two circles.

Since $E$ is on the radical axis of the two circles, and so is the midpoint of $BC$, it means that so is $C$, so $CA\cdot CM=CB\cdot CN=k^2$, and the inversion of pole $C$ and power $k^2$ invariates the two circles and turns $AB$, which is tangent to the two circles, into the circumcircle of $CMN$, which must then also be tangent to the two circles in $M,N$, the images of $A,B$ respectively.

Is not difficult to see that CE is the radical axis of the circles c(AEM) and c(BEN), i.e. CE pass through the second common point of the two circles. From this fact it follows that CM*CA=CN*CB  which in turn implies that ABNM is ciclic and finilly it holds that CMN is similar to ABC.
If O1 and O2 are the centers of c(A) and c(B) respectively let O3=C1M^C2N.
If O is the circumncer of ABC, by angle chasing, one can see that ABO is similar to MNO3. Then collecting those two fact we can conclude that O3 is the center of c(MNC) and also that this circle is tanget to c(A) and c(B).


geometry
rectangle
circumcircle
analytic geometry
trigonometry

You can solve it with complex numbers. Choose the circumcenter as origin and assume WLOG the circumradius as [tex]1[/tex]
Now we can write the coordinates of [tex]P, L[/tex] as:

[tex]L=ix(B-C)[/tex]
[tex]P=-iy(A-C)[/tex]

with [tex]x,y \in \mathbb{R}[/tex]

We have to prove that:

[tex]\dfrac{ix(B-C)-iy(A-C)}{2C}=i\dfrac{x(B\overline{C}-1)-y(A\overline{C}-1)}{2|C|^2}[/tex]

is real.
The condition concerning the areas of the rectangle is equivalent to:

[tex]y|A-C|^2=x|B-C|^2 [/tex]

[tex]\iff y(A-C)\overline{(A-C)}=x(B-C)\overline{(B-C)} [/tex]

After some algebra and considering that [tex]|A|^2=|B|^2=|C|^2=1 [/tex] we get

[tex] x(B\overline{C}-1)-y(A\overline{C}-1)=-\overline{x(B\overline{C}-1)-y(A\overline{C}-1)}[/tex]

Thus [tex]ix(B\overline{C}-1)-y(A\overline{C}-1)[/tex] is real
[tex]\Box[/tex]

A synthetic solution can be derived from the following simple fact:

Let ABC be a triangle and (O) its circumcircle. If D and F are the antipodes of A and C respectively and F = AE^DB, then AB/AC = FA/FD.

I will write later the rest of the proof.

[quote="Armo"]Excuse me , what is antipode?[/quote]

Excuse me. I have seen many this notation on the forum and I have thought that it was quite known.

BTW, if A is a point on a circle (O), the antipode D of A is the second end of the diameter through A. Or, in other words, is the opposite of the point A wrt O, the center of circle.

[quote="Michael"]Let $\triangle ABC$ be an acute triangle. Construct outwardly the rectangles $BCKL$ and $ACPQ$ such that they have the same area. Prove that the vertex $C$, the circumcentre of $\triangle ABC$ and the mid-point of segment $PL$ are collinear.[/quote]

This one is very easy:

Let O be the circumcenter of triangle ABC, and let the line CO meet the line PL at a point T.

Since the point O is the circumcenter of triangle ABC, the central angle theorem yields < COA = 2 < CBA = 2B. On the other hand, again because O is the circumcenter of triangle ABC, we have CO = AO, so that the triangle COA is isosceles, and thus its base angle < OCA must equal

$\measuredangle OCA=\frac{180^{\circ}-\measuredangle COA}{2}=\frac{180^{\circ}-2B}{2}=90^{\circ}-B$.

On the other hand, since ACPQ is a rectangle, < ACP = 90°. Thus, < PCT = 180° - < OCA - < ACP = 180° - (90° - B) - 90° = B.

After a condition of the problem, the rectangles BCKL and ACPQ have the same area. Denoting this area by t, we thus have $BC\cdot BL=t$ and $CA\cdot CP=t$. In other words, $a\cdot BL=t$ and $b\cdot CP=t$, so that $BL=\frac{t}{a}$ and $CP=\frac{t}{b}$.

By the sine law in triangle PCT, we have

$\frac{PT}{CP}=\frac{\sin\measuredangle PCT}{\sin\measuredangle PTC}$,

so that $PT=CP\cdot\frac{\sin\measuredangle PCT}{\sin\measuredangle PTC}$. Since $CP=\frac{t}{b}$ and < PCT = B, this becomes $PT=\frac{t}{b}\cdot\frac{\sin B}{\sin\measuredangle PTC}=\frac{t}{\sin\measuredangle PTC}:\frac{b}{\sin B}$. Similarly, $LT=\frac{t}{\sin\measuredangle LTC}:\frac{a}{\sin A}$. But sin < PTC = sin < LTC because of < PTC = 180° - < LTC, and $\frac{b}{\sin B}=\frac{a}{\sin A}$ by the sine law in triangle ABC. Thus, PT = LT. Hence, the point T is the midpoint of the segment PL. Since the point C, the circumcenter O of triangle ABC and the point T are collinear, we therefore can say that the point C, the circumcenter O of triangle ABC and the midpoint of the segment PL are collinear. And the problem is solved.

  Darij

[quote="Michael"]Hi!

Let $\triangle ABC$ be an acute triangle. Construct outwardly the rectangles $BCKL$ and $ACPQ$ such that they have the same area. Prove that the vertex $C$, the circumcentre of $\triangle ABC$ and the mid-point of segment $PL$ are collinear. :)

Michael[/quote]

Let O is the circumcenter of ABC,M is the intersection of OC with PK.
Then $A(OKC)=A(OCP)$ so KM/MP=1.

[quote="Virgil Nicula"]O.K., Hardsoul ! I didn't understand this problem. Maybe the middlepoint of the $PK$ or $QL$ is collinearly with the points $C$ and $O$ ! What proved Darij ? $OC\parallel KL\parallel BC$ ?[/quote]

In http://www.kalva.demon.co.uk/polish/pol02.html it is written as "the midpoint of PK lies on the line through C and the circumcenter" :?

[u]I note:[/u] the area $[XYZ]$ of a triangle $XYZ$; the circumcircle $C(O,R)$ of the triangle $ABC$; the middlepoint $M$ of the segment $[PK]$; $BL=CK=x$;
$AQ=CP=y$ ($ax=by$).

$m(\widehat {OCP})=180^{\circ}-B\Longrightarrow$ $[OCP]=\frac 12 yR\sin B=\frac14 by\Longrightarrow$
$[OAQ]=[OCP]=\frac 14 by.$

$m(\widehat {OCK})=180^{\circ}-A\Longrightarrow$ $[OCK]=\frac 12 xR\sin A=\frac 14 ax\Longrightarrow$
$[OBL]=[OCK]=\frac 14 ax.$

Therefore ($ax=by$), $[OCP]=[OCK]\Longrightarrow \overline {\underline {\left| \ M\in OC\ \right| }}.$

Let $ABC$ be a triangle with circum center $O$. rectangles $ACDE$and$BCFG$, have equal area.

Prove that:

                $O,C,M$ are collinear.

(which $M$ is the midpoint of $DF$.)

I want to use inversion ,but I don’t invert all the diagram. just a part of it.

ok lets solve the problem:

consider an inversion with center $C$ ,and radius $CA.CD$ or $CB.CF$.

$A,B$changes to$A',B'$.($CA'=CD$  ,   $CB'=CF$)

circle$(O)$changes to line$A'B'$. and$CO'=2CH$

(which$H$is the intersection of$A'B'$and$CO$)

only this.

Remember that points $G,E$ aren’t important. and I don’t use them in inversion

And after inversion circles $CA'B'$ and $CB'G'$ and $CF'G'$ and .$CA'E'$ and $CE'D'$ don’t help us

So  I don’t draw them in my inversive diagram.

Then problem will change to this one:

$A'B'C$ is a triangle. $CD$ is perpendicular to $CA'$. and $CD=CA'$

$CF$ is perpendicular to $CB'$. and $CF=CB'$

$CH$ is the altitude of triangle $CA'B'$. and $M$. is the mid point of $FD$

prove that $H,C,M$ are collinear.

Of course this problem is easier than the first.

Lets solve this one:

Suppose $HC$ intersects $DF$ at $M'$.

Its clear that these facts are true:

$\angle CB'H=\angle FCM'=T$

$\angle CA'H=\angle DCM'=K$

then area of triangles $CA'B'$ and $CDF$ is equal and we have:

$CM'.CD.sinK + CM'.CF.sinT= \frac{1}{2} (CA'.A'B'.sinK + CB'.A'B'.sinT)$

And :$CM'(CD.sinK+ CF.sinT)= \frac{1}{2} A'B'(CD.sinK+CF.sinT)$

Then : $CM'=\frac{1}{2}A'B'$

And if $S$ is midpoint of $A'B'$, triangles $CA'S$and$CDM$       and also      triangles    $CB'S$and$CFM$

Are equal to each other.

Then :$DM'=TS=MF$

And $M=M'$

 we done…

Inversion again showed its self very nice… ;)

Let $X=BK\cap AQ$. It is clear that $X\in (ABC)$ and that $O\in XC$. So, let $M=XC\cap PL$, we want to prove $M$ is midpoint of $PL$. We'll call $\angle BCO=\alpha$ and $\angle OCA=\beta$, therefore, $\angle LCM=90-\alpha$ and $\angle PCM=90-\beta$. The problem clearly states that $BC\cdot CL=AC\cdot CP$. Therefore, $\frac{CP}{CL}=\frac{BC}{AC}$.
Using generalized bisector theorem, we want to prove that:
\[CP\cdot \sin (90-\beta)=CL\cdot \sin (90-\alpha)\]
Which is equivalent to proving:
\[\frac{BC}{AC}=\frac{\cos \alpha}{\cos \beta}\].
Let the orthogonal projections of $O$ on sides $AC$ and $BC$ be called $D$ and $E$, respectively (which are the midpoints of $AC$ and $BC$). We have that $\cos \alpha=\frac{EC}{OC}$ and $cos \beta=\frac{DC}{OC}$. So then, $\frac{\cos \alpha}{\cos \beta}=\frac{EC}{DC}=\frac{BC/2}{AC/2}$, and the proof follows.

[quote=Christian Hirsch]
Now we can write the coordinates of $P, L$ as:

$L=ix(B-C)$
$P=-iy(A-C)$

with $x,y \in \mathbb{R}$[/quote]
Why?


algebra
polynomial
Vieta
analytic geometry
quadratics
symmetry
function

Find the value of $ c$ for which the roots of $ x^3\minus{}6x^2\minus{}24x\plus{}c \equal{} 0$ form an arithmetic progression.

[hide="For advanced people"]
Try to give hints or sketch the solution rather than giving it away! [/hide]

[hide="For beginners"]
Let's start from the beginning. For those who have just started math, definitely read and learn [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas]Vieta's Formulas[/url].  This is a very important concept that helps you to understand the relationship between the roots and the coefficients.  Now, after you read this, try to find the "middle" root. Why does this help us? Think about it..[/hide]

As usual, please hide your answers! I'm going to try to post some problems that cover basic yet crucial stuffs to motivate young problem solvers!

Some hints:
[hide="Hint 1"]Denote the polynomial's roots as $ r, s,$ and $ t$. Using Vieta's formulas, what sorts of equations can you get?[/hide][hide="Hint 2"]Using the fact that $ r, s,$ and $ t$ form an arithmetic progression can help you find $ s$.[/hide][hide="Rough Solution Sketch"]Find the value of $ r\plus{}s\plus{}t$ with Vieta's, then use the fact that they form an arithmetic progression to find $ s$. Then plug in the value of $ s$ into the other formulas derived from Vieta's, and solve for $ c$.[/hide]

Well done. Anyone wants to use his tip and finish this problem off?

I'll leave Brut3Forc3's method (the intended method) for others. I'd like to offer an alternate approach

[hide="Background and Setup"]
- An arbitrary cubic is symmetric about some point, which we will call the center-point.
- If the centerpoint has coordinates $ (h,k)$, then the cubic can be written in the form $ y \equal{} a(x \minus{} h)^3 \plus{} b(x \minus{} h) \plus{} k$, where we have eliminated the quadratic term. This form of the cubic makes the point-symmetry obvious.
- For monic cubics, $ a \equal{} 1$.
- The line $ y \equal{} k$ is the unique horizontal line that intersects the cubic equation in three evenly-spaced points.[/hide]
[hide="Solution"]In this case, we have three evenly spaced roots, meaning that the horizontal line $ y \equal{} 0$ intersects the curve in three evenly spaced points. Thus, the x-axis is the line $ y \equal{} k$, and $ k \equal{} 0$. Note that this means that $ y$ is divisible by $ (x \minus{} h)$.

$ \minus{} 3hx^2 \equal{} \minus{} 6x^2$ by term pairing, so $ h \equal{} 2$, and $ y \equal{} (x \minus{} 2)^2 \plus{} b(x \minus{} 2)$. Dividing the original form by $ (x \minus{} 2)$ via polynomial/synthetic division yeilds a remainder of $ 64 \minus{} c$. We know the remainder must be also zero, hence $ 64 \minus{} c \equal{} 0 \implies \boxed{c \equal{} 64}$[/hide]

Hmm, I never actually saw this approach! Cool  :)

how do you know that there is only one such line?

it is easy to prove with the vieta method, but you are trying to circumvent that, so what is your proof?

The idea was to provide a mental image of a cubic by which, on a contest, one can utilize this rather intuitive property :)

[u]Demonstration:[/u]

By translation/scaling it suffices to look at $ y \equal{} x^3 \minus{} kx$. It is easy to show that the function increases to some local maximum, decreases to a local minimum, and then increases.

Suppose the line $ y \equal{} j$, with $ j \equal{} 0$, intersects the line at x-values $ a,b,c$, which are evenly spaced because it is an odd function. Let $ a < b < c$. We want to show that this will not happen again as we vary the value of $ j$. If we increase $ j$, then $ a$ and $ c$ increase since the function is increasing, and $ b$ decreases, so $ b \minus{} a$ decreases and $ c \minus{} b$ increases, and they cannot be equal again. Similarly if you decrease $ j$, then $ b \minus{} a$ increases and $ c \minus{} b$ decreases, and again, the two cannot be equal.

This obviously holds only if the function is not monotone...

[i]-- I dunno, I tend to think in terms of pictures - when I saw the problem, I immediately divided by $ x \minus{} 2$, though I'm sure immediately setting the roots to $ 2,\; 2 \pm t$ and using Vieta's and multiplying would be nearly as fast[/i]  :maybe:

[hide="Vieta solution"]
Let the roots be $ p$,$ q$,$ r$. Since they are in an arith. progression, we have $ p$,$ q$, and $ 2q\minus{}p$. We have by Vieta's
$ p\plus{}q\plus{}(2q\minus{}p)\equal{}6$, so $ 3q\equal{}6$ and $ q\equal{}2$.
Also by Vieta's, we have $ pq\plus{}q(2q\minus{}p)\plus{}p(2q\minus{}p)\equal{}\minus{}24$, so subbing in 2 for q gives $ p^2\minus{}4p\minus{}32\equal{}0$, from which $ p\equal{}8$.
So since by Vieta's $ \minus{}c\equal{}pq(2q\minus{}p)$, we have $ c\equal{}64$. [/hide]

[quote="The Zuton Force"]- An arbitrary cubic is symmetric about some point, which we will call the center-point.[/quote]

I believe this point for a cubic of the form $ ax^2\plus{}bx^2\plus{}cx\plus{}d\equal{}0$ is given by $ \minus{}\frac{b}{3a}$, and is the inflection point about which it will change from concave down to concave up (or vice-versa for negative a).

Can anyone confirm that? Would be nice, I hadn't realized the extension of this property until now.

Yes, that's right. This is because if we let $ y\equal{}x \plus{} \frac{b}{3a}$ and write the cubic in terms of $ y$, we get $ a(y^3\plus{}py\plus{}q)$ for some $ p,q$ -- we call this a [i]depressed cubic[/i], which is an odd function in $ y$ because the squared term is gone.

Why does b-a decrease when c-b increases? (and vice versa).

[img]http://img.photobucket.com/albums/v194/mukmasterxps/cubic.png[/img]

Let $ a,b,c$ be the solutions to $ x^3\minus{}kx\equal{}j$. $ \text{sign} \left( \frac{da}{dj} \right) \equal{} \minus{} \text{sign} \left( \frac{db}{dj} \right) \equal{} \text{sign} \left( \frac{dc}{dj} \right)$.

If $ j$ increases, then $ a$ increases and $ b$ decreases, so $ \minus{}a$ decreases, and thus $ b\minus{}a$ decreases. $ c$ increases, and $ \minus{}b$ increases, so $ c\minus{}b$ increases.

If $ j$ decreases, we have exactly the opposite.

[quote="The Zuton Force"]

Let $ a,b,c$ be the solutions to $ x^3 \minus{} kx \equal{} j$. $ \text{sign} \left( \frac {da}{dj} \right) \equal{} \minus{} \text{sign} \left( \frac {db}{dj} \right) \equal{} \text{sign} \left( \frac {dc}{dj} \right)$.

[/quote]
There are still problems with this solution...Okay so if you increase j a bit, you have a correct, but what if you increase j a lot, and that pattern does not follow anymore?

Your idea is nice visually, but there is a very clear difference between a visual solution and a rigorous proof. I think that you can actually prove this using rolle's theorem.

But obviously this takes a lot of effort to prove (i.e. calculus)


induction
Galois Theory
superior algebra
superior algebra unsolved

Let $ K\equal{}K_0 \subset K_1 \subset ... \subset K_n \subset L$ be a sequence of finite extensions of fields. Prove that L/K is a separable extension if and only if for every i=1,...,n-1 extension $ K_{i\plus{}1}/K_i$ is separable.

Use induction and for $ E \leq K \leq F$ finite extensions, $ F/E$ separable iff $ F/K$ and $ K/E$ are separable.

RHS => LHS follows directly from definition.
LHS => RHS uses number of $ E$-embeddings of $ F$ into a normal extension of $ E$ is exactly $ [F: E]$.

Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.

My name is Soarer and I come aus Hk. I speak Chinesish, English and little german. I like math. Do not know if I correctly


Find the accelerations in the following pictures

acceleration of block A is 2g/7 upward and of block B is g/7 downwards....

plz reply and tell whether it is correct or not. :wink:

[quote="Jaswinder"]acceleration of block A is 2g/7 upward and of block B is g/7 downwards....
[/quote]
 :maybe:
I guess I'm wrong then- g/13,2g/13  :(
5g-T=5a
T-4g=2X4a
=> g=13a
=> a=g/13

[quote="nuclear_alchemist"][quote="Jaswinder"]yes certainly

u r wrong  :P

u haven't used the constraint relations properly

 :)[/quote]
Please elaborate your solution then :)[/quote]
Tension in the string is $ T$ throughout.
If acceleration of $ 5kg$ block is $ a$, the acceleration of $ 4kg$ block is $ 2a$.
Hence,
$ 40 - 2T = 2a(4kg) \\
\& 2T - 50 = a(5kg)$
Solve linearly, you get $ a= \frac{10}{7}$.
And I think it is high time Jaswinder stops mocking at people and blowing his own trumpet. :furious:


trigonometry
calculus
integration
modular arithmetic
induction
algebra
polynomial

[hide="Solution"] Set $ x_n \equal{} 2^{2^n}$ and $ k_0 \equal{} x_0^2 \plus{} x_0 \plus{} 1, k_{n \plus{} 1} \equal{} x_n^2 \minus{} x_n \plus{} 1$.  Verify that $ k_0 k_1 ... k_n \minus{} 1 \equal{} x_n(x_n \plus{} 1)$ by induction.  The key identity here is $ (x^2 \plus{} x \plus{} 1)(x^2 \minus{} x \plus{} 1) \equal{} x^4 \plus{} x^2 \plus{} 1$, which is found in Engel.

To prove that $ \gcd(k_i, k_j) \equal{} 1, i < j$, it helps to know some lemmas about cyclotomic polynomials.  A simpler argument:

[b]Case:[/b]  $ i \equal{} 0$.  Note that $ (x_{j \minus{} 1} \plus{} 1)k_j \equal{} x_j^3 \plus{} 1 \equal{} 8^{2^j} \plus{} 1 \equiv 2 \bmod 7$.

[b]Case:[/b]  $ i > 0$.  Then $ k_i | x_{i \plus{} 1}^3 \minus{} 1$ while $ k_j | x_j^3 \plus{} 1$.  Since $ x_j$ is a power of $ x_{i \plus{} 1}$, it follows that $ x_j^3 \plus{} 1 \equiv 2 \bmod k_i$. [/hide]

We need $ k_0k_1\cdots k_n \equal{} a(a \plus{} 1) \plus{} 1 \equal{} a^2 \plus{} a \plus{} 1$.

Start with $ a^2 \plus{} a \plus{} 1 \equal{} (a \minus{} \varepsilon)(a \minus{} \overline\varepsilon)$, where $ \varepsilon \equal{} \cos \frac {2\pi}{3} \plus{} i \sin \frac {2\pi}{3}$. Then we would like to find $ b$ such that
\[ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1) \equal{} c^2 \plus{} c \plus{} 1,
\]
so we think that we need $ (a \minus{} \varepsilon)(b \minus{} \varepsilon) \equal{} ab \plus{} \overline\varepsilon \minus{} (a \plus{} b)\varepsilon$, so we need $ a \plus{} b \equal{} 0$, and so $ b \equal{} \minus{} a$, and we get
\[ (a^2 \plus{} a \plus{} 1) (( \minus{} a)^2 \plus{} ( \minus{} a) \plus{} 1) \equal{} a^4 \plus{} a^2 \plus{} 1
\]
and it's easy to see that $ a^2 \plus{} a \plus{} 1$ and $ a^2 \minus{} a \plus{} 1$ are coprime (as odd integers). Which implies that $ a^{2^n} \plus{} a^{2^{n \minus{} 1}} \plus{} 1$ will contain at least $ n \plus{} 1$ different prime factors, so we are done.

[hide="My Solution"] Lemma. Infinitely many primes $ p$ exist such that $ p$ divides some term in the sequence $ f(a)=a^{2}+a+1$ for positive integral $ a$.

Proof. Assume there are finitely many $ p$. Clearly, there is at least one, since $ f(1)=3$. Then, multiply them all together, and you get $ f(a)\equiv1$ modulo the product of all of them. But then $ f(a)$ has a prime factor other than those primes, contradiction.

end lemma

Take $ c>1$ primes as described in the lemma, and we can take $ c$ arbitrarily large. Then, since $ f(a)$ modulo $ p$ has period $ p$, for each of these primes, there's some residue $ a_{1}$ such that if $ a\equiv a_{1}\pmod{p}$, then $ p|a$. By the Chinese Remainder Theorem, we have an $ a$ divisible by all of the $ c$ primes.

We write $ a=p_{1}^{e_{1}}p{}^{2}{}^{e_{2}}\cdots p_{c}^{e_{c}}K$, where $ K$ is relatively prime to the rest of the stuff. Then take $ k_{0}=p_{1}^{e_{1}},k_{1}=p_{2}^{e_{2}},\ldots,k_{n-1}=p_{c-1}^{e_{c-1}},k_{n}=p_{c}^{e_{c}}K$.[/hide]

[quote="worthawholebean"]Prove that for each positive integer $ n$, there are pairwise relatively prime integers $ k_0,k_1,\ldots,k_n$, all strictly greater than $ 1$, such that $ k_0k_1\ldots k_n \minus{} 1$ is the product of two consecutive integers.[/quote]

That's a very old and fairly well-known problem.

It goes (at least) back to
Solomon W. Golomb:
On certain nonlinear recurring sequences.
Amer. Math. Monthly  70, (1963), 403-405.

[quote="CatalystOfNostalgia"][hide="My Solution"] Lemma. Infinitely many primes $ p$ exist such that $ p$ divides some term in the sequence $ f(a) = a^{2} + a + 1$ for positive integral $ a$.

Proof. Assume there are finitely many $ p$. Clearly, there is at least one, since $ f(1) = 3$. Then, multiply them all together, and you get $ f(a)\equiv1$ modulo the product of all of them. But then $ f(a)$ has a prime factor other than those primes, contradiction.

end lemma

Take $ c > 1$ primes as described in the lemma, and we can take $ c$ arbitrarily large. Then, since $ f(a)$ modulo $ p$ has period $ p$, for each of these primes, there's some residue $ a_{1}$ such that if $ a\equiv a_{1}\pmod{p}$, then $ p|a$. By the Chinese Remainder Theorem, we have an $ a$ divisible by all of the $ c$ primes.

We write $ a = p_{1}^{e_{1}}p{}^{2}{}^{e_{2}}\cdots p_{c}^{e_{c}}K$, where $ K$ is relatively prime to the rest of the stuff. Then take $ k_{0} = p_{1}^{e_{1}},k_{1} = p_{2}^{e_{2}},\ldots,k_{n - 1} = p_{c - 1}^{e_{c - 1}},k_{n} = p_{c}^{e_{c}}K$.[/hide][/quote]
My solution was similar, except it involved less thought.
[hide="Solution Using Dirichlet"]
[b]Lemma.[/b]  If $ p$ is prime such that $ p\equiv 1 \pmod{3}$, there exists a residue $ r$ such that $ p \mid r^2+r+1$.

[i]Proof.[/i] Let $ g$ be a multiplicative generator of the nonzero integers mod 3.  Set $ r \equiv g^{(p-1)/3}$.  Then $ r-1 \not\equiv 0 \pmod{p}$, but $ r^3-1 \equiv 0 \pmod{p}$, so $ \frac{r^3-1}{r-1} \equiv r^2 + r+1 \equiv 0 \pmod{p}$.  $ \blacksquare$

By Dirichlet's theorem, there are infinitely many primes congruent to 1 (mod 3).  Let $ p_0, \dotsc, p_n$ be $ (n+1)$ such primes, and let $ r_0, \dotsc, r_n$ be respective residues as described in the lemma.  By the Chinese remainder theorem, there is a positive integer $ a$ satisfying the equivalences
\[ a \equiv r_i \pmod{p_i} . \]
Then $ p_0 \dotsc p_n \mid a^2+a+1$.  Now, for $ 1 \le i \le n$, take $ k_i$ to be the greatest power of $ p_i$ that divides $ a^2 + a +1$, and let $ k_0 = (a^2+a+1)/(k_1 \dotsm k_n)$.  We are done.[/hide]

My (lame) solution

[hide]
Define $ f(a) \equal{} a^2 \plus{} a \plus{} 1$. It suffices to prove $ f(a)$ can have as many distinct prime factors as we want.

Lemma: There exist infinitely many primes $ p$ such that $ f(a)\equiv 0 \pmod p$ has solution.

Proof: We'll proceed with induction.
Base: $ p \equal{} 3$, then $ a\equiv 1\pmod 3$ is a solution.
Inductive hypothesis: suppose there exist $ m$ such primes $ p_1,...,p_m$. By CRT we can find a $ a \equal{} a'$ so that $ f(a')\equiv 0 \pmod {p_i}$ for $ i \equal{} 1,...,m$.
Now, we see that $ \gcd (f(a'), f(a' \plus{} 1)) \equal{} (a'^2 \plus{} a' \plus{} 1,a'^2 \plus{} 3a' \plus{} 3) \equal{} (2(a' \plus{} 1), a'(a' \plus{} 1) \plus{} 1) \equal{} 1$, so $ f(a' \plus{} 1)$ is not divisible by any of $ p_1,...,p_m$. But $ f(a' \plus{} 1) > 1$, so $ f(a' \plus{} 1)$ must has another prime factor $ p_{m \plus{} 1}$, i.e., $ f(a' \plus{} 1)\equiv 0 \pmod {p_{m \plus{} 1}}$. Hence, $ p_{m \plus{} 1}$ also satisfies the condition.

Now let's choose any subset of these primes, say $ p_1,...,p_m$, and find the solution to each of congruences $ f(a)\equiv 0 \pmod {p_i}$, say $ a\equiv a_i \pmod {p_i}$. If we set up these solution congruences simultaneously, by CRT, there exist an $ a \equal{} x$ solving them all. Then $ f(x)$ is divisible by $ m$ distinct prime factors $ p_1,...,p_m$. Since $ m$ can be arbitrarily large, it holds true for every positive integer $ n$.
[/hide]

Yes! I did that too. Like you must have 4k0k1...k(n-1)=s^2+3 so we take kn=s^2+2s+4.

[quote="MellowMelon"]Was anyone else stupid enough like me to overcomplicate the problem and show by induction how to explicitly construct $ k_0 k_1 \ldots k_n$ for each $ n$? I'm actually surprised I was able to get the whole thing down in one hour looking back on it.[/quote]

I actually proved a sequence $ k_i$ for which $ k_0...k_n\minus{}1$ is a product. It wasn't that bad. (Also, my sequence has k_0=3, so k_0-1=2=1*2 and so n=0 works in my sequence also.)

I'm not sure if this works, but:

let $ k_1k_2\cdots k_n\equal{}x^2\plus{}x\plus{}1$
So we want to find infinitely many primes $ p$ such that $ x^2\plus{}x\plus{}1\equiv0\pmod p$
Since $ x$ is not $ 1 \pmod p$ except when $ x\equal{}3$, we can rewrite the above equation as
$ x^3\equiv1\pmod p$
By FLT, $ x^{p\minus{}1}\equiv1\pmod p$
Since $ p$ is prime, we can take the roots of this and we see that the original equation must be true if $ 3$ divides $ p\minus{}1$, or $ p\equiv1\pmod3$. Since it is known that there are infinite primes $ 1\pmod3$, then it is proven.

Doesn't that work?

[quote="MellowMelon"]Was anyone else stupid enough like me to overcomplicate the problem and show by induction how to explicitly construct $ k_0 k_1 \ldots k_n$ for each $ n$? I'm actually surprised I was able to get the whole thing down in one hour looking back on it.[/quote]

I did worse; for some random reason I thought I had to show that each of the $ k_i$s had to be prime .. and in the last half an hour I realized my mistake and fixed it.

just in your opinions, whats needed to get 1 point in this problem?
if  you preceded by induction for example...

[quote="nat mc"]I'm not sure if this works, but:

let $ k_1k_2\cdots k_n \equal{} x^2 \plus{} x \plus{} 1$
So we want to find infinitely many primes $ p$ such that $ x^2 \plus{} x \plus{} 1\equiv0\pmod p$
Since $ x$ is not $ 1 \pmod p$ except when $ x \equal{} 3$, we can rewrite the above equation as
$ x^3\equiv1\pmod p$
By FLT, $ x^{p \minus{} 1}\equiv1\pmod p$
Since $ p$ is prime, we can take the roots of this and we see that the original equation must be true if $ 3$ divides $ p \minus{} 1$, or $ p\equiv1\pmod3$. Since it is known that there are infinite primes $ 1\pmod3$, then it is proven.

Doesn't that work?[/quote]
Can you take a root in a modulus?
What I'm unsure about (though I'm probably missing something obvious) is $ x^{p\minus{}1}$ being able to imply $ x^{3}$, because $ x^{\frac{p\minus{}1}{3}}$ could be 1 mod p, right?

[quote="MellowMelon"]Was anyone else stupid enough like me to overcomplicate the problem and show by induction how to explicitly construct $ k_0 k_1 \ldots k_n$ for each $ n$? I'm actually surprised I was able to get the whole thing down in one hour looking back on it.[/quote]

Well, that's what I did, and it didn't take too long :)

We proceed by induction to show a such construction.\\
Base Case: $n = 1$, Notice that $k_0 =3,k_1=7$ suffices since $$k_0k_1-1 = 21-1 = 20 = 4\cdot 5$$
Assume that for some positive integer $n-1$, we have pairwise coprime positive integers $k_i$ such that  $$k_0\cdots k_{n-1}-1 = a(a+1)$$ for some integer $a$, and that let $S:= k_0\cdots k_{n-1}$. Now we claim that for case $n$, the positive integers $$k_0,\cdots ,k_{n-1},S^3+(2a+1)S+1$$ works, where $k_i$ for $0\leq i\leq n-1$ is the same as the $k_i$s for the $n-1$ case. Notice that $$k_0\cdots k_{n-1}\cdot k_{n} = Sk_n = S^4+(2a+1)S^2+S = S^4+(2a+1)S^2+a(a+1)+1$$ $$\Rightarrow k_0\cdots k_n-1 = S^4+(2a+1)S^2+a(a+1) = (S^2+a)(S^2+a+1)$$ which are evidently two consecutive integers. Lastly, notice that $k_n$ is indeed coprime to $k_0,\cdots, k_{n-1}$ Note that $$\gcd(k_n,S) = \gcd(S^3+(2a+1)S+1,S) = 1$$ which also implies that $k_n$ is also coprime to all the divisors of $S$. And since $k_i$ for $0\leq i\leq n-1$ are divisors of $S$, this construction works and we are done.

We use induction. The base case is trivial. Suppose that $k_0k_1\cdots k_n-1=c(c+1)$. Then, choosing $k_{n+1}=c^2-c+1$ suffices. Notice that this is relatively primes to all of $k_0, k_1, \cdots, k_n$ because $k_0k_1\cdots k_n=c^2+c+1,$ and obviously $\gcd(c^2+c+1, c^2-c+1)=1$(because their gcd is max 2, and clearly both are odd). We have that $k_0k_1\cdots k_n(c^2-c+1)-1=c^2(c^2+1)$. We are done.

oops

First, note that the given condition is the same as saying that there always exists an integer $a$ such that $a^2 + a + 1$ has at least $n$ distinct prime factors, $p_i$ for $1 \le i \le n$. Therefore:
$$ a^2 + a + 1 \equiv a^2 + (p+1)a + 1 = 0 \pmod{p_i}.$$
Assume that $p_i > 2$. Then, we can write this as:
$$ \left(a+\frac{p_i+1}{2}\right)^2 - \left(\frac{p_i+1}{2}\right)^2 + 1 \equiv 0 \pmod{p_i} \Longrightarrow \left(a+\frac{p_i+1}{2}\right)^2 \equiv \left(\frac{p_i+1}{2}\right) \cdot \left(\frac{p_i-1}{2}\right) \pmod{p_i}.$$

This means that $a^2 + a + 1 \equiv 0 \pmod{p_i}$ is going to have a solution iff $\left(\frac{p_i+1}{2}\right)\left(\frac{p_i-1}{2}\right)$ is a QR $\mod{p_i}$. Thus:
$$ \left(\frac{(\frac{p_i+1}{2})(\frac{p_i-1}{2})}{p_i}\right) \cdot 1 = \left(\frac{(\frac{p_i+1}{2})(\frac{p_i-1}{2})}{p_i}\right) \cdot \left(\frac{4}{p_i}\right)  = \left(\frac{p_i^2 - 1}{p_i}\right) = \left(\frac{-1}{p_i}\right) = 1.$$
This is only true iff $p_i \equiv 1 \pmod{4}$, and by Dirchlet there are infinetly many primes $p_i$ which satisfies this. Therefore, there are infinetly many primes $p_i$ for which $a^2 + a + 1 \equiv 0 \pmod{p_i}$ has a solution. Let $a \equiv k_i \pmod{p_i}$ be the solution. By CRT, there is an $a$ which satisfies all $n$ congruencies, so such an $a$ must exist, we are done!!! $\Box$

It suffices to show that there exists a number of the form $m^2+m+1$ having at least $n+1$ prime divisors for any $n.$ Indeed, let $p_1, p_2, \dots, p_{n+1}$ denote the first $n+1$ primes equivalent to $1 \pmod 3.$ From Dirichlet's there are infinitely many such primes and thus all $n$ are attainable. For a given $p_i$ with primitive root $g,$ set
$$m \equiv g^{(p-1)/3} \pmod {p_i},$$
Whence by CRT there always exists such $m.$

Don't feel like writing a detailed solution for this silly problem.
Take $k_0=2^2+2+1$, and $k_i=2^{2^i}-2^{2^{i-1}}+1$ for all $i \geq 1$. Then doing Euclidean Algorithm stuff shows that all the terms are relatively prime. Further their product minus one is $2^{2^{n+1}}+2^{2^n}=2^{2^n}(2^{2^n}+1)$ which is indeed the product of 2 consecutive integers.

Why did so many people overcomplicate the problem - posting for storage.

Notice that we want to prove that $n^2+n+1$ has arbitrarily large numbers of prime divisors. \\
By Schur's Lemma, we have infinitely many prime divisors of $n^2+n+1$, take $k$ of them $p_1,..,p_k$, notice that if $$p_i \mid n^2+n+1$$
Then also $$p_i \mid (n+p_ik)^2+n+p_ik+1$$ and consequently, we can construct an $n \in \mathbb{N}$ with at least $k$ distinct prime divisors by $CRT$. $\blacksquare$ \\

We want to show that $a^2+a+1$ has an arbitrarily large number of prime divisors.
Take $k$ pairwise relatively prime numbers $q_1,q_2,\dots, q_k$ from this and let $p_i^2+p_i+1=q_i$, we see that $q_i\mid a^2+a+1$ for all $a\equiv p_i\pmod{q_i}$. Now CRT guarantees a solution. As $k$ grows arbitrarily large we are done.

Actually being pairwise relatively prime is not needed; we can just pick $k$ that is sufficiently large.

Note that by dirichlet's, there are infinitely many prime in the form $3k+1$. Let $p_0,p_1.\dots,p_n$ be distinct prime numbers in the form $3k+1$. Let $\omega_i$ be a primitive root modulo $p_i$. Then $p_i\mid \omega_i^{p-1}-1$ but $p_i\nmid \omega_i^{\frac{p-1}{3}}-1$ so $p_i\mid (w_i^{\frac{p-1}{3}})^2+w_i^{\frac{p_i-1}{3}}+1$. By CRT, there exist positive integers $x$ such that $x\equiv w_i^{\frac{p_i-1}{3}}\pmod{p_i}$ for all $i=0,1,\dots,n$. Hence $p_0p_1\dots p_n\mid x^2+x+1$ so $x^2+x+1$ has at least $n+1$ prime divisors, and we're done.

Edit: Oops, in fact we don’t need primitive root. Schur’s kill it: we have that there are infinite many prime $p$ dividing $x^2+x+1$ for some $x$ and then do the same. (Use CRT)

yay

Note that it sufficies to show that if $m$ is a positive integer, the number of distinct prime divisors of $m(m+1)+1$ is unbounded. This is because if $m(m+1)+1$ has at least $n+1$ distinct prime divisors, it is always possible to pick the $k_i$.\\

We can rewrite this as $$\frac{m^3-1}{m-1}.$$ Therefore, it is divisible by $p$ if $m^3\equiv 1\pmod p$ but $m\neq 1\pmod p.$ \\

Claim: If $p$ is a prime for which $p\equiv 1\pmod 3$, then there exists $m$ such that $m^3\equiv 1\pmod p$ and $m\neq 1\pmod p.$ Let $r$ be a primitive root mod $p$. We can then pick $$m=r^{(p-1)/3}.$$

Pick a set of primes that are 1 mod 3. For each of these chosen primes $p$, pick a residue for $m\pmod p$ so that $m^3\equiv 1\pmod p$ but $m\neq 1\pmod p$ (which is possible due to our claim and the Chinese Remainder Theorem). This will guarantee that $\frac{m^3-1}{m-1}$ is divisible by every prime that we picked, so the number of prime divisors of $\frac{m^3-1}{m-1}$ is unbounded as we can pick any number of primes. Therefore, we are done.

[quote=CyclicISLscelesTrapezoid]Let $f(x)=x^2+x+1$. Check that $f\left(x^2\right)=f(x)f(x-1)$ and $\gcd(f(x),f(x-1))=1$, so $f\left(x^2\right)$ has strictly more prime factors than $f(x)$ for large enough $x$, so we can guarantee that $f\left(x^{2^{n-1}}\right)$ has at least $n$ distinct prime factors, done.[/quote]

wait wow that is smart

Let $f(x)=x^2+x+1$. If we can prove that an infinite number of $p$ satisfy $p|f(x)$ for some $x$, then we are done by CRT.

Now I claim that all $p \equiv 1 \pmod{3}$ work. Note it suffices to show that $p | \frac{x^3-1}{x-1}$. Assume $x \neq 1$, then we need $p|x^3-1 \implies x^3 \equiv 1 \pmod{p}$. It suffices to show there exists some $x\neq 1$ that satisfies this. To construct, consider a generator $g$, and note $g^{\frac{p-1}{3}}$ works. Thus, an infinite number of different primes divide $f(x)$, so by CRT, we can always form such a construction.

[quote=worthawholebean]Prove that for each positive integer $ n$, there are pairwise relatively prime integers $ k_0,k_1,\ldots,k_n$, all strictly greater than $ 1$, such that $ k_0k_1\ldots k_n\minus{}1$ is the product of two consecutive integers.[/quote]

Take $k_0=3,k_1=7$ and $ k_0k_1\ldots k_n-1=a_n(a_n+1)$. Our objective is to construct $k_{n+1}$. Note that $$(a_n^2+a_n+1)k_{n+1}=a_{n+1}^2+a_{n+1}+1,$$ so it suffices to construct $a_{n+1}$ such that $a_n^2+a_n+1$ divides $a_{n+1}^2+a_{n+1}+1$ and their quotient is relatively prime with $ k_0k_1\ldots k_n$.

Firstly, let me prove that if $-3$ is quadratic residue modulo $p>3$, then $-3$ is also quadratic residue modulo $p^n$ for all $n$. Let $-3\equiv x_n^2\pmod{p^n}$ and pick $x_{n+1}\equiv \frac{x_n^p}{(-3)^\frac{p-1}{2}}\pmod{p^{n+1}}$, then$$x_{n+1}^2\equiv \frac{x_n^{2p}}{(-3)^{p-1}}\equiv\frac{(-3)^p}{(-3)^{p-1}}\equiv -3\pmod{p^{n+1}}$$as desired by the LTE lemma. $\square$

For convenience, let $x:=a_{n+1}, y:=a_n$ and $A=(2y+1)^2+3$. We need $y^2+y+1\mid x^2+x+1$ or $(2y+1)^2+3\mid (2x+1)^2+3$. Let $p_i$ be all prime divisors greater than $3$ of $A$. By the above lemma, there exists $b_i$ such that $p_i^{\nu_{p_i} (A)+1}\mid b_i^2+3$. Let $c_i=kb_i$, where $$k\equiv 1-p_i^{\nu_{p_i} (A)}\pmod{p_i^{\nu_{p_i} (A)+1}},$$ then $$c_i^2\equiv -3k^2\equiv -3+6p_i^{\nu_{p_i} (A)}\pmod{p_i^{\nu_{p_i} (A)+1}}.$$ Note that $A\equiv 3\pmod{9}$ or $3\nmid A$, so by the CRT, there exists an (infinitely) $x'$ such that$$\begin{cases}
    x'\equiv c_i\pmod{p_i^{\nu_{p_i} (A)+1}}\\
    x'\equiv 0\pmod{3}\\
    x'\equiv 1\pmod 2
\end{cases}$$Combining with the fact that $y^2+y+1$ is odd, we get $\nu_{p_i}(y^2+y+1)=\nu_{p_i}(x^2+x+1)$ and the conclusion follows.

[hide=s][b]Induction
[/b]
We have $k_0k_1...k_n=a_n^2+a_n+1$
$\, \, \, \, \, \, \, \,$and need $ k_0k_1...k_nk_{n+1}=a_{n+1}^2+a_{n+1}+1  ,$ where $\left ( k_{n+1},k_0k_1...k_n \right )=\left ( k_{n+1},a_n^2+a_n+1\right )=1, (1).$
Which means $\left (a_n^2+a_n+1  \right )k_{n+1}=a_{n+1}^2+a_{n+1}+1 \leftrightarrow k_{n+1}=\frac{a_{n+1}^2+a_{n+1}+1}{a_n^2+a_n+1}. $
From $(1),$ we need for any prime $p \mid a_n^2+a_n+1,$ $v_p\left ( k_{n+1} \right )=0 \leftrightarrow v_p\left (a_{n+1}^2+a_{n+1}+1  \right )=v_p\left (a_{n}^2+a_{n}+1  \right ).$
So we choose $a_{n+1}: a_{n+1}^2+a_{n+1}+1 \equiv a_{n}^2+a_{n}+1 \left ( \textrm{mod }\left ( a_n^2+a_n+1 \right )^2 \right ).$
Or $a_{n+1} \equiv a_{n} \left ( \textrm{mod }\left ( a_n^2+a_n+1 \right )^2 \right ).$

I think y'all are overcomplicate this prb  ;)

[b][i]Q.E.D[/i][/b][/hide]


Somehow I have the checkmark for Round 4 but no scores.  Could anyone please help me with this?

Please email usamts@usamts.org with your USAMTS ID#.


LaTeX
complex numbers
absolute value
real analysis
real analysis theorems

Dear…
I have a problem in MATLAB solution. I want to show my problem as an example:

-32^0.2=?
1- -32^0.2=-2
2- (-32)^(0.2)= 1.618033988749895 + 1.175570504584946i
3- -32^0.2=2

The first and second answers are MATLABs and the last one is manual, which are different. I understand that it refer to the solution.
How can I notify the MATLAB to solve it as one of them (For example third answer). However, my problem is not as easy as this example and I have a long equation that I can’t extract it and put it in parenthesis or out of parenthesis.
Thanks a lot.
A.Pashaei
(Amir_Pashaie@Yahoo.com)

[quote="JBL"]Not at all clear to me what your question is, but it is not the case that $ 2$ is a fifth root of $ \minus{} 32$, i.e., $ ( \minus{} 32)^{0.2} \equal{} 2$ is false.  Every nonzero complex number has five distinct fifth roots, and these fifth roots vary from each other by multiples of the fifth roots of $ 1$.  Thus, to compute all five values of $ a^{\frac {1}{5}}$ for a complex number $ a$ (such as $ \minus{} 32$), it is enough to know a single value and be able to compute one of the complex fifth roots of $ 1$.[/quote]

Dear JBL,
Sorry for my poor english, (because my language is not english)
During my calculation in MATLAB software, I found a complex number which result from such calculation that I explain if above. For example, $ (\minus{}0.8773636)^{0.2480474}$ ...
As answer, MATLAB give me a complex number, which I`m not sure about its accuracy. I searched about it and find a web page shows me that some software for solving as quick solution give a complex answer which is not true [url]http://www.purplemath.com/modules/exponent5.htm[/url].
I think the answer 2 is correct because if you write the equation in this form: $ {10} sqrt {(\minus{}32)}^{2}$ you'll have $ {10} sqrt{(1024)}\equal{}2$ I mean that root of $ {10}$ of $ {1024}$ is equal to $ {2}$.

I don't know which one is my answer and because my problem is in physics subject, I guess the complex answer is not true. But how can I notify the MATLAB to answer me NOT in complex form.

Again, sorry for my poor english but this problem is very important to me. So thanks for your attention.
A.Pashaie
Amir_Pashaie@Yahoo.com

To do roots in LaTeX, write \sqrt[10]{1024} = 2 surrounded by dollar signs to get $ \sqrt[10]{1024} \equal{} 2$.  The "exponent rule" $ \left(a^b\right)^c \equal{} a^{bc}$ is valid when $ a$ is a positive real number and $ b$ and $ c$ are real numbers, but it is [i]not[/i] valid in other cases: for example, just consider the value $ \left((\minus{}i)^2\right)^{\frac{1}{2}}$, where $ i : \equal{} \sqrt{\minus{}1}$.

(Note that your analysis also leads to the false conclusion that $ \minus{}1 \equal{} 1$ by starting with $ \minus{}1 \equal{} \sqrt{1}$ and squaring both sides.)

However, you've now asked about non-rational exponents, and those behave rather differently.  In this case, there are infinitely many complex numbers represented by an expression like $ a^b$.  Luckily, because the exponent is still real, these numbers all have the same magnitude (absolute value) -- thus, the thing closest to the meaning you want is probably to take the negative of the magnitude of whatever MATLAB gives you.


calculus
integration
function
real analysis
real analysis unsolved

Let $ f: [0. \infty) \minus{} > [0, \infty)$ be a increasing function with the property $ f(x) \geq \int_{0}^{x}f(2t^{2})dt ,\forall x \geq 0$.Prove that $ f \equal{} 0$.


A bus ticket has six digits on it. It's considered to be lucky if the sum of the first three digits equals to the sum of the last three.
Prove that the sum of all the lucky numbers is divisible by 13.

Let $R_{k}$ be the set of all 3 digit numbers (include leading 0) that sum to $k$.  Essentially, we require that

$\sum_{k}\sum_{x\in R_{k}, y\in R_{k}}1000x+y \equiv 0 \mod{13}$

$\Leftrightarrow \sum_{k}\sum_{x\in R_{k}}1001|R_{k}|x \equiv 0 \mod{13}$

$\Leftrightarrow 13* \Bigg( 77* \sum_{k}\sum_{x \in R_{k}}|R_{k}|x \Bigg) \equiv 0 \mod{13}$

where the result follows.


Chris you seem to be asking a lot of definition questions. Might I suggest you go to mathworld.wolfram.com and type your questions into their search. All of the questions you have asked are explained perfectly on that site. Come here if you need clarification on something, but it's kind of a pain for people to sit here and type out what you could have just looked up. I know because I typed out the formal definition of a limit only to then realize there were two posts and someone had ALREADY done it. It's kind of upsetting....

http://mathworld.wolfram.com/CevasTheorem.html


search
combinatorics unsolved
combinatorics

For any simple graph $ G$ with directed edges, we denote the sets $ N_1(v)$ and $ N_2(v)$ for each $ v \in V(G)$ as follows:
$ N_1(v)\equal{} \{ u: v \rightarrow u \}$ and $ N_2(v)\equal{} \{ u: v \rightarrow u \ or \ v \rightarrow w \rightarrow u \}$.
(Note: $ N_1(v)$ is all the vertices whose distance to $ v$ is 1, while $ N_2(v)$ is all the vertices whose distance to $ v$ is at most 2)
Prove that there exists a vertex $ v$ such that $ |N_2(v)| \geq 2|N_1(v)|$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1139615159&t=237602]This result[/url] is a special case, and I think it's easier to prove, although I really don't know.


algebra
system of equations
algebra unsolved

Find the solutions to the following system of equations:

\[ x\plus{}y\equal{}\sqrt{4z\minus{}1}\]
\[ y\plus{}z\equal{}\sqrt{4x\minus{}1}\]
\[ x\plus{}z\equal{}\sqrt{4y\minus{}1}\]

Let $ \frac 14 x\le y\le z$, then $ y\plus{}z\ge x\plus{}y\equal{}\sqrt{4z\minus{}1}\ge \sqrt{4x\minus{}1}\equal{}y\plus{}z$. Therefore $ x\equal{}y\equal{}z\equal{}\frac 12.$


Find coordinates of the focus for the following problems.

i.	y = x2 + 6x + 9
ii.	y = -x2 - 4x
iii.	y = -2x2 + 9x - 9
iv.	x = -2y2 + 7y - 6
v.	x = y2 - 2y

x2 means x to the power of 2
y2 means y to the power of 2

[hide]
1. First find the vertex $(\frac{-b}{2a},f(\frac{-b}{2a}))$
For $y=x^2+6x+9$, the vertex is $(3,0)$
Then find the distance between the focus and the vertex by writing the coefficient of the $x^2$ term like this: $\frac{1}{4p}$ where $p$ is a real number. So in this case, $1=\frac{1}{4p}$
$p=\frac{1}{4}$ The parabola opens up so the focus is at $(3,\frac{1}{4})$
I think this is right. I actually just googled "focus of parabola" and found this. :D You can use this method for the other parabolas.[/hide]

[hide]
2. $y=-x^2-4x$  vertex: $(-2,4)$
$p=-\frac{1}{4}$  focus: $(-2,3\frac{3}{4})$

I don't have much experience with these.
5. $x=y^2-2x$  vertex: $(-1,1)$ I think. :?
$p=\frac{1}{4}$  focus: $(-\frac{3}{4},1)$[/hide]
I don't feel like doing the others.  :roll:

[quote="0714446459923"][quote="DanK"]Yeah, and once you've rewritten them in the form $(y-k)^2=4p(x-h)$ or $(x-h)^2=4p(y-k)$, the foci are $(h+p,k)$ or $(h,k+p)$, respectively...[/quote]

Don't you have those backwards?[/quote]

I don't think they're backwards. The first one expands into $x=\frac{1}{4p}(y-k)^2+h$ so the focus is at $(h+p,k)$.

The parabola either opens up or down for the second one.  Therefore $h$ is the x-coordinate of the focus and the y-coordinate depends on $k$.  The opposite holds for the parabola opening sideways.

oops I misread those. I thought the squared was on the other side.


A point object of mass m is connected to a cylinder of radius $ R$ via a massless rope. At time $ t \equal{} 0$ the object is moving with an initial velocity $ v_0$ perpendicular to the rope, the rope has a length $ L_0$, and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds $ T_{max}$. Express your answers in terms of $ T_{max}$, $ m$, $ L_0$, $ R$, and $ R_0$.

What is the angular momentum of the object with respect to the axis of the cylinder at the instant that the rope breaks?

I think that angular momentum isn't conserved because an external torque has to prevent the cylinder from spinnning. Your thoughts?

I figured this out. 2 hints:

1) Express angular momentum in terms of the length of the rope, express tension in terms of the length of rope, do a substitution.

2) Energy is conserved, while angular momentum is not.

Why is energy conserved? Isn't work done as the mass becomes closer and closer to the cylinder with tension pulling on it? Or is that just internal work? How do we know there is no external work?

If we say that energy increases because work is done, [i]from where does the mass receive additional energy?[/i]
The definition of work also supports this: as the dot product between the tension and displacement vectors, work is zero for the mass whose displacement at every point is perpendicular to the tension in the rope.

External work may be involved in preventing the cylinder from rotating. I shouldn't make assumptions on this point, so can someone else answer the question: Why does an external torque, which prevents the cylinder from rotating, produce no effect on the $ \frac{1}{2}mv^2$ of the revolving mass?


Romanian Junior Balkan Team Selection Test V, Bucharest, 6 June 2007
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152598]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152606]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152607]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152608]Problem 4[/url]


Να αποδειχθεί ότι κάθε παλίνδρομος αριθμός που δεν είναι πολλαπλάσιο του 10 έχει πολαπλάσιο που είναι πάλι παλίνδρομος !

[b]Σχόλιο ![/b]
 Είναι από τη γερμανική ολυμπιάδα 2009.
Βάζω και την άσκηση στο πρωτότυπο.

[quote="maniopas"]Ερώτηση: αν κατάλαβα καλά παλίνδρομος είναι αυτός που έχει συμμετρικά τα ψηφία του;

Τότε προφανώς αν πάρουμε τον αριθμό 100...001 όπου τόσα μηδενικά όσα τα ψηφία του παλίνδρομου+1 και τον πολλαπλασιάσουμε με παλίνδρομο, έχουμε πάλι παλίνδρομο (αυτό θέλει φυσικά λίγο παραπάνω μπλα μπλα, αλλά αυτή πρέπει να είναι η λύση).

Γενικεύοντας, μπορούμε να αποδείξουμε την ύπαρξη άπειρων παλίνδρομων πολλαπλασίων παλίνδρομου:

[hide]
τα πολλαπλάσια με τους αριθμούς 100...001 για τους οποίους το πλήθος μηδενικών είναι μεγαλύτερο από τα ψηφία του παλίνδρομου
[/hide][/quote]

Αυτό είναι το νόημα, αλλά ο παλίνδρομος δεν έχει υποχρεωτικά τη μορφή που έγραψες. Τα ψηφία μπορεί να είναι τυχαία : πχ

    12355321 ή  151617716151

 Μπάμπης

[quote="maniopas"]Εννοώ ότι πολλαπλασιάζοντας τον παλίνδρομο με τον αριθμό 100...001 έχω πάλι παλίνδρομο. π.χ
12321*100001 = 1232112321[/quote]
Κι όμως, δεν ισχύει για κάθε παλίνδρομο. π.χ. 73837*10001=738443837

Ίσως να ισχύει μόνο γι' αυτούς που το πρώτο και το τελευταίο ψηφίο τους είναι 1.

Κώστας Βήττας.

Ισχύει όμως $ 73837\cdot 100001\equal{}7383773837$. Γενικότερα αν ο αρχικός παλίνδρομος αριθμός έχει $ n$ ψηφία τον πολλαπλασιάζουμε με τον $ 100...001$ που έχει $ n\plus{}1$ ψηφία και προκύπτει ο αρχικός αριθμός γραμμένος 2 φορές!

[quote="bilot4"]Ισχύει όμως $ 73837\cdot 100001 \equal{} 7383773837$. Γενικότερα αν ο αρχικός παλίνδρομος αριθμός έχει $ n$ ψηφία τον πολλαπλασιάζουμε με τον $ 100...001$ που έχει $ n \plus{} 1$ ψηφία και προκύπτει ο αρχικός αριθμός γραμμένος 2 φορές![/quote]

Αυτό ακριβώς εννοώ!


All help is greatly appreciated!

20. The square matrix X is such that $X^{3}=0$. Show that the inverse of the matrix $(I-X)$ is $I+X+X^{2}$.

I've had basic matrices, but I don't even know what it means that $X^{3}=0$; does it mean that the sum of all the elements of $X^{3}$ is 0?

Thought: does the answer have anything to do with the fact that, therefore, $I^{3}-X^{3}=I^{3}=(I-X)(I+X+X^{2})$? (It works for polynomials... but if it works for matrices, I don't understand why).

Thanks!

$X^{3}=0$ means that all entries of the matrix $X^{3}$ are equal to $0$.
You are right: the identity $I^{3}-X^{3}=(I-X)(I+X+X^{2})$ is the key to solution. You can verify this identity by multiplying out the terms on the right.

Thanks... but I still have questions:

Since $X^{3}$=0, does that mean $X=0$ necessarily?

Ah, and a typo correction: the I on the RHS should be I squared, but that doesn't seem to have tripped you up, which is good.

Well if $X=0$ then the solution is obvious, because $(I^{2}+IX+X^{2})(I-X)=I^{3}=I$, which makes an inverse by definition.

Hmm, I suppose there's no reason matrices shouldn't multiply out like polynomials... you have to watch commutative stuff, but otherwise it should work, right?

Aha! Actually it doesn't matter what X is, becuase if that factoring/expanding is correct, we have an inverse no matter what. Zing! But still, do we necessarily have X=0?

An example of a nonzero matrix $X$ with $X^{3}=0$:
$X=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$.


advanced fields
advanced fields unsolved

I was recently reading a book on nonstandard analysis, and I was sorta wondering, what is the cardinality of the hyperreals? It seems to me that it should be more than the reals, but I can't come up with a convincing argument.

If people want, I can give a definition of the hyperreals, but I don't have time at the moment. Just ask, and I'll get to it this evening or something.

Since there is a surjection from the collection of sequences of reals to the hyperreals (a preorder $\le$ is defined on the collection of sequences of reals, and then two sequences $a,b$ are identified if $a\le b,b\le a$; the mentioned surjection is the natural projection from the collection of sequences onto the equivalence classes that this identification creates), and the collection of sequences of reals has as many elements as the reals, there cannot be more hyperreals than reals in the cardinality sense. Conversely, since the reals can be embedded in the hyperreals, the latter set has at least as many elements as the reals. This means that the two cardinalities are equal.

Ah, I guess I was guilty of the sort of fallacious argument like "there are way fewer powers of 2 than integers, so the cardinality of the powers of two must be less" (which is more obviously wrong than my thought for the hyperreals, but same concept I guess).

Thanks!


im using TexnicCenter and my outpuf profile is "LaTex -> PDF".
It was working when i compile and run output. All of a sudden when i compile again, the process became extremely slow, and eventually get stuck at "Loading: Context support macro pdf". It wouldnt work even after i reinstall it..

USAMTS is going to be due soon, i would like this to work.. asap..   thanks

Try rebooting your computer and running a simple $\LaTeX$ document such as the following (copy paste into a blank .tex file) and see if it outputs a pdf.

[code]
\documentclass[11pt]{article}
\begin{document}
Hello world!
\end{document}
[/code]

interestingly it does work.. it compiles successfully and shows the pdf document when i press F5

but my solution.tex still doesnt, thanks for trying to figure out whats wrong

Then there is nothing wrong with your $\LaTeX$ system (Miktex + Texnic center) but rather with your solution.tex file. Make sure you have no bad commands, that you didn't modify anything that you weren't supposed to from the template file, and check it in the following way: first make a backup of the current file. Then start deleting content and trying to compile until it does so. When it does, you know that the content you just deleted had some sort of error in it. Check it again (look for strange undefined commands \blabla or missing { } or missing dollar signs $\$$ ). Good luck with your submission.

Thanks! There were problems in my source code, and they are fixed now.

I hope i get to finish with my amateur LaTex skills.

Again, thanks a lot ^^

Isaac

You're welcome. Good luck on the USAMTS ;)


algebra
polynomial
logarithms
algebra proposed

Let $P(x)$ be a polynomial with non-negative integer coefficients and some unknown degree.
You may choose a value and I will give the value of the polynomial in that point.
It is possible to entirely determine $P(x)$ by such a guessing in a finite number of steps?
If you think it is possible, how many steps do you really need?
Note that you don't know the degree of the polynomial! It isn't cue?

[quote="Myth"]We will call all natural numbers step by step and we will construct Lagrange interpolational polynomial each time. Let $P_n(x)$ is an interpolational polynomial for $P(1)$, ..., $P(n)$. There is a first moment s.t. $P_k\in \mathbb{N}_0[x]$ and $P(k+1)=P_k(k+1)$, $P(k+2)=P_k(k+2)$, ..., $P(k\cdot M)=P_k(k\cdot M)$, $M\in\mathbb{N}$ is greater than all coefficients of $P_k$, $M\geq 4$.
We claim that $P=P_k$. Indeed, if $P\neq P_k$ then $\deg P\geq k\cdot M-1$. But it means that $P(x)\geq x^{k\cdot M-1}$, in particular \[P(k\cdot M)\geq (k\cdot M)^{k\cdot M-1}>(k\cdot M)^{k+1}>P_k(k\cdot M).\] Contradiction.[/quote]

C'mon Myth you can do much better.
Please, think a little bit more!  ;)

It is correct. In fact the answer is YES.
But how many steps are required?
I mean a finite number of steps that will not depend on the degree of the polynomial !?!

[quote="Myth"]Oh, I see... My poor head...

Calculate $M=P(2)$. Then calculate $P(M+2)$. Then the representation of $P(M+2)$ base $M+2$ gives us the coefficients of $P$.[/quote]

In fact you can even find $P$ with [i]one[/i] guess by asking for $P$ evaluated at any transcendental number, say, $\pi$.  If you ask for $P(\pi)$, the integer part of that value gives a bound on the degree of the polynomial -- it can't be larger than $\ln(P(\pi))/\ln(\pi)$, since all of the coefficients are nonnegative.  It also gives a bound on the magnitude of each of the coefficients of $P$.  Hence there are, at most, a finite number of possibilities for $P$.  Then, since $\pi$ is transcendental, the value of $P(\pi)$ is transcendental, and so you can take the value of $P(\pi)$ out to as many digits as necessary until you can eliminate all but one of the possible polynomials.  Of course, that may take a very long time, but it's possible.


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Here's another I don't get... Let
$ G \equal{} \langle a,b| a^8 \equal{} 1, b^2 \equal{} a^4, bab^{\minus{}1} \equal{} a^{\minus{}1} \rangle$
What is $ |G|$ and $ Z(G)$?

I tried to create some permutations $ a$ and $ b$ that satisfy these relations, and for some reason I just cannot construct them when $ a$ has order 8. Maybe I keep making some mistake, but it seems like it is impossible if my calculations are right. But I can't see why this would be true directly from the relations either, so I doubt what I am doing.

it's the quotient of $ \mathbb{Z}/8 \rtimes \mathbb{Z}/4$ (with $ \mathbb{Z}/4 \to Aut(\mathbb{Z}/8), 1 \mapsto (x \mapsto \minus{} x)$) by the smallest normal subgroup $ U$ that contains $ b^2 a^4$ where $ \mathbb{Z}/8 \equal{} \langle b \rangle, \mathbb{Z}/4 \equal{} \langle a \rangle$. now you can determine $ U$ and then the structure of the quotient. probably there are better methods ...

I'm lazy ...  :lol:

[quote="calculation with gap"]
gap> f := FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> g:=f/[f.1^8,f.2^2 * f.1^4, f.2^(-1)*f.1*f.2*f.1];
<fp group on the generators [ f1, f2 ]>
gap> List(Elements(g));
[ <identity ...>, f1, f1^7, f2, f1^4*f2, f1^2, f1*f2, f1^5*f2, f1^6, f1^7*f2,
  f1^3*f2, f1^4, f1^3, f1^2*f2, f1^6*f2, f1^5 ]
gap> Order(g);
16
gap> Center(g);
Group([ f2^2 ])
gap> Order(Center(g));
2[/quote]

with the above approach this can be done manually, and we have $ |U|\equal{}2$.

I think the structure can be determined by introducing a concrete group operation, like I did on the other question.

From the relations all elements can be written $ a^k b^j$ where $ 0 \leq k < 8$ and $ j$ is 0 or 1. It's easy to see how multiplication should be done, easier than in the last exercise (although honestly I've spent 3 hours on this exercise, but you always write everything is easy afterwards). Because you just use $ ba^m \equal{} a^{8 \minus{} m}b$. This does give a group with 16 elements, and it's as free as possible on the relations, so this describes the group presentation.

I still really don't understand why if we suppose $ a \equal{} (1 2 3 4 5 6 7 8)$ I cannot find a $ b$ that satisfies the relations. Since $ b^2 \equal{} a^4$, $ b$ must be the product of two 4-cycles, but it seems like no $ b$ works! It's driving me insane

ok, you realize the group directly instead of going a long way round as me.  :)

by the way, the group is known as $ Q_{16}$: http://planetmath.org/encyclopedia/QuaternionGroups.html

concerning the permutation representation:

[quote="gap"]gap> IsomorphismPermGroup(g);
[ f1, f2 ] -> [ (1,2,6,13,12,16,9,3)(4,10,15,8,5,11,14,7),
  (1,4,12,5)(2,7,16,8)(3,10,13,11)(6,14,9,15) ][/quote]

Is there a way to see directly that I cannot embed this group in $ S_8$, so I can save time by not trying?

look at the link above "and cannot be represented faithfully except by a regular representation" (including a proof). although this is about vector space representations, every set representation yields such one by permutation matrices.


AMC
AIME
calculus
ratio
integration
trigonometry
complex numbers

Given
$x=\frac{\displaystyle\sum_{n=1}^{44}\cos(n)}{\displaystyle\sum_{n=1}^{44}\sin(n)}$

Find $\left\lfloor100x\right\rfloor$

I know the answer is 241. I also know a calculus solution that simply uses the fact that the ratios of the definite integrals are the same as the ratio of the sums, that is:

$x=\frac{\displaystyle\sum_{n=1}^{44}\cos(n)}{\displaystyle\sum_{n=1}^{44}\sin(n)}=\frac{\displaystyle\int_{1}^{44}\cos(n)\:dn}{\displaystyle\int_{1}^{44}\sin(n)\:dn}=\frac{\sin(44)-\sin(1)}{\cos(1)-\cos(44)} \approx 2.41421$

Can someone please tell me how to solve this without calculus?
Thanks

there are a few approaches... (excuse me if i make mistakes, its late)
$\sin{a}+sin{b}=2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$
This can be derived from sin(x+y)+sin(x-y), setting a=x+y, b=x-y
$\cos{a}+cos{b}=2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}$ (similar derivation)
Now notice that 1+44=2+43...=22+23

so the expression is equivalent to
$\frac{2\cos{(\frac{45}{2})}\sum^{21}_{n=0}{\cos{n}+.5)}}{2\sin{(\frac{45}{2})}\sum^{21}_{n=0}{\cos{n}+.5)}}$
and now u can use the half angle formulas to find that it is $1+\sqrt{2}$

I believe another method is to make the sum telescope by multiplying through by cos(0.5) using the fact that sin(x+y)-sin(x-y)=2cos(x)sin(y) and thus 2sin(1)cos(.5)=sin(1.5)-sin(0.5), 2sin(2)cos(.5)=sin(2.5)-sin(1.5) and similarly for the denominator

Or you could do it
By DeMoivres Theorem:

$(\cos(1^{\circ})+i\sin(1^{\circ}))^n=\cos(n^{\circ})+\sin(n^{\circ})$

Thus our ratio is the real part to complex part of

$\sum_{n=1}^{44} ((\cos(1^{\circ})+i\sin(1^{\circ}))^n$

$=\frac{(\cos(1^{\circ})+i\sin(1^{\circ}))^{45}-1}{\cos(1^{\circ})+i\sin(1^{\circ})-1}-1$

$=\frac{\cos(45^{\circ})+i\sin(45^{\circ})-1}{\cos(1^{\circ})+i\sin(1^{\circ})-1}-1$

it comes out messy though

thanks u guys for all the solns.

I really like that last one a lot, it's very clever. I should have thought of that.

Anyway, thanks.


Here is a nice inequality in Math & Young,Vietnam:for $u\geq v+1$ and $\{a_i\},\{b_i\}$ is positive number:
$\sum\frac{a_i^u}{b_i^v}\geq \frac{(\sum a_i)^u}{n^{u-v-1}(\sum b_i)^v}$
  I guess that it will be [i]still true or revesed sign [/i]for v<0 but with the new condition(I yet find out).In these problems :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=quite+nice+triangle&t=33403 and http://www.mathlinks.ro/Forum/topic-30732-20.html
  I think if we can find out the correct form of the generalization we can reached solutions of above 2 inequal immediately...

[quote="Megus"]it was discussed before:

http://www.mathlinks.ro/Forum/viewtopic.php?t=20547[/quote]
they are different, nobody thought of the case v < 0 in the old post.

[quote="keira_khtn"]The following inequaly made me to expect more this conjecture is also true for u=v:
$\sum(\frac{a}{b+c})^k\geq \frac{3}{2^k}$
  where $k\geq \frac23$[/quote]

We have the more general and diffickut : Fraction Inequality!

Try to seach the forum.


A tachyon is a hypothetical particle that is able to travel faster than light. Let us consider an inertial observer $ \Sigma$ that has a tachyon emitter located in the origin of the coordinate system. He sends a signal with speed $ v_{T}>c$ to a second inertial observer $ \Sigma'$ moving with speed $ v$ in the positive $ x$ axis direction. Suppose $ \Sigma'$ receives the tachyon when it is at a distance $ D$ from the origin (measured with respect to $ \Sigma$), and that he immediately emits the signal back to the origin with a speed $ v_{T}$ relative to $ \Sigma'$. Find the time of traveling of the tachyon and show that it is possible to receive the signal before it is sent.

I always got confused with relative speeds, so I hope you can help me... :(

I am not sure you can apply the formulea of special relativity here. Using velocity addition etc. I am getting the absurd answer $ \Delta t\equal{}\frac{D}{v_{T}}\plus{}\frac{D}{\frac{v\minus{}v_{T}}{1\minus{}{vv_{T}}/{c^{2}}}}$


x,y is real numbers

|x|/2 + |y| = 1

find the max,min of
f(x,y) = x^2 - xy + y^2

Hi airrafer,

I posted a reply to this in the Pe-Olympiad (14+) forum

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302758[/url]


geometry
trapezoid
geometric transformation
homothety
angle bisector
geometry unsolved

$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC \equal{} \angle KDA$.

[i]Proposed by S. Berlov[/i]

Such an easy problem  :wink:

Let $ N \equiv QB \cap AD.$ Then $ QM$ is the Q-median of  $ \triangle QBC.$ Since $ BC \parallel ND,$ then $ P$ is midpoint of  $ ND.$ The triangle $ \triangle NDK$ is isosceles. So if $ R \equiv KD \cap BC,$ then $ \triangle KBR$ is also isosceles $ \Longrightarrow$ $ \angle QBC \equal{} \angle BRK \equal{} \angle KDA.$

Let $B' = QB%Error. "capAD" is a bad command.
$  and $R = KD%Error. "capBC" is a bad command.
$.
Notice that we only need to prove that $\triangle KB'D$ is isosceles, because if this happens, $\triangle KBM$ is also isosceles(because $B'D // BC$), and so $\angle KBR=\angle QBC = \angle KRB = \angle CRD=\angle RDA=\angle KDA$.
But $BC // B'D$, and so there're an homothety that goes from $\triangle QBC$ to $\triangle QND$, and if $M$ is the midpoint of $BC$, $P$ is the midpoint of $B'D$, what implies that $\triangle KND$ is isosceles.

Q.E.D.

Angle $ADK$=Angle $DMC$=Angle $KXM$=Angle $KBM$.

This is just one situation. $X$ is the point of intersection between $DK$ and $BC$.

From [b]Luis's[/b] solution we get $KP$ external bisector of angle $\angle QKD$, hence the parallel to the bases of trapezoid through $K$ is the internal angle bisector of $\angle QKD$, done.

Best regards,
sunken rock