This is where you talk about the difficulty of problems, exam, or just anything that doesn't relate to the solutions about the exam here.

RESULTS
Interesting fact is that AMC 12 takes people to AIME if their score is above 100 or top 5%.  I received 10 answers and top 5 people were also the only ones who made over 100. Even if your score is below 100, do not worry. This was a difficult exam.

First place
SplashD: 132

Second place:
n^4+4: 115.5

Third place:
azjps: 109.5

Fourth place:
alkjash: 108

Fifth place:
andersonw: 103.5

[hide="Rest"]
mnmath: 96
perfect628: 93
aufha: 72
aufha's brother: 67.2
Siddhant: 64.5 [/hide]

The details will be posted but #13 is actually a freebie to everyone (I didn't add that to scores though) because although I knew the answer, I made wrong answer choice (I added wrong while I"m making answer choice;  :P ).

You make some uber-uber-uber-hard problems. :(  I won't even bother sending them in. :noo:

amen. i won't bother either. i solved the first 10. made some progress on like 5 other ones, but didn't complete. i bombed.  :(

Nevermind. I did fail though. And now everyone can see my horribly incorrect mess...

Wait... which blog are you referring to, perfect628? How were we supposed to submit out answers?

Anyway, that was fun! I probably did much, much worse than all of you.
Oh well, I guess I'll have to train for next year to qualify for the AIME :)

m+n...
Did any one feel like they were sometimes doing the first couple problems from the AIME instead of the first half of an AMC?

Yea, this was very hard. I only answered 16 of them, and probably made a few careless mistakes.

@aufha:yes

The link at the top was the blog for submissions...

Anyway, I think 17 was a good one, but was ruined by its display. It gave away the solution.

[quote="perfect628"]The blog wouldn't [i]let[/i] me submit...[/quote]

I'm approving them as they are posted (yes, the blog thinks that those replies are spams so I have to do them individually).

I solved 14,guessed on 3, left 8 blank.
...I suck.

i had the exact same thought. "some of these problems look like old aimes iv seen before" (particularly the one about tossing the coin ten times)

Yeah, there were a few AIME like problems there.

(I hope its not a problem if my submission was at :32? started at :17) That was tough... answered 16, didn't even bother to read from question 20+. I recall at least 4 or so from AIME 10+s, #12 was a 5+ on an AIME, #16 was an 11 on an AIME, #19 an 10 (i think) on another, so forth.

Dang... I don't read directions very well.  I accidentally pm'd my answers to Silverfalcon.

[quote="perfect628"]omg...I just realized that I missed 13 solely because I put 44, as it ended with 4. I got the correct product of $ 2^{22}$, then didn't even read all the answers...lol...w/e I failed in general anyway. Made about 4 dumb mistakes.[/quote]
could you explain how you got taht?

I only got 18, and that was only because I recognized an AIME problem.
It was uber-hard.

I thought the test was extremely hard. :(

What was the correct answer to 14? B?

Should've tried #21, just solved it right now in half a minute.  :roll:  I have a <20% chance of making "aime" now.

Edit: Apparently #14 was E  :lol:

Results are posted on the first thread!  :)

Oof... 94.5. That was absurdly difficult :)

30 points lower than my last practice... Anyway, my favorite problem was the coin one as I got it. At least two of the problems I remembered how to do because I'd seen them on past AIMEs.

Yea...this was not a measure of AMC ability...The only way to do well on this test is to have seen the problems before. Most of the problems are medium to hard AIME level. The problems are not particularly impossible in themselves, but the time limit makes it a bit ridiculous.

It is still good practice for the AIME tests though...

The only one past the MC section I specifically remembered was #13, which everyone got. :mad: It's from 360 Problems for Mathematical Contests, by Titu Andreescu(sp? I don't have the book here.)

Ironically enough, #11 was the only question I remembered distinctly, besides the MC, (I think a #25 on a previous mock AMC?) and also one of the questions I skipped.

EDIT: nvm, I was thinking of #25 [url=http://wangsblog.com/jeffrey/mockAMC.pdf]here[/url]

Yeah, I've been taking it over the past hour or so, skipping problems I was confident I knew how to do but were kind of computationally intensive, but going to the point where all I had to do was find the numeric answer. I didn't mind the difficulty too much because I wasn't really timing myself.

I thought it was kind of funny that the last 3 were geometry, perhaps an insight into the author's prowess.

I recognized more than half, which was interesting. I enjoyed 17, 21, and 24.

17 would've made a nice short intro NT class problem.

[quote="1=2"][quote="perfect628"]#13, which everyone got.[/quote]

So I am nobody? That makes me sad. :(

Could the next Mock AMC possibly be easier? :stretcher:[/quote]

I think he meant that #13 was given for free because there was a typo in the answer choices. (or I could be nobody either)

[quote="1=2"][quote="perfect628"]#13, which everyone got.[/quote]

So I am nobody? That makes me sad. :(

Could the next Mock AMC possibly be easier? :stretcher:[/quote]

Don't worry. Many of the problems in this exam were ones that I couldn't get and learned from solutions (thus, by sharing it again, I review for myself as well).

I won't say anything positive about your second question because well, mock AMCs in AoPS are [i]usually[/i] harder than the real ones.


"An A4-paper has the sides 29,7 cm and 21 cm. If you make a cylinder of it, what is the values that the height and radius can have?"

[hide="My answer"]$ 0 < r < 7.425$
$ 0 < h < 29.7$

Am I right?[/hide]

Why should the height change?

$ h \equal{} 29.7$ or $ 21$

And solving for radius just apply $ 2\pi r \equal{} 29.7$ or $ 21$ and solve for $ r$.

I hope I'm interpreting this correctly.

[quote="SnipedYou"]Why should the height change?[/quote]You have to cut the paper in order to get the top and bottom. Bigger circles, less height.

[quote="AstroPhys"]Recheck the range of the radius $ r$. Notice that the diameter of a circle is twice the radius.  :wink:[/quote]:oops_sign:


what does it look like- i havent seen the design

It's at the top of the forum, next to the BLOGS button
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Comm_TShirt.php]T-Shirt[/url]


geometry
algebra
polynomial
area of a triangle
Heron\u0027s formula
cyclic quadrilateral
Galois Theory

In geometry, Heron's formula states that the area $ A$ of a triangle whose sides have lengths $ a, b,$ and $ c$ is $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ where $ s$ is the semiperimeter of the triangle.
Brahmagupta's formula gives the area of a cyclic quadrilateral whose sides have lengths $ a, b, c, d$ as $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)}$.
There does not seem to be a "similar" formula for a cyclic polygons with five or more "sides". If this is true, is there a link with Galois theory (We also know from Galois that quintic polynomials are non-solvable)? Or just a coincidence?

Hi¨Portland,
your formula for $ S$ gives a dimension equal to $ 2.5$ and is therefore false.
In the true formula the term $ "s"$ does not exist. Then if you change $ d$ into $ 0$ the quadrilateral becomes a triangle and you obtain the Heron's formula. Thus a priori, Galois theory has nothing to do here.

No, there is a link- because the key to things is whether we can solve the polynomial equation for the radius/diameter of the circumcircle. It doesn't lead neatly to a formula, but it is a method.

Hi loupblanc,
Thanks for noticing the mistake, instead of
[quote="Portland"]Brahmagupta's formula gives the area of a cyclic quadrilateral whose sides have lengths $ a, b, c, d$ as $ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$.[/quote]
it should read $ A \equal{} \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$.
Hi jmerry,
Thanks. Would mind elaborating a little bit more?


[quote="236factorial"]The four interior angles of a paticular quadrilateral are in the ratio 2:4:4:5. What is the measure of the smallest interior angle in this quadrilateral?[/quote]

[hide]Say that half the measure of the smallest interior angle is $x$. Then...

$2x+4x+4x+5x=360$

$15x=360$

$x=24$

So the smallest angle is $48$[/hide]


geometric transformation
geometry
circumcircle
trapezoid
parallelogram
perpendicular bisector
geometry unsolved

Two circles, of different radius, with centres at $ B$ and $ C$, touch externally at $ A$. A common tangent, not through $ A$, touches the first circle at $ D$ and the second at $ E$. The line through $ A$ which is perpendicular to $ DE$ and the perpendicular bisector of $ BC$ meet at $ F$. Prove that $ BC\equal{}2AF$.

Let $ P \equiv DE \cap BC$ and $ Q$ the orthogonal projection of $ A$ on $ DE.$ Since $ P$ and $ A$ are the exsimilicenter and insimilicenter of $ (B) \sim (C),$ we have that $ (B,C,A,P) \equal{} \minus{} 1$ $ \Longrightarrow$ $ DE,QA$ bisects $ \angle BQC$ externally and internally $\Longrightarrow$ $ QA$ and the perpendicular bisector of $ BC$ meet on the circumcircle of $ \triangle BQC$ $ \Longrightarrow$ $ F \in \odot(BQC).$

In the right trapezoid $ DECB$ with bases $ DB,EC,$ we have

$ \overline{AQ} \equal{} \frac {\overline{BD} \cdot \overline{AC} \plus{} \overline {CE} \cdot \overline{AB}}{\overline{AB} \plus{} \overline{AC}} \equal{} \frac {2 \cdot \overline{AB} \cdot \overline{AC}}{\overline{AB} \plus{} \overline{AC}} \ \ (1)$

From power of $ A$ to $ \odot(BQC),$ we have  $ \overline{AF} \equal{} \frac {\overline{AB} \cdot \overline{AC}}{\overline{AQ}} \ \ (2)$

From $ (1)$ and $ (2)$ we obtain $ \overline{AF} \equal{} \frac {_1}{^2}(\overline{AB} \plus{} \overline{AC}) \equal{} \frac {_1}{^2}\overline{BC},$ as desired.

Let $ D,N,E$ be the projections of $ B,M,C$ respectivley on the common tangent.
Since $ BM \equal{} MC$, $ DN \equal{} NE$ . We know that the common tangent at $ A$ to the circles bisects $ DE$. Since $ N$ bisects $ DE$, we conclude that $ AN$ is the tangent at $ A$ and hence, $ AN//FM$.
$ MN//FA$ since both are perpendicular to tangent $ DE$
Hence, $ ANMF$ is a parallelogram and $ AF \equal{} NM \equal{} \frac {1}{2}(BD \plus{} CE) \rightarrow BC \equal{} 2.AF$

[img]http://img269.imageshack.us/img269/7897/imageiz.png[/img]

Let $ P$ be the midpoint of $ BC$ and $ Q$ is its foot of perpendicular to line $ CE$. Clearly $ \triangle PAF$ and $ \triangle CQP$ are similar, and since $ AP\equal{}\frac{r_2\minus{}r_1}2\equal{}CQ$, they are congruent. So $ AF\equal{}PC\equal{}\frac12BC$. Q.E.D.


What is $ n^i$ when n is any natural number and $ i\equal{}\sqrt{\minus{}1}$?

You need the [url=http://en.wikipedia.org/wiki/Complex_exponential]complex exponential[/url], in particular the formula $ z^{\alpha}\equal{}e^{\alpha\log{z}}\equal{}e^{\alpha(\text{Log}|z|\plus{}i\text{Arg }{z})}e^{\alpha 2k\pi i}$, where $ k$ is an integer, and $ \alpha\in\mathbb{C}$.  The reason it has this odd form is because the [url=http://en.wikipedia.org/wiki/Complex_logarithm]complex logarithm[/url] has different branches and so is multi-valued in the complex plane.

Wow. This is a little complex to belong here.

[quote="JRav"]You need the [url=http://en.wikipedia.org/wiki/Complex_exponential]complex exponential[/url], in particular the formula $ z^{\alpha} \equal{} e^{\alpha\log{z}} \equal{} e^{\alpha(\text{Log}|z| \plus{} i\text{Arg }{z})}e^{\alpha 2k\pi i}$, where $ k$ is an integer, and $ \alpha\in\mathbb{C}$.  The reason it has this odd form is because the [url=http://en.wikipedia.org/wiki/Complex_logarithm]complex logarithm[/url] has different branches and so is multi-valued in the complex plane.[/quote]
All I can say is  :huh: .

Moderators, please move this to a different place. Thanks.

I believe that the moderator for this forum is no longer active...

You can get an admin (Valentin Vornicu, rrusczyk, etc.) to move it

It's actually not that complicated if you decipher everything.  $ e$ just the base of the natural logarithm (if you don't know what these are, I suggest you look them up; they're not hard to understand).  $ \text{Log}|z|$ just means the value of the logarithm (base $ e$) of the modulus of $ z$.  In this case, $ z\equal{}n\in\mathbb{N}$, so $ |z|\equal{}n$.  $ \alpha$ is just whatever complex number you're raising it to, so in this case, $ \alpha\equal{}i$.  $ \text{Arg }z$ is the argument of $ z$.  Remember that since a complex number is just an ordered pair of real numbers $ z\equal{}(a,b)\equal{}a\plus{}ib$, we can think of a complex number as a vector in the complex plane.  Since $ z\in\mathbb{N}$, the imaginary component is 0 and the real component is $ n$, so then $ z\equal{}(n,0)$.  This is of course just a point on the real (x)-axis.  The argument of a complex number is the angle formed by the vector and the x-axis so we get that for any purely real number (like we do here), the argument is just 0 since it's directly on the axis.  However, as you should recall, the angles repeat, so $ 0\equal{}\pm 2\pi\equal{}\pm 4\pi\cdots$.  That's why we add in that corrective $ 2k\pi i$ term, to account for the multiple angles.

Now everything is deciphered so you just have to plug in the numbers.

[quote="JRav"]It's actually not that complicated if you decipher everything.  $ e$ just the base of the natural logarithm (if you don't know what these are, I suggest you look them up; they're not hard to understand).  $ \text{Log}|z|$ just means the value of the logarithm (base $ e$) of the modulus of $ z$.  In this case, $ z \equal{} n\in\mathbb{N}$, so $ |z| \equal{} n$.  $ \alpha$ is just whatever complex number you're raising it to, so in this case, $ \alpha \equal{} i$.  $ \text{Arg }z$ is the argument of $ z$.  Remember that since a complex number is just an ordered pair of real numbers $ z \equal{} (a,b) \equal{} a \plus{} ib$, we can think of a complex number as a vector in the complex plane.  Since $ z\in\mathbb{N}$, the imaginary component is 0 and the real component is $ n$, so then $ z \equal{} (n,0)$.  This is of course just a point on the real (x)-axis.  The argument of a complex number is the angle formed by the vector and the x-axis so we get that for any purely real number (like we do here), the argument is just 0 since it's directly on the axis.  However, as you should recall, the angles repeat, so $ 0 \equal{} \pm 2\pi \equal{} \pm 4\pi\cdots$.  That's why we add in that corrective $ 2k\pi i$ term, to account for the multiple angles.

Now everything is deciphered so you just have to plug in the numbers.[/quote]

Indeed.


linear algebra
matrix
combinatorics unsolved
combinatorics

There is a $2n \times 2n$ array (matrix)  consisting of $0's$ and $1's$ and there are exactly $3n$ zeroes. Show that it is possible to remove all the zeroes by deleting some $n$ rows and some $n$ columns.

For each row $ R$ of the given matrix let us denote by $ N(R)$ the number of its zeros. Let $ R_1,R_2,\dots,R_{2n}$ be the rows sorted according to the number of zeros, i.e.

\[ N(R_1) \geq N(R_2) \geq \dots \geq N(R_n) \geq N(R_{n \plus{} 1}) \geq \dots \geq N(R_{2n})\]
and note that

\[ \sum\limits_{i \equal{} 1}^{2n} N\left(R_i\right) \equal{} 3n \qquad \qquad (*)\]
We can have the following three cases:

(i)   If $ N\left(R_n\right) \equal{} 0$ then $ N\left(R_i\right) \equal{} 0$ for each $ i > n$ and (*) implies that

\[ \sum\limits_{i \equal{} 1}^{n} N\left(R_i\right) \equal{} 3n > 2n\]

(ii) If $ N\left(R_n\right) \equal{} 1$ then $ N\left(R_i\right) \leq 1$ for each $ i > n$. Thus

\[ \sum\limits_{i \equal{} n}^{2n} N\left(R_i\right)\leq n \plus{} 1\]
and from (*) it follow that

\[ \sum\limits_{i \equal{} 1}^{n \minus{} 1} N\left(R_i\right)\geq 2n \minus{} 1 \qquad \Rightarrow \qquad \sum\limits_{i \equal{} 1}^{n} N\left(R_i\right) \geq 2n\]

(iii) If $ N\left(R_n\right) \geq 2$ then $ N\left(R_i\right) \geq 2$ for each $ i < n$, so

\[ \sum\limits_{i \equal{} 1}^{n} N\left(R_i\right)\geq 2n\]

Hence in each case it is possible to find $ n$ rows $ R_1,R_2,\dots,R_{n}$ containing at least $ 2n$ zeros. Obviously we can select $ n$ columns $ C_1,C_2,\dots,C_{n}$ containing the remaining $ n$ zeros.   Thus, by deleting the rows $ R_1,R_2,\dots,R_{n}$ and the columns $ C_1,C_2,\dots,C_{n}$ it is possible to remove all the zeros of the given matrix.

This complete the proof.

You need not complicate it so much.

Delete the $ n$ rows containing maximal number of zeros. So, we can say that there can be at most $ n$ zeros  in the remaining $ n$ rows. This is because if there are more than $ n$ zeros, say $ \ge n\plus{}1$, then since there are only $ n$ rows, two zeros must be in the same row. Since we have deleted $ n$ rows containing maximal number of zeros, all those $ n$ rows constitute $ 2n$ zeros. These along with $ n\plus{}1$ zeros (remaining) add up to $ 3n\plus{}1$ zeros $ > 3n$. Contradiction.
THis proves our claim and similarly we can do for the columns also.

Thus we have proved the statement.

[quote="Agr_94_Math"]You need not complicate it so much.

Delete the $ n$ rows containing maximal number of zeros. So, we can say that there can be at most $ n$ zeros  in the remaining $ n$ rows. This is because if there are more than $ n$ zeros, say $ \ge n \plus{} 1$, then since there are only $ n$ rows, two zeros must be in the same row. Since we have deleted $ n$ rows containing maximal number of zeros, all those $ n$ rows constitute $ 2n$ zeros. These along with $ n \plus{} 1$ zeros (remaining) add up to $ 3n \plus{} 1$ zeros $ > 3n$. Contradiction.
THis proves our claim and similarly we can do for the columns also.

Thus we have proved the statement.[/quote]

Dear Agr_94_Math, I think that the idea of your proof is substantially the same of the mine.
However our proof is written in a simpler way and I thank you for this .


In a processing plant, machines A, B, and C can be used to complete an order.  The three machines working together can complete an order in 9/5 hours.  If machine A is used alone for 1 hour, machine B can complete the order in 27/8 hours.  Machine C take three times as long as machine A to complete an order.  If machine C is used for 4 hours, how long must machine B be used to complete the order?

How can three machines take less time than one machine (A)?

I know the problem requires something to do with setting up rates and stuff, but I don't know how to start it. :(

[quote="chess64"]They work faster together. ;)[/quote]

I mean, how can it be [b]slower[/b]?  :blush:

Edit: never mind I just reread the question. strange how I missed all those details.  :?

This is what the solution key says, but I don't really understand.

---
Let a,b, c be the times for machines A, B, C to complete the order, respectively.

We have:

$\frac 1a + \frac 1b + \frac 1c = \frac 1{9/5}$
$\frac 1a + \frac{27/8}b = 1$
$c= 3a$

Using these:

$1 - \frac 1a = \frac{27}{8b}$
$\frac 8{27}(1 - \frac 1a) = \frac 1b$

Then:

$\frac 1a + \frac 8{27} - \frac 8{27a} + \frac 1{3a}= \frac{5}{9}$
$\frac 1a(\frac{28}{27}) = \frac 7{27}$

$a=4, b=\frac 92, c=12$

If x is the required time, then

$\frac 4{12} + \frac x{12} = 1, x=3$

Therefore, it will take 3 hours.

---

Could someone narrate the solution to me?

Check out my blog...

I've made a post about a problem just like this.  It's the first problem on the blog.
 :P

http://www.artofproblemsolving.com/Forum/weblog.php?w=53

[quote="tiredepartment"]

If x is the required time, then

$\frac 4{12} + \frac x{12} = 1$
[/quote]

I don't quite understand this part... can someone please explain?

[quote="prototypex24"]Try P=W/t, where P is the power of the machines, W is the amount of work needed to complete the order, and t is the time needed to complete the order[/quote]

Let $p_A, p_B$ and $p_C$ be the power of machines $A, B$ and $C$.

Let $W$ be the amount of work needed to complete the order.

Hence

$\boxed{\frac{9}{5} \cdot (p_A + p_B + p_C)=W}$ (I)

$\boxed{p_A = 3 \cdot p_C}$ (II) (Machine C take three times as long as machine A to complete an order)

Applying (II) in (I):

$\frac{9}{5} \cdot (4p_C+p_B)=W \Rightarrow \boxed{\frac{36p_C}{5}+\frac{9p_B}{5}=W}$ (III)

$\boxed{1 \cdot p_A + \frac{27}{8} \cdot p_B =W}$ (IV)    (machine A is used alone for 1 hour, machine B can complete the order in 27/8 hours)

Applying (II) in (IV):

$\boxed {3p_C + \frac{27}{8} \cdot p_B = W}$ (V)

Now taking (III) and (V):

$\frac{36p_C}{5}+\frac{9p_B}{5}=3p_C + \frac{27p_B}{8} \Rightarrow \boxed{p_C= \frac{3p_B}{8}}$ (VI)

Hence

$3p_C + \frac{27p_B}{8} = W \Leftrightarrow 3 \cdot \frac{3p_B}{8} + \frac{27p_B}{8} = W \Rightarrow \boxed{\frac{9p_B}{2} = W}$ (VII)

Let $x$ be the required time, then

$4 \cdot p_C + x \cdot p_B = W$ (If machine C is used for 4 hours, how long must machine B be used to complete the order)

Finally, applying (VI) and (VII), we get

$4 \cdot \frac{3p_B}{8} + x \cdot p_B = \frac{9p_B}{2} \Rightarrow x = 3$ hours.


calculus
integration
modular arithmetic
number theory
superior algebra
superior algebra unsolved

Show that the ring $ \mathbb{Z}[\sqrt{d}]\equal{}\mathbb{Z}\plus{}\mathbb{Z}\sqrt{d}$, for any squarefree rational integer $ d>1$, has infinitely many units.

Let $ \mathcal O$ be the integral closure  of $ \mathbb Z[\sqrt {d}]$ in $ \mathbb Q(\sqrt {d})$. Since $ \mathbb Z[\sqrt {d}]^*$ has finite index in $ \mathcal O^*$, the two unit groups have the same (finite) rank $ n$ (as $ \mathbb Z$-modules). Now apply Dirichlet's unit theorem to $ \mathcal O$ to see that $ n > 0$.

Thanks! I was wondering if there is more elementary solution, as this exercise is from the beginning of Neukirch's book "Algebraic number theory". I was wondering if there is a proof related to Pell's equations.

As Pell equations are exactly the same, there surely is.

$ x \plus{} y\sqrt{d}$ is a unit if it has an inverse in the ring
$ x \minus{} y\sqrt{d}$ will be an inverse if $ (x \plus{} y\sqrt{d})(x \minus{} y\sqrt{d}) \equal{} x^2 \minus{} dy^2 \equal{} 1$

This is Pell's equation which has an infinite number of solutions for given $ d$ so there are an infinite number of units in the ring

Specifically, we're looking for any elementary proof for Pell's equations.  I don't know of one that's entirely straightforward, but one method is to show that the partial fraction decomposition of $ \sqrt{d}$ yields rational approximations $ \frac{a}{b}$ to $ \sqrt{d}$ that are "good enough" to satisfy $ a^2 \minus{} db^2 \equal{} 1$.

Can anyone post more detailed sketch of that partial fraction decomposition method? It is new for me so I haven't found a proof for five days.

e.g. see p.64 of Baker's: A concise introduction to the theory of numbers
http://books.google.com/books?id=hAFfb8BBC0IC&pg=PA64

Refer to Joe Roberts' excellent "Elementary Number Theory: A Problem Oriented Approach" for a proof using pigeon-hole principle. It consists of four lemmas:

[b]Lemma 1[/b]. Let $ (x,y)$ such that $ |x \minus{} \sqrt D y| < \frac1y$ (there exist infinitely many such pairs by Dirichlet's theorem, which, by the way, admits a completely elementary proof). Thus
\[ |x^2 \minus{} Dy^2| < 1 \plus{} 2\sqrt D\]

[b]Lemma 2[/b]. There exists a positive integer $ k$ such that the equation $ x^2 \minus{} Dy^2 \equal{} k$ has infinitely many solutions with $ y > 0$.

[b]Lemma 3[/b]. Let $ k$ one of the numbers that satisfy lemma 2. There exist two pairs $ (x_1,y_1)\neq (x_2,y_2)$ such that $ x_i \equiv y_i \pmod k$, $ i \equal{} 1,2$.

[b]Lemma 4[/b]. Let $ (x_1,y_1)$ and $ (x_2,y_2)$ as in the previous lemma. If $ x_3 \plus{} y_3\sqrt D \equal{} (x_1 \plus{} y_1\sqrt D)(x_2 \minus{} y_2\sqrt D)$, $ x_3,y_3$ integers, then $ k \mid x_3$, $ k\mid y_3$ and $ x_3^2 \minus{} Dy_3^2 \equal{} k^2$.

Of well, after lemma 4 you can prove almost immediately that there is a solution to $ x^2 \minus{} Dy^2 \equal{} 1$.


Let be $ a,b,c,d$ the side lenghts of a convex quadrilateral such that $ \frac{a}{b\plus{}c\plus{}d}\plus{}\frac{b}{c\plus{}d\plus{}a}\plus{}\frac{c}{d\plus{}a\plus{}b}\plus{}\frac{d}{a\plus{}b\plus{}c}\le \frac{4}{3}$ . Prove that the given quadrilateral is a rhombus

[quote="alex2008"]Let be $ a,b,c,d$ the side lenghts of a convex quadrilateral such that $ \frac {a}{b \plus{} c \plus{} d} \plus{} \frac {b}{c \plus{} d \plus{} a} \plus{} \frac {c}{d \plus{} a \plus{} b} \plus{} \frac {d}{a \plus{} b \plus{} c}\le \frac {4}{3}$ . Prove that the given quadrilateral is a rhombus[/quote]

$ P\equal{}a\plus{}b\plus{}c\plus{}d$--->

$ \frac{a}{p\minus{}a}\plus{}\frac{b}{p\minus{}b}\plus{}\frac{c}{p\minus{}c}\plus{}\frac{d}{p\minus{}d}\equal{}\frac{p}{p\minus{}a}\plus{}\frac{p}{p\minus{}b}\plus{}\frac{p}{p\minus{}c}\plus{}\frac{p}{p\minus{}d}\minus{}4\geq 4\sqrt[4]{\frac{p}{p\minus{}a}*\frac{p}{p\minus{}b}*\frac{p}{p\minus{}c}*\frac{p}{p\minus{}d}}\minus{}4\equal{}\frac{4P}{\sqrt[4]{(P\minus{}a)(P\minus{}b)(P\minus{}c)(P\minus{}d)}}\minus{}4\geq\frac{4p}{\frac{(p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c)\plus{}(p\minus{}d)}{4}}\minus{}4\equal{}\frac{16p}{3p}\minus{}4\equal{}\frac{4}{3}$----->

$ \frac{4}{3}\geq\frac{a}{p\minus{}a}\plus{}\frac{b}{p\minus{}b}\plus{}\frac{c}{p\minus{}c}\plus{}\frac{d}{p\minus{}d}\geq\frac{4}{3}$---->

$ \frac{a}{p\minus{}a}\plus{}\frac{b}{p\minus{}b}\plus{}\frac{c}{p\minus{}c}\plus{}\frac{d}{p\minus{}d}\equal{}\frac{4}{3}$---> $ a\equal{}b\equal{}c\equal{}d$

[hide="A little bit shorter"]
Let the LHS be $ A$.
$ A\cdot(a(b\plus{}c\plus{}d)\plus{}b(a\plus{}c\plus{}d)\plus{}c(a\plus{}b\plus{}d)\plus{}d(a\plus{}b\plus{}c))\ge(a\plus{}b\plus{}c\plus{}d)^2$ by CSB
$ A\ge\frac{(a\plus{}b\plus{}c\plus{}d)^2}{2(ab\plus{}bc\plus{}ac\plus{}ad\plus{}bd\plus{}cd)}$
$ A\ge\frac{4}{3}$ because of AM-GM
So $ a\equal{}b\equal{}c\equal{}d$
[/hide]


Integer sequence
number theory
recurrence relation
Perfect Squares
Extremal combinatorics
IMO Shortlist

A finite sequence of integers $ a_0, a_1, \ldots, a_n$ is called quadratic if for each $ i$ in the set $ \{1,2 \ldots, n\}$ we have the equality $ |a_i \minus{} a_{i\minus{}1}| \equal{} i^2.$

a.) Prove that any two integers $ b$ and $ c,$ there exists a natural number $ n$ and a quadratic sequence with $ a_0 \equal{} b$ and $ a_n \equal{} c.$

b.) Find the smallest natural number $ n$ for which there exists a quadratic sequence with  $ a_0 \equal{} 0$ and $ a_n \equal{} 1996.$

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
first of all note that the existance of such a sequence with $ a_0,a_n$ is equivalent to the existance of numbers $ \epsilon_i$ for $ 1\leq i\leq n$ in which $ \epsilon_i\equal{}\pm 1$ and also:

$ a_n\minus{}a_0\equal{}\sum_{i\equal{}1}^n\epsilon_ii^2$

--------------------------------------------------
a.)first of all note that:

$ n^2\minus{}(n\plus{}1)^2\minus{}(n\plus{}2)^2\plus{}(n\plus{}3)^2\equal{}4$

now it's sufficient to show that the number $ c\minus{}b$ can be written as $ \sum_{i\equal{}1}^n \epsilon_ii^2$,we will prove this in the case $ c\minus{}b>0$ the other cases can be shown similarly:

let $ r$ be the remainder of $ c\minus{}b$ divided by $ 4$,now we have 4 cases:

$ 1)r\equal{}0\Rightarrow c\minus{}b\equal{}(1^2\minus{}2^2\minus{}3^2\plus{}4^2)\plus{}(5^2\minus{}6^2\minus{}7^2\plus{}8^2\plus{}\ldots$

in which the number of parenthesis are $ \frac{c\minus{}b}4$.

$ 2)r\equal{}1\Rightarrow c\minus{}b\equal{}1^2\plus{}(2^2\minus{}3^2\minus{}4^2\plus{}5^2)\plus{}(6^2\minus{}7^2\minus{}8^2\plus{}9^2)\plus{}\ldots$

there are $ \frac{c\minus{}b\minus{}1}4$ parenthesis.

$ 3)r\equal{}2\Rightarrow c\minus{}b\equal{}(\minus{}1^2\minus{}2^2\minus{}3^2\plus{}4^2)\plus{}(5^2\minus{}6^2\minus{}7^2\plus{}8^2)\plus{}(9^2\minus{}10^2\minus{}11^2\plus{}12^2)\plus{}\ldots$

there are $ 1\plus{}\frac{c\minus{}b\minus{}2}4$ parenthesis.

$ 4)r\equal{}3\Rightarrow c\minus{}b\equal{}\minus{}1^2\plus{}2^2\plus{}(3^2\minus{}4^2\minus{}5^2\plus{}6^2)\plus{}(7^2\minus{}8^2\minus{}9^2\plus{}10^2)\plus{}\ldots$

there are $ \frac{c\minus{}b\minus{}3}4$ parenthesis.

--------------------------------------------------
b.)first of all note that from the definition of $ a_n$ and the fact that $ a_0\equal{}0$,we obtain that:

1.)$ a_k\leq 1^2\plus{}2^2\plus{}\cdots \plus{}k^2\equal{}\frac{k(k\plus{}1)(2k\plus{}1)}6$

and

2.)$ a_k\equiv 1^2\plus{}2^2\plus{}\cdots \plus{}k^2\pmod 2$

now from 1.) we get that $ a_{17}\leq 1785$ hence $ n\geq 18$ and also by 2.) we get that $ a_{18}\equiv 2109\pmod 2$ but $ 1996$ is even,so $ n\neq 18$,so we must have $ n\geq 19$,now we will show that for $ n\equal{}19$ such sequence exists:

note that:

$ 1^2\plus{}2^2\plus{}\cdots \plus{}19^2\equal{}2470\equal{}1996\plus{}474$ so it's sufficient to write $ \frac{474}2\equal{}237$ as sum of some distinct perfect squares for example $ 237\equal{}14^2\plus{}5^2\plus{}4^2$,so we have:

$ 1996\equal{}1^2\plus{}2^2\plus{}3^2\minus{}4^2\minus{}5^2\plus{}6^2\plus{}7^2\plus{}\cdots \plus{}13^2\minus{}14^2\plus{}15^2\plus{}\cdots \plus{}19^2$

QED[/quote]


linear algebra
matrix
modular arithmetic
induction
trigonometry
inequalities
algebra

Let $ n > 1$ be an odd positive integer and $ A = (a_{ij})_{i, j = 1..n}$ be the $ n \times n$ matrix with
\[ a_{ij}= \begin{cases}2 & \text{if }i = j \\ 1 & \text{if }i-j \equiv \pm 2 \pmod n \\ 0 & \text{otherwise}\end{cases}.\]
Find $ \det A$.

It looks like the trick here is to realize that $ A=B^{2}$, where $ b_{ij}$ is 1 if $ i-j$ is either $ 1$ or $ -1$ modulo $ n$ and 0 otherwise.  (In other words, $ B$ is the adjacency matrix of the $ n-$cycle).

To determine the determinant of $ B$, we expand it out using definition of determinant as a sum over permutations
\[ det(B)=\sum_{\sigma}\prod_{i=1}^{n}b_{i \sigma(i)}sign(\sigma). \]

The only way a term in this sum will be nonzero is if $ \sigma(i)$ is either $ i+1$ or $ i-1$ for each $ i$.  If there is some $ j$ such that $ \sigma(j)=j+1$, then $ \sigma(j+2)$ must equal $ j+3$ for the term to be nonzero, and by induction $ \sigma(j+2k)=j+2k+1$ for any positive $ k$.  Since $ n$ is odd, each $ i$ can be written as $ j+2k$, so in fact we have $ \sigma(i)=i+1$ for all $ i$.

Therefore the only nonzero terms in the determinant correspond to the permutations $ \sigma(i)=i+1$ and $ \sigma(i)=i-1$.  These are both even permutations (as $ n$ is odd), so the determinant of $ B$ is 2, and the determinant of $ A$ is $ 2^{2}=4$.

or to observe that it's a circular determinant AND TO KNOW the formula for it...

On that note- all matrices of this form (circulants) can be diagonalized by the Fourier matrix $ F$, where the $ jk$ entry of $ F$ is $ \exp\left(jk\frac{2\pi i}{n}\right)$.
Let $ v_{j}$ be the $ j$th column; it's an eigenvector for $ A$. The associated eigenvalue is $ 2+\omega^{2j}+\overline{\omega}^{2j}= (\omega^{j}+\overline{\omega}^{j})^{2}=4\cos^{2}j\theta$, where $ \omega=e^{2\pi i/n}$ and $ \theta=\frac{2\pi}{n}$.

The numbers $ \cos j\theta$ are the roots of $ T_{n}(x)-1=0$, where $ T_{n}$ is the $ n$th Chebyshev polynomial $ T_{n}(x)=\cos(n\arccos x)$. The $ x^{n}$ coefficient of $ T_{n}$ is $ 2^{n-1}$ for $ n>0$; we have the recurrence relation $ T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x)$, with $ T_{0}=1$ and $ T_{1}(x)=x$. The constant term is zero when $ n$ is odd, $ 1$ when $ n$ is a multiple of 4, and $ -1$ when $ n\equiv 2\mod 4$.
The product of the zeros of $ T_{n}-1$ is thus $ (-1)^{n}\cdot\frac{a_{n}}{2^{n-1}}$, where $ a_{n}=\begin{cases}0&n\equiv 0\mod 4\\-1& n\equiv 1\mod 2\\-2&n\equiv 2\mod 4\end{cases}$
Squaring and multiplying by $ 4^{n}$, the determinant we want is $ 4$ when $ n$ is odd, $ 0$ when $ n$ is a multiple of $ 4$, and $ 16$ when $ n\equiv 2\mod 4$.


find all integers $ (a,b)$ satisfying the equation: $ 5a^{2}\minus{}b^{2}\equal{}4.$

see PEN L8. :)
to be complete,i must say that in L8 we had found all positive integer solutions and so all negative solutions.and of course none of (a,b) can be zero.


inequalities
complex analysis
function
integration
complex analysis unsolved

Let $f(z)$ be holomorphic function in the unit disk $U$ such that $f(0)=1$ and $|f(z)|<M$ in $U$. Show that in the disk $|z|\leq \frac{1}{M}$ the following inequality holds:$|f(z)-1|\leq M|z|$

Just  to avoid writing $\frac 1M$ in various expressions, let's reformulate the problem as follows. Let $\delta=\frac 1M$, $g=\delta f$. Then $|g|\leqslant 1$ in $U$, $g(0)=\delta$ and our task is to show that $|g(z)-\delta|\leqslant |z|$ when $|z|\leqslant \delta$.

Two approaches are possible.

The first one is to make the Moebius transformation that sends $\delta$ to $0$ and apply the Schwartz lemma. So, we have to consider the function $G(z)=\frac{g(z)-\delta}{1-\delta g(z)}$, which is also bounded by $1$ in $U$ and satisfies $G(0)=0$. Then $|G(z)|\leqslant |z|$ for all $z\in U$. Let $|z|<\delta$ and let $G(z)=\zeta$. Then $|g(z)-\delta|=|\zeta| \frac{1-\delta^2}{|1+\delta\zeta|}\leqslant |z|$ because $|\zeta|\leqslant |z|$ and $|1+\delta\zeta|\geqslant 1-\delta|\zeta|\geqslant 1-\delta^2$.

Another approach is to use the Taylor series $g(z)=\delta+\sum_{k>0}a_k z^k$ and the Parseval identity. We have $\delta^2+\sum_k|a_k|^2=\int_{\mathbb T}|g|^2\,dm\leqslant 1$ where $m$ is the Haar measure on the unit circumference $\mathbb T$. Therefore, $\sum_k|a_k|^2\leqslant 1-\delta^2$. Now, using Cauchy-Schwartz, we get $|g(z)-\delta|\leqslant \sum_k a_k |z|^k\leqslant \sqrt{1-\delta^2}\frac {|z|}{\sqrt{1-|z|^2}}\leqslant |z|$ (we used the identity $\sum_{k>0}|z|^{2k}=\frac{|z|^2}{1-|z|^2}$ here).


algebra
polynomial
function
domain
induction
modular arithmetic
Ring Theory

Let $ R$ be a commutative ring with $ 1$. Prove that $ p(x) \equal{} a_nx^n \plus{} \dots \plus{} a_1x \plus{} a_0 \in \text{U}\left(R\left[x\right]\right)$ iff $ a_0\in \text{U}\left(R\right)$ and $ a_i\ (i > 0)$ are nilpotent.

[i]Alternative formulation.[/i] Let $ R$ be a commutative ring with unity. Prove that a polynomial $ \sum_{i \equal{} 0}^n a_ix^i\in R\left[x\right]$ is a unit if and only if $ a_0 \in R^{\times}$ and $ a_1, \dots, a_n$ are nilpotent in $ R$.

It's been proved before on the forum, and it's easy to just use "brute force" :), but we could also prove it like this (all rings are commutative with unit, and a ring homomorphism is assumed to take $1$ onto $1$):

$\Rightarrow$

Consider the application $\phi: R[x]\to R,\ \phi(p)=p(0)$. This is a ring homomorphism, and if $p(x)=a_nx^n+\ldots+a_1x+a_0$ is invertible, then $\phi(p)$ must be invertible as well, i.e. $a_0$ is invertible.

Now let $\frak p$ be an arbitrary prime ideal of $R$. The canonical projection $\pi_\frak p: R[x]\to R/\frak p[x]$ is, again, a homomorphism, and hence takes $p$ onto an invertible element of $R/\frak p[x]$. However, $R/\frak p$ is a domain, so the invertible elements of $R/\frak p[x]$ are precisely the invertible elements of $R/\frak p$. This simply means that the projections of all $a_i,i>0$ are zero, i.e. all $a_i,i>0$ belong to $\frak p$. Since $\frak p$ was arbitrary, all $a_i,i>0$ belong to all prime ideals of $R$, and since the intersection of all prime ideals is the nilradical, we're done.

$\Leftarrow$

The hypothesis is now that $a_0$ is invertible, and all $a_i,i>0$ are nilpotent. Suppose $p(x)=a_nx^n+\ldots+a_1x+a_0$. Is not invertible. Then there exists a maximal ideal $\frak m$ of $R[x]$ which contains it. But since all $a_ix^i,i>0$ are nilpotent, all of them belong to $\frak m$, and this means that $a_0$ does too. This, however, is impossible, because $a_0$ is invertible in $R[x]$. We have a contradiction, and hence $p$ must indeed be an invertible element of $R[x]$.

@grobber: Yes, it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87441]here[/url]. But your proof for $\Leftarrow$ is more elegant. :-)

i have some elementary approach actually it suffices to prove $ a_n$ is nilpotent then by induction finishes,assume that $ g \equal{} b_0 \plus{} \dots \plus{} b_m\cdot x^m$ such that $ fg \equal{} 1$ for nilpotency of $ a_n$ it suffice to prove that every $ b_i$ are in $ I \equal{} \{ b\in R, a_n^t\cdot b \equal{} 0\text{ for some }t \}$.and it is easy by induction.

These are the straight forward and beautiful solutions using Zorn's lemma,
(though I did not know these posts) but some people are just not satisfied with that
and want everything constructive.
So here is a constructive proof for those who like that. (which I don't  :P )

Assume $ 1 \equal{} f(X)g(X)$ with $ f(X),g(X)\in R[X], f(X) \equal{} \sum_{k\ge 0}a_kX^k,g(X) \equal{} \sum_{k\ge 0}b_kX^k$.
The coefficient of $ X^0$ is $ 1 \equal{} a_0b_0$, so $ a_0,b_0\in R^{\times}$.
Assume $ a_n$ is not nilpotent for some $ n\ge 1$, and take $ n$ maximal.
The set $ J$ of all nilpotent elements of $ R$ forms an ideal of $ R$.

Define the ideal $ I_m: \equal{} J \plus{} \sum_{k\ge m}b_kR$ of $ R$ for each $ m\ge 0$.
Then $ I_m \equal{} J$ for $ m$ large enough, in which case we have $ a_n^k\not\in I_m$ for all $ k\in\mathbb{N}$.
Now take $ m\ge 0$ minimal with $ a_n^k\not\in I_m$ for all $ k\in\mathbb{N}$.

For $ m \equal{} 0$ we get the contradiction $ a_n\in R \equal{} b_0R\subseteq I_0$.
So $ m\ge 1$ and $ a_n^k\in I_{m \minus{} 1} \equal{} I_m \plus{} b_{m \minus{} 1}R$ for some $ k\in\mathbb{N}$.
But looking at the coefficient of $ X^{n \plus{} m \minus{} 1}$ in $ 1 \equal{} f(X)g(X)$ we get
$ 0\equiv a_nb_{m \minus{} 1}\pmod{I_m}$, which gives the contradiction
$ a_n^{k \plus{} 1}\in a_nI_m \plus{} a_nb_{m \minus{} 1}R\subseteq I_m$.

And conversely take $ f(X) \equal{} \sum_{k \equal{} 0}^na_kX^k\in R[X]$ with $ a_0\in R^\times$ and $ a_1,\ldots,a_n\in R$ nilpotent.
Then the set $ \tilde{J}$ of all nilpotent elements of $ R[X]$ forms an ideal of $ R[X]$
and $ a_1,\ldots,a_n\in\tilde{J}$ gives $ f(X) \equal{} a_0(1 \minus{} g(X))$ with $ g(X) \equal{} \minus{} a_0^{ \minus{} 1}\sum_{k \equal{} 1}^na_kX^k\in \tilde{J}$.
This gives $ g(X)^m \equal{} 0$ for some $ m\in \mathbb{N}$ and $ f(X)\in R[X]^\times$
with $ f(X)^{ \minus{} 1} \equal{} a_0^{ \minus{} 1}(1 \plus{} g(X) \plus{} g(X)^2 \plus{} \cdots \plus{} g(X)^{m \minus{} 1})$.


$n$ teams take part in a tournament. For any two teams $A$ and $B$, if the game is played at $A$'s town, then call this game as $A$'s "home game", as well as $B$'s "away game". Any two teams play against each other just twice, each having a "home game" respectively. Knowing that every team can be arranged more than one "away game" in one week (from Sunday to Saturday); but if one team has "home game" in certain week, then it cannot be arranged "away game" in that week.

If all the games can be arranged to finish in four weeks, find the maximum possible value of $n$.

DusT said
[quote]
A small unclarity!
Can a team play more than one game at home in one week?
If not, then the number of times a team can play away is limited by the number of times other can play at home!!![/quote]

I look through the official solutions, there is no limit on the number of games played at home in one week.

lets solve this one:
 I want to show there are at most 6 teams.

for $i=1,2,3,4$ let $A_i$ be the set of teams that play as away teams in week $i$, and let $H__i$ the set of teams that play as home, in week $i$.

We clearly have $A_i\cap H_i=\phi$, now we prove that if there are at least 6 teams, every team plays 2 weeks as away and 2 weeks as home.

Suposse, team $t_2$, plays weeks 1,2,3 as home, and week 4 as away. then every other team must play in week 4 as home in order to have a home game against $t_1$. So in the last week there are $n-1$ games.. so in total there are at most $|A_1||H_1|+|A_2||H_2|+|A_3||H_3|+n-1\leq 3n^2/4+n-1\leq n(n-1)$ for $n>7$, suppose $n=7$ and let $t_2,t_3...,t_7$ be the other teams. it is clear that they should have play the games between them in the weeks 1,2,3.  Now in each of this weeks there are at most $9$ games between these teams, so there are at most $27$ games between these teams, however we needed $30$ games between them, contradiction.

so in that case $n$ is at most $6$. Clearly we are left with the case where every team plays 2 weeks as home and 2 as Away. Look that we have $\binom{4}{2}=6$ ways to choose, the two weeks a team will play as home, so if there are 7 teams, 2 teams, play the same weeks as home and the same weeks as away, so they never get to ´play, because every game has an away team and a home team, so we get a contradiction.

It is easy to construct examples for $n=6$ I hope it's correct..

I think I can demonstrate 12 teams works? (I'm assuming only one game a day for each team.)

Let $\{a_1,a_2,...,a_6\}$ be the teams playing only home games in week 1; $\{b_1,b_2,...,b_6\}$ play only away games in week 1. Each $a_i$ plays all of the $b_j$, so 6 games for each team in week1.

In week 2 reverse the roles of home and away from week 1. Now all of the $a's$ have played all of the $b's$ as both home and away.

Now, the $a's$ and $b's$ play amongst themselves according to the same method above. Week 3:
$\{a_1,a_2,a_3\}$, $\{b_1,b_2,b_3\}$are playing only home games, and $\{a_4,a_5,a_6\}$, $\{b_4,b_5,b_6\}$ are playing only away games. Week 4 reverse the roles of week 3. So all 12 teams have played played every other team twice: once as home,and once as away. So, $n\ge 12$.

but under your scheme $a_1,a_2,a_3$ never play games between each other.
in fact there can't be more than 6 teams(i had the same result as Pascual2005).


Solve in $ \mathbb{R}_\plus{}^*$ the following equation :

$ \sqrt{x\plus{}\lfloor x\rfloor}\plus{}\sqrt{x\plus{}\{x\}}\equal{}\sqrt{x\plus{}\lfloor x\rfloor \cdot \{x\}}\plus{}\sqrt{x\plus{}1}$

It's clear that if $ \lfloor x\rfloor \equal{}1$, then equality is established. I am pretty sure that this is the only case, but I can't prove it.

By inspection notice that $ \lfloor x \rfloor \equal{} 0$ yields no solutions. For $ \lfloor x \rfloor \equal{} 1$ the equality holds.

So, to show there are no other solutions assume $ \lfloor x \rfloor > 1$

For simplicity let $ \lfloor x \rfloor \equal{} a, \{x\} \equal{} b$

squaring both sides gives

$ 2x \plus{} a \plus{} b \plus{} 2\sqrt {(x \plus{} a)(x \plus{} b)} \equal{} 2x \plus{} ab \plus{} 1 \plus{} 2\sqrt {(x \plus{} ab)(x \plus{} 1)}$

$ 2\sqrt {(x \plus{} a)(x \plus{} b)} \equal{} (a \minus{} 1)(b \minus{} 1) \plus{} 2\sqrt {(x \plus{} ab)(x \plus{} 1)}$

From our conditions we have $ (a \minus{} 1)(b \minus{} 1) < 0$

$ \therefore 2\sqrt {(x \plus{} a)(x \plus{} b)} < 2\sqrt {(x \plus{} ab)(x \plus{} 1)}$

$ \therefore x^2 \plus{} (a \plus{} b)x \plus{} ab < x^2 \plus{} (ab \plus{} 1)x \plus{} ab$

$ \therefore 0 < (a \minus{} 1)(b \minus{} 1)$ contradiction.

Hence the only solutions are $ \lfloor x \rfloor \equal{} 1$, with $ 0\le \{x\} < 1$


inequalities
logarithms
inequalities unsolved

:rotfl:
if$ x_1, x_2,...x_n$ positive real numbers prove inequality:

$ (\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}...\plus{}\frac{1}{x_n})(1\plus{}x_1)(1\plus{}x_2)...(1\plus{}x_n)\ge 2n^2$

I think $ n\geq2$.

First we prove that $ (1 \plus{} x)^n\geq2nx$ for all $ x > 0$ and positive integer $ n > 1$

By $ AM \minus{} GM$

$ 1 \plus{} (n \minus{} 1)\frac {x}{n \minus{} 1} \geq n.\sqrt [n]{\frac {x}{(n \minus{} 1)^{n \minus{} 1}}}$

Because of the sequence of $ (1 \plus{} \frac {1}{n})^n$ is increasing so $ (1 \plus{} \frac {1}{n \minus{} 1})^{n \minus{} 1} \geq2$ for $ n \geq2$ ($ n$ is a positive integer) $ \Longrightarrow$

$ 1 \plus{} (n \minus{} 1)\frac {x}{n \minus{} 1} \geq n.\sqrt [n]{\frac {x}{(n \minus{} 1)^{n \minus{} 1}}} \geq \sqrt [n]{2nx} \Longrightarrow$

$ (1 \plus{} x)^n \geq 2nx$

$ 1 \plus{} x_i \geq \sqrt [n]{2nx_i}$   ($ i \equal{} 1,2,...,n$)  $ \Longrightarrow$

$ \displaystyle\prod_{i \equal{} 1}^n(1 \plus{} x_i) \geq 2n.\sqrt [n]{\displaystyle\prod_{i \equal{} 1}^nx_i} \Longrightarrow$ By $ GM \minus{} HM$

$ \displaystyle\sum_{i \equal{} 1}^n\frac {1}{x_i} .\displaystyle\prod_{i \equal{} 1}^n(1 \plus{} x_i) \geq \frac {n}{\sqrt [n]{\displaystyle\prod_{i \equal{} 1}^nx_i}}.2n.\sqrt [n]{\displaystyle\prod_{i \equal{} 1}^nx_i} \equal{} 2n^2$ :wink:

[quote="fredi"]:rotfl:
if$ x_1, x_2,...x_n$ positive real numbers prove inequality:

$ (\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n})(1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_n)\ge 2n^2$[/quote]
Let $ t \equal{} \sqrt [n]{x_1x_2...x_n}$. By Hölder:
$ (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_n) \ge (1 \plus{} t)^n$
By AM-GM:
$ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n} \ge n \cdot t^{ \minus{} 1}$
So we just have to prove:
$ g(t) \equal{} (1 \plus{} t)^n \minus{} 2n \cdot t \ge 0$.
But $ g'(t) \equal{} 0 \iff t \equal{} \sqrt [n \minus{} 1]{2} \minus{} 1$ and $ g''(t) \ge 0$ so:
$ g(t) \ge g(\sqrt [n \minus{} 1]{2} \minus{} 1) \equal{} 2 \cdot \sqrt [n \minus{} 1]{2} \minus{} 2 (\sqrt [n \minus{} 1]{2} \minus{} 1) \cdot n \equal{} 2 \cdot (n \plus{} \sqrt [n \minus{} 1]{2} \minus{} \sqrt [n \minus{} 1]{2} \cdot n) \ge 0 \iff$
$ f(n) \equal{} n \plus{} 1 \minus{} \sqrt [n \minus{} 1]{2}n \ge 0$
$ f'(n) \equal{} 1 \minus{} \sqrt [n \minus{} 1]{2} \plus{} \sqrt [n \minus{} 1]{2} \cdot \ln(2) \cdot \frac {1}{(n \minus{} 1)^2}$
$ f''(n) \equal{} \sqrt [n \minus{} 1]{2} \cdot \ln(2) \cdot \frac {1}{(n \minus{} 1)^2} \minus{} \sqrt [n \minus{} 1]{2} \cdot \ln^2(2) \cdot \frac {1}{(n \minus{} 1)^4} \minus{} \sqrt [n \minus{} 1]{2} \cdot \ln(2) \cdot \frac {1}{(n \minus{} 1)^3} \ge 0 \iff$
$ 1 \ge \ln^2(2) \cdot \frac{1}{(n\minus{}1)^2} \plus{} \frac{1}{n\minus{}1}$. Which is clearly true for $ n \ge 2$. So $ f''(n) \ge 0$ and obviously $ f'(2) \ge 0$. So $ f'(n) \ge 0 \forall n \ge 2$. Thus since $ f(2) \ge 0$ we get $ f(n) \ge 0 \forall n \ge 2$. And we are done!

Equality only when $ x_1 \equal{} x_2 \equal{} 1$ and $ n\equal{}2$.


Do you know anything about Brocard's Theorem? Please tell me :)


function
number theory
totient function
number theory unsolved

$\phi(m)$ is Euler's Totient function right?

I don't know, I am a bit weak in Number Theory, but is it possible to conclude anything substantial about $P\mod m$ from the given data? And in terms of what are you trying to find it?

Let's see. If $m$ is prime to all $a_{i}$, all of the $a_{i}$s are counted in $\phi(m)$. As there are $\phi(m)$ of them, if they are all distinct, they cover all numbers less than $m$ and prime to it. That much we may conclude. But I still don't see how it helps us get an answer.

Okay. I give up.

Let $A=\{a_{1},\ldots,a_{\varphi(m)}\}$.
Then every $x\in A$ has an unique multiplicative inverse. If $x\neq x^{-1}\mod m$ then they drop out of $P$.
Hence we have
$P\equiv \prod_{x\in A}x \equiv \prod_{x^{2}\equiv 1\mod m}x \mod m$
But it is a very well known fact that $P\equiv-1\mod m$ iff $m$ has the form $1,2,4,p^{k},2p^{k}$ where $p$ is an odd prime and $k$ is a positive integer and $P\equiv 1\mod m$ else (see http://www.mathlinks.ro/Forum/viewtopic.php?p=258123#p258123 e.g.).


[quote="VANCHANH123"]$3^{x}+4^{2x-1}=7^{x}$[/quote]

It's rather easy to show that $f(x)=3^{x}+4^{2x-1}-7^{x}$ just have two zeros on $%Error. "mathlabb" is a bad command.
{R}$ :
$x=1$
$x=a$ with $a\in(0,1)$ (in fact $a$ near of $0.6859$

--
Patrick

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


inequalities
function
inequalities proposed

$ x_i > 0, n\ge3,$ then

$ (x_1^{n \minus{} 1} \plus{} n \minus{} 1)(x_2^{n \minus{} 1} \plus{} n \minus{} 1) \cdots (x_n^{n \minus{} 1} \plus{} n \minus{} 1) \ge n(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n )^{n \minus{} 1}$

and

$ \left( {x_1^2  \plus{} n \minus{} 1} \right)\left( {x_2^2  \plus{} n \minus{} 1} \right) \cdots \left( {x_n^2  \plus{} n \minus{} 1} \right) \ge n^{n \minus{} 2} \left( {x_1  \plus{} x_2  \plus{}  \cdots  \plus{} x_n } \right)^2$

are holds too.

but LBQ's ineq is stronger than mine...

[quote="xzlbq"]you ineq is nice!
best is:
$ 5\,bc \plus{} 5\,ca \plus{} 5\,ba \plus{} 4\,{a}^{2} \plus{} 4\,{b}^{2} \plus{} 4\,{c}^{2}\leq \left( {a}^{2} \plus{} 2 \right) \left( {b}^{2} \plus{} 2 \right) \left( {c}^{2} \plus{} 2 \right) .$----(*)
BQ

(*)<=>
$ \left( {b}^{2}{c}^{2} \plus{} 2\,{b}^{2} \plus{} 2\,{c}^{2} \right) {a}^{2} \plus{} \left( \minus{} 5\,c \minus{} 5\,b \right) a \minus{} 5\,bc \plus{} 2\,{b}^{2}{c}^{2} \plus{} 8\geq0.$

$ D \equal{} \minus{} 8\, \left( {c}^{2} \plus{} 2 \right) {c}^{2}{b}^{4} \plus{} 20\,c \left( {c}^{2} \plus{} 2 \right) {b}^{3} \plus{} \left( \minus{} 39 \minus{} 16\,{c}^{4} \minus{} 32\,{c}^{2} \right) {b}^{2} \plus{} 10\,c \left( 5 \plus{} 4\,{c}^{2} \right) b \minus{} 39\,{c}^{2}\le 0$

I have not a proof for this ineq.
BQ
BQ[/quote]

if the ineq holds when $ a,b,c$ be non-negative, then it's obvious holds for any reals $ a,b,c$, so we just need to prove that when $ a,b,c$ be non-negative.

$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \ge 4(a^2\plus{}b^2\plus{}c^2)\plus{}5(ab\plus{}bc\plus{}ca)$

$ \iff 2(ab\plus{}bc\plus{}ca)^2\minus{}4abc(a\plus{}b\plus{}c)\plus{}a^2b^2c^2\minus{}5(ab\plus{}bc\plus{}ca)\plus{}8 \ge 0$

$ \iff f(u) \equal{} \minus{}12w^3u\plus{}w^6\plus{}18v^4\minus{}15v^2\plus{}8 \ge 0,$ (setting $ a\plus{}b\plus{}c\equal{}3u, ab\plus{}bc\plus{}ca\equal{}3v^2, abc\equal{}w^3$)

that's a linear function of $ u$, so we just need to prove the following two case:

(1), $ a\equal{}0$;
(2), $ b\equal{}c$.

for (1), ineq becomes $ 2b^2c^2\plus{}8 \ge 5bc$, which is obvious true;

for (2), ineq becomes $ (b^4\plus{}4b^2)a^2\minus{}10ab\plus{}2b^4\minus{}5b^2\plus{}8 \ge 0$, $ \Delta \equal{} \minus{}4(2b^6\plus{}3b^4\minus{}12b^2\plus{}7)b^2$,

by Am-Gm, we have $ 3b^4\plus{}3 \ge 6b^2$ and $ 2b^6\plus{}2\plus{}2 \ge 6b^2$, then we get $ \Delta \le 0$, so we get the ineq holds.

the proof is completed.

[quote="daigiaga1994"][quote="akashram"]a,b,c positive reals

[a^2+2] [b^2+2] [c^2+2]   >= 9[ab+bc+ca][/quote]
[b]An stronger Problem: [/b]
Let $ a,b,c$ be positive real numbers. Prove that
 $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 3(a \plus{} b \plus{} c)^2$[/quote]

that's in 3# ineq when $ n\equal{}3$.

[quote="kuing"][quote="daigiaga1994"][quote="akashram"]a,b,c positive reals

[a^2+2] [b^2+2] [c^2+2]   >= 9[ab+bc+ca][/quote]
[b]An stronger Problem: [/b]
Let $ a,b,c$ be positive real numbers. Prove that
 $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 3(a \plus{} b \plus{} c)^2$[/quote]

that's in 3# ineq when $ n \equal{} 3$.[/quote]
Yes  :P
But, In your solution. I don't understand this paragraph:
[quote]that's a linear function of u, so we just need to prove the following two case:

(1), a=0;
(2), b=c.  [/quote]
Can you explain it more clearly, please ?
Thank you !

[quote="daigiaga1994"]
But, In your solution. I don't understand this paragraph:
[quote]that's a linear function of u, so we just need to prove the following two case:

(1), a=0;
(2), b=c.  [/quote]
Can you explain it more clearly, please ?
Thank you ![/quote]

I use uvw method, about this method, you can see here:
http://www.mathlinks.ro/viewtopic.php?t=278791


can someone tell me what the frequency of f(x) = 3sin(4x + pi) + 7 is?

isnt' the frequency always equal to B in f(x) = Asin(Bx + C) + D?

The period is equal to $\frac{2\pi}{B}$, so the frequency is $\frac{B}{2\pi}$.


abstract algebra
algebra
function
domain
calculus
integration
Ring Theory

we say the ring R is an Artinian ring,if we have:{R be an abelian ring and when every decreasing chain of ideals in R will be stoped or if $ %Error. "display" is a bad command.
\ I_1$   $ %Error. "display" is a bad command.
\supseteq$   $ %Error. "display" is a bad command.
\ I_2$    $ %Error. "display" is a bad command.
\supseteq$  $ %Error. "display" is a bad command.
\ I_3$   $ %Error. "display" is a bad command.
\supseteq$ ...     then exist n $ %Error. "display" is a bad command.
\in$  N so that  $ %Error. "display" is a bad command.
\ I_n$ =$ %Error. "display" is a bad command.
\ I_{n + 1}$ =...}.problem: if R be an abelian domain and be with unity  and be an Artinian ring,show that R is a field.should R be with unity?

Irrelevant terminology comment: the adjective is "Artin" or "Artinian," (after [url=http://turnbull.mcs.st-and.ac.uk/~history/Biographies/Artin.html]Emil Artin[/url]) not "arity."

if $ R$ is unital, pick $ 0 \neq x$  and consider $ (x) \supseteq (x^2) \supseteq (x^3) \supseteq ...$. then $ (x^n)\equal{}(x^{n\plus{}1})$ for some $ n$. because $ R$ integral, this shows that $ x$ is a unit.


[quote="Linh.NDD"]Prove that:
$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\geq \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}$
Where $a,b,c$ are positive real numbers and $abc=1$[/quote]
Let $a+b+c=3u,$ $ab+ac+bc=3v^{2}$ and $abc=w^{3}.$ Hence, your inequality equivalent to
$9u^{2}v^{2}w-5u^{2}w^{3}+3uv^{4}+6uv^{2}w^{2}-8uw^{4}-v^{4}w-v^{2}w^{3}-3w^{5}\geq0,$ which triviality.


Let $ p_{(x)}$ the number of primes less than or equal to $ x$. Prove that $ p_{(n)} < 2 \plus{} n/3$ for all positive integers $ n$.

[hide] Find the number of multiple of 2,3 from 1 to n.[/hide]

[b]TTTSpahn[/b]'s argument is not complete, although a small extension will do.

As for the non-completeness of his argument, e.g., $ n\equal{}6k\plus{}2$ has

$ 3k\plus{}1$ multiples of $ 2$
$ 2k$ multliples of $ 3$
less $ k$ multiples of $ 6$ to avoid double-counting
less primes $ 2$ and $ 3$ yields

at least $ 4k\minus{}1$ non-primes or equivalently, less than (or equal to) $ 2k\plus{}3$ primes, but $ 2k\plus{}3\not < 2\plus{}\frac{6k\plus{}2}{3}$

Can't we make it stronger to:
$ p(n)<\equal{}\frac{4\plus{}n}{3}$


Here is an interesting problem that I'm struggling on.

The digital sum $D(n)$ of a positive integer $n$ is defined recursively by:

$D(n) = n$ if $1 \leq n \leq 9$
$D(n) = D(a_0+a_1+...a_m)$ if n > 9
where $a_0,a_1,...a_m$ are all digits of $n$ expressed in base 10. (For example, $D(989) = D(26) = D(8)$). Prove that $D(n \times 1234) = D(n)$ for all positive integers of $n$.

I first thought of induction and tried 1, which worked as:

$D(1234n) = D(1234 \times 1) = D(1234) = D(1+2+3+4) = D(10) = D(1) = D(1)$.

But I'm not sure how to go from here using induction. Help!

[hide]$a + b + c + d = 10$

$D(n(1000a + 100b + 10c + d))=D(n(999a + 99b + 9c +10))$

The digital sum of a multiple of 9 will always be a multiple of 9 and hence 9 is the digital sum of any multiple of 9. $n(999a + 99b + 9c + 9 + 1)$ is $n$ more than a multiple of $9$ so

$D(n(999a + 99b + 9c + 10)) = D(9+n) = D(10 + n - 1) = D(1+n -1) = D(n)$[/hide]


Show that there are infinitely many pairs of positive integers (m, n) such that

$ \frac{m+1}{n}+\frac{n+1}{m}$

is a positive integer

[hide="Solution"]
I'm going to use $ (a,b)$ instead of $ (m,n)$ since that's what i did my work in.
\[ \frac{a+1}{b}+\frac{b+1}{a}=\frac{a^{2}+b^{2}+a+b}{ab}\qquad \cdots \qquad \textbf{(1)}\]
We wish to prove there infinitely many solutions that satisfy $ ab | a^{2}+b^{2}+a+b$. This is equivalent to showing the same result for $ ab | (a+b)(a+b+1)$. Consider the following factorization:
\[{ \frac{(a+b)(a+b+1)}{ab}=a^{2}+(2b+1)a+(b^{2}+b)}{ab}=\frac{a+(2b+1)}{b}+\frac{b+1}{a}\]
We deduce that $ a | b+1$, or $ b=ka-1$. Substituting this into $ \textbf{(1)}$ yields
\[ \frac{a^{2}+b^{2}+a+b}{ab}=\frac{(k+1)a-(k-1)}{ka-1}\]
Consider the substitution $ a=k-2$.
\[ \frac{(k+1)(k-2)-(k-1)}{k(k-2)-1}=\frac{k^{2}-2k-1}{k^{2}-2k-1}=1 \]
Therefore, there are infinitely many ordered pairs $ (a,b)$ that satisfy the equation.

 [/hide]

[quote="NeverOddOrEven"][hide=Solution]We deduce that $ a | b+1$, or $ b=ka-1$. Substituting this into $ \textbf{(1)}$ yields
$ \frac{a^{2}+b^{2}+a+b}{ab}=\frac{(k+1)a-(k-1)}{ka-1}$[/quote]

 :wink:

[quote="NeverOddOrEven"][hide="Solution"]
We deduce that $ a | b+1$, or $ b=ka-1$. Substituting this into $ \textbf{(1)}$ yields
\[ \frac{a^{2}+b^{2}+a+b}{ab}=\frac{(k+1)a-(k-1)}{ka-1}\]
Consider the substitution $ a=k-2$.
\[ \frac{(k+1)(k-2)-(k-1)}{k(k-2)-1}=\frac{k^{2}-2k-1}{k^{2}-2k-1}=1 \]
Therefore, there are infinitely many ordered pairs $ (a,b)$ that satisfy the equation.

 [/hide][/quote]

Yeah.. i dont get why $ b+1/a$  :blush:  could you please explain?

@Kurt Godel

gahh...funny how that one power screws up my whole solution  :rotfl: .

@riddler

If you look at the previous before I conjectured that $ a|b+1$, I wanted the second part to be an integer so I could try to show that the first part would also be an integer. However, it turns out that that's not the way to go i don't think

It's odd how these problems about hyperbolas have gotten so much attention since IMO 1988 Q6.  Anyways, the problem is a simple consequence of what I believe has been called "Viete jumping."  Okay, suppose that $ \frac{x+1}{y}+\frac{y+1}{x}=z$ for some positive integer $ z$.  For this problem, just suppose $ z=3$.  Then we have a solution $ (x_{1},y_{1})=(3,2)$ that turns out to equal $ 3$, just as planned.  We recursively will now produce a different solution.

Suppose that for some natural number $ n$ we have a solution $ (x_{n}, y_{n})$ where $ x_{n}>y_{n}$.  Notice that $ \frac{x+1}{y}+\frac{y+1}{x}=z=3$ is equivalent to the hyperbola $ x^{2}+y^{2}-3xy+x+y=0$.  Now, fix our $ x_{n}$.  Then the equation above becomes $ y^{2}-(3x_{n}-1)y+x_{n}^{2}+x_{n}=0$ meaning the two roots of the quadratic $ r_{1}$ and $ r_{2}$ sum up to $ 3x_{n}-1$.  But one root is $ r_{1}=y_{n}$, and so the other root is an integer, and better yet, $ r_{2}=3x_{n}-1-y_{n}>2x_{n}-1$ is a positive integer bigger than $ x_{n}$.  So then $ (x_{n},y_{n})$ being a solution implies that $ (x_{n}, r_{2}=3x_{n}-1-y_{n})$ is also a solution.  But because the original equation is symmetric about the line $ y=x$, we know that $ (x,y)$ being a solution implies that $ (y,x)$ is a solution.  Thus $ (x_{n+1},y_{n+1})=(3x_{n}-1-y_{n}, x_{n})$ is a solution, and $ x_{n+1}>x_{n}$.  Thus we have a new solution, and we can continue this way so that we have an infinite number of solutions.


Good job VA people!!! Nice job on 9th place.  Especially congrats to SeungIn  :D   that Wingardium Leviosa part was hilarious.   :rotfl:

What were the VA's individual rankings?

it sucked.

i heard i was one problem away from countdown?

edit:  nvm i was 2 away

Sorry to hear that.   :(   I hope the overall experience at Denver was fantastic though.  The nats trip isn't just about the competition.  Its about the experience.

Do you know what ranks Weina and Mitchell got?

not sure how they did.

actually 2 questions is a big difference. i thoguht i was one away...so yeah.

yawn.

its 9:30 here

Chill.  Think about how 13th place felt.

you mean alan chou? anyway, yuqing what was ur rank?

[quote="infinity4ever"]Chill.  Think about how 13th place felt.[/quote]

chill about what?

all i said is that i was far away from countdown and originally, i was told that i was very close.

ever consider actually reading posts before you answer them?

at cognos: 30th

Sorry, you sounded pissed when you said "it sucked"

no harsh feelings.

LOL so how were the problems?  What was the subject that they "honed in" on this year?  Heard it was Counting & Probability...  mehh

Nats is in Orlando next year!!!   :rotfl:  :rotfl:  :rotfl:

Wow, I know that no one has replied to this since May, but I got 88th at National MATHCOUNTS.

[quote="infinity4ever"]  What was the subject that they "honed in" on this year?  [/quote]

$ \equal{}$

$ \textbf{my defeat.}$

Especially since HALF of target was C and P. I hope that just means they do less of it next yr at nats. I mean I almost halved my average practice test score....


Prove that for all positive numbers a,b,c we have:

 a)  $ \frac {2a}{(b \plus{} c)^2} \plus{} \frac {2b}{(c \plus{} a)^2} \plus{} \frac {2c}{(a \plus{} b)^2}\ge\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}$

 b) $ \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}$

a)
Normalize $ a+b+c=3$:

$ \sum{\frac {2a}{(b + c)^2}-\frac {1}{b + c}}\geq 0$
$ \Longleftrightarrow \sum{\frac {2a}{(3-a)^2}-\frac {1}{3-a}\geq 0}$
$ \Longleftrightarrow \sum{\frac {2a}{(3-a)^2}-\frac {1}{3-a}-\frac{3}{4}(a-1)\geq 0}$
$ \Longleftrightarrow \sum{\frac {3(5-a)(a-1)^2}{4(3-a)^2}\geq 0}$

[quote="Marius Mainea"]Prove that for all positive numbers a,b,c we have:

 a)  $ \frac {2a}{(b \plus{} c)^2} \plus{} \frac {2b}{(c \plus{} a)^2} \plus{} \frac {2c}{(a \plus{} b)^2}\ge\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}$

[/quote]
The inequality is equivalent to:

$ \frac{2(a^3\plus{}abc)}{b\plus{}c}\plus{}\frac{2(b^3\plus{}abc)}{a\plus{}c}\plus{}\frac{2(c^3\plus{}abc)}{a\plus{}b}\plus{}a^2\plus{}b^2\plus{}c^2 \geq 3(ab\plus{}ac\plus{}bc),$

It suffices to prove that:

$ \frac{a^3\plus{}abc}{b\plus{}c}\plus{}\frac{b^3\plus{}abc}{a\plus{}c}\plus{}\frac{c^3\plus{}abc}{a\plus{}b} \geq ab\plus{}ac\plus{}bc,$

$ \Longleftrightarrow \frac{a^3\plus{}abc}{b\plus{}c}\minus{}\frac{ab\plus{}ac}{2}\plus{}\frac{b^3\plus{}abc}{a\plus{}c}\minus{}\frac{ab\plus{}bc}{2}\plus{}\frac{c^3\plus{}abc}{a\plus{}b}\minus{}\frac{ac\plus{}bc}{2} \geq 0,$

$ \Longleftrightarrow \frac{2a^3\minus{}ab^2\minus{}ac^2}{b\plus{}c}\plus{}\frac{2b^3\minus{}ba^2\minus{}bc^2}{a\plus{}c}\plus{}\frac{2c^3\minus{}cb^2\minus{}ca^2}{a\plus{}b} \geq 0,$

$ \Longleftrightarrow \sum \frac{(a\minus{}b)^2(a\plus{}b)(a\plus{}b\plus{}c)}{(b\plus{}c)(a\plus{}c)} \ge 0$

which is true.

[quote="Marius Mainea"]Prove that for all positive numbers a,b,c we have:

 b) $ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}\le\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}$[/quote]
Try $ a\equal{}2$ and $ b\equal{}c\equal{}1.$ :wink:

I think there is a typo error. Because, even if the ineq is flipped, it doesn't seem to hold for $ a\equal{}0$, $ b\equal{}c\equal{}1$.
Will you plz check your problem Marius Mainea! Thank u!
 :)

Plz give a little more detail, Math pro. It is still hard to guess for me. Are u using Karamata?

[quote="Marius Mainea"]Prove that for all positive numbers a,b,c we have:

 b) $ \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}$[/quote]
It's still wrong. Try $ a\equal{}1$ and $ b\equal{}c\equal{}5.$ :wink:

[quote="arqady"][quote="Marius Mainea"]Prove that for all positive numbers a,b,c we have:

 b) $ \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}$[/quote]
It's still wrong. Try $ a \equal{} 1$ and $ b \equal{} c \equal{} 5.$ :wink:[/quote]

 But $ \frac {4a^2}{(b \plus{} c)^3} \plus{} \frac {4b^2}{(c \plus{} a)^3} \plus{} \frac {4c^2}{(a \plus{} b)^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}$ :?:

For a:
WLOG assume $ a\leq b\leq c$.  Then
$ \displaystyle \frac{1}{(b\plus{}c)^2}\leq \frac{1}{(c\plus{}a)^2}\leq \frac{1}{(a\plus{}b)^2}$.
So we can now apply Chebyshev's Inequality to get
$ \displaystyle \frac{2a}{(b\plus{}c)^2}\plus{}\frac{2b}{(c\plus{}a)^2}\plus{}\frac{2c}{(a\plus{}b)^2}\geq \frac{2}{3}(a\plus{}b\plus{}c)\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right)$.
But the RHS of the above inequality can be written as
$ \displaystyle \frac{1}{3}((a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a))\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right)$.
Then we simply apply Chebyshev's Inequality a second time to get
$ \displaystyle \frac{1}{3}((a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a))\left(\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\plus{}\frac{1}{(a\plus{}b)^2}\right)\geq \frac{a\plus{}b}{(a\plus{}b)^2}\plus{}\frac{b\plus{}c}{(b\plus{}c)^2}\plus{}\frac{c\plus{}a}{(c\plus{}a)^2}$
$ \equal{}\frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}$.
Putting these two inequalities together we achieve the desired result.

a) Assume $ a \geq b \geq c$. Then we have $ \frac {a}{b \plus{} c} \geq \frac {b}{a \plus{} c} \geq \frac {c}{a \plus{} b}$, and $ \frac {1}{b \plus{} c} \geq \frac {1}{a \plus{} c} \geq \frac {1}{a \plus{} b}$

Now by Chebyshev we have:

$ LHS \geq \frac {2}{3}(\frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b})(\frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}) \geq \frac {2}{3}\frac {3}{2}(\frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}) \equal{} \frac {1}{b \plus{} c} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{a \plus{} b}$

Because
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \geq \frac {3}{2}$ (Nesbit's Inequality)

[quote="Sunkern_sunflora"]Will it be
\[ \frac {4a^2}{(b \plus{} c)^3} \plus{} \frac {4b^2}{(c \plus{} a)^3} \plus{} \frac {4c^2}{(a \plus{} b)^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2} \\
\geq \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3} ??
\]
:maybe:[/quote]

No,
$ \frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2} \\
\geq \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3}$ is not true for $ a \equal{} b \equal{} 1$ and $ c \equal{} 2$

But the inequality
$ \frac {4a^2}{(b \plus{} c)^3} \plus{} \frac {4b^2}{(c \plus{} a)^3} \plus{} \frac {4c^2}{(a \plus{} b)^3}\ge\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{c^2 \plus{} a^2} \plus{} \frac {c}{a^2 \plus{} b^2}$
is true for all positive reals

Proof: for positive reals $ x,y$ we have $ x^2 \plus{} y^2 \geq 2xy \Leftrightarrow 2(x^2 \plus{} y^2) \geq (x \plus{} y)^2 \Leftrightarrow (x \plus{} y)^3 \leq (x \plus{} y)2(x^2 \plus{} y^2)$, so we have:

$ \sum_{cyclic}{\frac {4a^2}{(b \plus{} c)^3}} \geq \sum_{cyclic}{\frac {4a^2}{2(b \plus{} c)(b^2 \plus{} c^2)}} \equal{} 2\sum_{cyclic}{\frac {a}{b \plus{} c}\frac {a}{b^2 \plus{} c^2}}$

Assume
$ a \geq b \geq c$ so we have $ \frac {a}{b \plus{} c} \geq \frac {b}{a \plus{} c} \geq \frac {c}{b \plus{} a}$ and $ \frac {a}{b^2 \plus{} c^2} \geq \frac {b}{a^2 \plus{} c^2} \geq \frac {c}{b^2 \plus{} a^2}$

By Chebyshev inequality we have now
$ 2\sum_{cyclic}{\frac {a}{b \plus{} c}\frac {a}{b^2 \plus{} c^2}} \geq \frac {2}{3}(\frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{b \plus{} a})(\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} c^2} \plus{} \frac {c}{b^2 \plus{} a^2}) \geq \frac {2}{3}\frac {3}{2}(\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} c^2} \plus{} \frac {c}{b^2 \plus{} a^2}) \equal{} (\frac {a}{b^2 \plus{} c^2} \plus{} \frac {b}{a^2 \plus{} c^2} \plus{} \frac {c}{b^2 \plus{} a^2}) \equal{} RHS$ and we are done.


AMC
AIME
geometry
AIME I
Pythagorean Theorem

Two squares side 1 are placed so that their centers coincide. The area inside both squares is an octagon. One side of the octagon is 43/99. Find its area.

Anyone have any ideas?

Problem (at least the version from kalva) doesn't specify so.

When I took this AIME, I solved it via a tedious, slow process of repeated applications of Pythagorean to find the dimensions of the corner triangles of one square (the part of one square not included in the octagon in question) and subtracted the area of those triangles from 1. It worked, but is not a recommended solution. I'll see if I can find a more elegant answer now...

EDIT (1): I don't think the octagon is regular.

Regular octagon would be too easy, methinks...

Hmm...what's with the misusage of "me"? Is gollum that popular? :)

Anyways, I'm clueless as to how to solve this when the octagon is irregular...any hints?

'methinks' has nothing to do with Gollum.

And no, it's definitely not regular...

Hmm, ok, I have solved it now. The corner triangles that are not part of the overlapping octagon are all congruent, so it is suffice to find the area of one of them, multiply it by 8, subtract that from the combined area of 2 of the squares, and divide by two to find the area of the octagon. Since the corner triangles are right, we are given the hypotenuse 43/99. We can express one leg as x and the other as 1 - 43/99 - x, and solve for x through the Pythagorean Theorem.

That's how I did it the first time, but I believe there is a much more elegant and much faster solution somewhere... (I think) the solution the AIME booklet gives is better, if you happen to have that.

[quote]That's how I did it the first time, but I believe there is a much more elegant and much faster solution somewhere[/quote]

Hmm, well, I don't think it was that bad. I mean, it was pretty much a noting of congruency, a single application of the Pythagorean Theorem, and the addition/subtraction of areas.

It's a matter of personal preference... I try to stay away from problems involving Pythagorean w/ fractions or multiple-digit numbers because I have a tendency to make mistakes and have to check several times, which does take me a while. I suppose it isn't that bad.

I'm looking at the answer booklet right now. A spoiler:

[hide] Use the circle in which both squares are inscribed to prove that the octagon is equilateral. Then find the distance from the center of the circle to a side of a square. Areas follow. [/hide]

I believe that solution is faster... though it would probably take more time just to discover that approach.


binomial coefficients
number theory unsolved
number theory

Show that for any $n\in\mathbb{N}$ the number of binomial co-efficients $\binom{n}{k}$ which leave remainder 1 when divided by 3 are greater than the no of binomial coefficients which leave remainder 2 when divided by 3.

Conider  the base-3 representation of n and let there are exactly N 2-s in it. We shall use Lucas's theorem. The only binomial coefficient $\equiv 2(mod 3 )$ is $\binom{2}{1}$. From here the number of binomial coefficients $\equiv 2$ is $\sum_{k-odd}\binom{N}{k}2^{N-k}$ and the others are $\sum_{k-even}\binom{N}{k}2^{N-k}$ and so the difference is exactly :
$\sum\binom{N}{k}(-1)^{k}2^{N-k}= 1$

sorry i didnt completely understand your argument. can you explain a little more?

[quote] and the others are$\sum_{k-even}\binom{N}{k}2^{N-k}$[/quote]

the others include the ones which are 0 mod 3 also. how do you conclude from the last step? sorry if this is a stupid question.

I am sorry i didn't write it very good. The others are only $\equiv 1(mod 3)$.
You use that $\binom{2}{k}$ is 1 or 2 (mod 3) depending on k. The product of some 2-s (mod 3) is 1 iff k is even and 2 iff k is odd.


In Triangle ABC, BC is perpendicular to AC. Angle A measures 30 degrees. Point D lies on Segment AC. If Angle BDC measures 45 degrees and AD=12, What is the length of BC?

Determine the volume of the solid formed by revolving the region in the xy plane bounded by the triangle with vertices 2,0   10,0   and 5,4  about the x axis.

Suppose the area of triangle ABC is 10 cm and the area of Triangle ADE which ABC is in has an area of 50 cm. BC is parallel to DE, and AF is perpendicular to Bc. The distance between BC and DE is 5 cm. What is the length of AF?

Solution #2:

[hide]

When revolved around the x axis, you get two cones.
One cone has a height of 3 and a base radius of 4.
The other cone has a height of 5 and a base radius of 4.

Then V = (1/3)(4^2)(pi) [ 5 + 3]

V = 128pi/3[/hide]

I'll put up the others later. My hand is sore.

For number 3, you need to mention if AF is on BC or DE. I assume it is on BC.

Solution #3:

[hide]
ABC similar to ADE

So, the base and altitude grow at the same ratio; r.

Hence,

A_1= 1/2rbh = 10 = 1/2(BC)(AF)
A_2 = 1/2bh = 1/2 (DE) (AF + 5) = 50
A_2/A_1 = 5 = r^2

(AF)/(AF+5) = r = (5)^(1/2)

Just use some simple algebra now and tada!

[/hide]

on number 2, how do you determine the heights? and how did they change? I get the radius part and how you plugged in the volume of a cone formula.

I think i am missing something.  :(

The heights are:
10-5 and 5-2

Draw it out on graph paper.
You have a "double ended" cone:
<>

solution to-1-
[quote="now a ranger"]In Triangle ABC, BC is perpendicular to AC. Angle A measures 30 degrees. Point D lies on Segment AC. If Angle BDC measures 45 degrees and AD=12, What is the length of BC? [/quote]
If we denote  CD as $a$ then $BC=a$ since BCD is isosceles right triangle
then by speacial triangle 30-60-90  $a=\frac{12}{\sqrt{3}-1}=6(\sqrt{3}+1)=BC$


I thought it would take a while before the solution to this problem is posted up, so am asking it here:

If a + b + c = 1 with a, b, c positive reals, prove

\sum (ab)^(5/4) < 1/4.

a solution would be available [url=http://www.mathlinks.ro/diff/e1r5p3.pdf]here[/url]

Also Cauchy Schwartz is a solution:

[sum ab^(5/4)]^2<=(sum sqrt(ab))(sum(a^2b^2))
and we know that sum(sqrt(ab))<=sum((a+b)/2)=1 we have to prove in the hypotesis that a+b+c=1 that sum(a^2b^2)<1/16

now.. it's easy to prove that sum(ab)<=(9abc+1)/4
and now denote t=abc and we get from the mean ineqs that 1/27>=t.
so from sum(ab)<=(9abc+1)/4 we get that:
sum(a^2b^2)=(sum(ab))^2-2t<=-2t+alpha<=1/16

alpha=(1+18t+81t^2)/16.

cheers! I hope it's okay.. it's kinda ugly but it's another solution! :D


function
combinatorics unsolved
combinatorics

How many ways to choose n numbers in 1,2,2,3,3,3,...,n,n,...,n

Are the n numbers you have to choose distinct?

If they are distinct, then the answer is 1.

If they aren't, the answer is $ mCn$, where m = $ 1^{2}\plus{}2^{2}\plus{}3^{2}\plus{}...\plus{}n^{2}$
= $ n(n\plus{}1)(2n\plus{}1)/6$

I think that the problem is not so easy. Note that, for example, there are $ n$ ways to choose $ \{1,n,n,\ldots,n\}$, so $ \binom{m}{n}$ is overcounting. (this word doesn't exist, no?)

In the case $ n\equal{}1$ the answer is 1. In case $ n\equal{}2$, there are two posibilities. In the case $ n\equal{}3$, the answer is 6...I think it's $ n!$...Let me see...

Nice thought, but for $ n \equal{} 4$ the answer is 20.  The sequence is (I believe) [url=http://www.research.att.com/~njas/sequences/A000707]A000707[/url] in the OEIS, which doesn't appear to have any simple expression associated with it.

As for a fast way of calculation the number is equal to the coefficient of $ x^{n}$ in $ \prod_{i\equal{}1}^{n}(1\plus{}x\plus{}x^{2}\plus{}\cdots\plus{}x^{i})$
more generally the number of r combinations from the multi set $ \{a_{i}\cdot m_{i}\}$ is the coefficient of $ x^{r}$ in $ \prod_{i\equal{}1}^{n}(1\plus{}x\plus{}x^{2}\plus{}\cdots\plus{}x^{a_{i}})$  :wink:

I note my problem in the case $ n\equal{}4$ late...But now I think it may be $ \binom{2n\minus{}2}{n\minus{}1}$...anyway, just am idea because I have no proof of anything else than the generating function of Albanian Eagle...But i belive that there must be a closed formula


In the data set {4,16,7,x,19,12,13,y,12,16,17,z,13}, 0 is less than or equal to x, x is less than y, y is less than z, and z is less than or equal to 20. The mean is less than the unique mode, which is less than the median. hat is the maximum possible value of x+y+z?

[hide=" click to reveal hidden content"]

Not entirely sure about this but as no-one else has responded to it, I'll give it a go:

Having played around with some numbers, I get $ 0, 12$ and $ 14$ for $ x, y$ and $ z$, in no particular oder.

This gives a mean of about $ 11\frac{12}{13}$, a unique mode of $ 12$ and a median of $ 13$ and a value of $ 0\plus{}12\plus{}14\equal{}26$ for $ x\plus{}y\plus{}z$.

[/hide]


Let $ A, B, C, D$ be four points in the complex plane $ a, b, c, d$ be the coordinates of these four points respectively. What is the coordinate of the intersection of $ AB$ and $ CD$, assuming that they do intersect. Find the nicest form as possible.

UGLY

dude you can just like do cartesian coordinates:  $ z=x+yi$; $ (x,y)=\left(\frac{z+\overline{z}}{2},\frac{z-\overline{z}}{2}\right)$.

P.S. You got that complex number book...shouldn't that cover this?


Which of the following is an equation of a line that is parallel to, and a distance of four units from, the line y= 3/4 x +6?

a) y= 3/4 x +1
b) y= 3/4 x -1
c)  y= 3/4
d)  y= 3/4 x +2
e)  y= 3/4 x -2/3

What's the best way to solve this?

Is the distance between lines same thing as the distance between the y intercepts (you both assumed that)?

 :wink:

Draw a picture, and apply basic trigonometry.

Well, you're both incorrect, since that's not how you measure distance from a line to another line...

Let's consider the line $ y\equal{}\minus{}\frac{4}{3}x\plus{}6$ Which is the perp line to the original line passing through $ (0,6)$

Let the point on the line we are finding intersect this line at $ (a,\minus{}\frac{4}{3}a\plus{}6)$. We have the distance from this point to $ (0,6)$ is $ 4$, so we have $ a^2\plus{}(\frac{4}{3}a)^2\equal{}16$ So $ \frac{25}{9}a^2\equal{}16$ So

$ a\equal{}\pm 12/5$. So we need to see which line has either $ (12/5, \frac{14}{5})$ or $ (\minus{}12/5, \frac{46}{5})$

The first one has the first of these, so the answer is a).

http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html

easier solution if you use this


algebra
polynomial
Galois Theory
group theory
number theory
superior algebra
superior algebra unsolved

Maybe this can be on another forum, but i dont know any elementary solution yet.

 Given $f(x) \in Z[x]$, irreducible,
a) Prove that there is a prime number $q$ such that $f$ haven´t got roots $mod\ q$.
b) Prove that in fact there are infinite such prime numers $q$.

PD: Put $f(x)=x^p-p$ for an imo problem.

Both are corollaries to Chebatorev's density theorem (when we find a permutation of the roots with no fixed points). But we need that $f(x)$ has degree $2$ or higher ;)

So we want to prove that there is a ($\infty$ many) prime $q$ such that $f(x)$ is never divisible by $q$.
I can prove it for degree exactly two (where it is simple), but for higher degrees it sounds funny to me to have a general proof without harder things...

Edit: nonsense removed ;)

$x^4+1$  is irreducible in $Z[x]$ but is reducible $mod\ p$ for every prime number $p$.

I somehow assumed that every Galois group of order has a cycle of length $n$, the degree of the polynomial.  :blush:
Se we need a permutation without fixed points now.

Well, the group is transitive (when regarded as a permutation group on the set of roots of our polynomial), and any transitive permutation group on an $n$-set has an element with no fixed point. This follows from the formula for the number of orbits: $\frac 1{|G|}\sum_{g\in G}|\mbox{Fix}\ g|$, where $\mbox{Fix}\ g$ is the set of fixed elements of $g$. This must be equal to $1$ (the action is transitive, so the number of orbits is $1$), but if $|\mbox{Fix}\ g|\ge 1,\ \forall g\in G$, then, since $|\mbox{Fix}\ 1|=n>1$, the expression would yield a value $>1$.

If you're only interested in the IMO problem and its generalization though, I believe I saw it (the generalization) in the Number Theory section, where it may or may not have received an elementary solution. I can't remember right now :).

I saw it long time ago but with an extra hypotesis (the degree of the polynomial being prime). ¿Is this what you are talking about?

 ¿Does anybody knows some elementary proof in this direction? I think i can manage to prove that in the imo problem there is an infinite number of prime number but only because of the special form of the polynomial.

Yes, that was what I was talking about. I actually thought that your $p$ was supposed to be a prime.


I was reading through some topics and someone said that they wanted to make USAMO by getting first in the state on AIME. Is that possible?

http://www.unl.edu/amc/e-exams/e0-usajmo/usajmo.shtml
Selection Protocol, Item 6

also:

Frequently Asked Questions USAMO #5, Item 5 :?:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, Mathematics Competitions


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

For $ R\equal{}\mathbb{Z}$, let $ S\equal{}\{2^m: m\geq 0\}\subset R$. Is $ R$ a subring of $ S^{\minus{}1}R$?

[quote="NcG"]For $ R \equal{} \mathbb{Z}$, let $ S \equal{} \{2^m: m\geq 0\}\subset R$. Is $ R$ a subring of $ S^{ \minus{} 1}R$?[/quote]

Hello.
Each element $ r$ of $ R$ can be written $ s^{ \minus{} 1}r$ with $ s \equal{} 1$; but $ 1 \in S$, since $ 1 \equal{} 2^0$. Thus $ R$ is contained in $ S^{ \minus{} 1}R$. Now, the multiplication in $ R$ and in $ S^{ \minus{} 1}R$ are both induced by the multiplication in $ \mathbb{Q}$, thus $ R$ is a subring of $ S^{ \minus{} 1}R$. (I assumed known that $ S^{ \minus{} 1}R$ is a subring of $ \mathbb{Q}$.)
Friendly,
P.

I guess that the problem was to be read as: "Is the ring homomorphism $ R\to S^{\minus{}1}R$ is injective?" owk

In general, if $ S$ is multiplicatively closed in $ R$, then the kernel of the natural map $ R\to S^{\minus{}1}R$ is $ \cup_{s\in S}(0: s)$. It's fun to prove that the last is an ideal without knowing that it is the kernel of a ring map.

The reason for the kernel being that is:$ S^{\minus{}1}R\ni\frac a1\equal{}0\Leftrightarrow \exists s\in S$ such that $ as\equal{}0$


[quote="mathexact"]can anybody help with that????

can calculate and design the area of the region bounded by the curves
y(x)=3x+0,5 and g(x)=(1-x^2)^1/2  and the positive axes x,y[/quote]
I'm assuming you meant $ y(x)\equal{}3x\plus{}0.5$.

First find the intersection points and then integrate the difference of the curves from these intersection points.

yes bθt i have to find thε area bounded by the curves and the positive axes...so what restrictions that have to take??

Well when you get to solve where they intersect you find that the x-value is approximately $ 0.162$. So these two curves intersect after $ 0$. So you integrate their difference from $ 0$ to $ 0.162$.


Find all functions $ f : Q\rightarrow Q$ that satisfy $ f (x\plus{}y) \equal{} f (x)\plus{} f (y)\plus{}xy$.

According to my book, first you must guess sth like $ f(x)\equal{}ax^{2}\plus{}bx\plus{}c$ and the calculate:
$ f(x)\equal{}\frac{1}{2}x^{2}\plus{}c$. Unfortunately answers there lack the essential proof of the fact that this one is the unique function :wink:

if we take $ g: Q\to Q$ such that $ g(x)\equal{}2f(x)\minus{}x^2$
then $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ so $ g(px)\equal{}pg(x)$
gives $ g(\frac{p}{q})\equal{}p.g(\frac{1}{q})$ and $ g(1)\equal{}p.g(\frac{1}{q})$
so $ g(\frac{p}{q})\equal{}g(1)\frac{p}{q}$
then $ \exists a\in\mathbb{Q},\forall x\in Q: \ g(x)\equal{}2ax \Rightarrow f(x)\equal{}\frac{1}{2}x^2\plus{}ax$


#25) A regular tetrahedron is inscribed in a sphere. If the volume of the sphere is 864 pi, what is the total surface area of the tetrahedron?

Hint
[hide]Look at the center of the sphere in a drawing. What does that remind you of? yea...pull those nifty formulae out of your pocket :wink: [/hide]

eek im stuck at 1152/pie equals r cubed. eek duno how to simplify.. sad for me, for circles radii

hmm how do u do that, am i doing it wrong?

[hide="Directions"]Start from a tetrahedron. Draw its altitude and find what $R^{2}-(H-R)^{2}$ equals to. Then, with the help of the formula for $H$, you can derive the formula connecting $a$ and $R$, where $a$ is the tetrahedron's edge.[/hide]


Prove that for any point P in the interior of the triangle ABC,

    PA*a  +PB*b   +PC*c   Is bigger or equal than 4(ABC).

Hola, Quién eres???????????

Nunca te creí capaz de hacer algo similar........

Mmmmmmmmmmmmmmmmmmmmm

Sale, nos vemos en el entrenamiento

De hecho vas bien, la transformaste adecuadamente

Se ve más bonito así:

$ PA\cdot{a} \plus{} PB\cdot{b} \plus{} PC\cdot{c}\geq 4\cdot{[ABC]}$

Que onda, Saludos .

              De paso que te pongan la respuesta porque no me ha salido,

           Saludos, Campero

    Nos vemos en el entrenamiento


integration
real analysis
function
real analysis unsolved

If $\mu(E)=\int_{E}fd\lambda$, show that $\int gd\mu=\int gfd\lambda$ for all measurable $g$. Measure space is $[0,\infty)$, Borel sets and Lebesgue measure.

You do this by showing it for integrable characteristic functions first, then for integrable simple functions, and eventually for integrable functions.


$(a,m)=1$ and $m=2\ or\ 4\ or\ p^k or\ 2p^k$ Prove that if $x^n\equiv a$ has solution it has exactly$(n,\varphi (m))$ solutions.(

I gave the generalisation here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=formula&t=15033
Prove: $\mathbb{Z}_{p^k}^*$ is isomorphic to $\mathbb{Z}_{\phi(p^k)}$ for odd primes $p$ and natural $k$. Also $\mathbb{Z}_{2^k}^*$ is isomorphic to $\mathbb{Z}_{2^{k-2}} \times \mathbb{Z}_2$. Then all follows from chinese remainder theorem.


A rectangle 12 inches long and 5 inches wide is inscribed in a circle. How many square inches are in the area of the region inside the circle but outside the rectangle? Use 3.14 as an approximation for $ \pi$, and express your answer as a decimal to the nearest tenth.

The diagonal of the rectangle is a diameter of length 13, so the radius of the circle is 6.5.  Then the area of the circle is $ (3.14)\times (6.5)^{2}\equal{}132.665$.  The area of the rectangle is $ 12\times 5\equal{}60$, so the area in between is 72.665.  To the nearest tenth, this is $ \boxed{72.7}$.

so the rectangle's diagonal is 13 inches.

and the rectangle's diagonal is the circle's diameter.

so the circle's radius is 6.5, and the area is 132.7 (since this is a target problem, we are allowed to use calc)

and the area of the rectangle is 5*12=60

so the area of the region inside the circle but outside the rectangle is 132.7-60=72.7

answer : 72.7

edit: blahh beaten


trigonometry
ARML
function
analytic geometry
graphing lines
slope

[hide]
naturally, I would look to see them both in the 4th quadrant (and as high as possible)

to find a lower bound, I set

$y= 2\pi$ and solved for the largest value of x  in the range

$2 \sin x (1)+\sin x+1 =-\frac{1}{2}$
$3 \sin x =-\frac{3}{2}$
$x \in \frac{7\pi}{6}, \frac{11\pi}{6}$

taking the larger result, we have
$x = \frac{11\pi}{6}$

this gives us a lower bound of
$2\pi+\frac{11\pi}{6}= \frac{23\pi}{6}$

now we'll analyze the LHS function in the range $(\frac{11\pi}{6}, 2\pi)$

as x gets larger, the function's value increases
as y gets smaller, the function's value also increases
BUT
the function will increase with x more rapidly than it will decrease with y because the of the slope of the unit circle in that interval
this means that making x larger and making y smaller will only serve to decrease the sum $x+y$

so the greatest sum is $\frac{23\pi}{6}$
[/hide]

[quote="da ban man"]Uh, this was like an ARML problem form a couple years ago..

I found this on the internet but there was no answer or solutions page to go along with it.  I got an answer of $\frac{23\pi}{6}$ but I dont know if thats correct or not..

the problem is

For $x$ and $y$ in radian measure with $0\leq x,y,\leq 2\pi$ compute the largest possible value of the sum $x+y$ if $2\sin{x}\cos{y}+\sin{x}+\cos{y}=-\frac{1}{2}$[/quote]

[hide="solution"]$(2\sin x+1)(\cos y+frac{1}{2})=0$

or

$(2\sin x+1)(2\cos y+1)=0$

if $\sin x=\frac{-1}{2}$, $x=\frac{11\pi}{6}$

if $\cos y=\frac{-1}{2}$, $y=\frac{10\pi}{6}$

since $x$ is greater than $y$, choose $2\sin x+1$ to equal 0.   This lets $y$ be any value so choose $2\pi$

Therefore $2\pi+\frac{11\pi}{6}=\boxed{\frac{23\pi}{6}}$[/hide]


(a) Name the product that is obtained from benzene after the following sequence of steps:

1) Reaction of benzene with succinic anhydride in the presence of AlCl3.
2) Hydrolysis.
3) Reflux with hydrochloric acid and amalgamated zinc.
4) Thionyl chloride, reflux.
5) AlCl3.
6) Hydrolysis.

(b) Steps 4-6 can be substituted by a single one. What is it?

(c) Show how the product obtained from the above sequence of reactions can be converted into an isomer that have the same number of secondary carbons.

[hide="Hint"]
$ C_{6}H_{6}+succinic\, anhydride+AlCl_{3}\longrightarrow C_{6}H_{5}C(=O)CH_{2}CH_{2}CO_{2}H$

(after hydrolysis).
[/hide]


One little tip:  drop in a \displaystyle at the start of that TeX code and you get:

[tex]\displaystyle\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}.....[/tex]

[quote="Rep123max"]Actually, heres the link http://www.sosmath.com/CBB/docs/howtolatex.pdf (you might have to be a free member to get it, not sure)[/quote]

[quote="mathfanatic"]The Not So Short Introduction to LaTeX2ε may be accessed at [url]http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf[/url][/quote]

Thank you!

Experiment:

[tex]\displaystyle \mbox{For all n, }(n+1)^2 - n^2 = n + n + 1[/tex]

[tex]\displaystyle\hbox{This is an hbox, and } \mbox{this is an mbox.}[/tex]

Are they the same thing?

[tex]\displaystyle \sum_{i=1}^{\infty} \frac{1}{i^{2}} = \frac{\pi^{2}}{6}[/tex]

To do combinations, you can use \choose:

\displaystyle{5\choose 3} gives:

[tex]\displaystyle{5 \choose 3}[/tex]

My TeX compiler gives me a warning when I do \choose. It prefers \binom{a}{b}.

[quote="Dark_Shadow"][tex]\displaystyle \sum_{x=0}^{n} \left(\begin{array}{cc}n\\x\end{array}\right) = {2^n}[/tex][/quote]
oops, i used a 1x2 matrice to do it.

[tex]\displaystyle \sum_{x=0}^{n} {n\choose x}= {2^n}[/tex]

Hi,

the most important tutorial you need to have is:

- http://www.ctan.org/tex-archive/info/companion-rev/ch8.pdf

There you can read that \binom {n}{k} requires the amsmath packagge.
Please browse the CTAN server it is really a valuable source. These links, standard resources on LaTex, can be recommended too:

- http://www.latex-project.org/intro.html
- http://www.tug.org.in/tutorials.html


geometry
3D geometry
sphere
function
geometric transformation
rotation
topology

Let $ S^2$ be the sphere in $ \mathbb{R}^3$. Let $ f: S^2\rightarrow \mathbb{R}$ a continuous function. Show that there exists an equilateral triangle in $ S^2$ such that $ f$ takes the same value in the vertices of the triangle.

Sounds interesting, but I have no idea.  :maybe: What about the others?  :roll:

i haven't worked out all of the details, but i think a trick like the sandwich theorem or the temperature-pressure-humidity problem should work.

pick a point $ x$ on the sphere, and name $ x',x''$ the two other point on the same maximal circumference such that the triangle $ xx'x''$ is equilater (and maximal). every function of $ x,x',x''$ which is symmetric in $ x',x''$ is a well-defined function of $ x$ only. so the trick should ask us to find such a symmetric function with nice properties.
now, my guess is that such an $ F: S^2\to\mathbb{R}^2$ may be
$ F(x) \equal{} (f(x')\plus{}f(x'')\minus{}2f(x), 2f(x')f(x'')\minus{}f(x)f(x')\minus{}f(x)f(x''))$:
if you're wondering where the hell this comes from, just play a bit with the elementary symmetric functions of $ f(x)\minus{}f(x'), f(x)\minus{}f(x'')$ (that we want to be both zero) and notice that this is a function (the only one?) that satisfies $ F(x)\plus{}F(x')\plus{}F(x'')\equal{}(0,0)$ (and of course that satisfies $ F(x)\equal{}(0,0)$ iff $ f(x)\equal{}f(x')\equal{}f(x'')$).
my guess (that i've not been able to prove, yet) is that if we assume that there's no equilater triangle $ xx'x''$ satisfying our thesis then we'd have a nontrivial map $ S^2\to S^1$ (which doesn't seem to be obvious at this point).

i don't have much time right now, if none fills in the details/concludes (as you're warmly invited to do :D ), i could think about it later.

I'm not sure exactly how to proceed, but this does look likely to work. However, one thing that you need to be careful of is that given a point $ x$ there is more than one maximal equilateral triangle (Consider rotating the sphere around an axis through $ x$. There's no way to tell which one of these rotations you should choose). So this function isn't quite well-defined in terms of $ x$.

lol... this just makes my previous post kind-of-nonsense... no comment, really.

- after 30 minutes break or so -

thinking about it once more... you can think of "my" map $ F$ as a map from $ X \equal{} T_1S^2$, the unit tangent bundle to $ S^2$, to $ \mathbb{R}^2$. we should know everything about this $ S^1$ bundle over $ S^2$, but i don't know if it's enough.
anyway, we should know something more about $ [X;S^1] \equal{} H^1(X;\mathbb{Z})$ to (try to) conclude.

You can calculate $ H^1(X)$ from $ \pi_1(X)$; from the universal coefficient theorem and the Hurewicz theorem, they should be isomorphic if $ \pi_1$ is abelian. Since $ S^1 \to X \to S^2$ is a fibration, there's an exact sequence $ \pi_2(S^2) \stackrel{\partial}{\to} \pi_1(S^1) \to \pi_1(X) \to 0.$, so $ \pi_1(S^1) \equal{} \mathrm{coker}(\partial)$.

Unfortunately, I don't remember the specifics of how the map $ \partial$ is defined at the moment... I suspect that $ \partial$ is either an isomorphism or the zero map, but I have nothing backing it up right now. I'll be coming back to this later after I do some review :).

you're right, it's either an isomorphism or the zero map, and both cases can occur: for example you can consider the trivial fibration and the hopf fibration.

anyway, always wondering (or maybe wandering) along the lines of my previous posts, we can also consider the involution $ v_x\mapsto \minus{}v_x$ and see where it takes, if it takes anywhere.
maybe considering a two-point orbit instead of a three-point one makes things easier. but, as usual, i could just be talking nonsense.


Show that the sum with n variables

\[\sum_{\mathbb{Z}_{+}}\frac{1}{(\lambda_{1}\cdots\lambda_{n})\cdot(\lambda_{1}+\ldots+\lambda_{n})}= n!\zeta(n+1)\]

where $\{\lambda_{1},\ldots,\lambda_{n}\}\in\mathbb{Z}_{+}$ .

It was posted before, but I don't remember where.


[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamowinners.html]Winners/Honorable Mention[/url]

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-results.html]Scores[/url]

Congrats to the winners and HM - I see a lot of AoPS Community members and participants in our classes there :)

Best of luck on the TST.

I got a 9.  So I accomplished my goal of not getting 0 :) .

That's almost exactly what I got on my first USAMO (adjusted for scoring changes - it was out of 100 then).  In fact, it's a little higher. . .

Wow I got a 4... Apparently 2 for #1, 2 for #4...

Wow, I got an 18.  Really surprised by that.

Being a first-time USAMO taker, I really have no clue what they gave me credit for and what they took off for though.

Will we ever get to see our point distributions for each problem?

You can ask Zuming if you're going to MOP, and he might look it up for you.

Otherwise, it's just a guessing game.

Does anyone else think the percentiles are a little bit messed up on the score table?  Mine says 44.9 for the 10-14 range but I calculated and I'm pretty sure it should be 55.9.  In one place it goes from 44.9-44.0 above it and in another 67.9-67.0 above it.  Do you think this might be a typo or something?

 :?  :?

Yeah, for some reason they seem to be a bit lower than they should.
Some of the stuff in the middle got shifted a bit lower.

With all the craziness probably going on in the AMC office these past few days, I guess some typos and stuff are to be expected.

Shloma, if you are who I think you are... with 99% certainty

I got a 12, but I completely misread question 4... ugg how come 4.5 hours wasn't enough for me to realize that table legs can have length 0? I always misread at least one question in any given competition...

Congrats on Mathcounts too, BTW

Even if you misread question 4, you could still get 6 pts for doing everything else properly. At least, I would give a 6 to anyone who did the problem correctly but didn't realize leg length 0 was permitted and thus had an incorrect (but CORRECT for the way they did it) answer. It's not like that assumption trivializes the problem.

Give him credit though, he is (presumably) in 8th grade, and 12 certainly isn't a bad score.

this previously is what kind of a post you get when bubala is acting like an idiot...

[quote="seungjun"]I'm an 8th grader, too. I only got a 2! I need to work on proof-writing skills.[/quote]

The Pre-Olympiad forum is a great place to do some of that work.

Who is shloma? If he really made CD at Nats, then I can tell Xantos Guin that he wasn't the #1 Penguin!

Indeed he made the countdown round at Nationals, and won the written test.  And judging by the tie he was sporting, he was clearly the top penguin in the contest.

Shloma is Sergei? I thought patricia, karlanna, and andrew were the only aopsers on the site...

Zhou Fan, a winner of the USAMO, came from the high school I am going tommorrow. I really want to meet him and learn how he studies.


floor function
number theory proposed
number theory

Prove that the values of $ \{\sqrt n\}$ are distinct in $ (0,1)$ as $ n$ ranges through the non-square positive integers. Here $ \{x\}\equal{}x\minus{}\lfloor x\rfloor$ denotes the fractional part of $ x$.

[hide]From this for once I thought that the values of $ \{\sqrt n\}$ cover all irrational numbers in $ (0,1)$ as $ n$ ranges through the non-square positive integers. But then I thought that this doesn't necessarily imply that. But is this true that the values of $ \{\sqrt n\}$ cover all irrational numbers in $ (0,1)$ as $ n$ ranges through the non-square positive integers?[/hide]

Suppose exist $ m,n$ is not perfect square and n>m satisfy
$ \{\sqrt {m}\} \equal{} \{\sqrt {n}\}$ then
$ \sqrt {n} \minus{} \sqrt {m} \equal{} [\sqrt {n}] \minus{} [\sqrt {m}]$
But we have a result :
If $ x,y\in N$ satisfy $ \sqrt {x} \minus{} \sqrt{y}$ is a integer then $ x,y$ are perfect square.
So this problem claim.

It would be helpful if you posted a complete solution. Thank you.

As there are uncountably many irrational numbers in $ (0,1)$, they are not all covered by $ \{ \sqrt n \}$.

And as a completion was asked for:
Let $ \sqrt x \minus{} \sqrt y$ be rational. Then by squaring, $ \sqrt {xy}$ is rational, thus $ xy\equal{}a^2$ for some rational $ a$.
Thus we want that $ \sqrt x \minus{} \sqrt y \equal{} \sqrt{x} \cdot ( 1 \minus{} \frac ax)$ is rational, thus $ x$ is rational (or $ \frac ax\equal{}1$ implying $ x\equal{}y$ which we exclude). As $ x$ is rational, $ y\equal{}\frac{a^2}x$ is, too.

Thank you $ \zeta x$. That was my solution too. And I think I have found another answer to my problem. The transcendental numbers aren't covered. :)

Other solution.$ x>y$
From $ (\sqrt{x}\plus{}\sqrt{y})(\sqrt{x}\minus{}\sqrt{y})\equal{}x\minus{}y$ is a rational number then if $ \sqrt{x}\minus{}\sqrt{y}$ is rational then $ \sqrt{x}\plus{}\sqrt{y}$ is too.
So $ \sqrt{x}\plus{}\sqrt{y}\equal{}a,\sqrt{x}\minus{}\sqrt{y}\equal{}b$ where $ a,b\in Q$ so
$ \sqrt{x}\equal{}\frac{a\plus{}b}{2}$
It means that $ x,y$ are perfect rational.